(→Define a DT LTI System) |
(→b) Response of Signal in Question 1) |
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From Question 1: | From Question 1: | ||
| + | Unfortunately, I did not not read ahead and make a DT signal for Parts 1/2. Instead of going back and redoing my work, I am going to "steal" the work of a Mr. Collin Phillips (my apologies and gratitude to Mr. Collin Phillips). | ||
| − | * <math> \,\ | + | According to his work: |
| − | + | * <math> \,\ X[n] = 3cos(3\pi n + \pi) </math> | |
| − | * <math> \,\ | + | * <math> \,\ a_0 = 0 </math> |
| − | * <math> \,\ | + | * <math> \,\ a_1 = -3 </math> |
| − | + | * <math> \,\ a_k = 0 </math> elsewhere | |
| − | + | * <math> \,\ N = 2 </math> | |
| − | * <math>\ | + | |
| − | + | ||
| − | * <math> \,\ | + | |
Revision as of 17:33, 25 September 2008
Define a DT LTI System
$ \,\ x[n] = 5*u[n-5] + 6*u[n+6] $
a) h[n] and H(z)
We obtain $ h[n] $ by finding the response of $ x[n] $ to the unit impulse response ($ \delta[n] $).
$ \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] $
$ \,\ H[z] = \sum_{m=-\infty}^\infty h[m] * Z $($ -m $)
$ \,\ H[z] = \sum_{m=-\infty}^{\infty} (5*\delta[n-5] + 6*\delta[n+6]) * Z $($ -m $)
By the sifting property, this sum equals:
$ \,\ H[z] = 5*Z $-5$ \,\ + 6*Z $6
b) Response of Signal in Question 1
From Question 1: Unfortunately, I did not not read ahead and make a DT signal for Parts 1/2. Instead of going back and redoing my work, I am going to "steal" the work of a Mr. Collin Phillips (my apologies and gratitude to Mr. Collin Phillips).
According to his work:
- $ \,\ X[n] = 3cos(3\pi n + \pi) $
- $ \,\ a_0 = 0 $
- $ \,\ a_1 = -3 $
- $ \,\ a_k = 0 $ elsewhere
- $ \,\ N = 2 $
