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<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math> | <math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math> | ||
| + | |||
| + | From Question 1: | ||
| + | <math>x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br> | ||
| + | With this expression we can conclude:<br> | ||
| + | <math>a_1 = 3\,</math><br> | ||
| + | <math>a_{-1} = 3\,</math><br> | ||
| + | <math>a_2 = 4\,</math><br> | ||
| + | <math>a_{-2} = -4\,</math><br> | ||
Revision as of 08:20, 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}\, $
From Question 1:
$ x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $
