Revision as of 16:21, 26 September 2008 by Apdelanc (Talk)

Defining an LTI System

For an input x(t), let the LTI system be defined as:

$ \ y(t)=0.5 x(t-5) u(t) $

Computing the Impulse Response and System Function

Inputting a delta into the system yields:

$ \ y(t)=h(t)=0.5 \delta(t-5) u(t) $

The System Function is defined by:

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

Now computing the actual response:

$ H(s)=\int_{-\infty}^{\infty} 0.5 \delta(t-5) u(t) e^{-st}\,dt\, $

which is turns into:


$ H(s)=\int_{0}^{\infty} 0.5 \delta(t-5) e^{-st}\,dt\, $

Now using the sifting property of the delta function we obtain:

$ \ H(s)= 0.5 e^{-5s} $

Remember$ \ s= jw $

Computing the Response of the Signal from Q1 using H(s)

When a periodic signal represented as a linear combination of complex exponentials is inputted into a LTI system the output is

$ \sigma $

Alumni Liaison

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

Francisco Blanco-Silva