(New page: ==CT LTI Impulse Response== The system <math>y(t) = 35x(t) - 2x(t+5)</math> is LTI Let's compute h(t): <math>x(t) = \delta(t)</math> <math>h(t) = 35\delta(t) - 2\delta(t+5)</math> ==R...) |
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<math>H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | <math>H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | ||
| − | <math>H(j\omega)=\int_{-\infty}^\infty 35\delta(\tau) - 2\delta(\tau+5) | + | <math>H(j\omega)=\int_{-\infty}^\infty (35\delta(\tau) - 2\delta(\tau+5))e^{-j\omega\tau}d\tau</math> |
| + | |||
| + | Using the sifting property we see that we get <math>35e^{0}-2e^{5 j\omega}</math> | ||
Revision as of 18:30, 25 September 2008
CT LTI Impulse Response
The system $ y(t) = 35x(t) - 2x(t+5) $ is LTI
Let's compute h(t):
$ x(t) = \delta(t) $
$ h(t) = 35\delta(t) - 2\delta(t+5) $
Response to My Function From Part 1
The function I chose for part 1 was
$ x(t) = sin(14t)+(1+3j)cos(2t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2} $
So lets compute the system's response.
Since the equation is already expressed as a sum of complex exponentials, we need only multiply the input signal by H(t).
Let's calculate it:
$ H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
$ H(j\omega)=\int_{-\infty}^\infty (35\delta(\tau) - 2\delta(\tau+5))e^{-j\omega\tau}d\tau $
Using the sifting property we see that we get $ 35e^{0}-2e^{5 j\omega} $
