Solution to Chapter 4 Problems

Solution to Chapter 4 Problems Problem 4.1 1) Since F[sinc(400t)]= 1 modulation index 400 ( f 400 β f = k f max[ m(t) ] W Hence, the modulated signal is ), the bandwidth of the message signal is W = 00 and the resulting u(t) = A cos(πf c t + πk f = k f 10 W = 6 k f = 10 = 100 cos(πf c t ++π100 m(τ)dτ) sinc(400τ)dτ) ) The maximum frequency deviation of the modulated signal is f max = β f W = 6 00 = 100 3) Since the modulated signal is essentially a sinusoidal signal with amplitude A = 100,wehave P = A = 5000 4) Using Carson s rule, the effective bandwidth of the modulated signal can be approximated by B c = (β f + 1)W = (6 + 1)00 = 800 Hz Problem 4. 1) The maximum phase deviation of the PM signal is The phase of the FM modulated signal is φ(t) = πk f = φ max = k p max[ m(t) ] = k p m(τ)dτ = πk f m(τ)dτ 0 πk f 0 τdτ = πk f t 0 t<1 πk f + πk f 1 dτ = πk f + πk f (t 1) 1 t< πk f + πk f πk f dτ = 3πk f πk f (t ) t<3 πk f 3 t 8

The maximum value of φ(t) is achieved for t = and is equal to 3πk f. Thus, the desired relation between k p and k f is k p = 3πk f ) The instantaneous frequency for the PM modulated signal is f i (t) = f c + 1 d π dt φ(t) = f c + 1 π k d p dt m(t) For the m(t) given in Fig. P-4., the maximum value of d m(t) is achieved for t in [0, 1] and it is equal to dt one. Hence, max(f i (t)) = f c + 1 π For the FM signal f i (t) = f c + k f m(t). Thus, the maximum instantaneous frequency is max(f i (t)) = f c + k f = f c + 1 Problem 4.3 For an angle modulated signal we have x(t) = A c cos(πf c t + φ(t)), therefore The lowpass equivalent of the signal is x l (t) = A c e jφ(t) with Envelope A c and phase π(t) and in phase an quadrature components A c cos(φ(t)) and A c sin(φ(t)), respectively. Hence we have the following A c envelope A c envelope k p m(t) phase t PM A c cos ( k p m(t) ) πk f m(τ) dτ FM ) phase in-phase comp. A c sin ( k p m(t) ) A c cos (πk f t m(τ) dτ in-phase comp. ) quadrature comp. A c sin (πk f t m(τ) dτ quadrature comp. Problem 4.4 1) Since an angle modulated signal is essentially a sinusoidal signal with constant amplitude, we have P = A c The same result is obtained if we use the expansion P = 100 = 5000 u(t) = A c J n (β) cos(π(f c + nf m )t) along with the identity J0 (β) + Jn (β) = 1 n=1 83

) The maximum phase deviation is φ max = max 4 sin(000πt) =4 3) The instantaneous frequency is f i = f c + 1 d π dt φ(t) = f c + 4 π cos(000πt)000π = f c + 4000 cos(000πt) Hence, the maximum frequency deviation is f max = max f i f c =4000 4) The angle modulated signal can be interpreted both as a PM and an FM signal. It is a PM signal with phase deviation constant k p = 4 and message signal m(t) = sin(000πt) and it is an FM signal with frequency deviation constant k f = 4000 and message signal m(t) = cos(000πt). Problem 4.5 The modulated signal can be written as u(t) = A c J n (β) cos(π(f c + nf m )t) The power in the frequency component f = f c + kf m is P k = A c J n (β). Hence, the power in the carrier is P carrier = A c J 0 (β) and in order to be zero the modulation index β should be one of the roots of J 0(x). The smallest root of J 0 (x) is found from tables to be equal.404. Thus, β min =.404 Problem 4.6 1) If the output of the narrowband FM modulator is, u(t) = A cos(πf 0 t + φ(t)) then the output of the upper frequency multiplier ( n 1 )is u 1 (t) = A cos(πn 1 f 0 t + n 1 φ(t)) After mixing with the output of the second frequency multiplier u (t) = A cos(πn f 0 t)we obtain the signal y(t) = A cos(πn 1 f 0 t + n 1 φ(t))cos(πn f 0 t) = A (cos(π(n 1 + n )f 0 + n 1 φ(t)) + cos(π(n 1 n )f 0 + n 1 φ(t))) 84

The bandwidth of the signal is W = 15 KHz, so the maximum frequency deviation is f = β f W = 0.1 15 = 1.5 KHz. In order to achieve a frequency deviation of f = 75 KHz at the output of the wideband modulator, the frequency multiplier n 1 should be equal to n 1 = f f = 75 1.5 = 50 Using an up-converter the frequency modulated signal is given by y(t) = A cos(π(n 1 + n )f 0 + n 1 φ(t)) Since the carrier frequency f c = (n 1 + n )f 0 is 104 MHz, n should be such that (n 1 + n )100 = 104 10 3 n 1 + n = 1040 or n = 990 ) The maximum allowable drift (d f ) of the 100 khz oscillator should be such that (n 1 + n )d f = d f = =.0019 Hz 1040 Problem 4.7 The modulated PM signal is given by u(t) = A c cos(πf c t + k p m(t)) = A c Re [ e jπfct e ] jk pm(t) = A c Re [ e jπfct e jm(t)] The signal e jm(t) is periodic with period T m = 1 f m and Fourier series expansion c n = 1 Tm e jm(t) e jπnfmt dt T m Hence, and = 1 T m 0 Tm 0 e j e jπnf mt dt + 1 e j = e jπnf mt T m jπnf m = ( 1)n 1 j(e j e j ) = πn e jm(t) = l= u(t) = A c Re [ [ e jπfct e jm(t)] = A c Re = A c l= Tm 0 T m Tm Tm e j e jπnf mt dt e j e jπnf mt T m jπnf m { 0 n = l T m sin(1) n = l + 1 π(l+1) π(l + 1) sin(1)ejπlf mt e jπf ct l= sin(1) π(l + 1) cos(π(f c + lf m )t + φ l ) 85 Tm ] π(l + 1) sin(1)ejπlf mt

where φ l = 0 for l 0 and φ l = π for negative values of l. Problem 4.8 1) The instantaneous frequency is given by f i (t) = f c + 1 d π dt φ(t) = f c + 1 π 100m(t) A plot of f i (t) is given in the next figure f c f i (t)... f c + 500 π...... f c 500 π 0 t ) The peak frequency deviation is given by f max = k f max[ m(t) ] = 100 π 5 = 50 π Problem 4.9 1) The modulation index is β = k f max[ m(t) ] f m = f max 0 103 = = f m 10 4 The modulated signal u(t) has the form u(t) = = A c J n (β) cos(π(f c + nf m )t + φ n ) 100J n () cos(π(10 8 + n10 4 )t + φ n ) The power of the unmodulated carrier signal is P = 100 = 5000. The power in the frequency component f = f c + k10 4 is P fc +kf m = 100 J k () The next table shows the values of J k (), the frequency f c + kf m, the amplitude 100J k () and the power P fc +kf m for various values of k. 86

Index k J k () Frequency Hz Amplitude 100J k () Power P fc +kf m 0.39 10 8.39 50.63 1.5767 10 8 + 10 4 57.67 1663.1.358 10 8 + 10 4 35.8 6.46 3.189 10 8 + 3 10 4 1.89 83.13 4.0340 10 8 + 4 10 4 3.40 5.7785 As it is observed from the table the signal components that have a power level greater than 500 (= 10% of the power of the unmodulated signal) are those with frequencies 10 8 + 10 4 and 10 8 + 10 4. Since J n (β) = J n (β) it is conceivable that the signal components with frequency 108 10 4 and 10 8 10 4 will satisfy the condition of minimum power level. Hence, there are four signal components that have a power of at least 10% of the power of the unmodulated signal. The components with frequencies 10 8 + 10 4, 10 8 10 4 have an amplitude equal to 57.67, whereas the signal components with frequencies 10 8 + 10 4, 10 8 10 4 have an amplitude equal to 35.8. ) Using Carson s rule, the approximate bandwidth of the FM signal is B c = (β + 1)f m = ( + 1)10 4 = 6 10 4 Hz Problem 4.10 1) β p = k p max[ m(t) ] = 1.5 = 3 β f = k f max[ m(t) ] = 3000 = 6 f m 1000 ) Using Carson s rule we obtain B PM = (β p + 1)f m = 8 1000 = 8000 B FM = (β f + 1)f m = 14 1000 = 14000 3) The PM modulated signal can be written as u(t) = AJ n (β p ) cos(π(10 6 + n10 3 )t) The next figure shows the amplitude of the spectrum for positive frequencies and for these components whose frequencies lie in the interval [10 6 4 10 3, 10 6 + 4 10 3 ]. Note that J 0 (3) =.601, J 1 (3) = 0.3391, J (3) = 0.4861, J 3 (3) = 0.3091 and J 4 (3) = 0.130. 87

0 10 6 8 10 3 AJ... (3)... 10 3 AJ 4 (3) In the case of the FM modulated signal u(t) = A cos(πf c t + β f sin(000πt)) = AJ n (6) cos(π(10 6 + n10 3 )t + φ n ) The next figure shows the amplitude of the spectrum for positive frequencies and for these components whose frequencies lie in the interval [10 6 7 10 3, 10 6 7 10 3 ]. The values of J n (6) for n = 0,...,7 are given in the following table. n 0 1 3 4 5 6 7 J n (6).1506 -.767 -.49.1148.3578.361.458.196 10 6... AJ 5 (6) f 14 10 3 4) If the amplitude of m(t) is decreased by a factor of two, then m(t) = cos(π10 3 t) and The bandwidth is determined using Carson s rule as β p = k p max[ m(t) ] = 1.5 β f = k f max[ m(t) ] = 3000 f m 1000 = 3 B PM = (β p + 1)f m = 5 1000 = 5000 B FM = (β f + 1)f m = 8 1000 = 8000 The amplitude spectrum of the PM and FM modulated signals is plotted in the next figure for positive frequencies. Only those frequency components lying in the previous derived bandwidth are plotted. Note that J 0 (1.5) =.5118, J 1 (1.5) =.5579 and J (1.5) =.31. 88

AJ 1 (1.5) AJ (1.5) 10 6 5 10 3 AJ (3) AJ 4 (3) 10 6 8 10 3 5) If the frequency of m(t) is increased by a factor of two, then m(t) = cos(π 10 3 t) and The bandwidth is determined using Carson s rule as β p = k p max[ m(t) ] = 1.5 = 3 β f = k f max[ m(t) ] = 3000 = 3 f m 000 B PM = (β p + 1)f m = 8 000 = 16000 B FM = (β f + 1)f m = 8 000 = 16000 The amplitude spectrum of the PM and FM modulated signals is plotted in the next figure for positive frequencies. Only those frequency components lying in the previous derived bandwidth are plotted. Note that doubling the frequency has no effect on the number of harmonics in the bandwidth of the PM signal, whereas it decreases the number of harmonics in the bandwidth of the FM signal from 14 to 8. 89

10 3 10 6 AJ (3) AJ 4 (3) 16 10 3 Problem 4.11 1) The PM modulated signal is u(t) = 100 cos(πf c t + π cos(π1000t)) = ( π ) 100J n cos(π(10 8 + n10 3 )t) The next table tabulates J n (β) for β = π and n = 0,...,4. n 0 1 3 4 J n (β).470.5668.497.0690.0140 The total power of the modulated signal is P tot = 100 = 5000. To find the effective bandwidth of the signal we calculate the index k such that k n= k 100 ( π ) J n 0.99 5000 k n= k J n ( π ) 0.99 By trial end error we find that the smallest index k is. Hence the effective bandwidth is B eff = 4 10 3 = 4000 In the the next figure we sketch the magnitude spectrum for the positive frequencies. 90

10 3 100 J 1( π ) 10 8 ) Using Carson s rule, the approximate bandwidth of the PM signal is B PM = (β p + 1)f m = ( π + 1)1000 = 5141.6 As it is observed, Carson s rule overestimates the effective bandwidth allowing in this way some margin for the missing harmonics. Problem 4.1 1) Assuming that u(t) is an FM signal it can be written as u(t) = 100 cos(πf c t + πk f α cos(πf m τ)dτ) = 100 cos(πf c t + k f α f m sin(πf m t)) Thus, the modulation index is β f = k f α f m = 4 and the bandwidth of the transmitted signal B FM = (β f + 1)f m = 10 KHz ) If we double the frequency, then u(t) = 100 cos(πf c t + 4 sin(πf m t)) Using the same argument as before we find that β f = 4 and B FM = (β f + 1)f m = 0 KHz 3) If the signal u(t) is PM modulated, then β p = φ max = max[4 sin(πf m t)]=4 The bandwidth of the modulated signal is B PM = (β p + 1)f m = 10 KHz 4) If f m is doubled, then β p = φ max remains unchanged whereas B PM = (β p + 1)f m = 0 KHz 91

Problem 4.13 1) If the signal m(t) = m 1 (t) + m (t) DSB modulates the carrier A c cos(πf c t) the result is the signal u(t) = A c m(t) cos(πf c t) = A c (m 1 (t) + m (t)) cos(πf c t) = A c m 1 (t) cos(πf c t) + A c m (t) cos(πf c t) = u 1 (t) + u (t) where u 1 (t) and u (t) are the DSB modulated signals corresponding to the message signals m 1 (t) and m (t). Hence, AM modulation satisfies the superposition principle. ) If m(t) frequency modulates a carrier A c cos(πf c t) the result is u(t) = A c cos(πf c t + πk f (m 1 (τ) + m (τ))dτ) = A c cos(πf c t + πk f m 1 (τ)dτ) +A c cos(πf c t + πk f m (τ)dτ) = u 1 (t) + u (t) where the inequality follows from the nonlinearity of the cosine function. Hence, angle modulation is not a linear modulation method. Problem 4.14 The transfer function of the FM discriminator is Thus, H(s) = R R + Ls + 1 Cs = R L s s + R L s + 1 LC H(f) = 4π ( ) R L f ( 1 LC 4π f ) + 4π ( R L ) f As it is observed H(f) 1 with equality if 1 f = π LC Since this filter is to be used as a slope detector, we require that the frequency content of the signal, which is [80 6, 80 + 6] MHz, to fall inside the region over which H(f) is almost linear. Such a region can be considered the interval [f 10,f 90 ], where f 10 is the frequency such that H(f 10 ) =10% max[ H(f) ] and f 90 is the frequency such that H(f 10 ) =90% max[ H(f) ]. 9

With max[ H(f) =1, f 10 = 74 10 6 and f 90 = 86 10 6, we obtain the system of equations 4π f 10 4π f 90 Solving this system, we obtain 50 103 + πf 10 [1 0.1 ] 1 1 L LC = 0 50 103 + πf 90 [1 0.9 ] 1 1 L LC = 0 L = 14.98 mh C = 0.018013 pf Problem 4.15 The case of φ(t) = β cos(πf m t) has been treated in the text, the modulated signal is u(t) = = A c J n (β) cos(π(f c + nf m )) 100J n (5) cos(π(10 3 + n10)) The following table shows the values of J n (5) for n = 0,...,5. n 0 1 3 4 5 J n (5) -.178 -.38.047.365.391.61 In the next figure we plot the magnitude and the phase spectrum for frequencies in the range [950, 1050] Hz. Note that J n (β) = J n (β) if n is even and J n (β) = J n (β) if n is odd. U(f) 950... 1000 1050 100 J 4 (5) U(f) π... 950 1000 1050 93

The Fourier Series expansion of e jβ sin(πf mt) is c n = f m 5 4fm = 1 π 1 4fm π 0 = e j nπ Jn (β) e jβ sin(πf mt) e jπnf mt dt e jβ cos u jnu e j nπ du Hence, [ ] u(t) = A c Re c n e jπfct e jπnf mt = A c Re [ e jπ(f c+nf m )t+ nπ ] The magnitude and the phase spectra of u(t) for β = 5 and frequencies in the interval [950, 1000] Hz are shown in the next figure. Note that the phase spectrum has been plotted modulo π in the interval ( π, π]. U(f) U(f) π 950...... 1000 1050 100 J 4 (5) π π....... 950 1000 1050 Problem 4.16 The frequency deviation is given by f d (t) = f i (t) f c = k f m(t) whereas the phase deviation is obtained from φ d (t) = πk f m(τ)dτ In the next figure we plot the frequency and the phase deviation when m(t) is as in Fig. P-4.16 with k f = 5. 94

f d (t) 50... 5... 1 3 4 5 6 5 50... t φ d (t) 50π... 5π............... 5 1 3 4 6 5π................. t Problem 4.17 Using Carson s rule we obtain B c = (β + 1)W = ( k f max[ m(t) ] W 000 k f = 10 + 1)W = 000 k f = 100 000 k f = 1000 Problem 4.18 The modulation index is β = k f max[ m(t) ] 10 10 = = 1.5 f m 8 The output of the FM modulator can be written as u(t) = 10 cos(π000t + πk f = 10 cos(π8τ)dτ) 10J n (1.5) cos(π(000 + n8)t + φ n ) At the output of the BPF only the signal components with frequencies in the interval [000 3, 000 + 3] will be present. These components are the terms of u(t) for which n = 4,...,4. The power of the output signal is then 10 4 J 0 (1.5) + 10 J n (1.5) = 50 0.630 = 13.15 n=1 Since the total transmitted power is P tot = 10 = 50, the power at the output of the bandpass filter is only 6.30% of the transmitted power. Problem 4.19 95

1) The instantaneous frequency is f i (t) = f c + k f m 1 (t) The maximum of f i (t) is max[f i (t)] =max[f c + k f m 1 (t)] =10 6 + 5 10 5 = 1.5 MHz ) The phase of the PM modulated signal is φ(t) = k p m 1 (t) and the instantaneous frequency f i (t) = f c + 1 π d dt φ(t) = f c + k p π d dt m 1(t) The maximum of f i (t) is achieved for t in [0, 1] where d dt m 1(t) = 1. Hence, max[f i (t)] =10 6 + 3 π. 3) The maximum value of m (t) = sinc( 10 4 t) is 1 and it is achieved for t = 0. Hence, Since, F[sinc( 10 4 t)]= 1 Carson s rule, we obtain max[f i (t)] =max[f c + k f m (t)] =10 6 + 10 3 = 1.001 MHz 10 4 ( f 10 4 ) the bandwidth of the message signal is W = 10 4. Thus, using B = ( k f max[ m(t) ] W + 1)W = KHz Problem 4.0 Since 88 MHz <f c < 108 MHz and f c f c =f IF if f IF <f LO we conclude that in order for the image frequency f c to fall outside the interval [88, 108] MHZ, the minimum frequency f IF is such that f IF = 108 88 f IF = 10 MHz If f IF = 10 MHz, then the range of f LO is [88 + 10, 108 + 10] =[98, 118] MHz. 96