Lecture 15

Lecture 15 - ECE 3101 Signals and Systems Reading...

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Lecture 15, 3/17/11 ECE 3101, Copyright P. B. Luh 1 ECE 3101 Signals and Systems Reading Assignment: Sections 5.1 and 5.2 Problem Set 6: Due Thursday next week Exam. 2, Tuesday next week, end of Ch. 3; 6 cheat sheets Last Time: Stability. If for every bounded u(t), y(t) is always bounded BIBOS < - 0 dt ) t ( h All the poles of H(s) have strictly negative real parts
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Lecture 15, 3/17/11 ECE 3101, Copyright P. B. Luh 2 Steady state response to e j ϖ t , assuming BIBOS h(t) or H(s) U(s) Y(s) u(t) = cos ϖ t Y ZS (s) = H(s)U(s) u(t) = e j ϖ t y ss (t) = H(j ϖ )e j ϖ t y ss (t) = |H(j ϖ )| cos( ϖ t+ H(j ϖ )) u(t) = sin ϖ t y ss (t) = |H(j ϖ )| sin( ϖ t+ H(j ϖ )) H(j ϖ ): Frequency response function , a function of ϖ u(t) = Σ c n e j ϖ n t y ss (t) = Σ c n H(j ϖ n )e j ϖ n t The Z -Transform: F(z) = Σ k from 0 to f[k]z -k Examples
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3 Today: Sample of problems for Exam 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] Convolution: h[k] f[k] H(z) F(z) Next Lecture: Sections 5.1 and 5.2 ; 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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4 3. Evaluate the convolution a. For 0 t < 2: f 2 (t) 3 6 2 f 1 (t) 2 4 -2 t t f 2 (- τ ) 3 -6 2 f 1 ( τ ) 2 4 -2 τ τ - τ = t 0 2 1 dt ) t ( f ) ( f ) t ( c t 6 dt 2 3 ) t ( c t 0 = = b. For 2 t < 4: t 6 24 dt 2 3 dt 2 3 ) t ( c t 2 2 0 - = - = c. For 4 t < 6: 0 dt 2 3 dt 2 3 ) t ( c 4 2 2 0 = - = d. For 6 t < 8: t 6 36 dt 2 3 dt 2 3 ) t ( c 4 2 2 6 t - = - = - e. For 8 t < 10: 60 t 6 dt 2 3 ) t ( c 4 6 t - = - = - e. For t > 10: c(t) = 0
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5 4. Compute h[k] for the following system y[k+2] – 2y[k+1] + y[k] = u[k+1] – u[k] Identify the coefficients: y[k+2] + a 1 y[k+1] + a 2 y[k] = b 0 u[k+2] + b 1 u[k+1] + b 2 u[k] a 1 = -2, a 2 = 1, b 0 = 0, b 1 = 1, b 2 = -1 h[k] = (b 2 /a 2 ) δ [k] + y c [k]U[k] To find the natural modes: Characteristic polynomial γ 2 – 2 γ + 1 = ( γ – 1) 2 = 0. A double root at γ = 1 Natural modes: γ k and k γ k ; or 1 and k in view that γ = 1 To find h[0] and h[1]: k = -2: h[0] – 2h[-1] + h[-2] = δ [-1] – δ [-2] = 0 h[0] = 0 k = -1: h[1] – 2h[0] + h[-1] = δ [0] – δ [-1] h[1] = 1 h[k] = (b 2 /a 2 ) δ [k] + y c [k]U[k] = - δ [k] + [c 1 + c 2 k]U[k]
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Lecture 15, 3/17/11 ECE 3101, Copyright P. B. Luh 6 Find the convolutions of the following (1) directly, and
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This note was uploaded on 03/29/2011 for the course ECE 3101 taught by Professor Luh during the Spring '11 term at UConn.

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Lecture 15 - ECE 3101 Signals and Systems Reading...

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