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We can get <math>h(t)=\delta(t)\,</math> | We can get <math>h(t)=\delta(t)\,</math> | ||
<br> | <br> | ||
| − | <math>y(t) = | + | <math>y(t) = \int^{\infty}_{-\infty} \delta(t) dt\,</math><br> |
| − | <math> | + | <br> |
| + | <math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau}</math> | ||
==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 07:04, 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}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau} $
