This preview shows page 1. Sign up to view the full content.
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 335 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 nonzero over a range of its argument from 0 to 4. It is convolved with
a function which is nonzero over a range of its argument from 3 to 1. What is the
nonzero range of the convolution of the two?
Imagine any two functions with finite nonzero 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 fullwave rectified sinusoid. 1 t t
(h) g( t) = sinc(2 t) ∗ comb rect 2 4
2 Solutions 336 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
 Spring '09
 ATOUSA

Click to edit the document details