| Line 25: | Line 25: | ||
<math>H(j\omega)=\int_{-\infty}^\infty (35\delta(\tau) - 2\delta(\tau+5))e^{-j\omega\tau}d\tau</math> | <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> | + | Using the sifting property we see that we get |
| + | |||
| + | <math>35e^{0}-2e^{5 j\omega} = 35 - 2e^{5j\omega}</math> | ||
| + | |||
| + | So since y(t) = x(t)*H(t), we can now solve for y(t): | ||
| + | |||
| + | <math>y(t) = (35 -2e^{5j\omega})*(\frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2})</math> | ||
Latest revision as of 17:48, 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} = 35 - 2e^{5j\omega} $
So since y(t) = x(t)*H(t), we can now solve for y(t):
$ y(t) = (35 -2e^{5j\omega})*(\frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2}) $
