Modern Digital And Analog Communications Systems B P Lathi ...

Chapter 22.1-

1 Let us denote the signal in question by q(t) and its energy by Ey . For parts (a) and (b)

/ 2"1 1 f

2r

Eg — Jsin

2t dt — - J

dt - -J

cos 2t dt = it + 0 = it

(c) Eg ~ f sin2Idl = ~ f dt-\ ( cos 2t dt = tr + 0 = tr

J2» 1 Jiw 1 JlK

(d) = y(2 sin 0

2dt = 4

Jdt~\J

cos2t dt = 4[5t + 0] = 4ir

Sign change and time shift do not affect the signal energy. Doubling the signal quadruples its energy. In the

same wav we can show that the energy of kg(t) is k 1Eg.

2.1-

2 (a) /•;, - /*( 1 ?di = 2. Eg = /J (1 )~dt + /*(- l)2dt = 2

= J (2

)

2dt = 4. £*-* - ^

(2}3dt = 4

Therefore £**„ - Ez ~ Eg.

r2r,.r r7r l"*/ 2 /•» /-a»/2 /•2»

(bl Eg = j(\)

3dt+J (

— l)2dt = 2rr. £„ = y j

(-ifdt^J

(1 )dt+

^- 1 )

dt = 2tr

,»/2 rW2 y-2.

£z^,= / (2)2</t + / (0)

2t/t + /

(-l)2d( = 47T

JO Jir/2 J%n/2

Similarly, we can show that £*.„ = 4rr Therefore £**„ = £„ + Ev . We are tempted to conclude that £,. ±1) =

Ez — Eg in general. Let us see.

fw/A r* fn

(c) £x= / (1 )

2<f<+

/(-l)

2c/« = 7T Ey = I (1 )

2dt = 77

,/o */ir/4 *'0

£*+„ = J(2

)

2./f + J (0)

2dt = 7T £*- „ = J

(0

)

2df

J(~-)

2dt. = 3tr

Therefore, in general E*±j, / Ez + £«

2.1-3

. = — / C 2cos

2(wot + 0)dt. = £- f |1 + cos (2w0 ' + 29)] dt

~o J0

z-»2 ch) rTa 1 r>2

= / dl 4. / cos (2-uo* + 29) dt = —- |To + 0] =2^0 [Jo Jo

27°£!2

2.1-4 This problem is identical to Example 2.2b. except that w, i- a* In this case, the third integral in Pg (see p. 19

is not zero. This integral is given by

~ lim2C,r2 r

r-4^ T1 -T/2

= limC1C2 '

rT/2

/T—*x. T L' -T/2

• r rT/- Irn

J—- / cos(9j — 62) dt 4- / cos(2-vjt + 0\ + 6j) dt

f [J-T/2 J-r/i

=. lim — [7 ’ cos(9i — 9j)| 4 0 — £162 cos(9i — O2)

T— *x. 7

1

Therefore

p9- £±. + Qi + C1C2 cos(0i - #2)

2.1-5

-u; {l'fdt - 64/7 (a) P ffdt = 64/7

2

(b) P2y = Ij

2

(Ot'fdt = 4(64/7) = 256/7 («) P«. = J)

2dt = 64c

2/

7

Sign change of a signal does not affect its power. Multiplication of a signal by a constant - incieases the po

bv a factor c2

.

2 .1-6

(a) P, -if* Ja

- f c-t dt=\ (1-a ’

 Ps= ^L (£)rft

5

P“ = 7 S. Tn[i)g

‘{t)d1 =

J-r,*i?JkDkD rf di

^ .> are finite because the integrands are periodic signals

The integrals of the cross-product terms ( hc" *J, d The remaining terms (.* = r) yield

(made up of sinusoids). These terms, when divided by T - 00. yield

/

T/2 n

T/2 kmm k~rr.

, (., P„.«, of . sinusoid of amplitude C is C*/2 IE,. (2-6.)) «rik. of iU fr«,u«ncv (at # °> and ph«.

JXtZ, 1 r,“. to fh. sum of tit* powet, o, .h.— |E,. «.«S)|. Thefts, in

this case P =^ = 178'

,. p _ | Ui =51.

(c) (10 + 2 sin 3f) cos 10/ = lOcos 10/ + sin 13/ - sin 3/. Hence from Eq. (2.6) ^^ 5

(d) 10 cos 5/cos lOt = 5 (cos 5/ + cos lot. Hence from Eq. (2.6b) P - +4^

- 25_

<«» ^ - d/O «./«> - «/>

2.2-1 For a real a

r_3o

‘ dt = oo

rT/2

For imaginary e. let a — j* Then

oc

P„ = IlifiT

r/2 , r7'/ 2

- I (rjx

')(r~'xt

)dt = lint — /

dt = 1

}_ T/;t-~ t J-t/2

2

*w

Fig. S2.3-2

Clearly, if o is real. r~“ is neither energy not power signal,

power 1.

However, if a is imaginary, it is a power signal with

g3 {i) = .<?(' - 1) + .9t( t ~ !)• 9s(0 = 9( < - 1 ) + -,J'f f + 1 ^ 9*{f)-a(t O.o) + ffi(f . 0.5)

The signal can he obtained by (i) delaying ,U) by 1 «~«l *- 1 »<">

by h factor 2 (replace t with t/2). (iii) then multiply with 1 o. Thus gs {t) - l-Offlj '

2.3-2 All the signals are shown in Fig S2.3-2.

2.3-3 All the signals are shown in Fig. S2.3-3

Fig. 52.3-3

2 J'4

- £>(>.« - r.. *<•

»

/>-»’* *

=£ led - Til’d/ - J »! (*)dr > £. E.I.,1 - £w«'>]’di -

;£»V)d* ~ F'J°

£ tot- - »]’ d. - i£ ,V) d.e **,-> - £«'/•))’ d. -£«’(') dl - as,

Eoe(t) - J[ag(J)ftl* - J 8~V)* = 0

" E*

2.4-

1 Using the fact that g(r)f‘(r) = g(0)h[r). we have

(.)„ (hlPM (.)).(/) W-p('-l) i.)A«- + 3| (f) ge{w) (use L‘ Hbptlals rule)

2.4-

2 In these problems remember ,ba. impute ><*> I. located a, a . 0. Thus, an impute *<—) » '»»'“* a. r I.

and so on. . .

(a) The impulse is located at r = l and g(r) at r = Ms gU). Therefore

3

L q(r)f<(t - r)»/r = g(t)

(b) The impulse *(t) is at r = 0 and q(1 - r) at r = 0 is s(t)- Therefore

fi[r)g(t - t) dr = ff(t)LUsing similar arguments, we obtain

(c) 1 (d) 0 (e) rs

(f) 5 (g)p(-l) (h) -e

2.4-3 Letting nf = .r. we obtain (for n > 0)

f dt = - f 0)

Similarly for (at)dt. — ~|d(0) =

jjjj J<t>(t)6(t)d1

Therefore

»<*•» - w‘">

2.5-

1 Trivial Take the derivative of |e;3 with respect to r. and equate it to zero.

2.5-

2 (a) In this case Ex = JQ'

lit = 1. and

= jrj

si'M'M' =\ J0

td, ~ 0 b

(b) Thus, git, * 0.5.r«). and the error c«) - t - 0.5 over (0 < i < 1). and zero outs.de this interval Also E„

and £, tt he energy of the error) are

E9 = Jg2 (t)dt - J

t2 dt- 1/3 and Et ~ J

— 0.5)2

rff = 1/12

The error (f - 0.5) is orthogonal to r(t) because

J(t — 05)(l)</f = 0

Note that £ = r*£x + £#. To explain these result, in terms of vector concepts we observe from Fig. 2.15

ihat the error vector e is orthogonal to the component «. Because of this orthogonally, the length-square of

Xe^ o^(O) ^ equal to the sum of the square of the lengths of cx and e [sum of the energ.es of «r(f) and

f(0l-

2.5-3 In this case Eg = J0‘ g

2(l)dt = f0t2dt - 1/3. and

c =4~ f r(t)g{t)dt = 3 ftdt= ll

Ey Jo Jo

Thus. r(t) * 1.5,(/). and the error r(l) = r(t) - 1.5»(f) = 1 - 1-W over (0 < t < 1). and zero outside this

interval Also £, (the energy of the error) is £e = /0 (1- 1-51) dt _ 1/4.

.5-4 (a) In this case Ez = j^sin7 2ntdt = 0.5, and

— ~ J «(()»(')* = djjf (sin 2irt<l( — -1/-

(b ) Thus. »(,) - -(l/*).r((). and (h. .no, .(.) - < + (•/>)»» to. o«r (0 < . < D- »d -» .«•

interval. Also £aand £, (the energy of the error) are

4

Eg = f g, (t)dt = jf’ t*Jt = 1/3 and £, = [f - (l/*)sin 2trf.]

2(ft = f

'

The error \t + (1 /tt) sin 2itl\ is orthogonal to x (/.Jbecause

Jsin 2trf[< + (l/tr)sin 2irt]dt = 0

Note tho. e. - **£a + TO -pH.L.— in ~the error vector e is orthogonal to the component ex. Because of this orthogon y, *

, ,y,^ equal to the sum of the square of the lengths of ex and e [sum of the energies of «(t) and

2.5-5 (a) If r(f) and u(t) are orthogonal, then we can show the energy of x(t) ± y(t) is Ex + Ev

rnv

P \x(t)±y(t)\1dt =p \r(t)\

3dt +p \y(t)\

2 dt±j

J\t)v(t)dt± jj^t)y{t)dt

= J|.r(f)|

5dt + J

|«(<)t*df

( 1 )

(2 )

The last result follows from the fact, that because of orthogonality the two integrals ®f 1

r(,)y‘(t) and r'(t)y(t) are zero [see Eq. (2.40)}. Thus the energy of x(t) + y(t) is equal to that of x(t) V (t) if

(b) Using similar argument, we can show that the energy of c,r(t) + n,,(t) is equal to that of ''i*(f) “ if

.r(/) and >/(/) are orthogonal. This energy is given by |m| £* +M Ey(c) If :(() = r(t) ± y(t). then it follows from Eq. (1) in the above derivation that

E, = Ex + Ev ± (Exy + Eyx)

2.5-6 g,(2. -1). ga (-1.2). g.t(0. -2). *4(1.2). gs(2.1), and «a(3,0). From Fig. S2.5-6. we sec that pairs (gs-ge).

(gl . g4 )and (gj.gs) aie orthogonal. We can verify this also analytically.

gvg«=(0x3) + (-2x0) = 0

gi g4 = (2xi)+(-lx2) = 0

g3 gs = (-l x 2) + (2 x 1 )= 0

We can show that the corresponding signal pairs are also orthogonal.

P {-rMZx,[t)\di = Q

„i (t)q*(tult =P |2r,(f) - ra(0][*i(0 + ^(t))dt = 0

g2 lt)gi (i) dt = P + 2x a (/)][2r,(f) + x3 (t)\dt = 0

J -'V-

2 .6-1

In deriving these results, we used the fact that = f^xfrtjdt - 1 and T,(t)xs (t.) dt 0

We shall compute , „ using Eq. (2.48) for each of the 4 cases. Let us first compute the energies of all the signals.

Ex = jsin

22irt dt = 0.5

In the same way we find = Em = Ets = Egt = 0.5.

Using Eq. (2.48). the correlation coefficients for four cases are found as

2trt sin 4trt tit = 0 (2)(1) ,

1

f sin 21 '

v/to sxo s) J0

I f 0.707 sin 2nt<it = 0 (4)’ s) Jor{t) and

2.8-1 Here To = 2. so that o = 2tr/2 = tr. and

Vg(t) = <*o +^ cosnirf + bn sinnrt.

n= l

whore

1

/ (sin 2*/)(-sin 2vt)dt = —

1

(3) i n I v.iviuu mum -» v»/ /^r:v ’v/to.sHO.s) J0 V<0!

Signals r(f) and 32 (f) provide the maximum protection against noise.

J— I / 0.707 sin 2»rtdt-j 0.S) (° S)

[70 J0.5

707 sin 2irtdt = 1.414/tt

- 1 < t < I

no -i /:Usinmrf dt = 0

Therefore1 4 A' (-1)"

3 n-1

n= 1

cos turf - 1 < f < 1

Figure S2.8-1 shows q\l) — t2

for all I and the corresponding Fourier series representing q(t

/

o\er I .1)

The power of q(l) is

Moreover, from Farseval s theorem [Eq. (2.90)j

Pg — C0 + / 9 ^3 ) 2 2—

t

l rr*n 2 ) 9 it4 ^ ” 4 9 90 5

," n«l V 7 " =1

(b) If the ;V-term Fourier series is denoted by r(t), then

V— 1

1 4 ( — l)n

n=l

The power Pz is required 10 be 99%P9 = 0.198. Therefore

p. = i +iri = 0.1989+

w 4 L* n*

- 1 < 1 < 1

6

For tt = 1. P, = 0.1111; for .V = 2. Px = 0.19323. For N = 3. Px = 0.19837, which is greater than 0.198.

Thus, .V = 3.

2.8-2 Here T0 = 2rr. so that u>0 = 2ff/2rr = 1. and

g (r) = a0 + cosnt + fe„ sinnt - tt < t < tr

where

oo = — f tdt — 0 , a„ = f2* J-r 2* J-rt cos nt dt — 0,

2 r . ,,, 2(-irhn = _ J

^tsinr>tdt = —

Therefore

— ir < t < ir.»(*) =2(-l)"

+1 ^-BinT)tn*l

Figure S2.8-2 shows g(t) = t for all t and the corresponding Fourier series to represent g(t) over (-it, tr).

1*1^>i Loil]

t-*

Fig. S2.8-2

The power of <j(t'j is

s £<•)’* -TMoreover, fioni Parseval s theorem [Eq. (2.90)]

1 tr‘

(b) If the .Y-tetm Fourier series is denoted by r(t). then

s

.T(t) = 2(-l)”+ 1 ^isinn W t

nxl

The power Px is required to be 0.90 x t = 0.37T2

. Therefore

- n < > < n

*-*£*-"*

For .V = 1. Px » 2; for A’ = 2. Pz = 2.5. for N = 5, P, = 2.927. which is less than 0.3tt2

. For N - 6. Px -

2.9825. which is greater than 0.3jt2

. Thus. N = 6

2.8-3 Recall that

flO ;

[To/2(la)

/.9(0*

J -To,'2

[To/i

1 g(t) cos njjot dt(lb)

J-T0 /

2

[To12(lc)

1 g(l) sin dt

J-To/2

/

Recall also that cos n-»l is an even function and sin Wjs an odd^

t. then g(i) cosW is also an even function and fl (t)sinW » an odd Junction

2 fT° /2 MM (

2a )

-t.J.,mno :

4 rTon

°""jo1q(t

)cos n*iQtAt (

2b)

(2c)

hr, = 0

Similarly, if *(*) is an odd function of t. then g(t) cos nwot is an odd function of t and g(t) sin r,^ t is an even

function of t. Therefore

ao = «» = 0(3a)

bn= ±f°'\mnn.0Ut MObserve that , because of symmetry, the intention retired to compute the coefficients need be performed over

only half the period.

2.8-4 (a) To = 4. -vo = ^ = f - Because of even symmetry, all sine terms are zero.

*X,

.<?(*)= «o + Y* a " cos \Tl

nss 1

n0 = 0 (by inspection)

«- - J [/* cos(f')

d1 -

1

cos (tO '"]= ^ s,nT

Therefoie. the Fouriei series for q{t) is

4 / at 1 3trt 1 5trf 1 rnr_7*t \

q(t) = - (cosy - gCOS— + -cos-y7cos

2 j

Here K = 0. and we allow C„ to take negative values. Figure S2.8-4a shows the plot of C„.

(b) To = lOtr. ^ = 3 . Because of even symmetry, all the sine terms are zero.

qii) = 00 + £«„ cos (Jf) +ftt.sin (t')

». = 1

n 0 = - (bv inspection)5

«. - -k£- (?•)* - & (H) (?•)L A !“n ("«'

)

. f sin f -tl dt = 0 (integrand is an odd function of t)

n10* J_„ V 5 )

Here b„ = 0. and we allow C„ to take negative values. Note that Cn = a„ for n = 0, 1, 2. 3,

shows the plot of Cn-

(c) Tq = 2tt . wo — 1*

9 (t) = oo + ^OnCosnt + b„sinnt with no = 0.5 (by inspection)

narl

—

£

cosn"" =0 -',n=

-X ^ S1

, 1

sin ntdtITT)

and

q(t) = 0.5 - ^(sinf + ^

sin 2/ + -sm3f + 4sin4M-

)

= 0.5 + i [cos (' + £) + \cos (2/ + §) + jcos (3* + 0 )

+ ']

Figure S2.8-4b

8

Figure S2.8-4d shows the plot of C„ and fi„.

(e) To = 3. uJo = 2ir/3.

fin2 2mr 3 ,

27rn 2nn:

3 J0iCOS— tJ'=

2^ [cOS— + -T

2vn.C ——

—

3^0 3 2

Therefore Co = g and

_ 3 4tr 2n 2 ”2jtti 4jtt<

"2rrn

c-- 53^[V a+— - 2c“— —

i= 6. -<o

2 /',

2nir, 3 . , .

2nn 2nn 2nn=

3 J0(sin—<d,= 27^'sm—--T cosT

(f) To -

!2+ ~t

rr/3 . 110

and fin = tan"/ 4ir 2n 2 ”2jtti 4jtt<

"

2rrn

\/2 + -—~ 2cos- 3

= 0.5 (by inspection). Even symmetry; 6n

2sh cos - sin 2^n \

+ 2iii sin^ -1 )

, ( ^fcoVcos^

= 0.

«n4 r

1

, . nrr

ij./if

2

3

6_ir2 n 2

>s~ tit+ J

(2 - t) cos Zj-t tit

nrr 2r)7r"|

lS__ C0S _r j

6 / jr 2 1 5ir ] 7t \

1 * 0 5 ~2(cos 3' *

9cos,r ' +

20cos— f t - cos— f +

J

2.8-5

Observe that even harmonics vanish The reason is that if the dc (0.5) is subtracted from <7 (f). the resulting

function has half-wave symmetry. (See Prob. 2 8-6). Figure S2.8-4f shows the plot of Cn .

An even function qr n) and an odd function 9o(0 have the property that

Se(() — 9e(~0 and Co(t) = -flo (—0 !1 '

F.vety signal </(t) can be expressed as a sum of even and odd components because

9(0 = \ i.o(0 + 2 i.9(0- 9(-0]

#v«»n

From the definitions in Eq. (1). it can be seen that the first component on the right-hand side is an even

function, while the second component is odd. This is readily seen from the fact that replacing t by —1 in the

first component yields the same function. The same maneuver in the second component yields the negative of

that component.

To find the odd and the even components of g{t) — u(t). we have

9(0 - 9«(0 + 9o(0

where {from Eq. (1)]

<iA<) = 5 M0 + «(-/)] = |and

9o(0 = 3 MO - «(-0 ! = ^sgn(f)

The even and odd components of the signal v(l) are shown in Fig. S2.8-5a.

Similarly, to find the odd and the even components of g(t) = e.~at

n(t), we have

9(0 = 9» (0 + 9o(0

where

9,(0 = i[e— «(fJ + e«u(-‘)]

and

10

ftOFig. S2.8-JS

The even and odd components of the signal c~tttu(t) are shown in Fig. S28-5b.

For g(f) = r''. we have

ejl = <j«(f) + flo(0

where

and

««(') = 3 K + r' J<

]* cos '

Soft) « ^[e

jt - <•"'*] * jsin t

2.8-6 (a) For half wave symmetry

and

and

<7(0 = -<7 (t ± y )

Toi* 7*0/3 /*To

= -1 [° g(t)cosn^otdt = — /

g(t)cosn«>otdt + »(f)co«W«ftTo

_/0JO

./o •'To/2

Let r = / - 7o/2 in the second integral. This gives

jf

r° /2

S(0 cos two* dt + Jg (* + y )

cos nu'o (t + y )<i*j

- JL]^J

a'

g(f) cosn^ot dt 4- J— <7(t)[- cosnwor) cfaj

,i.f r\To L/o .

fl(f) cosnwof tit

In a similar way we can show that

4 fTo/2

l,r = — jg(t) sin nuiot <lt

To Jo

(b) (i) Jo = 8. -o = j. no = 0 (by inspection). Half wave symmetry. Hence

11

2.9-

4 / ntr rift.

ntr \ ,= ^(cosT + T smT _1)

(nod)

Therefore

Similarly

4 /r)7T.

T17T ,>I Sin — 1

|

(n odd)r>3 it

3 V 2 2 /

flu “^,-^t(¥ + 0

n = 1 , 5 . 9 . 13 .-

n = 3 , 7 . 11 , 15 ,

1 f Lsm^Lt <11 =L 2 4

Hi coslil^ « -J-sinf^)fe"~

2 n 2tr2 l

Sm2 T 2 / n 2

tr2 V 2 /

and

nir „ ,•

n7r,

a„ cos— * + on sin— f

4.9(0 = Hn*l.S.5.-

(ii) To = 2ir. wo = 1 no = 0 (by inspection). Half wave symmetry Hence

0 (f) s n„ cosnf + sin til

n= l ,1

,S .

n,, = — Jr~ t,i0

cos nt ilt

_ £ [tr j^n 2 + 0.01

,- tr/tO

(-0 1 cosnf + nsinnf)

1

(n odd)

n 2 + 0.01

o

(01)n 3 + 0.01

107r{n'J + 001)

-*/10 -D =

(-0 . 1 )

0.0465

n 2 + 0.01

and

* Joe-,/l0

sin nf </f

-t/JO_ -(— 0.1 sinnf — n cosnt) (n odd)

ir|n 2 + 0.01 Jo

2n:(e

-w/10

“ (n 2 + 0.01)

1 (a): To = 4. wo = tr/2. Also Do = 0 {by inspection)

- 1 )=

1.461n

n 2 +0,01

- 5J-£ ^r-^W, dt = |n< > 1

(b) To = 10*. wo = 2tr/10fr = 1/5

g(,)= D„r J S\ where D„ =T^t /

* ^ <H =lint

(~ 2j S'" t)" "

12

-9 1

-J I Up

Fig. S2.9-1

(c>

</(i)~ Do + ^2 D”r'" t

- Where, by inspection D0 = 0.5

r»= -ti

Dn = — r = T^-' sothat l^-l = 2^r- and zDn = {l2ir JQ 2jt 2irn 2*71 l -i

(d) To = ir. wo = 2 and £>„ = 0

(e) 7'o = 3.^o —

.9(0

Therefore

= where On =5^

rft = 4^ C 3+ *)

l

\

lOnl - ,,.-2

_3_

4w 2 n :j

*sin 222 \

/') -L4lr3n2

-1 i( 1

cos aS1" -t \

/7T71

-2COS-3- and iDn — tan| , r 2ftn ,

2wnsin 25a - 1 /

V2 + ~9~ 3 3 ^COS -5- + 3

(f) To = 6. *>0 = *73 Do = 0.5

Dn = -o

*(t)® 0.5+ ^n*s-rtw

,1 ,1,

/-a 1 3 ( nit 2*n \

J <

1 + ay + l'-‘*•“

+ 1(-. + 2K^<"J = ^5('“T- C“~i

H • Cna -

t1 - _U0 15 8 0-

Co)

|J>*»

TV5t *

i_i_— ..-*t i-

J m. CS «Z

-8 -3

-<s

t 5 s

n4 ajL>

2.9-2

n

1

1

)3 -Ml l"* n

« > - a? 1 ^

For a compact trigonometric form, all terms must

reason, we rewrite 9(f) as

uO i .**

Fig. S2.9-2

q( 1) = 3 cost + sin ^5t “ " 2cos

(8t ~

3 )

have cosine form and amplitudes must be positive. For this

9(f) = 3cosr + cos(5t-|-|)+2cos(8t- *)

as 3 cos I 4- cos ^5t - + 2 cos (b* - —)

Figure S2.9-2a shows amplitude and phase spectra.

“ p:_ 09 q_2a we plot the exponential spectra as shown in rig.

(b) By inspection of the trigonometric spectra in Fig. S2.9 2a, we plot in p

S2.9-2b. By inspection of exponential spectra in Fig. S2.9-2a, we

,(„ .y + +iy-*> * + ['*-*’ +

. 5,,. *(I,-*) r-« + (.->») + I.- + (j''*) + (''*)”

-jSl

14

2.9-3 (a)

g[t) = 2 + 2cos(2< - n) + cos(3t - |)

= 2-2 cos 21 + sin 3t

(b) The exponential spectra are shown in Fig. S2.9-3.

(c) By inspection of exponential spectra

o(t) = 2 + [„<*-> + + 5[e**'*’

= 2 + 2cos(2t - it) + cos ^3t -

+ r-r(«-f)]

(d) Observe that the two expressions (trigonometric and exponential Fourier series) are equivalen

Fig. S2.9-3

2.9-4

D„ - —rTo/2

.J-Ta /2f(t) cos riuiot dt

/To/3

To/2

/(f) sin nu/of dl

If n't) is even the second term on the right-hand side is zero because its integrand is an odd function of t.

H«n«. D. rl.ll iTZiJ" If ,m is odd. ,h= first tsm on the righ.-h.nd sid. is soro boco.se » -»«»*

is an odd function of f. Hence. D n is imaginary.

15

Chapter 3

3.1-1dt

If ,(» is an even function of .'*)-»-* *• « odd function of

<7(t)cos is an even function of t. and the first integral is twice the integral

when fl(f) is even

/,nu

(1)G(w) = 2 / fl(t)coswtdt

Jo

Similar argument shows that when g(t) is odd

CM = -2j f 9 (t)*in~>tdt(2)

If g{t) is also real (in addition to being even), the integral (1) is real. Moreover from (1)

G(-~) = 2

yg(t)cos*/tdt = GM

Hence CM is real and even function of Similar arguments can be used to prove the rest of the properties.

3 . 1-2

.(.) - ££ CM-"' - s /~ *-•

„ J- |C(j)!cos[-' + #,Mi'<-' + J j|GMI»n!.-M- #»(-)[ J-’

and therefore

9 (C) = — f |G'(w)j cos[u;t + 0*(<*>)] da?

* Jo

For 9 (c) = r-‘»(t), CM = Therefore |C(u»)| = 1/V^T^ and *„M = - tan_1 <?) Hence

3.1-3

Therefore

and

n ii— cos 1

— tan

Jo v'w 2 4- o? 1

GM = r aMc-’^dt*f ~r*j

G*M = r^dtj -~*j

G'(—->) = -*X311<

O)^4

16

3.1-4 (a)

G(u-) =£ r-atc-J*‘ di = jf e

_0w+a)‘ = j _ f,-(j-'+°)'r

jvj + a.

(b)

3.1-5 (a)

1 — e-(jus-a)T

jul — aG(^) = JT

dt = Jr.-

(}w -a)di =

4 _ 2f~ JU’ - 2r"j2w

G(-0 = i:Ar-'^dt +

J^2r

' Jwtdt =

(b)

G(u>) . J°

dt +J'VJ“‘ dt = ^ [cos^-r + wr sinwr - 1]

This result could also be derived by observing that g(t.) is an even function. Therefore from the result .n Prob.

3.1-1

3.1-6 (a)

2 f ,2 .

Gp) = - / f cos -it dt = s '

ry0Tu;

cos ujt + sin U>T - 1]

i r 2

"0 / .2,2

(wpt2 ~ 2)sin-upf + 2uiq/ cos-upt

ir1'

a.

r-n—

f

i

1

-2. -I

A_

czz 1

- 1 [o 1

2 o

Fig. S3.1-0

(b) The derivation can be simplified by observing that G(~) ran be expressed as a sum of two gate functions

GiU) and Gj(*j) as shown in Fig. S3. 1-6. Therefore

a{,) = h JjGiM +

G

^)Viut^ = i {//“" ,U+L *** * sin 2/ + sin t

trf

3.1-T (a)

s(0 = j_ r'\

2* J-*/

2

*/2

,/wl, ^ w/2{jf cos a; + sin^)..,/2

2tr ( 1 - ts )

1

5t(l - t2)

cos(?)

(b)

<?(0_ J_ f

0

G(^)r i~,tf/wU = — [ [ G{~) cos *Jtdu/ + j f G(w) Sin attdw

2 it J2ir j_y_^ d- wo

Because G(w) is even function, the second integral on the right-hand side vanishes. Also the integrand of the

first term is an even function. Therefore

17

.1-8 (a)

17(0 =

1 f"0 yj 1 rcostiv-MwsinU]= — / COS t.u/ cL) = -9

JT yo UÔ ltU>0 L 1 J

nât 2(cos wot + wot sin wot - M

rJ‘*,(t‘ t° ) duj

“'0

0

*rju/(t-ep)

(2n)j(t - to)

U/Q

-“'0

8inu>o(t -_to) = i±sinc[u,o(t _ <o) ]

rr(t - to) *

(b)

i r° /u'°

<?(t) = — Jjr dw + J

-jcJW< dw

1 JW «

0_ 1 r

jw.*'°

= 1 - COS wot

2tr/r

-ô 2trt* o

1.2-

1 Figure S3 2-1 shows the plots of various functions. The function in part (a) is a gate function centered at the

origin and of width 2. The function in part (b) can be expressed as A (*£75)This is a triangle pulse centered

at the origin and of width 100/3. The function in part (c) is a gate function rect(|) delayed by 10. In other

words it is a gate pulse centered at t = 10 and of width 8. The function in part (d) is a sine pulse centered

at the origin and the first zero occurring at Sf = if, that is at w = 5. The function in part (e) is a sine pulse

sincf*) delayed bv 10*. For the sine pulse sine(f ), the first zero occurs at f = ft. that is at u; - 5tr. Ther t 1

the function is a sine pulse centered at w = 10tt and its zeros spaced at intervals of Sir as shown >n the fig.

S3.2-le. The function in part (f) is a product of a gate pulse (centered at the origin) of width 10* and a sine

pulse (also centered at the origin) with zeros spaced at intervals of 5ir. This results in the sine pulse truncated

beyond the interval ±5tr (|t| > 5ir) as shown in Fig. f. .......

5.2-

2 The function rect (f - 5) is centered at t = 5. has a width of unity, and its value over this interval is u»U». Hence

-jutdt = --

i

4.3

—>5w

_ _Lir --'4 - 5*' _ r -Js 1

/*’om ‘L

(f)-"= sine

i

18

3.2-3

.9(0 = — /°

2* Jio-

10+w WwlrJ“‘ dw =

2* (,/w)

10+ir

10— *

1 rfl

j(io+w)t _ ^(to-ion

j 2ttw*•

llOl

— [2j sin —f)= sinc(irf)fi

,,0‘

j2?rw3.2-

4 Observe that

result.

3.2-

5 Observe that

1 + Sgn(t) = 2u(t). Adding pah* 7 and 12 in Table 3.1 and then dividing by 2 yields the desired

cos(^ + #) = I[^^> + r-^0,+,)]

_ ir>Vu'0 ' + Ie--,9e"JW0 ‘

2 2

Fourier transform of the above equation yields the desired result.

3.3-1 (a)

,. 1

iy(t) <=> irt>(ui) + —0(0

Clw)

Application of duality property yields

irAlt) + — *=* 2ttu(-w)

' V1•“l/ 2wj(-u>)

6(0

3h +^]

>'(-*’)

Application of Eq. (3.28) yields

£(-t) - — <=> «(•*')jtrt

But Mf ! is an even function, that is — MO- an<*

+ -4) <=> «(*')2‘ trr

(b)

cos wo( •*==* rr[A(«*; t wo) + ^(w wq)|^

5/C*) G?(u>)


ir[A(/ + wo) + *(f - wo)] •*==> 2tr cos (-wow) = 2 tt cos (wow)

O'(t) 2»0(-w)

Setting wo = T yields

(c)

k{1 + T) + f>(t - T) <=> 2 cos Tw

sin wof <=f

G(u/)

sin

s(t)


jn^t + Wo) - t>{t ~ wo)j <=> 2tr Sin(-wow) = -2tr sin(w0w)W ' v——

'

G(e) 2

Setting wo = T yields

19

b{l + T) - b(t - T) <=> 2j sin Tu>

3.3-2 Fig. (b) jji(f) = fl(-f) and

GjP) = G(-w) * - 1)

Lu

Fig. (c) fl2 (0 = g(t - 1) + 9i(* - 1)- Therefore

G,(-») = [Gp) + G 1 (-»)]e"J-’ = [Gp) + G( -u.)]e-

2-(cos»j + wsinu> — 1)

Fig. (d) fl2(0 = flU - 1) + .?»(' + J )

C4 (-) = CP)*"*' + G(-w)r*-

= -L(2 - 2 cos.*,] = sin2

|= sine

2

(f

)

Fig. (e) 94(0 = fl(* “ ?) + flit' + £)• and

G4W = GO)r-'u/S + G,(w)r^2

. . . rlu,/2

- juirr'" - 1] + " V

= 35'2^ sin

f]=sinc(^)

Fig. (f) 95(f) can be obtained in three steps: (i) time-expanding 9(0 b> a factor 2 00 then deia>ing '

-

JLl mi and multiplying it by 15 [we may int.reh.ng, the tmqu.nc. to, .up. W and (n)l. lb. «.« «P(time-expansion by a factor 2) yields

/ (5)2G(2w) = - J2^ " l >

Second step of time delay of 2 secs, yields

/ (42)—^ - .2-'”' - i)e-

~

The third step of multiplying the resulting signal by 15 yields

3.3-3 (a)

(t)rect Tsinc (!y)

rect(^)— (f)p±j»Ti2

20

and

GU) = TsincwT/ai

Of)>= 2jTsinc

,/4 . a /^T\= T Sm (“)

wT\ .utT

s,nT

(b) Prom Fig. S3.3-3b we verify that

<t(t) = sin f u{f) + sin((. - ir)«(t - tr)

Note that sin(t - ir)u(f - tr) is sin t»i(«) delayed by tr. Now. sintu(t) <=* - 1) - + 1 ))+ i--3 and

sin(t - - ir) «=» {^-[AU - 1) - *U + 1)1 +j _35^

p J

Therefore

guo = {Jim-3 - d - *(-» + D] +

Recall that g(.r)*(.r - to) = .«(.t0 )/>(t - to). Therefore *U ± 1)0 + c~ , ’r“') = 0- and

1

GU) =1 -^ 2

(1 + r‘ J *-v)

(c) From Fig. S3.3-3c we verify that

q{ 1) = cost [»/(t) - « (f - = cosfti(f) - cost t, (f - -)

But sinft - f )= - cost. Therefore

<t(0 = cost i/(0 + sin ^t -//

(t -

GU; = |(*U -!) + «*• + !))+ + - 1) " «(*• + »)) + 7T^2}

2

Also because jj(t)A(.t — .to) — j?(to)A(t to).

^•±l)r‘ J~/J = *U± Dr*'*'2 = ±j/>(*± 1)

Therefore

(d)

GU) = J*’

-j»w/22

1 - U/J

1 - W2 1 - -t2[j-*' + <

-JWU//2I

9(0 = r-“‘|«(0 - «(* - r)l= r - T )

= c-0,(t) - c-o7V 8 (, - T,

rr(f - T)

GU) = -1

-a7-j^T

jjj + o jui + a Jut + O(i-

-(o+.i^)T]

3.3-4 From time-shifting property

q(t ± T) GU)c ±J

Therefore

g(t + T) + s(t - T) <=> GUy-T + GU)'"^T = 2GU) cos utT

\\e can use this result to derive transforms of signals in Fig. P3.3-4.

(a) Heie p(t) is a gate pulse as shown in Fig S3.3-4a.

21

9(0 = rect <= 2sinc(u>)

Also T = 3. The signal in Fig. P3.3-4a is g(t + 3) + g(t - 3), and

g(t + 3) + g{t - 3) <=*• 4 sinc(u>) cos 3ui

(b) Here g(i

)

is a triangular pulse shown in Fig. S3.3-4b. From the Table 3.1 (pair 19)

.9(0 = A(i) <=> sine2

Also r = 3. The signal in Fig. P3.3-4b is g(t + 3) + g{t - 3), and

g(t + 3) + g(t - 3) 4-a 2 sine2

(|)cos3ut

Fig. S3.3-4

3.3-5 Frequency-shifting propeny states that

Therefore

g(t)r±, 'J°‘ *=f G(u> T- -^o)

q(1) sin - y-IfllOe'"0 ' + .9(0^“^') - jj\

G^ ~ -*) + GU ~ ^o)i

Time-siiifting property states that

git ± T) <==> G(-')ed

Theiefoie

Q(f - T) - g(t - T) <=> G(o>)cj'*T - C(-)r J~T = 2jG(w)sin»T

— ! ff (/ + r)-g(f-T)) <=> GV)sinTv12./

‘

The signal in Fig. P3.3-5 is fl(f + 3) - »(/ - 3) where

g(t) = rect «=> 2sinc(v)

Therefore

4. 3) _ 9 (/ - 3) <=> 2j[2sinc(;*>)sin3w] = 4j sinc(-) sin 3->

3.3-6 Fig. (a) The signal g(t) in this case is a triangle pulse A(£) (Fig. S3.3-6) multiplied by cos lOt.

9(0 = A cos 10/

Also from Table 3.1 (pair 19) A(£)— trsinc=(^) From the modulation property (3.35), it follows that

9 (/) = a(—)cosl0f*=* ^jsme r2

+ S)nc2 JJ

The Fourier transform in this case is a real function and we need only the amplitude spectrum in this case as

shown in Fig. S3.3-6a.

Fig. (b) The »(.) here i, .he ,™. H .he.igh.lh, Fig. (.) dd*jd by 2.. Fta» >i™ ehif.ing prep.r.y-

its Fourier transform is the same as in part (a) multiplied by <* €l€ °,e

22

The Fourier transform in this case is the same as that in part (a) multiplied by e This multiplying factor

represents a linear phase spectrum -2 trot. Thus we have an amplitude spectrum [same as in part fa)j as well as

a linear phase spectrum ZG(v) = -2*w as shown in Fig. S3.3-6b. the amplitude spectrum in this case as shown

in Fig P3.3-6b

Note: In the above solution, we first multiplied the triangle pulse A(y;) by cos 10/ and then delayed the tesult

by 27t. This means the signal in Fig. (h) is expressed as iM^j^cos 10(/ - 2*).

We could have interchanged the operation in this particular case, that is. the triangle pulse A(^) is first delayed

by 2r and then the result is multiplied by cos 10/. In this alternate procedure, the signal in Fig. (b) is expressed

as A ( “ T"— )cos 10/ ..

This interchange of operation is permissible here only because the sinusoid cos 10/ executes tntegial numbei o

cycles in the interval 2tr Because of this both the expressions are equivalent since cos 10(/ - 2tr) = cos 10/.

Fig. (c) In this case the signal is identical to that in Fig. b. except that the basic pulse is rect(^) tnstsad o.

a triangle pulse A(£). Now

rect(27)

<=> 2/r sinc(/rut)

Using the same argument as for part (b). we obtain

C(ut) = tr{sinc[tr(ut + 10)) + sinc[7r(w - 10)]}rJ ~

3.3-7 (a)

/tj — 4 \ /lu + 4 \G(-t) = rect (—2” j

+ rect ~)

Also

isinc(f) <=> rectj

Therefore

2g(t) = — sinc(/)cos4/

7T

(b)

OM . a (2ii) + A(^i)

Also

23

Therefore

isinc2 (t) <=> A

g(t) = — sine2(I) cos 4t

IT

3.3-8 From the frequency convolution property, we obtain

g2{t) <=> — G(-v) * C(u>)

The width property of convolution states that if t.'i(x) * cj(.t) = y(x), then the width ot y(x) is equal to the

sum of the widths of ri(.r) and C2 (r). Hence, the width of G(u>) * G(*>) is twice the width of G(ui). Repeated

application of this argument shows that the bandwidth of gn(t) is nB Hz (n times the bandwidth of g{t)).

3.3-9 (a)

(b)

<?(•") - r'^

,dt ^-^- C08 ^T] = J

Z sin2 {T)

T/2\g(t) = rect (^)—(^)

and

rect (1) «=> Tsinc|(f)

rect|

'/±T/2\

,T J

•t=> Tsinc

cu )= Tsinc

|

\rJwTI'

i - c'^T/2]

/uT\ .-tT

= 2jl sine 1 — 1 sin2

,/ 4 . 2— em— 9111

sjJ V 2 )

(c)

^ =r A(t + T) - 2#(t) + A(i - T)r/f

The Fourier transform of this equation yields

ju>G(uj) = c-'wT - 2 + r~i

‘-'T = —2[1 - cos wT] = -4sin2

Therefore

C(w)=^sin 2

(^)

3.3-10

A basic demodulator is shown in Fig. S3.3-10a. The product of the modulated signal 9(f)cos~o< with 2cos^0 f

yields

9 (/ ) cos u.‘ot X 2 cos

-

2,9(«)cas2-t0 t = 9 (/)[l + cos2^0 <| = <?(0 + 9(f)cos2*o<

The product contains the desired g(t) (whose spectrum is centered at a) = 0) and the unwanted signal g(t)caa 2wo*

with spectrum i[C(^ + 2^o!-i'C(a;-2^o!. which is centered at ±2u>0 . The two spectra are nonoverlapping because

24

Fig. S3.4-1

-&zr

• Itdir

tvTT to-*

CrJc^^dO

MlW CO .

3.4-1

If < _’o (See Fig S3.3-l0b). We can suppress the unwanted signal by passing the product through a lowpas„

filter as shown in Fig S3.3-10a

O’i(w) = sinctjgggj) and Gi(u>) - 1

Figure S3. 4-1 shows G\(~). H\(~) and Now

Yib:)»GiMHi(u)

Vj(ui) = Gj(w)//3(ui)

The spectra YiM and V2P) are also shown in Fig. S3.4-1. Because y(t) « wt(*)va(0. t»» frequency conv°lution

property yieldsY<» = *(-/)•*(*)- From the width property ofconvo^ he band id

y(j) is the sum of bandwidths of and Y*J). Because the bandwidths of Vi(tf) and YaM are 10 kHz.

kHz. respectively, the bandwidth of Y(j0 is 15 kHz.

3.5-1*(<*>)«

Using pair 22 (Table 3.1) and time-shifting property, we get

MO = 1r-0-to)a /«fc

y/Titk

This is noncausal Hence the filter is unrealizable. Also

|In \H (u-')ll

J + 1

kj1

J1 + 1

dui = oc

25

Figure S3.5-1

H— -H. «.» isnoncausal .... .h^tor.— Sj»« ,,(<_)» •_*-» '”"'*£as shown in the adjacent figure. Choosing to — 3v2fc, n(0)

to = 3v^f is a reasonable choice to make the filter approximately realizable.

3 -5'22 x 10

8->wi0

~u,2 + lO 10

^

From pair 3. Table 3.1 and time-shifting property, we get

,,(,) = e-10i|-‘o1

The impulse response is noncausal. and the filter is unrealizable.

t.

The exponential delays to 1.8% at 4 times constants. Hence to - 4/a = 4 x 105 - 40/is is a reasonable choice

to make this filtei approximately realizable

3.5-3 From the results in Example 3 16

\HM\

Also H 10) = 1. Hence if is the frequency where the amplitude response drops to 0.95. then

10s

|tf(u>l)| = = 0.95= Wi = 328.684

y/u? + 10‘ 2

Moreover, the time delay is given by (see Example 3.16)

. r<i (Q) = i = 10' 6

w J + a2

If is the frequency where the time delay drops to 0.98% of its value at w - 0. then

= J2l°io iS

= 0,98 x 10-6 W3 = 142,857

We select the smaller ofw, and that is u, = 142,857, where both the specifications are satisfied. This yields

a frequency of 22.736.4 Hz.

3.5-4 There is a typo in this example. The time delay tolerance should be' 1 of^ sj 01 * 10». Let

Tire band of Aj = 2000 centered at « = 10s represents the frequency range from 0.99 x 10

us consider the gains and the time delays at the band edges. From Example 3.16

!«(-•)! = .nr rf ~ ZPTJ a ~ 10

At the edges of the band

26

l03

IHI0.99 » 10®)! - ~ 10

1

« 10~1

’ “d ‘“ 'I=

The s.i„ variation ove, the band i. only 1 .»%. SWMr. »•M «" «">* “"* “ 'h* “

= 9.901 x 10'

,MM « 10®) = ">'i “(1 01 - 10‘> *

The time deley variation over the bend ii <1%. Hence the ttenetniseton rnny

^ find

= 0.01 alt - 107)

Fig. S3.6-1

3.6-1

rU) = G(w)rect (~) <

- j{wfo+*c »m wT> )

*G(„)rect (^) U-./1: sin ^TK

This follows from the fart that r1 * 1 + t when * < 1- Moreover. Gp)rert («*g) = GU) hecause GU) i*

bandlimited to B Hz. Hence

YU) = nU)r~ J*t0 -jkGU) si" +-Tr-J“'°

Moreover, we can show that (see Ptob. 3.3-5)

lr\g(t + T) - g(t - T)] «=> GU)

Hence

I/(f) = g(t - t0 ) + |l»(t - to - T) - g(t ~*o + T)}

Figure S3.6-1 shows g(t) and ?/(/)

3.C.2 Rectth that the ttanafe, function of an ideal time deiay of T seconds is Hence, the uansf.r functton o,

the equalizer in Fig. P3.6-2 is

-jwil .-;2».At , -

HvfU) = °o + «i*3 +«!'• +• +a„r

Ideally, we require the equalizer to have

[h1‘

l

W) ] desired

1

1 + c,r

1 — or'

,wAt

J-Af ,2 -jZwAt

+ Q r.1 -jSutAI

a c + ••• + ••

The equalizer in Fig. P3.6-2 approximates this expression if we select a0 1- a, a. «2 a .

(-1)"q u.

27

3.7-1

Letting i = ^5and consequently dt =

1 » r -x1/!, ^ - -1—

E“=

27^75 J-J 2^2** 2(ty/H

Also from pair 22 (Table 3.1)

G(u>) = c

Letting <tu? = and consequently du/ = j^dx

Viir 1

3.7-2 Consider a signal

1 1 r -.VJ, V2tr L_^=2^772 L'

T ~2wi7\/2

<j(f) = sinc(A't) and G(w) = rect(2^)

= Jsine

2(ki)dt = -^ J [

r6Ct(2*-)]

d.j

3.7-3 Recall that

Therefore

flj(0_ _L G 2(~V

W< andJ

r, . (»)»<<)*- 5;/_»(<) [/^ox “

«/ —X -»

1

dt dui -*JGl(-w)C?2(^')^

Interchanging the roles of *(«) and «(#) in the above development, we can show that

P <n(t)92(t)dt = ~ JGip)G2(-w)dv

.. u • 17.1 if we identify 01U) = sinc(2irBf - rntr) and .9a(«)

3.7-4 In the generalized Parseval s theorem in

sine (2irBt - nit), then

Gi(a' )=jB

rect and C2(w) =2B

rect(475 )

r""

Therefore

Jgi(i)<l2 (t)di = ^ [

rect(4*5)]

But rect (^g) = 1 for kl < 2tr B. and is 0 otherwise. Hence

2,

r 20 rfw

28

1r2wB f 0

In evaluating the integral, we used the fact that r±J2wk - 1 when k is an integer

3.7-5 Application of duality property [Eq. (3.24)] to pair 3 (Table 3.1) yields

n ^ r»

n — m

2a

t* + a*2trr

The signal energy is given by

Eg = - r ^ = 4* f e' 2aw^ ^" dO do

The energy contained within the band (0 to W )is

fwEw = 4ir / r-

2au<hj = — [1 -

doa

If Eu = 0.99Ea . then

—2aW >

_ 0 012.3025 . ,

0.366 uIV = — rad/s Hz

a a

3-7-6, f 9

2(f)^ 4^.) then the outpul rU.) = AU)NM. where Jf (-») is the lowpass filter transfer function (Fig.

S3. 7-6). Because this filter band A/ 0. we may express it as an impulse function of area 47rA/. Thus.

H(*>) a [4 tt A/]A(—’) and >'(.*>) « (4tt.4p)A/]ft(w) *= [4trA(0)A/;M~)

Here we used the property q(r)hlr) = g(0)*(.r) [Eq. (1.23a)]. This yields

»/(f) = 2A(0)A/

Next because q2(t) <==> A(v). we have

Ap) = J <?

2(f)c 1*>t dt so that A(0) = r ,•(«>*-

J -flw

Hence. »/(<) = 2E9 A/.

40.f

«5-»-

Fig. S3.7-6

3.8-1 Let 9 (f) = r/i(f) + <?2 (0- Then

where

*«(r)1 /

,T/2

= lim — / [ffi(i) + ,92 (t)ji.9 i(f +t) +ff2(t + r)]d(

r— 7 J_ 77 2

= 7£9) (t) 4- ^siys(r ) 4" (T )

^ry(T)1 A

772

lim - /T(t)j/(t + r)dt

T-*oti i J - 772

29

3 .8-

lf we let g\(1) = Ci cospit + #i) and <?a(0 = C3 cospat + d2 ). then

/T/2

C,Ca cosPit + #1 ) cosPa* +^ dt

T/3 _

According to the argument used in Example 2.2b. the integral on the right-hand side is zero 9 ,n (

Using the same argument, we have i(T )= 0. Therefore

Ha (r) = = % COSiJlT + ^2 COSU,2T

This result can be extended to a sum of any number of sinusoids as long as the frequency of each

distinct , hence, if

g (l) =£ C„ «*(*-** + #«) then H a (r) =£% C0S^°T

n — 1

Moreover, for go(0 = Co, 72bo(t )= ant^

/T/a

ngo9 l(r)= Urn i

/CoCicosP,t + wir + ei)rft = 0

T— >». i J -T/2

Thus, we can generalize the result as follows. If

and

<,(0 = Co +^ C„ cos(n~of + then 7Za (r) - Cl +^ 2

nx 1

Ok_i

Syp) = 2*Cfa(~) + | 5^_ n~‘0) + A(U; + nW0)

'

cos njjQT

figure S3 8-2a show, the waveforms r„> and - r, for e < fa* U, T -NT, 0. .ho “'v/2 pulses in ,h< waveform of dura, ion 7". The .... onto the product .<(<),« O » •'« '1-

shown in Fig. S3.8-2b. Therefore

. rT/a

H x [t)= lim = /r(f)r(f - r)rft

i j_7-/a

1 JV /T* \ WI _ JL'j T < ^= "2 U Tj 2

2 \2 Tt)1T

! < y

for a < w < n. there » no ove„.phew puis...-Wl - ° SfcBut on the average, only half pulses overlap. Hence. *«(T )

ycomponents, as shown in Fig.

transoms of„W - -M-

Hence

S.(w) = ^ aincJ

(~) + st(-0

where s,W is the Fourier Worm of the periodic triangle function, shown in Fig. S3.8-2d W, find the

exponential Fourier series for this periodic signal to be

w,(r) = Cnfinu’T'Jb = Y

n* — nu

Using Eq. (2.80). we find D„ = ^sinc 2(^). Hence, according to Eq. (3.41)

S2P) = | ^ sine2

(y)AP ~ =

l7T>= - -X.

30

3 .8-

(C) T7(t)

(b) x'(f)

-n

2 '*' _ _ / dul = —* Jo *

and y (t)

A(uJ - \)d*i = - andIt

= - f —i— dw = i

* Jo uJ +l 4

y { )” Jo S + 1 * ^0 2 2rr

» The ideal differentiator transfer function is j-u. Hence, the transfer function of the entire system is

HW = (7^)°'w) = j^T and

~ )= hj r6Ct

= rect (l)^TT du,= -/0 dTi**’** (1_

4)=0 °.06831

32

4.2-1 (i) For ni(t) = cos 1000'

V’dsb-sc^) rri(t) cos 10,000 / = cos 1000/ cos 10, OOOt

x [cos 9000t + cos 1 1 . 000'j

LSB USB

(ii) Foi mi(/) = '2 cos 1000/ + cos 2000/

^d=b « c (0 = ">(/) cos 10. OOOt = [2 cos lOOOt + cos 2000/] cos 10. OOOt

= cos 9000/ + cos 11. 000/ + [cos 8000/ + cos 12. 000/]

= Icos 9000/ + i cos 8000/] + [cos 11. OOOt + ^cos 12. OOOt]

1 2 .*

(iii) For mU) = cos 1000/ cos 3000/

^nsBtc (f) = m(/)cos 10.000/ = i [cos 2000f + cos 4000/jcos 10. OOOt

= - [cos 8000/ + cos 12. OOOt] + i[cos6000f + cos 14,000/]

2 2

= - [cos 8000 / + cos 6000tj + j [

cos 1 2. 000 / + cos 14 , 000/)

LSB USB

This information is summarized in a table below. Figure S4.2-1 shows various spectra.

Fig. S4.2-2

c

USB

Fig. S4.2-3

LSB frequency USB frequency|

9000 11.000|

9000

8000

8000 12,000|

6000 14.CKK)

9000 and 11,000

9000 and 11,000

8000 and 12.000

8000 and 12.000

6000 and 14,000

4.2-

2 The relevant plots are shown in Fig. S4 2-2.

4.2-

3 The relevant plots are shown in Fig. S4.2-3.

4.2-

4 (a) The signal at point b is

f/a (i) = rn(t) cos* *jc t

= 7/1 (f f— COSU/c^ "i" *7 C06 3-Vct

The term 2,„(0 cos~-c f is the desired modulated signal, whose spectrum is centered at ±u»c/ The remaining

,erm JtsdUw * <he unwanted^term, which JETS^"» H*"“-

*

h" ,>s“” "”k! “

desired with the output ^rn(t) cosu)ct-

(b) Figure S4.2-4 shows the spectra at points b and c.

(c) The minimum usable value of-* is 2trS in order to avoid spectral folding at dc.

(d)

in (t) cos2 wc t = [1 + cos 2a;c t|

= im(t)+ im(t) cos 2wc t

The signal at point b consists of the baseband signal |f»(0 and a modulated signal

carrier frequency 2we not the desired value *e . Both the components will be suppressed by the filter, whose

center center frequency is wt . Hence, this system will not do the desired job.

(e) The reader may verify that the identity for cosn^t contains a term cosu;ct when n is odd. This

when n is even. Hence, the system works for a carrier cosn

u-c t only when n is odd.

Fig. S4.2-4

4.2-5 Wc use the ring modulator shown in Fig. 4.6 with the carrier frequency fc = 100 kHz Pc - 200 t x 10a

). and

the output bandpass filter centered at fc = 300 kHz. The output MO is found in Eq. (4.«b) as

„,(t) = i fm(t) cos S>et - |m(r)cos 3<1-J + im(t)cos 5

w

et + •

•

]

The output bandpass filter suppresses all the terms except the one centered at 300 kHz (corresponding to the

carrier 3

i

c f). Hence, the filter output is

-4,/(*) = — m(r.) cos 3

3ir

This is the desired output km{t) cos uie t. with *• = -4/3tr.

4 2-6 The resistance of each diode is , ohms while conducting, and oo when off. When the carrier A cos^J is positive

ZSXLto* (during ,h. entire positive cycle), end »'h.n the c.rrie, Sjgm,dU ri ne the entire negative half cycle). Thus, during the positive half cycle, the voltage TfpW) aPPc* s ™2S3 theSo”* During the .eg.tive half cycle, the output volte,. » ««ro

;

Therefore, .he drod.s », es

a gate in the circuit that is basically a voltage divider with a gain 2R/(R + r). The output is therefore

onr0 (l) = = w(t)w(l )

n + r

The period of «>(*) is To = 2rr/o>c . Hence, from Eq. (2.75)

35

!!•(>) = i^

^COS Wc' - ^cos 3wC t + g

cos *]

The output r0 (t) is

<•0(0 - ^.OMO - [5 + I (“* - 5~^ ' + 5“• 5“‘' + -

)]

(a) If .e p.v, the output 0.( 1 ) through . b.ndb.nd filter (centered .< .,.). the Biter .uPPre*«£ "t„,(f)cos nor-ct for all n # 1, leaving only the modulated term ^^rn^cos^c m

(b) Th‘ seine

1

circuit c.n be ueed „ . demodul.tor It .. « . b»ep». Biter .. the output, in .hi.« the input

is <p(t) ~ Tn(l) cosset and the output is rn(t ).

4.2-7 From the results in Prob. 4.2-6. the output co(f) = hni{t) cos u/d, where k = ln the present caSe '

m(t) = sin(^r t + ft). Hence, the output is

ro(t) = ksm(wet + 6)co&ujc i = ^[sinfl + sin(2^ct + 0))

The lowpass filter suppresses the sinusoid of frequency 2u,c and transmits only the dc term * sine.

at G>>arf (E>

-<©K 5K f©K <51

s

Of©

Fig. S4.2-8

channel b.ndwidth mua, be a, Leaf 30.000 ,.d/s (Bom

'shows’lhe receiver to recover ...i(t) and m,«) from the received modulated signal.

4.2-9 (a) S4.2-9 shows the output signal spectrum >'(-’)•.

. hi h frequencies

b Observe that W) is the same as J»/(w) with the frequency^ £l«Scy -Act™™,

are shifted to lower freq.tencies and vice versa. Thus, the scrambler in F.g, P4.2-9 tnverts the frequency

36

Y^) s«^*nes«vf

fFig. S4.2-9

M- 1

-04;

^ 30 *

- /c -

Fig. S4.3-2

To get back the original spectrum J17U-). wo need to invert the spectrum V(*>) once again. This can be done by

passing the scrambled signal >/(/) through the same scrambler.

4.2-10 We use the ring modulator shown in Fig. 4.6 except, that the input is ,,,(/)cos(2*)10b

/ insr^d of rn(t). The

carrier frequency is 200 kHz [*, - (400»)10\]. and the output bandpass filter is centered at 400 kHz T -

output r,(t) is found in Eq. (4 7b) as

,,(t) = j»n i) eos(2?r)10S

f j't'olO = ^(Ocos^lO6

*[cos (400*)10’t - |

cos 3<400»)l0»t + \cos 5(400tr)l0\ + ]

The product of the terms" (-1/3) cos 3(400-r)10’t and (4/*)m(0cos(2rr)106

t yields the desired term

cos (800rr)10Y whose spectrum is centered at 400 kHz. It alone passes throug P

(centered at 400 kHz). All the other terms arc suppressed. The desired output is

iy(t) = rn(t) cos (8Q0n) 10* t

07T

4.3-1 QaU) — \A ‘r rn(f)] cosu-vt. Hence.

gh {t) = \A + rn(f)] COS2u>e <

= + m(t)j + + w(t)]cos2u>cf

The first term is a lowpass signal because its spectrum is centered at u; = 0. The lowpass filter allows this term

"p« to Tupp^J <1* JLd term. who* ,-c.rom i. «n«r«d .. Hoc. <ho ottptt of ,h, lowp«»

filter is

j,(t) = A + rn(t)

When this signal is passed through a dc block, the dc term A is suppressed yielding the output n.(f). Tins

shows that the system can demodulate AM signal regardless of the value of A. I h.s is a synchronous or col,

demodulation.

37

4.3-2

(a) ,i = 0.5

(b) /t = 1.0

(c) /t = 2.0

fd) /! = OC

l/ip _ 10=> A = 20T “

Atrip _ 10

=> A = 10A

rn p __10

=>.4 = 5

>4 Ar/ip 10

=> A = 0>1 A

This means that // = oo represents the DSB-SC case. Figure S4.3-2 shows various waveform

4.3-3 (a) According to Eq. (4.10a), the carrier amplitude is .4 = mp/,i = 10/0.8 = 12 8. The camei powei is

pr = ,42/2 = 78.125.

Fig. S4.3-3

(b) The sideband power is m 2 (t)/2. Because of symmetry of amplitude values e q_

-' J m(f)

I,, (, )mav be computed by averaging the signal energy over a quarter cycle only Over a quarter cv

be represented as w(t) = 40>/T0 (see Fig. S4.3-3) Hence.

»"*(0 L. [

T° 4

[12I12

dt = 33.34

To/4 J0L To J

The sideband power is

P - mli

H

= 16.67J » rt

The efficiency is

V ~P, 16.67

+_

78.125+ 16.67x 100 = 19.66%

4.3-4 From Fig S4.3-4 i, i. *« .« d.

for demodulating AM signal using envelope detector is .4 -r m(f) > 0 for al .

27Tt

Fig. S4.3-5

4 3-5 When an input to a DSB-Sc generator is »»(*), the corresponding output is m(t)^by^dding4

A+

", <“ the corresponding output will be \A + »«)] «**. This is precse.y theAMs.gn^ Thus, by adchng

a dc of value .4 to the baseband signall we can gyrate> AM usmg a DS' ^ ,

f wg ^ tw0ti,« rnnvprse is eenerallv not true. However, we can generate UbU-at> using a- 8

identical AM s.nfrston in > balanced scheme shown in F*. S4.3-S to cancsl not the earner component.

1.3-6 When an input to a DSB-SCWM is -«> <tnsw.t. the

input is \A r«(/)]cos-*f. the corresponding output will A + m(t). By blocking

output, we can demodulate the AM signal using a DSB-SC demodulator.. , .

(,s

,

The converse unfortunate*, is not true. This is because, when an mput to an AM demodulator rn ,

)j3S c

.

coni.po„di”5 output is |U» envelop, of ,.,(.)! Hence, unlew »«) > 0 for .H < » <* ••

demodulate DSB-SC signal using an AM demodulator.

Fig. S4.3-7

4.3-7 Observe that ,n2(1) = -4

5for all t. Hence, the time average of m 2

(t.) is also A 2. Thus

= A 2 _ ™ (0 = i_

The carrier amplitude is A = ,n p/„ = m, « A. Hence Pc = A2/2. The total power is P, = P, + P. - The

power efficiency is

„ = £ x 100 = x 100 = 0.5' Pt

A2

The AM signal for /i = 1 is shown in Fig. S4.3-7.

4.3-8 The signal r.t point a is [.4 + »,(t)|cos wct. The signal at point b is

, .42 + 2.4rn(t) + , „„„ <j ,

r(t) = 1.4 + m(0l cosset = § * l + COS ’

The low-pass filter suppresses the term containing cos 2»*,«. Hence, the signal at point, c is

A 2 + 2Am(t) + m 2(l) _ A2

«‘(0 = —2 2

14^ +

Usually. m(t)/A « 1 for most of the time. Only when ™(f) is near its peak, this condition is violated

the output at point d is

Hence.

39

»/(() sb — + Arn(l.)

do .era, V/J. yield,., the oa.pa, *10 From *» •*“' ” “*

A blocking capacitor will suppress

that the distortion component is m {t)/2.

4.4-1 In Fig. 4.14. when the carrier is cos |(Aw)t + A] or sin l(Aw)t + *]. we have

nit) = 2[ir«»(0 M.’el : + n»a(0 sinwctjcos [(wc + Aw)t + A)

= 2m, (0 cos we t cos [(we + Aw)t + »} + 2n«a(0 sinwctcos [(w« + Aw)t + «

= rn i (t>{oos [(Aw)f + *1 + cos [(2we + Aw)t + b]} + rn 2 (t){sin !(2wc + Aw)f + 1

Similarly

.Tj(f') = "'i(t){sin|(JJc + ie)/ - ») + sin |(id)< + #1) + ».<*><«• K'WX + ‘I

-“ + a*'H + 6,1

After nit) and x»(t) are passed through lowpass filter, the outputs are

= r»j(t) cos [(Aw)t + A) - «< 2(0 sin IM* + A]

m '

a (t) = mi(f)«in [(Aw)t + *1 + m 2 (f)cos |(Aw)f + b)

, . ,... ... . I However, to generate the SSB signals

4.5-1 To generate a DSB-SC signal from "<(<)• we tnultip y

™

(

.it£ 2cosw c f.

This also avoids the nuisance

of the same relative magnitude, it is convergent to multiply () We suppress the USB spectrum

of the fractions 1/2. and yields the DSB-SC spectrum A/(w--*)+J

*

lh. Spectrum. we suppress

(above w, and below -we )to obtain the LSB

S4.M a. b and c show the three

the LSB spectrum (between -we and wc )from the DSB-SC spectrum

(a)From Fig. a. we can express *u.(0 = cos900f and *«b (0 =^os ll0° t

_ 2 cos l300 ,

(b)From Fig. b. we can express „ t..d) = 2 cos 70°t + c°s^ and , B (- co

(c)From Fig. c. we can express ,lf.(0 - il«.400t + cos600t] and *VSB (0 - 3^ 140

4.5-2 d UJ <t) - Tri(t)cosWet + m„(f)sinw c t

V^vsbIO = rr)(t)cosw e t - rr/(,(f)sinwe < and ^lsb! > v

° c«X, + .7. WO, 1000/ - cos£1000 - 100)/ —900,

,00, ,06 1000/ - .in 100/ .in 1000, . co.<1000 + 100), = =» 1 100/

(b) ,„(/) . co. 100/ .2 cos 30® and m»(<) = "in 100/ + 2.in300/. H.nc,

. (co. 100, + 2=0.300,) co. 1000, + (.in 100, + 2.1.300,) .in 1000, - co.900, + 2=0.700,

,.(„ - (cos 100, + 2CO. 300,) cca 1000, - (.in 100, + 2.1.300,) .1.1000, - co. 1100, + 2co.,300,

(cZ, - co. 100, co. 300, = 0.3 cos .00, + 0.5 cos 600, and -.(,) - 0 .5 .in.00, + 0.5.i»600,. Hence.

,„.(„ . (0.3 cos 400, + 0.3 co. 600,) co.1000, + (03.in.00. + 0.3.in600„ .in ,000, - 0 3CO..00, +0.3c«600,

dlia (,) = (0.5 cos 400, + 0.5CO6600,) cos 1000, - (0.3 sin 400/ + 0.5 .in 600,) .in 1000, — 0.5 cos 1.00, + O.Scos 1600,

40

A«u # 0 in SSB

Aw «u—-w

Fig. S4.5-5

f m and Fic S4 5-3b shows the corresponding DSB-SC spectrum

4.5-3 (a) Figure S4.V3a shows ihe spectrum of m(t) and Fig.

'2 hi 1 1 ' cos <OrO®C>Trt

obtained by suppressing the USB spectrum

(hi Figure S4.5-3C shows -he corresponding ^ by suppressing the LSB spectrum.

(c) Figure S4.5-3d shows the corresponding ^SB spec r

from Table 3.1 (pan 16) and the

We now find the inverse Fourier transforms of the LSB and U.B sp

fret|uency shifting property as

^ (/) s* 1000 sine {lOOOtri) cos 9000ttf

Vico (/)« 1000 sine (lOOOtrf) cos 11. OOOtrf

4.5.4 ® (“> "* " * B"b*rt ,ra"Sf°"”'r “

H{*i) — -j sgn (•*>)

„ „ apply a. .h. inpp. of .1. Hilb.,. W <*» °f *te ” < ') "

VP) = - l-jJl/U) sgn Mll-JSgn (-»)]= -

f f (t) is —m(f) To show that the energies of »«(0 and '" h( ai

This shows that the Hilbert transform of »u(0 »

equal, we have

£— = r™*(»*-5r£.i*,MI’*'

»/ - *Xi

42

4.5-5 The incoming SSB signal at the receiver is given by [Eq. (4.17b)]

flsb(0 = m(<) cosu/cf + ma(<)sin u

>

ct

Let the local carrier be cos [(wc + Au>)t + A], The prodeuct of the incoming signal and the local carrier is e d (t)

given by

ra(t) = *Lsa (f) [(^e+ Aw)« + A]

= 2[t»(t)coswjc « + ma(t) sin ta»ct] cos [(we + Au>)t + A]

The lowpass filter suppresses the sum frequency component centered at the frequency (2ujc + Au/). and passes

only the difference frequency component centered at the frequency Au>. Hence, the filter output c0 ( )is given

by

c0 (f) = m(f)cos(Au>)< + A) - tnft(t)sin(Aui)t + A)

Observe that if both Au and A are zero, the output is given by

p.0 (t) = m(t)

as expected. If only A = 0. then the output is given by

ro(l )= m(f-) cos(Aut)f - rru(t)sin(Aui)t

This is an USB signal corresponding to a carrier frequency Au as shown in Fig. S4.5-5b^ This spectrum is the

same as the spectrum MU) with each frequency component shifted by a frequency Au This changes the soun

of an audio signal slightly. For voice signals, the frequency shift within ±20 Hz is considered tolerable. Mos

svstems. however.lestrict the shift to ±2 Hz.

(b) When only A-u = 0. the lowpass filter output is

r0 {f) = m(t) cos A - rris(f) sin A

We now show that this is a phase distortion, where each frequency component of MU) is shifted in phase by

amount A. The Fourier transform of this equation yields

EoU) = MU) cos A - MhU) sin A

But from Eq. (4.14b)

MhU) = - jsgn U)MU

)

-{*—jMU)MU)

w > 0

it! < 0

and

( MU)***"

{ MU)f~J *

u> > 0

u; < 0

It follows that the amplitude spectrum of co(0 is MU)- the same as that for w(0- But the phase of each

component is shifted bv A. Phase distortion generally is not a serious problem with voice signals because h

human ear is somewhat insensitive to phase distortion. Such distortion may change the quality of speech, but

the voice is still intelligible. In video signals and data transmission, however, phase distortion may be intolerable.

4 .5-6 We showed in prob. 4.5-4 that the Hilbert transform of m h (t) is -rri(t). Hence, rn h (t) [instead of m(f)j is

applied at the input in Fig. 4.20. the USB output, is

tj(i) = nih(l) cos

—

rn(i') sin

= rn(t) cos [u;c l + ^ j+ r»/i(f) sin +

^

)

43

C-C.7p f

*• .

»""" —»

/1 . .

2-

-2.- < ± 2_ 4

f M*Fig. S4.6-1

Thus, if we apply m*(f) at the input of the Fig. 4.20. the USB output is an LSB signal corresponding to ».(#).

The carrier also acquires a phase shift ir/2. Similarly, we can show that if we apply n, K{t) at * he input of le

Fig. 4 20. the LSB output would be an USB signal corresponding to rn(t) (with a carrier phase shifted by it 12).

4.6*1 From Eq (4.20)

Wop) -1

P| < 2rBW,p + *lc) + W,p - «.

;

e )

Figure S4.6-U shows W.p - -»e )and H,p +«.)• Figure S4.6-lb shows the reciprocal, which is Hop).

4.8-

1 A station can be heard a. its allocated frequency 1500 kHz as well as at its image frequency. The frequencies

are 2hr Hz apart. In the present case. /IF = 455 kHz. Hence, the image frequency is 2 x 455 - 910 kHz apait

Therefore, the station will also be heard if the receiver is tuned to frequency 1500-910 _ 590 kHz f°

this is as follows. When the receiver is tuned to 590 kHz. the local oscillator frequency « /*> « 590 + 455 - 045

kHz. Now this frequency /lo is multiplied with the incoming signal of frequency fc - 1500 kHz. The ou p

vields the two modulated signals whose carrier frequencies are the sum and difference frequencies, w ic

«

-

1500 -c 1045 = 2545 kHz and 1500 - 1045 = 455 kHz. The sum earner is suppressed, but the difference carnei

passes through, and the station is received.

4.8-

2 The local oscillator generates frequencies in the range 1+8=9 MHz to 30+8-38 MHr "’hen

is 10MHz Jvo = 10 + 8 = 18 MHz. Now. if there is a station at 18 + 8 = 26 MHz. it will beat (nux) with

Ao I ,8 MHz to produce two signals centered at 26 + 18 = 44 MHz and at 26 - 18 = 8 MH*. The sum

component is suppressed by the IF filter, but the difference component, which is centered at 8 MHz. passes

through the IF filter.

(P

44

Chapter 5

Fig. S5.1-1

5.1-

1 In this case fc = 10 MHz. m p = 1 and m'p = 8000.

For FM :

A/ = k,w P!2* = 2tr X ]05/2tr = 10

sHz. Also fc = 10

T. Hence. (/,)-»« = 10 + 10 = 10.1 MHz. and

(

~

107 - 10

s = 9.9 MHz. The carrier frequency

increases linearly from 9.9 MHz to 10.1 MHz over a quarter (rising) cycle of duration n seconds. For the next a

seconds when w(f) = 1. the carrier frequency remains at 10.1 MHz. Over the next quarter (the falling) c>c e o

duration «. the carrier frequency decreases linearly from 10.1 MHz to 9.9 MHz., and over the last qua.tei cycle

when ,„(f) = -1. the carrier frequency remains at 9.9 MHz. This cycles repeats periodically with the period 4a

seconds as shown in Fig. S5 1-la.

For PM : , , „ „ w„A r _ n- = 50- x 8000/2* = 2 x 10

s Hz. Also fc = 107

. Hence. (/.)m« =10 +2x10 - J0.2 MHz.

and l f )m = 107 - 2 x 10

s = 9.8 MHz. Figure S5.1-lb shows m(t). We conclude that the frequency remains at

10 2 MHz over the (rising) quarter cycle, where rri(t) = 8000. For the next a seconds, m(f) - 0. and the carnet

frequency remains at 10 MHz. Over the next a seconds, where m(t) = -8000. the carrier frequency rem ,

9.8 MHz. Over the last quarter cycle rh(t) = 0 again, and the carrier frequency remains at 10 MHz. This cycles

repeats periodically with the period 4a seconds as shown in Fig. S5.1-1.

5.1-

2 In this case fc = 1 MHz. w p = 1 and vi'p = 2000.

Af = kttnJ2it = 2O,000t/2 tr = 104Hz. Also fc = 1 MHz. Hence, (/.)m« = 10* +- 10

4 = 101 MHz. and

(/ )min = 106 - 10

4 = 0.99 MHz. The carrier frequency rises linearly from 0.99 MHz to 1.01 MHz over the cycle

(oveTthe interval < t < ^). Then instantaneously, the carrier frequency falls to 0^ MHz and ^arts

rising linearly to 10.01MHz over the next cycle. The cycle repeats periodically with penod 10 as sho»n in

Fig. S5.1-2a.

For PM : ,

Here, because m{t) has jump discontinuities, we shall use a direct approach. For convenience, we select the

origin for *.(f) as shown in Fig. S5.1-2. Over the interval to if. we can express the message s.gnal as

rn (/) = 2000f. Hence.

^(t) = cos[2tr(10)6t+|m(0]

— ccs 2Jt(l0)*t +- |2000tj

r--- cos [2tr(10)6

f +- lOOOirt] = cos [2rr (10s

+- 500) f]

At the discontinuity, the amount

the carrier frequency is constant

of jump is u, d = 2. Hence, the phase discontinuity is kpm d = Therefore,

throughout at 10® +- 500 Hz. But. at. the points of discontinuities, there is a

45

5.1-3

5.2-1

5.2-

5.2-:

5.2-

5.2-

Fig. S5.1-2

phase discontinuity of * radians as shown in Fig. S5.1-2b. In this case, we

is a discontinuity of the amount 2. For k

,

> *, the phase discont.nu.ty

ambiguity in demodulation.

must maintain kr < ir because there

will be higher than 2* giving rise to

(a) *pm(0 = A cos [u».f + V«(0) = 10 cosilO, 000/ + kpm(t))

W, „e .iv„, rh.r »,„(.) - 10 a, (13.0000 with A, - WOO. Cb»rlj. m(.) - 3. over ,he inter*! HI < 1-

(b) r'FM(0 = -4 COS ^ + k,f rr.(a) da = 10 cos 10.000/ + kj

Jt»(a) da

j

Therefore k, j‘ rn(a)da = 1000Jm(a) do = 3000/

3/ = Jm(a) doHence rn(/.) = 3

In this case kf = lOOOtr and *> — 1. For

...(,) = 2 cos 100' + 18 cos 2000-1 end ”<(!) “ —200 sin 1001 - 36.000* sin 2000-1

Therefore . 20 end - 36.000. + 200. Abo the braebend IpAl b.od.ld.h B - 2000./2. - 1 kHz.

Fz PM ;

'“> “ 38 06366

kHz.

V,EM (,) = 10 cos(+c / + o.l sin 20007:/). Here, the baseband signal bandwidth 2? = 20007r/27t = 1000 Hz. - so,

U7,(t) = u>c + 200/T cos 2000tr/

Therefore. A- = 200 tt and A/ = 100 Hz and Bem « 2(A/ + B) « 2(100 + 1000) = 2.2 kHz.

- It

)

= 5 cos(wc / + 20 sin 1000*/ + 10 sin 2000 trt).

Here, the baseband signal bandwidth B = 2000/7/2* = 1000 Hz. Also,

u.\(/) = uic + 20, 000/r cos 1000/r/. + 20,000* cos 2000*/

Therefore. A+ = 20,000*+20.000* = 40,000* and A/ = 20 kHz and Bem = 2(A/+B) = 2(20. 000+ 1000)

kHz.

I The baseband signal bandwidth B = 3 x 1000 = 3000 Hz._ , „

For FM : A/ = = iS^7

1 = 15-951 kHz !lnd Bfw = + S) ~ 37 831 kH

For PM • A/ =kp™'r = = 31831 kHz and Bpm = 2(A/ + B) = 66 662 kHz.

i The baseband signal bandwidth B — 5 x 1000 = 5000 Hz’

2/2 + 51-14 kHzFor FM : A/ =

k)

^r ~ 200^— = 1 kHz and Bfm = 2(A/ + B) = 2(2 + 5) - 14 kHz.

46

For PM : To find BPM. we observe from Fig. S5.1-2 that vFM (t) is essentially a sequence of sinusoidal pulses

of width T = lO'1 seconds and of frequency fr. = 1 MHz. Such a pulse and its spectrum are depicted in Figs.

3.22c and d. respectively. The bandwidth of the pulse, as seen from Fig. 3.22d, is 4ir/T rad/s or 2/T Hz Hence.

Bpm = 2 kHz.

5.2-6 (a) For FM : A/ = = 22±™£z*± = 100 kHz and the baseband signal bandwidth B -—Therefore

Bfm = 2(A/ + B) = 202 kHz

For PM : A/ =^ = = 10 kHz and BPM = 2(A/ + B) = 2(10+1) = 22 kHz.

(b) m{t) = 2 sin 2000* r. and B = 2000?r/2tr = 1 kHz. Also mp = 2 and m'p = 4000tt.

For FM : A/ =^ - *»-<fps2 = 200 kHz. and

Bfm = 2(A/ + B) = 2(200 + 1) = 402 kHz

For PM : A/ = ^# = ^ig2te = 20 kHz and BPM = 2(A/ + B) = 2(20 + 1) = 42 kHz.

(c) m(t.) = sin 4000*/, and B = 4000tr/2?r = 2 kHz. Also w p = 1 and m'p = 4000tr.

For FM : A/ = = 2°?-'^L— = 100 kHz, and

Bfm = 2(A/ + B) = 2(100 + 2) = 204 kHz

For PM A / = = iP-jW = 20 kHz and BPm = 2(A/ + B) - 2(20 + 2) = 44 kHz.

(d) Doubling the amplitude of in It) roughly doubles the bandwidth of both FM and PM. Doubling the frequency

of /n(fi [expanding the spectrum A/(ui) by a factor 2] has hardly any effect on the FM bandwidth. However, it

roughly doubles the bandwidth of PM. indicating that PM spectrum is sensitive to the shape of the baseband

spectrum. FM spectrum is relatively insensitive to the nature of the spectrum Af(~)-

5 .2-7 From pair 22(Table 3.1). we obtain c' 1 ’ «-» The spectrum A/(ui) = a

which decays rapidly. Its 3 dB bandwidth is 1.178 rad/s=0.187 Hz. This is an extremely small bandwidth

compared to A/.3

Also Mt) = -2fr-*'». The spectrum of m(f) is A/V) = J*A/(->) = jy/*»r* >4. This spectrum also decays

rapidly away from the origin, and its bandwidth can also be assumed to be negligible compared to A/.

For FM: A/ = = 222&pl « 3 kHz and Bfm ft 2A/ = 2 x 3 = 6 kHz.

For PM : To find iii'p . we set the derivative of rri(f) •= -2fc"‘ /s equal to zero. This yields

,,,(/) = -2c-,S/a

+ 4/.ac-

,S/s = 0 t =V2

and rrij, = = 0-858. and

_ tgt - sooo^.Q.sss _ 3 432 kHz and BPW ft 2(A/) = 2(3.432) = 6.864 kHz.

5.3-1 The block diagram of the design is shown in Fig. S5.3-1.

X 1%s\Bf f

£= 1-&35N ’

/\C loRfitlUX 60

-iSL50H*T

£- q*ifi

(0 10-245 M«e

Fig. S5.3-1

47

5 .3-2 The block diagram of the design is shown in Fig. S5.3-2.

(a) ^pm(I) = ^ cos [u'ct + kPm{t)}

When this v’pnUO *s passed through an idea) FM demodulator, the output is kpfn(t) This signal. \ h " pas

lough an tdea! iterator, yields 1^(0- Hence. FM demodulator followed by an .dea

demodulator. However, if »(*) has a discontinuity. m(t) = oo at the po,nt(s) of d.scont.nutty, and the system

will fail.

(b) i£Fm(0 = cos u#'c^ *4" lc f Jm(a)do

5 .4

-

2

5 .4

-

3

When this signal „m(0 is passed through an ideal PM demodulator, the output is*,£m(o) **

signal is passed through an ideal differentiator, the output is k/m(t). Hence. PM demodula . >

ideal differentiator, acts as FM demodulator regardless of whether m(t) has jump discontinuities

Figure S5.4-2 shows the waveforms at points b. c. d. and e. The figure is self explanatory.

From Eq (5.30). the Laplace transform of the phase error MO

>

s K‘ven b>'

+ 7i0e ' (s)

For (9,(r) = kt2

. ©,(*) = and

, , ,2*

0<(s) “SVS + AKH(*)\

The steady-state phase error [Eq. (5 33)] is

2khm (9,(t) = Jim »©.(*) = + AK)

Hence, the incoming signal cannot be tracked. If

n + oH{s) = then ©«(*) =

= oo

2k

, 3 >]

and 2k Ik

t

ljm = Jim *©«(•-) = Jim + AK[s + ~) “ Ako

Hence, the incoming signal can be tracked within a constant phase 2k/Aka radians. Now, if

s* + n* + fc- - -

H(s) = then 0*( s)—

s 2 .i +—

and 2k.-

(

lim «,(/) = Jim «©«(*) - Jim ^ + AK^ + 0 , + /,)

0

1„ this case, the incoming signal can he tracked with zero phase error.

48

I*-TS—4*-%—>1

Chapter 6

0.1-2

0.1-1 The bandwidths of gi(t) and g3 (t) are 100 kHz and 150 kHz, respectively. Therefore the Nyquist sampling rates

for gi(t) is 200 kHz and for g3 {>) is 300 kHz., ... , 2,,-,

Also gi3 U) <=> ±qiU) * giM, and from the width property of convolution the bandwidth of ffi. W' >*

the bandwidth ofP 9,(0 and that of »*(*) » three times the bandwidth of g3 (t) (se also Prob. 4.3-10). Similar y

the bandwidth of ffl (/)9a (t) is the sum of the bandwidth of *,(#) and n (t). Therefore the Nyquist rate for fli (')

is 400 kHz. for <j23(f) is 900 kHz. for gi(t)g3 (t) is 500 kHz.

(a)

sinclOOtrf) <=> O.Olrect (3357)

1 he bandwidth of this signal is 100 7r rad/s or 50 Hz. The Nyquist rate is 100 Hz (samples/sec).

(b)

sinc2(100rrf) e*=> O.OlA(jj^)

The bandwidth of this signal is 200 tr rad/s or 100 Hz. The Nyquist rate is 200 Hz (samples/sec).

(c)

sinc(lOOrf) -t- sine (50irt) <==> O.Olrect 0.01 (jfc) + 0.02rect

The bandwidth of the first term on the right-hand side is 50 Hz and the second term is 25 Hz Clearly the

bandwidth of the composite signal is the higher of the two, that is. 100 Hz. The . vquist rate is

(samples/sec).

(d)

sinc(lOOTt) + 3sinc2(60trt)

The bandwidth of rect(^

0.01 rect( 555; ) + ^5 A(3^)

240* <is 50 Hz and that of A(rfb) is 60 Hz. The bandwidth of the sum is the higher of

the two. that is. 60 Hz. The Nyquist sampling rate is 120 Hz.

(•)

sinc(50trr) <=* 0.02 rect( 333^)

sinc(lOOirf) <=s 0.01 rectijg^-)

The two signals have bandwidths 25 Hz and 50 Hz respectively. The spectrum of the product of two signals is

1 /2n times the convolution of their spectra. From width property of the convolution the width of the convolu

signal is the sum of the widths of the signals convolved. Therefore, the bandwidth of tinc(50»f)amc(100irf) is

25 + 50 = 75 Hz. The Nyquist rate is 150 Hz.

6 1-3 The pulse train is a periodic signal with fundamental frequency 2B Hz. Hence, ui, — 2ir('2B) — AnB. The peiiod

is To =1/2B. It is an even function of t. Hence, the Fourier series for the pulse train can be expressed as

Using Eqs (2 72). we obtain

PT. (0 = Co +^ Cn cos ru.t

no

1 1/1601

4'

0/•1/1SB

2i( n7! \

/On = Cn = 7TTTo J

/ cos mj,t dt. =-1/16B

— sinT)7T [ 4 )

fcn =0

Hence.

g(t) = hOpt.V)

= ^ (f) +E^ sin (T) fl(,)cosW

50

-0.2

Fig. S6.1-4

For ,{*) = sinc2(5ir/) (Fig. Sfl.Ma). the spectra is GM = 0.2A(*) (Fig- S6.1-4b). The bandwidth of this

signal is 5 Hz (TO* rad/s). Consequently, the Nyquist rate is 10 Hz, that is. we must sample the s.gnal at a rate

no less than 10 samples/s. The Nyquist interval is T = 1/2B = 0T second.

Recall that the sampled signal spectrum consists of (1 /T)G(w) = ^ *(*) repeatmg penod.cally w.th a penod

equal to the sampling frequency /, Hz. We present this information in the following Table for three samphng

rates' /, = 5 Hz (undersampling). 10 Hz (Nyquist rate), and 20 Hz (oversamphng).

sampling frequency sampling interval T ~~±G(u)[

comments|

Thz 0.2 A (afe) I

Undersampling

iolH 0T 2A(^)|

Nyquist Rate" ]

20lh 0.05 4A(jfe) |

Oversampling-]

In the first case (undersampling), the sampling rate is 5 Hz (5 samples/aec) and the ^pea.s

every 5 Hz (10* rad/sec.). The successive spectra overlap, as shown in Fig. S6.1-4d_and the speciru O )

not recove, able from £(-•) that is. g(t) cannot be reconstructed from tts samples

sampled signal is passed through an ideal lowpass filter of bandwidth 5 Hz. the output spectrum

0

1

5 Hz 0.2

10 Hz 0.1

20 Hz 0.05

end the output signal '» lOsinc (20.1). which isnot'he <«•“! £ph3he Nyquist sampling t.te of 10 Hs ("*»-<•) T^f bfiiS“om 5(w) using a. Waal lowp~arepetitions of yGM repeating every 10 Hz. ence, (.

) Finally in the last case of oversampling

filter of bandwidth 5 Hz (Fig. S6.1-4f)_The output is,10s,nc Srt) - FmaU^m^^ (repeating every

(sampling rate 20 Hz), the spectrum CM consists of nonoveHapp g Pcan £ from ®(w)

2^ SKiSiJSi^ »

Thi.sd.Ls is analyzed fully in Problem 3 <-V“'here 30

kHz, 5 kHz. and 15 kHz, respectively. Hence, the Nyquist rates lor me

kHz. respectively._ ... _ q-y Th js acts ^ the input to

0 1-6 (a) When the input to this filter is *(<). the output of the summe ( )

the integrator. And, h(t). the output of the integrator is:

6.1-5

Hi) = J|/i(r) - b(r - T)] rfr = v{t) - «(# - T) = reel (^)

The impulse response /,(t) is shown in Fig. S6.1-6a.

(b) The transfer function of this circuit is:

and

ju>TJ2

Observe that the filter is a lowpass filter of

araa-output I fstaircase approximation of the input assU in F,g. S6.1-6c.

The amplitude response of the filter is shown in Fig. S6.1-6b

bandwidth 2ir/T rad/s or 1/T Hz.

t e*#

6 . 1-7 (a) Figure S6.1-7a shows the si§n.a* "co“,ruc* io

Jl1 ardre'^nîn^nstanf . ^he^hVightôf the* triangle is equal

roTlu-Lampie val ue* 'The'resuh hig^sfgn aTconsist s' of straight line segments joining the sample tops.

(b) The transfer function of this circuit is:/u>T\

//M = ^{^(0} = î^(or)| =T8inC \T) l . ,

2«ssgs

ITfiSmS' rfTS'SS? S S^Si'fcfo »u.d (Sizable). Such a delay would „«« -he

area' uudet'pM -P« - «• *«<*' “

, /'[„(.) - 2„(, - T) - u(r - 2T)|tlr = (..(1) - 2(1 - T)»(l - T) + (I - 2T)u{t -2T) - 4(—

)

shown in Fig S6.1-7b.

6 1-8 Assume a signal „(f) that is simultaneously timelimited and bandlimited. Let ff(w) = 0 for M > 2*B. Therefore

w (^.jicci( 4“'

g . ) = QU) for B' > B. Therefore from the time-convolution proper > (

q[i) = g(i) * i2B'sinc(2»rB'f)l

= 2 B'g(t) * sinc(2?rB't)

52

Figure S6.1-7

B— *(0 i. timelimiwi. .«) - 0 b>:|»| > T. B«|

,(0 k o(“ Sl'.l

is not timelimited. It is impossible to obtain a t.me-l.m.ted s.gnal from the convolution

with a non-timelimited signal.

fl.2-1 (a) Since 128 = 27

. we need 7 bits/character.

(b)For 100,000 characters/second . we need 700 kbit.s/second.

(a) 8 bits /character and 800 kbits/second.

6.2-2

6.2-3

(a) The bandwidth is 15 kHz. The Nyquist rate is 30 kHz.

(b) 65536 = 216

. so that 16 binary digits are needed to encode each sample.

(c) 30000 x 16 = 480000 bits/s.

(d) 44100 x 16 = 705600 bits/s.

(a) The Nyquist rate is 2 x 4.5 x 106 = 9 MHz. The actual sampling rate = 1.2 x 9 = 10.8 MHz.

(bl 1024 = 210

. so that 10 bits or binary pulses are needed to encode each samp' ' f . ^ run a /—

(c) 10.8 x 106

x 10 = 108 x 106or 108 Mbits/.-..

6.2-4 If rn p is the peak sample amplitude, then

,(0.2)(t« P ) _ ru

quantization error <jqq jq

Because the maximum quantization error is ^ — 4/

’r _ Til,

500

500

,it follows that

L - 500

f n , ; _ tto _ 9® This requires a 9-bit binary code per sample The

Because L should be a power of 2. we chooM L - 512 - 2 . 1 9 signal has 2400

Nyquist rate is 2 x 1000 = 2000 Hz. 20% ^ ' j™0^ ,L 9 x 2400 = 21.6 kbits/second.

sampies/second, and each sample is encoded yf * v 21 6 = 108 kBits/second data bits. Framing

Five such signals arex 0.005 = 540 bits, yielding a total of 108540

and synchronization requires additional 0.5% blts tnat

bits/second. The minimum transmission bandwidth is ™ "

6.2-5 Nyquist rate for each signal is 200 Hz.

The sampling rate /, = 2 x Nyquist rate = 400 Hz

Total number of samples for 10 signals = 400 x 10 = 4000 samples/second.

, 0 .25m, _ m.Quantization error < ,(/,

-4S0

Moreover, quantization error - 2 — ji l «oo, ,

Because L is a power of 2. we select L = 512 = 2*. that is, 9 bits/sample.

Therefore, the minimum bit rate = 9 x 4000 = 36 kbits/second.

Tlie minimum cable bandwidth is 36/2—18 kHz.

For a sinusoid, =?? = 0.5. The SNR = 47 dB =50119. From Eq. (6.16)

m.6.2-6

So _ 3L2'jfVl _ 3l 2(0.5) = 50119 =>1=182.8

No

Because L is a power of 2. we select I « 256 = 28

. The SNR for this value of L is

So _ a ) = 3(256)* (0.5) = 98304 = 49.43 dB

No

53

6.2-7 For this periodic tn(t). each quarter cycle takes onby averaging its energy

contributes identical energy. Consequently, we can compute the power^or^^^ = ^/Tq where A is

over a quarter cycle The equation of the first quar er eyesquared value (energy averaged over

the peak amplitude and To is the period of rn(f). The power or the mean squared

a quarter cycle) is

i fTa/A

( iA \* * - -L=(ft) 3

Hence. i1 — Jl

—:r /cici 9NR of 47 dB is a ratio of 50119- is

The rest of the solution is identical to that of Prob. 6.2-6. From Eq. (

So _ 2/2 T" = 3I

2(l/3) = 50119 => L = 223-87

A’o" "i?

B«,u„ I is , po«s, of 2. «. *1«< L - 25« - a'. The SNR for this v,h,e of L is

So _ ^L ' = 3(256)2(1 /3) = 65536 = 48.16 dB

So T"p

6.2-8 Here „ = 100 and the SNR = 45 dB= 31,622.77. From Eq- (6. 18)

So 31/31,622.77 /. = 473 83

So~

(In 101 )2

, „ i r — m 9 — 9® The SNR for this value of L is

Because L is a power of 2. we select L — ol2 —21So _ 3(512)

3

_ 3gg»>2 84 = 45.67 dBSo (In 101) ?

,. i ~ io v in6 '! = 3 x 106 Hz. Moreover. L = 256

6.2-9 (a) Nyquist rate = 2 x 106Hz. The actual sampling rate is 1.5 x (2 x 10 )

and ii = 255. From Eq. (6.18)

So 3I* = 3(256)2

^ 6394 = 33 Qg dBNo " |ln(/-+ l)!

s (In 256) 2

ZTvtZZ 7'r«d each sample is encoded by 8 bits (L = 256). Hence, the transmiss.on rate ,s

8 x 3 x 106 = 24 Mbits/second

Nyquist rate), then for the same transmission rate

™-— *- *• - io24 -,he

new SNR is

So 3L1 3(1024)2

_

% =[ln(,f"+~ (In 256)=

Clearly, the SNR is increased by more than 10 dB.

.= 102300 = 50.1 dB

Clearly, the arm is increase ^

6.2-10 Equation (6.23) shows that increasing n 20%.

by 12 dB (front 30 to 42) can be accomplished by increasing n trom iu

6.4-1 (a) From Eq. (6.33)<rf. thftt _ (ffKM^OOO) 0.0785

.4 mux =— 60 that 1 “2tr x 800

54

(b>(0.0785)

2(350Q) = n % 10

-4

(3)(64000)

(c) Here So = = 0.5. and

(d) For uniform distribution

2

So

No

0.5

1.12 x 10-= 4.46 x 10

So = =-jj

21 =^

, Soso that — =

0.333 = 2.94 x 103

No 1.12 x 10-4

(e) For on-off signaling with a bit rate 64 kHz, we need a bandwidth of 128 kHz. For a bipolar case, we need a

bandwidth of 64 kHz

Chapter 7

7.2-1 For full width rect pulse p(t) = rect

^j

For polar signaling [see Eq. (7.12)]

S

For on-off case [see Eq. (7.18b)]

P(eo) = Tb sine

-•(?)Sy{(o)^^-‘Tb smz

i = 0. Hence,

»*•(?)[•» _v(--¥l

But sinc2

^] = 0 for forall **0, and = 1 for n = 0.

Sy(o) *^smc2 +

For bipolar case [Eq. (7.20b)]

. . \P(*f . 2 (coTb \

5>) = LX-sm^ 2 J

= Tb sine2

sin2

The PSDs of the three cases are shown in Fig. S7.2-1. From these spectra, we find the bandwidths for all

three cases to be Rb Hz.

The bandwidths for the three cases, when half-width pulses are used, are as follows:

S^3bS&ssssss==bs=“bipolar case because the term .in

2 (^)i» S,(a.) daannin.. it. b»dwi*b.

56

From Fig. S7.2-2, it is clear that the bandwidth is— rad / s or 2Rb Hz.

7.2-3 For differential code (Fig. 7.17)

To compute K„we observe that there ate four possible 2-bit sequences 11. 00. 01 and 10. which are

equally likely. The product at,,oM for the fust two combinations is I and ts -1 for the last two

combinations. Hence,

Similarly, we can show that Rn = 0 n > 1 Hence,

%1-i (a) Fig. S7.2-4 shows the duobinary pulse train yfr) for the sequence 1110001 101001010.

(b) To compute Bo. we observe that on the average, half the pulses have o, =0and the mmaining half

have a* — 1 or —1. Hence,

*0= lim 4f(±l)2 +f(0)]4

To determine B| ,we need to compute0*0,. t

There.*,fourpossible equ^lIM>

11,10,01,00. Since bit 0 is encoded by no pulse(ui -0),the product of »»n».l - or

these sequences. This meanson dte avor*.Hcombin«ions have -Oand only - combinations

57

htVe nonzero akaM . Because of*0 duubuuuy rule. tb. bit *<.«»« .1« only be encoded by two

consecutive pulses of the same polarity (both positive or both negative).

This means o* endo**. are i and 1 or -I end -1 respectively. b-.—Wd-'-

these— combinations have akak+ \= 1. Therefore,

possible combinations of three bits m sequence.and/or the last bit 0.

-- --—--

combination 101 are of opposite polarity yielding akak^ .= -I. Thus on the average, ak *+2

— terms,-l for— terms, and 0 for~ terms. Hence,

m a similar way we can show that^ - 0 » > 1 ,and from Eq. (7.10c), we obtain

S,(«) = + cosair6 )=—jr" cos

{(o>Tb

'

l 2 J

yrO1 I 1 0 0 o 1 1 • * 9

mnu0 10»°

_JL_JL-n—

For half-width pulse P{t) = rect(2l / 7*)-

SJ,H = ^-sinc

2

^] cos2

(^)

From Fig. S7.2-4 we observe that the bandwidth is approximately Rb I 2 Hz.

7.3-1 From Eq. (7.32)

4000 =(l+r)6000 1

r ~3

58

7.3-

2 Quantization error S 0.0 \mp => L £ 1 002. J-t

•

j

(a) Because I is a power of 2, we select L * 128 = 2

(b) Ibis requires 7 bit code per sample. NWaisl«e=2 x2000.4lcHsfbrMchsign.l. lb. sanplm*

rate /, = 125 x 4000 = 5 kHZ.

Eight signals require 8 x 5000 * 40,000 samples/sec.

Bit rate * 40,000 x 7 = 280 kbits/s. From Eq. (7.32)

(\+r)Rb 12 x280x 103_

Bt2 ' 2

7.3-

3 (a) BT =2Rb =»Rb • IS kbits/s.

(b) Bt - => Rb = 3 kbits/s.

(c) Bt * Hence, 3000 =— Rb => Rb = 4J kbits/s.

(d) Bt = Rb => Rb = ‘S kbits/s.

(e) Bt = /?6 => - 3 kbits/s.

7.3-

4 (a) Comparison of P{co) with that in Fig. 7.12 shows that this P(eo) does satisfy the Nyquist criterion with

<ob = 2/r x 106 and r = 1. The excess bandwidth ax = n x 10

6.

(b) From Table 3. 1 ,we find

/>{f) = sinc2 (>rx!0

6/)

From part (a), we have cob = 2n x 106 and = 10

6. Hence, 7* = 10 . Observe that

p(n Tb)<*\ n * 0

«0 n*0

Hence /*(r) satisfies Eq. (7.36).

(c) the pulse transmission rate is— * Rb = 106

bits/s.

7.3-

5 In this case = 1 MHz. Hence, we can transmit data at a rate Rb - 2 MHz.

Also, Bt * 12 MHz. Hence, from Eq. (7.32)

12 x 106 - * 10&

)^ r = ®-2

7.3_6 /2 = 700 kHz. Also, -^- = 500 kHz and /, = 700 - 500 * 200 kHz.

HenCe,r-^i

.0.4 «.<l/, -500 - 20°. 30° Mb.

7.3-

7 To obtain the inverse transform of /(»),we derive the dual of Eq. (3.35) as follows:

g(r - r)o G(co)e'jTa>

and g(/ + T)o G(o>)eyT"

Hence,

g(i + T) + g(t - T)o 2G(a>) cos7©

Now, P{eo) in Eq. (7.34a) can be expressed as

*•>4 rea(^r)

+M^]“fe)59

(2)

7.3-8

7.3-9

Hence,

Using Pair 17 (Table 3.1) and Eq.(l) above, we obtain

P(t) - *fcsinc (2*Rbt)+&.smc [***(' + 2^]

+ ^-smc |2*^/ J

« /lysine (2/ri^/)+^sinc (2»Jty + >r)+^sinc (2ff^/-^)j

f sin(2jr/Zfe?) l sm(2xRbt + x).[;

1 M***4' -*) 1

=/?A

|_

2^^/+2 2xRbt+* 2 2nRbt-n J

* Rb sin(2nRbt)

= J?4 sin(2ff^t)j

1/2 1/2

(2ff*4f + ») (2* fy-*).

k —

2Rh cosx RbtimxRb t _ Rb cosx Rb t

sinc/ «

£

_ . _ 5 i AD.

2xRbt(\-4Rb2t2)\

inxRb t _ Rb cos* ^

2/rRt,t(\-4Rb2t2

)I-***

2'3

I,--""1*

Kt U»*w l

1 ( °> 1 1 Y

p(t) * sine (^^r) + sinc

sin/r/y(

sin(/r^-ff)

fffyf xRbt-x

s\nxRbt sinxRbt sin xR/,1

xRbt~ xRbt-*~ xRbA'-Rb*)

n,e Nyquist interval is Ts = -p - Tb . The Nyquist samples are p (±nTb )for n = 0, 1, 2,

RbFrom Eq. (7.16), it follows that

p{0) » p(Tb )= 1 and p{±nr0) = 0 for all other n.

Hence, from„ „ Rb 1

Eq. (6.10) with Ts = Tb , and B = -£ =—

.

p(r)*sinc tf^t+sinc

sin

‘

r i YxRb

-

sin sin <r/y

*~ xRbt-x xRbt{l-Rbt)

60

The Fourier transform of Eq. (1) above yields

. J_rwf_2_'l [«/*'**•

Rt, [2xRb )^

ja>/2Rt

-jullRt,

7.3-10 (.) No^rbec.u«the!raple».lu«ofa««n,.po^^s^^ eV“"mto0fimS "‘1

the sample values ofopposite polarities are separated by^ the first digit is 1. The

(b) The first sample value is 1 because there is no pulse before this digit Hence uie &(b) The first sample

detected sequence is

uoooiooiioiioioo

7.3-10 Hie lint ample vel»« is 1. indicata, tha d,. ^ “ IS’“^ ''

The duobinary rule is violated over the digits shown by underbracket.

12000 -200 -20 2 0 0 - 2 0 2 2 0 -2

Following are possible cornea ample values in place ofIhe 4 ondertnadcel values:: S 2 0 -2. or2j» -

-2j-S.

„ 0ooj” 2 0 0 0. These ample values lepmsen. Ae following 4 digit aqnence. 1 100, or 1000, or

0100, or 1010. Hence the 4 possible correct digit sequences are

1101001001X|XjXjX4lllOO

where x^xjx, is any of the four possible sequences 1100, 1000, 0100, or 1010.

7.4-1 S * 101010100000111

From example 7.2

T » (l© D3 © D5 © Db © D9 ® D10 © Z)M © Dn ©£>,3 ©D15 ©.~)s

K = (l®£>3 ©D5

)r

7- = 101110001101001

R = 101010100000111= S

7.4-2 5=101010100000111

T « (l © D2 © Z>4 © D6 ® Z>

8 © Di0 © Dn © Z),4

©...)S

R * (l© Z>2)r (see Fig. S7.4-2)

T = 100010000000110

R

=

10101C100000111 =

S

Fig. S7.4-2

7.4-3 S = 101010100000111

T = (l<BD<BD2 <BD*®D1 ®D*®D9 ®DU <BDU )s

/? = (l®Z>®03)7' (see Fig. S7.4-3)

r= lioiiiioiooioii

R = 101010100000111 = S

61

7.5-1 From Eq. (7.45), we obtain

’*-1

Co

«1

1 0.3 - 0.07"-1

'o'-0528"

0.1 1 0.3 1 = 1.07

-0.002 0.1 1 0 -0.113

7.6-1 (a) -2- * 5

<*n

(i) For polar case Pt 2(5) = 257 x 107

(ii) For on-off case Pt » 2(5 / 2) = 0.0062

1

(iii) For bipolar case Pt — 152(5 1 2) * 0.0093 1

5

In the following discussion, we assume = .4, the pulse amplitude.

(b) Energy of each pulse isf^, - A1Tf, / 2 and there are pulses/second for polar case and f

pulses/second for on-off and bipolar case. Hence, the received powers are

Fl -i!2L^.ii-1222^.u25Kio-#“2^2 2

. - ^*5-x =— = 05625x1 O'

6

2 2 4

*b

^on-off

'Dipolar_

.^ jSL x^- =— = 05625 x 10"^

2 2 4

(c) For on-off case:

We require /*(e) * 257 x 107 * 2(^p Hence,

For bipolar case:

/^ * 5.075

Hence

and

7.6-2 For on-off case:

A = Ap = 5.075 x 2a* = 0.003045

251x10“*

Pt - 10-6 =>-^-*4.75

2tr„

62

<t„ = 10"3 =>Ap 2 (4.75)(2 x 10"3)= 9S x 1<T

3

For on-off case, half the pulses are zero, and for half-width rectangular pulses, the transmitted power is:

/ v

Si4

(9ixl0-3

)s 2236 x 10“* watts.

There is an attenuation of30 dB, or equivalently, a ratio of 1000 during transmission. Therefore

ST * 1000S, = 22.56 x 10-3

watts

7.6-3 For polar case:

Pt - 10-6

=g^-j=>£-= 4.75 => Ap = 4.75 x 10"3

For polar case with half-width rectangular pulse:

z-h c -(4ixlO-3

)

2« 1 128 x 10

-6watts

V ‘ J2

ST * (I000)(l 128 x lO^BK) = 11.28 x 10-3

watts

For bipolar case:LAI W(UV.

Pf = lo-6 = 1-5^“

j

=>^-» 4.835 and Ap = 4.835 x 2 x 10"3 = 9.67 x 10"3

For bipolar (or duobinary), half the pulses are zero and the receive powerS, for half-width rectangular

pulses is

2

s = * 1(9.67 x 10“3f * 23.38 x 10-6

watts' 4 4 ' ’

Sr *(1000)S, =23J8xlO"3watts

7.7-2 Sampling rate = 2 x 4000 x 125 = 10,000 Hz.

mpQuantization error =y m 0.001mp L =

Because I is a power of 2, we select L = 1024 = 210

. Hence, n - 10 bits/sample.

(a) Each 4-ary pulse conveys log2 4 = 2 bits of information. Hence, we needy = 5 4-ary pulses/sample,

and a total of5 x 10,000 = 50,0004-ary pulses/second. Therefore, the minimum transmission bandwidth is

50,000

1000

(c)

. 25kHz.

T2 2

7.7-3 (a) Each 8-ary pulse carries log2 8 = 3 bits of information. Hence, the bandwidth is reduced by a factor of

(b) The amplitudes ofthe 8 pulses used in this 8-ary scheme are ±AI 2, ±3-4/2, ±SA 12, and ± 1A 12.

Consider binary case using pulses ±A 12. Let the energy of each of these pulses (of amplitude ±AI 2)

be Eb . The power of this binary case is

Wintry* SbRb

Because the pulse energy is proportional to the square of die amplitude, die energy of a pulse ± 21S

k2Eb . Hence, the average energy ofthe 8 pulses in the 8-ary case above is

63

Eb

£<» = '

2(±i)! +

2[±|)

2

+2(±f)

+a(±!)

= 21Eb

Hence,

Therefore,

fy.xy = £flvX pulse rate = 21£j x = 7EbRb •

^tty - 7 binary

7.7-1 (a) M - 16. Each 16-ary pulse conveys the information oflog2 16 * 4 bits. Hence, we need

12 ’0(M) - 3Q00 16-ary pulses/second.

43000

Minimum transmission bandwidth = —r~ = 1 500 Hz.

(b) From Eq. (7.32), we have Rb =— Br . Hence,

3000 *— Bt => Bt = 1 800 Hz.

(.) For polar signaling, *m^^*****<**«* 11» P"* <*

amplitude— has energy

^2

The power P is given by f = EbRb *— *i>=~

8 8

(b) The energy of a pulse ± is A2£j . Hence the average energy of the M-ary pulse is

EM =—[2£4 + 2(±3)2+2(±5)

2+ +2[±(M- >)f

E»]

M L

M-2

2£i (2*+o

2

M *«

o

Af2 -1

Bb

Each M-ary pulse conveys the information oflog2 M bits. Hence we require only^MM_ary

pulses/second. The power PM is given by

"idPu ’

log, M “3108, M * 24 logi M 24 log, M

7.7-S Each sample requires 8 bits(256 - 2*) . Hence: 24,000 x 8 - 192,000 bta/*c.

Bt = 30 kHz

R = _2_ Bt = —(30,000) = 50,000 M-ary pulses/sec.

1+r 12

We have available up to 5o!oo0 M-ary pulses/second to transmit 192,000 bits/sec. Hence, each pulse must

( 192,000^ai,

..j

transmit

:

• 3.84 bits.

64

7.8-1

=e choose 4 bits/pulse

=> AS = 16 is the smallest acceptable value

(a) Baseband polar signal at a rate of IMbits/sec

and using full width pulses has BW = 1MHz . PSK

doubles theBW to 2MHz.

(b) FSK can be viewed as a sum of2 ASK signals.

Each ASK signal BW = 7 MHz. The first ASK signal

occupies a band fgQ i 1 MHz, and die second ASK

signal occupies a band fe\± 1 MHz. Hence, the

bandwidth is 2 MHz + 100 kHz = 2.1 MHz.

PSD

7.8-2

7.8-3

(a) A baseband polar signal at a rate 1 Mbits/sec using Nyquist criterion pulses at r = 02 has a

BW = llld Rb mH x 106 = 6.0 x 10

5 Hz.

PSK doubles BW to 1.2 MHz.

(b) Similar to Prob. 7.8-1

.

B1PFSK =0.6 MHz + 0.6 MHz +100 kHz

SHfa* = 13 MHz

log2 M * 2 for AT = 4.

We need to transmit only 03 x 1064-ary pulses/sec

(a) BW is reduced by a factor of 2.

BWVsk = 1 MHz

(b) In FSK, there are four center (carrier) frequencies

/cl , fa • fa > *nd/c4 . each seParated by 100 kHz

Since ASK signal occupies bandfc ± 03 MHz, the total

bandwidth is

03 MHz + 0.5 MHz + 100 kHz + 100 kHz + 100 kHz = 13 MHz.

PSD ***%

c, -?c-

Jt 3,00 ij

Fig. S7.8-3

7.9-1

65

nt.tO•»

\—‘ft; 'T

w**)

—

*vw

—

Fig. (b)

/

-ft * Uioo cobi^/itc

Fig. S7.9-1

7.9-2

Either figure (a) or (b) yields the same result.

mi (/) has 8400 samples/sec.

ffl2 (t), «j(f). mA (t) each has 2800 samples/sec.

Hence, there are a total of 16,800 samples/sec.

Fira, we combine «,(<). -* ^W wiU,,comm.«W^of™0^^

™

slgoe, it now multiplexed with with .c—or tpeed of2.00 touuiont/tec, „e,dm6 dte ompu.

5600 samples/sec.

66

r* M'S ^4-s kbls

Fig S7.9-3

Chapter 8 Exercises

8 . 1-1 If a plesiochronous network operates from Cesium beam clock which is accurate to ± 3 parts „in 1012 , how long will it take for a DS3 signal transmitted between two networks to become out :

of sync if a 1/4 bit length time error results in desynchronization?

Solution: A DS3 bit is transmitted in 1/(44.736- 106)= 2.235336-10^ sec. In the worst case,

both network clocks will be out of synchronization by 6 parts in 101.

2.235336- 10*®/(6- 1012

) = 3922.27 sec/bit or 980.57 sec/ Vi bit

8.1-2 For the bit stream 01 1100101001 11 101 1001 draw an AMI waveform.

Solution:

8 . 1-3 For the following waveforms, determine if each is a valid AMI format

for a DS1 signal. If not, explain why not.

Solution: No. 16 0’s violation

Solution: No. bi-polar violation

Solution: No. l’s density violation

Solution: Yes

68

8.1-4 a) You have received the following sequence of ESF framipp pattern sequence bits

...00110010110010110...

Is this a legitimate framing bit sequence in order to maintain

synchronization between the T1 transmitter and receiver?

Yes NoIf yes, why? If no, why not?

Solution: No. The bit sequence 001 1 cannot be in an ESF framing bit sequence.

b) The following T1 AMI signal is received:

n

Is this an acceptable T1 signal?

Yes Noa. If yes, explain.

. , .

b. If no, explain why not (what, if any, DS1 standards are violated) and

draw on the figure the AMI waveform which would be transmitted by the DSU

Solution: No. 16 0’s violation. The 16 0’s will be replaced by a pattern of 1 s by the

DSU.

8.1-5 Tie signal 110100000000000000001 is received by die DSUin

8ZS format. Draw the output of the DSU for this signal? The

ihow the bit stream which is substituted by the DSU.

a T1 data stream which uses a

first 1 is already drawn.

8 .1.6 T-l synchronization at two distant locations is controlled by separate crystal controlled

oscillators which differ in freqoency by 125 parts per million If the termmj equipm

<

maintain sync in how many complete D4 superframes will the faster oscillator have generated (at

most) one more time slot (8-bit) than the slower oscillator ? Circle the correct answer.

69

a) 5

b) 10

c) 15

e) None of the above - if "none", what is the number of D4 superframes before an extra time slot

&£^)lteflSer oscillator will generate 125 10-61.544 106 = 193 bits per second more

than the slower oscillator. This is one frame/sec = 24.125 time slot^ Hence, a time slot

difference will be generated in 1/24.125 = 0.04164498 frames or 0.0034704 superframes.

8 . 1-7 Two plesiochronous digital networks, A and B, utilize Cesium beam clocks accurate to 3

parts in 1013

. The networks are operated by independent long distance companies and are

synchronized to each other by means of a UTC signal.

a If a company leases a T1 line with D4 framing which is terminated at one end m

network A and at the other end in network B, how often must the networks be resync d to

each other to avoid a framing bit error in the customers T1 signal in the worst case {You

may assume a framing bit error occurs when the two networks are out of sync by > 1/2 o

a T1 "bit time".} . ,.

Solution: ATI bit time is 1/(1.544. 106

) = 6.47668- 10'7 sec/bit. In the worst case, the

two clocks would be off by 2-3 = 6 parts in 1013

or 6- 1013

errored bits per bit transmitted.

Hence, 6.47668 10'7 sec/bit / 6- 1013

errored bits per bit = 1 .07945 10 seconds per errored

bit or 5.39723- 106seconds per errored half-bit.

b. UTC operates via GPS satellites which are approximately 23,000 miles above the Earth.

How long, in terms of T1 bits, will a correction signal take to be transmitted to

the network switches?

Solution: The speed of light is approximately 1 86000 miles/sec.

23000x2 » 46000miles up and down. 46000/186000 = 0.247 sec

0.247x1544000 s 381850 bits

70

Chapter 10

10. 1-1/ , JV 13+13 1

(a) /’(red card) = — = -

(b) />(black queen) =— =—

(c) /’(picture card) =

(“> ^-5-5. 20 5

(e) /»(« ^ 5) *

52“73

10.1-2 (a) S = 4 occurs as (1,1,2),(1,2,1),(2,1,1)-There are a total of6 * 6 x 6 = 21 6 outcomes.

Hence,^ = 4)

«

lsO,2,6) (1,3,5), (1,4,4), (1,5,3), (1,6,2), )

;5 3W5%I)’

(3,1,5), (3,2,4), (3,3,3), (3,4,2), (3,5,1), (4,1,4), (4,2,3), (4,3,2), (4,4,1), (5,1,3), (5,2,2),(b) S = 9 occurs as

c(5 ,3 , 1 ), (6, 1 ,2), (6 ,2, 1 )

P(5 =9)«^

(c) S = 1 5 occurs as (3,6,6), (4,5,6), (4,6,5), (5,4,6), (5,5,5), (5,6,4), (6,3,6), (6,4,5), (6,5,4), (6,6,3)

10.1-3 Note: There is a typo in this problem. The probability that the number, appears should be ki not k,

1 = £*/ = * +2* + 3* + 4* + 5* + 6* = 21* =o * *—,=l

P(/) =— (I = 1,2, 3, 4, 5,6)/ 2i

10.1-4 We can draw 2 items out of 5 in 20 ways as follows: 0,0,. 0,0,. 0,P,. 0,Pi. »A, 0A, Wt. 0,OI•

OMpTÔ,,“ 0„ PiOj, P,P„ W„ PA. PA, P.Pi All diese outcomes »e equally likely w,0l

^rto, fl U«,fiU.2AUOÛ03,UO! fiU flo,Uf1oJ U fl o3U^ 1

Ufi oJUfioJ

Hence, P(E\) =12

20

3

5

(ii) This event £2 * /’iÛ/^/’i

Hence, />(£2)

=^ = '^

(Hi) This event £3 = 0j02 UO1O3 (J020i U0203 UO3O1 U0 302

Hence, £(£3) =20

" Jo

(iv) This event £4 * £2 U £3 and bo* E2&E3 an disjoint.

Hence, P(EA )= P{E2 )+ P{E3 ) =^

= 1

71

10.1-5

10.1-6

10.1-7

10.1-8

Let x0 be the event that the fust chip is oscillator and be the event that the first chip is PLL. Also,

let jto2and represent events that the second chip drawn is an oscillator and a PLL, respectively. Then

P{ 1 osc and 1 PLL) = p(*0),xh )

+ P(*i\)

=/>(JCo,)/,(x/

i|xo,)+^fiM*o2h)

(BMHHUsing the notation in the solution of Prob. 10.1-5, we find:

(a)

(b) ^o2K) = f

(a) We can have (*§) ways of getting two l*s and eight 0’s in 10 digits

P (two r s and eight 0’ s) *= 45(0J)1

(0.J)' » 45(0.5) -

(b) Piat least four 0’s) = 1 - [/•(exactly one 0)]+[ /(exactlytwo 0’s)]+[/>(exactly three 0’s)]

/>(two 0’s) =(1

2°)(0-5)

,0“^P (three 0’s) -

()(0-5)

10 * -j—

( 5 45 120 'l 849

P(at least four O' s) = 1 - 1— +— +—

J J024

(a) Total ways of drawing 6 balls out of49 are

(4|) = _i£L = 13,983,816V 6

^ 6143!

Hence, Prob(matching all 6 numbers) *^ 3983816

(b) To match exactly 5 number means we pick 5 of the chosen 6 numbers and the last number can be

picked from the remaining 43 number,. We cer, choose 5 number, of our d in ( f)• 6 ways and can choose

number ou. of43 *(<?) - 43 ways. Hence. w, have 43. 6 combirmrions in wh.ch exactly 5 numbem

match. Thus,43x6 5

P (matching exactly 5 numbers) = 139g3gl6= 1-845 x 10

(e) To match exactly 4 numbers means we pick 4 out of the chosen « number in ( t)- 1 5 ways and choose

2 ou, of dr. remaining 43 numbers in(*J) = 903 way,. Thu, there are 15 * 903way, of picking exanly 4

numbers out of6 and. .

15x903 4

P (matching exactly 4 numbers)- ]39g3gl6

- 9 686

72

(d) Similarly, we can pick three numbers exactly in(f)(4j)

= 20 * 12341 *246820 ways. Hence,

p (matching exactly 3 numbers)- “ 001 765

10.1*9 (a) Let/ represent the system failure. Then

/>(/) = (l-O.Ol)10 = 0.90438

/>(/) = 1- /»(/) = 0.0956

(b) P(f) « 0.99 and P(f) = 0.01

If the probability of failure ofa subsystem st

is p, then

p[f) - (1 - pf or 0.99 - (1 - pf 0.001004510.1-

10 If/ represents the system failure andfu andfL represent the failure ofthe upper and the lower paths.

respectively, in the system, then:

(a) P{f) - P(fufl) - HfuWl.) =[p(/w)]

2

P(/u) = 1 - P{/«) = l-(l-O.Ol)10 = 0.0956

and

P(f) = (0.0956)2= 0.009143

Reliability is P[f)= 1 - P(f) = 0.9908

(b) />(/) = 0.999

P(f)= 1-0.999 = 0.001

/>(/„)- Voiom - 0.0316

P{fu) (1 - P)W - 1 - 003 16 => P = 0.003206

10.1-

1 1 Let P be the probability of failure of a subsystem (jj or rj)

.

TheSystem ^lils'if the upper and lower branches fail simultaneously. The probability of any branch not

failing is2

(1 _/>)(!_/>) = (l- P)2

. Hence, the probability of any branch failing is 1 - (1 - P) .

Clearly, Pj , the probability of the system failure is Pf * [l “ 0 ~ p? J1 “0 "

]

s 4/>2 P<< '

For the system in Fig. b: .. ..

We may consider this system as a cascade of two subsystems x, and x2 ,where x, is the parallel combmatio

ofjj and r, and x, is the parallel combination ofs2 ands2. Let Pf (x,) be the probability of failure ofx..

Then

r/M-r/M *'7

The system functions if neither s, nors; frils. Hence, the feobability of system not failing

is (l - P2)(l - P2

)•Therefore, the probability of system failing is

Pf = l-(l-P2){l-P2

)= 2P2 -PA s2P2 P«l

Hence the system in Fig. a has twice the probability of failure of the system in Fig. b.

73

10.1-12 There are(52

)= 2598960 ways of getting 5 cards out of 52 cards. Number ofways of drawing 5 cards of

the same suit (of 13 cards) isf^3)* 1287. There are 4 suits. Hence there are 4x1287 ways of getting a

flush. Therefore,

/’(flush) =4x1287

2598960= 1.9808 x 10"3

10.1-13 Sum of 4 can be obtained as (1,3), (2,2) and (3,1). The two dice outcomes are independent. Let x, be the

outcome of the regular die and x5be the outcome of irregular die.

?x\xi 0 *3)* ?x\ 0)^*2 0)

=g*3

=jg

^,x, (2J) - (2)^3 (2) - ^« o = 0

Thtrefor, W).Similarly,

W)-P.,.,(ub *U(M)+ p + ^,(4.1)

10.1-14 B=AB\jA eB

P(B) = P{A)P{B\A)+ P(AC)P(E\AC

)

10.1-

15 (a) Two l’s and three 0’s in a sequence of 5 digits can occur in (2)= 10 ways. The probability one such

sequence is

/> = (0.8)2(02)

3 = 0.00512

Since the event can occur in 10 ways, its probability is

10x0.00512 = 0.0512

(b) Three l’s occur with probability (^)(0i)3(02)

2 = 0.2048

Four l’s occur with probability( 4

}(0i)4(0.2)

1 = 0.4096

Five l ’s occur with probability( 5

j(08)S(02)° = 03277

Hence, the probability of at least three l’s occuring is

P = 02048 + 0.4096 + 03277 = 0.942

1

10.1-

16 Prob(no more than 3 error) = P{no error) + /*(l error)+ P{2 error) + P(3 error)

-<- p.r-iH/iii- p.f -(D^o -

s(l-1007‘,)-.K>0/',(l-99P,)*4950P,I(I-98f,,)-.l61700P,

3(l-97/>,)

74

10.1-17 Error can occur in 10 ways. Consider case of error over the first link

/^(correct detection over every link) = (1 ~ /*i)0 ” ^t) -0 “ ^lo)

/»£ =i-pc = i-0-^X 1 -^)

_ i-[i-(/»j + /^+...+/\0 )+ highcr order terms]

sPl+ P2 +...+Plo Pi«'

10.1-

18 *«) - i(!,)P,'(\-P.t‘ - lO'l’ll - rf -‘VO-

;=

3

sl0/»e3(l-/>

«)2

,/>.« 1

10.1-

19 (a) ^(success in 1 trial)»^ = 0.1

(b) /’(success in 5 trials) = 1 - /’(failure in all 5 trials)

•'-WhWSts

Pj - Prob(failure in 1* trial) -9/10

Pj - Prob(failure in 2nd

trial) = 8/9

Similarly, Pfi

=7/8, Pfi

=6/7, and/»/5=5/6

Hence, /’{success in 5 trials) - 1 “(^XfX^X «)“

10.1-10 Lets be the event of drawing the shot straw and the «(*) denote the event that/th person in the sequence

draws the short straw.

Now, /*i(-^)— 0.1

n.M - ProbO" person does not draw the short straw) * Pwb(2- person draws the short straw)

-MmiHS)- 01

"Uneidter I" nor 2“ person draw, the short straw) x ProbO1* person draws the short straw)

-[i-AM-aWjHfiKHSimilarly, /^(x) = Ps{*) *•••“ ^o(x)

~ 01

10.1-21 All digits are generated independently

(a) P(all 10 digits are 0) = (03)‘°

(b) There are ('§) ways of arranging eight l’s and two 0’s. Hence,

/height l’s and two 0’s)-(1

°)(0.7)S(0J)

2

(c) /»(at least five 0’s)=/^five 0’s)+/^six 0’s)+...+/\ten0’s)

= (

1

j)(O.7)S{01)

5+(*6)(0.7)

4(0.3)* + (*7}(8-7)

3(b^)’ + (*1){0^)

2(0J)* + ('o){®.^K®^)

9 + (0-^)10

10.2-1 /y (0)= /xy(l.°) + ^xy(0»°)

= />x(,)

/>y|x(

0l1)+ ^(Oj^yjxW

= 0.6 x 0.1 +0.4[l - /*y |x(l|0)] = 0.06+032 = 038

/y(l) = 1 - /y(°) = 062

75

... VMKU) (i -P')Q10.2-

2 (a) ” {\-Q)Pt +{\-Pe)Q

(b) ^x|y(°l1)= 1

- /,xly(

1l1)

(note that Py(l) and Py (0) are derived in Example 10.10)

10.2-

3 (a) P(x * l) = £^xe~xdx =

^

(b) Prob(-l < x <; 2) = £"**X* +lo\

xe-Xdx =

(c) Prob(x 5 -2) - “~f10.2-

4

Fig. S10.2-4

Since this is a half-wave rectifier, y assumes only positive values. So P{y < 0) - 0.

Hence, Fy (y)

= 0 (fory < 0) and P[y < 0+

)

- • Hence, Fy[o

)* ~

10.2-5 x is a gaussian r.v. with mean 4 and <rx - 3

Hence,

(a) P(x ^ 4) * * C(0) = 03

m n* 2 o) -e^) - 1 -e(|) - ' -<m»I7‘ - 0 !l0!3

(c) P(xZ -2) • - 1 - fi(2)'- 1 -0.02275 = 0.9773

Fig. S10.2-5

102-6 (a) From the sketch it is obvious that x is not gaussian. However, it ista unilateral(rectified) version of

Gaussian PDF. Hence, we can use the expression of Gaussian r.v. with a multiplier of .

For a gaussian r.v.

whhffya4

(b) Hence, (i) P(x * l) = 2P{y * 1)= = 05026

(H) p(l < x £ 2) = 2 P{1 < y 5 2) *2^)-12(f)]

=W 856

(c) Ifwe take a Gaussian random variable y

*w-asr^"Fig. SI0.2-6

py(y) **

and rectify y (all negative of y multipled by -1), the resulting variable is the desired random variable x.

76

10.2-7 The volume V under Pxy (x,y) must be unity.

A.2

Px(x)=jpxy(x'y)4y

y

But y = -x + 1 and the limits on y are 0 to 1 - x . Therefore,

l~x2dy = 2J

1 *•

0

f2(1—>) Oiyil

Px(x) = lo~

X24' = 2y|

Similarly, py (y) =

2(1-*) 0SxS4

0 otherwise

otherwise

/, X P*y(x>y) 2

Px\y(Ay)- 2(1 -y)

V\-y OsySl

0 otherwise

Fig. S10.2-7

1/1—x OSxSl

Similarly, />y |

x (.y|-*)-

0 otherwise

Clearly x and y are not independent.

10.2-8

—(x2 +>

2y2 . . ,

.

Pxy {x,y) = xye"

' u(x)u{y)

-x2 /2

(a)

Similarly,

and

px (x) = fixye^^ttxfr'= "

u(x)

py {y)

= ye~y 12

’Ky)

(b) From results in (a), it is obvious that x and y are independent.

« X 1 r«4ax2 +by

2-2ay)hM

10.2-9 Px{x) = i^Pxy(x'y)& = 71177l-~e**

1

re-^t2U

J*r-

1s - ' m£y2ftb

/ \1 —y^ /2tf

Similarly we can show thatpy (y)= c

Therefore/

1

\Pxy (

x>y) _ rrcAx

«y)h

‘|y(xW"

px(x) ~1l2xM

, , s Pxy(x>y)

77

10.2-10 K

But

jd* = 1

nJ-e^wW

=

^!Ze'x2

[lZ>e~y2-10,

‘fy,

j*. «.4

„ 1 3Hence, a =— .

-V4

,xW . *rjy<‘W)

‘» -^D"W - 'c '/”e'5'

!''' *

. _ |T -3>2/4

Because of symmetry ofPxy (x'>0 w*^ tespect to x and y. py (y) ^^

, , X PxyM _ 1/{* +,y+T

)

Px|y(;tW = ~ r_

/4

PyM V»

and

...'*W'1F 7

Since Pxy (x,y) * px (x)Py(p). * «»d Y are not dependent.

10.2-11 P« = P^P*! 1)* pM°)**(0)

If the optimum threshold iso ,then

j \ / i

__

jP, _ 1 L-{4p-a

)

2/W p^i). e-(^^f/2g2»p

r (0) =0 Fig. SI0.2-11

da 2nan[

Hence, e^f^1

,„(,) . *(.)

?ctlO

And a =2^„

Px(0)

Px (»).

. . 1 -(x-ifna10.3-1 x = 2, <rx =3, P*\x

)-~£j2ne

78

10.3-2 PxM = ^M*"W

Because of even symmetry ofpx (x), x = 0 and

x2 = 2^x2

px (x)dx = 2j"x2 ^xe'xdx

=J^°

x3e~xdx = 3! = 6

x2 = a 2 + x

2 -a\ = 3! = 6

10J-3

Fig. S10.3-2

y * J_*ypyb)^- +

J"

- * f® ye~y2,2a*dy = 0399CT

y2 -

= -L- r yV^'V^

+ r-4-ye->W <*2-00

<rv2ff

'e~>1>2°\

5 as2x

a 2 = y2 -(y

)

2 =y _ (°-399^2 = °-3408ct2

x2 = <t«

2

* o«3-4 .x2=

j

0°VpxW*

=

2C x2 /32a=2x42 *

Because * . ^ ?-*.’-<*f -’T'fe)*

10.3-5 The area of the triangle must be 1 . Hence K = - and px(x) *-(x + l) -15x53

^-

i

+ *)*-i(V

+

£

16ff-32

P*C-x. 5

ir«1.2Z 14l.il —«[< 3 4 3J 3

10*3*4 x-Zx,P.(x,).ip)4i(3)4^(4) +^{3)^(«)+^P)4

7.|4/>x(x,).4(4)+4(9)*i(16)+4(25).4(3<)*A(49)+

Fig. S10.3-5

79

10 3-7 I" - 1—

=

Fxne~^'1°'dx For n odd, the integrand is an odd function ofx. Therefore x -0.

ax y2/c

For n even, we find from tables

x" =(lX3K5)-(»-«)^ " even

a n odd

10.3-8 Let x, be the outcomes of the rth die. Then,

1 + 2 + 3 + 4 + 5 + 6 _ 2 j= 1,2, .... 101 6 2

—\2 + 21 + i2 + 4

2 + 52 +6 _ 91

xi

=5 6

2 2 /— \2 35<7,

» x.-(x<)

Ifx is a RV representing the sum, then

X*Xj +X2+—+X10 = 1

.2<rx*< +< +" +ax

1o -<a-Tk2 = <r

2 +x2 = —~ + (35)2 = 1254.167

10.4-1 px (x) = ^<*(x) +^{x ~ 3

)

, \1 -~«2 /8

y = x + n

Py (>-) = WVnW = + " 3)]*Fig. S10.4-1

„ * .->»/• + » e^'8

4^» 4V2«-

10.4-2

and

px (x) = 0.4<J(x) + 0.6£(x - 3)

1 -i?ll

, \1 ,-y2 /g +

3c-(y-3)

2/8

\0>[2n

10.4-3 ^(xJ^e^x-O +O-e^x + O. Pn(") = /,<y(w - ,)

+(1-2W ,, + ‘)

pyW=[e^-i) +0-e)^ + 0]*[M3'-i) +0-^(>' +1)]

= (P + Q-2PQ)S(y)+ PQS(y-2)+{\- P){\-Q)S{y + 2)

80

10.5-

2 When y = #,x + K2 Hence, y = Kxx + Kj

a\ - K\o\ and trxy - (x-x)(y~F) = (x - x)(#ix + K2- *

1* -*) = #,<rx . Hence,

Px = , Kxo\ / Kx

a\ = 1 if ATj is positive. If #, is negative, * K\°l « negative.

y axay

But o-

x and <xyare both positive. Hence, pxy = -l

10.5-

3 “S 9rf« = 0 Similarly, 5 = 0

. xy .SS .1*3- i

J

0

J'sin2M9><«= 0

2 2

Hence, crxy = x y = 0 and x, y are uncorrelated. But x + y =1.

Hence, x and y are not independent.

10.6-1 In this case

#11 = #22 ~ ^33 =mk *

#12 = #21 #23 “ ^32 “ ^01 = 0325#m

#13 “ #31 = #02 = 0362Pm

#03 = 0308fm

Substituting these values in Eq. (10.86) yields: a, = 1.1025, a2 = -03883, a3

From Eq. (10.87), we obtain

7 = [l- (0.825a, +0362a2 + 0308a3 )]#m = 03753#m

Hence, the SNR improvement is

-0.0779

81

Chapter 1

1

11 1-1 This is clearly a non-stationary process. For example,

amplitudes of all sample functions are zero at same

instants (one is shown with a dotted lind). Hence, the

statistics clearly depend on/.

11.1-2 Ensemble statistics varies with I. This can be seen by

finding

^ = AcoiaX + 0) = A J^°°cos(fflr + 0)p(a>)deo

- jL. f

,00Cos(<ar + B\ia . This is a function of /.

"100/ J°

Hence, the process is non-stationary.

Fig.SlU-2

11.1-3 This is clearly a non-stationary process since its

statistics depend on /. For example, at t = 0, the

amplitudes of all sample functions is b. This is not

the case at other values of /.

Fig. SI 1.1-3

11.1-4 x(/) = acos(fflf + 0)

^ = acos(<vf + 0) = acos(«af + 0) = cos(a* + 0) j_Aap»{o)da

=[cos(<b/ + 6)/2A\ j*

Aada = 0

_

*x (,b /2 )- a

2 ccs(/*i+0)cos(a*2+0) - «os(/uf1

+^)cos(/ar2 + *)a2

= CO$(<U/| + d)cos(fflf2 + 0)

2

* ^-cos(aX| + ^)cos(fiJ/2 + 0)

\ _

stA

A A

Fig. SI 1.1-4

t>aco

83

(e) The process is not ergodic. Time means of each sample function is different and is not equal to the

ensemble mean.

(f) x2 = *x (0)

= -^

11.2-

1 (a), (d), and (e) are valid PSDs. Others are not valid PSDs. PSD is always a real, non-negative and even

function ofa. Processes in (b), (c), (0, and (g) violate these conditions.11.2-

2 (a) Letx(t) = x, and x(r + r)=x2 Then,

(xi±x2 )

2 =^I + x22 +2x^7s0, x,

2 +x22 *±2^

But, x,x2 = Rx(t) and xj2 * x2

2 - flx (0) Hence, /?x(0)^j/ix (r)|

(b) Rx {r) = x(/)x(/ + r), Hm /?x (r)= Hmx(<)x(< + r)

As t -> oo, x(t) and x(t + r) become independent, so Urn,Rx(r) = x(t)x(t + r) = (x)(x) *11.2-

3 /?x (r)= 0 for r =±— and its Fourier transform Sx (e») is bandlimited to B Hz. Hence, rtx (r)isa

2Bwaveform bandlimited to B Hz and according to Eq. 6.10b

*x(r)= £ «x^sinc(2*Br-/»). Since Rx^j* 0 for all » except « = 0.

Rx (t) - Rx (0) sine (2«Br) andSx (a)= -^rect^j . Hence, x(<) is a white process bandlimited

to B Hz.

11.2-4 ^x( r)= ^x,x2 0»l) + />

XlXj(” 1 “ 1)“ /X|Xj(“ ,» l)“ ^xjxj 0

But because of symmetry of 1 and 0,

/*,x 2 0* 0* fyxjH*-1) *nd P*i*2 (

-1, 0* ^x^O*" 1)

and Rx (t) = 2[/,

X]Xj (I, 1)- Px,X2 (l,-l)]

-2J4, 0)[v, »)-('- JW 1W)] - 2W»-'

Consider the casenT* < |rj < (n+ 1)7* . In this case, there are at least n nodes and a possibility of (n + 1)

nodes Prob[(w + l)nodes] =f

= J~~n

Prob(« nodes) « 1 - Prob[(« + l)nodes] = (n+l)-y*b

k *x

y„/yyy/J//A

^-nTfcH*—nTb

The event (x2 = ijxj = l) can occur if there are N nodes and no state change at any node or state change at

only 2 nodes or state change at only 4 nodes, etc.

Hence, PXJ

j

x,

(l|l) - Prob[(w+ l)nodes] Prob(state change at even number ofnodes) +

Prob(n nodes) Prob(State changes at eveen number of nodes)

The number ofways in which changes at K nodes out ofN nodes occur

(iio-[(r>6)><rWr>‘)Vr'

is(jf). Hence,

+

85

[( 0)(0-6)° (0.4)” + (5K°6)

2(0-4)""

2+-

]{

n+l~fb )

aid/?x (r) = 2/>Xl

|X2

(l|l)-l This yields

*x (r) = 1-1.2 \r\<Tb (n = 0)

= -0.44 + 0.24M Tb < |rj < 2Tb (n- 1)

Tb

*0.136-0.048^ 27* <)r)< 37* (« = 2)

Tb

and so on.

Fig. SI 1.2-4

The PSD can be found by differentiating *x (r) twice. The second derivative d2Rx /dr2

is a sequence of

impulses as shown in Fig. SI 1.2-4. From the time-differentiation property.

— i?,(r)^(/fl>)2 Sx (ffl) = -®2SxH- Hence, recalling that ^r-rjoe'^.we have

jf2*' ’ ' ’

-ta2Sx (co) = -^-[-2.4 + 1.44(eJ

®7* +e-^)-02SS{^ +e~''2*Tt

)+••••]

= -L[-2.4 +258cosa»r4 -0576cos2fi>rfc+0.1 152cos3wf6 + .]

Tband ,

5x (q,) ,—1— j^2.4- 2£^coi<aTb ~cos2a>7i +^cos3a>r6

-^cos4fl>rA+....Jj

11.2-5 Because Sm(co) is a white process bandlimited to B, /^(r) = *m(°) sinc(2Bt

)811(1

M^) 350’ " = ±1 ’ ±2’ ±3 "‘

This shows that x(f)x^/ +—j = = 0

Thus, all Nyquist sample are uncorrelated. Now, from Eq. 1 1-29,

wRo + lLRm co*na>0Tb

_ n=l

J?n = ikbk+n S 1 and where a * is the klh Nyquist sample.

Rq = a* = x2 = /?m(0)- Hence,

Sy (*>)

= = 2BRm (Q)\p(vy sincere—

11.2-6 For duobinary

Ptk (1)

= PMk (-1) = 025 and Ptk (0)= 05

«.-.I>o)!

r(-o2

r°!

(i)4

K, =a*a* +1 =£ '£*k*k+iP»k*k+\(*k*k+i)

o* «*+t

Because a* and a*+ ,each can take 3 values (0, 1, -1), the double sum on the right-hand side of the above

equation has 9 terms out ofwhich only 4 are nonzero. Thus,

*1 * (W)*W, O*1)+HX-O'W.(-^-O + (1K-0V*+i

OX-0 HX 1)

Because of duobinary rule, the neighboring pulses must have die same polarities. Hence,

*•*»*! 0’1)= R

*k (0V.K *4 (2)

=8

Similarly, /»,*.**,K"0 = £Hence - *1

Also “ 8 *a*+

2

In this case, we have the same four terms as before, buta* anda*+2 are the pulse strengths separated by one

time slot. Hence, by duobinary rule,

*•**2^“ F*b^ />

**+2l#*^ *4(4) 16

Similarly, *a*a*+2H ^

87

11.2-7

11.2-8

In a similar way, we can show that Ptk»k+2 O*-1

)= *•*»*1 ( ^ *

16

uILgalTill procedure, we can show that*„=0 for n * 2. Thus, from Eq. ( 1 1 .29) and noting that *„ is

,|2

an even function ofn,we obtain Sy(a>) =———jy

(1 + cos a>7J, )

J= ^

C0S^ 2 J

‘ Mf

)

TFor half-width rectangular pulse P{o)) -£sinc

i7=(0e + (-i)(i-e)=2e-i

/io »5-(1)

2e + (-i)

2(i-e)=i

Because all digits are independent,

Rn = a 4 a*+i - a*at+ i= (20- 1)

2Hence,

SyH =_lfHi2 r -I »

1 + 2(20-1) £cos«ft>7i

V*c l n

Approximate impulses by rectangular pulses each of height A and width e such that he 1 and e -» 0

(Fig.Sll.2-8a)

^x(j‘) = IS xlx2

/>x,x1 (

;rlx2)

Since x, mdx2 can take only two values A and 0, there will only be 4 terms in the summation, out of which

only one is nonzero (corresponding to Xj = A, x2 =* A ). Hence,

*x(r) =A^xjM “ A^x.W^lx,WSince there area pulses/second, pulses occupy Of fraction of time. Hence,

PX] (A) = a* and Rx (r) =aWX2 ,X| (*W *^x,W

Now, consider the range |r| < e. /»X2

j

Xj(/i)A) is Ae -**6*"

Prob (x2 )* K given that xj = A. This means x, lies on one ^

|

of the impulses. Mark off an interval ofr from the edge of ifc- . med—LI

—

this impulse (see fig. SI 1.2-8b). If Xj lies in the hatched

interval, x2 falls on the same pulse.

Hence,

PX2|x

,

(A)A) = Pretax, lie in the hatched region) » - 1 --

and f?x (r)= oA^l-^j

Since Rx {r) is an even function of r, *x( r) = a^

1_ ~j

In the limit ase-kO, Rx {t) becomes an impulse of strength a.

Rx (t) = ccS{t) M = 0.

When r > e, xj and x2 become independent. Hence,

Rx (r) = a2he »a2 |r|>0

Hence, Rx (r) = aS(r) + a2

€-T —< '4

3T

<RxCt) 4

t •

f©..

o

Fig. SI 1.2-8

88

11.2-9 In this case the autocorrelation function at r = Oremain same as in Prob 1M ***£ O^henever

x(f), x(r + r) are both nonzero, the product x(/)x(r + r) is equally likely to

*x(r)«0, r * 0 and Rx (r) = aS[r)

Shift. The first-order probability density of the process is p{x\t). It can be shown that isproce

stationary of order 2. Hence, p{x,t) can be expressed as /<*)• We have

Rx (r) - J1£*1*2PxiXl (*i .*2^1*2

(1)

To calculate pXJ(x2 |x,

-x^we observe that in r seconds (interval between x, «idx2 ), thereat two

mutually exclusive possibilities; either there may be no amplitude shift (x2 - xi ),

or there may an

amplitude shift(x2 * x,)- We can therefore express pX2 (x2 |x, = x{)*s

Px2(x2 |x, -*i)-fc2 (*al*, -*i. no amplitude shift)P(no amplitude shift) +

p (x2 |xj = x,, amplitude shift)/^amplitude shift)

t fc-K-'U

*s Flg.Sll.2-10

The number of amplitude shifts are given to have Poisson distribution. The probability of* shifts

in r seconds is given by

„h«« there are on the averege/f shifts pr seeonti. The protabiH^ ofno shifts is obviously»<r) .where

Po(r) *C_/Jr

The probability of amplitude shift * 1 - Po(r) = 1 - er

. Hence

|x, -Wtl-e-^felx.”.. no amplitude shift) +(l -e~^

r)/>*j (ril* I

-x„ amplitude shifty

when there is no shift, x, - x, and the probability density ofn, is eoneentrared at the single value x,.

This is obviously an impulse located at x2 * x j . Thus,

Px2(-X2 |

xi= no amplitude shift) = S(x2 -x,) '

whenever there are one or more shifts involved, in generel, x, . x„ Moreover, we are given that the

amplitudes before and after a shift are independent. Hence,

Px2 (*2lx i= *1 >

atnpltawte shift) = Px2 (X2J = P\x)

89

where p%2(x2 )

is the first-order probability density of the process. This is obviously p(x). Substituting

Eqs. (3) and (4) in Eq. (2), we get

Px2(x2 |xj =xx

)*e~fiT8(x2 -*i)+(l-«pt

)Px2 (xz)

* e”^r|<5(x2 - X|)+ [e^ — l)px2 (*2 )|

Substituting this equation in Eq. (1), we get

Rx (t) = e-;?r

j^o J^oX,x2 A>Xl

(x1)|<S(x2 -xj)+(e^

r-l)px

2(x2 )]£&i<*f2

=«"*[££Wx,

(

x,)5(x2 -*i>*,*2+nJlW^r -‘K (

x«)Px2(x2 )rfr,A2

]

* e~P

\\IxfPx^Xl)^ 1 + («* - 0JZo JCl Px

l

(

Xl )^ 1 -CoX2Px2 (

X2)t&2

]

=e-^p +(^-l)x2

]

where x and? are the mean and the mean-square value of the process. For a thermal noise x = 0 and

Eq. (5) becomes __

Rx(r)*x2e~fiT r>0

Since autocorrelation is an even function of r, we have

*x(r) = x2«_/W

and

SXW 2fix2

P2+a>

2

2

11.3-

1 For any real number a, (ax- y) SO

a2x2 +y2 -2axy SO

Therefore the discriminant of the quadratic in a must be non-positive. Hence,

(2xy)2<4x2

y2or (xy)

2< x

2y2

Now, identify x with x(r,)and y with y(/2 ) ,

and the result follows.

1

1.3-

2 *u (r)= u(/)u(f + r) * [x(r) + y(r)Ix(r + r) + y(< + r)]

« /?x (r) + Ry {r) + *xy (r) + Ryx(r) - *x(r) + Ry (r)

since x(r)and y(r) are independent.

Rv (r) = [2x(r) + 3y(/)l2x(/ + r) + 3y(r + r)]

*4/?x (r) + 9rty(r) since *xy(r) = *yx(r) 5=0

Rw {t) = [x(r) + y(t)l2x(/ + r) + 3y(r + rjj = 2Rx (r) + 3Ry (r)

Rm {r) - «u>(-r) - 2/?„(r) + 3/?y(r)

113-3 *xy( r) ABcos{o}qI + 4>) cos(muo(^ + r)+

n

+\

=^ (cosfnio' + »©o(' + f) + (» + 1)*]+ cos[na>0 (' + r) - a>0t + (n-

1)*]}

90

11.3-4

11.4-1

CO<^0^ + + 1)#] =^io^ + W = 0

Similarly, cos[»fi>o(, + f)_ ®o, +

(,,“ 1)^]

= 0 8,1(1 *xy(r)*°

x(0 - C0 +£Cn cosmu0 (f- b

)+0n

11=1

00

• C0 + £C„(nfi>of - «a>0b + 0n)

n«l

Since b is a r.v. unifomtl, distributed in the tta,e(0. T„). ».b .^ is . r.v. unifOmtly distribrned in ihe

™big^the^ngument in problem 1 1 .3-3, we observe dint nil haimonics ere incoherent. Hence.be

autocoirelaiion (taction of«.(r) ia the eon, of autottarelation Ibnclion of each tenn. Hence follows

result.

(a) S,(©) * 2KTR{and S2 (f») = 2KTR2 . p„n

Since the two sources are incoherent, the principle of superposition applies to the PS .

IfS^ (o») and 5^ Hare the PSD’s at the output terminals due toS^ru) and S2 {a>) respecivey,

(a>) = $iH and SrfH =|

w2(fl>)fS2H

where

R2 / jaC «2_

Ri+V je£ _ jaRlC + 1 = .

«iW *: ' s2 / tac-” $_ jk.

*2

R-, / icuC " /?2~

J?iO<uR2C+1)+ *2 ja>R\Ric + R

\+ R2

j«

2 :u0

s v0

TV-;

Ri,

c?

*l*i-.. *

/?, + r2 ;

5 v- *

i

• «L_QL__

(a) (b) (c)

Fig. SI 1.4-1

Similarly,

Ri RHiK03)

/?2 (yalR,c + l) + f?i

* jo)R\R2C+Rx+ R2

•f IfTPi ffn t \2KTR7R\“d

2CT*2(*i+*2)5WH =^h +^H -

^2*2*2^ +( *, + Rlf

, „ i/;<uc _ *i+*2(b) #H = ~ =

+ (Rl + *2)

jaC Ry + ^2

(J?i + J?2)2

7KTR\R2 2KTRxR2 {R\ + R2 )

= + *2 ’^l«22C2

+(^, + «2)2

which is the same as that found in part (a).

91

1 1.4-2 y(f) =£ " a)da

(f) = x(/)y(/ + r) = x(/)£ ^(o)x(/ + r - a)da

= f h(a)x(t)x(t + t -ajrfa = h(a)Rx(r - a)da = h(r)*

R

x (r)wdS^i”) = H{a>)Sx (o)

J-flO

1

. . jeoC 1

in Fig. 11.13. H{») «—

—

=-^71+jwC

and SnVo (©) = 2X77?/ (><aRC + 1)and 7?,^ (r) = 2X77? «- r/*c«(r)

11.4-3 (a) We have found /?x (r)of impulse noise in Prob. 1 1.2-8

R% (r) = a$(r) + ar2 ,and Sx (t») = or +2*a2

$(fl>)

Hence,

Sy(n>) = \H((of[a + 2/ia

2S{o))] = 2;ra

2j//(0)|

2^®)+

and 7?y(r) = ST 1

^®)] = a2\H{of + ah(r)*h(-r)

(b) h(t) = 2-e-,lr

u(t). = l* 1

inJ+ —

Fig. SI 1.4-3

11.5-1 n(/) = ncW C0S<uc, + wsW sina, cf _The PSD of«c (/) and *,(/) are identical. They are shown in Fig. SI 1.5-1- Also, n

1is the area under

5n (o>) , and is given by n2 «

2 y x 104 +“(y 1 = 1-25 x 10

4Ol

n2^orn2

jistheareaunderS„

t(®),andisgivenbyn

2 =n f=2^5(HXUI +yy x5000

j

1-25 10 ^

Fig. SI 1.5-1

92

1 1.5-2 We follow a procedure similar to that of the solution of Prob.

11.5-1 except that the center frequencies are different. For the

3 center frequencies S„c(<»)[or (®) ]

8,8 shown in Fig.

SI 1 .5-2. In all die three cases, the area under S„c(<u) is the

same, viz., 125 x 104

cAI. Thus in all 3 cases

I^ = n2 =i25xl04 J(

r . foBk. C*riW “f * ^15 fc- Ctf'H* -£re<j- Qc

Fig. SI 1.5-2

ii.M

Fig. SI 1.5-3

. , \SmH 9 + fiT - ~

—

11.5-4 (a)+ __L_ + 6 &u 2 +60 o> +1

<b)

(c) Tlie time constant is Hence,, retntnmble «loeoftimeslehym<|uimd to entice this filter

Vio

realizable is * 0.949 sec.

>10

(d) Noise power at the output ofthe filter is

. 1 p Sm{a)Sn (eo) p”—$— dco =—^-nrtan"1 =-4-

=2«-J-*Sm (<a) + Sn (<u) ©2 +10 2W10 VIO -« VIO

The signal power at the output and the input are identical

St * S0 *— f*6

g<tto = 1

12ar^-0O 9 + fl»

2

SNR=-^-*— = 1-054

93

4

11.5-5 (a)„ ,

, Sm(a) a>2__

H^ {6)) 'Sm(c) +Sn(a>) _L_ +a)2 +4

_a)

2 +64 l[n 5333 1

9a2 +96 eu2 + 10.67J

+ 432

eo2 +64

(b)

(c) The time constant ofthe filter is 0.306 sec.

A reasonable value oftime-delay required to make this filter realizable is 3 x 0306 = 0.91 8 sec

(d) Noise power at the output of die filter is

w _l_f® Sm(fl>)SnH . _ _L p £ -do) = 0544° ' In SmH + S„(a>) 2n ^ + 10.67)

The signal power is

S,=Se =

ia.

No

Fig. SI 1.5-5

94

Chapter 12

12.1*1N0

S,

y = 1000= L -=>5, = 0.08

12 .1-2

2xl0"8 x 4000

Also, //c(a>)= 10~3

. Hence, Sr =— = 8xl °4

5r , _L£ [2 x 8000>r] * 8 x 1

0

4 =c 0 * 1

0

£5

— oC ®< (|>*>

Fig. S12.1-2

.2 . _2

aa = 8000/r

5n„H - * 10I0

[ a.2

35 dB = 3 162 = -f- = yT2—: => So " 337 * 10

_x 10-7

3

iq~3

But s0 {t)•—«(0- Hence-

So = 121-«2(/) . 337 x 10'3 => «2

(/)* 215.7 x 10

9

a 2

Also, i? -—f fiU» »^ -mop = 215.7 x 109

2n J~eo x

Hence, 0 = 26.96 x 106 andSm (e>) * 26.96 x 10

6rectQ^j

\ e<x - j / ii2 . 1 f®#

/rJ°

26.96 __-i ©i tan —an a

r io6

]

U2+ «2J

^iCD

26.96

4a: 2.68 X 10“

Sr =I£ SmH^ = ^Joa26*96 x 10

6<to =

2656^0 . 6M5 x 109

12.2-1 (.) 30dB = 1000--^- =y=^ =^ri^^=>S<sl4xl0

‘

(b) From Eq. (12.7), N0 = cAlB = 10" 10(4000) = 4 x 10'7

(c) Sj = |//c(a»)j

2Sr and 10'8Sr = 4 x 10^* => ST = 4 x 10

4

95

12 2-2 ft) -^2_ = 1000 =— 7P =5-5/ =4x10'12.2-2 W „ ,wu

jufi io_l°x4000N„

*-7

(b) N0 =JUB = 10',0 x 8000 = 4x10'

(c) S, = |//c(®)|

2Sr = l<T

8Sr = 4 x 10-4

=> 5^ = 4 x 104

12.2-3 Let the signals m,(f) and m2 (f) be transmitted over the same band by carriers of the same

frequency (t»c ), but in pha* quadrature. nte two transmitted signals are ^t)coia> ct +

ft©5 rigp ,^005^

lpf

_lPf

*9

r%

Fig. S12.2-3

The bandpass noise over the channel is nc(/)cosn>er n^sin^t. Hence, the received signal is

y2mx(f) + n c

(f)]cosa>cf + [>/2m2 (t) + n$(')]sm°

c

1

Eliminating the high frequency terms, we get the output ofthe upper lowpass filter as m,(f) +^n c

(r)

Similarly, the output of the lower demodulator is m2 (r) +^ ns(0

ntesearesimilartotheoutputsobtainedforDSB.se on page 535. Hence, we havejf-

= y for both QAM

channels.

12.2-4 (a)Hence, mp >M

Sa m<b) TT-— 2N0 A2 +m2

_H»2

j. y1 *2-V

—£- +mV

2mP

where*-

m2

m2 SIX(c) For tone modulation*

2 =-£r = 2 andfor/r = l, -j*--—r = jmpr

. ST A2 +m2 f»p2 +m2

_,wg^ 1 \{ K2 »\

(d) Ratio -f-=—

ST m2 m m

96

2^

^2

12.2-

5 OFromProb. 122-4. For*r-l«dmg, »p •J--£— »

™ o fC Tî

and when// =1, ^ = =^

s„ (os)2

r0,) Whence

12.2-

6 For tone modulation, let m(r) * /*< cos©mf . For BSB-SC,

^XU»(/) * if2pAcosa>mt -co$a>ct

* -^=-[cos(r»c + + cos(<uc - <Wm),j

,, _Zil +Zdi = inland m=-^+^ = V2/*4 Hence, the peak poweri<-

4 4 2P

V2 V2

Sp =(V^f = 2^2

2<2 and^ = r -^ ^ where Si = ^Sp

For SSB-SC

^S$b(0 = m(/)cosoc/ + m*(/)sin/wc/

= pA cos<amf cosojf/ + /*4 sin <amr sin <uct * /*4 cos(<ac - “«)'

2 ,2 2 2 S0 _ Sj p2 A2

_S

f>

Sj = — and mp = /i4. Hence, Sp = m A

N ~ j\b 2JiB 2JiB

For AM^w (r) * /<(l + /«os<»mf)cos<Hcr

c _0,2

_ jl+fiidiS‘~2

+ 2 2 2

mp = .4(1 + //) and Sp * /<2(1 + ^)

2-

Hence,

Sp(2 + /.2

)

Si-4(l + „)

2 “d »* "A2 + m2

' "A2 *{SaH2)

mr =

*A2l2 A. =1f *

21

’

5,(2V)lAIB 1l2 + /r

2J 4(l + /i)

2<^J£

S„ ^pUnder best condition, ie., for// 1,

N<> 1(Ui5

Hence, for a given peak power (givenSp ) DSB-SC has 6dB superiority, and SSB-SC has 9dB supônty

over AM. These results are derived for tone modulation and forp = l(the case most favorable for A ).

12.2-7 For 4<r loading, mp •= 4am and the carrier amplitude A = mp = 4<rm (for // = !)• For Gaussian m(r),

m2 s o2m (assuming m-0)

Prob(£ * A) = r-ê-^ l2al dEn = e'^12^ = <

” fT_

Hence, -î- « - - 4.605 and S(=

2<t?

= 0.01

A2 +m2 \6*l*oh2 2 2

n -n

_ , _S±_ s IZfdZffi.I s —(4.605) = 9.79dBTherefore, r Thresh ^ 2JUB 8 JIB

J8

V

97

12 3-1 = 28dB = 63 1. Hence,

Nc -> iw

2(/)

Z£- = 63\ = ‘i02Y— jk r »A m.

= 3(2)27ttV(^m)

,631x9

Therefore, y =—jj— = 973-^3

(.) Also, r = -^L_ => s, = r^«® = 473.25 x 2 x 10" 10 x 1 5000 = 1.4197 x 10

lMB

m B.mJ-Hy^l.igZsL91;,, =20,000*(b) P

2kB 2k

B

30,000*

sa . -ahy„. (10-,

)

I

(20,000*)! - 4/r

2

(c)

12.3-2 mp = S. S 'T “d band',''dth *Y Hen“'

VsMou (2z*2/T0 )2 B7

_3kJ_

(So/»o)FM

=

3(4s/r0 )

2 4

,-3

12.3-3 m(<) = cos3<u0/ and mp = 1

For a maximum:_2 .

m(t) = 0. This yields cos2

<»0/ = 2sin Q>o'

or l-sin2ai 0/ = 2sin

2<u0f=>sin<u0/ = ^, cosav

and

S = |-3<u0 cos2 w0( sinfl>0/|

- 3®0(jj(^J

,2 2

I

(So /No)pM m (

3a>o) mp =

9fi?o _ = 225

&,/**)FM 3m‘>2

<3-’)

12.3-4 m(r) = aiC0S<al/ + fl2 C0Sfi, 2r >

«^=oi +a2

^) = -(a,fi)lsinfi>if + a2 a, 2sina>2 r )'

mP * °l®» + °2®2

(So/^o)^ 3m’p 3{o\0)\ +02^2)

<»2o|KT

3^a22

(

\2

i +^|,

02^2 ;

98

_0+*)

2

=3(1 + *y)

2

12.3-5 Error in this problem. There should be 4*2in the denominator (see below).

SA(m) = o>2Sm («>). Hence,

£ -

n

0s lh(2^ -r>v2sm(2w

From Eq. (12.42a)2— i D*M] <*

e”'Js.(VW

These results are true for a waveform m(r)

12.3-6

~2 ^^(///o)2*^ /o3^l + *

2*?Sm “ r 1

/• r —a^l + (///e)

2t oJ^, + x2

‘

t \

fon

“*(§I-/i

2* sin(£)

The definite integrals are found from integral tables.

—2

sin(*/2*) 2 (*l2k ) _ 1 ABm ~ fo

sin(3*/2k)fo

(3/r/2fc) 3°

Ks)

As k -* ®,

f> 1 ledn\tt.

<r a

.Ifl>l -a1 12a1

Hence, the normalized PSD s is--—e2O

,f W = 2«*,then^ =(W - - to2- M«0 » *•

’-41-band-W to W.

la

-W2/2ct

2I

I

,^).2..nd4^ = '-'‘WW -°» = W = 30to' S '°4K'’

/H£^ = 0.95=>W = 2.45<7, B = 0395c

/H= 0.9 =>W = 2.15<t, fl = 0342(7

pH

x = o 99 = 3.06tr2 >W2 =- PM superior

99

x = o 95 = 2.00a2 -W2 => PM and FM equal

x = 0 9Wl.

1.54a2 < W2 => FM superior

12.3-8 m(r) = aiCosfl>1f + fl2 cos®2, < 41,(1

m2= |^5m(/y/ = (a? +

)/ 2

a2 + a

2 2

Since B = A,PM is superior to FM if/2 >5 2

a\ +05

)

2__[<fo2

,

4ft]J a

2 + o2 2 2

a?/2 +alfi

=a? +fl2

*

| 1 2

§a- m tfrdL~lp

r . Since mp = 3a, in2 - a2 and 23.4 dB- 218.8 ,

218.8 = =-(2) r

12.3-9 (a)m2 3mP

y =218--~ = 164.1 . Also, r-nn* - 20(^ + 1)

74

„ 164.1,

...So /3Thresh

m_2^

I s7 -21

!

jl = lply = 1(7J21)2(164.1) = 2844 = 3453 dB (40 dB = 10,000)

- ’" 242i

Required increase in y = = 1-479 = 1.7 dB

12.3-10 From Eq. (12.40) p2 * -

(1) Tone modulation

(2) Gaussian with 3a- loading

(3) Gaussian with 4a - loading

where For tone modulation,

[l +(m2

/mj)j

^-KnbW- a47

fl2 1=^ = 0547P

3(l + l/9j

fl2 - if—-—W- 0J6P

3^1 + 1/16j

HL-,05mj,

.

m2 a2

For Gaussian modulation with 3a— loading,2=

/, ,2 "9

'

mp (3o7

100

2 _2m o. .. ill u _ 1

For Gaussian modulation with 4cr - loading, -y = 2-

1 <(4c)

12.3-1 1 Let us first analyze the L+R channel. In this case, the demodulator output signal, when passed through the

0-15 kHz(lowpass) filter, is given by (L + R) + n0 (t), where S„0(<a

) = ^-[see Eq. (12.33)].

When this signal is passed through the de-emphasis filterHd(at) -— .*e s'8nal is restored t0 (L

and the output noise power N'a is given by

---3

Let us now consider the (L-R) channel.

Let <yc=2*x 38,000 and =2*x2100.

The received signal is FM demodulated (Fig. 5.19c). The PSD ofthe noise at the output of the FM

demodulator isS„0H = Jla

2/ A2

[see Eq. (12.33)] The output ofthe FM demodulator is separated

into (L + R)' over 0-15 kHz and (L-R)' cos <oct over the band 38 ± 15 or 23 kHz to 53 kHz. Let us consider

the signal over this passband, where the noise can be expressed as nc (r)cos<ucr + n s(/)sino»cf. The signal

is (

L

- R)' cos co ct. Hence, the received signal is [(I - R)' + nc(/)jcosoc + n

s(/)sin©ef. This signal is

multiplied by 2 costu cr and then lowpass-filtered to yield the output (Z. -R) + nc (/). But

S„£(a) = Sn(o + <uc )

+ Sn (c )= + °>c)

2+ (® '

“

°>c)2

]

When this signal is passed through de-emphasis filter //„(«) = *e signal is restored to (L-R) and

the output noise power N£ is given by

s«M)da = J \"[(a+o,c)2 +

(a ~”c )

2

]^i^(dco

m2Jia,L f? +

"c-"?. tan"1— W = 2/rx 15,000

jiA2

[<*>\ " 1

.

Hence, the (L-R) channel is noisier than (L+R) channel by factor—2- given by

at: l “i »ij 71

B-f\ tan'

Substituting ZJ - 15,000, fe = 38.000, /, * 2100 in this equation yields:

= 166.16 = 222 dB.

101

12.4-1

12.4-2

12.4-3

L = M n =>n = logw L

— = 3Lrm2

(t)

= 3M2n

_2nP

m* 2

Vwp 7

= 55 dB = 316200

For uniform distributionfn

m2m„ J'"1

/-

1

:

3mP

- rA •

(a) 316200 = 3(2)2n m

2

1WP )

o\

Fig. S12.4-2

m i

,2n

Since n must be an integer, choose n = 10 and L = 1024

nil -^2- = 3(2)20 - = 1.048576 * 10

6 = 60.17 dB.

Na 3

Bpcm - * *>MH2 bip0l““f?

1^ „„ . „ men tte new bnndwidih of transmission is

(c) To increase the SNR by 6 dB, increase n by 1 ,that is n - 1 1. l nen tne new

22x45 = 99 MHz.

Sj = 2BnEp ,Ep = 2xlO

-S,

B = 4000, « = 8

Sj = 2x4000x8x2xl0~5 =128

v-^L- 128• = 2i6xl02

\Ali_2x625 xlO

-7 x4000

^JZj-eV32=7i69xl0“9

Bm = nB = 8 x 8000 = 64 kHz (assuming bipolar line code)

1 + 4i(2,6

-l)e(V32)

21 845 = 43.4 dB.

(b) If power is reduced by 10 dB, then y = 25.6, c(V32)- 0(1.79) = 0.0367 and

S0 _N0 l + 4(2

l6-l)e(>/32)

227 = 3.56 dB.

102

The table below gives SNR for various values ofn under the reduced power.

(d)

Hence, n = 3 yields the optimum SNR. The bandwidth in this case is Bm = 3 x 8000 * 24 kHz.

12.4-4 i — />£ = p (correct detection over all K links) + smaller order terms

s(l- Ptf~' (1 - Pi) = [!-(*- l)^Jl - Pi]s\- Pi -(K- 1)Pt

So PE = Pi+{K-')P*

(b) y = 25 dB = 3162, y = 23 dB = 199.5

Pt = 2(73162/8) = 2(6287) = 1.6 x 10" 10

Pi * 2(V199i/8 )= CK4-994

)= 3 x 10"7

p£ m 99 X 1.6x 10"10 +3 X 10~7 * 3.16 x 10-7

s /»;

12.4-5 |m| - mpm{m)dm - * y

m2 = = J^m2jfim =y

o-m*m2 -(mf-y_ 2 / 2m/

^[ln(l^)]2(crm

2/«l,

2) + (2H/Aimp ) + (l///

2)

4_

3(

28)

°"2/^ 6383—j ^(In 256)

2 ^,J + 0.0068^“- + L53xl0"

5

m2 255mP (255)

2 mP

12.5-1 As noted on Pg. (570), the optimum filters for DSB-SC and SSB-SC can be obtained from Eqs. (12.83a)

and (12.83b), provided we substitute ^[5m(ty + <w e )+ Sm(to - t»c )]

forS„(a>) in these equations. Let

5m(®)“7[Sm(" + ®e) +5m(®-®c)]

2 [(a> + a;c )

2+o 2

(a>-a>e )

2 +a2

J

a 2(a>

2 +<u2 +a 2

)a»3000»

(a>2+

2 ai^^xio5

We shall also require the power ofSm (ffl)

.

/ =2

We can simplify the evaluation of this integral by recognizing that the power of the modulated signal

m(/) cos coc t is half the power ofm(f). Hence,

Ilf* , . If* a 2, o _j (o

00 a/=±J_f Sm(twVtu =— f

—; rtf<u=— tan — = —2 2^“°°

m ' ”2jt^ <y

2 +a2 a o 4

We shall use the PDE system shown in Fig. 12.19

103

(a) For this system

(n>)/Sm(®)

i „ / \iJ

(3)

(4)

Because tf» and Sn(o) are constants, we have

. ,.,2 SrfilM »o

V

v£H_{hM

"d/qss* -s^)do)

where Sm(o>) is found in Eq. (1). Also from Eg. (12.83b)

ST yli/sm (a>) tfx/JsM

(b) The output signal is Gm(f ) . Hence, S„=G m (0

We have already found the power ofm(/) to be 2(a/4) * 2 . Hence

Gla (lQ-2

)

2(3000ff)

_ 3„So = 2~ ~

2 20

To find the output noise power N„, we observe drat the noise signs! With PSD5,(o>)

-

2xl ° *WSS*S

through the de-emphasis filter in Eq. (4) above. Hence, Sn (flj)the noise PSDat die otdput of

Hd (co) is

SnW-S.H"dH

Also, the output noise power i, nc(,)/V2 and N. * |s« E,. (ITdb)]. where n2 -n2

and

12.5-

2 Similar to Prob. 12.5-1

12.5-

3 The improvement ratio in FM is-

3nJ20

10JoS,(a)da>

B3m2 —,where

m2 = 2j0°°Sm(a>Hf = andJ

0WSn,H# -J ^ 0"^ 2

Hence, the improvement ratio is

B’W-L.lsUdB.H 3

104

Chapter 13

13.1-1

Ay '"K

% -t-*

*T>/| *S^D

f & t-

Fig. S13.1-1

Tb *Tb

For the integrate and dump filter (I&D), the output is the integral ofp(t). Hence, at t » Tb , p0{Tb )= ATb .

Ifwe apply S(t) at the input of this filter, the output h(t) = u(t) - u(t - Tb )

.

Hence,

and

and

,2plin) A2T? _ 2

This is exactly the value of

p

2for the matched filter.

13.1-2 The output p0{i) of this R-C filter is I

p0(t)^A[\-e-‘IKC

)

°*‘* Tb ^—= A(\-e-

T>lRC)e*-n)lRC t>Tb

. -f 4 P-t£

The maximum value ofp0 (t) is Ap ,which occurs at Tb .

Ap- Po{Tb)=A(\-e~

T>l*C)

2 _Lffn =

2fr'

2 )-*>\ +o2 R2C2 _ 4RC

.2nP

P ~Z2°n

4A2RC(\-e-T*l*C

)

2

4A2n l'-‘-T>IRC

fJi Tb/RC

K(f>

rh

103

We now maximize p2with respect to RC. Letting x = Tb{RC, we have

(>-«-)

JU

and

This gives

and

Hence,

eb2 2xe

~X{'-e-x)-(' -e~

Xf

etc

2xe~x = \-e~x

x = U6

or

or

l+2x = e

1 126

RC* Tb

pin =(0^16)2A2TbJi

Observe that for the matched filter,

22E

p 2A2TbPmtx ~

Ji Ji

The energy of />(/) is 7}, times the power ofp(r)

.

Hence,

A2{l- m2

)„ ,

a2t .AEp Tb +^f-Tb-^-Eb

/rTSimilarly, £, =

^

* = £*

£«-J?*MrfO‘*-J? -^(l-^Jcos^ct + ^Vsin <V

= Tj, + A 2m2 Tb

Hence,

and

4/<2 7*{l-/n

2

)8£*(l-m2

)

maxJU

2£*(l-m2

)

13.2-2 Let Cj be the cost of error when 1 is transmitted, andC0 be the cost of error when 0 is transmitted,

optimum threshold bea0 in Fig. SI 3.2-2. Then:

c, =Qo£ (€

lm “ 0“ Oo Qh?)

Q> = 0)1 P («h °) = Coi 0 — -J

Let the

106

The average cost ofan error is

IfPm (l) = JPm {0)= 0J5

C>

C = /m(l) C\ + ^(o) Co(i)

For optimum threshold dC/da0 = 0. Hence, to compute dC/da0 ,we

observe that

and

Hence,

0(x) = 1—i-f* e~y2^dy

i „-*V2—= =—s*edx fin

Hence,

and

But

Hence,

dC _ 1

dlfl0 2<t_V2™

Cio

—3— : 1

^

C,o«2<r" -Q)l*

(Ap+a0 f _(^p-flp)

2

2*2 = 0

2<r» 2<r*

and q„ = ^ In £si

l C10 )

0 2Ap CW .

^J1EP2

and Ap = Ep .

^ln|>l4 L

cioj

13.2-3 We follow the procedure in the solution of Prob. 1 3.2-2. The only difference is Pm (l) and Pm (0) are not

0.5. Hence,

c - pm (i) c, + />„,(<>) c0 = Pm(i) Cio 2 Pm^ C° 1 2\

^\ °n /

and

(Ap~aof (

AP+°o)

\

Hence,

dC1

da0 2<rn -j2n

In

/n(l) Cjq e ^ -Pm(0) <^01 •2d = 0

\pm(0)c0l-

_2a<>Ap O*_ \Pm(0)C0l-

./’mCOQoJPm0)CloJ2

~~ 0 7A<*n

2AP

107

But

13.5-1

13.5-2

Hence,

2 cAiEgand Ap = Ep

Pm(0) c01a°"

4ta

[/»m(l)C10 J

/It 1 ~(r+Ep) A®*

A'h)—

W

rVtoJ

a„V2/r

p(,K)—!

—

Jr-bffaon j2it

The thresholds are ± Epj

2

and

2

T“

-4JS'/

/£ Ae-p

Here, />(/)and 9(r)are identified with 3p(t) and p(t), respectively. Hence,

H{a) = [3P(-£u) - /»(-a>)]e'-,0'r* = 2P(-a)e~J<oTt

and

But multiplication of/i(/) by a constant does not affect die performance. Hence we shall choose h(t) to be

p{T^ -i) rather than 2p{7j -/). This will also halve the threshold to a0 =2Ep . This is shown in Fig.

SI 3.5-2. Also,

108

and

Pt ^EEEMJiEEE)

9E„ + EpThe energy/bit is Eb

—^ = 5£P Hence,

Pt

13” p"

"

s^

«

msm«sk,» nqufctt . mibimum b»dwi*h 128 kto But^pltode

modulation doubles the bandwidth.

HenceBt = 256 kHz

io-7=et

V- *J

rv—

7

£fc= 2.7x10'

5, = £4£A = 2.7 x 10-7

x 256,000 = 0.069H/

For A/ = 16

o =«WOO = 64kH2

rlog2 16

7 2(15) ( l24EbPtM~Pb\°i2

16-4x10 =J6 21^2550!

This yields Eb= 5.43 x 10

$ = £6£4 = 5.43 x 10-6

x 256,000 = 1391F

For M = 32256^000 =J12kHz

rlog2 32

-7 2(31) J 30gT**i ,082 32*5x10 = ^ Cly

1023jy

This yields £* = 1.719 x 10-5

St * EbRb * 1719 x 10-5

x 256,000 = 4.4IF

-813.5-4 For A/ = 2 and <3! * 2 x 10

This case is identical to MASK for M = 2

l°’7=e

[\

if

L

j^ £6=2 ‘7><10"7

Si = EbRb = 2.7 x 10'7 x 256,000 = 0.0691F

qt =2^6,0<

?2 x 2 = 256 kHz

109

For M = 16

PeM * 0°g2 16)^ " Apb a 4 X 107

4xl0"7 s2gj2ir x 4£t

25601. £

fc= 1.67 x 10

-6

In MPSK, the minimum

Hence,

Sj = E„Rb = 1.67 x 10"6 x 256,000 = 0.42751F

bandwidth is equal to the number of M-ary pulses/second.

256 ,000 ^ 641^

' log2 16

For M 32

PtM -0og2 32)Pb = 5xlO-7

5xlO“7 s20|

2*2(5Eb )

102401=o £*=524x10“

S. » EbRb = 524 x 10-6

x 256,000 = 134W

256^000 = 512kHzT

log2 32

110

Chapter 14

14. 1-3 1) (1,1.0) is -^Lr[l + >/2sinfiv]

2) (2,1, 1) is -^[2-V2sma>o' + V2cos£»o'] 2 + 2cos^a> 0f + 4 j

3) (3’ 2> 4)is

7^3 + 2>/2 sin a>o* --J^costoo1

.

^3+ 2 .9

1

cos(<tfor “ 1040

)]

4)

-I, -1, 1) is ~7^[“|_ ‘^sin<»of + >^ cosa,o,

j

-- + 2cos^o0,+

112

-u

I*

14.1-4

< 3,-3, 3, OFig. S14.1-4

b) The energy of each signal is:

c) F3 F4 = (-6-8 + 6+8 + 0) = 0. Hence, /3 (/) and/4 (r) are orthogonal.

14.2-1 Letx(/) = xIi

x(/ + l) = x2 x(f + 2) = x 3

We wish to determine

Fxix2 x 3 (*l»*2 **3 )

Since the process x(/) is Gaussian, x,, x2 , x 3 are jointly Gaussian with identical variance

£ It The covariance matrix is:

ax,

ffX,X2 °*1*30’x

1x 2

* ^xjxi= x

tx2 * x(') x(‘ + 1

)=

O'xjX! a\x2

CTX2XJAlso trX2x3

= crX3x2« = x(* + 0 x(' + 2

)* **(')

^XjXj °*3X2 <

.

x,x3- axjx,

= xi*3 = x(') x(' + 2) = R*&

so

113

14.3*1

1 _ 1 IA,i = A33 = 1—j and A 12 = A2 i

= A23 = A32 ~g3~

e6

_ 1

A >3* A31 = 0,

~~* 4

e

And-IIA*V>

<?' J

Fig. S14.3-1

p(C|m]) = Prob^ni < and P(C|mM )= Prob^nj >

2 j

p(C|m2 )= />(CH - = P(ClmM-l) = Prob

(ln

>l < f

)

Hence

. |-[p(m|)f(Cl«i)++p

(mMlacl

mM)]

The signal energies are

114

Hence the average pulse energy E is

7_i[£+2d+2d+.....+i*^

2 1" *(M2 -\W-^<2k+,) =

-

£ K -'K"b

log2 M 121og2 M

2(W-1)|

6(log2^)

,>,M “ W S^(W2

-l) Ji

M 4 4

Hence

Which agrees with the result in Eq. (13.52c)

14.3-2 /»(C|m1 )= P{C\mA )

« />(Clm5 )

= />(C|mg )

p(C\m2 )= />(C|m3 )

= />(CK) = ^<>»)

S. S*. *s

<X —:

"t. - • • • •

s5 ; S4 -*7

?(C|wi)*/^n, -e|, n2>y)

/*(cK)* ^Khf* ,!2 > y)

f(0 = ±[j-(ch)+ *ch)] -[

2_3Cfel

P,M .i-f(c).ie(xr)f5- 34ij]]

Fig. S14J-2

115

The average pulse energy E is

£* =log2 8

and

«1

This perfonnance is considerably better than MASK in Prob- 14.3-1, which yields

PeM = \.15Q '(““ftlfor tf-8V ^

14.3-3 In this case, constants ak ' s are same for k = 1,2,

in Fig. 14.8 with terms ak 's omitted.

Hence, the optimum receiver is the same as that

We now compare r-j|, r-J2 * r sM

Since r sk= -J~Er cos6k is the angle between t and sk ,

it is clear that we

which r has the smallest angle. In short, the detector is a phase comparator,

at the smallest angle with r.

are to pick that signal sk with

. It chooses that signal which is

14.3-4 Because of symmetry,

= />(C|m2 )

= = P[C\mM )

116

and

P(C)

P'M*

Here, M = 2N

Hence

and

14.3-5

Also

Hence

14.3-6

1-22£

JIN

NP{C1«|)

1 - P{C) = 1 -

. Hence, each symbol carries the information log2 M = Nbits

.

Eb = E/Ji

1-2 2

1

£jy

Jl ^(wp),

^ 2d z*(mi) 2

«4ln2 + 42d 2

/>(Cl«,) = P(C\m_,) = Prob.[ni > -(d - /i)] = 1

/’{Cjmo) = Prob(|n,|</i) = l-

/>(C) = 1 /’(Clm0 ) + ^/>(Clm, ) +£

P(Clm_,)

1 Jf-Jilni ] JV +

2 2dJj!/f J \ 2

, 2 d1

E » 0.5(0) + 025d2 + 025d

z *—

'2£-ln2

2/I/jl

'2£

+0+ ln2

2J£/Ji

/»(c)

=^[/’(ct»,,)+

/

>

(c1,«2 )]

= _L]

«JU iJ-«t »"(*/*)

117

14.3-7

and

1 7 *,«(*/») e^-ld)l*Abdq2dq\

~nJi h

~q'

_ <j2 +4d2 _5 dj

£=2 '2

d = >/04£ - V°-1£*

FI*. S14.3-7*

Note that

.t/ , - ——A. +

—

5l =--t\~2*2 ’ 229

22

^ .d,

5 . = _— d. +— ^

2

JJ—T#l~r#2 .J4 2 ^ 2

2

118

Fig. S14.3-7b

(c) /’(c|m5 )= Prob^noise originating from s5

remains within the square of side

j

w<^)

/>(f)m5 )= 4

0^

d

2V2<t„

_ -| f 4</

2

We also observe that £ ,the average energy is £ « - —

£ 0.4rf2 02</

2

Ji Ji

Therefore F(4”s)s4d

The decision region R2 for is shown in Fig. a and again in Fig. C-l. R2 can be expressed as the first

quadrant (horizontally hatched area in Fig. C-l) -Ax

. Thus

P(Cjm2 )= noise originating from s2 lie in R2

= £(noise lie in 1st quadrant) - /’(noise lie in A,)

- j\ _jj

~ £(noise originating from s2 lie in A\

)

But /’(noise tie in Ax ) = j[/>(noise lie within outer square) - £(noise lie within inner square)] (See

Fig. C-2)

119

Moreover, by symmetry

P(e\m2 )= P(fh) = *M"*) = P(*K)

Hence

PiC) = i-l4?(C|mi )= i[?(C|m

1 )+ P(Cl«2 )

+ P(Clm3 ) + P(C|m4 )]

l 6 i=l4

The decision region /?, for m, (see Figure) can be expressed as

= outer square of side d>[2-~ (outer square - inner square of side d)

= I outer square of side djl + 1inner square of side </

4120

Now />(Cj«j) = Prob(noise originating from mj lies in R

\

)

= - P(n lie in outer square) + ^P(n lie in inner square)

=Mw<£H fK)

Similarly P2 ,the decision region for m2 (see figure above) can be expressed as

R2 = outer square of side d^2 -— (outer square - inner square of side d)

and

2

= — outer square of side dV2 -— inner square of side d

2 2

P(Cjm2 )= noise originating from m2 lie in R2

f)

i['- 2C(^)] 4[- 2e(^)

m. id

- J'. - -

ft

The decision region P3for «3 can be expressed as

R2 = RA + Rb ~ Rc

and

P(C|m3 )= Prob(noise originating from m3 lie in Ry)

= P(noise in RA )+ P{noise in Rg

)- P(noise in Pc)

" \ n ' '

P{n, >0, hM+^N’ l

n2 l

<^)-^[

;a

(ln

i!- I«»2|<^)-'p(ln il*

a*

•rm4

*4-*i

*

r— ‘ l *' .< 1• • I

V' . ad

h££i i

121

tl|<N

The decision region *4 for m4 can be expressed as

and

p(C|oi<) - P(»i > -A "2 ’

For any practical scheme g(-)« I ,»! »•« 'xPress

[l-k&)fs\-2kQ()

Using this approximation, we havex

» 1 -^Ji]-

Hence

P(C) = i[p(clmO + J’tcK)* /’(chh ^cK)]

,3 r/ ^ 1 lof4=V-eUF^I

1'2 el^rJ"4elv^j WJ

Now

Therefore

And

so that

Therefore

Moreover

Hence

£, = d2,E2 - 2d2

, £3 * 4* 2-

£< “^ ’

£ = -(d2 +2d2 +4d2 +W2)*— d

4

'

. _I I* log2 16 4

£L =X = li^ljy

=401 16 ^

And

122

Contptoon ofto result wfth tobto* »«.3[E,d457))

approximately 1 .5 times the power of the system m Example 14.3 to ach.eve tne sam ^

14.3-9 If jj is transmitted, we have

bi = E+a + jE n, ,6_,=-£ +a-V£ n,

A2= fl = V£n2 .

b_2=a-Je n2

and

Note that

Hence

Similarly

Hence

bk =a + SE n k ,b-k =a-JEnk

p(C\mi )= Prob.(b

x>6_ h, i_2 ,

— bk , b.k )

bx>b.

ximplies £ + a + 4e n, > -£ +a-V£ ni or «!>->/£

*,>*2 implies £ + a + ^£n, >a + Vfn2orn2 <Vf+ni

6, > 6_2 implies £ + a + V£ n,> a - 41 n2 or > -(Vi + " i

)

A, > &2 and 6_2 implies - (n, +4e)< n2 < (n, + Je

)

bx> bk and 6_* implies - (n, + Je) < n^ < (nj + Je)

cncc

p(c]mi ) * Prob.(Z»i > b.\, b2 , 6-2. 63, 6-3 bk , 6_*)

. Prob.[n, > ~j£, |n2 | <(m VI). |nsl < ("I + &)• '•W < ("'* '^)\

Since n„ n2,-nk «« .11 intlependen, gaussian .toon. variables each wi* variance ^2 .

/.(CH,) -[/>(»,> -41) 4^1 < n, *t/EMM < »l *- HKI ‘ 4*

)]

I" J/JrmWr I dhi

. T e~n'/Ji

THaT-Jj

,-4- J•-"M*

[CSV*'*'*

1-2 dnt

JEni +

,w-«-4W- ,

-g5?jj!^ 4'-^)]” '

Alsolog2 2N

14.4-1 The on-off signal se, to iis—ES££5SZ*

«

K5g£»tS2C?£»*•- - F*

«

is-with half the energy of on-off or orthogonal signals.

123

Therefore

5[(/) = V20^2(0

s2 (r) = V? *i(r) s2 =>/5<t>1

jj(0--V5#iW *3 = ~Vs <t>

The vector o = j£s, = j[VlO «D1- y/lO *

,+ ^20 4>a ]

— «>2 •

Hence the minimum energy signal set is given by

i,(0 = = ^20

The optimum receiver - a suitable form - in this case would be that shown in Fig. 14.8a or b.

Fig. SI 4.4-2

1 4.4-3 To find the minimum energy set, we have 0 * j (5

l+ *2 + *3 + *4 )

- ~+l ~h

Hence the new minimum energy set is

lj = V3 + ^2 , s2 = + V3 ^2 , s3 = -V3 = ^i-V3^2

124

Note that all the four signals form vertices ofa square because (*i s2 ), (*2 h)> (*3 *4 )- and (s4 *0 are

orthogonal. The distance between these signal pairs is always ifl . This set is shown in Fig. SI 4.4 3a.

J1

1

fien^iJtg„

^ V

:

1

£ rj*ii

> V

. /

1

l

XQ. e-

*1\ - ' ^7 *

iifA

P±-1*3£)—*

—

i> 1

t ~0 V<3 53

UJ, ts.

q. . - -O' Fig. S14.4-3

14.4-4

Observing symmetry we obtain

P(C) * /’(Clm,) = - ^(<>3) * W*)* Z^nj > -V2 and nj > -V2

)

HMPtM = 1 - F(C) - 1 -[1 - 0(3.16)f

s 138 x 10-3

i y /

_ 1

lofislOVg

N ,. . * f iO TRe

s,L$ - 3

~~c

_ it

t-

H1

• 4*3^

Z*\b3

-fc-*

(d)T*E- nin.Kv^ ***&/

Fig. SI 4.4-4

125

P(C,.|[2 c(cK) fM-,)] - i[2 -jefro?) * i - 2C(7 0,)

]

= 1-jQ(7.07)

PeM =i-P(C)-^Q(7.07) = l03xlO-12

Also £l= £3“(^)= 4x,0

":>

£ -f-j—f+f—LSI = 2 x 10-3 £ = {(£l + £2+£3)=

3X, °

2

£2‘liovroj iToVToJ

3 3

Mean energy of the minimum energy set:

£min4(2 ’‘ 10'3+0+2 ’“°",*‘ :

5’< '0

"5

14.4-5 Tile use of Eq. (14.76) end siqiial rotation shows th“V*' ^^^“c^atfyMt.^rs'uIeMicssl

that in Prob. 14.4-4. Hence the minimum energy set is as shown in Fig.

to that in Prob. 14.3-5 with * -^ From the results in the solution of Prob. 14.3-5, we have

e =y = 10-3

Also, we are given s„(o>) *^ = 105

* Hence ’ ^ * 2 * 10

(a) From the solution of Prob. 14.3-5

PeM = ~6(7.02) + C(7.12)= 1.09 x 10

- *2

(b) and (c> identical to those in Prob. 14.4-4

14.4^ (a) The center of gravity of the signal set is (», + «a)/2

Hence, the minimum energy signal set is

The minimum energy signals are

*,(,)« 0.5-0.707 sin^-j

x2(0 = 0.707 sin ~ 03

a>0 = 2000a

(b) EX]=°

J

>,

^0i -0,707 sin dt * 0.4984 x 10"5

£ - E = 0.4984 x 10_s

. We are given = 5x 10x2 ^1

fl-dsR?- =2(4.465) = 0.41x10,-5

126

(c) We use Gram-Schmidt orthogonalization procedure in appendix C to obtain

yi(*)= Jl(0

.001

JV2sina>0'd*

»(<)-•£ •‘""o'--4—SSi

—

, 8 8

y\ (r) * 1 - cos2a>0/-— sin a>0t +-j

N if

«i(0*y

jj(/) = 7J7^(0 + = O138p2(0+ -o285>’l(0

127

Chapter 15

15-1.1 P\ = 0.4, Pj - 03, Pi — 0.2 and P4 - 0.1

H(m) = -{Pi log Pi + />2 log P2 + ^*3 log /*3 + /»4 log ft)

=1 .846 bits (source entropy)

There are 10‘ symbols/s. Hence, the rate of infonnation generation is 1*46 x 10* bits/s.

15.1-2 Infotmation/element = log2 10 = 332 bits.

Information/picture frame - 332 x 300,000 = 9.96 x 105

bits.

15.1-3 lnfoimation/word » log2 10000 * 133 bits.

Informetion content of 1000 words - l^' <“^“'i“2tobt 9 ,6x , 0

5 biK . obviously, It is no,

™sSS“n^rPXcome

pT“c'y% 1000 words, in genentl. Hence, • pieuue is worth 1000 wo,ds

iTvery much an underrating or understating the reality.

15.1-4 (a) Both options are equally likely. Hence,

/ = log(Jj) = lbit

(b) P{2 lanterns) = 0.1

1(2 lanterns) = log2 10= 3322 bits

15.1-5 (a) All 27 symbols equiprobable and P(xt )= •

//,(x) = 27(3^ log2 27) = 4.755 bits / symbol

(b) Using the probability table, we compute

ffw(x) = -I^log «*i) = 4127 bits/symbol

(c) Using Zipf s law, we compute entropy/word Hw(x)-

S727

//w(x)=-iwiogwr=l

8727_ _ £Miog(M) = 9.1353 bits / word.

r=l'

H/letter =1 1/82/5.5-2.14 bits/symbol.

Entropy obtained by Zipf s law is much closer to the real value than W,(x) or H2 (x).

128

Redundancy y = (100- rj) * 0%

7

15.2-2 //(m) * -£ log Pj = 2289 bits

i=l

=2-2g— = 1.4442 3 - aty units

log2 3

Message Probability Code Sl

nti 1/3 0 1/3 0 1/3

m2 1/3 1 1/3 1 1/3

m 31/9 20 1/9 20“ 1/3

m4 1/9 21 1/9 21

m51/27 220 9. 1/9 22J

m6 1/27 221

m7 1/27 222

J

0

1

2

i-ifiil -j(')4w+i(2)+

i(2)+3F(3)

3-ary digits

* 1.4442 3 -ary digits

Efficiency 77 =tf(m)

=L4442

xlQ0= 100o/o

L 1.4442

Redundancy y =(1-77)100 = 0%

129

1 5.2-3 H(m) = -S Ptlog P

t= 1-69 bits

i=l

Message Probability Code

mim2

m3

nu

0.5

0.3

0.1

0.1

0 0.5

1 0 0.3

1 1 0“(-v 0.2

1 *1

J

0.5

51 oT— 0.

llj

0

1

Efficiency rj = x J00 = -jj- x 100 = 992%

Redundancy y = (1 - >7)100 * 0.8%

For ternary coding, we need one dummy message of probability 0. Thus,


mim2

m3

m4

m 5

0.5

0.3

0.1

0.1

0

0 0.5

1 0.3

20*1—r 0.2

sT

0

1

2

L = 03(1) + 03(1) + 0.1(2) + 0.1(2) = 12 3 - ary digits

H{m) = 1.69 bits = -—- = 1.0663 3 -ary units

log2 3

Efficiency „ . SSL , 100 -^ x 100 » S8J6%

Redundancy y = (1 - ^)100 = 1 1.14%

15.2-4

Message

m,

m2

m3

numjm»m7

Probability

1/2

1/4

1/8

1/16

1/32

1/64

1/64

l = ZPiLim T2 3-ary digits

lo

From Problem 15.2-1, Him) =^ bit!

-

1242 3-aiyuniB

Efficiency 7 . * 100- 100- 94«%

Redundancy y - (1 - »7)100 = 5-37%

130

m,m2

m3

m5

m«in?

1/9

1/9

1/27

1/27

1/27

Si

"1/3 1

1/3 00

1/9 Oil

1/9 0100

1/27 01010“

0101 lOT-* 2/27 01011

OlOlllJ

63

%1/3 1 1/3

1/3 00 1/3

1/9 Oil f»2/9010 1/3 01J

1/9 0100*1 1 1/9 Oil

1/9 0101

J

l “ £ ^ “ 2 4074 binstfy digits

//(m) = 2289 bits (See Problem 152-2).

Efficiency „ = ^*100 = xlOO = 95.08%

Redundancy y =(1-t?)100 = 4.92%

15.2-6 (a) H(m) = 3(^ log 3)— 1385 bits

(b) Ternary Code


m,m 2

m3

1/3

1/3

1/3

0

1

2

I = I(l) + i(l) + -j(l) = l 3 -ary digits

1 <85H(m) * 1.585 bits = - = 1 3-ary un;t

log2 3

Efficiency n - ~~p * 100 = ,00°/#

Redundancy y =(1-^)1 00 = 0%

(c) Binary Code

Message Probability Code Si

m,

m2

m3

1/3

1/3

1/3

0001

2/3

1/3

l = j(l) + (2) j(2)- 1-667 binary digits

Efficiency tj = x 100 = x 1 00 = 95.08%

Redundancy y *=(1-/7)100 = 4.92%

131

(d) Second extension - binary code

•

^[(7{03)+(2)

(?)4)

] =

f

= 161 1 binary d5gits

H(m) = 1385 bits

Efficiency tj = —7-^ x 100 = x 100 = 98.39%L 1.6 1

1

Redundancy y * (1 - >7)100 = 1.61%

Message

mimi

ProbTT Code

001 *2/9

S|

01 TT»»

Si

1/9 0000 1/9 001 10

mim) 1/9 0001 1/9 0000 1/9 001

1/9 110 1/9 0001 1/9 0000

lUjinj 1/9 111 1/9 110 1/9 0001

mimi 1/9 100 1/9 111 1/9 no"

ntiiot

IBjIll;

TflylTlj

1/9

1/9

1/9

101

010-1

01 1J

1/9

1/9

100-1

I01_P1/9

2/9

001 f*2/9>1 r»2

I I 1/9 oooonj 1

nj 1/9 0001

15.4-1 (a) The channel matrix can be represented as shown in Fig. S15.4-1

P{y\) • P(y\\x\)P(.xx )+ P(y\\x2 )P(x2 )

(b)

_ 2 jl

+ J_ 2 =13

=3 3

+10 3 45

P{yi)^\- P{y\)-^

H{x) = />(*,) U>g^TT + ^JlogF(x,)

’ F(x2 )

1 2 3= — logj 3 + — logj — = 0.9 1 8 bits

P- 2-r 3

Fig. S15

To compute //(x|y) , we find

FO-,) 13

rto)-u-

PWn)p{yi) 32

F( 2,W)/>0'2 )

64

//(x|y,) = F(x,|yi)»oi + ^(X2l>’l) ,081

^X2 lFl)

10, 13 3 . 13ft_70

:— tog2— +— log2— = 0.779

1382

10 1362

3

H(x\y2 ) = F(xi|>|)log1

+ P(x2 \y2)\og1

P(x2 \y2 )

5, 32 54. 64

132

and

Thus,

Also,

H(x|y) « Piyx)H(x\yO+ P(yi)H{x\yi )

- 11(0.779) + 11(0.624) = 0.6687

/(x|y) = H(x) - W(x|y) = 0.91 8 -0.6687 = 024893 bits/ binit

H(y)»Z^) |o8/

1

Piyd

13 . 45 32

45°8

TT « *45

32= 0.8673 bits /symbol

//(y| X) = tf(y)-/(x|y) = 0J8673 - 02493 * 0.618 bits / symbol

-rn. _ .Lo-nal matriv p( 1J ,lr \ IKi. y / P)

l nc buoiui^i uibuia » vjnt -

'10 0

0 p l-p

0 l-p p

(p) *1 • 7\

'

y

)

f<3>j

yAlso, P(yi) = p, P(P2) P(yi) = Q

(<5) x3^- >3 '

P(yj\x,)P(xi) . •

Now we use “ obl,“ Fig. SI 5.4-2

P(x,\y,) matrix as y}

xi10 0

0 p l-p

|_0 1 -p p

1

//(x) = I /*(jc1)log—1— = -/’log P - 2gloge with (20 = 1- P)

syXj

)

P log P + (l-P) log(¥)] n(p)+o-p)

i

//(x|y) = II P(yj)P(xi\yj)'<>& P{x ,

lyj)• j

/> log 1

+^ log-^+ ( 1 - P) log

J

+ ^jd - P) !og— + p log-

O+20£2(p) = (l-P)n(P)

/(x|y) = H(x) - H(x|y) = fl(P)+(l - P)-0 -TO= fi(P) + (l-P)[l-n(P>3

Letting p = 20**0 or £2(p) = \o%P

/(x|y) = n(/’)+(l-/>Xl-log^)

— /(xiy)-O or 4[n</>

> + < l - /’>< 1 - ,oS^ = 0 - 71,151,168,15

4Pv dP

iL[p log P + (1 - P) -0 - P)M1 - P)(1 * log#] = 0

log P - log(l - P) + [1 - »og/?] = 0

133

Therefore\og— = -^°lP

PNote: -1 + log2 fi

—^2 2 + l°g2 P 9 lo^ 1

P P_ p = P— and 1 - P - ~—

r

JI7 T =*f + 2 P+2

so

C= MAT/(xjy)to,!

(00)

=

0 - f|« AO-V * ^ _3fifi

fa)

BSta ">k. M M > —y « 'K-lFig. S15.4-3

-ft*-B^CL

*

a;

n-A a' W2 Pi

rU Pi \ -Pi.

Hence, the channel matrix of the cascade is

\l-Px-Pl-2P\Pi P\ + Pi ~2P[Pl

[p, + P2 -2P1P 1-A“^"2^P

This result will prove everything in this problem.

, , with , . bom the above result it follows that the channel matrix *s^**~^

(a) With P\ Pi e>oftwo cascaded channels is M\Mi-

(b) We have already shown thal the c

,with the H* channel.

ret Consider a c^cade of . identica'^h^en^^-JtT^lhs denvcd in (b).

if «... ,s the channel m.mx of the fim * - chenneU ^ ^^ for^ * Wehjye

-*——*7 ‘^.7^ -— « --- «*r«;•

have, for a cascade of 3 channels

\-Pe .Q-Pti* + 3P'2Q-P')

= l-3Pe+ 3?e2 -Pe

3 +3/>r2 -3/»

e3

*1-3P,+*P?-*P?

we

and

/>£ =3Pe -6/>e2 +4/>

e

3

Now

134

3 fl -Pe Pt l3 [l-(3Pe -6/»e

2+4/i

3) 3Pe -6Pe

2 +4P,3

M"U J-^J

=[3P# -6Pe

2 +4P<3 l-(3Pe -6Pe

2 +4Pe3).

Clearly

P£ =3P,-6P,2 +4P, 3

which confirms the results in Example 10.7 for * = 3.

^Log-^a-Zilloe^j

(d) From Equation 15.25

C, = l-

wbere PE is the error probability of cascade of k identical channel

We have shown in Example 10.7 that

Pe- 1-

If kPe « 1, PE s kPe

and

C, = 1-\tft log^+0 - ) log

15.4-4 The channel matrix is

>1 >2 >3

X, 9 0 P

Xi0 4 P

<?= 1-P

Let

x, « 0 yi=0

x2 = 1 yi = 1

>3 = £

Also,

’(>1) » P(Xl,>l)+ P(x2 ,>i) - P(x1)P(>i!x,)+ P(x2 )PO-i|x2 )

[y2)‘ P(x,,>2 ) + P(x2 .>2 ) = P(xi)P(>2|x,)+P(x2)PCy2 |x2 )

P(y3 )= 1- P(>i)- PCP2)* 1- 9 * P

«ala)£££i>.i£.i'W,)

/-O',) «/2

'™w)—twT p ^

'!lW) P(«) P 2

135

15 .4-5

Therefore,

P{xl ,yi )= P(x\)P(yi\xi) = ^

P(xuy2)= P(x\)P(y2\*\) = °

P(xi,y2 ) = P(x{)P(y-}\x\) = |-

P(x2 >yi) ~ P(.x2 )P(y\\x

‘i )= 0

P{x2 ,yi) = />(*2)/>0'2lx2)=2

P(x2 ,yy) = />(jt2> />0'3lx2) = f

W(x) = -P(*,)l0g />(*!>- /’(*2) ,°8^X2)

~ui-i2 2

W(x|y) = SI /’(*/.>';) log

’

= i(0) + 0 + ^p + 0 +^x0 +^ = P2 2 2 2

/(x|y) = //(x)-Wx|y)

= 1-p bits/ symbols

H(x| z) - W(xly) = 11 P(x, , zk ) log—~ -11 } log

/>(x,|y7 )

Note that for cascaded channel, the output z depends only on y. Therefore, X J

By Bayes’ rule. Fi_ si 5.4-5

P(x,\yj,ik ) - /’(x/l^) r,g> MS-

„ Wry**)//(x|z)- //(x|y) =

/»(jc(|x*

)

r_ .. P(*i\yj»*k)

y x L x

„„ b. shown that the sunonaion ove, , of the «m> Inside the bn*ke, is nonnegative. Hen.., it Mows

**//(x|z)-W(x|y)2 0

From the relationship for /(xly)and /(x|z), it immediately follows that

/(x|y) £ /(x|z)

136

15.5-1

15.5-2

We have H(x) = J_^plog-Ur = M ~p\ogpdx and j Mpdx = 1

Thus,

dFF{x,p) = -p\ogp and— = -(1 + log/))

dp

d*4\(x,p) = /) and -r1 = 1

dp

Substituting these quantities in Equation 15.37, we have

-(1 + log /?) + «! =0^ p = ea

'1

and

Hence,

Also,

*r''TS

m* r0,)mTii

We have H(x) = -J" p logpdx, A =J" x/wfr, 1 = J*

dFF(x, /?)

= -/> log /> and — = -(l + log/>)

dp

di\h(x,p) =pxmd— = x

di-, ,

h (x,/>) = pand— = 1

Substituting these quantities in Equation 15.37, we have

-(l + log/») + aiX + a2 =0

p = ea =(e

ar2- 1

)eflr

»jr

Substituting this relationship in earlier constraints, we get

,a2 -l

or

Hence,

ea2~ l = -a.

and

Hence,

so

i/Kx) = -a,ea|Jt

a\ -—- and e®2”

1 = -«i = A

p(x) =1/^ x £ 0

0 x< 0

To obtain //(x)

137

H(x) = -J" p(x) logp(x)dx = -J*

/jj^-log^-- loge] dx

= log /if® P(x)& + îfj*xp(x)A

* log >< + loge = log(e/l)

15 5.3 Information pm picture f*m« - 9.96 x 10s

bits. (Sec Problem 15 . 1 ). For 30 pientre frames per second, we

need a channel with capacity C given by

C = 30x9.96* 10s = 2.988 x 10

7bits/sec.

But for a white Gaussian noise

C = fllog

We are given — = 50 db = 100,000 (Note: 100,000 = 50 db)

N

K)

Hence,B = 1.8 MHz

15.5-1 Consider a narrowband *4 where Af -+ 0 so that we may consider both signal noise power density to be

constant (bandlimited white) over the interval A/" . The signal and noise power Sand N respectively are

given by

S-2Ss(ia)£4 and N = 2S„(<w)A/

rs + A,1.Anc»S,(a)) + Sn (fli)

[tJ Sn (*>) J

The capacity of the channel over the entire band (/i,/2 ) is given by

n(») .

We now wish to maximize C where the constraint is that the signal power is constant.

f/2

Vi[

5n (fl

<4

2$ Ss(a>) df -S (a constant)

Using Equation 15.37, we obtain

or

Thus,

A log[Ûfôas, [

s„ Jas.

s + S„ = -— (a constant)s a

Ss(<u) + S„(a>) = ~^

This shows that to attain the maximum channel capacity, the signal power density + noise power density

must be a constant (white). Under this condition,

C= f/2 l0g[W^ni£0l# = |/2|og y-rWC

Vi8

[Sn (ffl) J

j/lL

aSn (tu)J

= (/2 -/i)log (-')'^ log

138

15.5-5

* Blog [5s(a) + Sn(®)]-J/f ,06 [5n(©)]#

In this problem, we use the results of Problem 15.5-4. Under the best possible conditions,

C = Blog [St(a>) + Sn (<u)] - \}}\

log [Sn (<w)]

constant

We shall now show that the integral § log [Sn (a>)] df is maximum when $„(*) = constant if the noise

is constrained to have a given mean square value (power). Thus, we wish to maximize

$ log [S„(®)]tf

under the constraint

2jflog [Sn (<u)]4f

= N (a constant)

Using Equation 15.37, we have

-r“(l°B‘Sn ) + flf—- = 0dS, 3Sn

or

and

— + a = 0

S„(<o) *— (a constant)

a

Thus, we have shown that for a noise with a given power, the integral

log [Sn(co)]df

is maximized when the noise is white. This shows that white Guassian noise is the worst possible kind of

noise.

139

I Chapter 16

(?)-(?)+(?)(?)*(?)J=0

2048*1 + 23 + 23x11 + 23x77 = 204816.1-

2 (a) There are(})

ways in which j positions can be chosen from n . But for a temery code, a digit can

be mistaken for two other digits. Hence the number of possible errors in j places is

("V3-1)' or 3" * 3* I ("ft*- 3"-* * f (j)2

y

'' j*0 j*

0

(b) (1 1,6) code for t = 2

35

^(y) + (!

1

)2+(y)22 = 1 + 22+220 = 243

This is satisfied exactly.

16.1-

3 For (1 8,7) code to correct up to 3 errors

Jm0

« 1 + 1®: + -I®-- + = 1 + 18 + 153 + 816 = 98817! 2! 16! 3! 15!

211 =2048

Hence

7=0

Clearly, there exists a possibility of 3 error correcting (18,7) code. Since the Hamming bound is

oversatisfied, this code could correct some 4 error patterns in addition to all patterns with up to 3 errors.

[>

16.2-

1 GH t = [/* P]L/m.

= p©p-0

16.2-

2 c — dG where d is a single digit (0 or l)

.

For d = 0

c = 0 [l 1 1] = [000]

For d = 1

c = l[l 1 1]-[1 1 1]

140

16.2-3

16.2-4

16.2-5

c = dG where d is a single digit (0 or l)

.

For d = 0

c = 0 [l 1 1 1 1] = [0 0 0 0 0]

For d = 1

c = 1 [1 1 1 1 1] = [1 1 1 11]

Hence in this code a digit repeats 5 times,

detection.

Such a code can correct up to two errors using majority rule for

0 is transmitted by [0 0 0] and 1 is transmitted by [l 11]

(a) This is clearly a systematic code with

10 0- • • 0 f T0 10- • 0 1

p =1

000 •

_jn

i

/*

Note that m = 1

(c) This is a parity check rode, if a single am,, ocean »ywhara in ihe coda woni. ihc pariiy is violated.

Therefore this code can detect a single error.

T(d) Equation (16.9a) in the text shows that cH = 0

.

Nowr~c@e

and

rH T =*(c®e)HT =cH t 9eHT *eH T

If there is no error e * 0 and

rH T = eH T =0Also

-E]-But since m * 1, lm - [l]

and

1

1

HT =

1

1

141

ll<km i, a single error in the received word r, e has a single 1 elenrnn. wid, all other elemenrs being 0.

Hence

rH T =eH T = 1 (for single error)

16.2-6

Data word Code word

0 0 0 o 0 0 0 0 0

0 0 1 1 1 0 0 0 1

0 1 o 1110 100 1 1 0 0 10 111 0 0 0 1110 1

1 0 1 10 110 0

1 1 o 10 0 1111 1 1 0 10 110

From this code we see the. die dismnee between any two code words is « least 3. Hence = 3

16.2-7

16 .2-8

Data word Code word

0 0 0 "0 6 0 0 o o

0 0 1 0 0 1 1 1 0

0 1 o 0 1 o 1 o 1

0 1 1 0 110 1110 0 1 0 0 0 1 1

1 0 1 10 110 1

1 1 o 110 1101 1 1 1110 0 0

Observe that dm

m

= 3

T *

H T « a 1 5 x 4 matrix with all distinct rows. One possible H is:

HT

ini 1

1110

110 1

1 1 oo

ion10 10

1001

00 1 1

0 111

ono0 101

1000

0 1 00

00 10

000 1

142

G = [/*J*] =

'l 00000000001 1 1 1

010000000001 110

001000000001 101

000100000001 100

0000 1000000 1 01 1

0000010000010100000001 000010010000000 1000001

1

0000000010001 1

1

000000000 100110

000000000010101

For 4 = 10111010101c = dG = [10111010101] <7-101110101011110

16.2-9 (a)

1 1 r

1 1 0100 111

T 1 0 1

0 10 110 & H t =1 00

00 1 1 0 1

Ik P.

0 1 0

00 1

(b)

Data word 1Code word

0 0 0 0 0 0 0 0 0

0 0 1 0 0 1 1 0 1

0 1 0 0 1 0 1 1 0

0 1 1 0 1 1 0 1 1

1 0 0 1 0 0 1 1 1

1 0 1 1 0 1 0 1 0

1 1 0 1 1 0 0 0 1

1 1 1 1 1 1 1 0 0

(c) The minimum distance between any two code words is 3. Hence, this is a single error correcting code.

Since there are 6 single errors and 7 syndromes, we can correct all single errors and one double error.

(d)

e

1 0 0

0 1 0

0 0 1

0 0 0

0 0 0

0 0 0

1 0 0

0000 0 0

0 0 0

1 0 0

0 1 0

0 0 1

1 0 0

s

1 1 1

1 1 o

1 0 1

1 0 0

0 1 0

0 0 1

0 1 1

s = eHT

143

(e)

r s *

101100

000110101010

110

110

000

c d

010000 111100 111

010000 010110 010

000000 101010 101

16.2-10 (a) done in Prob. 16.2-7

(b) HT =

0 1 f

1 0 1

1 1 0

100

0 1 0

00 1

six single errors

1 double error

e s

100000 Oil

010000 101

001000 110

000100 100

000010 010

000001 001

100100 111

16.2-11

fl 000 1 0 l

<?=['* p]=0 100111

00 100 11

jo 0 0 1 1 1 o

c “ dG

144

d c

0000 0000000

0001 00011100010 0010011

0011 0011101

0100 0100111 101

0101 0101001 1 1

1

0110 0110100 0 1 1

0111 0111010 ht = 1 1 0

1000 1000101 100

1001 1001011 010

1010 1010110 00 1

1011 1011000

1 100 1100010

1101 1101100

1110 1 1 10001

nil 1111111

s = eH T

e 5

0000001 00 1

0000010 010

0000100 100

0001000 1 1 0

0010000 Oil

0100000 1 1

1

1000000 10!

s-rH T where r = received code

c = r®,

c = corrected code

16.2-12 We observe that the syndrome for all the three 2-error patterns 100010. 010100 or 001001 have the same

syndrome namely 111 Since the decoding table specifies , = 111 for , = 100010 whenever^ 100010

occurs, it will be corrected. The other two patients will not be corrected. If for exampte, -0101

occurs ,**111 and we shall read from the decoding table e = 100010 and the error is not corrected.

Ifwe wish to correct the 2-error pattern 010100 (along with six single error patterns), the new decoding

table is identical to that in Table 16.3 except for the last entry which should be

e s

010100 111

16.2-13 From Eq. on P.737, for a simple error correcting code

2""**n + l or 2""*i« + l -*• n-85>log2(«+l)

This is satisfied for « ;> 12 . Choose n = 12 . This gives a (12, 8) code. H Tis chosen to have 12 distinct

rows of four elements with the last 4 rows forming an identity matrix. Hence,

145

90 1 1,

0 10 1

0 1100 1111 00 1

101010 11

1 10010000 1 0000 10000 1

G = [I*P]

G -

1000000000 1

1

01000000010100 10000001 1

0

00010000011100001000100100000100101000000010101100000001 1 100

Tile number ofnon-zero syndromes =16-1 = 15. mere ere 12 single error paneros. Hence we may be

able to correct 3 double-error patterns.

£ -

0000 oooooooooooo

0011 100000000000

0101 01 0000000000

0110 001000000000

0111 000100000000

1001 000010000000

1010 000001000000

1011 0000001 00000

1100 000000010000

1000 000000001000

0100 000000000100

0010 000000000010

0001 000000000001

nil 100000010000

1110 001000001000

1101 000000010001

16.2-14

Data word Code word

00 00000001 01101110 1011101 1 110101

The minimum distance between any two code words is dmi„ = 4 . Therefore, it can

patterns. Since the code oversatisfies Hamming bound it can also correct some 2-error

3-error patterns.

correct all 1 -error

and possibly some

146

(b)

i/r =

1 1 1 0‘

1011

1000

0100

00 1 0

0001.

and s = eH T

6 single error patterns

7 double-error patterns

2 triple-error patterns {

e s

100000 1110010000 1011001000 1000000100 0100000010 0010000001 00011 1 0000 0101101000 0110100100 1010100010 1 1 00100001 1 1 1

1

011000 0011010010 10010001 1 1 0111001101 1101

16.3-1 Systematic (7, 4) cyclic code

g(x) = x3 +x + l

For data 1111 d{x) = x3 + x

2 + x + l

x3(x

3 + X2 + X + l) = X

6 + xs +x4 + x

3

x3 +x2 +

1

X3 + x + l)x

6 +x5 +x4 +x3~~

' - 4 1

+ x +X

x3 +x

2

2

X3 +X + l

x2 + X + 1

c(x) = (x3 +x + l)(x

3 + x + l) = x6 + x

5 + x4 + x

3 +x2 +x + l

The code word is 11111111

147

For data 1110 </(*) = x3 + x

2 +x

x3 + x2

x3 + x + ljx6 +xs +x4

r6 +X4 +X 3

+ XJ

+ x 3 + x2

The code word is 1110100 nA similar procedure is used to find the remaining codes (see Table )

(b) From Table 1 it can be seen that the minimum distance between any two codes is 3.

single error correcting code.

d c

1 1 1 1 11111111110 11101001101 1 1010011100 1 1000101011 10 110001010 10100111001 10011101000 10001010111 01110100110 01 100010101 01011000100 01001110011 00111010010 00101100001 000101 1

0000 0000000

Table 1

(c) There are seven possible non-zero syndromes.

x3 + x + l

for * = 1000000 x3 + x + ljx*

x6 + x4 +x3

x4 + x3

X4 +x2

x3 +x2

+ x

+ x

+ X + 1

+ 1

5-101

Hence this is a

148

The remaining syndromes are shown in Table 2.

e s

1000000 1010100000 1110010000 1100001000 Oil0000100 1000000010 0100000001 001

Table 2

(d) The received data is 1101100

r(x) = x6 + x

5 +x3 +x 2

x 3 + x + l]x6 + x

3 + x3 + x

2

+ x4 + x

3

Xs + x

4 + x/

+ x3 + x2

x4 +x 3

+ x2 +x

X3 +x2 +x

s(x) = x2 + 1 X3

S = 1 0 1

From Table 2

,=1000000c=r® *= 1101100© 1000000- 0101100

Hence d - 0 1 0 1

16.3-2 g(x) = x" +x9 +x 7 +x6 +x 5 +x + l

c(x) = d(x)g(x)

x + 1

+ 1

and

and

<f, = 00001 11 10000, </,(x) = x7 +x6 +x5 + x

4

d2 = 101010101010,</2 (x) =

x

11 +x9 +x7 +x5 +x3 +x

C,(x) = rf,(x)*(x) = x18 +x 17 +X 13 +x 12 +Xn +x9 +X8 +x7 +X4

C =00001100011101110010000

c2 (x) = d2(x)g(x)= x32 + x

l' + x ' 7 +x l5 + x ,3 + x* + x5 + x4 + x3 + x2

c2=10001101010000100111110

+ x

149

16.3-3

16.3-4

16.3-4

x2 +1

x + l')jc3+Jt

2 +x + l

x + 1

x + 1

0

Hence x3 + x

2 + x + 1 = (x + l)(x2 +

1)- (* + ty* + + 1M* +

Hence xs + x

4 +x2 + 1 = (x + l)(x4 +x + l)

x2 + 1

x + 1

x + 1

0

Now dividing s4 +x+l by x+ 1 , we jet a remainder 1. Hence (*+t) is not a factor of (x4 +Jt+ *)'

Hie 2"d-order prime factors nol divisible by a. lure x! andi

2 +a + l Since (x4 * x +

1)is not divisible

by x’.w, tty dividing by (x’-x-l). TO also yields, remainder I . Hence x4+ xtl doesnothave

either a first or a second order factor. This means it cannot have a third order factor erther. Hence

X5 + X

4 + X2 + 1 “ (x + l)(x

4 + X + l)

rx4 +x + l

x + ljx5 +x4 +x2 +l

Xs +X4

Try dividing x7 + l by x + 1

: +1

JV2 +

1

x6 +xS +x4 +x 3 +x2 +x + l

xi+xi.

x6 + l

x6 + x

5

X5 + l

xs + x4

x4 + l

x4 +x 3

x3 + 1

x3 +xi

x2 + l

x2 +x

x + 1

X + 1

0

150

Therefore (x7 +

1)= (x + l)(x

6 + xs + x4 + x3 + x2 + x +1)

Now try dividing (x6 + x 5 + x

4 + x3 + x

2 x+1)

by (x + 1) . It does not divide. So try dividing by

(x2 + 1) . It does not divide. Try dividing by (x

2 + x + 1) . It does not divide. Next try dividing by

(x3 + lj . It does not divide either. Now try dividing by (x

3 + x + 1) . It divides. We find

(x6 + X5 + X

4 + x3 + x2 + X +1)= (x

3 + x + iXx3 + X

2 + l)

Since (x3 + x2 +

1)is not divisible by x or x + 1 (the only two first-order prime factors), it must be a

third-order prime factor. Hence

x7 + 1 - (x + l)(x

2 + X + l\x2 + X

2 + l)

16.3-6 For a single error correcting (7, 4) cyclic code with a generator polynomial

g(x) = x3 +x2 +1

A=4 n =

7

Hence

'l 1 0 1 0 0 O'

_0110100

c “ 00110100001101

Each code word is found by matrix multiplication c » dG’

c * [0 0 0 0]

c = [0 0 0 l]

1 101000'

011010000110100001101

1101000'

011010000110100001101

0000000

0001101

151

The remaining codes are found in a similar manner. See table below.

d c

0000 00000000001 00011010010 00110100011 00101110100 01101000101 01110010110 01011100111 01000011000 11010001001 11001011010 11100101011 11111111 100 10111001101 10100011110 10001 10

1111 1001011

16.3-7 g(x)^x 3 +x2 *

1

The desired form is

G’ =

1 000 • • 0hn h2 \hi i •*m,

0 100- • o *12 *22 *32• •

*FB2

0010- • • 0 *13 *23 *33'

0000 •

_J *u *2* *3k •

/* r

(kxk) [kxm)

The code is found by using c * dG

Proceeding with matrix multiplication, and noting that

0 + 0-0, 0 + l = l + 0 = l, l + l-o and 0x0-0, Oxl-lxO

we get

c15 =[1111]

10001100100011ooioi 1

1

0001 101

=[1111111]

C14 =[U10]C=[1110010]

and so on.

i, 1x1 = 1

132

d c

1111 11111111110 11100101101 11010001100 11001011011 10111001010 10100011001 10010111000 10001 1 o

0111 01110010110 01101000101 01011100100 0100011001 1 00110100010 00101110001 00011010000 0000000

These results agree with those of Table 16.5

16.3-8 (a)

*101 1 0 0 O'

0101100G -

0010110000 1 0 11

(b) The code is found by matrix multiplication. dG'

In general gix)= S\x

" *+ 82x

" * 1+ ’"8»-*+i

For this case gj = 1. g2 * •. S3= 8* ~ 1

Since A,* -ft, ^ =g3 . *» -ft. the fourth row is immediately found. Thus, so far we have

(7 =

0001 1 0 1

Next, to get row 3, use row 4 with one left shift.

00110100001 101

The i is eliminated by adding row 4 to row 3.

00101110001101

Next for row 2, use row 3 with 1 left shift.

153

01011100010111000 1 10 1

The 1 is eliminated by adding row 4 to row 2.

010001

1

00101110001101

Next for row 1, use row 2 with 1 left shift.

’10 0 01 10*

0 1000 1 1

0010111000 1101

This is the desired form.

c d

0000 00000000001 00010110010 0010110001 1 00111010100 01011000101 01001110110 01110100111 01100011000 10110001001 10100111010 10011101011 10001011 100 11101001101 11111111110 11000101 1 1

1

1101001

(c) All code words are at a minimum distance of 3 units. Hence this is a single error correcting code.

16.3-9 g(x) = x3 + x + 1 . Hence row 4 is 0 0 0 1 0 1 1.

G'

'1 0 1 1000*

010110000101100001011

Row 4 is ok.

Row 3 is left shift ofrow 4.

For row 2, left shift row 3.

And add row 1 to obtain row 2.

For row 1 , left shift row 2.

And add row 1 to obtain row 1

.

0 0 0 1 0 1 1 «- row 4

0 0 10110 «- row 3

01011000 100111 4- row 2

10011101 000 1 0 1 4- row 1

154

1 0 0 0 10 1

0 100111

00101100001011

16.4-1

16.5-1

The burst (of length 5) detection ability is obvious. The single error correcting i

as follows^ If in any segment of b digits a single error occurs, it will violate the parity in that segment.

H^ewe'l<>«M*«8«^*n,w*.ere,^ e,Tor

augmented segment. By checking which bit in the augmented segment violates die parity, we can

die wrong bit position exactly.

The code can correct any 3 bursts of length 10 or less. It can also correct any 3 random errors in each code

word.

16.7-1 PEu = kQ(pEb /Ji) = 120^2x9.12) = 9525 x 10"6

4-4

»(23)[e(V9lT65)] = 9.872 xlO"9

To achieve a value 9.872 x 10-9

for PEu .we need new value Eb/Ji say E'b !^ Thcn

This means Eb /Ji must be increased from 9.12 to 18.18 (nearly doubled).

Pec =(2kE±

nJ!

155

Modern Digital And Analog Communications Systems B P Lathi ...

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