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<math> y(t) = \int_{-\infty}^{\infty} e^{-j*w(t-k)} * 2\delta(\tau) *d\tau </math> | <math> y(t) = \int_{-\infty}^{\infty} e^{-j*w(t-k)} * 2\delta(\tau) *d\tau </math> | ||
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| + | <math> y(t) = e^{j*w*t} \int_{-\infty}^{\infty} e^{-j*w*k} * 2\delta(\tau) * d\tau </math> | ||
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| + | <math> H(s) = \int_{-\infty}^{\infty} e^{-j*w*k} * 2\delta(\tau) d\tau </math> | ||
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| + | <math> 2*1 = 2 </math> | ||
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| + | H(s) = 2 | ||
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| + | == Part B: The response to my question 1. == | ||
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| + | <math> x(t) = cos(2 * \pi * t) * cos(4 * \pi * t) </math> | ||
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| + | <math> x(t) = \frac{1}{2}(e^{2\pi t}+e^{-2\pi t}) + \frac{1}{2}(e^{4\pi t}+e^{-4\pi t}) </math> | ||
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| + | <math> y(t) = H(s)*e^{-kt} * x(t) </math> | ||
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| + | <math> == (2*e^{-kt} )*(\frac{1}{2}(e^{2\pi t}+e^{-2\pi t}) + \frac{1}{2}(e^{4\pi t}+e^{-4\pi t}) </math> | ||
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| + | <math> = \frac{1}{1}* e^{-kt}*e^{2\pi t}+e^{-2\pi t}) + e^{-kt}* e^{4\pi t}+e^{-4\pi t}) </math> | ||
Latest revision as of 16:41, 25 September 2008
CT LTI sytem
An example system would be:
y(t) = 2*x(t)
Part A: The unit impulse response and system function H(s)
The unit impulse response:
$ x(t) \to \delta(t) * h(t) = 2*\delta(t) $
The system function, H(s) derivation:
$ y(t) = \int_{-\infty}^{\infty} x(\tau) * h(\tau) *d\tau $
$ y(t) = \int_{-\infty}^{\infty} e^{-j*w(t-k)} * 2\delta(\tau) *d\tau $
$ y(t) = e^{j*w*t} \int_{-\infty}^{\infty} e^{-j*w*k} * 2\delta(\tau) * d\tau $
$ H(s) = \int_{-\infty}^{\infty} e^{-j*w*k} * 2\delta(\tau) d\tau $
$ 2*1 = 2 $
H(s) = 2
Part B: The response to my question 1.
$ x(t) = cos(2 * \pi * t) * cos(4 * \pi * t) $
$ x(t) = \frac{1}{2}(e^{2\pi t}+e^{-2\pi t}) + \frac{1}{2}(e^{4\pi t}+e^{-4\pi t}) $
$ y(t) = H(s)*e^{-kt} * x(t) $
$ == (2*e^{-kt} )*(\frac{1}{2}(e^{2\pi t}+e^{-2\pi t}) + \frac{1}{2}(e^{4\pi t}+e^{-4\pi t}) $
$ = \frac{1}{1}* e^{-kt}*e^{2\pi t}+e^{-2\pi t}) + e^{-kt}* e^{4\pi t}+e^{-4\pi t}) $
