Revision as of 20:33, 30 June 2008 by Magasser (Talk)

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)

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