(New page: == Part A == Consider the system: <math>y(t)=\int_{-\infty}^{\infty}3x(t-1)dt</math> The unit impulse response is then <math>h(t) =3u(t-1)</math> Using <math>H(s) = \int_{-\infty}^{\...) |
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== Part B == | == Part B == | ||
| − | + | The Fourier series coefficients in <math>x(t)=cos(3 \pi t) + sin(8 \pi t)</math> are: | |
| − | <math>a_{ | + | <math>a_{3} = \frac{1}{2}</math> |
| − | <math> | + | <math>a_{-3} = \frac{1}{2}</math> |
| − | + | <math>a_{8} = \frac{1}{2j}</math> | |
| − | + | <math>-a_{-8} = \frac{1}{2j}</math> | |
| + | |||
| + | For <math> k \neq [3,-3,8,-8], a_{k}</math> = 0 | ||
| + | |||
| + | The response of the input <math>x(t)</math> to the system <math>y(t)</math> using <math>H(s)</math> and the above coefficients is: | ||
<math>y(t)=\sum_{k=-\infty}^{\infty}a_{k}H(s)</math> | <math>y(t)=\sum_{k=-\infty}^{\infty}a_{k}H(s)</math> | ||
Latest revision as of 16:31, 26 September 2008
Part A
Consider the system:
$ y(t)=\int_{-\infty}^{\infty}3x(t-1)dt $
The unit impulse response is then $ h(t) =3u(t-1) $
Using $ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}dt $
we find that
$ H(s) = \int_{-\infty}^{\infty}3u(t-1)e^{-st}dt $
$ =\int_{1}^{\infty}3e^{-st}dt $
$ =(\frac{-3}{s}e^{-st})|_{1}^{\infty} $
$ =\frac{3}{s} $
Part B
The Fourier series coefficients in $ x(t)=cos(3 \pi t) + sin(8 \pi t) $ are:
$ a_{3} = \frac{1}{2} $
$ a_{-3} = \frac{1}{2} $
$ a_{8} = \frac{1}{2j} $
$ -a_{-8} = \frac{1}{2j} $
For $ k \neq [3,-3,8,-8], a_{k} $ = 0
The response of the input $ x(t) $ to the system $ y(t) $ using $ H(s) $ and the above coefficients is:
$ y(t)=\sum_{k=-\infty}^{\infty}a_{k}H(s) $
$ =\frac{1}{s} + \frac{1}{s} + \frac{1}{sj} - \frac{1}{sj} $
$ =\frac{2}{s} $
