(→h[n] and H(z)) |
(→h[n] and H(z)) |
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| Line 9: | Line 9: | ||
<math> \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] </math> | <math> \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] </math> | ||
| − | <br> | + | <br><br> |
<math> \,\ H[z] = \sum_{m=-\infty}^\infty h[m] * Z</math><sup>(<math>-m</math>)</sup><br><br> | <math> \,\ H[z] = \sum_{m=-\infty}^\infty h[m] * Z</math><sup>(<math>-m</math>)</sup><br><br> | ||
<math> \,\ H[z] = \sum_{m=-6}^{5}Z</math><sup>(<math>-m</math>)</sup> | <math> \,\ H[z] = \sum_{m=-6}^{5}Z</math><sup>(<math>-m</math>)</sup> | ||
Revision as of 16:20, 25 September 2008
Define a DT LTI System
$ \,\ x[n] = 5*u[n-5] + 6*u[n+6] $
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=-6}^{5}Z $($ -m $)
