Lecture 6

# Lecture 6 - ECE 3101 Signals and Systems Reading Assignment...

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Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 1 Reading Assignment: Sec. 2.4 Problem Set 2: Due next Tuesday Problem Set 3: Due Tuesday 2/22/11. 2.2-5, 2.3-2, 2.3-4, 2.4- 2, 2.4-10, 2.4-18 (a, c, e, and g), 2.4-37 Last Time: Time domain analysis of input/output descriptions Response caused by initial conditions: Zero input response Based on characteristic polynomial Response caused by input: Zero state response Based on linearity, time invariance, and response to δ (t) ( unit impulse response ) Total response: y(t) = y ZI (t) + y ZS (t) ~ Based on linearity ECE 3101 Signals and Systems

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Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 2 y (N) + a 1 y (N-1) + . . + a N-1 y (1) + a N y=b N-M u (M) + . . + b N u Q(D)y(t) = P(D)u(t) y ZI (t): Characteristic polynomial and characteristic modes Q( λ ) = λ N + a 1 λ N-1 + . . + a N-1 λ + a N = ( λ - λ 1 ) ( λ - λ 2 ) . . ( λ - λ N ) All roots are distinct: y 0 (t) = c 1 e λ 1 t + c 2 e λ 2 t + . . + c N e λ N t With r-repeated root ( λ - λ 1 ) r : c 11 e λ 1t + c 12 te λ 1 t + . .+ c 1r t r-1 e λ 1 t With complex roots α± j β : c e α t cos( β t+ θ ) Unit Impulse Response: h(t) = b 0 δ (t) + [P(D)y n (t)]U(t) y n (t): Unit impulse response to Q(D)w(t) = u(t), or zero input response to that system with y n (N-1) (0 + ) = 1; y n (0 + ) = y n (1) (0 + ) = . . = y n (N-2) (0 + ) = 0
Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 3 Today: Zero State Response: Convolution – linearity/time invariance Convolution of Continuous-Time Signals Direct integration Semi-graphical method Properties of Convolution Examples Next Time: Classical Solution of Differential Equations, Section 2.5 Time-Domain Analysis of D-T Systems Sections 3.6-3.8, 3.12

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Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 4 δ (t) Input System Output Zero State Response, y ZS (t): Response of a system to input, assuming zero initial conditions u(t) y ZS (t) = ? h(t) u(t) = u 0 (t) + u 1 (t) + . . + u m (t) + . . u 0 (t) y 0 (t) [u(0) Δτ ] h(t) u 1 (t) y 1 (t) [u( Δτ ) Δτ ] h(t- Δτ ) (linearity and time invariance) : y(t) = y 0 (t) + y 1 (t) + . . + y m (t) + . . (linearity) t u(t) u(m Δτ ) Δτ Δτ m Δτ u 0 (t) u 1 (t) u m (t) Zero State Response u(0) Δτ
Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 5 Major Result: y(t) = h(t) u(t) τ τ τ = τ τ τ t 0 t 0 d ) t ( h ) ( u d ) t ( u ) ( h δ (t) h(t) δ (t-m Δτ ) h(t-m Δτ ) [u(m Δτ ) Δτ ] δ (t-m Δτ ) [u(m Δτ ) Δτ ] h(t-m Δτ ) u m (t) y m (t) u(t) = u 0 (t) + u 1 (t) + . . + u m (t) + . . y(t) = y 0 (t) + y 1 (t) + . . + y m (t) + . . = Σ m y m (t) = Σ m [u(m Δτ ) Δτ ] h(t-m Δτ ) ~ When Δτ 0, this becomes Convolution . How to get it? t u(t) u(m Δτ ) Δτ Δτ m Δτ u 0 (t) u 1 (t) u m (t) τ τ τ = t 0 d ) t ( h ) ( u τ τ τ = t 0 d ) t ( u ) ( h ~ with a change of variable

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Lecture 6, Makeup, 2/10/11 ECE 3101, Copyright P. B. Luh 6 What does the above mean?
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Lecture 6 - ECE 3101 Signals and Systems Reading Assignment...

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