Line 8: Line 8:
 
<math>y(t) = \int^{\infty}_{-\infty} \delta(t) dt\,</math><br>
 
<math>y(t) = \int^{\infty}_{-\infty} \delta(t) dt\,</math><br>
 
<br>
 
<br>
<math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}</math><br>
+
<math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau</math><br>
 
<br>
 
<br>
<math>H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}</math><br>
+
<math>H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}d\tau</math><br>
 
<br>
 
<br>
<math>H(s)=\int_{0}^{\infty}e^{-s\tau}</math><br>
+
<math>H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau</math><br>
 
<br>
 
<br>
<math>
+
<math>H(s)=-e^{-s\tau}</math><br>
  
 
==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal==
 
==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal==

Revision as of 07:08, 26 September 2008

Obtain the input impulse response h(t) and the system function H(s) of your system

A very simple system:
$ y(t)=x(t)\, $ and $ x(t)=\delta(t) $

We can get $ h(t)=\delta(t)\, $
$ y(t) = \int^{\infty}_{-\infty} \delta(t) dt\, $

$ H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau $

$ H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}d\tau $

$ H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau $

$ H(s)=-e^{-s\tau} $

Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva