| Line 1: | Line 1: | ||
| + | == A == | ||
| + | |||
<font size="3">Let <math>y(t)=\int_{-\infty}^{\infty}2x(t)dt</math> | <font size="3">Let <math>y(t)=\int_{-\infty}^{\infty}2x(t)dt</math> | ||
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<math>=(\frac{-2}{s}e^{-st})|_{0}^{\infty}</math> | <math>=(\frac{-2}{s}e^{-st})|_{0}^{\infty}</math> | ||
| − | <math>=\frac{2}{s}</math> | + | <math>=\frac{2}{s}</math></font> |
| − | + | == B == | |
Revision as of 12:51, 24 September 2008
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} $
