(New page: Given the following system <math>\,s(t)=6x(t-2)-5x(t)\,</math> == Part A == Find the system's unit impulse response <math>\,h(t)\,</math> and system function <math>\,H(s)\,</math>. =...) |
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| − | Given the following system | + | Given the following LTI CT system |
<math>\,s(t)=6x(t-2)-5x(t)\,</math> | <math>\,s(t)=6x(t-2)-5x(t)\,</math> | ||
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Find the system's unit impulse response <math>\,h(t)\,</math> and system function <math>\,H(s)\,</math>. | Find the system's unit impulse response <math>\,h(t)\,</math> and system function <math>\,H(s)\,</math>. | ||
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| + | The unit impulse response is simply (plug a <math>\,\delta(t)\,</math> into the system) | ||
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| + | <math>\,h(t)=6\delta(t-2)-5\delta(t)\,</math> | ||
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| + | The system function is found using the following formula (for LTI systems) | ||
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| + | <math>\,H(s)=\int_{-\infty}^{\infty}h(t)e^{-st}dt\,</math> | ||
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| + | <math>\,H(s)=\int_{-\infty}^{\infty}(6\delta(t-2)-5\delta(t))e^{-st}dt\,</math> | ||
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| + | <math>\,H(s)=6\int_{-\infty}^{\infty}\delta(t-2)e^{-st}dt - 5\int_{-\infty}^{\infty}\delta(t)e^{-st}dt\,</math> | ||
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| + | using the sifting property | ||
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| + | <math>\,H(s)=6e^{-2s}-5e^{0}\,</math> | ||
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| + | <math>\,H(s)=6e^{-2s}-5\,</math> | ||
== Part B == | == Part B == | ||
| − | Compute the system's response to | + | Compute the system's response to (from problem 1 [[HW4.1 Jeff Kubascik_ECE301Fall2008mboutin]]) |
<math>\,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\,</math> | <math>\,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\,</math> | ||
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| + | Using the Fourier coefficients determined in problem 1, we can express the system's response to <math>\,x(t)\,</math> as | ||
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| + | <math>\,y(t)=\sum_{k=-\infty}^{\infty}a_kH(jkw_o)e^{jkw_ot}\,</math> | ||
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| + | <math>\,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})}\,</math> | ||
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| + | <math>\,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}}\,</math> | ||
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| + | <math>\,y(t)=-\frac{3\pi}{8}(6e^{j\frac{9\pi}{2}}-5)e^{-j\frac{9\pi}{4}} - \frac{3\pi}{8}(6e^{j\frac{3\pi}{2}}-5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6e^{-j\frac{3\pi}{2}}-5)e^{j\frac{3\pi}{4}} - \frac{3\pi}{8}(6e^{-j\frac{9\pi}{2}}-5)e^{j\frac{9\pi}{4}}\,</math> | ||
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| + | <math>\,y(t)=-\frac{3\pi}{8}(6j-5)e^{-j\frac{9\pi}{4}} - \frac{3\pi}{8}(-6j-5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6j-5)e^{j\frac{3\pi}{4}} - \frac{3\pi}{8}(-6j-5)e^{j\frac{9\pi}{4}}\,</math> | ||
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| + | <math>\,y(t)=-\frac{3\pi}{8}(6j-5)e^{-j\frac{9\pi}{4}} + \frac{3\pi}{8}(6j+5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6j-5)e^{j\frac{3\pi}{4}} + \frac{3\pi}{8}(6j+5)e^{j\frac{9\pi}{4}}\,</math> | ||
Latest revision as of 17:11, 25 September 2008
Given the following LTI CT 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}}\, $
$ \,y(t)=-\frac{3\pi}{8}(6e^{j\frac{9\pi}{2}}-5)e^{-j\frac{9\pi}{4}} - \frac{3\pi}{8}(6e^{j\frac{3\pi}{2}}-5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6e^{-j\frac{3\pi}{2}}-5)e^{j\frac{3\pi}{4}} - \frac{3\pi}{8}(6e^{-j\frac{9\pi}{2}}-5)e^{j\frac{9\pi}{4}}\, $
$ \,y(t)=-\frac{3\pi}{8}(6j-5)e^{-j\frac{9\pi}{4}} - \frac{3\pi}{8}(-6j-5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6j-5)e^{j\frac{3\pi}{4}} - \frac{3\pi}{8}(-6j-5)e^{j\frac{9\pi}{4}}\, $
$ \,y(t)=-\frac{3\pi}{8}(6j-5)e^{-j\frac{9\pi}{4}} + \frac{3\pi}{8}(6j+5)e^{-j\frac{3\pi}{4}} - \frac{3\pi}{8}(6j-5)e^{j\frac{3\pi}{4}} + \frac{3\pi}{8}(6j+5)e^{j\frac{9\pi}{4}}\, $
