Line 6: Line 6:
  
 
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>

Revision as of 08:04, 18 September 2008

Part B: The basics of linearity

$ e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} $

$ e^{-2jt} \rightarrow linear-system \rightarrow 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) $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett