| Line 7: | Line 7: | ||
h(t) = 10delta(t-1) | h(t) = 10delta(t-1) | ||
| − | <math>H(s) = \int_{-\infty}^{\infty} h(t)e^{-st}</math> | + | <math>H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}</math> |
| − | <math>H(s) = \int_{-\infty}^{\infty} 10delta(t-1)e^{-st}</math> | + | <math>H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st}</math> |
| − | <math>H(s) = 10\times\int_{-\infty}^{\infty} | + | <math>H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st}</math> |
Revision as of 11:00, 25 September 2008
CT LTI system:
y(t) = 10x(t-1)
plugging in delta(t) into the system we get:
h(t) = 10delta(t-1)
$ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st} $
$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $
$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $
