lecture10 - EE313 Linear Systems and Signals Fall 2010...

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EE313 Linear Systems and Signals Fall 2010 Initial conversion of content to PowerPoint by Dr. Wade C. Schwartzkopf Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin Difference Equations and Stability
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10 - 2 [ ] [ ] [ ] Input Response Impulse n x n h n y s = Example: Second-Order Equation y [ n +2] - 0.6 y [ n +1] - 0.16 y [ n ] = 5 x [ n +2] with y [-1] = 0 and y [-2] = 6.25 and x [ n ] = 4 - n u[ n ] Zero-input response Characteristic polynomial γ 2 - 0.6 γ - 0.16 = ( γ + 0.2) ( γ - 0.8) Characteristic equation ( γ + 0.2) ( γ - 0.8) = 0 Characteristic roots γ 1 = -0.2 and γ 2 = 0.8 Solution y 0 [ n ] = C 1 (-0.2) n +
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10 - 3 Example: Impulse Response h [ n +2] - 0.6 h [ n +1] - 0.16 h [ n ] = 5 δ [ n +2] with h [-1] = h [-2] = 0 because of causality General form of impulse response h [ n ] = ( b N / a N ) δ [ n ] + y 0 [ n ] u [ n ] Since a N = -0.16 and b N = 0 , h [ n ] = y 0 [ n ] u [ n ] = [ C 1 (-0.2) n + C
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This note was uploaded on 02/16/2011 for the course ECE 313 taught by Professor Evans during the Spring '11 term at University of Texas at Austin.

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lecture10 - EE313 Linear Systems and Signals Fall 2010...

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