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d)Find the output y(t) of the above LTI system when the input x(t) is <math> e^{j7t} </math> | d)Find the output y(t) of the above LTI system when the input x(t) is <math> e^{j7t} </math> | ||
| − | <math> y(t) = x(t) | + | <math> y(t) = x(t) * h(t) </math> (where * indicates convolution) |
| − | < | + | <math> x(t) = e^{j7t} </math> |
<math> h(t) = e^{-7t}u(t) </math> | <math> h(t) = e^{-7t}u(t) </math> | ||
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| − | <math> y(t) = \int_{-\infty^\infty e^{j7(t-\tau)}e^{-7\tau}u(\tau)d\tau </math> | + | <math> y(t) = \int_{-\infty}^\infty e^{j7(t-\tau)}e^{-7\tau}u(\tau)d\tau </math> |
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| − | <math> y(t) = \int_0^\infty e^{j7(t-\tau)}e^{-7\tau}d\tau | + | <math> y(t) = \int_0^\infty e^{j7(t-\tau)}e^{-7\tau}d\tau </math> |
Revision as of 19:59, 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} $ Numerator 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) * h(t) $ (where * indicates convolution)
$ 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) = \int_{-\infty}^\infty e^{j7(t-\tau)}e^{-7\tau}u(\tau)d\tau $
Drop the u(t) and change the integration limits.
$ y(t) = \int_0^\infty e^{j7(t-\tau)}e^{-7\tau}d\tau $
Simplify the exponentials.
$ y(t) = \int_0^\infty e^{-7\tau + j7t - j7\tau}d\tau $
$ y(t) = \frac{e^{-7\tau + j7\tau - j7\tau}}{-7-j7} \bigg|_0^\infty $
$ y(t) = \frac{e^{-\infty}}{-7-j7} + \frac{e^{j7t}}{7+7j} $
$ y(t) = \frac{e^{j7t}}{7+7j} $
