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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: | |
| − | |||
| − | y(t)=1 | + | <math>y(t)= \frac{1}{7}x(t)- \frac{1}{7}\frac{dy(t)}{dt}</math> |
| + | b) Show that the impulse response to this LTI system is given by <math>h(t)=e^{-7t}u(t)</math> | ||
| − | + | This means that <math>x(t) = \delta{(t)} </math> and <math> y(t) = e^{-7t}u(t)</math> | |
| − | |||
| − | e^ | + | <math> e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} </math> |
| − | Differentiating | + | |
| + | Differentiating <math>\frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} </math> requires use of the chain rule. | ||
This portion of the equation becomes: | This portion of the equation becomes: | ||
| − | |||
| − | -(1 | + | <math>\frac{1}{7}(-7)e^{-7t}u(t) - \frac{1}{7}\delta{(t) }e^{-7t} </math> |
| + | |||
| + | |||
| + | <math> \frac{-1}{7}\delta{(t) }e^{-7t} </math> is <math> \frac{-1}{7}\delta{(t) }e^{-7t} </math> evaluated at t=0 or <math> \frac{-1}{7}\delta{(t) }(1) </math> | ||
| + | |||
Plugging that back in yields: | Plugging that back in yields: | ||
| − | e^ | + | |
| + | <math> e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{-7}{7}e^{-7t}u(t) - \frac{1}{7}\delta{(t) } </math> | ||
| + | |||
This equation simplifies to: | This equation simplifies to: | ||
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| − | c) Find H(s) at s=jw for the LTI system with impuls response h(t)=e^ | + | c) Find H(s) at s=jw for the LTI system with impuls response <math>h(t)=e^{-7t}u(t)</math> |
| + | |||
H(s) = Integral(-infinity to infinity)h(t)*e^(-st)dt | H(s) = Integral(-infinity to infinity)h(t)*e^(-st)dt | ||
| + | |||
H(s) = Integral(-infinity to infinity)e^(-7t)u(t)*e^(-st)dt | H(s) = Integral(-infinity to infinity)e^(-7t)u(t)*e^(-st)dt | ||
| + | |||
Limits change to 0 to infinity and u(t) drops out. | Limits change to 0 to infinity and u(t) drops out. | ||
| + | |||
H(s) = Integral(0 to infinity)e^(-7t)*e^(-st)dt | H(s) = Integral(0 to infinity)e^(-7t)*e^(-st)dt | ||
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H(s) = Integral(0 to infinity)e^(-7t-st)dt | H(s) = Integral(0 to infinity)e^(-7t-st)dt | ||
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H(s) = e^(-7t-st)/(-7-s) evaluated from 0 to inifinity | H(s) = e^(-7t-st)/(-7-s) evaluated from 0 to inifinity | ||
| + | |||
H(s) = e^(-7*infinity-s*infinity)/(-7-s) - e^(0)/(-7-s) | H(s) = e^(-7*infinity-s*infinity)/(-7-s) - e^(0)/(-7-s) | ||
| + | |||
H(s) = e^(-7*infinity-s*infinity)/(-7-s) goes to 0 | H(s) = e^(-7*infinity-s*infinity)/(-7-s) goes to 0 | ||
| + | |||
H(s) = 1/(7+s) | H(s) = 1/(7+s) | ||
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H(jw) = 1/(7+jw) | H(jw) = 1/(7+jw) | ||
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Revision as of 20:33, 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 impuls response $ h(t)=e^{-7t}u(t) $
H(s) = Integral(-infinity to infinity)h(t)*e^(-st)dt
H(s) = Integral(-infinity to infinity)e^(-7t)u(t)*e^(-st)dt
Limits change to 0 to infinity and u(t) drops out.
H(s) = Integral(0 to infinity)e^(-7t)*e^(-st)dt
H(s) = Integral(0 to infinity)e^(-7t-st)dt
H(s) = e^(-7t-st)/(-7-s) evaluated from 0 to inifinity
H(s) = e^(-7*infinity-s*infinity)/(-7-s) - e^(0)/(-7-s)
H(s) = e^(-7*infinity-s*infinity)/(-7-s) goes to 0
H(s) = 1/(7+s)
H(jw) = 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)
