(→a) h[n] and H[z]) |
(→b) Response of Signal in Question 1) |
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<math> \,\ a_k </math> is only valid at <math> a_1 = -3 </math>. Therefore... | <math> \,\ a_k </math> is only valid at <math> a_1 = -3 </math>. Therefore... | ||
| + | |||
| + | <math> \,\ Z = e</math><sup><math> jk(2\pi /N) </math></sup> | ||
<math> \,\ y[n] = -3 * | <math> \,\ y[n] = -3 * | ||
Revision as of 18:16, 25 September 2008
Define a DT LTI System
$ \,\ y[n] = 5 * x[n-5] + 6 * x[n-6] $
a) h[n] and H[z]
In order to find h[n], we input $ x[n] = \delta [n] $ to y[n]. h[n] is then the unit impulse response.
$ \,\ x[n] = \delta [n] $
$ \,\ h[n] = 5 * \delta [n-5] + 6 * \delta [n+6] $
H[z] is the system's function, and is defined by:
$ \sum_{m = -\infty}^{\infty} h[m] * Z $-m
$ \,\ H[z] = \sum_{m = -\infty}^{\infty} ( 5 * \delta [n-5] + 6 * \delta [n+6] ) * Z $-m
This function will only be valid when n = 5 or n = -6 due to the $ \delta $s.
Therefore, the end result is:
$ \,\ H[z] = 5 * Z $5$ \,\ + 6 * Z $-6
b) Response of Signal in Question 1
From Question 1: Unfortunately, I did not 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 $
Response of the System $ \,\ y[n] = \sum_{k = <N>}^{\infty} a_k H[e $$ j2\pi k/N $$ \,\ ]e $$ jk(2\pi /N) n $
Because $ a_k $ only has one value, this shouldn't be that hard to calculate.
$ \,\ a_k $ is only valid at $ a_1 = -3 $. Therefore...
$ \,\ Z = e $$ jk(2\pi /N) $
$ \,\ y[n] = -3 * $
