ECE3101S11Lec160

ECE3101S11Lec160 - ECE 3101 Signals and Systems Reading...

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Lecture 16, 3/24/11 ECE 3101, Copyright P. B. Luh 1 ECE 3101 Signals and Systems Reading Assignment: Sections 5.1-5.3 Problem Set 6: Due today Problem Set 7: 2.6-1 (BIBOS only), 2.6-4, 4.5-1, 4.5-3, 4.6-13 (deriving H(s) for the three configurations only), 4.8-1, 5.1-2 (b, d, f, h, j, and l only), 5.1-4 ( b, d, f , and h). Due next Thursday Project 1: Due Tuesday 4/7/11 Last Time: The Z -Transform: F(z) = Σ k from 0 to f[k]z -k 0 0; δ [k] 1; δ [k-1] z -1 ; δ [k-m] z -m ; z z ] k [ U k γ - γ ; 1 z z ] k [ U - ( 29 2 1 z z ] k [ kU -
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Lecture 16, 3/24/11 ECE 3101, Copyright P. B. Luh 2 Properties of the Z -Transform • Linearity: a 1 f 1 [k] + a 2 f 2 [k] a 1 F 1 (z) + a 2 F 2 (z) Shifting to the right: f[k-1] z -1 F(z) + f[-1] f[k-m] z -m F(z) + z -m+1 f[-1] + z -m+2 f[-2] + . .+ f[-m], m > 0 f[k-m]u[k-m] z -m F(z), m > 0 Shifting to the left: f[k+1] zF(z) - zf[0] f[k+m] z m F(z) - z m f[0] - z m-1 f[1] - . .- zf[m-1], m > 0 Convolution: h[k] f[k] H(z) F(z) ; z z ] k [ U k γ - γ 2 1 k ) z ( z ] k [ U k γ - γ - m 1 m k ) z ( z ] k [ U )! 1 m ( ) 2 m k ( ) 1 k ( k γ - γ - + - - + -
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3 Multiplication by k: kf[k] -zF (1) (z) Scaling in the z-plane: γ k f[k]U[k] F(z/ γ ) • Partial Sum: Σ m from 0 to k f[m] (z/z-1)F(z) Today: • Initial Value Theorem: f[0] = lim z →∞ F(z) • Final Value Theorem: f[ ] = lim z 1 (z-1)F(z) if all the poles of (z-1)F(z) are strictly inside the unit circle Inverse Z -Transform Z -Transform Solution of Difference Eqs Next Time: Sections 5.3, 10.6, 3.10, and 5.5 Z -Transform Solution of Difference Eqs (Continued) State-Space Analysis of Discrete-Time Systems System Stability Frequency Response of Discrete-Time Systems
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Lecture 16, 3/24/11 4 Initial Value Theorem and Final Value Theorem f[k] F(z) ~ Obtain f[0] and f[ ] from F(z) Initial Value Theorem: f[0] = lim z →∞ F(z) Final Value Theorem: f[ ] = lim z 1 (z-1)F(z) if all the poles of (z-1)F(z) are strictly inside the unit circle + + + + + - - - k 2 1 z ] k [ f z ] 2 [ f z ] 1 [ f ] 0 [ f ) z ( F f(0 ) = lim sF(s) ] 0 [ f ) z ( F lim z = Example. γ - = z z ) z ( F 1 z z lim ] 0 [ f z = γ - = ? z z ) 1 z ( lim ] [ f 1 z = γ - - = γ γ = γ = γ < γ = 1 | | NA 1 , 1 | | NA 1 1 1 | | 0 ] [ f f[k] = γ k U[k] Proof of IVT:
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Lecture 16, 3/24/11 ECE 3101, Copyright P. B. Luh 5 Example. 2 ) z ( z ) z ( F γ - = 0 ) z ( z lim ] 0 [ f 2 z = γ - = ? ) z ( z ) 1 z ( lim ] [ f 2 1 z = γ - - = γ γ = γ = γ < γ = 1 | | NA 1 , 1 | | NA 1 NA 1 | | 0 ] [ f f[k] = k γ k-1 U[k], f[ ] doesn’t exist unless | γ | < 1
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Example. A discrete-time function f[k] has F(z) = . 1
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ECE3101S11Lec160 - ECE 3101 Signals and Systems Reading...

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