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| − | Exam 1, Problem 6 - Summer 2008 | + | == Exam 1, Problem 6 - Summer 2008 == |
a) By Inspection the linear constant-coefficient differential equation that describes the LTI system is: | a) By Inspection the linear constant-coefficient differential equation that describes the LTI system is: | ||
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| − | c) Find H(s) at s=jw for the LTI system with | + | c) Find H(s) at s=jw for the LTI system with impulse response <math>h(t)=e^{-7t}u(t)</math> |
| − | H(s) = | + | <math>H(s) = \int_{-\infty }^\infty h(t)e^{-st}dt</math> |
| − | H(s) = | + | <math>H(s) = \int_{-\infty }^\infty e^{-7t}u(t)e^{-st}dt</math> |
| − | Limits change to 0 to | + | Limits change to 0 to <math> \infty </math> and u(t) drops out. |
| − | H(s) = | + | <math>H(s) = \int_0^\infty e^(-7t)e^{-st}dt</math> |
| − | H(s) = | + | <math>H(s) = \int_0^\infty e^{-7t-st}dt</math> |
| − | H(s) = e^ | + | <math>H(s) = \frac{e^{-7t-st}}{-7-s} \bigg|_0^\infty</math> |
| − | H(s) = e^ | + | <math>H(s) = \frac{e^{-7\infty -s\infty}}{-7-s} - \frac{e^{0}}{-7-s} </math> |
| − | + | <math> \frac{e^{-7\infty -s\infty}}{-7-s} </math> goes to 0 | |
| − | H(s) = 1 | + | <math>H(s) = \frac{1}{7+s}</math> |
| − | H(jw) = 1 | + | <math>H(jw) = \frac{1}{7+jw}</math> |
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| − | d)Find the output y(t) of the above LTI system when the input x(t) is e^ | + | d)Find the output y(t) of the above LTI system when the input x(t) is <math> e^{j7t} </math> |
y(t) = x(t) convolved with h(t) | y(t) = x(t) convolved with h(t) | ||
Revision as of 20:48, 30 June 2008
Exam 1, Problem 6 - Summer 2008
a) By Inspection the linear constant-coefficient differential equation that describes the LTI system is:
$ y(t)= \frac{1}{7}x(t)- \frac{1}{7}\frac{dy(t)}{dt} $
b) Show that the impulse response to this LTI system is given by $ h(t)=e^{-7t}u(t) $
This means that $ x(t) = \delta{(t)} $ and $ y(t) = e^{-7t}u(t) $
$ e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} $
Differentiating $ \frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} $ requires use of the chain rule.
This portion of the equation becomes:
$ \frac{1}{7}(-7)e^{-7t}u(t) - \frac{1}{7}\delta{(t) }e^{-7t} $
$ \frac{-1}{7}\delta{(t) }e^{-7t} $ is $ \frac{-1}{7}\delta{(t) }e^{-7t} $ evaluated at t=0 or $ \frac{-1}{7}\delta{(t) }(1) $
Plugging that back in yields:
$ e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{-7}{7}e^{-7t}u(t) - \frac{1}{7}\delta{(t) } $
This equation simplifies to:
0=0 indicating that it is correct.
c) Find H(s) at s=jw for the LTI system with impulse response $ h(t)=e^{-7t}u(t) $
$ H(s) = \int_{-\infty }^\infty h(t)e^{-st}dt $
$ H(s) = \int_{-\infty }^\infty e^{-7t}u(t)e^{-st}dt $
Limits change to 0 to $ \infty $ and u(t) drops out.
$ H(s) = \int_0^\infty e^(-7t)e^{-st}dt $
$ H(s) = \int_0^\infty e^{-7t-st}dt $
$ H(s) = \frac{e^{-7t-st}}{-7-s} \bigg|_0^\infty $
$ H(s) = \frac{e^{-7\infty -s\infty}}{-7-s} - \frac{e^{0}}{-7-s} $
$ \frac{e^{-7\infty -s\infty}}{-7-s} $ goes to 0
$ H(s) = \frac{1}{7+s} $
$ H(jw) = \frac{1}{7+jw} $
d)Find the output y(t) of the above LTI system when the input x(t) is $ e^{j7t} $
y(t) = x(t) convolved with h(t)
x(t) = e^(j7t)
h(t) = e^(-7t)u(t)
Using the properties of convolution we will choose to convolve using x(t-tau) and h(tau) for simplicity.
y(t) = Integral(-infinity to infinity)e^(j7(t-tau))e^(-7tau)u(tau)dtau
Drop the u(t) and change the integration limits.
y(t) = Integral(0 to infinity)e^(j7(t-tau))e^(-7tau)dtau
Simplify the exponentials.
y(t) = Integral(0 to infinity)e^(-7tau + j7t - j7tau)dtau
y(t) = (e^(-7tau + j7tau - j7tau))/(-7-j7) evaluated from 0 to inifity
y(t) = e^(-infinity)/(-7-j7) + e^(j7t)/(7+7j)
y(t) = e^(j7t)/(7+7j)
