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<math>X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br> | <math>X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br> | ||
<br> | <br> | ||
| + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math> | ||
| + | |||
| + | <math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math> | ||
Revision as of 08:18, 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)=-se^{-s\tau}|_0^\infty \, $
$ H(s)=-s(e^{-\infty} - e^{0})\, $
$ H(s)=s\, $
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
Signal defined in Question 1:
$ X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\, $
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\, $
