(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(s) = \int^{\infty}_{- \infty} h(t) e^{- s t} dt</math>
 +
 
 +
<math>H(s) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- s t}</math>
 +
 
 +
<math>H(s) = 3 e^{0} + 7 e^{- 3 s} </math>
 +
 
 +
<math>H(s) = 3 + 7 e^{- 3 s}</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>

Revision as of 08:44, 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(s) = \int^{\infty}_{- \infty} h(t) e^{- s t} dt $

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

$ H(s) = 3 e^{0} + 7 e^{- 3 s} $

$ H(s) = 3 + 7 e^{- 3 s} $

The signal used in question 1:

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

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Ryne Rayburn