Principles of Communication 5Ed R. E. Ziemer%2C William H. Tr Solution

8/10/2019 Principles of Communication 5Ed R. E. Ziemer%2C William H. Tr Solution

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2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY

, Hz , Hz

0 2 4 6 0 2 4 6

10

5

/4

/8

Single-sided amplitude Single-sided phase, rad.

Figure 2.1:

Problem 2.3(a) Not periodic.(b) Periodic. To nd the period, note that

62

= n1f 0 and 20

2 = n2f 0

Therefore103

= n2n1

Hence, take n1 = 3, n2 = 10 , and f 0 = 1 Hz.(c) Periodic. Using a similar procedure as used in (b), we nd that n1 = 2, n2 = 7, andf 0 = 1 Hz.(d) Periodic. Using a similar procedure as used in (b), we nd that n1 = 2 , n2 = 3 , n 3 = 11 ,and f 0 = 1 Hz.

Problem 2.4(a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6Hz, and the phase spectrum consists of a single line of height - / 6 radians at frequency 6Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of 6 and -6 Hz, and the double-sided phase spectrum consists of a line of height - / 6 radiansat frequency 6 Hz and a line of height / 6 at frequency -6 radians Hz.

(b) Write the signal asxb(t) = 3 cos(12 t / 2) + 4cos(16 t)

From this it is seen that the single-sided amplitude spectrum consists of lines of height 3and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists


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2.1. PROBLEM SOLUTIONS 3

, Hz

0 2 4 6

/4

/8

Double-sided phase, rad.

, Hz

-6 -4 -2 0 2 4 6

5

Double-sided amplitude

- /8

-/4

-6 -4 -2

Figure 2.2:


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of a line of height - / 2 radians at frequency 6 Hz. The double-sided amplitude spectrum

consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of height 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrumconsists of a line of height - / 2 radians at frequency 6 Hz and a line of height / 2 radiansat frequency -6 Hz.

Problem 2.5(a) This function has area

Area =Z

1sin( t/ )( t/ ) 2 dt

=

Z sin( u)( u) 2

du = 1

A sketch shows that no matter how small is, the area is still 1. With 0, the centrallobe of the function becomes narrower and higher. Thus, in the limit, it approximates adelta function.(b) The area for the function is

Area =Z

1 exp(t/ )u (t) dt =

Z 0

exp(u)du = 1

A sketch shows that no matter how small is, the area is still 1. With 0, the functionbecomes narrower and higher. Thus, in the limit, it approximates a delta function.(c) Area = R 1 (1 |t | / ) dt = R 11 (t) dt = 1. As 0, the function becomes narrowerand higher, so it approximates a delta function in the limit.Problem 2.6(a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9.

Problem 2.7(a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively.The waveform of part (c) is a periodic train of impulses extending from - to spacedby 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). Thewaveform of part (e) is a doubly-in nite train of square pulses, each of which is one unithigh and one unit wide, centered at , 6, 3, 0, 3, 6, . Waveform (f) is a raisedcosine of minimum and maximum amplitudes 0 and 2, respectively.


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Problem 2.8

(a) The result is

x(t) = Re e j 6 t+ 6Re e j (12 t / 2) = Re he j 6 t + 6 e j (12 t / 2)i(b) The result is

x(t) = 12

e j 6 t + 12

e j 6 t + 3 e j (12 t / 2) + 3 e j (12 t / 2)

(c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequenciesof 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height

/ 2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height3, 1/2, 1/2, and 3 at frequencies of 6, 3, 3, and 6 Hz, respectively. The double-sidedphase spectrum consists of lines of height / 2 and / 2 at frequencies of 6 and 6 Hz,respectively.Problem 2.9(a) Power. Since it is a periodic signal, we obtain

P 1 = 1T 0 Z T 00 4sin2 (8 t + / 4) dt = 1T 0 Z T 00 2 [1cos(16 t + / 2)] dt = 2 W

where T 0 = 1/ 8 s is the period.(b) Energy. The energy is

E 2 = Z e2 tu2(t)dt = Z 0 e2 t dt = 12(c) Energy. The energy is

E 3 = Z e2 t u2(t)dt = Z 0

e2 t dt =

12

(d) Neither energy or power.

E 4 = limT Z T

T dt

( 2 + t2)1/ 4 = P 4 = 0 since limT

1T R T T dt( 2 + t 2 ) 1 / 4 = 0.(e) Energy. Since it is the sum of x1(t) andx2(t), its energy is the sum of the energies of these two signals, or E 5 = 1 / .


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(f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we use

P 6 = limT

12T Z

T

0sin2 (5 t) dt = lim

T 1

2T Z T

0

12

[1cos(10 t)] dt

= limT

12T T 2 12 sin(20 t)20 T 0 = 14 W

Problem 2.10(a) Power. Since the signal is periodic with period / , we have

P = Z

/

0A2 |sin(t + )|2 dt =

Z /

0A

2

2 {1 cos[2(t + )]} dt = A

2

2

(b) Neither. The energy calculation gives

E = limT Z T T (A )

2 dt + jt jt

= limT Z T T (A )

2 dt 2 + t2

The power calculation gives

P = limT

12T

Z T

T

(A )2 dt 2 + t2 = limT

(A )2

2T ln

1 +

p 1 + T 2/ 2

1 +

p 1 + T 2/ 2

! = 0

(c) Energy:

E = Z 0 A2t4 exp(2t/ ) dt = 34A2 5 (use table of integrals)(d) Energy:

E = 2 Z / 20 22dt + Z / 2 12dt! = 5 Problem 2.11(a) This is a periodic train of boxcars, each 3 units in width and centered at multiples of 6:

P a = 16Z 33 2t3dt = 16Z

1.5

1.5dt =

12

W


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(b) This is a periodic train of unit-high isoceles triangles, each 4 units wide and centered

at multiples of 5:

P b = 15Z 2.52.5 2t2dt = 25Z

2

0 1 t22 dt = 25 231 t2320 = 415 W(c) This is a backward train of sawtooths (right triangles with the right angle on the left),each 2 units wide and spaced by 3 units:

P c = 13Z 20 1 t22 dt = 13 231 t23

2

0

= 29

W

(d) This is a full-wave recti ed cosine wave of period 1/5 (the width of each cosine pulse):

P d = 5Z 1/ 101/ 10 |cos(5 t)|2 dt = 2 5Z 1/ 10

0 12 + 12 cos(10 t)dt = 12 WProblem 2.12(a) E = , P = ; (b) E = 5 J, P = 0 W; (c) E = , P = 49 W; (d) E = , P = 2 W.Problem 2.13(a) The energy is

E = Z 6

6 cos2

(6 t) dt = 2Z 6

0 12 +

12 cos (12 t)dt = 6 J(b) The energy is

E = Z he| t | / 3 cos(12 t)i2 dt = 2Z 0 e2t/ 312 + 12 cos(24 t)dtwhere the last integral follows by the eveness of the integrand of the rst one. Use a tableof de nte integrals to obtain

E = Z 0 e2t/ 3dt + Z 0 e2t/ 3 cos(24 t) dt = 32 + 2/ 3(2/ 3)2 + (24 )2Since the result is nite, this is an energy signal.(c) The energy is

E = Z {2 [u (t) u (t 7)]}2 dt = Z 7

04dt = 28 J


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Since the result is nite, this is an energy signal.

(d) Note that Z tu ( ) d = r (t) = 0, t < 0t, t 0which is called the unit ramp. The energy is

E = Z [r (t) 2r (t 10) + r (t 20)]2 dt = 2Z 10

0 t102 dt = 203 Jwhere the last integral follows because the integrand is a symmetrical triangle about t = 10.Since the result is nite, this is an energy signal.

Problem 2.14(a) Expand the integrand, integrate term by term, and simplify making use of the orthog-onality property of the orthonormal functions.(b) Add and subtract the quantity suggested right above (2.34) and simplify.(c) These are unit-high rectangular pulses of width T/ 4. They are centered at t =T/ 8, 3T/ 8, 5T/ 8, and 7T/ 8. Since they are spaced by T/ 4, they are adjacent to eachother and ll the interval [0, T ].(d) Using the expression for the generalized Fourier series coe ffi cients, we nd that X 1 =1/ 8, X 2 = 3/ 8, X 3 = 5/ 8, and X 4 = 7 / 8. Also, cn = T/ 4. Thus, the ramp signal isapproximated by

t

T =

1

81 (t) +

3

82 (t) +

5

83 (t) +

7

84 (t) , 0

t

T

where the n (t)s are given in part (c).(e) These are unit-high rectangular pulses of width T / 2 and centered at t = T / 4 and 3T/ 4.We nd that X 1 = 1/ 4 and X 2 = 3 / 4.(f) To compute the ISE, we use

N = Z T |x (t)|2 dt N Xn =1 cn X 2nNote that

R T |x (t)|

2 dt =

R T 0 (t/T )

2 dt = T / 3. Hence, for (d),ISEd = T 3

T 4

164 +

964 +

2564 +

4964

= 5.208 103T .

For (e), ISE e = T 3 T 2 116 + 916 = 2.083 10

2T .


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Problem 2.15

(a) The Fourier coeffi

cients are (note that the period = 12

2 0 )

X 1 = X 1 = 14

; X 0 = 12

All other coeffi cients are zero.(b) The Fourier coe ffi cients for this case are

X 1 = X 1 = 12

(1 + j )

All other coeffi cients are zero.(c) The Fourier coe ffi cients for this case are (note that the period is 22 0 )

X 2 = X 2 = 18; X 1 = X 1 = 14; X 0 = 14All other coeffi cients are zero.(d) The Fourier coe ffi cients for this case are

X 3 = X 3 = 18

; X 1 = X 1 = 38

All other coeffi cients are zero.

Problem 2.16The expansion interval is T 0 = 4 and the Fourier coe ffi cients are

X n = 14Z 2

22t2e jn ( / 2) t dt = 2

4Z 2

02t2 cosn t2 dtwhich follows by the eveness of the integrand. Let u = n t/ 2 to obtain the form

X n = 2 2n3Z n0 u2 cos u du = 16(n )2 (1)nIf n is odd, the Fourier coe ffi cients are zero as is evident from the eveness of the functionbeing represented. If n = 0 , the integral for the coe ffi cients is

X 0 = 14

Z 2

22t2dt =

83

The Fourier series is therefore

x (t) = 83

+Xn = , n 6=0 (1)n 16n e jn ( / 2) t


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Problem 2.17Parts (a) through (c) were discussed in the text. For (d), break the integral for x (t) upinto a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry contition.

Problem 2.18

This is a matter of integration. Only the solution for part (b) will be given here. Theintegral for the Fourier coe ffi cients is (note that the period really is T 0/ 2)

X n = A

T 0 Z T 0 / 20 sin(0t) e jn 0 t dt=

Ae jn 0 t

0T 0 (1 n2) [ jn sin(0t) + cos ( 0t)]

T 0 / 2

0

= A1 + e jn 0T 0 (1 n2) , n 6= 1

For n = 1, the integral is

X 1 = AT 0 Z T 0 / 20 sin(0t)[cos( jn 0t) j sin( jn 0t)] dt = jA4 = X 1

This is the same result as given in Table 2.1.

Problem 2.19(a) Use Parsevals theorem to get

P |nf 0 | 1/ =N

Xn = N |X n |2 =N

Xn = N A T 0 2 sinc2 (nf 0 )

where N is an appropriately chosen limit on the sum. We are given that only frequencesfor which |nf 0 | 1/ are to be included. This is the same as requiring that |n| 1/ f 0 =T 0/ = 2 . Also, for a pulse train, P total = A2 /T 0 and, in this case, P total = A2/ 2. Thus

P |nf 0 | 1/ P total

= 2A2

2

Xn = 2A22

sinc2 (nf 0 )

= 1

2

2

Xn =

2

sinc2 (nf 0 )

= 1

21 + 2sinc2 (1/ 2) + sinc2 (1)=

12"1 + 222# = 0.91


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(b) In this case, |n| 5, P total = A2/ 5, andP |nf 0 | 1/

P total=

15

5

Xn = 5 sinc2 (n/ 5)=

15n1 + 2h(0.9355)2 + (0 .7568)2 + (0 .5046)2 + (0 .2339)2io= 0 .90

Problem 2.20(a) The integral for Y n is

Y n = 1T 0 Z T 0 y (t) e jn 0 t dt = 1T 0 Z

T 0

0x (t t0) e jn 0 t dt

Let t0 = t t0, which results in

Y n = 1T 0 Z T 0 t 0t 0 xt0e jn 0 t 0 dt0e jn 0 t 0 = X n e jn 0 t 0(b) Note that

y (t) = A cos 0t = A sin(0t + / 2) = A sin[0 (t + / 20)]

Thus, t0 in the theorem proved in part (a) here is / 20. By Eulers theorem, a sine wavecan be expressed as sin(0t) =

12 j

e j 0 t 12 j

e j 0 t

Its Fourier coe ffi cients are therefore X 1 = 12 j and X 1 = 12 j . According to the theoremproved in part (a), we multiply these by the factore jn 0 t 0 = e jn 0 ( / 2 0 ) = e jn / 2

For n = 1, we obtainY 1 =

12 j

e j / 2 = 12

For n = 1, we obtainY 1 =

12 j e j

/ 2 =

12

which gives the Fourier series representation of a cosine wave as

y (t) = 12

e j 0 t + 12

e j 0 t = cos 0t


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We could have written down this Fourier representation directly by using Eulers theorem.

Problem 2.21(a) Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0to obtain the series

1 = + 425 2

+ 49 2

+ 4 2

+ 4 2

+ 49 2

+ 425 2

+

Multiply both sides by 2

8 to get the series in given in the problem. Therefore, its sum is 2

8 .(b) Use the Fourier series of a square wave (specialize the Fourier series of a pulse train)with A = 1 and t = 0 to obtain the series

1 = 4 1 13 + 15

Multiply both sides by 4 to get the series in the problem statement. Hence, the sum is

4 .

Problem 2.22(a) In the expression for the Fourier series of a pulse train (Table 2.1), let t0 = T 0/ 8 and = T 0/ 4 to get

X n = A4

sincn4exp j nf 04 (b) The amplitude spectrum is the same as for part (a) except that X 0 = 3A4 . Note that

this can be viewed as having a sinc-function envelope with zeros at multiples of 43T 0 . Thephase spectrum can be obtained from that of part (a) by adding a phase shift of fornegative frequencies and subtracting for postitive frequencies (or vice versa).

Problem 2.23(a) There is no line at dc; otherwise it looks like a squarewave spectrum.(b) Note that

xA (t) = K dxB (t)

dtwhere K is a suitably chosen constant. The relationship between spectral components istherefore

X An = K ( jn 0) X

Bn

where the superscript A refers to xA (t) and B refers to xB (t).


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Problem 2.24

(a) This is the right half of a triangle waveform of width and height A, or A(1 t/ ).Therefore, the Fourier transform is

X 1 (f ) = AZ 0 (1 t/ ) e j 2 f t dt=

A j 2 f 1 1 j 2 f 1 e j 2 f

where a table of integrals has been used.(b) Since x2 (t) = x1 (t) we have, by the time reversal theorem, that

X 2 (f ) = X 1 (f ) = X 1 (f )= A

j 2 f 1 + 1 j 2 f 1 e j 2 f

(c) Since x3 (t) = x1 (t) x2 (t) we have, after some simpli cation, that

X 3 (f ) = X 1 (f ) X 2 (f )=

jA f

sinc (2f )

(d) Since x4 (t) = x1 (t) + x2 (t) we have, after some simpli cation, that

X 4 (f ) = X 1 (f ) + X 2 (f )

= A sin2 ( f )

( f )2

= A sinc2 (f )

This is the expected result, since x4 (t) is really a triangle function.

Problem 2.25(a) Using a table of Fourier transforms and the time reversal theorem, the Fourierr transformof the given signal is

X (f ) = 1

+ j 2 f 1

j 2 f

Note that x (t) sgn(t) in the limit as 0. Taking the limit of the above Fouriertransform as 0, we deduce that

F [sgn (t)] = 1

j 2 f 1

j 2 f =

1 j f


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(b) Using the given relationship between the unit step and the signum function and the

linearity property of the Fourier transform, we obtain

F [u (t)] = 1

2F [sgn (t)] +

12

F [1]

= 1 j2 f

+ 12

(f )

(c) The same result as obtained in part (b) is obtained.

Problem 2.26(a) Two di ff erentiations give

d2x1 (t)dt2 =

d (t)dt (t 2) + (t 3)

Application of the di ff erentiation theorem of Fourierr transforms gives

( j2 f )2 X 1 (f ) = ( j 2 f )(1) 1 e j 4 f + 1 e j 6 f

where the time delay theorem and the Fourier transform of a unit impulse have been used.Dividing both sides by ( j 2 f )2, we obtain

X 1 (f ) = 1

j 2 f e j 4 f e j 6

f

( j 2 f )2 =

1 j 2 f

e j 5 f j 2 f

sinc (2f )

(b) Two di ff erentiations give

d2x2 (t)dt2

= (t) 2 (t 1) + (t 2)Application of the di ff erentiation theorem gives

( j 2 f )2 X 2 (f ) = 1 2e j 2 f + e j 4 f

Dividing both sides by ( j 2 f )2, we obtain

X 2 (f ) = 12e j 2

f + e j 4 f

( j 2 f )2 = sinc

2 (f ) e j 2 f

(c) Two di ff erentiations give

d2x3 (t)dt2

= (t) (t 1) (t 2) + (t 3)


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Application of the di ff erentiation theorem gives

( j2 f )2 X 3 (f ) = 1 e j 2 f e j 4

f + e j 6 f


X 3 (f ) = 1e j 2

f e j 4 f + e j 6 f

( j 2 f )2

(d) Two di ff erentiations give

d2x4 (t)dt2

= 2 (t 1/ 2) 2 (t 1) 2d (t 2)

dt

Application of the di ff erentiation theorem gives( j2 f )2 X 4 (f ) = 2 sinc (f ) e j f 2e

j 2 f 2 ( j 2 f ) e

j 4 f


X 4 (f ) = 2e j 2 f + ( j 2 f ) e j 4 f sinc (f ) e j

f

2 ( f )2

Problem 2.27(a) This is an odd signal, so its Fourier transform is odd and purely imaginary.(b) This is an even signal, so its Fourier transform is even and purely real.(c) This is an odd signal, so its Fourier transform is odd and purely imaginary.(d) This signal is neither even nor odd signal, so its Fourier transform is complex.(e) This is an even signal, so its Fourier transform is even and purely real.(f) This signal is even, so its Fourier transform is real and even.

Problem 2.28(a) Using superposition, time delay, and the Fourier transform of an impulse, we obtain

X 1 (f ) = e j 16 t + 2 + e j 16 t = 4cos 2 (6 t)

The Fourier transform is even and real because the signal is even.(b) Using superposition, time delay, and the Fourierr transform of an impulse, we obtain

X 2 (f ) = e j 12 t

e j 12 t = 2 j sin(12 f )

The Fourier transform is odd and imaginary because the signal is odd.


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(c) The Fourier transform is

X 3 (f ) =4

Xn =0 n2 + 1e j 4 nf It is complex because the signal is neither even nor odd.

Problem 2.29(a) The Fourier transform of this signal is

X 1 (f ) = 2 (1/ 3)

1 + (2 f / 3)2 =

2/ 31 + [f / (3/ 2 )]2

Thus, the energy spectral density is

G1 (f ) = 2/ 31 + [f / (3/ 2 )]22(b) The Fourier transform of this signal is

X 2 (f ) = 23

f 30Thus, the energy spectral density is

X 2 (f ) = 49

2 f 30 = 49 f 30(c) The Fourier transform of this signal is

X 3 (f ) = 45

sincf 5so the energy spectral density is

G3 (f ) = 1625

sinc2

f 5

(d) The Fourier transform of this signal isX 4 (f ) =

25sincf 205 + sincf + 205


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so the energy spectral density is

G4 (f ) = 425sincf 205 + sincf + 205

2

Problem 2.30(a) Use the transform pair

x1 (t) = e t u (t) 1

+ j 2 f

Using Rayleighs energy theorem, we obtain the integral relationship

Z |X 1 (f )|2 df = Z df 2 + (2 f )2 df = Z |x1 (t)|

2 dt = Z 0 e2 t dt = 1

2

(b) Use the transform pair

x2 (t) = 1

t sinc ( f ) = X 2 (f )Rayleighs energy theorem gives

Z |X 2 (f )|2 df = Z sinc2 ( f ) df = Z |x2 (t)|2 dt= Z

1 2 2

t dt = Z

/ 2

/ 2dt 2 =

1

(c) Use the transform pair

x3 (t) = e | t | 2

2 + (2 f )2

The desired integral, by Rayleighs energy theorem, is

I 3 = Z |X 3 (f )|2 df = Z 1 2 + (2 f )22

df

= 1(2 )2 Z

e2 | t |

dt = 12 2 Z

0 e

2 tdt =

14 3

(d) Use the transform pair1

t sinc2 ( f )


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The desired integral, by Rayleighs energy theorem, is

I 4 = Z |X 4 (f )|2 df = Z sinc4 ( f ) df

= 1

2 Z 2 (t/ ) dt = 2 2 Z

0[1(t/ )]

2 dt

= 2

Z 10 [1u]2 du = 23 Problem 2.31(a) The convolution operation gives

y1 (t) =0, t 1/ 21 1 e (t +1 / 2), 1/ 2 < t + 1/ 21 e (t 1/ 2) e (t +1 / 2), t > + 1/ 2(b) The convolution of these two signals gives

y2 (t) = (t) + tr (t)

where tr (t) is a trapezoidal function given by

tr (t) =

0, t < 3/ 2 or t > 3/ 21, 1/ 2 t 1/ 23/ 2 + t, 3/ 2 t < 1/ 23/ 2 t, 1/ 2 t < 3/ 2

(c) The convolution results in

y3 (t) = Z e | | ( t) d = Z t+1 / 2

t1/ 2e | | d

Sketches of the integrand for various values of t gives the following cases:

y3 (t) = R t+1 / 2t1/ 2 e d , t 1/ 2R 0t1/ 2 e d + R t+1 / 20 e d , 1/ 2 < t 1/ 2R

t+1 / 2t1/ 2

e d , t > 1/ 2

Integration of these three cases gives

y3 (t) =

1 e (t+1 / 2) e (t1/ 2), t 1/ 21 e (t1/ 2) e (t+1 / 2), 1/ 2 < t 1/ 21 e (t1/ 2) e (t+1 / 2), t > 1/ 2


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(d) The convolution gives

y4 (t) = Z t

x ( ) d

Problem 2.32(a) Using the convolution and time delay theorems, we obtain

Y 1 (f ) = F e t u (t) (t )= F e t u (t)F [ (t )]= 1 + j 2 f

sinc (f ) e j 2 f

(b) The superposition and convolution theorems give

Y 2 (f ) = F {[ (t/ 2) + (t)] (t)}

= [2sinc (2f ) + sinc (f )] sinc (f )

(c) By the convolution theorem

Y 3 (f ) = F he | t | (t)i=

2 2 + (2 f )2

sinc (f )

(d) By the convolution theorem (note, also, that the integration theorem can be applieddirectly)

Y 4 (f ) = F [x (t)u (t)]

= X (f ) 1 j 2 f + 12 (f )=

X (f ) j 2 f

+ 12

X (0) (f )

Problem 2.33(a) The normalized inband energy is

E 1 (|f | W )E total

= 2

tan 12 W


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(b) The result is

E 1 (|f | W )E total = 2 Z W

0sinc2 (u) du

The integration must be carried out numerically.

Problem 2.34(a) By the modulation theorem

X (f ) = AT 0

4 sinc(f f 0) T 02 + sinc(f + f 0) T 02 =

AT 04 sinc12 f f 0 1+ sinc12 f f 0 + 1

(b) Use the superposition and modulation theorems to get

X (f ) = AT 0

4 sinc f 2f 0+ 12sinc 12 f f 0 2+ sinc12 f f 0 + 2Problem 2.35Combine the exponents of the two factors in the integrand of the Fourier transform integral,complete the square, and use the given de nite integral.

Problem 2.36

Consider the development below:

x (t)

x (t) = Z x ( ) x (t ) d = Z x ( ) x (t + ) d where = has been substituted. Rename variables to obtain

R ( ) = limT

12T Z T T x ( ) x (t + ) d

Problem 2.37

The result is an even triangular wave with zero average value of period T 0. It makes nodiff erence whether the original square wave is even or odd or neither.


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Problem 2.38

Fourier transform both sides of the diff

erential equation using the diff

erentiation theoremof Fourier transforms to get

[ j 2 f + a]Y (f ) = [ j 2 bf + c]X (f )

Therefore, the transfer function is

H (f ) = Y (f )X (f )

= c + j 2 bf a + j 2 f

The amplitude response function is

|H (f )| = qc2 + (2 bf )2

qa2 + (2 f )2and the phase response is

arg[H (f )] = tan 12 bf c tan 12 f a Amplitude and phase responses for various values of the constants are plotted below.

Problem 2.39(a) The nd the unit impulse response, write H (f ) as

H (f ) = 1 5

5 + j 2 f

Inverse Fourier transforming gives

h (t) = (t) 5e5t u (t)

(b) Use the transform pair

Ae t u (t) A

+ j 2 f and the time delay theorem to nd the unit impulse as

h (t) = 25

e 815 (t3) u (t 3)


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Problem 2.40

Use the transform pair for a sinc function to

nd that

Y (f ) = f 2B f 2W (a) If W < B , it follows that

Y (f ) = f 2W because f 2B = 1 throughout the region where f 2W is nonzero.(b) If W > B , it follows that

Y (f ) =

f

2Bbecause f 2W = 1 throughout the region where f 2B is nonzero.Problem 2.41(a) Replace the capacitors with 1/j C which is their ac-equivalent impedance. Call the junction of the input resistor, feedback resistor, and capacitors 1. Call the junction at thepositive input of the operational ampli er 2. Call the junction at the negative input of theoperational ampli er 3. Write down the KCL equations at these three junctions. Use theconstraint equation for the operational ampli er, which is V 2 = V 3, and the de nitions for0, Q, and K to get the given transfer function.(d) Combinations of components giving

RC = 2 .3 104 seconds

andRaRb

= 2 .5757

will work.

Problem 2.42(a) By long division

H (f ) = 1

R1/LR 1 + R 2

L + j 2 f Using the transforms of a delta function and a one-sided exponential, we obtain

h (t) = (t) R1L

expR1 + R2L tu (t)


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(b) Substituting the ac-equivalent impedance for the inductor and using voltage division,

the transfer function isH (f ) =

R2R1 + R2

j 2 fLR 1 R 2

R 1 + R 2 + j2 fL=

R2R1 + R2 1 (R1 k R2) /L(R1 k R2) /L + j2 f

Therefore, the impulse response is

h (t) = R2

R1 + R2 (t) R1R2(R1 + R2) L exp R1R2(R1 + R2) L tu (t)Problem 2.43The Payley-Wiener criterion gives the integral

I = Z f 2

1 + f 2df

which does not converge. Hence, the given function is not suitable as the transfer functionof a causal LTI system.

Problem 2.44(a) The condition for stability is

Z |h1 (t)| dt = Z |exp( t)cos(2 f 0t) u (t)| dt=

Z 0

exp(

t) |cos(2 f 0t)| dt 0A sketch is shown in Figure 2.4.


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(b) It follows that

x2 (t) = 34x (t) + 34 j bx (t)exp( j 2 f 0t) X 2 (f ) = 34 [1 + sgn (f f 0)] X (f f 0)= 0, f < f 032 X (f f 0) , f > f 0

A sketch is shown in Figure 2.4.(c) This case has the same spectrum as part (a), except that it is shifted right by W Hz.That is,

x3 (t) = 34x (t) +

14 j bx (t)exp( j 2 Wt) X 3 (f ) = 34 + 14sgn (f W )X (f W )

A sketch is shown in Figure 2.4.(d) For this signal

x4 (t) = 34x (t) 14 j bx (t)exp( j Wt) X 4 (f ) = 34 14sgn (f W/ 2)X (f W/ 2)

A sketch is shown in Figure 2.4.

Problem 2.66(a) The spectrum is

X p (f ) = X (f ) + j [ j sgn (f )] X (f ) = [1 + sgn (f )] X (f )The Fourier transform of x (t) is

X (f ) = 12

f f 02W + 12 f + f 02W Thus,

X p (f ) = f f 0

2W if f 0 > 2W (b) The complex envelope is de ned byx p (t) = x (t) e j 2

f 0 t


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, Hz

-W 0 W

2 A

A

1( f )

, Hz

0 f 0 f 0 - W

3 /22( f )

, Hz

0 W 2W

2 A A

3( f )

Hz

- W/2 0 W /2 3 W /2

2 A A

4( f )

Figure 2.4:


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Therefore

x (t) = x p (t) e j 2 f 0 t

HenceF [x (t)] = F [x p (t)]|f f + f 0 =

f f 02W f f + f 0 = f 2W (c) The complex envelope is

x (t) = F 1 f 2W = 2W sinc (2Wt)Problem 2.67

For t < / 2, the output is zero. For |t| / 2, the result isy (t) =

/ 2

q 2 + (2 f )2 ncos[2 (f 0 + f ) t ] e (t+ / 2) cos [2 (f 0 + f ) t + ]o

For t > / 2, the result is

y (t) = ( / 2) e t

q 2 + (2 f )2

ne / 2 cos[2 (f 0 + f ) t ] e

/ 2 cos[2 (f 0 + f ) t + ]oIn the above equations, is given by = tan

12 f 2.2 Computer Exercises

Computer Exercise 2.1% ce2_1: Computes generalized Fourier series coe ffi cients for exponentially

% decaying signal, exp(-t)u(t), or sinewave, sin(2*pi*t)%tau = input(Enter approximating pulse width: );type_wave = input(Enter waveform type: 1 = decaying exponential; 2 = sinewave );if type_wave == 1


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2.2. COMPUTER EXERCISES 35

t_max = input(Enter duration of exponential: );

elseif type_wave == 2t_max = 1;endclf a = [];t = 0:.01:t_max;cn = tau;n_max = oor(t_max/tau)for n = 1:n_max

LL = (n-1)*tau;UL = n*tau;if type_wave == 1

a(n) = (1/cn)*quad(@integrand_exp, LL, UL);elseif type_wave == 2

a(n) = (1/cn)*quad(@integrand_sine, LL, UL);end

enddisp( )disp(c_n)disp(cn)disp( )disp(Expansion coe ffi cients (approximating pulses not normalized):)disp(a)disp( )x_approx = zeros(size(t));for n = 1:n_max

x_approx = x_approx + a(n)*phi(t/tau,n-1);endif type_wave == 1

MSE = quad(@integrand_exp2, 0, t_max) - cn*a*a;elseif type_wave == 2

MSE = quad(@integrand_sine2, 0, t_max) - cn*a*a;enddisp(Mean-squared error:)

disp(MSE)disp( )if type_wave == 1

plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(t),ylabel(x(t); x_a_p_p_r_o_x(t))


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hold

plot(t, exp(-t))title([Approximation of exp(-t) over [0, ,num2str(t_max),] withcontiguous rectangular pulses of width ,num2str(tau)])

elseif type_wave == 2plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(t),

ylabel(x(t); x_a_p_p_r_o_x(t))holdplot(t, sin(2*pi*t))title([Approximation of sin(2*pi*t) over [0, ,num2str(t_max),] with

contiguous rectangular pulses of width ,num2str(tau)])end

% Decaying exponential function for ce_1%function z = integrand_exp(t)z = exp(-t);

% Sine function for ce_1%function z = integrand_sine(t)z = sin(2*pi*t);

% Decaying exponential squared function for ce_1%

function z = integrand_exp2(t)z = exp(-2*t);

% Sin^2 function for ce_1%function z = integrand_sine2(t)z = (sin(2*pi*t)).^2;

A typical run follows:

>> ce2_1Enter approximating pulse width: 0.125Enter waveform type: 1 = decaying exponential; 2 = sinewave: 2

n_max =8c_n0.1250Expansion coe ffi cients (approximating pulses not normalized):


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Figure 2.5:

Columns 1 through 70.3729 0.9003 0.9003 0.3729 -0.3729 -0.9003 -0.9003Column 8-0.3729Mean-squared error:0.0252

Computer Exercise 2.2% ce2_1: Computes generalized Fourier series coe ffi cients for exponentially% decaying signal, exp(-t)u(t), or sinewave, sin(2*pi*t)%tau = input(Enter approximating pulse width: );type_wave = input(Enter waveform type: 1 = decaying exponential; 2 = sinewave );if type_wave == 1

t_max = input(Enter duration of exponential: );elseif type_wave == 2

t_max = 1;endclf a = [];t = 0:.01:t_max;


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cn = tau;

n_max =

oor(t_max/tau)for n = 1:n_maxLL = (n-1)*tau;UL = n*tau;if type_wave == 1

a(n) = (1/cn)*quad(@integrand_exp, LL, UL);elseif type_wave == 2

a(n) = (1/cn)*quad(@integrand_sine, LL, UL);end

enddisp( )disp(c_n)disp(cn)disp( )disp(Expansion coe ffi cients (approximating pulses not normalized):)disp(a)disp( )x_approx = zeros(size(t));for n = 1:n_max

x_approx = x_approx + a(n)*phi(t/tau,n-1);endif type_wave == 1

MSE = quad(@integrand_exp2, 0, t_max) - cn*a*a;elseif type_wave == 2

MSE = quad(@integrand_sine2, 0, t_max) - cn*a*a;enddisp(Mean-squared error:)disp(MSE)disp( )if type_wave == 1


holdplot(t, exp(-t))

title([Approximation of exp(-t) over [0, ,num2str(t_max),] withcontiguous rectangular pulses of width ,num2str(tau)])elseif type_wave == 2



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hold

plot(t, sin(2*pi*t))title([Approximation of sin(2*pi*t) over [0, ,num2str(t_max),] withcontiguous rectangular pulses of width ,num2str(tau)])

end

% Decaying exponential function for ce_1%function z = integrand_exp(t)z = exp(-t);

% Sine function for ce_1%function z = integrand_sine(t)z = sin(2*pi*t);

% Decaying exponential squared function for ce_1%function z = integrand_exp2(t)z = exp(-2*t);

% Sin^2 function for ce_1%function z = integrand_sine2(t)z = (sin(2*pi*t)).^2;

A typical run follows:>> ce2_2Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: 1

Computer Exercise 2.3% ce2_4.m: FFT plotting of line spectra for half-recti ed, full-recti ed sinewave, square

wave,% and triangular waveforms%clf

I_wave = input(Enter type of waveform: 1 = positive squarewave; 2 = 0-dc leveltriangular; 3 = half-rect. sine; 4 = full-wave sine );T = 2;del_t = 0.001;t = 0:del_t:T;


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Figure 2.6:

L = length(t);fs = (L-1)/T;del_f = 1/T;n = [0 1 2 3 4 5 6 7 8 9];if I_wave == 1x = pls_fn(2*(t-T/4)/T);X_th = abs(0.5*sinc(n/2));disp( )disp( 0 - 1 level squarewave)elseif I_wave == 2x = 2*trgl_fn(2*(t-T/2)/T)-1;X_th = 4./(pi^2*n.^2);X_th(1) = 0; % Set n = even coe cients to zero (odd indexed because of MATLAB)X_th(3) = 0;X_th(5) = 0;X_th(7) = 0;X_th(9) = 0;

disp( )disp( 0-dc level triangular wave)elseif I_wave == 3x = sin(2*pi*t/T).*pls_fn(2*(t-T/4)/T);X_th = abs(1./(pi*(1-n.^2)));


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X_th(2) = 0.25;

X_th(4) = 0; % Set n = odd coe

cients to zero (even indexed because of MATLAB)X_th(6) = 0;X_th(8) = 0;X_th(10) = 0;disp( )disp( Half-recti ed sinewave)elseif I_wave == 4x = abs(sin(2*pi*t/T));X_th = abs(2./(pi*(1-n.^2)));X_th(2) = 0; % Set n = odd coe cients to zero (even indexed because of MATLAB)X_th(4) = 0;X_th(6) = 0;X_th(8) = 0;X_th(10) = 0;disp( )disp( Full-recti ed sinewave)endX = 0.5* ff t(x)*del_t; % Multiply by 0.5 because of 1/T_0 with T_0 = 2f = 0:del_f:fs;Y = abs(X(1:10));Z = [n Y X_th];disp(Magnetude of the Fourier coe fficients);disp( )disp( n FFT Theory);disp( __________________________)disp( )disp(Z);subplot(2,1,1),plot(t, x), xlabel(t), ylabel(x(t))subplot(2,1,2),plot(f, abs(X),o),axis([0 10 0 1]),...xlabel(n), ylabel( |X_n |)


>> ce2_3Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect.

sine; 4 = full-wave sine 3Half-recti ed sinewaveMagnetude of the Fourier coe ffi cientsn FFT Theory__________________________


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Figure 2.7:

0 0.3183 0.31831.0000 0.2501 0.25002.0000 0.1062 0.10613.0000 0.0001 04.0000 0.0212 0.02125.0000 0.0001 06.0000 0.0091 0.00917.0000 0.0000 08.0000 0.0051 0.00519.0000 0.0000 0

Computer Exercise 2.4Make the time window long compared with the pulse width.

Computer Exercise 2.5% ce2_5.m: Finding the energy ratio in a preset bandwidth

%I_wave = input(Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 =half-rect. sine; 4 = raised cosine. );

tau = input(Enter pulse width );per_cent = input(Enter percent total desired energy: );


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clf

T = 20;f = [];G = [];del_t = 0.01;t = 0:del_t:T;L = length(t);fs = (L-1)/T;del_f = 1/T;if I_wave == 1

x = pls_fn((t-tau/2)/tau);elseif I_wave == 2

x = trgl_fn(2*(t-tau/2)/tau);elseif I_wave == 3

x = sin(pi*t/tau).*pls_fn((t-tau/2)/tau);elseif I_wave == 4

x = abs(sin(pi*t/tau).^2.*pls_fn((t-tau/2)/tau));endX = ff t(x)*del_t;f1 = 0:del_f*tau:fs*tau;G1 = X.*conj(X);NN = oor(length(G1)/2);G = G1(1:NN);ff = f1(1:NN);f = f1(1:NN+1);E_tot = sum(G);E_f = cumsum(G);E_W = [0 E_f]/E_tot;test = E_W - per_cent/100;L_test = length(test);k = 1;while test(k) < = 0

k = k+1;endB = k*del_f;

if I_wave == 2tau1 = tau/2;elsetau1 = tau;end


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subplot(3,1,1),plot(t/tau, x), xlabel(t/ \ tau), ylabel(x(t)), axis([0 2 0 1.2])

if I_wave == 1title([Energy containment bandwidth for square pulse of width ,num2str(tau), seconds])

elseif I_wave == 2title([Energy containment bandwidth for triangular pulse of width ,

num2str(tau), seconds])elseif I_wave == 3

title([Energy containment bandwidth for half-sine pulse of width ,num2str(tau), seconds])

elseif I_wave == 4title([Energy containment bandwidth for raised cosine pulse of width ,

num2str(tau), seconds])endsubplot(3,1,2),semilogy( ff *tau1, abs(G./max(G))), xlabel(f \ tau), ylabel(G), ...

axis([0 10 -inf 1])subplot(3,1,3),plot(f*tau1, E_W,), xlabel(f \ tau), ylabel(E_W), axis([0 4 0 1.2])legend([num2str(per_cent), bandwidth % X (pulse width) = , num2str(B*tau)],4)


>> ce2_5Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4

= raised cosine. 3Enter pulse width 2

Enter percent total desired energy: 95

Computer Exercise 2.6The program for this exercise is similar to that for Computer Exercise 2.5, except that thewaveform is used in the energy calculation.

Computer Exercise 2.7Use Computer Example 2.2 as a pattern for the solution (note that in earlier printings of the book Computer Excercise 2.2 should be Computer Example 2.2).


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Figure 2.8:


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Chapter 3

Basic M odulation Techniques

3.1 P roblems

P roblem 3.1The demodulated output, in general, is

yD (t) = Lp {xc (t)2cos[! ct + " (t)]}

where Lp {} denotes the lowpass portion of the argument. With

xc (t) = Acm (t)cos[! ct + # 0]

the demodulated output becomes

yD (t) = Lp {2Acm (t)cos[! ct + #0 ]cos[! ct + " (t)]}

Performing the indicated multiplication and taking the lowpass portion yields

yD (t) = Acm (t)cos[" (t) ! # 0 ]If " (t) = " 0 (a constant), the demodulated output becomes

yD (t) = Acm (t)cos[" 0 ! # 0 ]Letting Ac = 1 gives the error

$ (t) = m (t) [1 ! cos(" 0 ! # 0)]The mean-square error is

!$2 (t) = Dm2 (t) [1 ! cos(" 0 ! # 0 )]2E1


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2 CHAPT ER 3. BASIC MODULATION TECHNIQUES

where hi denotes the time-average value. Since the term [1! cos(" 0 ! # 0)] is a constant,we have !$2 (t) = !m2 (t)[1! cos(" 0 ! # 0 )]

2

Note that for "0 = # 0 , the demodulation carrier is phase coherent with the original modu-lation carrier, and the error is zero. For " (t) = ! 0 t we have the demodulated output

yD (t) = Acm (t)cos( ! 0 t ! # 0 )Letting Ac = 1 , for convenience, gives the error

$ (t) = m (t) [1 ! cos(! 0 t ! # 0)]giving the mean-square error

!$2 (t) = Dm2 (t) [1 ! cos(! 0 t ! # 0 )]2EIn many cases, the average of a product is the product of the averages. (We will say moreabout this in Chapters 4 and 5). For this case

!$2 (t) = !m2 (t)D[1! cos(! 0 t ! # 0 )]2ENote that 1 ! cos(! 0 t ! # 0 ) is periodic. Taking the average over an integer number of periods yields

D[1! cos(!0 t ! # 0)]

2

E = !1 ! 2cos(!0 t ! # 0 ) + cos

2(! 0 t ! # 0 )= 1 + 12 = 32

Thus

!$2 (t) = 32 !m2 (t)P roblem 3.2Multiplying the AM signal

xc (t) = Ac [1 + am n (t)]cos ! ct

by xc (t) = Ac [1 + am n (t)]cos ! ct and lowpass ltering to remove the double frequency(2! c) term yields

yD (t) = Ac [1 + am n (t)]cos " (t)


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3.1. PROBLEMS 3

( )C x t

( ) D y t CR

Figure 3.1:

For negligible demodulation phase error, "(t) " 0, this becomesyD (t) = Ac + Acam n (t)

The dc component can be removed resulting in Acam n (t), which is a signal proportionalto the message, m (t). This process is not generally used in AM since the reason for usingAM is to avoid the necessity for coherent demodulation.

P roblem 3.3A full-wave recti er takes the form shown in Figure 3.1. The waveforms are shown inFigure 3.2, with the half-wave recti er on top and the full-wave recti er on the bottom.The message signal is the envelopes. Decreasing exponentials can be drawn from the peaksof the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-waverecti ed xc (t) de nes the message better than the half-wave recti ed xc (t) since the carrierfrequency is e! ectively doubled.

P roblem 3.4

Part !m2n (t) a = 0.4 a = 0.6 a = 1a 1/ 3 E ff = 5 , 1% E ff = 10 .7% E ff = 25%b 1/ 3 E ff = 5 .1% E ff = 10 .7% E ff = 25%c 1 E ff = 13 .8% E ff = 26 .5% E ff = 50%

P roblem 3.5By inspection, the normalized message signal is as shown in Figure 3.3.Thus

mn (t) = 2T

t, 0 # t #T 2


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

Figure 3.2:

t

-1

( )nm t

T

1

0

Figure 3.3:


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3.1. PROBLEMS 5

and

!m2n (t) = 2T Z

T/ 2

0 2T t

2

dt = 2T

2T

2 13

T 2

3

= 13

Also

Ac [1 + a] = 40Ac [1! a] = 10

This yields1 + a1 ! a

= 4010

= 4

or

1 + a = 4 ! 4a5a = 3Thus

a = 0.6

Since the index is 0.6, we can write

Ac [1 + 0.6] = 40

This givesAc =

401.6

= 25

This carrier power isP c =

12

A2c = 12

(25)2 = 312 .5 Watts

The e " ciency is

E ff = (0.6)2131 + (0 .6)213 =

0.363.36

= 0 .107 = 10.7%

ThusP sb

P c + P sb= 0 .107

where P sb represents the power in the sidebands and P c represents the power in the carrier.The above expression can be written

P sb = 0.107 + 0 .107P sb

This givesP sb =

0.1071.0 ! 0.107

P c = 97.48 Watts


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P roblem 3.6For the rst signal

% = T 6

, mn (t) = m (t) and E ff = 81 .25%

and for the second signal

% = 5T

6 , mn (t) =

15

m (t) and E ff = 14 .77%

P roblem 3.7(a) The rst step in the solution to this problem is to plot m (t), or use a root- ndingalgorithm in order to determine the minimum value of m (t). We nd that the minimumof m (t) = ! 11.9523, and the minimum falls at t = 0.0352 and t = 0.0648. Thus thenormalized message signal is

mn (t) = 1

11.9523 [9cos20&t ! 7cos60&t]

With the given value of c (t) and the index a, we have

xc (t) = 100 [1 + 0 .5mn (t)]cos200&t

This yields

xc (t) = !14.6415cos140&t + 18 .8428 cos180&t+100cos200 &t+18 .8248cos220&t ! 14.6415 cos260&t

We will need this later to plot the spectrum.(b) The value of !m2n (t) is

!m2n (t)

=

1

11.9523

2

12

h(9)2 + (7) 2

i = 0.455

(c)This gives the e " ciency

E = (0.5)2 (0.455)1 + (0 .5)2 (0.455)

= 10 .213%


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3.1. PROBLEMS 7

70 90 100 110 130-130 110 100 90 -70

AB

5050

f

BA

BA

BA

Figure 3.4:

(d) The two-sided amplitude spectrum is shown in Figure 3.4.where

A = 14.4615

2 = 7 .2307

andB =

18.82482

= 9 .4124

The phase spectrum results by noting that A is negative and all other terms are positive.(e) By inspection the signal is of the form

xc(t) = a(t)c(t)

where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap.Thus the Hilbert transform of c(t) is

bxc(t) = a(t) bc(t)in which c(t) = 100cos 200 &t. This the envelope ise(t) = 100qa2 (t)cos2 200&t + a2 (t)sin 2 200&t = 100a(t)

wherea(t) = [1 + 0 .5mn (t)]

P roblem 3.8


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(a) mn (t) = 116 [9cos20&t + 7cos60 &t]

(b) !m2n (t) = 116

2

12

h(9)2

+ (7)2

i = 0.2539(c) E ff = 0.25(0 .2539)1+0 .25(0 .2539) = 0 .05969 = 5 .969%(d) The expression for xc (t) isxc (t) = 100 1 + 132 (9cos20&t + cos 60&t)cos200&t

= 10 .9375cos140&t + 14 .0625 cos 180&t+100cos200 &t+14 .0625 cos220&t + 10 .9375cos260&t

Note that all terms are positive so the phase spectrum is everywhere zero. The amplitudespectrum is identical to that shown in the previous problem except that

A = 1

2 (10.9375) = 5 .46875

B = 1

2 (14.0625) = 7 .03125

(e) As in the previous problem, the signal is of the form

xc(t) = a(t)c(t)

where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap.Thus the Hilbert transform of c(t) is

bxc(t) = a(t) bc(t)where c(t) = 100cos 200 &t . This the envelope ise(t) = 100qa2 (t)cos2 200&t + a2 (t)sin 2 200&t

= 100 a(t) = 100 1 + 132 (9cos20&t + cos 60&t)P roblem 3.9The modulator output

xc (t) = 40 cos 2 &(200) t + 4cos 2 &(180) t + 4cos 2 &(220) t

can be writtenxc (t) = [40 + 8cos 2 & (20) t]cos2&(200) t


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3.1. PROBLEMS 9

or

xc (t) = 40 1 + 840 cos 2& (20) tcos2&(200) tBy inspection, the modulation index isa =

840

= 0 .2

Since the component at 200 Hertz represents the carrier, the carrier power is

P c = 12

(40)2 = 800 Watts

The components at 180 and 220 Hertz are sideband terms. Thus the sideband power is

P sb = 12

(4)2 + 12

(4)2 = 16 Watts

Thus, the e " ciency is

E ff = P sb

P c + P sb=

16800 + 16

= 0 .0196 = 1.96%

P roblem 3.10

A = 14.14 B = 8.16 a = 1.1547

P roblem 3.11The modulator output

xc (t) = 20 cos 2 &(150) t + 6cos 2 &(160) t + 6cos 2 &(140) t

isxc (t) = 20 1 + 1220 cos 2& (10) tcos2&(150) t

Thus, the modulation index, a, isa = 12

20 = 0 .6

The carrier power isP c =

12

(20)2 = 200 Watts


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and the sideband power is

P sb = 12 (6)2 + 1

2 (6)2 = 36 Watts

Thus, the e " ciency isE ff =

36200 + 36

= 0 .1525

P roblem 3.12(a) By plotting m (t) or by using a root- nding algorithm we see that the minimum valueof m (t) is M = ! 3.432. Thus

mn (t) = 0 .5828cos (2&

f m t) + 0 .2914cos(4&

f m t) + 0 .5828 cos(10&

f m t)The AM signal is

xc (t) = Ac [1 + 0.7mn (t)]cos2&f ct= 0 .2040Ac cos 2&(f c ! 5f m ) t

+0 .1020Ac cos 2&(f c ! 2f m ) t+0 .2040Ac cos 2&(f c ! f m ) t+ Ac cos 2&f ct+0 .2040Ac cos 2&(f c + f m ) t+0 .1020Ac cos 2&(f c + 2 f m ) t+0 .2040Ac cos 2&(f c + 5 f m ) t

The spectrum is drawn from the expression for xc (t). It contains 14 discrete componentsas shown

Comp Freq Amp Comp Freq Amp1 ! f c ! 5f m 0.102Ac 8 f c ! 5f m 0.102Ac2 ! f c ! 2f m 0.051Ac 9 f c ! 2f m 0.051Ac3 ! f c ! f m 0.102Ac 10 f c ! f m 0.102Ac4 ! f c 0.5Ac 11 f c 0.5Ac5 ! f c + f m 0.102Ac 12 f c + f m 0.102Ac6

!f c + 2 f m 0.051Ac 13 f c + 2 f m 0.051Ac

7 ! f c + 5 f m 0.102Ac 14 f c + 5 f m 0.102Ac(b) The e " ciency is 15.8%.

P roblem 3.13


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3.1. PROBLEMS 11

f

Filter Characteristic

( ){ }2m t ( ){ }m t

2 f c f c f W c + f W c 0 W 2W

Figure 3.5:

(a) From Figure 3.75x (t) = m (t) + cos ! ct

With the given relationship between x (t) and y (t) we can write

y (t) = 4 {m (t) + cos ! ct} + 10 {m (t) + cos ! ct}2

which can be written

y (t) = 4 m (t) + 4cos ! ct + 10 m2 (t) + 20 m (t)cos ! ct + 5 + 5cos 2 ! ct

The equation for y (t) is more conveniently expressed

y (t) = 5 + 4 m (t) + 10 m2 (t) + 4 [1 + 5 m (t)] cos ! ct + 5cos 2 ! ct

(b) The spectrum illustrating the terms involved in y (t) is shown in Figure 3.5. The centerfrequency of the lter is f c and the bandwidth must be greater than or equal to 2W . Inaddition, f c ! W > 2W or f c > 3W , and f c + W < 2f c. The last inequality states thatf c > W , which is redundant since we know that f c > 3W .(c) From the de nition of m (t) we have

m (t) = Mmn (t)

so thatg (t) = 4 [1 + 5 Mm n (t)]cos ! ct

It follows thata = 0 .8 = 5M


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Thus

M = 0.8

5 = 0 .16(d) This method of forming a DSB signal avoids the need for a multiplier.

P roblem 3.14

H U (f ) = 1 !12

sgn (f + f c)

xUS B (t) = 12

Acm (t)cos ! ct !12

Ac bm (t)sin ! ctP roblem 3.15For the USB SSB case, the modulator output is a sinusoid of frequency f c + f m , while forthe LSB SSB case, the modulator output is a sinusoid of frequency of f c ! f m .P roblem 3.16Using Figure 3.13 and the phases given in the problem statement, the modulator outputbecomes

xc (t) = A$

2 cos[(! c ! ! 1 ) t + #]

+A (1 ! $)

2 cos[(! c + ! 1 ) t + "1 ]

+B

2 cos [(!

c + ! 2) t + "2 ]

Multiplying xc (t) by 4cos ! ct and lowpass ltering yields the demodulator output

yD (t) = A$ cos(! 1 ! # ) + A (1 ! $)cos( ! 1 t + "1 ) + B cos(! 2 t + "2 )For the sum of the rst two terms to equal the desired output with perhaps a time delay," 1 must equal ! # . This gives

yD (t) = A cos(! 1 t + #) + B cos(! 2 t + "2 )

which we can write

yD

(t) = A cos ! 1

t +

"1

! 1+ B cos ! 2

t +

"2

! 2For no distortion, yD (t) must be of the form m (t ! % ). Thus" 1! 1

= "2! 2


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3.1. PROBLEMS 13

so that

" 2 = ! 2! 1 " 1

Thus, the phase must be linear. The fact that # = ! " 1 tells us that the lter phase responsemust have odd symmetry about the carrier frequency.P roblem 3.17We assume that the VSB waveform is given by

xc (t) = 1

2A$ cos(! c ! ! 1) t

+12

A (1 ! $)cos( ! c + ! 1) t

+12B cos(! c + ! 2 ) t

We let y (t) be xc (t) plus a carrier. Thus

y (t) = xc (t) + K cos ! ct

It can be shown that y (t) can be written

y (t) = y1 (t)cos ! c (t) + y2 (t)sin ! ct

where

y1 (t) = A

2 cos ! 1 t +

B

2 cos! 2 t + K

y2 (t) = A$ + A2sin ! 1 t ! B2 sin ! 2 tAlso

y (t) = R (t)cos( ! ct + ")

where R (t) is the envelope and is therefore the output of an envelope detector. It followsthat

R (t) = qy21 (t) + y22 (t)For K large, R (t) = |y1 (t)| , which is 12 m (t)+ K , where K is a dc bias. Thus if the detectoris ac coupled, K is removed and the output y (t) is m (t) scaled by 12 .

P roblem 3.18The required gure appears in Figure 3.6.

P roblem 3.19


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1

DesiredSignal

1 2

LocalOscillator

2 IF 1 22

Signal atMixer Output

1 22

ImageSignal

2 13 2 2

Image Signalat Mixer Output

Figure 3.6:


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3.1. PROBLEMS 15

Since high-side tuning is used, the local oscillator frequency is

f LO = f i + f IF

where f i , the carrier frequency of the input signal, varies between 5 and 25 MHz . Theratio is

R = f IF + 25

f IF + 5where f IF is the I F frequency expressed in M Hz . We make the following table

f IF ,MHz R0.4 4.700.5 4.630.7 4.51

1.0 4.331.5 4.082.0 3.86

A plot of R as a function of f IF is the required plot.

P roblem 3.20For high-side tuning we have

f LO = f i + f IF = 1120 + 455 = 1575 kHzf IMAGE = f i + 2 f IF = 1120 + 910 = 2030 kHz

For low-side tuning we have

f LO = f i ! f IF = 1120 ! 455 = 665 kHzf IMAGE = f i ! 2f IF = 1120 ! 910 = 210 kHz

P roblem 3.21For high-side tuning

f LO = f i + f IF = 1120 + 2500 = 3620 kHzf IMAGE = f i + 2 f IF = 1120 + 5000 = 6120 kHz

For low-side tuning

f LO = f i ! f IF = 1120 ! 2500 = ! 1380 kHzf LO = 1380 kHz

f IMAGE = f i ! 2f IF = 1120 ! 5000 = ! 3880 kHzf LO = 3880 kHz


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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

0

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

0

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

0

1

Figure 3.7:

In the preceding development for low-side tuning, recall that the spectra are symmetrical

about f = 0 .P roblem 3.22By de nition

xc (t) = Ac cos [! ct + k pm (t)] = Ac cos [! ct + k pu(t ! t0 )]The waveforms for the three values of k p are shown in Figure 3.7. The top pane is fork p = &, the middle pane is for k p = ! &/ 2, and the bottom pane is for k p = &/ 4.P roblem 3.23Let # (t) = ' cos ! m t where ! m = 2&f m . This gives

xc

2 (t) = Ac Re

ne j ! c t e j " cos ! m t

oExpanding a Fourier series givese j " cos ! m t =

!

Xn= "! C n e jn ! m t


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3.1. PROBLEMS 17

where

C n = !

m2& Z # / ! m

" # / ! me j " cos ! m t e

"

jn ! m t dt

With x = ! m t, the Fourier coe " cients become

C n = 12&Z #" # e j " cos x e" jnx dx

Since cosx = sin x + #2 C n = 12&Z #" # e j [" sin (x+ !2 ) " nx ]dxWith x = x + #2 , the preceding becomes

C n = 12&Z 3# / 2

" # / 2e j [" sin y" ny + n

!

2 ]dy

This gives

C n = e jn !

2 12&Z #" # e j [" sin y " ny ]dywhere the limits have been adjusted by recognizing that the integrand is periodic withperiod 2&. Thus

C n = e jn !

2 J n (' )

and

xc2 (t) = Ac Re(e j ! c t

!

Xn= "! J n (' ) e

j (n ! m t+ n !2 )

)Taking the real part yieldsxc2 (t) = Ac

!

Xn = "! J n (' )cosh(! c + n! m ) t + n&2 iThe amplitude spectrum is therefore the same as in the preceding problem. The phasespectrum is the same as in the preceding problem except that n#2 is added to each term.

P roblem 3.24Since sin(x) = cos( x !

#2 ) we can write

xc3 (t) = Ac sin(! ct + ' sin ! m t) = Ac cos! ct ! &2 + ' sin ! m twhich is

xc3(t) = Ac Rene j ( ! c t " # / 2) e j sin ! m to


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Since

e j sin ! m t =

!

Xn = "! J n (' )e jn ! m t

we have

xc3 (t) = Ac Re(e j ( ! c t " # / 2) !Xn = "! J n ( ' )e jn ! m t)Thus

xc3 (t) = Ac!

Xn= "! J n ( ' ) Rene j ( ! c t+ n ! m t " # / 2)oThus

xc3 (t) = Ac

!

Xn = "! J n (' )cosh(!

c + n ! m ) t !&

2iNote that the amplitude spectrum of xc3 (t) is identical to the amplitude spectrum for bothxc1 (t) and xc2(t). The phase spectrum of xc3 (t) is formed from the spectrum of xc1 (t) byadding ! &/ 2 to each term.For xc4 (t) we write

xc4 (t) = Ac sin(! ct + ' cos ! m t) = Ac cos! ct ! &2 + ' cos! m tUsing the result of the preceding problem we write

xc4(t) = Ac Re(e j ( ! c t " # / 2)

!

Xn = "! J n (' ) e

j (n ! m t+ n !2 )

)This givesxc4(t) = Ac

!

Xn = "! J n (' ) Rene j ( ! c t+ n ! m t " !2 + n !2 )oThus

xc4 (t) = Ac!

Xn = "! J n ( ' )cosh(! c + n! m ) t + &2 (n ! 1)iCompared to xc1(t), xc2(t) and xc3 (t), we see that the only di ! erence is in the phase spec-trum.

P roblem 3.25From the problem statement

xc (t) = Ac cos [2& (40) t + 10 sin (2 &(5) t)]


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3.1. PROBLEMS 19

The spectrum is determined using (3.101). Since f c = 40 and f m = 5 (it is important to

note that f c is an integer multiple of f m ) there is considerable overlap of the portion of thespectrum centered about f = ! 40 with the portion of the spectrum centered about f = 40for large ' (such as 10). The terms in the spectral plot, normalized with respect to Ac,are given in Table 3.1 for f # 0 (positive frequency terms and the term at dc). The termsresulting from the portion of the spectrum centered at f = 40 are denoted S 40 and the termsresulting from the portion of the spectrum centered at f = ! 40 are denoted S " 40 . Giventhe percision of the Table of Bessel coe " cients (Page 131 in the textbook), we have overlapfor |f | # 45. Where terms overlap, they must be summed. The sum is denoted S T in Table3.1. In developing the Table 3.1 be sure to remember that J " n (' ) = ! J n (' ) for odd n.The power must now be determined in order to nd Ac. With the exception of the dc term,the power at each frequency is S 2T A

2c / 2 (the power for negative frequencies is equal to the

power at positive frequencies and so the positive frequency power can simply be doubled togive the total power with the dc term excepted). The power at dc is S 2T A2c = (0 .636)2 A2c .Carring out these operations with the aid of Table 3.1 gives the total power

0.8036A2c = 40

which is

Ac = r 400.8036 = 7 .0552The waveform xc(t) is illustrated in the top pane of Figure 3.8. The spectrum, normalizedwith respect to Ac, is illustrated in the bottom frame. The fact that xc (t) has a nonzero

dc value is obvious.

P roblem 3.26We are given J 0 (3) = ! 0.2601 and J 1 (3) = ! 0.3391. Using

J n +1 (' ) = 2n

' J n (' ) ! J n " 1 (' )

with ' = 3 we have

jn+1 (3) = 23

nJ n (3) ! J n " 1 (3)With n = 1,

J 2 (3) = 2

3J 1 (3) ! J 0 (3)

= 2

3 (0.3391) + 0 .2601 = 0.4862


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Table 3.1: Data for Problem 3.25

f S 40 S " 40 S T 125 0.001 0 0.001120 0.002 0 0.002115 0.005 0 0.005110 0.012 0 0.012

105 0.029 0 0.029100 0.063 0 0.06395 0.123 0 0.12390 0.207 0 0.20785 0.292 0 0.29280 0.318 0 0.31875 0.219 0 0.21970 !0.014 0 ! 0.01465 !0.234 0 ! 0.23460 !0.220 0 ! 0.22055 0.058 0 0.05850 0.255 0 0.25545 0.043 0.001 0.04440 !0.246 0.002 ! 0.24435 !0.043 0.005 ! 0.03830 0.255 0.012 0.06725 !0.058 0.029 ! 0.02920 !0.220 0.063 ! 0.15715 0.234 0.123 0.35710 !0.014 0.207 ! 0.1935 !0.217 0.292 0.0750 0.318 0.318 0.636


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3.1. PROBLEMS 21

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1

-0.5

0

0.5

1

Time

x c ( t )

-150 -100 -50 0 50 100 1500

0.2

0.4

0.6

0.8

Frequency

M a g n i t u d e

Figure 3.8:

With n = 2,

J 3 (3) = 4

3J 2 (3) ! J 1 (3)

= 4

3 (0.04862) + 0 .3391 = 0.3091

Finally, with n = 3 we have

J 4 (3) = 2 J 3 (3) ! J 2 (3)= 2 (0 .3091) + 0 .4862 = 0 .1320

P roblem 3.27The amplitude and phase spectra follow directly from the table of Fourier-Bessel coe " cients.The single-sided magnitude and phase spectra are shown in Figure 3.9. The magnitudespectrum is plotted assuming Ac = 1.


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600 700 800 900 1000 1100 1200 1300 14000

0.1

0.2

0.3

0.4

Frequency

M a g n i t u d

e

600 700 800 900 1000 1100 1200 1300 1400-4

-3

-2

-1

0

Frequency

P h a s e

Figure 3.9:


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3.1. PROBLEMS 23

P roblem 3.28

The modulated signal can be written

xc (t) = Re h10 + 3e j 2# (20) t + 5 e" j 2# (20) tie j 2# (100) tWe will concentrate on the term in brackets, which is the complex envelope described inChapter 2. Denoting the complex envelope by exc (t), we can writeexc (t) = [10 + 3cos2 &(20t) + 5 cos 2 &(20) t]+ j [3sin2&(20t) ! 5sin2&(20) t]

= [10 + 8 cos 2 &(20t)] ! j [2sin2&(20) t]

It follows from the de

nition of xc (t) that

exc (t) = R (t) e j $ ( t )

Thus

R2 (t) = [10 + 8cos2 & (20t)]2 + 4sin 2 2&(20) t= 134 + 160 cos 2 & (20) t + 30 cos 2 &(40) t

This givesR (t) =

p 134 + 160cos 2& (20) t + 30 cos 2 &(40) t

Also# (t) = tan " 1 !2sin2& (20) t

10 + 8cos 2&(20) t

P roblem 3.29Since sin x = cosx ! #2, we can write

xc (t) = Re ne j 200 # th5e" j 40 # t + 10 + 3 e j (40 # t " !2 )ioor

xc (t) = Re

a (t) e j 200 # t

where a (t) = (5cos40 &t + 10 + 3sin 40 &t)+ j (5sin40&t ! 3cos40&t)


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Thus if we write xc (t) as

xc (t) = R (t) cos (200&t + # (t))

the envelope, R (t), is given by

R (t) = h(5cos40&t + 10 + 3 sin40 &t)2 + (5sin40 &t ! 3cos40&t)2i12

and the phase deviation, # (t), is given by

# (t) = tan " 1 5sin40&t ! 3cos40&t

10 + 5cos 40&t + 3sin40 &t

The envelope, R (t), can be simpli ed and expressed in several di ! erent forms.

P roblem 3.30(a) Since the carrier frequency is 1000 Hertz, the general form of xc (t) is

xc (t) = Ac cos [2&(1000) t + # (t)]

The phase deviation, # (t), is therefore given by

# (t) = 20 t2 rad

The frequency deviation is

d#dt = 40 t rad/sec

or1

2&d#dt

= 20

& t Hertz

(b) The phase deviation is

# (t) = 2 &(500) t2 ! 2& (1000) t radand the frequency deviation is

d#

dt = 4&(500) t

!2&(1000) rad/sec

= 2000 &(t ! 1) rad/secor

12&

d#dt

= 1000( t ! 1) Hertz


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3.1. PROBLEMS 25

(c) The phase deviation is

# (t) = 2 &(100) t = 200&t rad

and the frequency deviation is

d#dt

= 200& rad/sec

or1

2&d#dt

= 100 Hertz

which should be obvious from the expression for xc (t).(d) The phase deviation is

# (t) = 200 &t + 10$ t radand the phase deviation is

d#dt

= 200& + 12

(10) t "12 = 200& +

5$ t rad/sec

or1

2&d#dt

= 100 + 52&$ t Hertz

P roblem 3.31(a) The phase deviation is

# (t) = 2 &(30)Z t0 (8)dt = 480&t, t # 4The maximum phase deviation is # (4) = 480 &(4) = 1920 &. The required plot is simply

# (t) = "#$

0, t < 0480&t, 0 # t < 41920&, t %4

(b) The frequency deviation, in Hz, is

12&

d#dt

= 30 m(t) =0, t < 0

240, 0 # t < 40 t %4


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The required sketch follows simply.

(c) The peak frequency deviation is 8f d = 8 (30) = 240 Hertz(d) The peak phase deviation is

2& (30)Z 40 (8) dt = 2&(30) (8)(4) = 1920 & radThe modulator output power is

P = 12

A2c = 12

(100)2 = 5000 Watts

P roblem 3.32(a) The message signal is

m (t) ="%%#%%$

0, t < 4t ! 4, 4 # t # 68 ! t 6 # t # 80, t > 8

The phase deviation is

# (t) = 2 &f dZ t4 (t ! 4) dt= 60 &(! 4) ( t ! 4) + 30 &

t2 ! 16

= 30 &t2

! 8t + 16, 4 # t # 6# (t) = 2 &f dZ t6 (8 ! t) dt + # (6)

= 120 & + 60&(8)( t ! 6) ! 30&t2 ! 36= 120 &! 30&t2 ! 16t + 60, 6 # t # 8Also

# (t) = 240 &, t > 8# (t) = 0 , t < 4

The sketches follow immediately from the equations.(b) The frequency deviation in Hertz is 30m(t).(c) The peak phase deviation = 240 & rad.(d) The peak frequency deviation = 120 Hertz.


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3.1. PROBLEMS 27

(e) The carrier power is, assuming a su " ciently high carrier frequency,

P = 12

A2c = 12

(100)2 = 5000 Watts

P roblem 3.33The frequency deviation in Hertz is the plot shown in Fig. 3.76 with the ordinate valuesmultiplied by 25.The phase deviation in radians is given

# (t) = 2 &f dZ t m (( ) d( = 50&Z t m (( ) d(For 0 # t # 1, we have # (t) = 50 &Z t0 2( d( = 50&t2For 1 # t # 2

# (t) = #(1) + 50 &Z t1 (5 ! ( ) d( = 50& + 250&(t ! 1) ! 25&t2 ! 1= !175& + 250&t ! 25&t

2

For 2 # t # 3# (t) = # (2) + 50 &

Z t

23d( = 225& + 150&(t ! 2)

For 3 # t # 4# (t) = # (3) + 50 &Z t3 2d( = 375& + 100&(t ! 3)

Finally, for t > 4 we recognize that # (t) = # (4) = 475 &. The required gure results byplotting these curves.

P roblem 3.34The frequency deviation in Hertz is the plot shown in Fig. 3.77 with the ordinate valuesmultiplied by 10. The phase deviation is given by

# (t) = 2 &f dZ t

m (( ) d( = 20&Z t

m (( ) d(

For 0 # t # 1, we have# (t) = 20 &Z t0 ( d( = 10&t2


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For 1 # t # 2

# (t) = #(1) + 20 &Z t

1(( ! 2) d( = 10& + 10&t2 ! 1! 40&(t ! 1)

= 10 &t2 ! 4t + 4 = 10&(t ! 2)2For 2 # t # 4

# (t) = #(2) + 20 &Z t2 (6 ! 2( )d( = 0 + 20 &(6)( t ! 2) ! 20&t2 ! 4= !20&(t2 ! 6t + 8)Finally, for t > 4 we recognize that # (t) = # (4) = 0 . The required gure follows by plottingthese expressions.

P roblem 3.35The frequency deviation in Hertz is the plot shown in Fig. 3.78 with the ordinate valuesmultiplied by 5. The phase deviation in radians is given by

# (t) = 2 &f dZ t m (( ) d( = 10&Z t m (( ) d(For 0 # t # 1, we have

# (t) = 10 &

Z t

0

(

!2( )d( =

!10&t2

For 1 # t # 2

# (t) = # (1) + 10 &Z t1 2d( = ! 10& + 20&(t ! 1) = 10 &(2t ! 3)For 2 # t # 2.5

# (t) = #(2) + 10 &Z t2 (10 ! 4( )d( = 10& + 10&(10) ( t ! 2) ! 10&(2)t2 ! 4= 10 &(! 2t

2 + 10 t ! 11)

For 2.5 # t # 3# (t) = #(2.5) ! 10&Z t2.5 2d( = 15&! 20&(t ! 2.5)

= 10 &(! 2t + 5 .5)


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3.1. PROBLEMS 29

For 3 # t # 4

# (t) = #(3) + 10 &Z t

3(2( ! 8) d( = 5& + 10&(t

2

! 9) ! 10&(8)( t ! 3)= 10 &(t2 ! 8t + 15 .5)

Finally, for t > 4 we recognize that # (t) = #(4) = ! 5&. The required gure follows byplotting these expressions.P roblem 3.36(a) The peak deviation is (12.5)(4) = 50 and f m = 10. Thus, the modulation index is5010 = 5.(b) The magnitude spectrum is a Fourier-Bessel spectrum with ' = 5 . The n = 0 term fallsat 1000 Hz and the spacing between components is 10 Hz. The sketch is that of Figure 3.24in the text.(c) Since ' is not 1, this is not narrowband FM. The bandwidth exceeds 2f m .(d) For phase modulation, k p (4) = 5 or k p = 1 .25.

P roblem 3.37The results are given in the following table:

Part f d D = 5f d/W B = 2 ( D + 1) W a 20 0.004 50.2 kHzb 200 0.04 52 kHzc 2000 0.4 70 kHzd 20000 4 250 kHz

P roblem 3.38From

xc (t) = Ac!

Xn = "! J n (' )cos( ! c + ! m ) twe obtain

!x2

c (t)

=

1

2A2

c

!

Xn ="!

J 2n (' )

We also know that (assuming that xc(t) does not have a signi cant dc component - seeProblem 3.24)

!x2c (t) = !A2c cos2 [! ct + # (t)]


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which, assuming that ! c 1 so that xc (t) has no dc component, is

!x2c (t) = 12A2cThis gives12

A2c = 12

A2c!

Xn = "! J 2n (' )from which

!

Xn = "! J 2n (' ) = 1P roblem 3.39Since

J n (' ) = 12&Z #" # e" j (nx " " sin x) dx = 12&Z #" # e j ( " sin x " nx ) dx

we can write

J n (' ) = 12&Z #" # cos( ' sin x ! nx ) dx + j 12&Z #" # sin( ' sin x ! nx ) dx

The imaginary part of J n (' ) is zero, since the integrand is an odd function of x and thelimits (! &, &) are even. Thus

J n (' ) = 12&Z

#

" # cos(' sin x ! nx ) dxSince the integrand is even

J n (' ) = 1&Z #0 cos(' sin x ! nx ) dx

which is the rst required result. With the change of variables ) = &! x, we haveJ n (' ) =

1&Z #0 cos[' sin (&! ) ) ! n (&! ) )] (! 1) d)

= 1

&Z #

0cos[' sin (&

!) )

!n& + n ) ]d)

Since sin(&! ) ) = sin ) , we can write

J n (' ) = 1&Z #0 cos[' sin ) + n) ! n&]d)


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3.1. PROBLEMS 31

Using the identity

cos(u ! * ) = cos u cos * + sin u sin * withu = ' sin ) + n)

and* = n&

yields

J n (' ) = 1

&Z #0 cos[' sin ) + n ) ]cos(n&) d)+

1

&Z #

0sin[' sin ) + n ) ]sin(n&) d)

Since sin(n&) = 0 for all n, the second integral in the preceding expression is zero. Also

cos(n&) = ( ! 1)n

ThusJ n (' ) = ( ! 1)

n 1&Z #0 cos[' sin ) + n ) ]d)

HoweverJ " n (' ) =

1&Z #0 cos[' sin ) + n) ]d)

ThusJ n (' ) = ( ! 1)

n J " n (' )

or equivalentlyJ " n (' ) = ( ! 1)

n J n (' )

P roblem 3.40(a) Peak frequency deviation = 80 Hz(b) # (t) = 8sin(20 &t)(c) ' = 8

(d) P i = 50 Watts , P 0 = 16.76 Watts(e) The spectrum of the input signal is a Fourier_Bessel spectrum with ' = 8. The n = 0term is at the carrier frequency of 500 Hz and the spacing between components is 10 Hz.The output spectrum consistes of the n = 0 term and three terms each side of the n = 0term. Thus the output spectrum has terms at 470, 480, 490, 500, 510, 520 and 530 Hz.


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Figure 3.10:


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3.1. PROBLEMS 33

P roblem 3.41

The required spectra are given in Figure 3.10. The modulation indices are, from top tobottom, ' = 0 .5, ' = 1 , ' = 2 , ' = 5, and ' = 10.

P roblem 3.42We wish to nd k such that

P r = J 20 (10) + 2k

Xn =1 J 20 (10) %0.80This gives k = 9, yielding a power ratio of P r = 0.8747. The bandwidth is therefore

B = 2kf m = 2 (9) (150) = 2700 Hz

For P r %0.9, we have k = 10 for a power ratio of 0.9603. This givesB = 2kf m = 2 (10) (150) = 3000 Hz

P roblem 3.43From the given data, we have

f c1 = 110 kHz f d1 = 0 .05 f d2 = n (0.05) = 20

This givesn = 20

0.05 = 400

andf c1 = n (100) kHz = 44 MHz

The two permissible local oscillator frequencies are

f 0 . 1 = 100 ! 44 = 56 MHzf 0 . 2 = 100 + 44 = 144 MHz

The center frequency of the bandpass lter must be f c = 100 MHz and the bandwidth is

B = 2 ( D + 1) W = 2 (20 + 1) (10) 103

or B = 420 kHz


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P roblem 3.44

For the circuit shownH (f ) =

E (f )X (f )

= R

R + j 2&fL + 1 j 2# fC or

H (f ) = 1

1 + j2&f % L ! 12# f % C where

% L = L

R =

10" 3

103 = 10 " 6 ,

% C = RC =

103

10" 9

= 10 " 6

A plot of the amplitude response shows that the linear region extends from approximately54 kHz to118kHz. Thus an appropriate carrier frequency is

f c = 118 + 54

2 = 86 kHz

The slope of the operating characteristic at the operating point is measured from the am-plitude response. The result is

K D &= 810" 6P roblem 3.45We can solve this problem by determining the peak of the amplitude response characteristic.This peak falls at

f p = 1

2&$ LC It is clear that f p > 100 MHz. Let f p = 150 MHz and let C = 0 .00110" 12. This gives

L = 1

(2&)2 f 2 p C = 1 .12610" 3

We nd the value of R by trial and error using plots of the amplitude response. Anappropriate value for R is found to be 1 M ! . With these values, the discriminator constant

is approximately K D " 8.510" 9

P roblem 3.46


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3.1. PROBLEMS 35

For A i = Ac we can write, from (3.176),

xr (t) = Ac [cos! ct + cos ( ! c + ! i ) t]

which isxr (t) = Ac [(1 + cos ! i t)cos ! ct ! sin ! i t sin ! ct]

This yieldsxr (t) = R (t)cos[! ct + + (t)]

where+ (t) = tan " 1 sin ! i t1 + cos ! i t = tan " 1 tan ! i t2 = ! i t2

This gives

yD (t) = 12& ddt 2&f i t2 = 12f iFor Ai = ! Ac, we get

+ (t) = tan " 1 ! sin ! i t1 ! cos ! i t = ! tan " 1 sin ! i t1 + cos ! i tSince

sin x1 ! cos x

= cot x2

= tan &2 ! x2we have

+ (t) = tan " 1

h! tan

&2 !

x2

i =

x2 !

&2

ThusyD (t) =

12&

ddt 2&f i t2 ! &2 = 12f i

Finally, for Ai Ac we see from the phasor diagram that

+ (t) " " (t) = ! i tand

yD (t) = K D2&

ddt

(2&f i t) = f i

P roblem 3.47From Example 3.5, m (t) = Au (t). Thus

# (t) = kf Z t Au (( ) d( = Akf t, t %0


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( )t

( )t

( )t f T

Ak

K

0 t

Figure 3.11:

Also" (s) =

AK T kf s2 (s + K T )

Taking the inverse transform gives

" (t) = Akf t + 1K T e" K T t ! 1K T u (t)The phase error is + (t) = # (t) ! " (t). This gives

+ (t) = Akf K T 1 ! e" K T t

u (t)The maximum phase error occurs as t ' ( and is clearly Akf /K T . Thus we require0.2 =

Akf K T

andK T = 5 Akf

The VCO constant is contained in K T . The required sketch follows.

P roblem 3.48For m (t) = A cos ! m t and

# (t) = Akf Z t cos ! m ( d( = Akf ! m sin ! m t


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3.1. PROBLEMS 37

and

# (s) = Akf s2 + ! 2m

The VCO output phase is

" (s) = # (s) K T s + K T

= Akf K T

(s + K T ) ( s2 + ! 2m )

Using partial fraction expansion, " (s) can be expressed as

" (s) = Akf K T K T 2 + ! 2m 1s + K T ! ss2 + ! 2m + K T s2 + ! 2m

This gives, for t %0" (t) = Akf K T

K T 2 + ! 2m e" K T t ! cos ! m t + K T ! m sin ! m tThe rst term is the transient response. For large K T , only the third term is signi cant.Thus,

e& (t) = 1K &

d"dt

= Akf K 2T

K & K 2T + ! 2m cos ! m tAlso, since K T is large, K 2T + ! 2m ! K 2T . This givese& (t) "

Akf K &

cos ! m t

and we see that e& (t) is proportional to m (t). If kf = K & , we have

e& (t) " m (t)

P roblem 3.49The Costas PLL is shown in Figure 3.57. The output of the top multiplier is

m (t)cos ! ct [2cos(! ct + ")] = m (t)cos " + m (t)cos(2! ct + ")

which, after lowpass ltering, is m (t)cos " . The quadrature multiplier output is

m (t)cos ! ct [2sin(! ct + ")] = m (t)sin " + m (t)sin(2 ! ct + ")

which, after lowpass ltering, is m (t)sin " . The multiplication of the lowpass lter outputsis

m (t)cos "m (t)sin " = m2 (t)sin2"


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A

d dt /

Figure 3.12:

( )0e t

( )ve t

PhaseDetector

Loop Filter and Amplifier VCO

BandpassFilter

Figure 3.13:

as indicated. Note that with the assumed input m (t)cos ! ct and VCO output 2 cos (! ct + "),the phase error is " . Thus the VCO is de ned by

d"dt

= d+dt

= ! K & e& (t)This is shown below.

Since the d'dt intersection is on a portion of the curve with negative slope, the point Aat the origin is a stable operating point. Thus the loop locks with zero phase error and zerofrequency error.

P roblem 3.50With x (t) = A2&f 0 t, we desire e0 (t) = A cos2&73f 0 t. Assume that the VCO output is apulse train with frequency 13 f o. The pulse should be narrow so that the seventh harmonicis relatively large. The spectrum of the VCO output consists of components separated by


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3.1. PROBLEMS 39

073

f 013

f

Bandpass filter passband

This component (the fundamental)

tracks the input signal

f

0

Figure 3.14:

13 f 0 with an envelope of sinc(% f ), where % is the pulse width. The center frequency of thebandpass lter is 73 f 0 and the bandwidth is on the order of

13 f 0 as shown.

P roblem 3.51The phase plane is de ned by

+ = $ ! ! K t sin + (t)at + = 0, + = + ss , the steady-state phase error. Thus

+ ss = sin" 1$ !K t = sin " 1 $ !2&(100)

For $ ! = 2& (30)

+ ss = sin" 1 30100 = 17.46 degrees

For $ ! = 2& (50)

+ ss = sin " 1 50100 = 30 degreesFor $ ! = 2& (80)+ ss = sin

" 1 80100 = 53.13 defrees


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For $ ! = ! 2&(80)+ ss = sin " 1! 80100 = ! 53.13 degreesFor $ ! = 2&(120) , there is no stable operating point and the frequency error and the phase

error oscillate (PLL slips cycles continually).

P roblem 3.52From (3.228)

" (s)# (s)

= K t F (s)s + K tF (s)

=K t s+ as+ (

s + K t s+ as+ (which is " (s)

# (s) = K t (s + a)

s (s + $) + K t (s + a) = K t (s + a)

s2 + ( K t + $) s + K t a

Therefores2 + 2 ,! n s + ! 2n = s

2 + ( K t + $) s + K t a

This gives! n = p K t aand, =

K t + $2$ K t a

P roblem 3.53Since ! n = 2&(100) we have

p K t a = 2&(100)orK t a = 4&2 104Since

- = 1$ 2 =

K t + $2$ K ta =

1.1K t2 (2&) (100)

we haveK t = $ 2 (2&) (100)1.1 = 807.8

Thus$ = 0 .1K t = 80.78


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3.1. PROBLEMS 41

( ) dt ( ) PAM t x t PWM ( )

x t 2 ( )

Pulse train

x t 1( )

Figure 3.15:

anda =

4&2104

807.8 = 488.7P roblem 3.54From (3.247), we see that the phase deviation is reduced by 1+ 12# K D K & . The VCO constantis 25 Hz/volt . Thus

K & = 2 & (25) rad/s/volt

we may then writeD1D2

= 50.4

= 12 .5 = 1 + 12&

K D (2&)(25)

which gives1 + 25K D = 12 .5

ThusK D = 0 .46

P roblem 3.55A system converting PWM to PAM can be realized as illustrated in Figure 3.15.The operation should be clear from the waveforms shown in Figure 3.16.The integrator can be realized by using a capacitor since, for a capacitor,

* (t) = 1C Z i (t) dt

Multiplication by the pulse train can be realized by sampling the capacitor voltage. Thus,the simple circuit is as illustrated in Figure 3.17.


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T 2T 3T 4T 0

x t PWM ( )

t

T 2T 3T 4T 0

t 1( )

t

T 2T 3T 4T 0

x t 2 ( )

t

T 2T 3T 4T 0

x t PAM ( )

t

Figure 3.16:


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3.1. PROBLEMS 43

switch closesat t nT =to restartintegration

sampling switch isclosed for nT t nT < B . For 0 x B , the result is (A/b )1 e

bx

;(c) The mean isE [X ] =

1b1 bB e

bB

1 ebB(d) The mean-square is

E X 2

= 2A

b 1b2 ebB

b2 (1 + bB)

AB 2

b ebB

where A must be substituted from part (a.).

(e) For the variance, subtract the result of part (c) squared from the result of part (d).Problem 4.22(a) The integral evaluates as follows:

E X 2n

= Z

x2n

ex2/ 2 2 2 2 dx = 2 Z 0 2 2y

2n ey2 dy =

2n+1 2n

Z 0 y2n eydywhere y =

Principles of Communication 5Ed R. E. Ziemer%2C William H. Tr Solution

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Transcript of Principles of Communication 5Ed R. E. Ziemer%2C William H. Tr Solution