Line 7: Line 7:
 
The input, cos(2t) is equal to <math>\frac{1}{2}(e^{j2t} + e^{-j2t})</math>
 
The input, cos(2t) is equal to <math>\frac{1}{2}(e^{j2t} + e^{-j2t})</math>
  
From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t)</math>,
+
From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t)</math>,  
 +
where <math>a = b = \frac{1}{2}</math>
  
 
The response to cos(2t) is <math>\frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math>
 
The response to cos(2t) is <math>\frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math>

Revision as of 08:10, 18 September 2008

Part B: The basics of linearity

$ x_1(t) = e^{2jt} \rightarrow linear-system \rightarrow y_1(t) = te^{-2jt} $

$ x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} $

The input, cos(2t) is equal to $ \frac{1}{2}(e^{j2t} + e^{-j2t}) $

From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t) $, where $ a = b = \frac{1}{2} $

The response to cos(2t) is $ \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood