EECS Signals & Systems University of California, Berkeley: Fall 7 Ramchandran November, 7 Homework 8 Solutions Problem OWN 8.47 (Effects from loss of synchronization.) In this problem we assume that c > M and π > c + M. Let G c (e j ) represent the Fourier transform of cos( c n+θ c ) and G d (e j ) represent the Fourier transform of cos( d n+θ d ), which are shown below. G c (e j ) G d (e j ) (πe jθc ) (πe jθc ) (πe jθ d ) (πe jθ d ) c c d d Y (e j ) = X(e j( θ) )G c (e jθ )dθ π π Y (e j ) e jθc ejθc π c c π W(e j ) = Y (e j( θ) )G d (e jθ )dθ π π (a) If =, then d = c. Therefore, W(e j ) is as shown below. π c W(e j ) cos(θ d θ c ) 4 e j(θc+θ d) 4 ej(θc+θ d) M c π (b) If we pass W(e j ) from Figure through the LPF R(e j ) = cos(θ d θ c )X(e j ) = cos( θ)x(e j ) and r[n] = cos( θ)x[n]. If θ = π/, then r[n] =. (c) In this case, W(e j ) is as shown below. If w > M +, then R(e j ) = X(ej( ) ) + X(ej(+ ) ) for π < π. Therefore, r[n] = x[n] cos( n).
W(e j ) 4 e jθc 4 4 4 ejθc π ( d + c) d + c π Problem OWN Problem 8.6. (Asynchronous demodulation.) First let s solve for the Fourier transform of y(t). Y (j) = ejθc X(j( c )) + e jθc X(j( + c )) + Aπ ( e jθc δ( c ) + e jθc δ( + c ) ) Let s define w (t) to be the output after y(t) is multiplied by cos( c t) and w (t) to be the output after y(t) is multiplied by sin( c t). Also, let z (t) be the output after w (t) is passed through the low-pass filter (LPF) and z (t) be the output after w (t) is passed through the LPF. W (j) = Y (j( c)) + Y (j( + c)) = 4 ejθc( X(j( c )) + Aπδ( c ) ) + + e 4 (e jθc jθc ) ( X(j) + Aπδ() ) + 4 e jθc( X(j( + c )) + Aπδ( + c ) ) W (j) = j Y (j( c)) j Y (j( + c)) = 4j ejθc( X(j( c )) + Aπδ( c ) ) + e 4j (e jθc jθc ) ( X(j) + Aπδ() ) 4j e jθc( X(j( + c )) + Aπδ( + c ) ) Assuming the LPF has a gain of, Z (j) = (e jθc + e jθc ) ( X(j) + Aπδ() ) = cos(θ c ) ( X(j) + Aπδ() ) Z (j) = e j (e jθc jθc ) ( X(j) + Aπδ() ) = sin(θ c ) ( X(j) + Aπδ() ) Thus, z (t) = cos(θ c ) ( x(t) + A ) z (t) = sin(θ c ) ( x(t) + A )
Figure : OWN Problem 8.9 r(t) = = z (t) + z (t) (x(t) + A) cos θ c + (x(t) + A) sin θ c = x(t) + A Problem 3 OWN Problem 8.9. (Single-sideband amplitude modulation.) (a) The sketches in the Figure show S(j) and R(j). (b) In Figure we show how P(j) may be obtained by considering the outputs of the various stages of Figure P8.8(c). From the sketch for P(j), it is clear that P(j) = S(j). (c) In Figure we show the results of demodulation on both s(t) and r(t). It is clear that x(t) is recovered in both cases. Problem 4 OWN Problem 8.4 (Quadrature multiplexing.) If we approach this problem analytically in the time domain, we see that: r(t) = x (t)cos( c t) + x (t)sin( c t) R(j) = X (j( c )) + X (j( + c )) + j X (j( c )) j X (j( + c )) 3
Let z (t) = r(t)cos( c t) and z (t) = r(t)sin( c t). z (t) = r(t)cos( c t) = x (t)cos ( c t) + x (t)sin( c t)cos( c t) + cos(c t) sin(c t) + sin() = x (t) + x (t) + cos(c t) sin(c t = x (t) + x (t) ) z (t) = r(t)sin( c t) Thus, after the LPF with gain. = x (t)cos( c t)sin( c t) + x (t)sin ( c t) sin(c t) + sin() cos(c t) = x (t) + x (t) sin(c t) cos(c t = x (t) + x (t) ) y (t) = x (t) y (t) = x (t) Let s approach this problem graphically now and in the frequency domain. Let X (j) and X (j) be as shown in Figure. Then R(j) is as shown in Figure. The overlapping regions in the figure need to be summed. When r(t) is multiplied by cos c t, in the vicinity of = we get ( X (j) + j X (j) + X (j) j ) X (j) = X (j). Therefore the first lowpass filter output is equal to x (t). When r(t) is multiplied by sin c t, in the vicinity of = we get ( j X (j) + j ( X (j) X (j) j )) X (j) Therefore the second lowpass filter output is equal to x (t). Problem 5 OWN Problem 8.3 (Intersymbol spacing.) = X (j ). (a) p() = π = π = π π T π T [ P(j)d + sin ( + cos( T ) ) d ( T T )] π T π T = T 4
Figure : OWN Problem 8.4 (b) Since P(j) satisfies eq/ (8.8), we know that it must have zero-crossings every T. Therefore, p(kt ) =, for k = ±, ±, Problem 6 (Signal transmission system.) (a) c > z + x. This is because we need to avoid aliasing. Q(j) B A - c c (b) C = and c z > f > x. R(j) A B A 4 - c c c c Problem 7 OWN Problem 8.39 (FSK.) (a) There are two possible cases. Case : b(t) = m (t). D = cos ( t)dt cos( t)cos( t)dt 5
Case : b(t) = m (t). D = cos ( t)dt cos( t)cos( t)dt Both D and D are maximum when cos( t)cos( t)dt =. (b) cos( t)cos( t)dt = = (cos(( + )t) + cos(( )t)) dt [ sin(( + )t) ( + ) = sin(( + )T) ( + ) + sin(( )t) ( ) + sin(( )T) ( ) ] T Thus for any choice of and,, we can always find T so that cos( t)cos( t)dt =. 6