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<math> | <math> | ||
| − | \ w(t) = 5v(t) | + | \ w(t) = 5v(t-2) |
</math> | </math> | ||
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<math> | <math> | ||
| − | \ L[h(t)] = \int_{-\infty}^{+\infty} 5\delta(t)e^{-st}\, dt | + | \ L[h(t)] = \int_{-\infty}^{+\infty} 5\delta(t-2)e^{-st}\, dt |
</math> | </math> | ||
<math> | <math> | ||
| − | \ L[h(t)] = 5\int_{-\infty}^{+\infty} \delta(t)e^{-st}\, dt | + | \ L[h(t)] = 5\int_{-\infty}^{+\infty} \delta(t-2)e^{-st}\, dt |
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[h(t)] = 5\int_{-\infty}^{0^-} \delta(t-2)e^{-st}\, dt + 5\int_{0^-}^{+\infty} \delta(t-2)e^{-st}\, dt | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[h(t)] = 5\int_{0^-}^{+\infty} \delta(t-2)e^{-st}\, dt | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[h(t)] = 5\int_{2}^{+\infty} \delta(t)e^{-st}\, dt | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[h(t)] = 5\left [ e^{-(s)(\infty)} - e^{(-s)(2)} \right ] \quad | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[h(t)] = 5e^{-2s} | ||
| + | </math> | ||
| + | |||
| + | The response of this system to my signal y(t) = cos(t) from HW4.1 yields the output | ||
| + | |||
| + | <math> | ||
| + | \ L[cos(t)] = \int_{-\infty}^{+\infty} cos(t)e^{-st}\, dt | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \ L[cos(t)] = \frac{s}{s^2 + 1} | ||
</math> | </math> | ||
Latest revision as of 16:43, 26 September 2008
A continuous-time Linear Time-Invariant (LTI) system defined for the purpose of this page will be
$ \ w(t) = 5v(t-2) $
where v(t) is an input signal dependent on the parameter of time.
The unit impulse response of the system would then simply be
$ \ w(t) = 5\delta(t) $
and the system function H(s) of the system, where
$ \ s = j\omega $
can be determined by taking the Laplace Transform of the system's unit impulse response, h(t).
$ \ L[h(t)] = \int_{-\infty}^{+\infty} h(t)e^{-st}\, dt $
$ \ L[h(t)] = \int_{-\infty}^{+\infty} 5\delta(t-2)e^{-st}\, dt $
$ \ L[h(t)] = 5\int_{-\infty}^{+\infty} \delta(t-2)e^{-st}\, dt $
$ \ L[h(t)] = 5\int_{-\infty}^{0^-} \delta(t-2)e^{-st}\, dt + 5\int_{0^-}^{+\infty} \delta(t-2)e^{-st}\, dt $
$ \ L[h(t)] = 5\int_{0^-}^{+\infty} \delta(t-2)e^{-st}\, dt $
$ \ L[h(t)] = 5\int_{2}^{+\infty} \delta(t)e^{-st}\, dt $
$ \ L[h(t)] = 5\left [ e^{-(s)(\infty)} - e^{(-s)(2)} \right ] \quad $
$ \ L[h(t)] = 5e^{-2s} $
The response of this system to my signal y(t) = cos(t) from HW4.1 yields the output
$ \ L[cos(t)] = \int_{-\infty}^{+\infty} cos(t)e^{-st}\, dt $
$ \ L[cos(t)] = \frac{s}{s^2 + 1} $
