(HW 4.1 Response)
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The convolution of the input signal and the system.
 
The convolution of the input signal and the system.
  
<math>7e^{-4jt}\frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + 7e^{-4jt}\frac{1}{2}(e^{j3\pi t} + 7e^{-4jt}e^{-j3\pi t}) + e^{j2\pi t}</math>
+
<math>7e^{-4jt}\frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + 7e^{-4jt}\frac{1}{2}(e^{j3\pi t} +e^{-j3\pi t}) + 7e^{-4jt}e^{j2\pi t}</math>

Revision as of 17:36, 26 September 2008

The CT LTI Signal

$ y(t) = 7x(t-4) \! $

The Unit Impulse Response

$ y(t) = 7x(t-4) \! $

$ h(t) = 7\delta(t-4) \! $

$ H(s) = 7e^{-4s} \! $

HW 4.1 Response

HW 4.1 signal

$ \frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + \frac{1}{2}(e^{j3\pi t} + e^{-j3\pi t}) + e^{j2\pi t} $

The response is equal to

The convolution of the input signal and the system.

$ 7e^{-4jt}\frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + 7e^{-4jt}\frac{1}{2}(e^{j3\pi t} +e^{-j3\pi t}) + 7e^{-4jt}e^{j2\pi t} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn