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Harold Eaton
6 years ago
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1 : Introduction to Communication Systems Spring 2017 Solution to Homework Assignment #5: 1 For a message signal m t = 2 cos 1000t + 9 cos 2000πt 1-a Write expressions (do not sketch for φ /0 t and φ 10 t when A = 10 ω 4 = 10 5 k 7 = 1000π and k 8 = 1. For determining φ 10 t use the indefinite integral of m t ; that is take the value of the integral at t = to be 0. (6 points ( ò- t 6 ( t pò ( a ( pa da - t ( = cos + ( j t A w t k m a da FM c f = 10cos cos cos 2000 æ 6 é 2 9 ùö = 10 cosç 10 t p ê- sin ( 1 000t - sin ( 2 000pt p ú è ë ûø 6 ( t p ( t ( pt = 10 cos 10-2 sin sin j ( ( 6 t ( t ( pt ( t = Acos w t+ k m( t PM c p = 10cos cos cos b Estimate the bandwidth of φ /0 t and φ 10 t. (10 points ( ( ( p mt = 2cos 1000t + 9cos 2000 t Þ m p = 11 ìb= 2000p 2p = 1kHz \ í Þ BFM = 2( D f + B = 2( = 13kHz î D f = kfmp 2p = 5.5kHz ( ( ( mt =-2000sin 1000 t psin 2000pt Þ m p = p ìb= 2000p 2p = 1kHz æ 1 ö \ í Þ BPM = 2( D f + B = » 20.64kHz f kpmp 2p 1p 9kHz ç î D = = + èp ø

2 2 An angle-modulated signal with carrier frequency ω 4 = 2π 10 5 is described by the equation φ <0 t = 10 cos ω 4 t sin 2000πt 2-a Find the power of the modulated signal. (2 points P <0 = AB C = 50 Watts 2-b Find the frequency deviation f. (5 points φ t = ω 4 t sin 2000πt ω t = GH I GI f t = ω t = ω π cos 2000πt rad/s 2π = f 4 + CJJK f = 100 Hz 2-c Find the phase deviation φ. (2 points φ t = ω 4 t sin 2000πt φ = 0.1 rad 2-d Estimate the bandwidth of φ <0 t. (5 points Baseband signal bandwidth: B = CJJJK = 1000 Hz Therefore: B <0 = 2 B + f = = 2.2 khz 3 Design (the block diagram of an Armstrong indirect FM modulator to generate an FM carrier with a carrier frequency of 96 MHz and Df = 20 khz. A narrowband FM generator with f c = 200 khz and adjustable Df in the range of 9 to 19 Hz is available. The stockroom also has an oscillator with adjustable frequency in the range of 9 to 10 MHz. There are bandpass filters with any center frequency and only frequency doublers are available. (20 points f c1 =200 khz Df c1 =9 to 19 Hz M 1=2 m1 f c2 Df c2 fc3 f c3 Df c3 f c4 =96 MHz M2=2m2 Df c4 =20 khz Crystal Oscillator 9 to 10 MHz First note that given the available components a one-stage modulator design is not possible (show. So we move to the two-state modulator. Total required number of multiplier can be obtained as follow: M = f 4 O f 4P f 4 O f 4P QRS M f 4 O f 4PQUV M

3 2 Y = M = M Z M Z = 2 Y P 2 Y B Y P\Y B ^_`B m = m Z + m C Therefore total number of required doublers is 11 and f 4P = 7 bo = CJ ZJc 9.77 Hz. 0 C PP f 4B = f 4P 2 Y P f 4O = f 4c 2 Y B 1. f 4c = f 4B + f fg f 4O = f 4P 2 Y P + f fg 2 Y B = 2 Y P\Y B f 4P + 2 Y Bf fg 96 = 2 ZZ Y Bf fg = 2 Y Bf fg Multiplication of two positive terms cannot be negative. 2. f 4c = f 4B f fg f 4O = f 4P 2 Y P f fg 2 Y B = 2 Y P\Y B f 4P 2 Y Bf fg 96 = 2 ZZ Y Bf fg = 2 Y Bf fg = 2 Y Bf fg hzh.5 2 Y B hzh Y B ^_`B 4.97 m C 5.12 ü. Thereby m C = 5 f fg = hzh.5 C l 9.8 MHz and m Z = f 4c = f fg f 4B f 4O = f fg f 4P 2 Y P 2 Y B = 2 Y Bf fg 2 Y P\Y B f 4P 96 = 2 Y Bf fg 2 ZZ = 2 Y Bf fg = 2 Y Bf fg ojo.5 2 Y B ojo Y B ^_`B 5.66 m C 5.81 f c1 =200 khz Df c1 =9.77 Hz M 1=64=2 6 f c2 =12.8 MHz Df c2 =625 Hz Mixer f c2 - f f c3 f c3 =3 MHz Df c3 =625 Hz f c4 =96 MHz M 2=32=2 5 Df c4 =20 khz Crystal Oscillator flo=9.8 MHz 4 Design an Armstrong indirect FM modulator in block diagram to generate an FM signal with carrier 96.3 MHz and Df = khz. A narrowband FM generator with f c = 150 khz and Df = 10 Hz is available. Only a limited number of frequency doublers are available as frequency multipliers. In addition an oscillator with adjustable frequency from 13 to 14 MHz is also available for mixing along with bandpass filters of any specification. (20 points

4 f c1 =150 khz Df c1 =10 Hz M 1 =2 m1 f c2 Df c2 fc3 f c3 Df c3 f c4 =96.3 MHz M2=2m2 Df c4 =20.48 khz Crystal Oscillator 13 to 14 MHz First note that given the available components a one-stage modulator design is not possible (show. So we move to the two-state modulator. Total required number of multiplier is: M = f 4 O = h = 2048 = 2 ZZ f 4P 10 2 Y = M = M Z M Z = 2 Y P 2 Y B Therefore total number of required doublers is 11. f 4B = f 4P 2 Y P f 4O = f 4c 2 Y B 1. f 4c = f 4B + f fg f 4O = f 4P 2 Y P + f fg 2 Y B = 2 Y P\Y B f 4P + 2 Y Bf fg 96.3 = 2 ZZ Y Bf fg = 2 Y Bf fg Multiplication of two positive terms cannot become negative. 2. f 4c = f 4B f fg f 4O = f 4P 2 Y P f fg 2 Y B = 2 Y P\Y B f 4P 2 Y Bf fg 96.3 = 2 ZZ Y Bf fg = 2 Y Bf fg = 2 Y Bf p4 CZJ.q 2 Y B CZJ.q Y B ^_`B 3.91 m C 4.02 ü. Thereby m C = 4 f fg = CZJ.q C O = MHz and m Z = f 4c = f fg f 4B f 4O = f fg f 4P 2 Y P 2 Y B = 2 Y Bf fg 2 Y P\Y B f 4P 96.3 = 2 Y Bf fg 2 ZZ = 2 Y Bf fg = 2 Y Bf fg rjh.o 2 Y B rjh.o Y B ^_`B 4.85 m C 4.96 f c1 =150 khz Df c1 =10 Hz M 1 =128=2 7 f c2 =19.2 MHz Df c2 =1280 Hz Mixer f c2 - f f c3 f c3 =6.02 MHz Df c3 =1280 Hz f c4 =96.3 MHz M 2 =16=2 4 Df c4 =20.48 khz Crystal Oscillator f LO=13.18 MHz

5 5 Let s t be an angle-modulated signal that receiver obtains s t = 2 cos 10 t πt + 2 sin 2000πt + 0.3π 3π cos 100t 5-a Find the bandwidth of this FM signal. (5 points The baseband bandwidth of this angel-modulated signal is = CJJJK φ v t = 10 t πt + 2 sin 2000πt + 0.3π 3π cos 100t ω v t = GH U I GI = 1 KHz. Also = 10 t π π cos 2000πt + 0.3π + 300π sin 100t Therefore ω = 4300π and f = rhjjk B 10 = 2 B + f = 6.3 khz. = 2.15 khz. Thus 5-b If s t is sent to an (ideal envelope detector find the detector output signal. (5 points Since s t is an angel-modulated signal with a constant amplitude of 2 the output of an ideal envelope detector would be just constant. 5-c If s t is first differentiated before the envelope detector find the detector output signal. (5 points s t = t π π cos 2000πt + 0.3π + 300π sin 100t sin 10 t πt + 2 sin 2000πt + 0.3π 3π cos 100t Thus the output of the ideal envelope detector would be 2 10 t π π cos 2000πt + 0.3π + 300π sin 100t 5-d Explain which detector output can be processed to yield the message signal m t and find the message signal m t if k 7 = 200π. (5 points Clearly it is necessary to first differentiate the signal to obtain the message. If k 7 = 200π then m t = 4000π cos 2000πt + 0.3π + 300π sin 100t 200π = 20π cos 2000πt + 0.3π sin 100t

6 6 A transmitter transmits an AM signal with a carrier frequency of 1530 khz. When an inexpensive radio receiver (which has a poor selectivity in its RF-stage bandpass filter is tuned to 1530 khz the signal is heard loud and clear. This same single is also heard (not as strongly at another dial setting. State with reasons at what frequency you will hear this station. The IF frequency is 455 khz. (10 points Note that the image station is always 2f 1 = 910 khz away from the target station. The current station at f 4 = 1530 khz is also heard when we tune to a different station with carrier frequency f 4. So the current station should be the image station for the station at f 4. This means that at the carrier frequency of f 4 = = 620 khz (and f fg = f 4 + f 1 = 1075 khz f 4 will be also heard owing to a poor RF-stage bandpass filter.

Problems from the 3 rd edition

(2.1-1) Find the energies of the signals: a) sin t, 0 t π b) sin t, 0 t π c) 2 sin t, 0 t π d) sin (t-2π), 2π t 4π Problems from the 3 rd edition Comment on the effect on energy of sign change, time shifting