Chapter 2

2.1-1 Let us denote the signal in question by g j t ) and its energy by E,. For parts (a) and (b)

4 r 4 r 4 r

(e) ~ ~ - ~ ~ s i n ~ r d r = f ~ ~ d t - ~ / ~ ~ s ~ ~ d t = n + O = ~

2 n

(d) E. = 1 ( ls in t12 dt = 4 [i 12" dt - ; 1. cos 2 dt ] = Iln + 01 = 4n

Sign change and time shift do not affect the signal energy. Doubling the signal quadr~iples its energy. In the same way we can sllow that the energy of kp(t) is k2&.

Sirnilally. we can show that E x - , = 4n Therefore Ex*, = E:, + E,. Mre are tempted to conclude thar C,*, = E, - t', in general. Let us see.

Therefore. in general Eli, # El $ E,

2.1-4 This prob!em is identical to Example 2.2b. except that ;u.i f a. J n this case. the third integral in Po (see p. 19 is not zero. This integral is given by

Cl Cz T/?

= T-,r. lim T [ ~ T , 2 c o l i ~ l - O 2 ) 0 t + c o s ( 2 ~ l t + (9, + 8 2 ) rlt I

Therefore

2.1-5 2

( ~ ' ) ~ d t = 64/7 (a) p-y 5 f 12( - t3 )2d t = 64/7

1 ' 2 ? 2 (b) Pz. = J 2 ( 2 t ) dt = 4(64/i) = 256/7 (e) Peg = $ /_2(ct')2dt = 64r?/l

Sign change of a signal does not affect its power. hfultiplication of a signal by a constant r . increases the power by a factor 2.

w2 (t) rlt = - dt = 0.5

T;2 T/2 r~

Py = !kn= ; J ,q(t)g4 i t ) dt I z D ~ D * , ~ J ( - A - - ~ ) ~ dt

T i 2 - 7 / 2 k=m rrm

The integrals of the cross-product terms (when k # r ) are finite because the integrands are periodic signals Onnde up of sinusoids!. These terms. when dh~ided by T - oo. yield zero. The remaining terms ( k = r . ) yield

k e n :

2.1-8 (a) Power of a sinusoid of amplitude C is c 2 / 2 [Eq. (2.6a)l regardless of its frequency (4, # O j and phase. Tllerrfore. in this case P = (10)'/2 = 50. (b) Power of a sum of sinusoids is equal to the sum of the powers of the sinusoids IEq. (2.6b)j. Therefore. in this case P = + = 178.

(c) (10 + 2 sin 3t) cos 10t = locos lint 0 sin 131 - sin 3,. Hence from Eq. (2.6b) P = + 4 + 4 = 51.

(d) locos St cos lot = 5(cos St + cos 1 3 . Hence from Eq. (2.6b) P = a f + e& = 25. (E) IOsin i tcos 101 = 5(sin 151 - sin S t . Hence from Eq. (2.6b) P = % + = 23. ( f ) rtat C0S40i = 4 [ p-~(a+-ro)t + cJ("-"o)']. Using the result in Prob. 2.1-7. we obtain P = (1/4) + (1 /4 ) = l j 2 .

2.2-1 Foi a real n

T / 2

P, = l i~n - (c-")~ rlt = lim - FOI irnegina~ y o. let n = j r . Then

( p J f ' ) ( r - ' f t ) d t = lim - tlt = 1

Fig. S2.3-2

CIlearly. if o is veal. r-"' is neither energy not power signal. However. if a is imaginary, it is a power signal with POH'CI. 1.

The signal g s ( t l can be obtained by (i) delaying g ( t ) by 1 second (replace t with t - 1). (ii) then time-expanding by n factor 2 (replace t with 1,.'2). (iiij then ri~ultiply with 1.5. Thus p s ( t ) = 1 . 5 g ( i - 1).

2.3-2 All the signals are showri in Fig. 52.3-2

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

Fig. 52.99

3.4-1 Using the fact that g ( r ) h ( . t ) = g(O)O(r). we haw

( a ) 0 (b) $A(&) ( c ) i h ( t ) (d) - A t - I ) is) h h ( ; + 3) ( f ) k d ( u ) (UX L' Hirpital's ride)

1.4-2 In these problems ternember that impulse h ( n : ) is located at 3. = 0. Thus. an impulse 6 ( t - :) is located at 7 = t and so on. (a) The impulse is located at : = t and g17) at s = t is g ( t ) . 'Therefore

(b) The impulse b(r) is at T = 0 and g(t - s ) at r = 0 is g ( t ) . Therefore

Using similar arguments. we obtain ( c ) I (d) 0 (e) r v f ) 5 (g) g ( - 1 ) (h) -c2

2.4-3 Letting n t = 2.. we obtain (for n > 0)

Similarly for n < 0 . we show that this integral is - 3 @ ( 0 ) . Therefore

2.5-1 Trl\.lal Take the derivative of lei2 with respect to c and equate it to zero.

2.5-2 (a) In this case El = $ dt = 1 . and

(b) f111rs. q t t ) e 0.5 .7(!) . and the error r ( t j = t - 0.5 over (0 5 t 5 1). and zero outside this intc~.vai -Also E, and E . (the energy of the el rorj are

Eg = 1' g 2 ( t j d t = 1' t 2 d + = l j 3 and E. - (I - 0.5)'df = 1 / 1 2 1' 'I'hc error ( 1 - O..5) is orthogonal to . r ( t ) hecause

Sote that E, = r 2 ~ , + E,. To explain these results in terms of vector cotacepts we observe from Fig. 2.15 that the e r ~ o r \-ector e is orthogonal to the component 12. Because of this orthogonality, the length-square of g jo)e:gy of g ( t ) l is equal to I he sum of the square of the lengths of EX and e isurn of the energies of r , r ( t ) and :

r ( t ) l .

2.5-3 I n this case E, = g 2 ( t ) d t = ,/: t2rlt = 1/3. and

Thus. r ( t ) 2' 1 .5g ( t ) , and the error ~ ( t ) = ~ ( t ) - 1.5g(t) = 1 -- 1.5t over (0 5 t 5 1). and zero outside this inte~v:rl. Also Ec (the cnerglv of the error) is E, = S,'(l - 1.5t)'dt = 1/4.

2.5-4 (a) In this case E, = J, sin2 2 s t $t = 0 .5 . and

(h) Thus. p(t) zz - ( l / n ) r ( t ) . and the error ~ ( t ) = t + ( I / t ) s i n 2ni over (0 _< t 5 1). and zero outside this itlterval. Also E, and Ee (the cnergy of the error) we

I

I' 1 1 Eg = jdl q 2 ( t ) l t = & t 2 dt = 1/3 and El = It - ( I / r ) sin ?ntl2 dt = - - - 3 2x2

The error [ t + ( I /s) sin 2st) is orthogonal to ~ ( t ) because

l1 sin 2rrlt + ( l / r ) sin 2r t ] Jt = 0

Note that Eg = v 2 ~ 1 + EC. To explain these results in terms of vector concepts we observe from Fig. 2.15 that the error vector e is orthogonal to the component cx. &cause of this orthogonality, the lengoli of f [energy of g ( t ) i is equal to the sum of the square of the lengths of cx and e [sum of the energies of cr ( t ) and c(t)j .

2.5-5 (a) If r ( t ) and ~ j ( t ) are orrhogo~lal, rhen we can show the energy of z ( t ) f y ( t ) is Er + E,.

The last result follows from the fact. that beca\isc of orthogonality, the two integrals of the cross products . r f t ) ! / * ( t ) and s g ( t ) ? , ( t ) are zero [see Eq. (2.40)]. Thus the energy of + ( t ) + ! / ( t ) is equal to that of r ( t ) - :/it) i f ~ ( t j and y ( i ) are orthogonal. ( b ) Using similrr argument. we can show that the energy of clt ( t ) + r : 2 y ( t ) is equal to that of pi+(!) - rx? / ! t ) if . n j t ) and y ( t ) are orthogorial. This energy is given by I C ~ ( ~ E , + ( c . 2 / 2 ~ , .

( c ) I f i ( t ) = ~ ( t ) f ? l ( t j , then it follows from Eq. (1) in the above derivation that

2.5-6 g1(2.-1). gz( -1 .2 ) . g?(O. - 2 ) . g d ( l . 2 ) . gs(2. I), and g6(3,0). From Fig. S2.5-6. we sec that pairs (g3.g6). (g,. g4) and (g2. g5 1 ale orthogonal. Ct'e can verify this also nnalytically.

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

J- .% J - .r

L

[ 1 m ( t ) q 5 ( t ) t 1 t = 1 * 1 - ~ ~ ( t ) + 2 = 2 ( t ) ] l 2 r , ( t ) + r 2 ( i ) l t i t = 0

In deriving these results. we used the fact that S-L tTdt 5 J-L ~ $ ( t ) d t = 1 and j;: r l ( t ) r ? ( t ) d t = O 2.6-1

We shall compute V , using Eq. (2.48) for each of the 4 cases. Let us first c0mput.e the energies of all the signals 1

& = 1 sin22rtdt = 0.5

I:I the same way we find E,, = E,, = E,, = E,, = 0.5. ('sing Eq. (2.48). the correlation coefficients for four cases are found as

* 1. sin 2r t sin 4nl tlt = 0 (2)

I 0.8 I (3) 0.707 sin 2nl tit = 0 4 - [ 0.707sin 2rt df - O.7Otsin = 1.4141. . .

Signals ~ ( t ) and p2(t) provide the maximum protection against noise.

2.8-1 Here To = 2. so that 40 = 2x12 = r. and

Therefore

n s l

F1gul.e S2 6-1 shows ~ ( t ; = t 2 for all t and the corresponding Fourrer serles representing q ( t ) over I - 1. 1)

Fig. S2.6-1

The power of ( t ) is

hlor.eove~. from Parseval's theorem (Eq. (2.90)]

(b) If the 'V-tern~ Fourier series is denoted by x ( 1 ) . then

The power Pz is required to he 99%Pg = 0.198. Therefore

For N = 1. P. = 0.1111: for .V = 2. P. = 0.19323. For N = 3. P. = 0.19837. which is greater than 0.198. Thus. N = 3.

2.8-2 Here To = 2n. so that wo = 2n/2n = 1. and

Clu

g(t) = a. + En, cosnt + h, sinni - n < t l n

where

Therefore 1

q ( i ) = ? ( - 1 ) " " ~ ; ~ n n t - s < t < n n3 1

Figure 52.8-2 shows g(t) = t for all t and the corrmponding Fourier series to reprewnt g(t) over (-n. n).

Fig. S2.8-2

The power of q f t ) is

I\lo~.eovet. f~on l Porseval's theorenr [Eq. (2.90)]

( b ) I f the S-rerm For~riel- series is denoted by ~ ( t ) . then

The power P. is required to be 0.93 x $ = 0.3n2. Therefore

F ~ ~ . .y = 1. pz = 2; for s = 2. pi = 2.5. for N = 5 , p. = 2.927. which is less than 0.3n2 For N = 6. PZ = 2.9625. which is greater tllarl 0.3n2. Thus. N = 6.

2.8-3 Recdl that

nn = - g(t) cos nvrot d t /"I2 -To12 .

b,, = - q(1) sin n;ot dt ( l c )

Recall also that cos ndnt is an e m function and sin nwot is an odd function o f t . If g(t) is a n even function of t. then ~ ( ~ ) c o s ndot is also an even function and g(t)sin nuot is an odd function of t. Therefore (e hint)

To/2 nn = - g( t ) cos n s o t RI.

To .

Similarly. if g(t) is an odd function of t. then g(t)cos nwot is an odd function of t and g(t)sin nwot is an even function of t . Therefore

T o l l q ( t ) sin r u o t dt b n = $ L . (3b)

Observe that. because of symmetry. the integration r e ~ u i r e d t o compute the coefficients need be performed over - only half the period.

2.8-4 ( a ) TO = 4. JO = = 3. Because of even symmetry, all sine terms are zero.

no = 0 (by inspection)

4 . 177t .,, = 2 4 [l' cos (yt) lit - [eos (yt) d t ] =

T h e ~ e f o ~ e . the Foul.ier series for g [ t j is

Here tjn = 0. and we allow C, to take negative values. Figure S2.8-4a shows the plot of C,. (b) To = 10:. i o = = 4 . Because of even symmetry. all the sine terms are zero.

1 no = - (by inspect in^^)

5

1,. = 1 /" sin ( t t ) dt = 0 (inlegrand ir an odd function of t ) Ion -"

H e ~ e I ) , = 0. and we allow Cn to take negative values. Note that Cn = n, for n = 0, 1. 2. 3, . . .. Figure S2.84h shows t tle plot of C, . (c) To= 22n.CV.O = 1.

2.

(I(t) = no + En. cosnt + hn sinnt with a0 = 0.5 (by inrpection) n r l

and

Fig. S2.8-4

Tllr reason foi vanishing of the cosines terms is that when 0.5 (the dc componel~t) is subtracted from ) ( t ) . the mnaining function has odd symmetry. Hence. the Fourier writs would contain dc and sine terms only. F~gcre S2.8-4c shows tlie plot of C, and 8,. (d) To = n. ..ro = 2 and g ( t ) = tt. no = 0 (by inspertion). o n = 0 ( 7 1 > 0) because of odd symmetry.

Figure S2.8-4d shows the plot of C, and 0,.

Therefore Co = and

( f ) To = 6. d o = 713. 110 = 0.5 (by inspection). Even symmetry; b, - 0.

2x11 4rir 2 x n - 2 cos - - - sin - 3 3 J

d

Ohsr~.ve 11121 even h;r~.monics vanish. The reason is that if t,he dc (0.5) is subtracted frorn r l ( t ) . the re:;k~lrlt~g ' -

function has half-wavc symmetry. (See Prob. 2.8-6). Figure SZ.8-4f shows the plot of C,,.

cos p - sin 9 and 0, = tan

c o s v + % s i n + - 1 ,

2.8-5 An cven fut~ction g,rt) and at1 odd function go(#) have the propel.ty that

s,.ft) = qe( - t ) and o d t ) = -go(-t)

Evet \. sig11;il , q ( t ) can be expressed as a suln of even and od? components hecallse !,(I) = 4 [ .d l ) 4- d - t ) ] + 4 1.9(t) - d-1);

- ' - I

cvcn odd

From the definitions in Eq. (1). i t can be seen that the first componellt on the right-hand s ide is st1 evsn f~~nc t ion . while the second component is odd. This is readily seen from the fact that replacing t by - f in the first component yields the same function. The same maneuver in the serond component yields the negatis? of that component. To find the odd and the (?\.en compoilents of g ( t ) = u(t ) . we have

u.he~.t. [fro111 Eq. ( I ) ]

and

The even and odd compo~lents of the signal ~ ( 1 ) ale shown in Fig. 52 8-5a. Sln~llarly. to find the odd and the even conlponellts of y(t) = c - " ' l ~ ( t ) , we have

g(t) = ! l e f t ) + o o ( f )

whe~c.

g,(t) = f [ ~ - ~ ' r l ( t ) + ra'1l(-f)]

arid

~ . b > Fig. S2.8-5

T h e evrn and odd components of the sigl~al r - " t ~ ( t ) are shou-n in Fig. S2.8-5b. For gft) = r " . we Iiave

P" = . 9 C ( f ) + go(f.)

where

,(,*(I) = f [PI' + .-ji] = cos t

and

go(t) = 4 [CJ' - C - J ' ] = j sin t

2.8-6 (a) For half wave symmetry

Lct 7 = t - To/2 in the second integral. This gi\*es

g ( t ) cos nuot dr + 1"'" .I (. + $1 cos n.0 ( r + 3) d r ]

q ( t ) c o s n ~ o t dt + /0T0'2 -,(T)[- cosnwox] drr: I

In a similar way we can show that

q ( t ) sin ndot tlt To

(b) (i) To = 8. ;,-, = 3. no = 0 (by inspection). Half wave symmetry. Hence

4 RB n~ . n r = - ( + - n - - 1) (n odd) nZn2 2 2

4 nn nn =- ,,znz ( i n - 1) tn odd)

Therefore

( - 1) n = 1,5.9.13 . . . . 0" = { -* (y + 1) n=3 ,7 .11 ,15 , . . .

Similarly

2 t n A 4 nn n7r 4 b n = i l - d n - t d t = - ( ~ i n - - ~ ~ ~ - ) ~ - s i n ( ~ ) 2 4 nzx2 2 2 2 n2nz (nodd)

and Y

nn nn g(t) = an cos - t + b, sin - t

4 4

(ii) To = 27. d o = 1. no = 0 (by inspection). Half wave symmetry. Hence

and

= ;? 1- r-'rlOsinnt dt

(-0.1 sinnt - n cosnt)

2n 1.461 n - (p--rlo - 1) = - - (n2 + 0.01) 112 $- 0.0)

2.9-1 (a): ro = 4.do = a/?. Also Do = 0 (by inspection).

Fig. 52.9-1

q ( f ) = Do + Dndn'. when. by inspection Du = 0.5

(d) To = n. 40 = 2 and Dn = O

(e) fi = 3 . ~ 0 = + ... 1

= J where Dn = 5 1 t P-'l0.'& =

Fig. S2.9-2

q ( t ) = 3cost +sin

Fol a compact trigonometric form. all terlns must have corine form and mplitudes must be pmitive For this

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

(b) By inspection of the trigonometric spectra in Fig. S2.9-2a, we plot the exponential spectra s shown in Fig. S2.9-2b. By inspection of exponential spectra in Fig. S2.9-la, we obtain

(b) The exponential spectra are shown in Fig. S2.9-3. (c) By inspection of exponential spectra

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

Fig. S2.9-3

f ( t ) cos ndot dt - j f ( t ) sin nuot dl

I f 9 ! t ) is fverl. the second term on the right-hand side is zero because its integrond is an odd function of t . Hence. D,, is real. In contrast. if g ( ~ ) is odd. the first term on the right-hand side is zero because its iqregranrl is an odd function o f t . Hence. Dn is imaginary.

Chapter 3

If g ( t ) is on even function of 1 . g ( t ) s i n ~ t is an odd function of t , and the second integral vanishes. Moreover, q(t)co3&t is an even f u n c t i o ~ ~ of t . and the filst integral is twice the int.egrd over the interval 0 t o oc. Thus when g ( l ) is even

'U

G(4) = 2 6 q ( t ) cos wt dl

Similar argument shows that when g( t ) is odd

If q f t ) is also real (in addition to being even), the integral (1) is real. Moreover from ( I )

Hence G;;) is ~.eal and even furlction of J. Sin~ilar arguments can be used t o yl,ove the I.esr of the properties.

Since !C(;)/ is an cven ful~ction and B,(J) is an odd function of LJ. the integrand in the second integral is an odd function of ;. and therefore vanishes. Moreover the ir~tegrand in the first integral is an even function of d. and therefore

For g ( 1 ) = r-"ir(t) . G ( J ) = &. Therefore IG(w)l = l /JGf and B,(w) = -tan-'($). Hence

Therefore

and

3.1-4 (a)

(b)

3.1-5 (a)

(b) 0 2

r - ~ ~ ~ dt + lT L C - J W ~ dt = - ka ST + WI sin 4 7 - 11 ~ ( u ) = l7 -; 7 7 ' 2

This lrsult could also be derived by observing that g(t) is an even function. Therefore from the result in Pmb. 3.1-1

3.1-6 (a)

1 -O 1 r'-I (&t2 - 2) sin ~ o t A 2 ~ 0 1 cos JC

= s /*o u2,rdt (id. = - - [ -&2t2 - 2 , . ~ t + 2

2~ ( ~ f ) ' x i7

Fig. S3.1.0

(h) The derivation can be simplified by olsrrrirrg that C(;) ca11 be expresvd aa a sum nf two gate hlncrions G I (Y.) nnd G2!d j R S sllou,n in Fig. S3.1-6. Therefore

3.1-7 (a)

,,.1ut = { j t cosd + ~ i n d ) f / , 2 / ~

2n(l - tz)

= - c06: r ( 1 - tZ)

Because G(u) is even function. the second integral on the right-hand side vanishes. Also the i~~tegrand of the first term is an even f1111ctio11. Therefore

1 [COS tu 3; sin LJ " .(,) = ~ C O s t a 6 - , = -

TWO 10

= A ( c o s ; o t + uot sin uot - I] lry'otZ

3.1-8 (a)

Fig. 53.2-1

3.2-1 Figure S3.2-1 shows the plots of various functions. The funct,ion in part (a) is a gate function centered at the

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

at the origin and of width 100/3. The fuliction in part (c) is a gate function rect (6) delayed by 10. It] other words it is a gate pulse centered a t t = 10 and of width 8. The f~inction in part (d) is a sinc pulse cen te ld at the origin and the first zero occurring a t 9 = r , that is a t w = 5. The function in part (e) is a sinc pulse sine(%) delayed by 107. For the sinc pulse s inr(g) . the first zero occurs at : = a. that is at w = 5n. Therefore the function is a sinc pulse centered at j, = 10n and its zeros spaced a t intervals of 5x as shown in the fig. S3.2-le. The function in part ( f ) is a product of a gate pulse (centered at the origin) of width 10n and a sinc pulse (also centered at the origin) with zeros spaced a t intervals of .5n. This resrllts in the sinc pulse truncated beyovol~d tibe interval k5n ( I t ) 2 5n) as shown in Fig. f.

3.2-2 The function rect ( t - 5) is centered at t = 5. has n width of unity. and its value over this interval is uait?.. Hence

- I & p-3% - - L k 1 ~ 1 2 - ,, - jwI21 - [2 j sin ;] , j ~ jw -

,,.a 1 ot = -12j sin r t ) = s i n c ( ~ t ) e j ' ~ j2r4

3.2-4 Observe that 1 + sgn(t) = 21i(t). Adding pails 7 and 12 in Table 3.1 and then dividing by 2 yields the desired result.

3.2-5 Observe thai

Fourier transform of the above equation yields the desired result. 3.3-1 (a)

I rr ( t ) ~3 nA(w) + - v 13 9 ( : ) -

C ( w )

Application of duality property yields

Application of Eq. (3.28) yields

Bul b i t ! is an even function. that is h ( - t ) = O(t). and

Application of duality property yields

n lb(t + do) + 6 ( t - wo)] e 29r cos (-wow) = 2r Cos (wod) \ I - -

~ ( t ) a r g ( - d )

Setting = T yields

Application of duality property yields

Setting i o = 'r yields

h( t + T) - b ( t - T ) 2 j sin Td 3.3-2 Fig. (b) g l ( t ) = g ( - t ) and

1 G , ( d ) = G ( - - d ) = + j ~ r - ' ~ - 11

642

Fig. (r) g 2 ( t ) ;: g ( t - 1 ) + gl ( t - 1). Therefore

Fig. (e) g s ( t ) = g ( t - 3 ) + g r ( t + 4). and

I ;r: = -;i ?-J sin -1 = sinc 2

Fig. (I) .g .~( t ' j can be obtained in three steps: ( i ) time-expmdiilg g ( t j by a factor 2 [ii) then delaying i t by 2 seconds ( i i i ) and mrltiplying i t by 1.5 [we may interchange the sequence for steps (i) and (ii)!. The first step

'

(time-expansion by a factor 2) yields

I f (i) - 2 G ( b ) = -(r324 2 3 - j 2 4 J 2 d - 1 i

Second step of time delay of 2 secs. yields

t - 2 1 1 j (+ a lI?""d - ,,?d"2d - 1)'- = -!I 2 i 2 - j?; - , . '=*I

?.he third step of multiplying the resulting signal by 1.5 yields

3 g S ( t ) = 1 . 5 j (9) - D ( l - , j 2 d - r - l 2 * )

Fig. S3.3.9

3.3-3 (a)

rect (;) - Tainc (q) t 7' /2

YIX~ (-1 - Tsinc ($1 P ~ J w T / ?

idT - 2jTsinc ( ) sin - 2

= "sin' (f ) d

(b) From Fig. S3.3-3b we verify that,

h(t) = sin t u( t ) -t sin(t - n)u.(t - A)

Note that sin(t - *)u(i - x ) is sin t n ( t ) delayed by r. Now. sint.rr(t) e $ [ b ( c ~ , - 1) - b ( d + I)] + &y and

The1 efore

n 1 C ( 4 ) = {?; [h(d - 1) - 6(.*1.+ I ) ] + -

I - d = ](1 + c--'"-')

Recall that q(.r)b(t - r o ) = q(rroib(r - T O ) . Therefore h ( 4 f 1)(1 + c-'"") = 0. and

1 C(r') = -(I + c-'"") 1 - i '

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

g i t ) = c o s t [ 7 1 ( t ) - 1 1 ( t - j ) ] = c a t i ( r ) - c o s t a

But sin1 t - 5 j = - cost. Therefore

.Also because g ( . r ) h ( ~ - l o ) = g( . ro )h(~ - Y O )

3.3-4 From time-shifting property

q(t i T) L.. C ( L ~ ' ) C * ' ~ ~

Theref01 e

q(t i T) + q ( t - T) ~ ( d ) r ' ~ " l + c ( J ) ~ - - J ' ~ = 2C(d) cos JT

\Ye can use this result to derivr t~ ansforms of signals in Fig. P3.3-4. (a) Heie q( t) IS a gate pulse as sliown in Fig S3.3-4a.

g(t) = rect (i) - 2 sinc(w)

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

g(t + 3) + g(l - 3) (=$ 4 sinc(u) cos 3w

(b) Here g ( t ) is a triaxtgular pulse shown in Fig. 53.3-4b. From the Table 3.1 (pair 19)

,(,.I= A (i) er sinc2 (i) Also 7 = 3. The signal in Fig. P3.3-4b is g(t + 3) + g(t - 3). and

p(t + 3) + p(t - 3) - 2sinc2 (j) cos3w

Fig. 53.5-4

3.3-5 Frequency-shift ilrg propel ty states that

g( t ) f* luot CsS C(.J J1 do)

l'lre~ efore

1 1 q ( f ) sin ~ . o t - - Iq i t l ~ ' Y O t + R ( t ) ~ - J W ~ t ] = -IC(" - jO) + G(i - J O ) j

2 I 2 j

Time-stlift ing propel.ty st a: es that

qjl -C 7) - C(*.)C*'"~

Thcl.efoi e

q ( t - 7) - g ( t - T) c(u)F'"~ - ~ ( ~ ) r - . " ' ~ = 2jC(u;) sin

illld

I ->! t - T) - p(t - T)] (=, G'(rl)sin TJ 2.i

The signal in Fig. P3.3-5 is g(r + 3) - g(t - 3) where

j ( t ) = rect (;) - ~s inc(2)

There forc

g(t + 3) - g(t - 3) 2j[2sinc(~) sin 3 4 = 43' sinc(4) sill 3 . ~

8.3-6 Fig. (a) The signal p( t l in this case is a triangle pulse A(&) (Fig. S3.3-6) multiplied by cos lot.

g(t) = A (&) cos 101

Also from Table 3.1 (pair 19) A(&) n s i n c 2 ( y ) From the modulation property (3.35), it follows that

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

shown in Fig. S3.3-6a.

Fig. (b) The signal g(t) here is the same as the signal in Fig. (a) delayed by 2r. From time shifting property. its Fburier. transform is the same as in part (a) multiplied by ~ - j ~ ( ~ " ) . Therefore

Fig. S3.3-6

Th" Fourier transform in this case is the sa& as that in part (a) multiplied by P - ~ ' ' ' ~ ' ' . This multiplyicg iactor represents a linear phnse spectrkrm -2nd. Thus r e halve an amplitude spectrum [same as in part (a)j as \rtell as a linear phase spectrum LC(&) = -2nd i1.5 shown in Fig. S3.3-6b. the amplitude spectrum in this c s e a5 shov:n in Fig S3.3-6h. Sotc: I n the above solution. we first mult.iplied the triangle pulse A(&) by cos 101 and then delayed the result by 2 ~ . This means the signal in Fig. (h) is expre~sed ns ~(%$)cos 10(t - 2n). \\P could have interchanged the operation in this particular care. that is. the triangle pulse A ( & ) is f i s t delayed bv 2% and then the result is multiplied by cos lot. In this alternate procedure. the signal in Fig. (b) is espressed as A(+)COS lot This ~ n ~ e l c h a n g e of o p e l n t i o ~ ~ is permissible here only because the nintuoid cos lot exccutes integ~al n~rmber of cvcles in the rnterval 2% Because of this both the expressions are equivdent since cos 10(t - 2n) = cos 1Qt. Fig. ( c ) In this case the signal is identical to that in Fig. b. except that the basic pulse is rect (&) instead of a triangle pulse A { + j . SOH.

rect (t) - ~nsinc(-)

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

3.3-7 (a)

A lsu

Therefore

w + 4 C ( J ) = rect (q) + rect ( T )

1 -sinc(t) 4 rect lr

Also

3.3-8 From the frequency convolution property, we obtain

'The width property of co~lvolution states that if cl(z) c2(z) = y(+), then the width of ~ ( x ) is equal to the sum of the widths of r. ,(r) and t2 ( r ) . Hence, the width of G ( w ) * G ( w ) is twice the width of G(w). Repeated application of this argument shows that the bandwidth of gn(t) is nB Hz (n times the bandwidth of g(l)).

3.3-9 (a)

mt (f) - Tsinc ( f ) t i TI-

rect ( T z ) - Tsinc ($) P*~~'!'

and

&IT = 2jTsinc (q) sin -1- - " .in' ( f ) -

J

T5e Fourier transform of this equation yields

'Therefore

3.3-10 A basrc demodulator is shown in Fig. S3.3-10a. The product of the modulated signal q(t)cosdot with 2 c o s ~ o t y ~elds

q(t)cosdot x 2 c o s ~ ~ t = 3.41t)cos2~ot = g(i)[l + cos2;otI = q(tj + q(t)cos2;ot

The product contains the desired q(i) (whosespectrum is centered a t s = 0) and the unwanted signal q ( t ) c 0 ~ 2 ~ * 0 t wit I1 spect 1 urn 4 [ c [ ~ - r 2 & 0 ] + C ( w - ZJO]. which is centered at f Zlo. The two spectra are nono\er lapping because

Fig. S3.S-10

Fig. 53.4-1

I\. < d o (See Fig. S3.3-lob). We can suppress the unwanted signal by passing the product through a lowpasc filter as shown in Fig S3.3-10a.

3.4-1

Cl(;)=sinc(&) and C . ( u ) = l

Figure S3.4-1 shows GI(;). Gz(;). Hi(;) and Hz(&). NOW

The spectra Y,(J) and Y2(ur) are also shown in Fig. S3.4-1. Because u(t) = !,l(t)ua(t), the frequency convolution property yields Y ( J ) = ti (ur) * Yz(d). From the width property of convolution, it follows that the bandwidth of )I(;) is the sum of bandwidths of )i(d) and Yz(d) Because the bandwidths of Yl( i~) and Y2(d) are 10 kHz. 5 kHz. respectively. the bandwidth of Y(ur) is 15 kHz.

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

1 I , ( , ) - ,.,-(t-f0)~/4k m This is noncausal. Hence the filter is unrealizable. Also

Figure S3.5-1

Hence the filter is noncausal and therefore unrealizable. Since h(t) is a Gaussian function delayed by to. it looks as shown in the adjacent figure. Choosing t o = 3m, h(0) = ,?-4.s == 0.011 or 1.1% of it.s peak value. Hence to = 3 f i is a reasonable choice to make the filter approximately realizable.

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

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

Figure S3.5-2

The exponential delays to 1.8% nt 4 times constants. Hence to = 4/a = 4 x lo-' = 40,'s is a reasonable choice to make this liltel app~.mi~nately realizable

3.5-3 From the results il l Example 3.16

Also H (0) = 1 Heilcc if is the frequency where the amplitude response drops to 0.95. then

hlorwvel-. the time delay is given by (see Example 3.16)

R t*(a) = - 1 * t'#(O) = - = 1 0 - ~

ur2 + a2 a

If 4 2 is I he frequency where the time delay drops to 0.98% of its value at u = 0. then

We select the smaller of 41 and ~ 2 . that is w = 142,857, where both the specificat ions are satisfied. This yields n frequency of 22.736.4 Hz.

3.5-4 There is a typo in this example. The time delay tolerance rhould be 4% instead of 1%. The band of ALI = 2000 centered at 2 = lo5 represents t.he frequency range from 0.99 x 10' to 1.01 x 10'. Let us consider the gains and the time delays at the band edges. From Exmnple 3.16

At the edges of the band

iH(O.99 x lo5)! = = 10.1 x lo-', and lH(l.01 x lo')\ = 5* = 9.901 x lo-'

The gain variation over the band is only 1.99%. Similarly, we find the time delays at the band edges as

The time delay variation over the band is 4%. Hence, the transmission may be considered distortionless. The sigrlnl is transmitted with a gain and time delay at the center of the hand, that is at u = lo5. We also find I ~ ( 1 0 ' ) a 0.01 and tr(105) 1- -'. Hence, if g ( t ) is the input. the corresponding output is

Fig. S3.81

b ' b j = G(4j rect (5) - ~ ~ u f o + k * ~ n WT)

= C(d) rect (&) 11 - . jk sin J T ] ~ - ~ ~ ~ ~ ~

This fo1lou.s from the far1 that r" z 1 + r when 7 < 1. Moreover, C ( ~ ) r e c t (a) = C ( d ) hecause G(4) ia handlimited to B Hz. Hence

).(i) = C ( J ) V - ~ ~ ' O - jkC(4)sin &T r-'"O

.\loreover. we can show that (see Prob. 3.3-5)

1 -Ig(t + T ) - g( t - T)] u G(4) sin4T 2.i

Hence

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

3.0-2 Recall that the transfer function of an ideal time delay of T seconds is r- jwT. Hence. the tralsfer fu~~ction of the equalizer in Fig. P3.6-2 is

Ideally. we require the equalizer to have

The ccl~lalize~ in Fig. P3.6-2 approxitnatcs this expression if we select a0 = 1. a, = -0. nl - . . . , an = ( - 1 ) " ~ " .

Letting $ = 3 and consequently tit = % d l

Also from pair 22 (Table 3.1)

G ( w ) = C -,0w2/2

Letring nr. = j; and consequentb d* = &Jr

3.7-2 Consider a signal

g ( t ) = sinc(kt) and G(J) = -re' k

3.7-3 Rerall that

Interchanging the roles of g l ( t ) and g2( t ) in the above development. we can show that

3.7-4 In the paeralized Parwval'r theorem in Prob. 3.7-3. if we identify g ~ ( t ) = sinc ( 2 r B t - mn) and g2(t) = sinc (2nBt - nn), then

There fore

But rect (A) = 1 for 1-1 5 27rB and is 0 ot i~errir . Hence

28

In evaluating the integral. we ulrd the fact that c*J21k = 1 when k is an integer.

3.7-5 Application of dua1it.y property [Eq. (3.24)) t o pair 3 (Table 3.1) yields

The signal energy is given by

The energy contained within the band ( 0 to H' ) is

3'7-6 If q 2 ( t ) - .4(d). then the o u t p u ~ l ' f i ) = A(&)H(d ) . where H ( L ) is the lowlpass filter transfer function (Fig S3.7-6). Becalise this filter band A f - 0. we may express it as an impulse function of area 4xA. f . Thus,

H ( L ) ;= I4nA f ] A ( , . ) and Y ( 4 ) zz [4n.4(d)A j ] b ( d ) = [4nA(O)Aj16(;)

Here we used the property g ( r ) b ( r ) = p(O)b(+) IEq. (1.23a)l. This yields

~ l r t ) = 2A(O)Af

Sext , because nz( t ) A ( J ) . we have

Hence. ~ ( t ) = 2E,A j .

Fig. 53.7-6

3.8-1 Let g ( t ) = m ( t ) + g2 ( t ) . Then

where

I f we let g ~ ( f ) = C1 c o s ( ~ 1 t + 81) wd gz(t) = C2ccw(yt + 02). then

According to the argument used in Example 2.2b. the integral on the right-hand side is zero. Hence. Rg,,, (1) = 0. Csing the same argument. we have 'R,,, ( r ) = 0. Therefore

This result can be extended to a sum of any number of sinusoids as long as the frequency af each sinusoid is distinct. hence, if

1U

g ( t ) = c,. m ( n 4 0 t + 8,) " c:

then n . ( r ) = - ~ ~ c m n r n t

Moreover. for po(t) = Co, R,, ( r ) = ~ 8 , and

Thus. ate can generalize the result as follows. If

1- ,a.

q ( t ) = CO + cn cos(niot + 8") then Rg(r) = c,' + $cosnror

and %

sy (i) = ~.C,'D(;) + i C:jb(i - n.do) + O(J + nio)] n = l

3.8-2 Fig111.e S3.8-2a show?s the waveforms ~ ( t ) and ~ ( r - r ) for t .: TA/2. Let T = NTb. On the average. there are 9 / 2 pulses in the waveform of duration T. The area under the product r( t)=(t - r ) is .V/2 times (3 - 7) as shown ill Fig. S3.8-2b. Therefore

For I / T I 5 Tb. there b no overlap between pulses, and R,(r) = 0. For Tb 5 171 q, pulws again overlap.

Rut on the average. only half pulses overlap. Hence. R l ( r ) repeats every Tb WLCCO~&, but only with half the magnitude. as shown in Fig. S3.8-2c. We can expregs R,(T) as a sum of two components. as shown ill Fig. S3.8-2d. Thus. R,(T) = R I ( r ) + R2(1). The PSD is the sum of the Fourier transforms of 'R.1(( and li2(+). Hence

T b . 2 + s.(J) = 5 s1nc ( ) + where S2(i) i s the Fourier trar~sform of the periodic triangle function, shown in Fig. S3.8-2d. We find the exponential Fourier series for this periodic signal to be

Using Eq. (2.80). we find Dn = &sinc2i?f). Hence, according to Eq. (3.41)

Fig. SS.8-2

N MALy K I< 1U rulr+

1 2 ( 1 ) = 6 X = 05 and ? i2 ( t ) = 1 m d d -- 2

- 1 1

@I . 2 ( ( ) = ; 1 & = I n and - = i l 1 & u = - 4

3.8-4 The ideal differentiator transfer function is j ~ . Hence, the transfer function of the entire system is 2

ju 1 = ( ) : = and I H ( ~ ) I ' =

r*MC 1 1 2 ~ ~ ~ y e c t , ( ~ ) c i u = + ~ d a = ; 2- ( t ) =

n

Chapter

4.2-1 (i) For r ,~(t) = cos 1C00t

~ ~ s ~ . ~ ~ 0 ) = r" (t) cos 10, OOOt = cos 10001 cos 10, OOOt

LSB -

USB

( i i ) FOI 1 1 , ( 1 ) = 2 cos lOOOt + cos 20001

(t) = 111 ( t ) cos 10. OOOL = I2 cos lOOOt + cos 200011 cos 10.0001 1 = ros 90001 + cos 11.000t + - [cos 80001 + cm 12. OOOtj 2

.1 1 = (cos 90001 + T; cas 80001) + [cos 11,000t + ;; cos 12.000~) L * L

* d \ # - LSB USB

( i i i F a r r t ! I ) = cos 1000t cos 30001

1 $, ,sB.+c(~) = ul(t)ros 10.OOOt = 5[cos2000t -k cos4000tj cos 10.M)Ot

1 1 = - [cos 8000t + cos 1 2. OOOt j + - [coa 60001 + cos 14,0001] 2 2

This illforlnation is summarized in n table below. Figure S4.2-I shows various spectra.

~ ~ d ~ j ~ t ~ d s $no[ s \ ; e h m 5 - * %

i 0- - (f,CO4' - 9CO 8 I . qcoc ( t . b @ d

Fig. 84.2-1

Fig. 54.2-2

Fig. 54.2-3

4.2-3 The relevant plots a1.e show*n in Fig. S4.2-3. 4.2-4 (a) The signal at point b is

case

i

ii

i i i

4.2-2 The relevant plots are shown in Fig. S4.2-2.

Baseband frequenq

1000 1000 2000 2000 4000

DSR frequency I LSB frequency 1 USB frequency [ 9000 and 11,000 1 9000 1 11.000 1 9000 and 11,000 1 9000 I 11,000 8000 a d 12.000 1 8000 1 12,000 8000 and 12.000 ( 8000 I 12,000 6000 and 14.000 1 6000 I 14,000

The t,esrn { tn ( t ) cosdCt is the desired modulated signal, whose spectrum is centered at *wc. The renlaining term frrr(t)cos3wCt is the unwanted term, which represents the modulated signal with carrier frequency 5, with spectrum centered at f L y C as shown in Fig. S4.2-4. The bandpass filter centered at f we alloars to pass the desired term $ r r r ( f ) c o s ~ ~ t . blrt s t ~ p p r w m the unwanted term fm(i)cos3wct,. Hence. this system works as desired wit], tlle outpi~t Qrn( t ) cosuct. (b) Figure S4.2-4 shows the spectra at points b and c. (c) The minimun~ usable value of dc is 2 ~ 8 in order to avoid spectral folding at dc. (dl

The signal at point b consists of the baseband signal i rn( t ) and a modulated signd anr(t) c0~2wd, which has a carrier frequency 2wC . not the desired value u,. Both the components will be suppressed by the filter. whose center center frequency is J,. Hence, this system will not do the desired job. (c) The readel may verify that the idcntity for cosndct contains a term cosw,t when n is odd. This is not true wl~cn n is even. Hence. the system works for a carrier cosn u,t only when n is odd.

Fig. S4.2-4

4.2-5 Wc use the ring modulato~ shown in Fig. 4.6 with the carrier frequency j, = 100 kHz (;tc = 200n x 10"). and tllo output bandpass filter. centered at f, = 300 kHz. The output r>,(t) is found in Eq. (4.7b) as

Tlle output bandpass filter suppresses all the terms except the one centered at 300 kHz [corresponding to the carrier 32, t) . Hence. the filter output is

This is the desired output krrr(t)ros;~~t. with k = -4/3r.

4.2-6 'I'hc resistance of each diode is I. ohms while conducting, and oo when off. When the carrier A cosdr/,t is positive. the diodes conduct (during the entire positive half cycle). and when the carrier is negative the diodes are open (during the elltire nego!ive half cycle). Thus, during the positive half cycle. the voltage &$(I) appears across each of thr resistors It. During the negative half cycle. the output voltage is zero. Therefore, the diodes act as a gate in the circuit that is hasirally a voltage divider with a gain 2R/(R + 1. ) . The output is therefore

The period of tr~(t) is To = 2z/dc. Hence. from Eq. (2.75)

Tlie out pi11 co(t) is

2R 2 R 2 1 1 ~ . ( r ) = -tt-(t ),n(t) = -~n(t) [+ + (cos u.t - - cos 3uct + - u* + - . . R + R + r 3 5

(a) If we pass the output co(t) through ti bandhand filter (centered a t us). the filter suppresses the signal nr(t) and rrt(1)cos ndct for all n # 1. leaving only the modulated term * j m ( t ) ~ ~ s u ~ t intact. kjence. the system acts ,as a modulator.

(b) T h e same circuit can h e used as a demodulator if we use a b e p a s s filter a t the output. in this case, the input is @ ( t ) = r t r ( t ) c o s ~ ~ t and the output is ~*jrrt(t).

4.2-7 From the results in Prob. 4.2-6. the output eo(f) = km(t)cosw,:t, where k = ,*,. In the present case. ,

rrr(t) = s i n ( ~ , t + 0 ) . Hence, the output is

The lowpass filte~ supplesses the sinusoid of frequency 2w, and transmits only r he dc term $ sin 0.

- co~c sk wk rm 03- a4

1 C m, (4) - I

cos (7-t

, , & ,- yf> Fig. 54.2-8

4.2-8 (a) Fig. 54.2-8 shows the signals at points a. b. and c. (b) R.orn the spectrum af point c. it is clear that the channel bandwidth must be at least 30.0M rmd/s (from 5000 to 3.5.000 rad/s. j. (c) Fig. S4.2-8 shows the receiver to Iscovcr r r r ~ ( t ) and m d t ) from the received modulated signal.

4.2-9 (a) 54.2-9 showr the ou tp t~ t signal spectl.um t'(;). (h) Observe t!lat Y ( J ) is the same as A](;) with the frequency spectrum inverted. that is. the high frequencies are shifted to lower freq~lencies and vice versa. 'rhuu. t.he scrambler in Fig. P4.2-9 inverts the freq~rency spectrum.

Fig. 54.2-9

Fig. S4.3-2

l o get back the original spectrlrrn $41 (*,I. u10 need to invert the spectrum Y ( z ) once agaill. 'This can he donc by pasing the sc~.nmb!ed signal y(t) through the same scrambler.

4.2-10 \Ye use the ring modulator shown in Fig. 4.6. except, that the input is rr1(t~co.s(2r)lO~t instead of t r t ( f ; . TILe car.~.ic-r fr,eq\~ency is 200 kHz [dc., = (40Orj10'tj. and the output bandpass filter is centered at 400 kHz. T ~ I P output l . , ( t , ! is found in Eq. (4.7b) as

4 [

1 1 1 , ) = 1 ~ o s ( 2 ) l ~ t ) = 1 1 ( t ) c o s ( ) 0 ~ cos (100n)10~1- -50s 3 3@00r)j$t + - 5 cos 5(400r)10't A 1 J

The prodltct of the terms (-113) cos 3(400r)103t and (4/r)rn(t) cos(2*)10~t yields the desired term - &rrr(t) cos (800n)l~". ulhoac spectrum is centered at 400 kHz. I t alone passes through the bandpass filter (centered at 400 kHz). All the other terms arc slippremed. The desired output is

4.3-1 O o ( t ) = [ A -C rrt(t)] cosi,t. Hence.

1 1 = -j.4 + nr(t)] + Z[A + rrr(t)] cos2act

2 The f i s t term is a lowynss sigl~al because its spectrum is centered at u = 0. The 1owp.w filter allows this term to pass. hut suppresses the second term. whose spectrum is centered at &b,. Hence the output of the lowpass filrer is

?,(t) = A + 711 ( t !

When this signal is gassed thl~ough a dc block, the clc tenn A is suppressed yielding the output rrt (t). This shows that the system can demodulate AM signal regardless of tlre value of A. This is a syncl~ronous or collerellt demodt~lation.

This means that / c = 3a represents the DSB-SC case. Figure S4.3-2 shows various waveforms.

4.3-3 (a) According to Eq. (4.10a), the carrier amplitude is -4 = rnpllt = 10/0.8 = 12.8. The carrier power is PC = .4'/2 = 78.125.

I met)

(h) The s~debarld power is r r i 2 ( t ) / 2 . Because of syrnrn~tly of amplitude values every quartpr cycle, the power of r ~ d ( t ) may be computed by averaging the signal energy over a quarter cycle only. Over a quarter cycle m(t) car1 be lepresented as irt(t) = 40t /To (see Fig. S4 3-3) Hence.

The sideband power is

The efficiency is

4.3-4 From Fig. S4.3-4 it is clear that the envelope of the signal 111 (t) C O S J , ~ is ( rrr ( t ) l . The signal ( A + rr~(t)! cos&,i is identical to ni(t) cos ~ e t with m ( t ) replaced by A + r r r ( t ) . Hence. using the previous argument, it is deal that ~ t s envclope is IA + m(1)~. Now. if A + m ( t ) > 0 for dl t , then -4 + r n ( t ) = IA + in ( t ) i . The~efore. the co!ldition for demodulating A11 signal using envelope detector is -4 r rn( t ) > 0 for all t.

Fig. 54-54

38

4.9-1 When an input to a DSB-Sc generator is T I I ( ~ ) . the corraponding output is m ( t ) coswct. Clearly, if the input is .4 + r t r ( 1 ) . the corresponding output upill be 1.4 + m(t)] cosdct. This is precisely the Ahl signal. Thus. by adcling I\ dc of value .4 to the baseband signal rrr(t), we can generate AM signal using a DSR-SC generator. 'rht converse is generally not true. Iiowever. we can generate DSB-SC using AM generators if we two ider~tical A M generators in a balanced scheme shown in Fig. S4.3-5 to cancel out the carrier component.

4.3-6 \Vhen an input to a DSR-SC demodulator is m ( f ) cosri,t. the corresponding ot.1tpu1 is 7n( t ) . Clearly. if the input is ( A + rrr ( t ) ! cosdct. the corre~ponding output, will A + rr,( t) . By blocking the dc component A from this out put. we call dei~~odulatc the -4 31 signal using a DSB-SC demodulator. The converse. unfo~~unately. is not true. This is because. when an input to an AM demodulator is m(l ) COSJ,~.

the corresponding outprtt is I rn(t ) j [the envelope of r n ( t ) J . Hence. unless nc(i) 2 0 for all t , it is not possible to dernodufate DSB-SC signal using an A31 demodulator.

Fig. S1.37

4.3-7 Obselvr t l ~ r ? ~ r , r 2 ( t ) = A? fot. all t . Hence. rhe time average of r r r 2 ( t ) is also A2. Thus

The carrier amplitude is A = rrrp/,r = mp = A. Hence PC = A2/2. The total power is R = I', .t P. = A'. The power efficiency is

'T'f~e ,411 signal for / I = 1 is shown in Fig. S4.3-7.

4.3-8 'Zh. signal ~t point a is 1.4 + t ~ ~ ( t j ] cos wct. The siqnal at point b is

.4' + 2 A r n ( t ) + nr2(t) r ( t ) = 1.4 + r n ( t ) f 2 c ~ Z ~ c t = ( 1 + cos 2 4 )

2

The lowpass filter suppresses t . 1 ~ term containing cos Let. Hence. the signal a t point. c is

L'sually. r n ( t ) f , 4 < 1 for most of the time. Only when t r r ( 1 ) is near its peak, this condition is violated. Hence. the output at point d is

A blocking capacitor will suppress the dc term .4*/2. yielding the output An~(t). From the signal l t l ( t ) , we see that the distortion component is rrr '(1)/2.

4.4-1 In Fig. 4.14. when the carrier is cos I(Ayl)t + b] or sin [(Au)t + 61, we have

rl(1) = 2j r r r l ( t ) coodct + m ~ ( t ) sinw,t]cos [(wc + Au)t + h]

= 2n1 I ( t ) cos wct cos I(&, + Aw)t + h] + 2nr 2(t) sin uct cos ( (w, + Aw)t + 61

= rill (t){cos [ (Aw)~ + A] + cos I(2uC + Au)t + b ] ) + rn2(t){sin ((2dc + Aw)t + 6) - sin [(Ayl)t + 61)

.4fte1 ~ , ( t ) and . 7 2 ( t ) are passed through lowpass filter. the outputs are , :

trl;(t) = rnI(t) cos l ( A ~ ) t + 6) - r112(l) sin [(Aw)t + h!

~rr;(t) = rrrl(t) sin [ (ALJ)~ + h] + r n 2 ( t ) cos [ ( A z ) ~ + h]

4.5-1 To generate a DSR-SC signal from nr ( f ) . we multiply m ( t ) with cosw,t. However. to generate the SSB signals of the same relative magnitude. i t is convenient to multiply rn (t) with 2cosdct. This also avoids the nuisat~ce of the fractions I j 2 , and yields the DSB-SC spectrum Af(u - c~,) + M ( j , .t dc). We suppress the CSB spectrum (above and below -J,) t.o obtain the LSB spectrum. Similarly. to obtain the USR spectrum. we suppress the LSB spectrum (between -4, and dc) from the DSB-SC spectrum. Figures S4.5-1 a. b and c shou. the three CiSCS.

(a)From Fig. a. use can express ~ , , , ( t ) = cos900t and y,,, (1) = cos 1100t.

(b)From Fig. b. ufe can express yL,,(t) = 2cos700t + ccw900r and jc,,,(f) = cos ll00t + 2cos 13001

!c)Flom Fig. c. we can express gL,,(t) = $[cos400t + cos6OOt] and q,,,(f) = ;[cos l4OOt + cos 16001:

4.5-2 qCss ( t ) = m ( t ) c o s ~ , t - mt,(t)sin wcf and 9,,,(t) = rrr(L) coswct + mh(t) sin ~ , t

(a) m ( t ) = cos 1001 and rrrh(t) = sin 1001. Hence, g,,, (1) = cos lOOt C08 lOOOt + sin loot sin lOOOt = cos(1000 - 100)t = cos 9001

qCSB (1) = cos lOOt cos lOOOt - sin lOOS sin lOOOt = cos(1000 + 100)t = cos 1 l O O t

(b) rrr ( t ) = cos 1001 + 2 cos 3001 and mh (t) = sin loot + 2 sin 300t. Hence,

;,,, (1) = (COS lOOt + 2 cos 300t) cos lOOOt + (sin lOOt + 2 sin 300t) sin lOOOt = cos 900t + 2 cos 720t

;, ,, f f ) = (cos lOOt + 2~0~3001) cos 10001 - (sin lOOt + 2sin 300t) sin lOOOt = cos 1100t + 2 cos 13001

(c) 111 ( t ) = cos 1001 cos 5Wt = 0.5 cos4M)t + 0.5 cos600t and rnh(t) = 0.5 sin4001 + 0.5 sin 600t. Hence.

g,,, (1) = (0.5 cos400t + 0.5 cos600t) cos lOOOt + (0.Ssin 4001. + 0.5sin600t) sin 10001 = 0.5 cos400t + 0.5~0~6001

p, ,, ( t ) = (0.5 cos400t + O.Scos6001) cos lOOOt - (O.5sin 400t + 0.5sin600t) sin 1000t = 0 .5~0s 1400t t 0 . 5 ~ 0 ~ 16001

, fl Tcfl" ""r. .T

-I00 loo L3 9 -/Joe -qm d., qoo 1toa

F;quzec not ~ Z J scale - Fig. S4.5-1

Fig. S4.5-3

Qo, f 0 in SSB (b)

&I- . ,

Fig. 54.5-5

4.5-3 Figure S4.5-3a shows the spectrum of n r ( t ) and Fig. 54.5-3b shows the correspor~ding DSB-SC' spec:trum 2 , , , : t ) c o ~ i4000nt

( h i Figure S4.5-3c shows I he cot-respol~ding LSB spectrum obtained by suppressing the CSB spectrum. (c j Figu1.e S4.5-3d shows the corresponding L'SB spectrum obtained by suppressing the LSB spectrum. \\'Q now find the i ~ ~ w r s e Fourier transforms of the LSB and USB spectra from Table 3.1 (pair 16) and the f r ~ c ~ ~ r e n r y shifting property as

G , . ~ ~ ( I ) = 1000 sinc (1000ni) cos 9000nl

cp, ,, ( t ) = 1000 sinc (1000rt) cos ll.000nl

4.5-4 Because Afh(w) = - j A I ( & ) sgn (w). the transfer function of a Hilbert transformer b

H (a) = - j sgn (J)

If wr apply rn h ( t ) at the input of the Hllbert. transformer. Y (w). the spectrum of the outp~lt signal ! / ( t ) is

This sllows that the Hllbert transform of r r r h ( t ) is -1r>( t ) . To show that the energies of r r c ( t ) and t r r h 0 j ale ; equal. we liave

4.5-5 The incoming SSB signal at the receiver is given by (Eq. (4.1 7b)]

pLED (L) = m(i) cosuct + m h ( l ) sin wet

Let the local carrier be caq [(wc + Aw)t + 6 ) . T h e prodeuct of the incoming signal and the local carrier is cd(t) . given by

rd(1) = yLsB(f) COS [(we + A U ) ~ + b]

= 2[711(i) cosur,t + n r h ( t . ) sinw,t] cos [(u, + Aw)i + b]

The lowpass filter suppresses the sum f~equency component centered a t the frequency ( b e + A d ) . and passes only tile difference frequency component centered a t the frequency Aur. Hence, the filter output no(t) is given by

Observe that if both Ai and A are zero. the output is given by

as expected. If only 6 = 0. then the output is given by

This is an I'SB signal corresponding t o a carrier frequency Ad as shown in Fig. S4.5-5b. This spectrum is the same as the spectrum .\I (d) with each frequency component shifted by a frequency A,.. This changes the sound of an audio signal slightly. For voice signals. the frequency shift within *20 IIz is coilsidered tolerable. Most L'S systems. howeve1 .testlict the shift to f 2 Hz. (b ) \Vhen only AJ = 0. the lowpass filter output is

N'e now shou* that this is a phase distortion, where each frequency component of .I/(&) is shifted in phase by a m o ~ ~ n t 6 . The Fourier ~ransform of this equation yields

But from Eq. ( 4 . 13b )

Iif ( d ) (.'A W > o Eo(rl) =

Af ( J ) r-J" d < O

It follows that the amplitude spectrum of ro( t ) is A~(J). the same ns that for rr)(f) But the phase of each component is shifted by h. Phase distortion generally is not a serious problem with voice signals, because the human ear i s 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 nrh ( l ) is -m( t ) . Hence, if m h ( t ) /instead of ~ n ( t ) ] is applied at the input ill Fig. 4.20. t h e L'SB output. is

I ) = I ) 0 - I ( ) sin d C

= rrl(t1 cos (uct + i) + ruh ( t ) sin (dc t + i)

Fig. 54.81

Thus. if we apply 7 n h ( t ) at the input of the Fig. 4.20. the USE) output is an LSB signal corresponding to n r ( f > . The cairie~ also acquires a phase shift n/2. Similarly. we con show that if we apply ~ t , h ( t j at the Input of the Fig. 420, the LSB output w.ould be an USB signal corresponding to rn(t) (with a carrier phase shifted by n/2).

4.6-1 Fro113 Eq. (4.20)

Figure S4.6-la shows H , ( & - A=) and H I ( & ++a Figure S4.6-lb shows the reciprocal. which is Ha(&).

4.8-1 A station can be heard at its allocated frequency 1500 kHz as well as at its image frequency. The two frequencies are 2 f l ~ HZ apart. In the present case. fir = 455 kHz. Hence. the image frequency is 2 x 455 = 910 kHz apart. 'I'herrlore. the station will also be heard if the receiver is tuned to frequency 1500-910 = 390 kHz. The reascjn for this is as follows. When the receiver is tuned to 590 kHz. the local oscillator frequency is f ~ o = 590+455 = 1015 kHz. Now this frequency f ~ o is ~nultiplied with the incoming signal of frequency fc = 1500 kIJz. The outpul i'

yields the two modulated sigl~als whose carrier frequencies are the sum and difference friquencies. which arc 1500 + 1045 = 2545 kHz and 1500 - 1045 = 455 kHz. The sum carrier is suppressed. but the difference carrier ' '

ptsses 1 I~rough. arid the st ntion is receit~ed. J . ' ,

4.8-2 The local oscillato~ genelates f~rquencies in the range 1+8=9 MHz to 30+8=38 MHz IVhen the reccive~ setting is 1OhlHz. f ~ o = 10 + 8 = 18 MHz. Now. if there is a station a t 18 + 8 = 26 MHz. it svill beat (mix) with fLo = 18 MHz to produce two signals centered at 26 + 18 = 44 MHz and at 26 - 18 = 8 MHz. The sum component is suppressed by the IF filter. but the difference component. which is centeted at 8 MHz. passes th~oilgh the IF filte~.

Chapter 5

Fig. S5.1-I

5.1-1 I n this case f, = 10 ?.IHz. rn, = 1 and m6 = 8000.

For Fhll :

Af = k,rtrpi/2x = 2n x 105/2r = 10' Hz.. Also f, = 10'. Hence. (f8)m-x = l o 7 + 1 ~ " 10.1 MHz. and /I,),,, = 10' - lo5 = 9.9 MHz. The carrier frequency incremes linearly from 9.9 MHz to 10.1 MHz over a quarter (rising) cycle of duration n seconds. For the next n seconds. whcn l i l ( f j = 1. the carrier frequency remains at 110.1 MHz. Over the next quarter (the falling) cycle of duration n. the carrier frequency decreases linearly from 10.1 MHz to 9.9 MHz., and over the last quartel cycle, when 111 ( t ) = - 1. the carrier frequency remains at 9.9 MHz. This cycles repeats periodically with the period 4n seconds as show11 in Fig. S5.1-la. For PSI : Jj = X-,rri,b/lx = 50:: x 8000,'2r = 2 x IO"H Also j, = 10'. Hence. (f , )max = 10' + 2 x lo5 = 10.2 MHz. and (f, jm;, = 10' - 2 x 10' = 9.8 MHz. Figure S5.1-1 b shows m(t). We conclude that the frequency remains at 10.2 5111~ over the (rising) quarter cycle. where rir(f) = 8000. For the next n seconds, rn(t) = 0. and the carrier frequency remains at 10 MHz. Over the next Q .seconds, where rh(t) = -8000. the carriel. frequency remains at 9.8 31 Hz. Over tlie last quarter cycle tir ( t ) = 0 again, and the carrier frequency remains at 10 MHz. This cycles repeats pe~.iodically with the period 4n seconds as shown in Fig. SS.1-I.

5.1-2 111 this case fc = 1 MHz. nr, = 1 and rrr; = 2000.

For Fhl : Af = k,ir1,/2n = 20,0007/27r = 10' Hz. A h j, = I MHz. Hence. (I,),, = lo6 + 10' = 1.01 MHz. and (/,)man = lo6 - lo4 = 0.99 MHz. The carrier frequency r i m linearly from 0.99 MHz to 1.01 MHz over the cycle (over the inrersd -q c t c q). Then instantaneously. the carrier frequency falls to 0.99 MHz and starts rising linearly to 10.01 MHz over the next cycle. The cycle repeats periodically with period lo-' as shoal1 in Fig. S5.1-2a. For PM : Here. beca~rse nr(t) has jump discontinuities, we shall use a direct approach. For convenience. we select the origin for rrl(t) as shown ill Fig. SS.l-2. Over the interval to $. we can express the message signal as 711 (1) = 20001. Hence.

7: cos [2n( l0 )~1+ 1000rt] = cos [an (loR + 500) I]

At the discontinuity. the amount of jump is rnd = 2. Hence. the phase discontinuity is k,md = r;. Therefore, the carrier frequency is constant throughout at lo6 + 500 Hz. But. at the points of discontinuities. there is a

Fig. 95.1-2

p11a.w discontinuity of rr radians as shown in Fig. S5.1-2b. In this case, we must maintain kp < n because there is a discontinuity of the an~ount 2. For kp > r, the phase discontinuity will be higher than 217 giving rise to ambiguity in demodulntion.

W e are given that ~ P J I ( ~ ) = 10 cos ( 1 3 . 0 1 ) with kp = 1000. Clearly. m(t) = 3t over the interval 11) 5 I .

Therefore kf 1' ,,'(a) ,la = 1000 nc (a) do = 30001 r Hence 3t = It m ( a ) d o - m(r) = 3

5.2-1 In this case kf = 1OOn and k, = 1. For

r r~( t ) = 2 cos lOOt + 18 cos2000xt and r i i ( t ) = -200 sin 100t - 36. OOOx sin 2000rt

Therefore ntp = 20 and rrrb = 36. OOOn + 200. Also the baseband signal bandwidth B = 2000%/2n = 1 kHz. For F M : : A J = kf rnP/2n = 10.000. and B F ~ . ~ = 2 ( A f + B ) = 2(20.000 + 1000) = 42 kIjz. For Phl : : AJ = kprr,;/2s = 18.000 + 9 Hz. and Bphr = ?(Af + B) = 2(18.031.83 + 1000) r- 38.06366 kHz.

5.2-2 g,,()) = 10 cos(ul,t + 0.1 sin 2000st). Here. the baseband signal bandwidth B = 2000n/2n = 1000 Hz. Also.

Tilerefore. AJ = 200a and A j = 100 Hz and BE^^ = 2 ( A f + B ) = 2(100 + 1000) = 2.2 kHz. I

5.2-3 ;,, ( 1 ) = 5 cos(dct + 20 sin lOOOnl + 10 sin2000nt). Here. the baseband signal bandwidth B = 2000n/2a = 1000 Hz. Also.

u, ( t ) = u, + 20, OOOn cos 1000nt + 20, OOOn cos 2000nl

Therefore. AJ = 20.000n+20. OOOn = 40, OOOn and A f = 20 kHz and B E ~ I = 2(A f +B) = 2(20.000+ 1000) = 12 kHz.

5.2-4 The bnseband signal bandwidt!~ U = 3 x 1000 = 3000 Hz. For Fbf : Af = = Jw = 15.951 k H z md BFhl = 2 ( A j + B ) = 37.831 kHz.

k m' For PM : Aj = = = 31.831 kHz and B ~ M = 2(Af + B ) = 66.662 kHz.

5.2-5 The baseband sigllal h n ~ ~ d w i d t l ~ B - 5 x 1000 = 5000 HZ. For F M : A f = r ~v. = 1 kHz and BFIl = 2 ( A f + B ) 2(2 + 5 ) = 14 kHz.

For P M : TO find B P M . we observe from Fig. S5.1-2 that tp,,(t) is essentially a sequence of sinusoidal pulses of width T = lo-' seconds and of frequency f, = 1 MHz. Such a pulse and its spectrum are depict.ed in Figs. 3 . 2 2 ~ and d. respectively. The bandwidth of the pulse, as seen from Fig. 3.22d, is 4 n l T rad/s or 2 / T Hz. Hence. Bphl = 2 kHz.

5.2-0 (a) For FM : S f = = 200.yx1 = 100 kHz and the baseband signal bandwidth B = 9 = I kHz. Isherefore

Bpu = 2 ( A j + B) = 20'2 kHz

F o r P M : Af=*Pt": zw - - >Ox = l o kHz and BpM = 2 ( A j + 23) = 2(10 + 1 ) = 22 kHz (b) rn ( f ) = 2 sin 2000nl. and B = 2000n/2n = 1 kHz. Also m, = 2 and tnb = 4000n.

For FM : A1 = = 20b.-x2 = 200 kHz. a d ?r

B w l = 2 ( A f + B) = 2(200 + 1) = 402 kHz

For P M : ~ f = ~ = ~ = ~ k ~ ~ . n d ~ p n = l ( A f + B ) = 2 ( 2 0 + 1 ) = 4 2 k H z . (c) rn(t) = sin 4OOC)nt, and B = 4000n/2x = 2 kHz. Also m , = 1 a ,d mi = 4 0 0 0 ~ . F o r F M : A f = = 2 0 b . y x 1 = 100 kHz, and

L m' For P M : Af = = = 20 kHz and Bphl = 2 ( A f + B) = 2(20 + 2 ) I 44 kHz.

(d ) Doubling the amplitude of rrr(t) roughly doubles the bandwidth of both FM and PM. Doubling the frequency of r r r ( t ! jexpanding the spectrum .\l(sr) by a factor 21 has hardly any effect on the Fhl bandwidth. However. it roughly doubles the bandwidth of Ph1. indicating that P3I spectrum is sendti\.e to the shape of the baseband sprr t lum. FYI spectrum is relative!y insensitive to the nature of the spectrum h i ( & ) .

5.2-7 From pair 22(lnble 3.11. wc obtain r - ~ ' - The spectrum A f ( w ) = is a Gaussian pulse. which decays iapidly. I t s 3 dB bandwidth is 1.178 rad/s=0.187 Hz. This is an extremely small bandwid~ l~ comya~.ed lo A f .

.Also ri ,( t j = - 2 t 1 - - ' ~ / ~ . 'The spectrum of r i ~ ( t ) is Al l (d) = j d h l ( d ) = jfid~-u214. This spectrum also deca:;s rapidly away from the origin. and its bandwidth can also be assumed t o be negligible compared to Li f . For FM : A f = = = 3 kHz and = 2 A f = 2 x 3 = 6 kHz.

F o r Phl : To find r r r ; , we set the derivative of , i t (+ ) = -2tc-"I2 equal to zero. This yields

and nth = r i r ( 5 ) = 0.858. and

af=*= s*\",0.Bs8 = 3.432 kHz and Bphl = 2 ( A f ) = 2(3.432) = 6.864 kHz.

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

Fig. S5.3-1

Fig. 96.344

5.3-2 The block diagram of the design is shown in Fig. S5.3-2. I

5.4-1 (a) ( ~ p ~ l ( t ) - A cos [& + kpm(t)l

When this psll(t) is p u r d through an ided FM demodulator. the output is kpm(t) This s@al. when p-d through an ideal integrator. yields k,rn(t). Hence. FM demodulator follow~ed by an ideal integrator acts a PM demodulator. how eve^. if m ( i ) has a discontinuity. m(t) r rn at the ~o in t ( s ) of discontinuity. and the system will fail.

(b) gFhr(t) = A cos ~ ; c f + kf ni(0) do [ I \Vhm this signal ,cFv(t) is passed thmugh an ideal PM demodulator, the output is k, It mlo)do When this '

signal is passed throagh an ideal differentiator. t.he output is kfm(t). Hence. PSf demodulator. followcd by an ideal differentiator. acts as Fhl demodulat~r regardless of whether ~ n ( t ) has lump discoecinl~ities or not.

5.4-2 Figule SJ.4-2 shows the rareforn~s at points b. c, d. and e. The figure is self explanatory.

5.4-3 From Eq (5.30). the Laplare T I ansform of the phase error & ( t ) is gi\.en by

2k ''('') = ~11.~ + A K H ( . + ) ]

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

Iim 0, ( t ) = l i~n see(*) = Ik = % t - -Y b -9 a(n + A K )

Hence. the incoming signal cannot be tracked. If

* + a 2k H(.s) = -. then 0 4 s ) = 3z + AKc:+m,l

2k 2k = - lim 8 , ( t ) = lim * ~ L ( . U ) = !yo a~ + AK(* + ,,) ~k~ 1 -U r -0

Ilence. the incoming signal can be tracked within a constant p h u . 2kjAka radians. !Vow, if

2 + n* i l l 2k H(s) = then 8 c ( s ) =

~2 '

In this rase. the incoming signal can he tracked with zero phase error.

I I r 1 r 1 1 4*

&to

Fig. SS.4-2

Chapter 6

0.1-1 The bandwidths of gl(t) and g2(t) are 100 kHz and 150 kHz, respectively. Therefore the Nyquist sampling rates for g1(t) is 200 kHz and for gp(t) is 300 kHz. Also g12(t) &pi (u) * gl (w), and from the width property of convolution the bandwidth of glz ( t ) is twice

'

the bandwidth of gl(t) and that of g2'(t) is three times the bandwidth of gn(t) (se also Prob. 4.3-10). Similarly the bandwidth of gl(t)g2(tj is the sum of the bandwidth of gl(t) and gz(t). Therefore the Nyquist rate for gi2(t) is 400 kHz. for gz3(t) is 900 kHz. for gl(t)g2(t) is 500 kHz.

0.1-2 (a)

'I he bandwidth of this signal is 100 n rad/s or 50 Hz. The Syquist rate is 100 Hz (samples/sec). (h)

sinc2(100st) +=+ O.OlA(?dii;)

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

sinc (100rt) +sine (50rt) - 0.Olrect 0.01 (&) + 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 IIz. The Syquist rate is 200 Hz ! san~ples/sec). (d l

The bandwidth of rert(&) is 50 Hz and that of A(&) is 60 Hz. The bandwidth of the sum is the higher of the two. that is. 60 Hz. The Syquist sampling rate is 120 IIa. (el

The two signals have bandwidths 25 Hz and 50 Hz respectively. The spectrum of the product of two signals is 1/23 times the convolution of their spectra. Rom width property of the convolution. the width of the convoluted

'

s i g ~ ~ a l is the sum of the widths of the signals convo!ved. Therefore. the bandwidth of sinc(SOnt)sinc(100*f) is 2.5 + 30 = 75 Hz. The Xyqi~ist rate is 130 Hz.

6.1-3 l'he pulse train is a periodic signal with fundamental frequency 2 8 HZ. Hence, w, = 2r(.LB) = 4nB. The period is To = 1/2B. It is an even function of t . Hence, the Fourier series for the pulse train can be expressed as

w

pr. ( t ) = CO + C C. c a n h r n r l

Csing Eqs. (2 72). we obtain

no = Co - - tit = - 4 ' a n = & = - c ~ s ~ . t d t = ~ s i n ( ~ ) . n n b n = 0

Hence.

lU 2 = I,(+) + C ; sin ( ) q ( t ) cos n r , t

4

1 r 3h.. -20 -5 5 20 F(Hz) -* . . , . Proctlcal Filter

! :

' -0.3 ' -0.1 '0 0.2 1+ -40A

Fig. S6.14

6.1-4 Foi g ( t ) = sinc2(5nt) (Fig. S6.1-4a). the spectrum is G(u) = 0.24(%) (Fig S6.1-4b). The bandwidth of t1:is signal is 5 Hz ( l o x rad/s). Consequently, the Xyquist rate is 10 Hz, that is. we must sample the signal at a rate JIO less than 10 samples/s. Tlle Nyquist interval is T = 1 / 2 8 = 0.1 second. Recall that the sampled signal spectrum consists of (l/T)C(u) = A(*) repeating periodically with a period cqual to the sampling frequency fa Hz. We present this information in the following Table for three sampling rates: f. = 5 Hz (undersampling). 10 Hz (Nyquist rate). and 20 Hz (owraampling).

In the first rase (undersmpling). the samplitlg rate is 5 Hz (5 samples/sec.), and the spectrum ; G ( J ) repeats evely 5 Hz (lox radjsec.). The successive spectra overlap, as sllown in Fig. S6.1-4d, and the spectrum G(w.) is not recoveiable from z(;). that 1s. g ( t ) cannot be reconstructed from its samples B ( t ) in Fig. S6.1-4c. If the sampled signal is passed through an ideal lowpass filter of bandwidth 5 Hz. the output spectrum is lect ( & J .

51

F saml)ling frequency j.

5 Hz

10 Hz 20 Hz

sampling interval T

0.2

0.1

0.05

f C(d) I comments 1 A (&-) I Undersampling I 2A (&) 4A (&)

Nyquist Rate

Oversampling

and the output signal is lOainc(20nt), which is not the desired &nal sinc2(5*f). In the second case. we use the Syquist sampling rate of 10 Hz (Fig. S6.1-4e). The spectrum G(w) consists of b3ck-to-back. nonoverlapping repetitions of +G(3) repeating every 10 Hz. Hence, G(y) can be recovered from C ( w ) using an ideal lowpass filter of bandwidth 5 Hz (Fig. S6.1-4f)-The output is 10si1lc2(5rt). Finally, in the last case of oversampling (sampling rate 20 Hz). the spectrum C(u) consists of nonoverlapping repetitions of +G(&) (repeating every 20 Hz) with empty band between successive cycles (Fig. S6.1-4h). Hence. C(w) can be recovered from z ( w ) using an ideal lowpass filter 01 even a practical lowpas filter (shown dotted in Fig. S6.1-4h). The output is 20sinc '(ant).

4 $ 6.1-5 This schcine is analyzed fully in Problem 3.4-1. where we found the bandwidths of !11(t), ?tz(t ,and!t(t to be 10 kHz, 5 kHz. and 15 kHz. respectively. Hence, the Nyquist rates for the three signals are 20 k z. 10 k z. and 30 kHz. respecttvely. i

0.1-6 (a) H'hen the input to this filter is h(t). the output of the summer is h(t) - b( t - T). This acts as the input to the tntegratol. And. Ir(tj. the output of the integrator is:

The impulse response Ir(t) is shown in Fig. S6.1-6a. (b) The transfer function of this circuit is:

H ( u ) = TsiIic ($) ~- -J~T"

and

The amplitude response of the filter is shown in Fig. S6.1-6b. Observe that the filter is a lowpass filter of ba~ldwidth 2r;T rad/s or 1!T Hz. The impulse response of the circuit is a rectangular pulse. When a sampled signal /s applied at the in ut. each

output 1s a staircase approximation of the input as s own in Fig. S6.1-6c. It sample generates a rectangular pulse at the output. ypor t ional to the corrnpondtng sample value. ence the

Figure SB. 1-6

6.1-7 (a) Figure S6.1-7a shows the si nal reconstruction from its samples usitlg the first-order hold ci1,cuit. Each sam Ie generates a triangle of r i a th 2T and centered at the sampling instant. .The height of the triangle is eqltnl to t {e sample \.;the. The resulting slgnnl consists of straight line segments join~ng the sample tops. (b) The traasfe~ fitnction of this circuit is: -

1 1 4 1 / J \ L / Because H ( J ) is positive for all J. it also represents the amplitude response. Fig. S6.1-7b shou-s the impulse response h(t) = A(&). The corresponding amplit.ude response H ( J ) and the ideal amplitude response (lowpgrs) required for signal reconstruction is rhowa in Fig. S6.1-7c. ( c ) A ininimum of T secs delay is r uired to make I r ( t ) causal (realizable). Such a delay would cause the recotistructed signal in Fi . SB.1-Ta tSbe delayed by 7 usr. (d) When thr input to t f e first filter is b ( t ) . then as shown in Pmb. 6.1-4. its output is a rectangular pu le p(t) = " ( 1 ) - ~ ( t - T) sliown in Fig. S6.1-4a. This pulse p ( t ) is applied to the input of the second ident~cal filter The output of the summer of the seco~ld filter is p(t) - p(t - 1') = t ~ ( t ) - 2u(i - T ) + u ( t - ST). which is applied to the integrator. The output Ir(t) of the integrator is the area under p ( t ) - p( t - T). which. as T -

sl~o\vn in Fig S6.1-fb.

6.1-8 Assume n signal g ( t ) that is rimult.aneously timelimited and bandlimited. Let g ( u ) = 0 for Iwl > 2nB. Therefore .q(;jrect (& ) = q ( ~ ) fot B' > B. Therefore from the time-convolutio~~ property (3.43)

Figure S6.1-7

Because g ( t ) is timelimited. g ( t ) = 0 for It( > T. But q ( t ) is equal t.o canvolution of g ( t ) with s inc(2n~ ' t ) which is not timelimited. It is impossible to obtain a time-limited signal from the convolution of a time-limited signal with a non-timelimited signal.

6.2-1 (a) Since 128 = 2'. we need 7 bits/charact.er. (b)For 100,000 characters/second . we need 700 kbit,s/second. (a) 8 bits/charact.er and 800 kbitsjsecond.

0.2-2 (a) The bandwidth is 15 kHz. The Nyquist rate is 30 kHz. (b) 6.5536 = 216. so that 16 binary digits are needed to encode each sample. (c) 30000 x 16 = 480000 bits/$. (d) 44100 x 16 = 705600 bitsls.

0.2-3 (a) The Nyquist rate is 2 x 4.5 x lo6 = 9 MHz. The actual sampling rate = 1.2 x 9 = 10.8 MHz. (b) 1024 .= 2''. so that 10 bits or binary pulses are nreded ro encode each sample. (c) 10.8 x lo6 x 10 = 108 x lo6 01. 108 Mbits/n.

6.2-4 If nr, is the peak sample a~nplitude. then

quantization error < (0.2)(1"P) = 3 I 0 0 500

%eca~tse tlie ~naxirnurn quar~tization error is 9 = % = T, it follows that

Because L sl~ould be a power of 2, we choose L = 512 = 29. This requires a Qbit binary code per sample The Nyquist rare is 2 x 1000 = 2000 Hz. 20%, abo\-e this rat,e is 2000 x 1.2 = 2400 Hz. Thus. each signal h i ~ s 2400 saniples/second. and each sample is encoded by 9 bits. Therefore. each signal uses 9 x 2400 = 21.6 kbits/second. Five such signals are n~ultiplcxed. hence. we need a total of 5 x 21.6 = 108 kBits/acond data bits. Framing and synchronization requires additional 0.5% bits. that is. 108,000 x 0.005 = 540 bits, yielding a totd of 108540 bits/second. The minimum transll~ission bandwidth is = 54.27 kHz.

6.2-5 Nyquist rate for each signd is 200 Hz. The sampling rate J, = 2 x Nyquist rate = 400 Hz Total number of samples for 10 signals = 400 x 10 = 4000 sampla/second.

O.2Sm Quantization error + = 3 hloreover. quantizatio,, error = 9 = 2 = = 3 ===+ L = 400 Because L is n power of 2. we select L = 512 = 29. that is. 9 bits/sample. Therefore: the minimum bit rate = 9 x 4000 = 36 kbit.s/second. T11e miilimum cable bandwidth is 36/'2=18 kHz.

zv 6.2-6 For a sinusoid. = 0.5. Tlte Sh'R = 47 dB -501 19. From Eq. (6.16)

Because L is a power of 2. we select L = 256 = 2'. The SNR for this value of L is - So 2 r , i Z ( t ) - = 31- - = 3(25612(0.5) = 98304 = 49.43 dB N o Ill ,

- C

Fig. 56.2-7

6.2-7 For this periodic r n ( t ) . each quarter cycle takes on the same net of amplitude values. Hence, each quarter cycle contributes identical energy. Consequently, we can compute tohe power for this signal by averaging its energy over a quarter cycle. The equation of the first quarter cycle aa shown in Fig. S6.2-7 is nr(t) = 4A/To. where A is the peak amplitude and To is the period of m(t). The power or the mean squared value (energy averaged over a quarter cycle) is

Hence. = = 4 . m~

The rest of the solution is idelltical to that of Prob. 6.2-6. From Eq. (6.16). SNR of 47 dB is a ratio of 50119. is

Because L is a power of 2. we select L = 2.56 = 2'. The SNR for this value of L is

6.2-8 Here I r = 100 and the SSR = 43 dB= 31,622.77. From Eq. (6.18)

So 3 ~ 2 - = - = 31,622.77 N o (In 101)2

Berause L is a power of 2. we select L = 512 = 2'. The SSR for this value of L is

6.2-9 (a) Syquist rate = 2 x lo6 Ha. The actual sampling rate is 1.3 x (2 x lo6) = 3 x 10' Hz. Moreover. L = 256 and 11 = 235. From Eq. (6.18)

(b) If we reduce the sampling rate and increase the d u e of L so that the same number of bits/second is maintained. we can improve t.he SNR (because of increased L) with the same bandwidth. In part (a). the sampling rate is 3 x lo6 Hz and each sample is encoded by 8 bits (L = 256). Ilence. the transmission rate is 8 x 3 x lo6 = 24 hfbits/second. If we reduce the sampling rate to 2.4 x lo6 (20% above the Nyquist rate). then for the same t.ransmission rate (24 Xlbitsls). we car1 have (24 x 106)/(2.4 x 10') = 10 bits/sample. This results in L = 2'' = 1024. Hence. the new SNR is

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

6.2-10 Equation (6.23) shows that increasing n by one bit incre- the SNR by 6 dB. Hence, an increase in the SNR bv 12 dB (from 30 to 42) can be accomplished by increasing n from 10 to 12, that is increasing by 20%.

6.4-1 (a j F I O I ~ Eq. (6.33) "'" so that 1 = 9 mcr = - (~)(64*00') _4 ,, = 0,0783 Y' 2ir x 800

n 2 ~ (0.0785)*(3~00) = 1.12 (b) = -- = 3js (3)(64000)

(c) Here So = = 0.5. and So - = O" = 4.46 x 10.' No 1.12 x lo-"

(d) For uniform distribution

7 1 , 1 So SO = v,, ( t ) = - = - SO that - = 0.333 = 2.94 x 10"

3 3 N~ 1.12 x 10-4

l 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 bandwidt h of 64 kHz.

Chapter 7

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

t

For polar signaling [see Eq. (7.1211

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

2nn = 0 for t-i~ = - for all n # 0 , and = 1 for orn = 0. Hence,

Tb

For bipolar case [Eq. (7.20b)l -

= $ sinc2 (9) sin2 (7) The PSDs of the thee ELK. are shown in Fig. S7.2-1. From these s p f 4 we find the bandwidths for all three cases to be Rb Hz. The bandwidths for the three sun, when half-width pulses arc used, are as follows: Polar and on-OR 2 Rb Hz; bipolar: & H z Clearly, for polar and on-off cases the bandwidth is halved when full-width pulvs are used. However, for the bipolar case, the bandwidth nmainr unchanged. The puke shape has only a minor influence in the

bipolar case because the term sin2 in Sy (m) dctamincs its bandwidth.

Fig. S7.2-1

Fig. S7.2-2

P(t) = rect I;] - - rect [$I - and

= +she (+) . /~GM + $sinc 2 r - ~ a ~ " ( 4 1

= jq sinc (9) sin (F) lp(Q$

S,(d= = q, sinc2 (F) sin2 (9) 4n

From Fig. 57.2-2, it is clear that the bandwidth is - Tb rad 1 s or 2% HZ.

7.2-3 For differential code (Fig. 7.17) l N 2 N

= N+m lirn -[-(I) N 2 +-(-I)'] 2 = 1

To compute RI, we observe that there arc four possible 2-bit squences 11.00.01, and 10, which are

equally likely. The product ~ k a k , ~ for the fm two combinations is 1 and is -1 for the last two combinations. Hence,

1 N N Rl = N+m lh -[-(I)+$-l)]=O N 2

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

7.24 (a) Fig. S7.2-4 shows the duobinuy pulse train At) for the sequence 1110001101001010.

(b) To compute &, we observe that on the average, half the pulses have a& = 0 and the remaining half have ak = 1 or - 1. Hence,

1 2 N =- & = N-+m h -[-(*I) N 2 +T(o)] 2

To determine Rl , we need to computeakak+l. There are four possible equally likely sequences of two bits:

11,10,01, 00. Since bit 0 is encoded by no pulse(ak = O).the product of akak+1' Ofor the last three of 3N N

these sequences. This means on the average - 4 combinations have a&ak+, = 0 and only - combinations 4

have nonzero aka&+]. Because of the duobinary rule, the bit sequence 11 can only be encoded by two consecutive pubes of the same polarity (both positive or both negative). This means ok andak+, arc 1 and 1 or -1 md - 1 respectively. In either case akok+i = 1. Thus,

N these - combinations have akak+l = 1. Therefore,

4

To compute R2 in a similar way, we need to observe the product okok+2. For this we need to observe all possible combinations of three bits in sequence. There are eight equally likely combinations: 11 1,101, 110,loO, 01 l,OlO,oOl, and 000. The last six combinations have either the first andlor the last bit 0. Hence, okok+z = 0 for dl these six combinations. The fvst two combinations an the only ones which yield nonzero okak+2. Using the duobinary mlt, the first combination is encoded by three pulses of the same polarity (all positive or negative). Thus ak are 1 and 1 or - 1 and - I , respectively, yielding

akak+2 = 1. Similarly, because ofthe duobinary rule, the fvst and the third pulses in the second bit combination 101 are of opposite polarity yielding akok+2,= -1. Thus on the average, okak+2 = 1 for N N 3N - terms,-1 for - tmns, and 0 for - terms. Hence, 8 8 4

In a similar way we can show that R,, = 0 n > 1, and from Eq. (7.10c), we obtain

Fig, S7.24

For half-width pulse P ( t ) .: rect(21 I q).

From Fig. S7.24 we observe that the bandwidth is approximately Rb / 2 Hz.

7.3-1 From Eq. (7.32)

4000 = (1 + r)6000 1

a r t - 2 3

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

P(r) = Rbsbc ( 2 x ~ ~ t ) + & S b C 2

Hence,

sin zRbt sin(* &t - x) s-+

XRbt %Rbt-A

1 7.3-9 TheNyquistintcrvalirT,=-=%. The~yqoiannplcr~.p(tn~)fan=0.1,2,......

Rb From Eq. (7.16), it follows that

p(0) = p(q) = 1 and ~ ( f n ~ ~ ) = 0 for all othcrn.

1 Hence. ha. Eq. (6.10) with T, = 5, and B =$= -, 2 2%

sin rrR t sin n b t sin x&t b- - - x Rb X Rbr - X x Rbf(l - Rbl)

The Fourier transform of Eq. (1) above yields

7.3-30 (a) No error because the sample value, of the m e polarities arc separated by even number of lnor and the sample values of opposite polarities are separated by odd number of zeros. (b) The fmt sample value is I because there is no pulse before this digit. Hence the first digit is I. The detected sequence is

110(10100110110100

73-10 The fvJt sample value is 1, indicating that the transmissions starts with a positive pulse, that is, first digit I. The duobmary rule is vioiated over the digits shown by underbracket.

Following arc possible correct sample values in place of the 4 underbracket values: 2 2 0 -2, or 2 0 -2 -2, or 0 0 0 -2, or 2 0 0 0. These sample values represent the following 4 digit sequence: 1100, or 1000, or 0100, or 1010. Hence the 4 possible comct digit sequences arc

where x t x l x , ~ is any of the four possible sequences 1100,1000,0100, or 1010.

7.4-1 S = 101010100000111 From example 7.2

T r (1Q D3 Q D5 Q D6 Q D9 Q Dl0 Q D" @ Dl2 Q Dl3 $ D"Q ...) s

T = 101110001101001 R = 101010100000111 = S

7.4-2 S = 101010100000lll

T = ( I @ D ~ @ D ~ Q D ~ @ D ~ @ D ' ~ @ D ' ~ @ D ' * ~ , . . ) S

R = (1 @ 02)r (we Fig. S7.4-2) T

Fig. S7.4-2

7.5-1 From Eq. (7.43, we obtain

1 4.002 0.1 -0.1 13

(i) For polar case P, = ~ ( 5 ) = 2.87 x lo-'

(ii) For on-off case P, = e(5 12) = 0.0062 1

(iii) For bipolar case 4 = lSQ(5 / 2) = 0.0093 15

In the following discussion, we assume Ap = A, the pulse amplitude.

R& (b) Energy of each pulse is E, = A' 5 I2 and there are Rb puldsecond for polar case and - 2

pulses/second for on-off and bipolar case. Hence, the received powers are

(c) For on-off case:

We require P(E) = 267 x lo-' = Q( A~ 1 2 0 ~ ) . Hence,

Ap/2un =5andAp = IOu, =0.003

For bipolar case:

Hence

and

7.6-2 For onoff case:

on . lo5 2 (4.7~)(2 loJ) - 95 x lo-'

For on-off case, half the pulses an zero, and for half-width rectangular pulses, the aansmitted power is:

= 2256 x 1 od watts. 4

There is an attenuation of 3 0 d ~ , 6r equivalently, a ratio of 1000 during transmission. Therefore

ST = lOOOS, = 2256 x 10" watts

7.6-3 For polar case:

For polar case with half-width rectungular pulse:

For bipolar case:

For bipolar (or duobinary), half the pulses arc zero and the receive power St for half-width rectangular pulses is

ST = (1 OOO)Si = 2338 x 1 o - ~ watts

7.72 Sampling rate = 2 x 4000 x 125 = 10,000 Hz.

m~ Quantizstion error = - = 0.00 Imp a L = I000 L

Because L is a power of 2, we select L = 1024 = 2". Hence, n = 10 bitslsample. 10

(a) Each Cary pulse conveys log2 4 = 2 bits of information. Hence, we need- = 5 4-ary pulses/semple, 2

and a total of5 x 10,000 = 50,000 4-ary pulscs/second. Therefore, the minimum transmission bandwidth is

7.7-3 (a) Each 8-ary pulse carrier logz 8 = 3 bits of information. Hmce, the bandwidth is reduced by a factor of 3. @) The amplitudes of the 8 pulses used in this 8-ary scheme are f A 12, f 3A 1 2, f 5 A / 2, and i 7A 12. Consider binary ccrse using pulses f A 1 2. Let the energy of each of these pulses (of amplitude f A I 2 ) be Eb . The power of thb binary case is

hinry = EbRb

Because the pvhe energy is proportional m the square of the amplitude, the energy of a pulse t fi is 2

k2 E ~ . Hence, the average energy of the 8 pulses in the 8-ary case above is

Hence,

Therefore, %-my = 7 %iW

7.7-1 (a) M = 16. Each 16-ary pulse conveys the information oflog2 16 = 4 bits. Hence, we need

-= '2s000 3000 1 Gary pulrcdsecond. 4

Minimum transmission bandwidth = 3000 1 1500 Hz. 2

2 (b) From Eq. (7.32), we have 4 = - Br. Hence,

l + r

7.7-4 (a) For polar signaling, Rb bits/second requires a bandwidth of Rb Hz. The half-width rectangular pulse of A

amplitude - has mergy 2

A ~ T /i2 The power P is given by P = EbRb = -A R~ = -

8 8 2 (b) The energy of a pulse f - is k Eb. Hence the average mergy of the M-ary pulse is

2

E M = -!-FLi A4 +2(i312 +2(*)'+.....+2[*(~ - I)]' Ed]

R Each M-ary pulw sonveys the information of logl M bits. Hence we require only -A M-zvy

Jog2 M pulsedsecond. The power PM is given by

7.7-5 Each sample requires 8 bits (256 = 2') . Hence: 24,000 x 8 = 192,000 bidsec. BT = 30 kHz

2 R=- 2 Br = -(30,000) = 50,000 M-ary puWsec.

I+r 12 We have available up to 50,000 M-ary pulsedsccond to transmit 192,000 biWsec. Hence, each pulse must

transmit at least - = 3.84 bits. (ZE)

= choose 4 bits/pulsc M = 16 is the smallest acceptable value

7.8-1 (a) Baseband polar signal at a rate of IMbitdsec and using full width pulses has BW = IMHz . PSK doubles the B W to 2MHz.

(b) FSK can be viewed as a sum of 2 ASK signals. Each ASK signal BW = 2 MHz. The first ASK signal occupies a band fd f 1 MHz, and the second ASK signal occupies a band f,, f 1 MHz. Hence, the bandwidth is 2 MHz + 100kHz ~ 2 . 1 MHz.

7.8-2 (a) A baseband polar signal at a rate 1 Mbits/sec using Nyquist criterion pulses at r = 02 has a

PSK doubles BW to 1.2 MHz.

(b) Similar to Rob. 7.8-1. BWFsK = 0.6 MHz + 0.6 MHz + 100 HZ BWFsK = 13 MHz

# P s D FsCi 7.8-3 logz M = 2 for M = 4.

We need to tansmit only 05 x 1 o6 4-ary pulses/sec (a) B W is reduced by a factor of 2.

BWFsK = I MHz (b) In FSK, there are four center (carrier) Erequencics fcl, fez, fc3, udfc4, each sepmted by 100 kHz-

0 Since ASK signal occupies band 1, i 05 MHz, the total bandwidth is 05MHz+O.5MHz+100Wz+100~z+100Wz=13MHz. t . 3 M H t

Either figure (a) or (b) yields the same nsuIt. ml (t) has 8400 ~unpled~oc.

m2 ( t) , m3(t), nq (r) each bas 2800 sampledsec. Hence, there are a total of 16,800 sampledscc.

7.9-2 First, we combinem2(t), m3(t), and m,(r) with a commutator speed of 700 rotationsfsec. This combined

signal is now multiplexed with nr,(t) with a commutator speed of 2800 rotationdsec, yielding the output of 5600 samples/sec.

Fig S7.9-3

Chapter 8 Exercises I .

8.1- 1 If a plesiochronous network operates from Cesium beam clock which is accurate to * 3 p s N ;, in 1012, how long will it take for a DS3 signal transmitted between two networks to become out' 1 '

of sync if a 114 bit length time error results in desynchronization? * &

Solution: A DS3 bit is transmitted in ll(44.736.10~) = 2.235336.14' sec. In the wont case, both network clocks will be out of synchronization by 6 parts in 1012.

2.235336-10'81(6.10"2) = 3922.27 sectbit or 980.57 sect W bit

8.1-2 For the bit stream 01 1100101001 11 101 1001 draw an AM1 waveform. Solution:

Note that typically, for illustrative purposes, the waveform is given as

8.1-3 For the following waveforms, determine if each is a valid AM1 format for a DS 1 signal. If not, explain why not.

solution: No. 16 0's violation

I

2 -

~olution: No. bi-polar violation 1

Solution: No. 1 ' s density violation d.

solution: Yes

e.1-a a) You have received the following sequence of ESF framing att tern seauence bits

Is this alegitimate framing bit sequence in order to maintain synchronization between the T1 transmitter and receiver? Yes No If 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 AM1 signal is received:

Is this an acceptable T1 signal? Yes No a. If yes, explain. b. If no, explain why not (what, if any, DSl standards are violated) and

draw on the figure the AM1 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 The signal llOlOOOWOOOOOOOOOOO1 is received by the DSU in a T1 data stream which uses a BSZS format. Draw the output of the DSU for this signal? The first 1 is already drawn. Show the bit stream which is substituted by the DSU. Solution:

8.1-6 T-1 synchronization at two distant locations is controlled by separate crystal controlled oscillators which differ in frequency by 125 parts per million. If the terminal equipment doesn't maintain sync in how many complete D4 superframes will the faster oscillator have generated (at most) one more time slot (&bit) than the slower oscillator ? Circle the correct answer.

a) 5 b) 10 c) 15 d) 20 e) None of the above - if "none", what is the number of D4 superframes before an extra time slot is generated? Solution: e) The faster oscillator will generate 125.10~1.544.10~ = 193 bits per second more than the slower oscillator. This is one framelsec = 24.125 time slots. 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 10'). The networks an 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 in

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 1 112 of a T1 "bit time".) Solution: A Tl bit time is ll(1 .544.106 ) = 6.47668.10" seclbit. In the worst case, the two clocks would be off by 2.3 = 6 parts in 10') or 6.10"~ errored bits per bit transmitted. Hence, 6.47668.10.' sechit 1 6.10.') enorcd bits per bit = 1.07945.10~ seconds per enored bit or 5.39723.10~ seconds per errored half-bit.

b. UTC operates via GPS satellites which are approximately 23,000 miles above the Earth. How long, in terms of TI bits, will a correction signal take to be transmitted to the network switches? Solution: The speed of light is approximately 186000 miles/sec. 23000x2 = 46000miles up and down. 46000/186000 = 0.247 sec

0.247~1544000 z 381850 bits

Chapter 10

13+13 1 0 I -1 (a) P(red card) = - = -

52 2 I+1 1

(b) P( black queen) = - = - 52 26

12 3 (c) P(picture card) = - = -

52 13 4 1 (d) P(7)= -=-

52 13 20 5

(c) P ( n ~ 5 ) = - 5 - 52 13

10.1-2 (a) S = 4 occurs as (1,1,2),(1,2,1),(2,1,1). Thm ere a total of 6 x 6 x 6= 216 outcomes. 3 I

Hence, P(S = 4) = - = - 216 72

(b) S = 9occurs as (1,2,6), (1,3,5), (1,4,4), (1,5,3), (1,6,2), (2,1,6), (2,2,5), (2,3,4), (2.4.31, (2,5,2), (2,6,1), (3.1.51, (3,2,4), (3,3,3), (3,4,2), (3,5,1), (4,1,4), (4,2J), (4,3,2), (4,4,I), (5,1,3), (5,2,2), (5,3,1), (6,131, (62,l)

10.1-3 Note: There is a typo in this problem. The probability that the numberiappws should be ki not ki .

10.1-4 We can draw 2 items out of 5 in 20 ways as follows: 0102, 0103, OIPt, o1P2, 0201,0203, &PI, 02P2, 0301,0302, 03PI, 03P2, PIOl, P102, P103, PIPz, P201, P202, P203, P2P1. AII these outcomes arc equally likely with probability 1 DO. (i) ThiseventE, = 0 1 P l U ~ I ~ U O 2 ~ U O 2 ~ U O 3 ~ U 0 3 ~ U ~ 0 1 U ~ O ~ U ~ 0 3 U ~ 0 1 U ~ 0 ~ U ~ ~ 0 3

12 3 Hence, P(El) = - = -

20 5 (ii) 'Ihis event E2 = 4 4 U 4 4

4 I Hence, P(E2) = - = -

20 10 (iii) This event EJ = 0102 UO1o3 Uo2o1 U0203 U0301 Uo302

6 3 Hence, P(E3) = - = -

20 10 (iv) This event El = E2 U E3 and both E2& Ej are disjoint.

4 2 Hence, P ( E 4 ) = P( E ~ ) + p(E3) = - = -

10 5

10.14 Let xo, be the event that the firn chip is oscillator n d x q be the event hat the fmt chip is PLL. Also,

let xo2 and x 4 represent events that the second ship drawn is an aci l l l ta and a PLL, respctively. Then

10.16 Using the notation in the solution of Prob. 10.1-5, we fmd:

(a) p(xs l q ) =

10.1-7 (a) We can have(';) ways of getting two 1's and eight 0's in 10 digits

45 45 P (two 1's and eight 0's) = 4S(0$(0.5f = 45(05)'O = - = - 2'' 1024

(b) P(at least four 0's) - 1 - [~ (exac t l~ one 0)] + [ ~ ( e x a c t l ~ two OS)]+ [ ~ ( e x a c t l ~ three o's)]

10 5 P (one 0) = ('P)O~)'O = - = -

1024 512, 45

P (two 03) = - - 1024

120 P (three 0's) = ('10)(05)'~ = -

1024 5 45 120 849

~ ( a t ~ e a s t ~ o u r ~ ' s ~ = l - -+-+-)=- 512 1024 1024 1024

10.1-8 (a) Total ways of drawing 6 balls out of 49 are

1 Hence, Rob(matching all 6 numbers) = 13983816

(b) To match exactly 5 number means we pick 5 of the chosen 6 numbm and the last number can be

picked imn the remaining 43 numbers. We can ch- 5 numben of our 6 in ( f ) = 6 ways and can choose ,

one number out of 43 in (:)) = 43 ways. H a w , we have 43 r 6 combiiatiom in which exactly 5 numbers

match. Thus,

P (matching exactly 5 numbers) = 43x6 =1845x10d 13983816

(c) TO match exactly 4 numbers mums we pick 4 out of the chosen 6 number in ( f ) = 15 ways and choose

2 out of the remaining 43 numbers in ('3) = 903 ways. Thus t h m are 15 x 903 ways of picking exactly 4

numbers out of 6 and

P (matching exactly 4 numben) = 15x903 =9.686x104 13983816

(d) Similarly, m can pick three nwbm exactly h(f14i) = 20 x 12341 = 246820 ways. Hare ,

246820 P (match'mg exactly 3 numbers) = - = 0.01 765 13983816

10.1-9 (a) Let f represent the system failure. Then

P ( j ) = ( i - o . 0 1 ) ~ ~ - 0 . ~ 3 8 a

P( f ) = 1 - ~ ( j ) = 0.0956

(b) ~ ( ] ) = 0 . 9 9 a n d ~ ( f ) = 0 . 0 1

If the probability of failure of a subsystem s, is p, then P(j) = (I - P)ID or 0.99 = (1 - Py * P = 0.0010045

10.1-10 Iff represents the system failun and f , and f L represent the failure of the upper and the lower paths. respectively, in the system, then:

(a) P(f 1 = P ( f u f ~ ) = P ( f " ) P V d = [ ~ ( f " f P( 1,) = 1 - ~ ( j , ) = 1 - (1 - 0.01)'~ = 0.0956

and

P( f) = (0.0956)~ = 0.0009143

Reliability is ~ ( j ) = I - P ( / ) = 0.9908

10.1-1 1 Let P be the probability of fiilun of a subsystem (s, or q) . For the system in Fig. a: The system fails if the upper and lower branches fail simultaneously. q e probability of any branch not failing is

(1 - P)(1- P ) = (1 - P ) ~ . Hence, the probability of any branch failing is 1 - ( 1 - P ) ~ .

Clearly, PI . the probability of the system failure is P, = [I - (1 - P).' 11 - ( 1 - P)'] r 4 P2 P << I

For the system in Fig. b: We may consider this system as a cascade of two subsystems xi and x2 , whm x, is the parallel combination

of sl and sl and x, is the parallel combination of q and s2. Let PJ (xi) be the probability of failure ofx, . Then

= P (x2)=!J2 Pf (XI) f The system functions if neither xl nor x2 fails. Hence, the probability of system not failing

is (1 - ( 1 - P ) 7hmfon, the probability of system failing is

P / = I - ( ~ - P ~ ) ( ~ - P ~ ) = ~ P ~ - P ~ . ~ P ~ P<<1

Hence the system in Fig. a has twice the probability of failure of the system in Fig. b.

10.1-12 There arc (I:) = 2598960 ways of geaing 5 car& out of 52 cards. Nupbcr of ways of drawing 5 cards of

the same suit (of 13 cards) is(':) = 1287. There arc 4 sub. Hence there are4 r 1287 ways of getting a

flush. Therefore,

10.1-13 Sum of 4 can be obtained as (1,3), (2,2) and (3,l). The two dice outcomes are independent. Let xl be the outcome of the regular die and x, be the outcome of irregular die.

1 1 1 Therefore P, (4) = - + - o - 18 36 12

Similarly,

<(5) = (194) + (293) + e,, (312) + (491)

10.1-15 (a) Two 1's and three 0's in a sequence of 5 digits can occur in(:) = loways. The probability one such

sequence is

P (oa12 ( 0 2 ) ~ = 0.005 12 Since the event can occur in 10 ways, its probability is

10~0.00512 = 0.0512

(b) Three I 'r occur with probability(~)(0.E)~(02)~ = 02048

Four 1's occur with probabi1ity(~)(0l)'(02)' - 0.4096

Fiw 1 'r occur with (:)(0.8)~(02)~ = 03277

Hence, the probability of at least three 1's occuring is P = 02048 + 0.4096 + 03277 = 0.942 1

10.1-1 6 Prob(no more than 3 error) = P(no error) + ~ ( 1 error) + P(2 =or) + ~ ( 3 emr) - (I - p, ) lW +(p) Pe(l - p,)w + (p)~: (1 - + (:0°) P:(l- PJ9'

10.1-17 Error can occur in 10 ways. Consider case of mor over the first link correct detection over every link) = (1 - q)(1- 4). ..(I- Plo)

PE "1-PC = I - ( I - 4 x 1 - 4 ) ...( 1-90)

= 1-[I-(4 + 4 + . . ~ 4 ~ ) + h i ~ h e r ordertexms]

= p , + q + ...+qo p i e 1

1 10.1-19 (a) success in 1 trial) = - = 0.1

10 (b) P(success in 5 trials) = 1 - ~(fa i lun in all 5 trials)

= ' - Ph Ph Pf3 Pf4 Pfs PA = Prob(fii1ure in I ' trial) = 9 / 10

p/2 = Rob(failun in 2" trial) = 8 I 9

Similarly, Pj3 = 7/8, PI, = 617, and Pj, = 516

Hence, P(success in S trials) = 1 - 05 I0

10.1-20 Let x be the event of drawing the short straw and the 4(x) denote the event that ith penon in the sequence draws the short straw. Now, q (x) = 0.1

4 (x) = Rob(1 person does not draw the short straw) x Prob(2" person draws the short straw)

Similarly, P3(x) = Prob(neitha 1' nor 24 person draws the short straw) x ~ o b ( 3 ~ penon draws the shon straw)

Similarly, P4(x) = P5(x) =...= qo(x) = 0.1

10.1-21 All digits are generated independently

(a) P(all 10 digits arc 0) = (03y

(b) There are ( J) ways of armng*g eipht 1's md two 0's. Hence,

P(eight 1's and two 0's). ('~)(0.7)*(03)'

(c) P(at least five 0's) - P(five O's)+P(six O's)+. . . +P(m 0's)

= ('!)(0.7)~(03)~ + ('~)(0.7)* (0.3)~ + (110)(~.~)3(~3)7 + ('!)(0.7)~ ( 0 3 ) + ('$)(0.7)(03)~ + (03)"

- 10.2-2 (a) P,',~ (111) - - (1 - Pe)Q

Py(l) (1-Q)G+(l-8)Q

px,y (41) = 1 - q y (111)

(note that pY (I) and pY (0) arc derived in Uumple 10.10)

1 1 10.2-3 (a) P(x 2 I) = IfP-'& = - e

1 1 3 (b) Rob(-] <xs2)=C-5zex&+fLxe-x&=~---- 0 2 e 2e2

3 (c) Prob (x L -2) = r2 -!kex& - -

-Q 2 2e2

Fig. S10.2-4

Since this is a half-wave rectifier* y assumes only positive values. So ~ ( y < 0) = 0. 1 1

Hence, Fy ( y ) = 0 (for < 0) and < 0') - - 2 . Hence* F,, (o+) = - 2

10.2-5 x is a gaussian r.v. with mean 4 and a, = 3

Hence, 4-4

0-4 (b) p ( X , o ) = P ( T ) = ] - p ( ~ ) = 1 - 0 1 ) 9 ~ 7 6 1 ~ . ~ 8 3

-2 4 p(x 2 -2) = Q(+) 1 - Q(2) = 1 - 0.02275 = 0.9773

Fig. S103-5

10.24 (a) From the sketch it is obvious that x is not gaussian. However, it is a unilateral (rectified) version of Gaussian PDF. Hence, we can use the expression of Gaussian r.v. with a multiplier of 2. For a gaussian r.v.

1 (x2m w j & u y = 4 PAY) = ?~r; I

(b) Hence, (i) P ( x 2 I) = 2 ~ ( y 2 1) = 2 = 08026 43 (ii) P ( I C X S ~ ) = ~ ~ ( I < ~ L ~ ) = {(i) - (:)I = 0.1 856

(r) If we take a Gaussian random va&le y 1 -y2 132 Fig. S10.2-6 pY (y ) = xe

I .

and rectify y (all negative of y multipled by -I)* the resulting variable is the desired random variable x.

10.2-7 The volume V underpw(x, y) must be unity.

Y But y = -x + 1 and the limits on y are 0 to 1 - x . Therefore,

Ill-x MxSl

Similuly, pyl,(y(x) = 0 otherniK

Clearly x and y arc not independent.

and

(b) From results in (a), it is obvious that x and y are independent.

Fig. S10.2-7

1 -y2/2a Similarly we can show thatpy (y) = r e 2m

Then fore

Hence, K = - -(x2+m+y2) -2 " e-~2-rvdy= K G e -3x2 14 = E e - 3 X 2 14

px(x) = K L e d y = K e j,

and

Since p,, ( x , y ) L px (x )pY ( y ) , x and y are not independent.

10.2-1 1 Pe = ~ ( 4 1 ) ~ ~ ( 1 ) + P ( ~ O ) P ~ ( ~ ) If the optimum threshold is a , then

Fig. S10.2-11

2 2 - ( ~ ~ - a ) I Z U , ~ -(Ap+a) 12un

Hence, e P,(l) = e PI (0) I

px (0) And a = In[-] 2 A p Px(f )

1 10.3-2 pX ( x ) = -M~-M

2 Because of even symmeay of px(x) , 7i = 0 and

mx2 1 7i2 = 2JOe xzPx ( x ) m = 2J0 5xe-x&

= r x 3 e - ' & = 3! = 6 - X2 = q x 2 + ~ 2 =a: =3!=6 Fig. S10.3-2

I e -~212du(y ) . Therefore 10.3-3 P ~ ( Y ) = ~ ~ ( Y ) + ~

I -gaddy

- 1 "xe-x2~32m 8 1 6 ~ - 3 2 Because P = - = lo

2 2n Z n

1 1 10.3-5 The area of the triangle must be I. Hence K = - 2 and px(x) = -(x 8 + I) - I S x L 3

Fig. S10.3-5

- * xne-" For n odd, the is .D odd function of X. Thmf0re xn = 0. 10.3-7 xn = .x J4

For n even, we fmd h r n tables ( I X ~ M S ) - ( ~ - I ) ~ . n

0 n odd

10.3-8 Let xi be the outcomes of the ith die. Then, - xi 5 1+2+3+4+5+6 - 1 i = 1,2, ..., 10 -

6 2 2 - l2 +22 +32 +4* +52 +6 91 xz = =-

6 6 3 5

If x is a RV representing the sum, then - x = ill +K2+...+Xlo = 1

2 2 2 175 Ox = O X I + ux2 +...+ Oxlo

- 175 2 x2 = Q: + z2 = - 6 +(35) = 1254.167

1 1 10.4-1 pX ( x ) = - 6(x ) + - 6(x - 3)

2 2 1 -n2/8

Pn(n) = se y = x + n

1 1 -n2/8 pY(y ) = P X ( X ) ~ n ( n ) = [16(4 + Ib(X-3)]* -' 2 f i Fig. S10.4-I

. u - ~ ) ~ 1 8 ] , + ii16(. 2 - 2 2n

p, (x ) = 0 . 4 4 ~ ) + 0.66(~ - 3)

1 - n ' / ~ ~ n ( n ) = ~ ~ o

and 1 -y2/8 + 3 -(y-3)2~8 Fig. S10.4-2

p y ( v ) = - x e 1 6 .

10.4-4 PZ(Z)= PX (4 * P~(Y) Taking Fourier transform of both sides, we have

Taking inverse Fourier transform we get

It isclearthat L = K + ~ a n d u ~ = u ~ + <

2 10.61 For any real a, [n(x - 5i) - (y - T)] 2 0, m 020: + o: - 2aoW 2 0 . Hence, the discriminant of this

quadratic in a must be nonpositive, that is:

4 4 - 4 o : r : 5 0, that is, I - s l s l a Ids1

10.5-2 When y = Klx + K2 Hence, 7 = K15i + K2 2 2 2 0 , - K I o x a n d o x y = ( x - ~ ) ( ~ - ~ ) = ( x - P ) ( K ~ x + K ~ - K ~ ~ - K ) = K ~ o ~ . Hence,

2 OXy - ~ ~ o : I K l o : = 1 if KI is positive. If K I is negative, a, = K l o , is negative. PXY =O.Oy-

But a, and oy are both positive. Hence, p x y = -1

2n 1 2n 10.5-3 X = I, cos Op(8)ds = - cos 0 do = 0 Similarly, 7 = 0

2n 0 - 1- no = xy 5 cos~s in8= -sin28= ~ ~ r i n 2 6 ~ ( 8 ) d ~ = ~ ~ ~ s i n 2 8 d 8 = 0 2 2 4n --

Hence, o,, = x y - 0 and x, y are uncomlued. But x2 + y2 = I. Hence, x and y are not independent.

10.61 In this case - 2 R I I =Rz =R33 =mk =Pm

R12 = R21 R23 0 R32 = 0825Pm R13 P R3i = RO2 = 0562 Pm 4 3 = 0308Pm Substituting these values in Eq. (10.86) yields: a1 = 1.1025, a2 = -02883, a3 = -0.0779 From Eq. (10.87). we obtain - r2 = [I - (0.82k1 + 0 5 6 2 ~ ~ + 0 3 0 & ~ ) ] ~ ~ = 02753 P, Hence, the SNR improvement is

Chapter 11

11.1-1 This is clearly a non-stationary process. For example, I I

amplitudes of all sample hctions are zero at same instants (one is shown with a dotted tine. Hence, the statistics clearly depend on t.

I

Fig. S11.1-1

11.1-2 Ensemble statistics varies with t. This can be seen by finding

100 = ASO+ + e) = A Jo C O + ~ + e)p(a)d,~

= A J'@' co<a + ~pcu . his is a iinction of r . lo& 0

Hence, the process is non-stationary.

11.13 This is clearly a non-stationary process since its statistics depend on r . For example, at t = 0, the amplitudes of all sample fbnctions is b. This is not the case at other values of r.

Fig. S11.1-2

a = ~ ~ ~ ~ c o r ( t t # +8) dcu = -sin(u# + 8

100 1 00t 1'7 a = -[sin(l~t +8)-sin81

loot Using this result, we obtain

- - - - 11.1-6 x(t) = at + b = ar + b. But a = 0 Hence, x(t) = b

- - 2 2 Also, a - 0 , a2

11.1-7 (b) G= = O

(c)

- 1 (d) The process is W.S.S. Since x(t) = 0 and Rx (tl , t2) = - I #

3 al t* (e) The process is not ergodic since the time mean of each sample function is different from that of the other and it is

not equal r the ensemble mean (i = 0) - 1 (9 X ~ = R ~ O = ; .

Fig. Sll.1-7

(d) he process is W.S.S. Fig. Sll.1-8

(e) Thc process is not ergodic. Time means of each Jarnple function is different and is not equal to the ensemble mean. - 1 (f) x2 = ~ ( 0 ) - 5

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), (f), and (g) violate these conditions.

11.2-2 (a) Let x(r) = xl and x(t + r) = x2 Then, - - - 7 -

2 2 - (xl k x2)' = x12 +xZ2 +2x1r2 5 0, X I +x2 2 t 2~1x2 - - - But, xlx2 = RX(r) and x12 = x~~ = RX (0) Hence. ~ ~ ( 0 ) 2 l~,(ry

(b) ~ , ( r ) = x(r)x(r + r). lim RX(r) = lim x(t)x(t + r) ?-+a r-+a,

As r -+ a, x(t) and x(t + r) become independent, so lim ~ ~ ( r ) = x(t)x(t + r) = (ZXQ = ii' r-ba

n 11.2-3 RX(r) = 0 for r = f - and its Fourier transform &(m) is bandlimited to B Hz. Hence, R,(?) is a

28 waveform bandlimited to B Hz and according to Eq. 6. lob

RX(r) = ~ ~ ( $ ) r i n c (2nBr -n). Since R, = 0 for alln arceptn = 0, n=-s

Rx(o) r e c t ( 2 ) . Hence, x(t) is a white process bandlimited R, ( r) = Rx (0) sinc ( Z d r ) and Sx (o) ;. - 2B

to B Hz.

11.2-4 R, (r) = PXtX2 (1. 1) + Pxlx2 (-1.-1) - PxlxZ (-1, 1) - Pxlx2 (I,-]) But because of symmetry of 1 and 0,

P,,,, (1, 1) = PXIX2 (-19-1) a d Px,x2 (-19 1) = PXIX2 (19-1)

and Rx(r) = 2[px1x2 (1. 1) - px1x2 (1*-1)] L !

Consider the case n q < (4 < (n+ 1)G . In this we, there are at least n nodes and a possibility of (n + I) -nT r

nodes ~ o b ( ( n + l ) n o d e s ] = ~ = - - n Tb Tb ,.

The event (x2 = llxl = 1) can occur if there arc 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. Px2ixl (111) - R O I ) ] Rob(= cbanp m even number of nodes) + Rob(n nodes) Prob(State changes at eveen number of nodes)

m e number of ways in which changes at K nodcr out of N nodes occur is ( i ) Hence,

d R ) = P X x ( 1 - 1 This yields

and so on.

Fig. S11.2-4

The PSD can be found by differentiating ~ ~ ( r ) twice. The second derivative d2 ~ , / d r ~ is a sequence of 2

impulses as shown in Fig. S11.2-4. From the time-differentiation property.

d - Rx ( r ) u (jcu)' S. (a ) = -a2sX (u) . Hence, recalling that b(r - T) o i j W T , we have dr2

and

11.2-5 BecauseS,(ru) is a white process bandlimited to B , It,,,(?) = &(O) sinc (281) and

This shows that x(t) t + - = Rm - = 0 d ;B) I;) 'Ihus, all Nyquist sample are kcorrelated. Now, k m Eq. 1 129,

- R, = aka&+, = 0 n 2 1 and whenak is the AthNyquist sample. - - Rg =a: = x2 = R,,,(o). Hence,

11.24 For duobinary Pa, (1) = Pa, (-1) = 0 2 5 and Pa, (0) = 05

- RI = arak+~ = C C a k a k + ~ ~ ~ a ~ + , (akak+i)

Y Ok+l

Because a, and a&+, each can take 3 values (0,1, -I), the double sum on the tight-hand side of the above equation has 9 tenns out of which only 4 an nwero. 'I%us,

R~ (l~l)pakak+l (191)+(-1x-1)4kak+i (-1x-1)+(1M-1)p8kak+l (1x-1)-(-1~1)p&k8k+l (-')(') Because of duobinary rule, the neighboring pulses must have the same polarities. Hence,

- - Also R2 =akak+2

In this case, we have the same four terms as before, but ak and ak+2 are the pulse strengths separated by one time slot. Hence, by duobinary rule,

1 In a similar way, we can show that Pa,,,, ( 1 ~ 1 ) = P,~,,, (-1,l) = 16 Hence, R2 = 0 -. <.

Using a similar procedure, we can show that R, = 0 for n 2 2. Thus, from Eq. (1 1.29) and noting that R,, is

146' [ l an even function of n. we obtain Sy (0) = -- -(I + G 2

For half-width rectangular pulse P ( r ) E %sine (5) and S, ( r ) = +sinc2 2

- = - ( i ) 2 ~ + ( - i ) 2 ( ~ - Q) = i

Because all digits arc independent, - = akak+l = i k X k + l - ( 2 ~ - I)' Hence,

11.2-8 Approximate impulses by rectangular pulses each of height h and width & such that h& = 1 and & + 0 i

(Fig. S 1 1.2-8a)

Rx(r) = Z C X I X ~ P X , ~ ~ ( ~ 1 ~ 2 ) 3

XI 9 Since xl and x2 can uLe only two values h and 0,thae will only bc 4 terms in the summation, out of which only one is nonzero (corresponding to xl = h, x2 = h ). Hence,

Since there art a pulses/second, pulses occupy fraction of time. Hmce,

Now, consider the range (4 < E. Pxzlxl (Ah) is the +€I* -. 7

Rob(x2) = h, given that xl = h. This means xl lies on one t -- -I 7

h of the impulses. Mark off an interval of r from the edge of k, . this impulse (see fig. S11.2-8b). If XI lies in the hatched %- interval, x l fills on the same pulse. Hence,

6-r r Px2(x, (qh) = b b ( x I lie in dY hatched region) E - = 1 - -

6 E

and ~ ~ ( r ) = d ( l - ~ )

Since Rx(r) is an even function of r , ~ , ( r ) = &(I-!) It=

In the limit as E + 0, Rl(r) becomes an impulse of strength a.

R X ( r ) = d ( r ) I ~ = O . When t > 6, xl and x2 become independent. Hmce,

Px21x, (44 = px2 (h ) = a&

~ , ( r ) =a2h&=a I4>0

Hence, ~ ~ ( r ) = d ( r ) + 0 2 Fig. Sl 1.2-8

11.2-9 In this case the autocornlation fimction at r = Oremain same as in Prob 11.2-8. But fort > 0 whenever

~ ( t ) , x(t + r ) arc both nonzero, the product x(t)x(r + t ) is equally likely to be h2 and - h2. Hence,

Rx(s)=O, r+Oand&(r)=a6(r)

11.2-10 The process in this problem represent$ the model for the thermal nohe in conductors. A typical sample function of this process is shown in Fig. Sl12-10. The signal x(t) changes abNptly in amplitude at random instants. The average number of changes or shifts in mpliides arepper second, and the number of changes are Poisson-distributed. The amplitude aAer a shift is independent of the amplitude prior to the shift. The fa-order probability density of the process is ~(x;t). It can be shown that this process is

stationary of order 2. Htnce, p(x;t) can be expressed as p(x). We have

Z(r) - P , P , x l x 2 ~ x , , ( x l , ~ 2 ~ 1 ~ 2

= P_ j: ~1x2 PX, (x,)px2 ( ~ 2 1 ~ 1 - x1 Fl4 (1)

To calculate px2 (x2 (x = xl), we observe that in r seconds (interval between xl and x2 ), thm are: two

mutually exclusive possibilities; either t h m may be no amplitude shiA (x2 = XI), or thm may be an

amplitude shift (x2 t xl). We can therefore express fi2 (*lxl - xl ) as px, (x2 lx 1 = xL ) = px2 (SIX = 11, no mplitudl shift)P(no amplitude shifl) +

px2 (x2 1x = XI, amplitude shift)f(amplitude shift)

The number of amplitude s h i b are given to have Poisson distribution. The probability of k shifts in r seconds is given by

whm then are on the average Pshifls per second. The probability of no shih is obviously po(r) , where

Po(r) = i f l r

The probability of amplitude shift = I - po(r) = I - e'Br . Hence

px, (x2 lx 1 = xl ) - e-flrpx2 (x2 IX = XI, no amplitude shift) + (I - e-pr)Px2 (x2 IX I = 1 1 , amplitude shift)

(2) when there is no shift, x l = XI and the probability density of x2 is concentrated at the single value XI .

'his is obviously an impulse located at x2 .: xi . Thus,

P,(x21xI =XI , noamp~itude&ift)=b(~ -q) (3

whenever thm are one or more shifts involved, in general, x2 # xl. Moreover, we are given that the amplitudes before and after a shift an independent. Hence,

px2 ( ~ 2 1 x 1 ~ XI, amplitude shift) = px, (xZ) = Ax) (4)

where px2 (x2 ) is the fmt-orda probability density of the poceu. This is obviously ~ ( x ) . Substituting Eqs. (3) and (4) in Eq. (2), we get

px2 (x2PI = X I ) = e-brb(x2 - X I ) + (1 - e - p r ) ~ x 2 ( ~ 2 )

Substituting this equation in Eq. (I), we get

~ ( 7 ) = e+$' 4 4 p x,x2px, (xI)b(x2 - X I ) +(eflr - l)n2 ( ~ 2 ) b 1 & 2 -, -. - ,

- - where I and x2 arc the mean and the mean-square value of the process. For a thermal noise X = 0 and Eq. (5 ) becomes -

2 -81 r>O Rx(r )=x e

Since autocorrelation is an even function of r, we - have 2 -Arl Rx(r) = x e

and

2 1 1.3-1 For any real number a, (a - y ) 2 0

Therefore the discriminant of the quadratic in a must be non-positive. .Hence, - - -- (2i$ < qx2 - y2 or ( ~ r < x2 y2

Now, identify x with x(t,) and y with fit2), and the result follows.

since x(t) and y(t) arc independent.

~ " ( 3 = [2x(t) + 3y(t)I2x(t + r ) + 3y(t + r)]

=4RX(r)+9Ry(r) since Rv(r)=Rp(r)=O

R, ( r ) P [x(t) + Y(t)1[21(t + 3 + 3Y(t + r)] = 2 Rx ( r ) + 3 (7)

-

Similarly, cor(nso(r + r) - ro t + (n- 1141 = 0 and ~ ~ ( 7 ) = 0

2nt, Since b is a r.v. uniformly distributed in the range(O, q), tvob = - is a r.v. uniformly distributed in the

Tb range (0, ZR) . Using the argument in problem 1 1.3-3, we observe that all harmonics are incoherent. Hence the autocorrelation function of R,(r) is the sum of autocorrelation function of each term. Hence follows the result.

1 1.4-1 (a) .Y1 ( a ) = 2 KTR, and S2 (o) = 2 KTR2 Since the two sources are incoherent, the principle of superposition applies to the PSD. If S,,, ( a ) and Sq (a ) are the PSD's a the output terminals due toSl(@) and S2(o) respectively, then

s, (0) = pl (@)I' sl md sO2c4 = la (a]' s 2 ( 4 where

R2 I j d R2 R2 + I / j d

z jluR2C + 1 - - R2

H l ( 4 = t R2 R2 I j d R2 Rl(jaR2C+1)+R2 jORlR2C+RI+R2

R'* ~ ~ + l l j o C R1 + j ~ 2 ~ + I

(a

Similarly, (c)

Fig. Sll.4-1

ZKTR, R= and Sq(tv) = ZKTR~R?

Sq'"'t m 2 ~ ? # c 2 +(R, + ~ 2 ) ~ r 2 ~ f ~ $ c 2 +(Rl + ~ 2 ) ~

= ( 4 + ~ 2 ) ' 2KTRlR2 - ~ K T R ~ R ~ ( R I + R ~ ) - u2c2 R: R: + (Rl + ~~f ' RI +R2 m2 RfR:c2 + (RI + ~ 2 ) ~

which is the same as that found in pan (a).

- JQ h(a)x(t)x( t + r - a)da = h(a)~, ( r - a)& = yr ) . R~ ( r ) and sxy (0) = H ( r ) S . (o) -Q -QD

In Fig. 11.13, H(m)= jcuc = 1

1 j o R C + l R+- j e

ads,, (ru) = ~ K T R I ( ~ U R C + 1)and k 0 ( r ) = 2KTR e-""u(r)

11.4-3 (a) W e have found R,(r) of impulse noise in Rob. 1 12-8

~ , ( r ) - a 6 ( r ) + a 2 , a n d &(a) = a + 2 n r r 2 6 ( 0 ) Hence,

and Ry ( r ) = 9 - ' [ S Y ( u ) ] = a21 H(0n2 + ah(rk)tY - I )

Fig. Sll.4-3

The PSD of n,( t ) and %( t ) are identical. They are shown in Fig. S11.5- 1. Also, 2 is the area under ..

Fig. S11.5-1

11.5-2 We follow a procedure similar to that of the solution of Rob. 1 1.5-1 except tha! the center ikcquencies arc different. For the 3 center fbqucncier SQ (.)(or S~ (a) ] I shown in Fig.

S 1 1 5-2. In all the rhne cases, the area under Snc (a) is the

same, viz, 125 x lo4&. Thus in all 3 cases - - nf =nf =125x10~~

Fig. S11.S-2

-3 \b -. k . . . _

E # ~ H H @ ' 4 = - Fig. S11.5-3

srn(@) =&= 6 1 . 11.5-4 (a) Hw(@)= =- 6 S t ~ ( @ ) + ~ n ( @ ) _+6 6cu2+60 cv2+10

(c) The time constant is I . Hence, a reasonable value of time-delay required to make this filter 75 realizable is & = 0.949 scc.

(d) Noise at the output of the filter is " 3

The signal power at the output and the input are identical

11.5-5 (a)

(c) me time constant of the filter ii 0.306 sec. A reasonable value of time-delay required to make this filter realizable is 3 x 0306 = 0.91 8 sec.

(d) Noise power at the output of the filter is

The signal power is

Fig. Sl1.5-5

,.

Chapter 12

-3 'O m ct>+r~,,Ct)

i ~ + 4 4 4 - = --

-d. o(

L

Fig. S12.1-2

Sn0 (w) = s,, (UWH~ (el2 = 10- ( a = 8 0 0 0 ~

S si 12.2-1 (a) 3 0 d B = 1 0 0 0 = A = y = - = = ~ ~ = 4 x l 0 ~

No 10"~x4000

(b) From Eq. (12.71, No = JIB = 10-~~(4000) = 4 x 10" 2

(c) Si = ( ~ , ( a , ) ( ST and = 4 x lo4 ST = 4 x 10 4

12.2-3 Let the signals ml(t) and m2(t) be transmitted over the m e band by &m of the same

fkquency (or), but in phase quadrature. The two transmitted signals arc Ji[ml(t)c0soct + m2(t)sin WS] , ,

The bandpass noise over the channel is nc(t)cosert + n,(t)sinoct. Hence, the received signal is

[JTml (t) + nc(t)]cosoct + [&m2 (t) + n,(t)]sin act

1 Eliminating the high frequency term* we get the o u ~ u t of the upper lowpass filter as ml(t) + J;nc(t)

1 Similarly, the output of the lower demodulator is m2(t) + An,(t) i

These arc similar to the outputs obtained for DSB-U: on page 535. Hence, we h a v e s = 7 for both QAM * ; N o

channels.

-[.(IH..I=% 12.2-4 (a) p = Hence, mp = M

A A

rn P Y , I where K' = 3;;:

2 m~ S 1 . Y

(c) FOI tone modlt~ation r2 = -j- = 2 and for P = 1, = -Y a - "PI2

No 2+1 3

2 S p2 p (30.1'

12.2-5 (a) From Rob. 1224, = n r . F0r30- loading, m,, = 3a, andc2 = 7 = - = 9 No x + p m 2

Qm

s (05))' @) When p = 05, --B = Y

No 9+(05)~'- '36

12.26 For tone modulation, let m(r) = M coscu,r . For BSB-SC,

gDSB(r) = fi@cos@,t -c0s(uct

For SSB-SC +,(t) = m(r) coscu,t + mh(t) sin cuct

= jL( COSO~,,~ COSW,+ + p~ sin wmr sin @,r = p-4 cos(cuc - @,)t

p2 A2 si = - S Si p 2 A 2 Sp and ip=.u4. Hence, Sp=p2A2and9=y=-=-=-

2 No d B 2dB 2cUB

For AM 4AM(t) = ~ ( 1 + ~ C O S W , ~ ) C O S ~ ~ ~

n, = ~ ( I + p ) a n d Sp = A2(l+p)'. Hence, - S, = SP(~*>) and -=-y So m2 - - p2r2h & _ ( p2 [sp(2*p2)]

4(1+ PI2 No A2 +m2 A' +(p2~'/2) ( 2 + ~ ' 4(1+p)~JlB

s SP Under best condition, ie., forp = 1, A = - No 1(MIB

Hence, for a given peak power (given Sp ) DSB-SC has 6dB superiority, and SSB-SC has 9dE superiority

over AM. These results are derived for tone modulation and for p = I (the case most favMable for AM).

12.2-7 For40 loading, mp = 40, and the carrier amplitude A = mp = 40. (for p = I). For Gaussian m(t), - 2 m2 e CT, (assuming iii = 0 )

- A2 nrn 2 ,t2+m2 - - 16u~+:t4, =-.urn 17 2

Hence, - = 8--r = 4.605 and Si = 202, 0" 2 2 2

12.3-1 5 = 28dB = 63 1. Hence, No

631x9 Therefore, y = - = 47325

12

4 8 3 12.3-2 mp = B, m i = - and bandwidth = - . Hence,

T, To

12.3-3 m(t) = cos3 mot and mp = 1

2 2 4 t ) = -3m0 cos mot s i n r o t and y t ) = -3m0[m0 M mot cosmot - 2 0 , cosmot sin

For a maximum 2 4 t ) = 0. This cos2 aot = 2 sin mot

2 2 or 1-sin mor=2sin ~ , t *s inm, t=

and

I 2 m i = -3w0 COS a0t sinmot

123-4 m(t )=a lcoso~ t+a2coso2 t , mp = a1 +a2

t ) = ( a sin t a sin m ) , mb = alm1 + a p 2

(%/NO)pM = (2d)2m; 2 = &(al+a2)'

(SOINO)FM 3mb qa1al +a2w2)2

12.3-5 Wmr in this problem. T h m should be 4x2 in the denomlumr (see below). s,(,) = u2sm(o). Hence,

p -QD [W)I2d - P _ ~ h ( 2 # ) 4 ' P, 4x2f2sm(2a)df

From Eq. (12.42a)

Thew results am m e for a waveform m(t)

x fo?

0

X

2 k sin(%)

The definite integrals are found from integral tables.

Hence, the normalized PSD's is H e - @ 2 / 2 d

2u2 - - 3 If W = 2 d , then w2 = ( 2 4 ' = 2 g 3 - @ 2 2 " 2 d m = 2 2 . 1f p(W) is the power within the

2u band -W to W.

Aw) = 2r %-m212u2 0 2 u

P(w) p(oo) = 2, and - = 1 - e' w212u2 0.99 =$ W = 3.03u, B = 0.482~ dm)

P(w) - = 0.9 3 W t 2.150, B = 0.3420 d m )

w2 - x = 0.99 - , 3.wy2 > w2 => PM superior

3

w2 - x = 0.95 - = 2 . 0 0 ~ ~ = w2 a PM and FM equal

3 w2 -

x = 0.9 - = 154a2 < w2 , FM superior 3

12.3-8 m(t) = a1 cosolf + a2 coso2t, and 2

s,,,(f) -$[6(f-/,)+d(f 2 + fI)]+%[6(f -f2)+6(f +A)]

3(a?f? +dl?) Since B = f2, PM is superior to FM iff: >

a: +a; 9

2 2 1+x 2

that is, if ( a l k ) ' ( f ~ l h ) ~ + ~ <I 3 Or i f l+x ,, <--. 3 (a1 /a2 )2 + 1

= 164.1 . Also, y- = 20(8 + 1) Y '

2 (b) &, t B 2 ? = ;(yn-20) yn = 10,000 or (rn -20)' =- 12x107 ~ ~ 2 4 2 3

No 20 Y n l

12.340 From Eq. (12.40) f12 = - 'Il ++/mil J (1) Tone modulation

(2) Gaussian with 3u - loading d =i(-&)38-~3~ 1

(3) Gaussian with 4 u - loading d a A(-) 8 = 0.56 3 1+1/16

where For tone modulation,

- m2 u2 1

For Gaussian modulation with 3 s - loading, 7 = - = - m (30)~ 9

- m2 a2 1

For Gaussian modulation with 4a - loading, 7 = - = - 3 ( 4 l6

12.3-1 1 Let us fvst analyze the L+R channel. In this case, the demodulator output signal, when passed through the

0-1 5 W (lowpass) filter, b given by (L + R)' + no(?), wheresn0 ( a ) = Eg (12.3311.

When this signal is passed through the de-emphasis filter Hd ( a ) = ,the signal is restored to (L+R) J@+@l

and the output noise power NA is given by

Let us now consider the (L-R) channel. Let a, = 2x x 38,000 and aI = 2n ~2100 . The received signal is FM demodulated (Fig. 5.19~). The PSD of the noise at the output of the FM demodulator is Sno (a ) = J ~ B ~ I A' [see Eg. (12.33)] The output of the FM demodulator is separated

into (L + R)' over 0- 15 kHz and (L - A)' cosa,? over the band 38* 15 or 23 kHz to 53 kHz. Let w consider

the signal over this passband, when the noise can be expressed as nc(t)cosoct + n,(r)sin cu,t. The signal

is (L - R)' coso,r. Hence, the received signal is (L - R)' + nc(t) costu, + n,(r)sin ac t . This signal is C I multiplied by2 cosw,r and then lowpass-filtered to yield the output(l - R)' + n,(r). But

'"I When this signal is passed through de-emphasis filter Hd(a) = -, the signal is restored to (L-R) and I@+ *I

the output noise power N; is given by

N" Hence, the (L-R) channel is noisier than (L+R) channel by factor d g i v e n by

N;

Substituting B = 15,000, f, x 38,000, f1 .t 2 100 in this equation yields: A'" = 166.16 = 222 dB. N;

FO; uniform distribution

2n = 18.27 Since n must be an integer, choose n = 10 and L = 1024

Fig. S12.4-2

"

BpCM = 2nB = 90 MHz (assuming bipolar signaling) tc) To increase the SNR by 6 dB, increase n byl, that isn = 1 1. Then the new bandwidth of transmission is

. , B, = nB = 8 x 8000 = 64 kHz (assuming bipolar line d e )

where 2 x 625 x lo-'

S SO, 9- = '(2f6 (i) z 21845 = 43.4 d ~ .

NO 1 + 4(216 - l)P(JjI) . .

(b) If power is reduced by 10 dB, then? = 25.6, Q(G) = Q(1.79) = 0.0367 and

The table below gives SNR for various values ofn under the reduced power.

Hence, n = 3 yields the optimum SNR. 'The bandwidth in this case is Bm = 3 x 8000 = 24 kHz.

12.4-4 1 - PE = P (correct detection over d l K links) + smaller order tmns

=(I - P ~ ) ~ - ' ( ~ - P ; ) s [ I - ( K - I ) P ~ ~ I - P;]S I - P;-(K-1)pe

So PE = +(K - I)P, (b) y=25dB=316.2, y=23dB-1995

P, = o(Jm) = P(6287) = 1.6 x 10-lo

P; = Q( Jq) = ~(4.994) - 3 x lo-'

12.5-1 As noted on Pg. (570), the optimum filters for DSB-SC and SSB-SC can be obtained from Eqs. (12.83a) I

and (12.83b), provided we substitute-[sm(r + e c ) + Sm(m - uE)] forSI(~) in these equations. Let 2

We shall also require the power of Sm (o) . I

1 = -c5m(m)im 2%

We can simplifL the evaluation of this integral by recognizing that the power of the modulated signal m(r)coscucr is half the power of m(r). Hence,

We shall use the PDE system shown in Fig. 12.19 1 03

(a) For this system

Because H,(o) and ere constants, we have

where i,,,(cu) is found in Eq. (1). Also h m Eq. (12.83b)

(b) The output signal is ~ m ( t ) . Hence. So - c2n2(t) We have alnady found the power of m(t) to be 2(a/4) = 2 . Hmce

c2a (1 0" )' (3000~) 3, So e-= - -- 2 2 20

To fmd the output noise power No, we o b w e that the noise signal with PSD S n ( r ) = 2 x 1 0 ~ ~ a s s e s

through the de-emphasis filter &((u) in Eq. (4) above. Hence, S,, (a) the noise PSD at the output of fid(cu) is

- n2

- - 1 Also, the output noise power is n , ( t ) / f i and No = [a Eq. (12.4b11, where n: = n2 = -rSrt (@)da

2 X 0

A, 3%/20 i. 3n2 and No 1 ~. (a) ' fm 10jom t (@)dm

A

12.52 Similar to Rob. 12.5-1

12.5-3 The improvement ratio in FM is , , where

Hence, the improvement ratio is

Chapter 13

13.1-1

Fig. S13.1-1

For the integrate a d dump filter (I&D), the output is the integral of ~ ( r ) . hen^, at r = 5, p o ( q ) = 4%

If we apply S(t) at the input of this filter, the output h(t) = u(r) - u(t - 4 ) . Hence,

,+). q s e c ( F ) - j w 5 / 2

and

and

This is exactly the value ofp2 for the matched filter.

13.1-2 The output po(t) of this R-C filter is

(t) = ~ ( 1 - e-'IK) o r t s ~ - --, ,,,-. - 5 1 ~ ~ -(r-lb)/RC .A(I-e )@ t ' G ,

The maximum value of p,(t) is Ap, which occurs at Tb:

AP = = ~ ( 1 - e-%IRC)

and

We now maximize p2 with respect to RC. Letting x = G/RC, we have

and 2

+' - 2xe-'(l- e-') - (I - e-') --

2 = O

dx X

This gives zxe-= = I - e'X or 1+2x=eX

and

x z 126

Hence,

Observe that for the matched filter,

The energy of At) is Tb times the pow& of At).

Hence,

A2& Similarly, E9 = - =

2 Eb

E~ = 12 p(tk(t)dl= J: [ - A z ( ~ - m2)cm2 (~,j + A2m2 sin2 act 1 dt

Hence,

and

13.2-2 ~ e t cI be the cost of m r when 1 is transmitted, and Co be the cost of mot when 0 is transmitted. Let the optimum threshold bea, in Fig. S13.2-2. Then:

The average cost of an error is c = Prn(l) c, + P,(O) co

For optimum threshold &/aho = 0. Hence, to computedC/doo, we observe that

and

Hence,

Hence,

I

Fig. 513.2-2

and

But

Hence.

2 ~ E P 0, =- and Ap = E p .

2

13.23 We follow the procedure m the solution of Rob. 13.2-2. The only difference is ~ ~ ( 1 ) and ~ ~ ( 0 ) are not 0.5. Hence,

A -a, A +ao C = Prn(1) CI + Pm(0) CO = pm(1) c10 Q (&)+ Qn h ( 0 ) COI Q (e) Qn

and

I - (A,-%,' - dc 6''i;;bG Prn(1) CIO e '4 - *rn(o) COI e

Hence,

But

Hence,

Fig. S13.5.1

The thresholds are f E p / 2 and

13.5-2 Here, p(r) and q(t) an identified with 3p(r) and&), mpc.ctively. Hence,

H(@) = [ ~ P ( - w ) - P ( - ~ ) E - J ~ % = 2q-wkL.'W and

But multiplication ofh(t) by a constant does not affect the performance. Hence we shall choose h(t) to be

p ( 5 - t) rather than 2 p ( q - r). This will also halve the threshold to a. = 2 E,,. This is shown in Fig. S13.5-2. Also,

and

pe = {/-) = (J-]=

13.5-3 For M = 2, IA = 2 x lod For 256,000 bps the baseband transmission requires a minimum bandwidth 128 W But m p l i ~ d e modulation doubles the bandwidth. Hence

Br = 256 kHz

For M = 16

This yields Eb = 5.43 x lo4 Si = EbRb = 5.43 x lo4 x 256,000 = 139W

13.54 For M = 2 n d J = 2 x loJ

This case is identical to MASK for M = 2

10-7 = Eb = 2.7 x 10-7

For M = 16

In MPSK, the mmimum bandwidth is equal to the number of M-ary pulsedsecond. Hence,

For M = 32 PeM = (log2 32)Pb = s x 10"

Chapter 14

14.1-1 The following signals represent 2 sets of 5 mutually orthogonal signals.

Fig. S14.1-3

Fig. S14.1-4

b) The energy of each signal is:

C) F3 - F4 = (-6 - 8 + 6 + 8 + 0) = 0. Hence, f3 (t) and f4 (I) arc orthogonal.

14.2-1 Let x(t)=xl, x ( r+ l )=x2 x(t+2)=x3 We wish to determine

~ x ~ x ~ x ~ (XI 9 ~ 2 9x3)

Since the process x(r) is Gaussian, XI, xa , x3 arc jointly Gaussian with identical variance

* = o2 = o2 = ~ ~ ( 0 ) = I). TIIK covarinse matrix is: (0x1 x2 1)

and

And -CZa(j xixj

1 i j px,x2x3 (x1x2x3) =

(2 43'2 f i

Hence

and

Fig. S14.3-1

p(qrn2) = f ( q m 3 ) = ..... . = P ( C \ ~ ~ - ~ ) = Rob Inl\ c I 9

Hence PeM = 1 - P(C)

= 1 - p(ml ) p ( q m 1 ) + P(%) P ( C ~ ~ ~ ) + . . - . + P ( ~ M ) P ( c I ~ M 1)

Hence the average pulse energy E is

Also

Hence

Which agrees with the result in Eq. (1 3.52~)

and

Fig. S14.3-2

The average pulse energy is

and

This performance is considerably k n e r than MASK in Rob. 14.3-1, which yields

PeM = 1.75Q [,/?I for A4 = 8

14.3-3 In this case, constants ak 's arc same for k = 1, 2, ..... M. Hence, the optimum receiver is the same as that in Fig. 14.8 with terms a k 's omitted. We now compare r . q , r e g , ..... r . s ~ . Since r ask = f i r c0st9~ is the angle between r and sk , it is clear that we are to pick that signal sk with which r has the smallest angle. In short, the detector is a phase comparator. It chooses that signal which is at the smallest angle with r .

14.3-4 Because of symmetry, ~ ( q m , ) = p ( l m 2 ) =.....= p ( I r n M )

where M = 2N

Let

q = (+, T, ....., 5) 2

Then

Fig. S14.3-4

and

p ( c ) = p(Clm, )

Here, M = 2 . Hence, each symbol d e r the information log2 M = N bits .

Hence E* = E / d

and

14.3-5

Also

Hence

and

Fig. s14.3-7a

Note that d d d h=-h+T#2 Sl = --4, - 4 2 3

2 2 2

Fig. S14.3-7b

c ~ ( 4 ~ ) = Rob(noise originating Imm y remains within the square of side -

and

We also observe that E , the average energy is

Therefore

The decision region R2 for is shown in Fig. a and again in Fig. C-I. R2 can be expressed as the fvrt quadrant (horizontally hatched area in Fig. C-1) -Al . Thus

P(&) = noise originnting from q lie in R2

= P(noir lie in 1st quadrant)- P(noisc lie in AI ) 2

= [I - Q(&")] - P(noise originating fKnn s2 lie in -

1 But P(noise lie in A ~ ) = noise 4 lie within outer squaw) - P ( n o i ~ lie within inner square)] (See

Fig. C-2)

and

and

P(+""

Moreover, by symmetry

P(alm2) = ~ ( d m , ) = p ( l m 3 ) = ~ ( r l m 4 ) Hence

Fig. S14.3-7c

Fig. S14.3-8

The decision region RI for mi (we Figure) cm be exprrned as 3

RI = outer square of side d f i - - 4 (outer square - inner square of side d)

1 3 . - outer square of side d f i + - inns quvc of side d 4 4

120

Now p(clml) = Rob(noise originating horn ml lies in RI)

1 3 = - 4 ~ ( n lie in outer square) + - 4 P(n lie in inner square)

Similarly R2, the decision region for * (see figure above) can be expruxd as 1

R2 = outer squm of side d f i - - ( o w 2 squm - inner sqvve of side d)

1 1 = -outer square of side d f i --inner square of side d

2 2 and P(&) = noise originating fiom m2 lie in R2

The decision region R3 for m3 can be expressed as .

R3 =RA+RB-RC and

~ ( ~ l m ] ) = ~ob(noise originating Imm m3 lie in ~ 3 )

The decision region 4 for m4 can bc express4 as % = R A - R e

and I

p(qm4j = p(nl > d . nz > - ~ ) - - { P o ~ I I ~ 4 1n21'~)- {~ni~, 14 < '-)I f i

For any practical scheme P(.) << 1 , and we can expa,

[l - &a+)? = 1 - 2kQ(.)

Using this approximation, we have

p(+t) 1 - 4%)- 3 4 7 3

Hence

Now

Therefore

And

SO that

Therefore

Moreover

Hence

And

122

Cmpluison of this result with that in Example 14.3m.(14.57)] (7ows that this configuration rqukes approximately 1.5 times the power of the system in Example 14.3 to achieve the same perfonnance.

143-9 If sl is transmitted, we have

bk = o + G n k , b-k = * - f i n k and

~ ( ~ l r n ~ ) = prob.(bl > b-,, q, b 4 , ... bt , b-k ) Note that

bl >b-, i m p l i e s ~ + a + f i n l > - E i + a - f i n l or q >-6 bl >h i m p l i e s ~ + a + f i n l > a + f i n 2 or "I <*+"I

bl >b-2 i m p l i a b + a + f i n l >.-fin2 or > - ( f i + n l )

Hence

bl > and b-2 implies - (nl + 6) < n2 (nl + f i ) Similarly

bI > b k and^^ implies - ( n , + J F ) < n t c ( n l + 6 )

Hence f(clrnl) = Prob.(bl > b-I, b2, b 4 , b j , b-j,..- bk, b-k)

Since nl , nz ,-.. nk are all independent gaussian random variables each with variance d/2 ,

p(qrnl) = [ ~ ( n l > -6) P((*,(< nl + G ) ~ ( l n ~ l < nl +q ..- p(IntJ < nl +JT)]

Also

14.4-1 The on-off signal set and its minimum energy equivalent set art shown in Figs. (a) and (b), respectively. The minimum energy equivalent set of orthogonal signal set in Fig. (c) is also given by the set in Fig. (b). Hence, onsff (Fig. a) and orthogonal (Fig. c) have identicaI error probability. The set in Fig. (b) is polar with half the energy of on-off or orthogonal signals.

123

14.4-2 Here

Fig S14.4-1

Fig. S14.4-2

Therefore

1 14.4-3 To fmd the minimum energy set, we have a = -(sl 4 + sz + s j + s4) = -6 - h

Hence the new minimum energy set is

~ ~ ( t ) = f i 4 ~ ( t )

s 2 ( t ) = & h ( t )

s - 4 , )

S I = J 0 2

12 =&el

.=-fie 1 456 , e . e c r n . = - x s i = ~ [ f i ~ l - f i e l + f i e 2 ] = T ~ ~ 3 3 .

Hence the minimum energy signal set is given by

f i f i 2 0 f i ~ ( 1 ) = S~(I)--(~(I) = 4 41(1)-Th(t) = l o f i c ~ s ~ o t - - sin wot

3 3 6 f i 20JS sin wot

i3(t) = s ~ ( ~ ) - ~ ( ~ ( I ) = -6 +1(r)-T42(t) = -10ficosm01-- 3

The optimum receiver - a suitable form - in this case would be that shown in Fig. 1 4 . l or b.

* . .. \

R3 - - r o

s3

. ... 2 that all the four signals fonn vertices of a square because (i, i2) , (i2 it ) , (i3 i4) , and ( i4 il) arc

onhogonal. The distance bcwen these signal pairs is always 2 f i . This set is shown b Fig. S14.4-3a

The signal set is now rotated so a to yield a new set shown in Fig. SI4.4-3b. a

Observing symmetry we obtain

qc) = !(+I) = 444 = P(C1m3) = N m 4 )

= P(n1 >-fiandn2 >-q

Fig. S14.4-4

and

P(C) - f (2 p(+tl) + ~ ( q ~ ) ] = f [2 - ih(7.07) + i- 2~(7.07)]

Mean energy of the minimum energy set:

14-44 The use of Eq. (14.76) and signal mution shows that the m'himum energy set in this case is identical to that in Rob. 14.44. Hence the minimum energy set is as shown in Fig. 5144-4c. this situation is identical

to that in Rob. 14.3-5 with d = . From the results in the roluri" of Rob. 14.3-5, we have lo.crj

JJ ~ 1 ~ 0 , we u e given s,,(o) = - = 10" . Hence, d = 2 x 10" 2

(a) From the solution of Prob. 14.3-5 1 Pa = eQ(7.02) 2 + Q(7.12) = 1.09 x 10-l2

@) and (c) identical to those in Rob. 14.44

14.4-6 (a) me center of gravity of me signal set is ($1 + s2)/2 Hence, the minimum energy signal set is

X I = $1 - 2 2 2

The minimum energy signals ate

xl(r) = 03 - 0.707 sin

&, = E x , =0 .4984~10-~ . Wean given J l=5x10 -6

(c) We use Gram-Schmidt orthogonalintion procedure in appendix C to obtain

.00 1 I f i sin mot at

y2( t ) = f i s h m o t - 0 2J5 = J r ~ i n @ ~ t --

.oo 1 I

1 s1 ( t ) = - jl ( t ) =.03 I 6 j l ( t )

3 1.6 1 1 s2 ( t ) = - j 2 ( t ) + -91 ( t ) ~ . 0 1 3 8 ~ ~ ( ~ ) + . 0 2 8 ~ ~ 1 ( ~ )

72.7 35.1

Fig. S14.4-6

Chapter 15

H(m)=-(PI log4 + 4 l o g 6 + P3 log&+ ~g 10gP4) = 1.846 bits (source mmpy)

There are 10' symbolds. Hence, the rate of information generation is 1.846 x 10bbits/s.

15.13 Infonnationlelement = log2 10 = 3.32 bits.

Inforrnatiordpicture h e = 332 x 300,000 = 9.96 x 10' bits.

15.1 -3 Infonnatiordword = log2 10000 x 133 bits. Information content of 1000 words = 133 x 1000 = 13 f 00 bits.

The information per picture M e was f a d in Problem 15.1-2 to be 9.96 x 10'bits. Obviously, it is not possible to describe a picture completely by 1000 words, in general. Hence, a picture is worth 1000 words is very much an underrating or understating the reality.

15.1-4 (a) Both options are equally likely. Hence, I = log(&) = ! bit

(b) ~ ( 2 lanterns) = 0.1 l ( 2 lanterns) = logt 10 = 3.322 bits

15.1-5 (a) A11 27 symbols equiprobable and P(x,) = X,. H,(X) = 27(& log2 27) = 4.755 bits/ symbol

(b) Using the probability table, we compute 27

H,,,(x) = - P(xi) log P(xi) = 4.1 27 bits 1 symbol I-1

(c) Using Zipf s law, we compute entropylword H,(x). 8727

H,(x) = - 2 P(r) log P(r)

8727 = - log(+) = 9.1353 bits / word.

r=l

Entropy obtained by Zipf s law is much closer to the reai value than Hl (x) or H2 (x).

f 63 15.2-1 H(m) = C ?j log l'j = - bits

i= I 32

Message Probability Code SJ st S3 4 Ss

m, ID o ln o 1 12 o i n o in o in o

63 = - binary digits 32

H(m) x 100 = 100% Efficiency q = - L

Redundancy y = (1 00 - 7) = 0%

7 15.2-2 H(m) = -x 4 log 4 = 2289 bits

17.1

2289 = - = 1.4442 3 - ary units Jog2 3

Message Probability Code SI s2

ml 1 I3 0 113 0 113 0

13 =- 3-qdig i t s 9

= 1 A442 3 - ary digits

Redundancy y = (1 - q)100 = 0%

4 15.2-3 H(m) = -C log I j = 1.69 bits

i= 1

Message Probability Code SI S2

m I 0.5 o 0.5 0- 0.5 o

L = C pi Li = 031) + 03(2) + 0.1(3) + 0.1(3) = 1.7 binary digits

Redundancy y = (1 - q)100 = 0.8%

For ternary coding, we need one dummy message of probability 0. Thus,

Message Probability Code SI

m I 0.5 0 0.5 0

L = 031) + 03(l) + 0.1(2) + 0.1(2) = 12 3 - ary digits

1.69 H(m) = 1.69 bits = - = 1.0663 3 - ary units lot32 3

H(m) I 0663 Efficiency q=-x100=-xl00=88%6% L 12

Redundancy y = (1 - q)100 = 1 1.14%

Message Robability Code SI $2

m I I n o In o i n o

63 From Problem 15.2- 1, H(m) = - bits = 1242 3 - ary units 32

Redundancy y = (1 - q)100 = 537%

15.2-5 Message Probability Code SI sl S3

m I 1 /3 I 113 1 I 13 I 1 13 1 1 0 m2 1 0 00 113 00 10 00 1 6 00 1 d Ii' m3 1 I9 01 1 1/9 011 219 010 113 01 nh 1 I9 0100 119 0100 119 0100 119 011 ms 1127 01010 119 0101 nb 1/27 i n 7 1/27 0101 11

65 L = q L i = - = 2.4074 binary digits 27

H(m) = 2289 bits (See Problem 152 - 2).

Efficiency q = - 2289 H(m) x 100= x 100.: 95.080/o L 2.4074

Redundancy y = (1 - 7)) 00 = 4.92%

15.2-6 (a) H(m) = 3($log3) - 1585 bits

(b) Ternary Code

Message Probabiliw Code

1585 H(m) = I585 bits = - = 1 3 - ary unit

log2 3

H(m) x 100 = 100% Efficiency q = - L

Redundancy y .: (1 - 7)lOO = 0%

(c) Binary Code

Message Probability Code SI

m I 113 1 e 2 3 o

1 1 L = -(l) +(2)?(2) = 2 = 1.667 binary digits 3 3

H(m) 1585 Efficiency = - x 100 t - x 100 = 95.08% L 1.667

Redundancy y = (1 - q)100 = 4.92%

(d) Second extension - binary code

L = ~[(7)(+)(3) 2 + (2)(;(4)] = 5 = 1.6 1 1 binary digits

H(m) = 1585 bits

H(m) 1585 Emciency 7 = - x 100 = - X 100 = 98.39% L 1.61 1

Redundancy y = ( I - q)100 = 1.6 1%

Marrge Rob Code L h SI 4

maml 1/57 110 mlm, 119 111 mrm, 1/57 100 mml 1/57 101 mfi 1/57 010 m,m, 1/57 011

15.4-1 (a) The channel matrix can be represented as shown in Fig. S15.4-1

?= $- 1 1 P(x1 ) log - + P(x2 ) log -

P(XI ) P(x2 1 2 3 , -log2 3+ -log2 - = 0.918 bits Fig. S15.4-1 3 3 2

To compute H(xly), we find

and H(x)y) = PQl )H(xl~l) + PCvz)H(xlY2)

13 32 = ~ ( 0 . 7 7 9 ) + - (0.624) = 0.6687

45 45 n u s ,

I(xly) = H(x) - H(x1y) = 0.9 1 8 - 0.6687 = 024893 bits 1 binit

Also, H(y1x) = H(y) - I(xly) = 06673 - 02493 = 0.61 8 bits1 symbol

15.4-2 The chamel matrix P(Yl1xi) is

xi

P(Y, l ) m i to obuin Now we use P(x,ly, ) = C P(xj)P(y,lxi)

i

Fig. S15.62

P(x, 1 y, ) matrix as

1 H ( x ) = P(xi)log- = -flog P - 2QlogQ with (2Q = I - P)

P(x,

Letting P = Z ~ P ) or ~ ( p ) = logs

I(xly) = Q(P)+(l- PX1-logs) d d d p l ( ~ ~ ~ ) '0 or -[Q(P)+(I- dP PXI-logs)]= 0. This means

d -[flog P+(l - P)-(1- P)log(l- P)(I-IO~B)] - 0 dP

log P - log(1- P) +[I - log p] = 0

Therefore P

log- = -I + logs I - P

B Note: -1 + log2 P = - log2 2 + log2 = log2 - 2

15.4-3 Consider the cascade of 2 BSCS shown in Fig. S15.4-3. In this case Pdx(lll)=(I- Pl)(I-q)+Pjq - 1 - 5 -5 -244 P'x(Wl)=(I-q)q + 5(1-9)= 9+ 5 - 2 4 5

Hence, the channel matrix of the cascade is

1-4-~5-2qq q + q - z q q P, + q - 2 4 P I - q - q - 2 q p

k * B S a C a ~ e a d e OF . > &'

0 k-I BSCs

This result will prove everything in this problem.

r 4- I 4 m t r 4 M

(a) With 4 = 4 = P' , h m the above result it foUows that the channel matrix is indeed M' . @) We have a h d y shown that the chsnntt matrix of two cascaded channels is MI M2 . (c) Consider ti d e of k idmfical channels broken up as k - I channel cascaded with the k* channel. If M.., is the channel matrix of the first k - I channels in cascade, then from the results derived in part (b),

the channel matrix of the k cascaded channels is Mk = Mk,, M. This is valid for any k . We have

- I e

already proved it for k = 2, that & = M'. Using the process of induction it is clear that Mk = M k . We can verily these results from the dcvelojmcnt in Example 10.7. From the msults in Exmple 10.7, we have, for a cascade of 3 channels

8 Fig. 515.43 J

and

Now

I - Pe M 3 = [ p e

pe I=[ I - ( 3 ~ ~ - 6 c 2 + 4 ~ . ) ) 3pe - 6 ~ : + 4 ~ :

1-pe 3 4 - 6 ~ : + 4 c 3 1-(3Pe - 6 ~ : +4pe3) Clearly

PE = 3Pe - 6 ~ : +4pe3

I which confinns the results in Example 10.7 for k = 3.

(d) From Equation 15.25

where PE is the error probability of cascade of k identical channel.

We have shown in Example 10.7 that -

and

15.4-4 The channel manix is

Let

XI '0 y1 = o

Also,

Therefore, H(x) = - P(xl ) log P(q) - P(x2) log P(x2

= 1 - p bits / symbols

1 1 = C ~ C P < x i , Y J ~ z k ) l o g -CCCP(x, ,~, ,zk)log

X Y Z P(x,Izk) x y z PO, I Y , ~ ( x ~ I Y ,

= x ~ ~ P ( x , , ~ / , ~ k ) ~ o g p(x Iz ) x Y z i k

Note that for cascaded channel, the output z depends only on y. Therefore, X )' i )

p(zk1yjsxi)= p(zkl~j) - By Baycs' rule

Fig. S15.4-5

and

It can be shown that the summation over x of the term inside the bracket is nonnegative. Hence, it follows that

H(xlz)- WXIY) 2 0

From the relationship for /(xly) and I(xlz), it immediately follows that I(xIy) 2 I(xlz)

1 M 15.51 We have H(x) = J-$ p ~ o g - & = j-M -p log p& and j-cpfr = 1

P Thus,

dF F(x, p) = -p log p and - = -(I+ log p)

ap

Substituting these quantities in Equation 15.37, we have

-(l+Iogp)+al = : ~ = ~ = e ~ v - '

and

~ ' p m = j-:,eal-'& = 2 ~ ( e ~ l - ' ) = I

Hence, 1 ea~-l , - I

2M and p(x) = -

2M Also,

15.52 We have H ( x ) = -r logpfr, A =J;xp&, 1 = I;p& aF

F(x, p) = -p log p and - = -(I+ log p) ap

Substituting these quantities in Equation 15.37, we have

Substituting this relationship in earlier constraints, we get

and

Hence.

SO

To obtain H(x)

loge = 4 A J ~ p(x)dx + -% xp(x)& A

= log A + loge = log(eA)

15.5-3 Information per picture frame e: 9.96 x lo5 bits. (See Problem 15.1). For 30 picture frames per second, we need a channel with capacity C given by

But for a white Gaussian noise

S We are given - = 50 db = 100,000 (Note: 100,000 = 50 db) N

Hence, B = 1.8 MHz

15.54 Consider a narrowband 4 where -, 0 so that we may consider both signal noise power density to be constant (bandlimited white) over the interval L(I . The signal and noise power S and N respectively an i

given by s = 2s,(0)4 and N = 2sm(0)4f

The maxknum channel capacity over this band ly is given by

The capacity of the channel over the entire band (f,, f2) is given by

We now wish to maximize C where the constraint is that the signaI power is constant.

212 SJU) U) = s (a constant)

Using Equation 15.37, we obtain

or

Thus,

(a constant) s s + s n = -- 4

1 SS(0)+S,,(0) = --

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,

15.5-5 In this problem, we use the results of Problem 15.5-4. Under the best possible conditions,

c = BIog [Ss(w) + Sn(.)]- jf 10s [sn(.)] 4 constant

We shall now show that the integral 12 log [S,(@)] )I4 is maxim- when S,,(o) = constant if the noise

is constrained to have a given mean square value (power). Thus, we wish to maximize

under the constraint

21:: log (sn (u)] d j = N (a constant)

Using Equation 15.37, we have

and I

S,,(co) = -- (a constant) a

Thus, we have shown that for a noise with a given power, the integral

I; 10s [s. (011 d !

is maximized when the noise is white. This shows that white Guassian noise is the wont possible kind of noise.

f r Chapter 16 i

16.1-2 (a) There are (7) ways in which j positions can be chosen Imm n . But for a ternen' d e , a digit can

be mistaken for two other digits. Hmce the number of possible mrs in j places is

(b) (1 1,6) code for r = 2

35 r (:I) + (11)2 + (i1)22 = 1 + 22 + 220 = 243

This is satisfied exactly.

16.1-3 For (18,7) code to correct up to 3 errors

2" = 2048 Hence

Clearly, there exists a possibility of 3 error correcting (1 8,7) 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.

16.23 c = dG when d is a single digit (0 or 1) . Ford =0

c = 0 [ 1 11]=[000J F o r d = l

c - 1 [l 1 1]=[1 111

16.2-3 c = dG where d is a single digit (0 or 1) . Ford - 0

c=0[11111]=[00000] Ford = I

c = l [ l l l l l ] = [ l l l l l ] Hence in this code a digit repeats 5 times. Such a code can conect up to two errors using majority rule for detection.

16.2-4 0 is transmitted by 10 0 0) and 1 is transmitted by [l 1 I] (a) This is clearly a systematic code with

G=[I 1 11

16.2-5 (a) 1 0 0 . . - 0 1 0 1 0 . . . 0 1

Note that m 6 1

\ - I

Data word I Code word 0 0 0 1 0 0 0 0

(c) This is a parity check code. If a single error occun anywhere in the code word, the parity is violated. Therefore this code can detect a single cmr.

(d) Equation (16.9a) in the text shows that CH = 0 . Now

r = c @ e and

If then is no error = 0 and

~ H ~ = ~ H ~ = o Also

and

But since m = 1, I , = [I]

If then is a single error in the received word r, e has a single 1 dement with all other elements being 0. Hence r~ = eH = 1 (for single mor)

16.2-6

From this code we see that the distance between any two code w a d is at least 3. Hence dmh = 3 .

16.2-7

Data word Code word 0 0 0 0 0 0 0 0 0 0 0 l 1 l 0 0 0 1 O l O l l l O l O O l l O O l O l 1 1 0 0 0 1 1 1 0 1 1 0 1 l 0 l 1 0 0 1 l O l O O l l l l 1 l O l O l l O

Observe that dmh = 3

Data word

16.2-8 H is a 15 x 4 matrix with all distinct rows. One possible is:

Code word 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 0 0 1 0 0 0 l 1 1 0 1 1 0 1 1 0 . 1 l l O l l O l 1 0 1 1 1 1 1 1 0 0 0

16.2-10 (a) done in Rob. 162-7

six single errors

1 double error

s = r~ ' where r = received code c = r e e c = corrected code

16.2-12 We observe that the syndrome for all the three 2-mor patterns 100010, 010100, or OOlOOt have the same syndrome namely 11 1. Since the decoding table specifies s = 111 for e = 100010 whenever e = 100010 occurs, it will be comcted. Ihe other two pattern will not be cbmcted. If for example e = 010100 occurs, s = 11 1 and we shall read from the decoding table e = 100010 and the error is not comcted.

If we wish to correct the 2-error pattern OlOl00 (along with six single error patterns), the new decodhg table is identical to that in Table 16.3 except for the last entry which o;hould be

Z J_

010100 111

16.2-1 3 From Eq. on P.737, for a simple error correcting code

2n-k & n + l or 2 n - 8 ~ n + 1 + n - 8 ~ l o g ~ ( e + l )

This is satisfied for n 2 12 . Choose n = 12 . This gives a (12, 8) code. H' is chosen to have 12 distinct rows of four elements with the last 4 rows forming an identity matrix. Hence,

The number of non-zero syndromes = 16 - 1 = 15. Then are 12 single enor patterns. Hence we may be able to correct 3 double-error patterns.

The minimum distance between any two code words is dmi, = 4 . Therefore, it can correct all 1-error patterns. Since the code oversatisfies Hamming bound it can also comct some 2-enor and possibly some 3-error patterns.

The number of non-zero syndromes = 16 - 1 = 15. Then are 12 single enor patterns. Hence we may be able to correct 3 double-error patterns.

The minimum distance between any two code words is dmi, = 4 . Therefore, it can correct all 1-error patterns. Since the code oversatisfies Hamming bound it can also correct some 2-enor and possibly some 3-error patterns.

6 single enor patterns I 7 double-error patterns I

\

2 triple-error patterns {

16.3-1 Systematic (7, 4) cyclic code 3 g(x)=x + x + I

Fordata 1111 d ( x ) = x 3 + x 2 + x + 1

x3(x3 + x 2 + x + I) = x6 +xJ +x4 + x 3

, 1 +x4+x3

X 5

xS + x 3 + x 2

x3 +x2

x 3 + x + 1 2 x + x + l

C ( X ) = ( X ~ + X + I ) ( X ~ + X + I ) = X ~ + X ~ + X ~ + J ~ + X ~ + X + I

The code word is 11111111

X +x3+x2

X 2

The code word is 1 1 10100 A similar procedure is used to fmd the remaining codes (see Table I).

(b) From Table 1 it can be seen that the minimum distance between any two codes is 3. Hence this is a single error comcting code.

(c) There are seven possible non-zero syndromes. x 3 + x + l

for e=lOOOOOO 1 3 +x+1)x6

x6+x4 +x3

x4 +x3

x4 + x 2 + x

x3 +x2 +x

x3 +.+I

x2 + 1

The remaining syndromes are shown in Table 2.

1 1 1

O Table 2 0 1 1 1 0 0 0 1 0 0 0 1

(d) The received data is 1101 100

s(x) = x2 + 1 s = l O l

From Table 2

Hence d = 0 1 0 1

I . 7 6 5 4 d l = O O O O 1 l l l O O O O , d 1 ( x ) = x + X + X + X

d 2 = 1 0 1 0 1 0 1 0 1 0 1 0 , d 2 ( x ) = x " + x 9 + x 7 + x 5 + x 3 + x

C ~ ( X ) = ~ ~ ( X ) B ( X ) = X ~ ~ + X ~ ~ + X ' ~ + X ~ ~ + x l ' + x 9 + x 8 + x 7 + x 4 and

c ~ = 0 0 0 0 1 1 0 0 0 1 1 1 0 1 1 1 0 0 1 0 0 0 0

~ ~ ( x ) = d ~ ( x ) ~ ( x ) =x22 + X ' ~ + X ' ~ + x 1 ' + x J 3 + x 8 + x 5 + x 4 + x 3 + x 2 + x and

c ~ = 1 0 0 0 1 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 1 1 0

Hence x 3 + x 2 +x+l=(x+l)(x2 + l ) = ( x + l ) ( ~ + 1 ) ( ~ + 1 ) = ( x + 1 ) ~

4 x + x + l 16.3-4 x + l v Hence ~ ~ + x ~ + x ~ + l = ( x + l ) ( . ~ ~ + x + l )

5 4 X + X

Now try dividing x4 + x + I by x + 1 , we get a remainder I . Hence ( x + 1) is not a factor of (x4 +r + 1) .

The 2*-order prime factors not divisible by x + 1 arc x2 and x2 +x + 1 . Since (x4 + x + I) is not divisible

by x2 , we try dividing by (x2 + x + 1) . B i r also yields a remainder I . Hence x4 + x + 1 does not have

either a first or a second order factor. This means it cannot have a third order factor either. Hence

x'+x4 + x 2 + l = ( x + i ) ( x 4 + I + l )

16.3-4 Try dividing x7 + 1 by x + 1

n~refore (x' + I) = ( x + l)(x6 + xs + x4 + x3 + x2 + x + I)

Now by dividing (x6 + x5 + x4 + x3 + r2 + x + I ) by ( x + . It does not divide. So by dividing by

(x2 + I) . It does not divide. Try dividing by ( x2 + x + I) . It does not divide. Next try dividing by

( x3 + 1) . It docs not divide either. Now try dividing by (x3 + x + I) . It divides. We fmd

( x 6 + x 5 + x 4 +X3+x2+ I+ l ) r ( x~+x+1 ) (X3+x2+ l )

Since ( x 3 + x2 + 1 ) is not divisible by x or x + 1 (the only two fht-ordn prime factors), it must be a

third-order prime factor. Hence

x ' + l - ( ~ + l ) ( X ~ + X + l & ~ +X2+1 )

16.36 For a single error correcting (7, 4) cyclic code with a generator polynomial

,y (x )=x3+x2+1 k - 4 n = 7

Hence

Each code word is found by matrix multiplication c = dG'

The remaining codes are found in a similar manner. See table below.

0001101 0011010 O O l O I l l O l l O l d O 0111001 O l O l l l O 0100001 1101000 llO0IOl 1110010

, l l l l l l l ,1011100 l010001 ,1000110 1001011

The desired form is

1 0 0 0 . . . O h l l b l h l ' . 'h,,,,

0 1 0 0 . . - O h , 2 ~ h 3 2 + ' * % 1

The code is found by using c=dG

Roceeding with matrix multiplication, and noting that

we get

and so on.

These results agree with those of Table 16.5

16.3-8 (a) [ l o 1 1000)

(b) The code is found by matrix multiplication. c = dG'

In general B(X) = glxn-k + g2xn-k-1 +..-g, , -k+l

For this case gl -- I, g2 = 1, gj = 0, gs = 1 Since hlk = g2, bk = 83. bk = g4, the fourth row is immediately found. Thw, so far we have

Next, to get row 3, use row 4 with one left shift.

The 1 is eliminated by adding row 4 to row 3.

Next for row 2, use row 3 with 1 left shift.

T'he 1 is eliminated by adding row 4 to row 2.

Next for row 1, use row 2 with 1 left shift.

This is the desired form.

(c) A11 code words an at a minimum distance of 3 units. Hence this is a single m o r correcting code.

Row 4 is ok. Row 3 is left shift of row 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 .

16.61 The burst (of length 5 ) detection ability is obvious. The single emr conecting ability can be demonstrated as follows. If in any segment of b digits a single error occurs, it will violate the parity in that segment. Hence we locate the segment whm the aror exists. lhis error will also cause parity violation in the augmented segment. By checking which bit in the augmented segment violates the parity, we can locate tbe wrong bit position exactly.

16.5-1 The code can conect any 3 bunts of length 10 or less. It can also correct any 3 random mors in each code word.

To achieve a value 9.872 x for PE,, we need new value Eb JcAI say El Id . Then

Hence

and

This means Eb 14 must be increased from 9.12 to 18.18 (nearly doubled).