| Line 14: | Line 14: | ||
<math>H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau</math><br> | <math>H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau</math><br> | ||
<br> | <br> | ||
| − | <math>H(s)=-e^{-s\tau}</math><br> | + | <math>H(s)=-e^{-s\tau}\,</math><br> |
==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal== | ==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal== | ||
Revision as of 07:08, 26 September 2008
Obtain the input impulse response h(t) and the system function H(s) of your system
A very simple system:
$ y(t)=x(t)\, $ and $ x(t)=\delta(t) $
We can get $ h(t)=\delta(t)\, $
$ y(t) = \int^{\infty}_{-\infty} \delta(t) dt\, $
$ H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau $
$ H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}d\tau $
$ H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau $
$ H(s)=-e^{-s\tau}\, $
