(New page: Let us use the CT LTI system: <math>y(t) = x(3t) + x(t+3)</math> ---- The impulse response, h(t), of this system is computed using the following: <math>x(t) = \delta (t)</math> <math>...)
 
 
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Let us use the CT LTI system:
 
Let us use the CT LTI system:
  
<math>y(t) = x(3t) + x(t+3)</math>
+
<math>y(t) = 3x(t) + 7x(t+3)</math>
  
 
----
 
----
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<math>x(t) = \delta (t)</math>
 
<math>x(t) = \delta (t)</math>
  
<math>h(t) = \delta (3t) + \delta (t+3)</math>
+
<math>h(t) = 3\delta (t) + 7\delta (t+3)</math>
 +
 
 +
The system function, H(s) is:
 +
 
 +
<math>H(j w) = \int^{\infty}_{- \infty} h(t) e^{- j w t} dt</math>
 +
 
 +
<math>H(j w) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- j w t}</math>
 +
 
 +
<math>H(j w) = 3 e^{0} + 7 e^{- j 3 t} </math>
 +
 
 +
<math>H( jw) = 3 + 7 e^{- j 3 t}</math>
 +
 
 +
----
  
 
The signal used in question 1:
 
The signal used in question 1:
  
 
<math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math>
 
<math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math>
 +
 +
<math>= \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}</math>
 +
 +
The response, y(t) = H(jw)*x(t)
 +
 +
<math>y(t) = (3 + 7 e^{- j 3 t}) * (\frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}) </math>

Latest revision as of 08:55, 26 September 2008

Let us use the CT LTI system:

$ y(t) = 3x(t) + 7x(t+3) $


The impulse response, h(t), of this system is computed using the following:

$ x(t) = \delta (t) $

$ h(t) = 3\delta (t) + 7\delta (t+3) $

The system function, H(s) is:

$ H(j w) = \int^{\infty}_{- \infty} h(t) e^{- j w t} dt $

$ H(j w) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- j w t} $

$ H(j w) = 3 e^{0} + 7 e^{- j 3 t} $

$ H( jw) = 3 + 7 e^{- j 3 t} $


The signal used in question 1:

$ x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t} $

$ = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t} $

The response, y(t) = H(jw)*x(t)

$ y(t) = (3 + 7 e^{- j 3 t}) * (\frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}) $

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Questions/answers with a recent ECE grad

Ryne Rayburn