(New page: <font size="3">Let <math>x(t)=cos(4 \pi t) + sin(6 \pi t)</math> Then <math>h(t) =</math> </font>) |
|||
| Line 1: | Line 1: | ||
| − | <font size="3">Let <math> | + | <font size="3">Let <math>y(t)=\int_{-\infty}^{\infty}2x(t)dt</math> |
| + | |||
| + | Then <math>h(t) =2u(t)</math> | ||
| + | |||
| + | And <math>H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}dt</math> | ||
| + | |||
| + | <math>=\int_{-\infty}^{\infty}2u(t)e^{-st}dt</math> | ||
| + | |||
| + | <math>=\int_{0}^{\infty}2e^{-st}dt</math> | ||
| + | |||
| + | <math>=(\frac{-2}{s}e^{-st})|_{0}^{\infty}</math> | ||
| + | |||
| + | <math>=\frac{2}{s}</math> | ||
| − | |||
</font> | </font> | ||
Revision as of 11:50, 24 September 2008
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} $
