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<math>H(s)= \frac {5}{s}</math> | <math>H(s)= \frac {5}{s}</math> | ||
| + | |||
| + | |||
| + | == part B == | ||
| + | |||
| + | <math>h(t) = \sum_{-\infty} ^\infty a_k H(s) | ||
| + | </math> | ||
| + | |||
| + | since I got | ||
| + | |||
| + | <math>a_1 = \frac{3}{j}</math> | ||
| + | |||
| + | <math>a_2= \frac{-3}{j}</math> | ||
| + | |||
| + | <math>a_3 = 2 </math> | ||
| + | |||
| + | <math>a_4 = 2</math> | ||
| + | |||
| + | so, | ||
| + | |||
| + | <math> s + s + \frac{s}{j} -\frac{s}{j}= 2s</math> | ||
Latest revision as of 18:46, 25 September 2008
part A
$ e^{st}\rightarrow h(t)\rightarrow H(S)e^{st} $
$ H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt $
assume
$ h(t) = 5u(t-3) $
$ H(s) = 5\int_3^\infty e^{-st}dt $
$ H(s) = \frac{-5}{s} $
$ 4 to \infty $
$ H(s)= \frac {5}{s} $
part B
$ h(t) = \sum_{-\infty} ^\infty a_k H(s) $
since I got
$ a_1 = \frac{3}{j} $
$ a_2= \frac{-3}{j} $
$ a_3 = 2 $
$ a_4 = 2 $
so,
$ s + s + \frac{s}{j} -\frac{s}{j}= 2s $
