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Difference Equations Linear Systems and Signals Lecture 9 Spring 2008
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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] + f[n] Compute answer iteratively: y[0], y[1], … y[0] = ½ y[-1] + f[0] = ½ (16) + 0 = 8 y[1] = ½ (8) + (1) 2 = 5 y[2] = 6.5 y[3] = 12.25 y[4] = 22.125
![9 - 2 Iterative Solutions Example: y[n] - ½ y[n-1] = x[n] How many initial conditions do we need.](http://images.slideplayer.com./17/5381178/slides/slide_2.jpg "For x[n] = n 2 u[n] and y[-1] = 16, causal system, y[n] = ½ y[n-1] + f[n] Compute answer iteratively: y[0], y[1], … y[0] = ½ y[-1] + f[0] = ½ (16) + 0 = 8 y[1] = ½ (8) + (1) 2 = 5 y[2] = 6.5 y[3] = y[4] =")
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9 - 3 Stability Is the system stable? y[n] - ½ y[n-1] = x[n] Impulse response occurs when x[n] = [n] Values zero-state response y s [n] to input [n] are 1, 0.5, 0.25, …, for n = 0, 1, 2, … System appears to be stable. x[n] = n 2 u[n] is unbounded
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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
![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](http://images.slideplayer.com./17/5381178/slides/slide_4.jpg "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")
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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
![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 ) (½) n](http://images.slideplayer.com./17/5381178/slides/slide_5.jpg "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 ) (½) n")
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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) Lathi, p. 271
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9 - 7 Operator Notation E operator E{x[n]} = x[n + 1] E 2 {x[n]} = x[n + 2] … E k {x[n]} = x[n+ k] What is E -1 {x[n]}? E -1 {x[n]} = x[n - 1] E k {x[n]} = x[n + k] for any integer n First-order difference equation y[n + 1] + a 0 y[n] = x[n + 1] (E + a 0 ) y[n] = E{x[n]} Zero-input response: solve Q( ) = 0 = -a 0 y 0 [n] = C (-a 0 ) n u[n]
![9 - 7 Operator Notation E operator E{x[n]} = x[n + 1] E 2 {x[n]} = x[n + 2] … E k {x[n]} = x[n+ k] What is E -1 {x[n]}.](http://images.slideplayer.com./17/5381178/slides/slide_7.jpg "E -1 {x[n]} = x[n - 1] E k {x[n]} = x[n + k] for any integer n First-order difference equation y[n + 1] + a 0 y[n] = x[n + 1] (E + a 0 ) y[n] = E{x[n]} Zero-input response: solve Q( ) = 0 = -a 0 y 0 [n] = C (-a 0 ) n u[n].")
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9 - 8 Characteristic Modes Continuous Time Discrete TimeCase non-repeated roots repeated roots
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