(→h[n] and H(z)) |
(→Define a DT LTI System) |
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By the sifting property, this sum equals:<br> | By the sifting property, this sum equals:<br> | ||
<math> \,\ H[z] = 5*Z</math><sup>-5</sup><math> \,\ + 6*Z</math><sup>6</sup> | <math> \,\ H[z] = 5*Z</math><sup>-5</sup><math> \,\ + 6*Z</math><sup>6</sup> | ||
| + | |||
| + | |||
| + | === b) Response of Signal in Question 1 === | ||
| + | ---- | ||
| + | |||
| + | From Question 1: | ||
| + | |||
| + | * <math> \,\ x(t) = 5cos(2t) - 4sin(5t) </math><br><br> | ||
| + | * <math> \,\ T = 2\pi </math><br><br> | ||
| + | * <math> \,\ x(t) = \frac{5}{2} * e</math><sup>(j2t)</sup> <math> \,\ + \frac{5}{2} * e</math><sup>(-j2t)</sup><math> - \frac{4}{2j} * e</math><sup>(j5t)</sup> <math>+ \frac{4}{2j} * e</math><sup>(-j5t)</sup><br> | ||
| + | * <math> \,\ w_0 = 1 </math><br><br> | ||
| + | * <math>\mathbf{a_2} = \mathbf{\frac{5}{2}}</math><br><br> | ||
| + | * <math>\mathbf{a}</math><sub>-2</sub><math> = \mathbf{\frac{5}{2j}}</math><br><Br> | ||
| + | * <math>\mathbf{a_5} = \mathbf{-2}</math><br><br> | ||
| + | * <math>\mathbf{a}</math><sub>-5</sub><math> = \mathbf{\frac{2}{j}}</math><br><br> | ||
| + | * <math> \,\ a_x = 0 </math> elsewhere | ||
Revision as of 17:19, 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:
- $ \,\ x(t) = 5cos(2t) - 4sin(5t) $
- $ \,\ T = 2\pi $
- $ \,\ x(t) = \frac{5}{2} * e $(j2t) $ \,\ + \frac{5}{2} * e $(-j2t)$ - \frac{4}{2j} * e $(j5t) $ + \frac{4}{2j} * e $(-j5t)
- $ \,\ w_0 = 1 $
- $ \mathbf{a_2} = \mathbf{\frac{5}{2}} $
- $ \mathbf{a} $-2$ = \mathbf{\frac{5}{2j}} $
- $ \mathbf{a_5} = \mathbf{-2} $
- $ \mathbf{a} $-5$ = \mathbf{\frac{2}{j}} $
- $ \,\ a_x = 0 $ elsewhere
