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Lecture 14

# Lecture 14 - ECE 3101 Signals and Systems Reading...

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Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 1 ECE 3101 Signals and Systems Reading Assignment: Sections 2.6, 4.8 and 5.1 Problem Set 5: Due today Problem Set 6: Due Thursday next week Exam. 2, Tuesday next week, end of Ch. 3; 6 cheat sheets Last Time: Transformed Network Method: with Z(s) R, Ls, 1/Cs V(s) = Z(s)I(s) ~ Analyzed as a resistive circuit ! Block Diagrams: Connecting components to form a system

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Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 2 H 1 (s) H 2 (s) U(s) Y(s) H 1 (s) H 2 (s) U(s) Y(s) + H 1 (s) U(s) Y(s) + H 2 (s) Serial: H(s) = H 1 (s)H 2 (s) ; Parallel: H(s) = H 1 (s) + H 2 (s) Feedback: V(s) ); s ( U ) s ( H ) s ( H 1 1 ) s ( V 2 1 - = ) s ( U ) s ( H ) s ( H 1 ) s ( H ) s ( V ) s ( H ) s ( Y 2 1 1 1 - = = H(s)
Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 3 Today: Sections 2.6, 4.8 and 5.1 System Stability (Continued) Frequency Response to Sinusoidal Inputs Discrete-Time Systems: The Z -Transform Analysis Next Lecture: Sections 5.1 and 5.2

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Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 4 If for every bounded u(t), y(t) is always bounded BIBOS It is related to the above intuitive concept in that any “small” disturbance will remain “small” h(t) or H(s) U(s) Y(s) How to check stability? Is an integrator stable? BIBOS < - 0 dt ) t ( h All the poles of H(s) have strictly negative real parts Proof. Suppose that h(t) is absolutely integratable and |u(t)| M < ) t ( u * ) t ( h ) t ( y = τ τ - τ = - t 0 d ) t ( u ) ( h τ τ - τ = - t 0 d ) t ( u ) ( h ) t ( y τ τ - t 0 d ) ( h M τ τ - τ - t 0 d ) t ( u ) ( h
Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 5 The other direction: By contradiction Relating to poles of H(s): Recall that for a system described by y (N) + a 1 y (N-1) + .. + a N-1 y (1) + a N y = b 0 u (N) + b 1 u (N-1) + .. + b N-1 u (1) + b N u h(t) = b 0 δ (t) + [P(D)y n (t)]U(t), where y n (t): Unit impulse response to Q(D)w(t) = u(t) with y n (N-1) (0) = 1; y n (0) = y n (1) (0) = .. = y n (N-2) (0) = 0 Combination of e λ t , te λ t , t 2 e λ t ,…, t r-1 e λ t If there is a pole whose real part is not strictly negative, then < - 0 dt ) t ( h

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6 Example ) 3 s )( 2 s ( 1 ) s ( H + + = Poles: -2, -3 ~ BIBOS Example 1/s U(s) Y(s) 1/s 2 s 1 ) s ( H = Poles: 0, 0 ~ Not stable Can the system be stabilized by using output feedback as above? 1/s 2 U(s) Y(s) + a - 2 2 s 1 a 1 s 1 ) s ( H + = ; a s 1 2 + = Poles: < ± ± 0 a if | a | 0 a if a j ~ Not stable What to do? 1/s 2 U(s) Y(s) + s + a - 2 2 s 1 ) a s ( 1 s 1 ) s ( H + + = a s s 1 2 + + = PD controller
Lecture 14, 3/15/11 ECE 3101, Copyright P. B. Luh 7 Stable? a < 0: One positive pole and one negative pole. Not stable a = 0, Poles are at -1 and 0. Not stable 0 < a < 0.25: Two negative poles.

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