(New page: == Part A: Unit Impulse Response and System Function == <font size ="4"><math>h(t) = 9\delta(t)</math></font> so: <math>y(t) = \int^{\infty}_{-\infty} h(\tau) x(\tau) d\tau</math> <mat...) |
(→Part B: Response of the System) |
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<math>x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2})</math> | <math>x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2})</math> | ||
| − | So: | + | So with <math>y(t) = H(s)e^{-st}x(t)</math>, the final response is: |
| − | <math>y(t)</math> | + | <math>y(t) = 9e^{-st} + 9e^{-st}(\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + 9e^{-st}(e^{j8 \pi t} + e^{-j8 \pi t}) + 9e^{-st}(\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2})</math> |
Revision as of 17:52, 25 September 2008
Part A: Unit Impulse Response and System Function
$ h(t) = 9\delta(t) $
so:
$ y(t) = \int^{\infty}_{-\infty} h(\tau) x(\tau) d\tau $
$ y(t) = \int^{\infty}_{-\infty} 9\delta(\tau)e^{-j\omega_0(t - \tau)} d\tau $
$ y(t) = e^{j\omega_0 t} \int^{\infty}_{-\infty} 9\delta(\tau)e^{-j\omega_0 \tau} d\tau $
$ H(s) = 9e^{j\omega_0} $
$ H(s) = 9 $
Part B: Response of the System
$ x(t) = 1 + sin(8\pi t) + 2cos(8\pi t) + cos(16\pi t + \frac{\pi}{4}) $
$ x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $
So with $ y(t) = H(s)e^{-st}x(t) $, the final response is:
$ y(t) = 9e^{-st} + 9e^{-st}(\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + 9e^{-st}(e^{j8 \pi t} + e^{-j8 \pi t}) + 9e^{-st}(\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $
