Chap6StudentSolutions

X tn yt n xrn cos2fc n yd n lpf yf n cos2fc

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Unformatted text preview: pulse train, p( t) = is modulated by a signal, 1 t rect ∗ 4 comb( 4 t) w w x( t) = sinc( t) . Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response for (a) and w = 10 ms (b) w = 1 ms . This is similar to previous modulation exercises. Solutions 6-31 M. J. Roberts - 8/16/04 n t− 1 4 y( t) = sinc( t) ∑ rect w n =−∞ w ∞ ∞ Y( f ) = 4 ∑ sinc( 4 wk ) rect ( f − 4 k ) k =−∞ PAM Modulator Response (a) |Y( f )| x(t) and y(t) 4 100 -200 200 f Phase of Y( f ) -3 3 π t -200 -50 200 f -π Because of the scale, it is difficult to see what is really happening in the magnitude plot of the transform of the response. It consists of a large number of closely-spaced impulses. 1 n and the cutoff DT frequency of the 39. In the system below, x t [ n ] = sinc , Fc = 20 4 1 lowpass filter is . Plot the signals, x t [ n ] , y t [ n ] , y d [ n ] and y f [ n ] and the magnitudes 20 and phases of their DTFT’s. x t[n] yt [n] = xr[n] cos(2πFc n) yd [n] LPF yf [n] cos(2πFc n) This is a DT modulation system. The analysis is very similar to that used for CT modulation systems. 40. Repeat Exercise 22 but with an excitation, x( t) = rect (1000 t) ∗ 20 comb(20 t) . x sh ( t) = [rect (1000 t) ∗ 20 comb(20 t)] cos(2πf c t) Solutions 6-32 M. J. Roberts - 8/16/04 ∞ n x sh ( t) = cos(2πf c t) ∑ rect 1000 t − 20 n =−∞ 1 f f 1 X sh ( f ) = sinc comb ∗ δ ( f − f c ) + δ ( f + f c ) 1000 20 2 1000 [ X sh ( f ) = ∞ 1 f sinc ∑ δ ( f − 20 k ) ∗ δ ( f − f c ) + δ ( f + f c ) 1000 k =−∞ 8 [ 1∞ k X sh ( f ) = ∑ sinc δ ( f − f c − 20 k ) + δ ( f + f c − 20 k ) 4 8 k =−∞ [ 1 ∞ k f Y( f ) = ∑ sinc δ ( f − f c − 20 k ) + δ ( f + f c − 20 k ) rect 4 2B 8 k =−∞ [ ∞ f k rect ∑ sinc δ ( f − f c − 20 k ) 2 B k =−∞ 4 1 Y( f ) = ∞ 8 f k + rect ∑ sinc δ ( f + f c − 20 k ) 2 B k =−∞ 4 1 k Y( f ) = ∑ sinc δ ( f − f c − 20 k ) + 4 8 f c + 20 k < B Py = 1 2 k ∑ sinc + 4 64 f c + 20 k < B k sinc δ ( f +...
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