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== Define a CT LTI System == | == Define a CT LTI System == | ||
| − | <math>y(t)=2x(t)-3x(t- | + | <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- | + | 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- | + | <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- | + | <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, |
<br> | <br> | ||
<br> | <br> | ||
| − | <math>=2-3e^{- | + | <math>=2-3e^{-2s}\!</math> |
| + | |||
| + | |||
| + | == System Response == | ||
| + | The signal found in Question 1 was:<br> | ||
| + | <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> | ||
| + | <br> | ||
| + | To find the system's response to the signal, simply multiply it by the system function:<br> | ||
| + | <math>y(t)=H(jw)*x(t)\!</math><br> | ||
| + | <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> | ||
| + | <math>=[\frac{4}{j}e^{j3t}-\frac{4}{j}e^{-j3t}+(1+2)e^{j2t}+(1+2j)e^{-j2t}] | ||
Revision as of 11:44, 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}] $
