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<math>
 
<math>
 
\ L[h(t)] = 5\int_{0^-}^{+\infty} \delta(t-2)e^{-st}\, dt
 
\ 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>
  

Revision as of 16:20, 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)] = \left [ e^{-2s} \right ] \quad $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn