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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

2 9 - 2 Iterative Solutions Example: y[n] - ½ y[n-1] = x[n] How many initial conditions do we need? For x[n] = n 2 u[n] and y[-1] = 16, causal system, y[n] = ½ y[n-1] + x[n] Compute answer iteratively: y[0], y[1], … y[0] = ½ y[-1] + x[0] = ½ (16) + 0 = 8 y[1] = ½ (8) + (1) 2 = 5 y[2] = 6.5 y[3] = 12.25 y[4] = 22.125

3 9 - 3 Stability Is the system bounded-input bounded-output (BIBO) stable? y[n] - ½ y[n-1] = x[n] Impulse response is the output y[n] for input x[n] =  [n] Zero-state response y s [n] to input  [n] is 1, 0.5, 0.25, …, for n = 0, 1, 2, … System appears to be BIBO stable. x[n] = n 2 u[n] is unbounded in amplitude as n goes to infinity

4 9 - 4 Zero-Input, Zero-State Solutions Example: y[n] - ½ y[n-1] = x[n] with y 0 [-1] = 16 Zero-input solution y 0 [n] - ½ y 0 [n-1] = 0 y 0 [n] = ½ y 0 [n-1] y 0 [n] = 8 (½) n u[n] 8, 4, 2, 1, ½, … Zero-state solution y s [n] - ½ y s [n-1] = n 2 y s [0] = 0 y s [1] = (½) 0 + (1) 2 = 1 y s [2] = (½) 1 + (2) 2 = 4.5 Important identity

5 9 - 5 Complete Solution Closed-form solution of impulse response Algebraic solution (not numerical or iterative solution) y[n] = zero-input solution + zero-state solution y[n] = y 0 [n] + y s [n] Using Mathematica on example on previous slide Needs[ “DiscreteMath`Master`” ]; RSolve[ { y[n] - (1/2) y[n-1] == n^2, y[-1] == 16 }, y[n], n] Complete solution (for non-negative n) y[n] = -14 ( n + 2 ) + 2 ( n + 2 ) ( n + 3 ) + 22 + 2 (½) n

6 9 - 6 Forms of Difference Equations Anti-causal (advance operator form) Causality requires M  N Set M = N; delay input/output by N samples Iterative solution 2N initial conditions (N initial conditions if causal input)

7 9 - 7 Characteristic Modes Continuous Time Discrete TimeCase non-repeated roots repeated roots

8 9 - 8 Zero Input Response Linear combination of characteristic modes Type of RootsCharacteristic Polynomial and Zero-Input Response Distinct Repeated Complex Slide by Prof. Adnan Kavak

9 9 - 9 Distinct Roots

10 9 - 10 Repeated Roots


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