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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap08

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap08

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Unformatted text preview: Chapter 8: Case Studies for CT Systems Problem 8.1 : (a) The AM signal is given by ) 2 cos( )] 2 cos( 2 ) 2 sin( 3 1 [ ) ( 2 1 t f t f k t f k A t s c π π + π + = . To ensure that the envelope of s ( t ) ≥ 0 for all t )) 2 cos( 2 ) 2 sin( 3 1 ( 2 1 ≥ π + π + t f k t f k . Taking the worst case scenario, i.e., both sin(2 π f 1 t ) and cos(2 π f 2 t ) take their minimum values of − 1 at the same time, we get ) 2 3 1 ( ≥ − − k k or, 2 . ≤ k . (b) Expressing ) 2 cos( )] 2 cos( 2 ) 2 sin( 3 [ ) 2 cos( ) ( 2 1 t f t f k t f k t f A t s c c π π + π + π = , or, ). ) ( 2 cos( ) ) ( 2 cos( ) ) ( 2 sin( ) ) ( 2 sin( ) 2 cos( ) ( 2 2 1 2 3 1 2 3 t f f Ak t f f Ak t f f t f f t f A t s c c c Ak c Ak c − π + + π + − π + + π + π = Assuming f 1 ≠ f 2 ≠ f c , the power in the modulated signal is composed of 2 2 1 carrier in the Power A = and 2 4 13 2 ) ( 2 ) 2 / 3 ( ) ( 2 2 signal modulating in the Power 2 2 Ak Ak Ak = × + × = . Hence, the ratio of power lost in the carrier and the total power = 2 5 . 6 1 1 k + . (c) The power spectrum of s ( t ) is given by [ ] [ ] [ ] [ ] [ ] . ) ( 2 ( ) ( 2 ( ) ( 2 ( ) ( 2 ( ) ( 2 ( ) ( 2 ( ) ( 2 ( ) ( 2 ( ) 2 ( ) 2 ( ) ( 2 2 2 2 1 1 2 3 1 1 2 3 f f f f Ak f f f f Ak f f f f j f f f f j f f A f S c c c c c c Ak c c Ak c c − π − ω δ + − π + ω δ π + + π − ω δ + + π + ω δ π + − π − ω δ − − π + ω δ + + π − ω δ − + π + ω δ + π + ω δ + π − ω δ π = π π For f 1 = 10 kHz, f 2 = 20 kHz, and f c = 50 kHz, the power spectrum is shown in Fig. S8.1 (a). (d) Signal x ( t ) can be reconstructed using the synchronous detector shown in Fig. S8.1(b). The information signal x ( t ) can be extracted from the output of the above system by using an amplifier with a gain of 2/ A 2 and removing the dc offset. ▌ 2 Chapter 8 kradians/s in ω ) ( ω S 10 20 30 40 50 60 70 80 90 (×2 π ) − 10 − 20 − 30 − 40 − 50 − 60 − 70 − 80 − 90 π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak kradians/s in ω ) ( ω S 10 20 30 40 50 60 70 80 90 (×2 π ) − 10 − 20 − 30 − 40 − 50 − 60 − 70 − 80 − 90 π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak π A π Ak 5 . 1 π Ak π Ak 5 . 1 π Ak (a) x ( t ) Lowpass filter with cutoff frequency of 50kHz A cos(2 π f c t ) with f c = 50kHz modulated signal s ( t ) + 0.5 A 2( 1 + kx ( t )) x ( t ) Lowpass filter with cutoff frequency of 50kHz A cos(2 π f c t ) with f c = 50kHz modulated signal s ( t ) modulated signal s ( t ) + 0.5 A 2( 1 + kx ( t )) (b) Figure S8.1: (a) Spectrum for the AM modulated signal s ( t ) in Problem 8.1(c). (b) synchronous detector in Problem 8.1(d)....
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Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap08

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