| Line 16: | Line 16: | ||
<math> y(t) = \sum_{k = -\infty}^{\infty} a_k H(jkw) (sin(4\pi t) + sin(6\pi t)) \!</math> | <math> y(t) = \sum_{k = -\infty}^{\infty} a_k H(jkw) (sin(4\pi t) + sin(6\pi t)) \!</math> | ||
| + | |||
| + | <math> y(t) = \sum_{k = -\infty}^{\infty} a_k 4e^{-jw} (sin(4\pi t) + sin(6\pi t)) \!</math> | ||
| + | |||
| + | from before: | ||
| + | |||
| + | <math> a_5 = \frac{-1}{4}, a_-5 = \frac{-1}{4},a_1 = \frac{1}{4},a_-1 = \frac{1}{4}</math> | ||
| + | |||
| + | |||
| + | <math>y(t) = \frac{1}{4}4e^{-jw} - \frac{1}{4}4e^{-jw} + \frac{1}{4}4e^{-jw} - \frac{1}{4}4e^{-jw}</math> | ||
| + | |||
| + | <math>\ y(t) = 0</math> | ||
Latest revision as of 14:19, 26 September 2008
get h(t), H(s), and H(jw)
$ \ y(t) = 4x(t-1) $
$ \ h(t) = 4d(t-1) $
$ \ H(s) = \int^{\infty}_{-\infty} h(t)e^{-st}dt $
$ \ H(s) = \int^{\infty}_{-\infty} 4d(t-1)e^{-st}dt $
$ \ H(s) = 4e^{-s} $
$ \ H(jw) = 4e^{-jw} $
get the response of H(s) to signal proposed in previous question
$ y(t) = \sum_{k = -\infty}^{\infty} a_k H(jkw) (sin(4\pi t) + sin(6\pi t)) \! $
$ y(t) = \sum_{k = -\infty}^{\infty} a_k 4e^{-jw} (sin(4\pi t) + sin(6\pi t)) \! $
from before:
$ a_5 = \frac{-1}{4}, a_-5 = \frac{-1}{4},a_1 = \frac{1}{4},a_-1 = \frac{1}{4} $
$ y(t) = \frac{1}{4}4e^{-jw} - \frac{1}{4}4e^{-jw} + \frac{1}{4}4e^{-jw} - \frac{1}{4}4e^{-jw} $
$ \ y(t) = 0 $
