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

Revision as of 19:28, 18 July 2008

The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic

MAY BE: X(jw) is periodic only if x(t) is periodic

The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic

Alternate Solutions

Problem 6 - Using Frequency Domain

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-== Exam 1, Problem 6 - Summer 2008 ==
+The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic
-a) By Inspection the linear constant-coefficient differential equation that describes the LTI system is:
+MAY BE: X(jw) is periodic only if x(t) is periodic
-<math>y(t)= \frac{1}{7}x(t)- \frac{1}{7}\frac{dy(t)}{dt}</math>
+The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic
-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>
-<math> e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} </math>
-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:
-<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:
-<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:
-=0 indicating that it is correct.
-c) Find H(s) at s=jw for the LTI system with impulse response <math>h(t)=e^{-7t}u(t)</math>
-<math>H(s) = \int_{-\infty }^\infty h(t)e^{-st}dt</math>
-<math>H(s) = \int_{-\infty }^\infty e^{-7t}u(t)e^{-st}dt</math>
-Limits change to 0 to <math> \infty </math> and u(t) drops out.
-<math>H(s) = \int_0^\infty e^(-7t)e^{-st}dt</math>
-<math>H(s) = \int_0^\infty e^{-7t-st}dt</math>
-<math>H(s) = \frac{e^{-7t-st}}{-7-s} \bigg|_0^\infty</math>
-<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> Numerator goes to 0
-<math>H(s) = \frac{1}{7+s}</math>
-<math>H(jw) = \frac{1}{7+jw}</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) * h(t) </math> (where * indicates convolution)
-<math> x(t) = e^{j7t} </math>
-<math> h(t) = e^{-7t}u(t) </math>
-Using the properties of convolution we will choose to convolve using <math> x(t-\tau) </math> and <math> h(\tau) </math> for simplicity.
-<math> y(t) = \int_{-\infty}^\infty e^{j7(t-\tau)}e^{-7\tau}u(\tau)d\tau </math>
-Drop the u(t) and change the integration limits.
-<math> y(t) = \int_0^\infty e^{j7(t-\tau)}e^{-7\tau}d\tau </math>
-Simplify the exponentials.
-<math> y(t) = \int_0^\infty e^{-7\tau + j7t - j7\tau}d\tau </math>
-<math> y(t) = \frac{e^{-7\tau + j7\tau - j7\tau}}{-7-j7} \bigg|_0^\infty </math>
-<math> y(t) = \frac{e^{-\infty}}{-7-j7} + \frac{e^{j7t}}{7+7j} </math>
-<math> y(t) = \frac{e^{j7t}}{7+7j} </math>
 ==Alternate Solutions==
 Problem 6 - Using Frequency Domain

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

Revision as of 19:28, 18 July 2008

Alternate Solutions

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