(New page: Consider the following CT LTI system defined by: <math>y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\,</math> the impulse response is... <math>h(t)=\int_{-\infty}^{t} e^{-(t-\ta...)
 
(System Response to Q.1)
 
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==Impulse Response==
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Consider the following CT LTI system defined by:
 
Consider the following CT LTI system defined by:
  
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<math>h(t)= e^{-t}u(t)\,</math>
 
<math>h(t)= e^{-t}u(t)\,</math>
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==System Function==
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The system function is...
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<math>H(s)=\int_{-\infty}^{\infty} h(t)e^{-st}\,dt\,</math>
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Where <math>s=j\omega\,</math>
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for this system....
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<math>H(s)=\int_{-\infty}^{\infty} e^{-t}u(t)e^{-st}\,dt\,</math>
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<math>H(s)=\int_{0}^{\infty} e^{-t}e^{-st}\,dt\,</math>
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<math>H(s)=\int_{0}^{\infty} e^{-(s+1)t}\,dt\,</math>
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<math>H(s)=\frac{-1}{s+1}|e^{-(s+1)t}|_0^{\infty}\,=\frac{1}{s+1}\,</math>
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== System Response to Q.1 ==
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The input in Q1 was not CT, but if it was it would have been:
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<math>x(t)=sin(3\pi t)\,</math>
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The systems response to this signal would be...
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<math>x(t)=sin(3\pi t)\,=\frac{-j}{2}(e^{j3\pi t}-e^{-j3\pi t})</math>
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<math>y(t)=\frac{-j}{2}(H(j3\pi)e^{j3\pi t}-H(-j3\pi)e^{-j3\pi t})</math>
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The systems response to this signal by direct computation would be...
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<math>y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\,</math>
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<math>y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}sin(3\pi\tau)\,d\tau\,</math>
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<math>y(t)=\int_{-\infty}^{t} e^{-t}e^{\tau}sin(3\pi\tau)\,d\tau\,</math>
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<math>y(t)=e^{-t}\int_{-\infty}^{t}e^{\tau}sin(3\pi\tau)\,d\tau\,</math>
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from the table of Integrals...
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<math>y(t)=e^{-t}[\frac{e^\tau}{1^2+(3\pi)^2}(sin(3\pi\tau)-3\pi cos(3\pi\tau))]_{-\infty}^t\,</math>
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<math>y(t)=\frac{1}{1+9\pi^2}(sin(3\pi t)-3\pi cos(3\pi t))\,</math>

Latest revision as of 07:23, 25 September 2008

Impulse Response

Consider the following CT LTI system defined by:

$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\, $

the impulse response is...


$ h(t)=\int_{-\infty}^{t} e^{-(t-\tau)}\delta(\tau)\,d\tau\, = e^{-(t-\tau)} |_{ \tau=0}= e^{-t} $

but this will diverge when t is less than 0 so...


$ h(t)= e^{-t}u(t)\, $

System Function

The system function is...


$ H(s)=\int_{-\infty}^{\infty} h(t)e^{-st}\,dt\, $


Where $ s=j\omega\, $

for this system....

$ H(s)=\int_{-\infty}^{\infty} e^{-t}u(t)e^{-st}\,dt\, $


$ H(s)=\int_{0}^{\infty} e^{-t}e^{-st}\,dt\, $

$ H(s)=\int_{0}^{\infty} e^{-(s+1)t}\,dt\, $

$ H(s)=\frac{-1}{s+1}|e^{-(s+1)t}|_0^{\infty}\,=\frac{1}{s+1}\, $

System Response to Q.1

The input in Q1 was not CT, but if it was it would have been:

$ x(t)=sin(3\pi t)\, $

The systems response to this signal would be...


$ x(t)=sin(3\pi t)\,=\frac{-j}{2}(e^{j3\pi t}-e^{-j3\pi t}) $


$ y(t)=\frac{-j}{2}(H(j3\pi)e^{j3\pi t}-H(-j3\pi)e^{-j3\pi t}) $



The systems response to this signal by direct computation would be...

$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\, $

$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}sin(3\pi\tau)\,d\tau\, $ $ y(t)=\int_{-\infty}^{t} e^{-t}e^{\tau}sin(3\pi\tau)\,d\tau\, $

$ y(t)=e^{-t}\int_{-\infty}^{t}e^{\tau}sin(3\pi\tau)\,d\tau\, $

from the table of Integrals...


$ y(t)=e^{-t}[\frac{e^\tau}{1^2+(3\pi)^2}(sin(3\pi\tau)-3\pi cos(3\pi\tau))]_{-\infty}^t\, $

$ y(t)=\frac{1}{1+9\pi^2}(sin(3\pi t)-3\pi cos(3\pi t))\, $

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