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<math>H(s)=\int_{-\infty}^{\infty} 0.5 \delta(t-5) u(t) e^{-st}\,dt\,</math> | <math>H(s)=\int_{-\infty}^{\infty} 0.5 \delta(t-5) u(t) e^{-st}\,dt\,</math> | ||
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Latest revision as of 17:50, 26 September 2008
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 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 exponential is inputted into a LTI system the output is
$ \ \Sigma a_{k} e^{jkwt} \longrightarrow sys \longrightarrow \Sigma H(jkw) a_{k} e^{jkwt} $
Therefore using this fact the system's output to the input of $ \ cos(2 \pi t/3) sin(2 \pi t/9) $ is
$ \ y(t)= \frac{j}{8} e^{-10 \frac{2 \pi}{9}} e^{2 \frac{2 \pi}{9}t} -\frac{j}{8} e^{10 \frac{2 \pi}{9}}e^{-2 \frac{2 \pi}{9}t} - \frac{j}{8} e^{-20 \frac{2 \pi}{9} e^{4 \frac{2 \pi}{9}t}} + \frac{j}{8} e^{20 \frac{2 \pi}{9} e^{-4 \frac{2 \pi}{9}t}} $
