(System Function)
(System Function)
Line 12: Line 12:
 
<br>
 
<br>
 
<br>
 
<br>
<math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]e^{st}dt</math>
+
<math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]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> which by the sifting property,  
+
<math>=\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{-st}dt</math> which by the sifting property,  
 
<br>
 
<br>
 
<br>
 
<br>
<math>=2-3e^{4s}\!</math>
+
<math>=2-3e^{-4s}\!</math>

Revision as of 10:12, 25 September 2008

Define a CT LTI System

$ y(t)=2x(t)-3x(t-4)\! $


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-4)\! $


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-4)]e^{-st}dt $

$ =\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{-st}dt $ which by the sifting property,

$ =2-3e^{-4s}\! $

Alumni Liaison

EISL lab graduate

Mu Qiao