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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 627 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 628 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.
 Spring '09
 ATOUSA

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