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<math> y(t) = H(s)*e^{-kt} * x(t) </math>
 
<math> y(t) = H(s)*e^{-kt} * x(t) </math>
  
<math> == (5*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>
+
<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>

Revision as of 16:38, 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}) $

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EISL lab graduate

Mu Qiao