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== Define a CT LTI System ==
 
== Define a CT LTI System ==
  
<math>y(t)=2x(t)-3x(t-4)\!</math>
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<math>y(t)=2x(t)-3x(t-2)\!</math>
  
  
 
== Unit Impulse Response ==
 
== Unit Impulse Response ==
The unit impulse response is simply the systems response to an input <math>\delta(t)\!</math>.  Thus, in our case, the unit impulse response is simply <math>h(t)=2\delta(t)-3\delta(t-4)\!</math>
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The unit impulse response is simply the systems response to an input <math>\delta(t)\!</math>.  Thus, in our case, the unit impulse response is simply <math>h(t)=2\delta(t)-3\delta(t-2)\!</math>
  
  
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<br>
 
<br>
 
<br>
 
<br>
<math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]e^{st}dt</math>
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<math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-2)]e^{-st}dt</math>
 
<br>
 
<br>
 
<br>
 
<br>
<math>=\int_{-\infty}^{+\infty}2\delta(t)e^{st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{st}dt</math>
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<math>=\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-2)e^{-st}dt</math> which by the sifting property,
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<br>
 +
<br>
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<math>=2-3e^{-2s}\!</math>
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 +
 
 +
== System Response ==
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The signal found in Question 1 was:<br>
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<math>x(t)=\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t}</math>
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<br>
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To find the system's response to the signal, simply multiply it by the system function:<br>
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<math>y(t)=H(jw)*x(t)\!</math><br>
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<math>y(t)=(2-3e^{-2jw})*[\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t}]</math><br>
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<math>=\frac{4}{j}e^{j3t}-\frac{4}{j}e^{-j3t}+(1+2)e^{j2t}+(1+2j)e^{-j2t}-\frac{6}{j}e^{3j(t-2)}+\frac{6}{j}e^{-3j(t-2)}-\frac{3+6j}{2}(e^{2j(t-2)})-\frac{3+6j}{2}(e^{-2j(t-2)}

Latest revision as of 10:52, 25 September 2008

Define a CT LTI System

$ y(t)=2x(t)-3x(t-2)\! $


Unit Impulse Response

The unit impulse response is simply the systems response to an input $ \delta(t)\! $. Thus, in our case, the unit impulse response is simply $ h(t)=2\delta(t)-3\delta(t-2)\! $


System Function

To find the system function $ H(s)\! $ we use the formula:
$ H(s)=\int_{-\infty}^{+\infty} h(t)e^{st}dt $ where $ s=j\omega\! $.

$ H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-2)]e^{-st}dt $

$ =\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-2)e^{-st}dt $ which by the sifting property,

$ =2-3e^{-2s}\! $


System Response

The signal found in Question 1 was:
$ x(t)=\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t} $
To find the system's response to the signal, simply multiply it by the system function:
$ y(t)=H(jw)*x(t)\! $
$ y(t)=(2-3e^{-2jw})*[\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t}] $
$ =\frac{4}{j}e^{j3t}-\frac{4}{j}e^{-j3t}+(1+2)e^{j2t}+(1+2j)e^{-j2t}-\frac{6}{j}e^{3j(t-2)}+\frac{6}{j}e^{-3j(t-2)}-\frac{3+6j}{2}(e^{2j(t-2)})-\frac{3+6j}{2}(e^{-2j(t-2)} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang