Difference between revisions of "HW4.3 Jeff Kubascik ECE301Fall2008mboutin" - Rhea

Revision as of 16:25, 25 September 2008

Given the following system

$ \,s(t)=6x(t-2)-5x(t)\, $

Part A

Find the system's unit impulse response $ \,h(t)\, $ and system function $ \,H(s)\, $.

The unit impulse response is simply (plug a $ \,\delta(t)\, $ into the system)

$ \,h(t)=6\delta(t-2)-5\delta(t)\, $

The system function is found using the following formula (for LTI systems)

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

$ \,H(s)=\int_{-\infty}^{\infty}(6\delta(t-2)-5\delta(t))e^{-st}dt\, $

$ \,H(s)=6\int_{-\infty}^{\infty}\delta(t-2)e^{-st}dt - 5\int_{-\infty}^{\infty}\delta(t)e^{-st}dt\, $

using the sifting property

$ \,H(s)=6e^{-2s}-5e^{0}\, $

$ \,H(s)=6e^{-2s}-5\, $

Part B

Compute the system's response to (from problem 1 HW4.1 Jeff Kubascik_ECE301Fall2008mboutin)

$ \,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\, $

Using the Fourier coefficients determined in problem 1, we can express the system's response to $ \,x(t)\, $ as

$ \,y(t)=\sum_{k=-\infty}^{\infty}a_kH(jkw_o)e^{jkw_ot}\, $

$ \,y(t)=-\frac{3\pi}{8}H(j(-3)(\frac{3\pi}{4}))e^{j(-3)(\frac{3\pi}{4})} - \frac{3\pi}{8}H(j(-1)(\frac{3\pi}{4}))e^{j(-1)(\frac{3\pi}{4})} - \frac{3\pi}{8}H(j(1)(\frac{3\pi}{4}))e^{j(1)(\frac{3\pi}{4})} - \frac{3\pi}{8}H(j(3)(\frac{3\pi}{4}))e^{j(3)(\frac{3\pi}{4})}\, $

$ \,y(t)=-\frac{3\pi}{8}H(-j\frac{9\pi}{4})e^{-j\frac{9\pi}{4}} - \frac{3\pi}{8}H(-j\frac{3\pi}{4})e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}H(j\frac{3\pi}{4})e^{j\frac{3\pi}{4}} - \frac{3\pi}{8}H(j\frac{9\pi}{4})e^{j\frac{9\pi}{4}}\, $