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== CT LTI sytem ==
  
== 3 ==
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An example system would be:
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y(t) = 2*x(t)
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== Part A: The unit impulse response and system function H(s) ==
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The unit impulse response:
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<math> x(t) \to  \delta(t) * h(t) = 2*\delta(t) </math>
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The system function, H(s) derivation:
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<math> y(t) = \int_{-\infty}^{\infty} x(\tau) * h(\tau) *d\tau </math>
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<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 17:41, 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}) $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett