# lecture10 - EE313 Linear Systems and Signals Spring 2009...

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EE313 Linear Systems and Signals Spring 2009 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 +
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 In general, from Lathi (3.49), 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 01/22/2012 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.

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lecture10 - EE313 Linear Systems and Signals Spring 2009...

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