{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chap3StudentSolutions

# Change the pulse duration to 02 s and the pulse

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online