Revision as of 12:53, 24 September 2008 by Mjwhitta (Talk)

A

Let $ y(t)=\int_{-\infty}^{\infty}2x(t)dt $

Then $ h(t) =2u(t) $

And $ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}dt $

$ =\int_{-\infty}^{\infty}2u(t)e^{-st}dt $

$ =\int_{0}^{\infty}2e^{-st}dt $

$ =(\frac{-2}{s}e^{-st})|_{0}^{\infty} $

$ =\frac{2}{s} $

B

Let $ x(t)=cos(4 \pi t) + sin(6 \pi t) $ with Fourier series coefficients are as follows:

$ a_{4} = a_{-4} = \frac{1}{2} $

$ a_{6} = -a_{-6} = \frac{1}{2j} $

All other $ a_{k} $ values are 0

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood