Difference between revisions of "Problem 6 Old Kiwi" - Rhea

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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} $

Alternate Solutions

Problem 6 - Using Frequency Domain