Chap3StudentSolutions

Change the pulse duration to 02 s and the pulse

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Unformatted text preview: 8/16/04 ∞ m n 7 7 y[ n ] = h[ n ] ∗ x[ n ] = ∑ m u[ m] u[ n − m] = ∑ m 8 8 m =−∞ m =0 m 1 − r N , r ≠1 with respect to r, Differentiating ∑ r n = 1 − r n =0 N , r =1 N −1 N −1 ∑ nr n −1 = n =0 N −1 r ∑ nr n =0 N −1 (1 − r)(− Nr N −1) − (1 − r N )(−1) , r ≠1 (1 − r) 2 n −1 − Nr N −1 + Nr N + 1 − r N =r , r ≠1 (1 − r) 2 ∑ nr n = r n =0 Nr N −1 ( r − 1) + 1 − r N , r ≠1 (1 − r) 2 7 (n + 1) 8 7 y[ n ] = 8 n 7 7 − 1 + 1 − 8 8 1 − 7 8 n +1 2 u[ n ] n n +1 7 1 7 y[ n ] = 56 ( n + 1) − + 1 − u[ n ] 8 8 8 7 n n y[ n ] = 56 1 − + 1 u[ n ] 8 8 x[n] Excitation 1 -5 60 h[n] n Impulse Response 3 -5 60 y[n] n Response 50 -5 60 Solutions 3-35 n M. J. Roberts - 8/16/04 n (b) x[ n ] = u[ n ] , 4 3 h[ n ] = δ [ n ] − − u[ n ] 4 7 50. A CT function is non-zero over a range of its argument from 0 to 4. It is convolved with a function which is non-zero over a range of its argument from -3 to -1. What is the non-zero range of the convolution of the two? Imagine any two functions with finite non-zero width and convolve. 51. What function convolved with −2cos( t) would produce 6sin( t) ? Think of a sine as a shifted cosine. There are multiple correct answers to this exercise. 52. Sketch these functions. (a) 1 1 g( t) = 3 cos(10πt) ∗ 4δ t + = 12 cos10π t + = 12 cos(10πt + π ) = −12 cos(10πt) 10 10 g(t) 12 -0.5 0.5 t -12 (b) g( t) = tri(2 t) ∗ comb( t) t (c) g( t) = [tri(2 t) − rect ( t − 1)] ∗ comb 2 t t (d) g( t) = tri comb( t) ∗ comb 8 4 1 t (e) g ( t ) = sinc ( 4 t ) ∗ comb 2 2 The result should look like a Dirichlet function. It is a Dirichlet function written in a different form. (f) g( t) = e −2 t u( t) ∗ (g) 1 t − 2 t comb 4 − comb 4 4 This result looks like a full-wave rectified sinusoid. 1 t t (h) g( t) = sinc(2 t) ∗ comb rect 2 4 2 Solutions 3-36 M. J. Roberts - 8/16/04 53. Find the signal power of these signals. (a) t x( t) = rect ( t) ∗ comb 4 ∞ t rect ( t) ∗ comb = 4 ∑ rect ( t − 4 n ) 4 n =−∞ This is a periodic signal whose period, T, is 4. Between -T/2 and +T/2, there is one rectangle whose height is 4 and whose width is 1. Therefore, between -T/2 and +T/2, the square of the signal is [4 rect (t)] (b) 2 T 2 1 = 16 rect ( t) and P = T 2 16 2 ∫T 16 rect (t)dt = 4 −∫ rect (t)dt = 4 2 2 2 − t x( t) = tri( t) ∗ comb 4 2 Remember, the square of a triangle function is not triangular. 54. A rectangular voltage pulse which begins at t = 0, is 2 seconds wide and has a height of 0.5 V drives an RC lowpass filter in which R = 10 kΩ and C = 100 µF . (a) (b) (c) (d) Sketch the voltage across the capacitor versus time. Change the pulse duration to 0.2 s and the pulse height to 5 V and repeat. Change the pulse duration to 2 ms and the pulse height to 500 V and repeat. Change the pulse duration to 2 µs and the pulse height to 500 kV and repeat. The solutions in this problem approach the impulse response of the system. 55. Write the differential equation for the voltage, vC ( t) , in the circuit below for time, t > 0,...
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