Chap6StudentSolutions

# E 2 just as there is no phase shift in the frequency

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Unformatted text preview: y 2 ( t) = y1 (− t) ∗ h( t) in the time domain or Y2 ( jω ) = Y1* ( jω ) H( jω ) in the frequency domain. (This uses the CTFT property, F * g ( −t ) ← G * ( jω ) .) The final signal is y 2 (− t) in the time domain and Y2 ( jω ) in the → frequency domain. Therefore the final signal in the frequency domain is [ * Y2 ( jω ) = Y1* ( jω ) H( jω ) = Y1 ( jω ) H* ( jω ) = X( jω ) H( jω ) H* ( jω ) * and in the time domain this is y 2 ( t) = x( t) ∗ h( t) ∗ h(− t) . So the effective transfer function is Solutions 6-27 M. J. Roberts - 8/16/04 1 H( jω ) H* ( jω ) = and, using e − a t ←F → 1+ j 1 ω ω 1− j ωc ωc = 1 ω 1+ ωc 2 2a , the effective impulse response is a + ω2 2 h( t) ∗ h(− t) = ω c e −ω c t u( t) ∗ ω c eω c t u(− t) = ω c −ω c t . e 2 Just as there is no phase shift in the frequency domain, the effective impulse response, h(− t) ∗ h( t) is symmetrical about t = 0 which also means it creates no time delay when it is convolved with the excitation. 35. Repeat Exercise 18 but with the second cos(2πf c t) replaced by sin(2πf c t) . Similar to Exercise 18. 36. In the system below, x t ( t) = sinc( t) , f c = 10 and the cutoff frequency of the lowpass filter is 1 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases of their CTFT’s. This is a single sideband system. Analysis is similar to several previous communication system exercises. The biggest difference is the addition of the filter which removes a sideband before transmission. yt (t) = x r(t) |H( f )| x t(t) fm -fc fm fc yd (t) f LPF cos(2πfct) cos(2πfct) Yt ( f ) = 1 f + 10.25 f − 10.25 rect 0.5 + rect 0.5 2 1 t y t ( t) = sinc cos(20.5πt) 2 2 1 t y d ( t) = sinc cos(20.5πt) cos(20πt) 2 2 Yd ( f ) = 1 rect (2( f − 20.25)) + rect (2( f + 0.25)) 4 + rect (2( f − 0.25)) + rect (2( f + 20.25)) Solutions 6-28 yf (t) M. J. Roberts - 8/16/04 Demodulated Carrier |Yd( f )| yd(t) 0.5 0.5 -20 20 f Phase of Yd( f ) -4 4 π t -20 20 -0.25 f -π Demodulated and Filtered Carrier |Yf( f )| yf(t) 0.25 0.25 -2 2 f Phase of Yf( f ) -4 4 π t -2 -0.125 2 f -π 37. A quadrature modulator modulates a sine carrier, sin(20πt) , with a signal, x1 (t ) = sinc( t) , and a cosine carrier, cos(20πt) , with a signal, x 2 (t ) = rect ( t) . The quadrature π demodulator has a phase error making its local oscillators be sin 20πt − and 6 π cos 20πt − . Plot the two demodulated and filtered signals, x1...
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## This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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