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lecture10

# 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 + C (0.8) 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 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 2 (0.8) n ] u [ n ] Discrete-time version of slides 5-9 and 5-10 ] [ ] [ ] [ n u n m n y = ( 29 ] [ ] 1 [ ] [ ] 1 [ ] [ ] [ ] 1 [ ] 1 [ ] 1 [ ] 1 [ n n m n u n m n n u n m n u n m n y δ δ - - - = - - = - - = - balances impulsive events at origin

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10 - 4 Example: Impulse Response Need two values of h [ n ]
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