ece_350_lecture4_sp10_v5

ece_350_lecture4_sp10_v5 - ECE 350 Lecture 4 Lecture...

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ECE 350 D. van Alphen 1 ECE 350 – Lecture 4 Lecture Overview • Convolution Tables • Interconnected Systems Example • Total Response of a Linear, Time-Invariant System • Some MATLAB Tools for Time-Domain Analysis • System Stability • Invertible Systems • Impulse Response – Multipath Channel
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ECE 350 D. van Alphen 2 No. x 1 (t) x 2 (t) x 1 (t) * x 2 (t) 1x ( t ) δ (t – T) x(t – T) 2e λ t u(t) u(t) 3 u(t) u(t) t u(t) 4e λ 1 t u(t) e λ 2 t u(t) ( λ 1 ≠λ 2 ) ………… ) t ( u e 1 t λ λ Convolution Tables (Lathi Table 2.1, p. 177) ) t ( u e e 2 1 t t 2 1 λ λ λ λ
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ECE 350 D. van Alphen 3 Convolution Tables – An Example Find the zero-state response for a system with impulse response: h(t) = e -3t u(t) if the input is x(t) = u(t). Note: The zero-state response when the input is the step function, u(t), is called the step response of the system. Solution: We want: y(t) = x(t) * h(t) = u(t) * [e -3t u(t)] Using entry #2, with λ = -3: _____ __________ ) t ( u e 1 ) t ( y t = λ = λ
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ECE 350 D. van Alphen 4 Interconnected Systems: In-class Example Find the impulse response for the over-all system below, given that: h 1 (t) = u(t) h 2 (t) = 3 u(t - 1) h 3 (t) = -2 δ (t – 1) h 4 (t) = e -2t u(t) h 1 (t) h 2 (t) Σ h 4 (t) h 3 (t)
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ECE 350 D. van Alphen 5 Total Response of LTI Systems Recall: For an LTI system with impulse response h(t) and input x(t) Total Response = Zero-Input Response + Zero-State Response = Example below: From Lecture 2, pp. 20 – 21, we obtained the diff. eq: ) t ( h ) t ( x e c t N 1 k k k + λ = (assumes distinct roots for characteristic equation) ± x(t) 1 H 3 Ω ½F + v C (t) - y(t) (D 2 +3D+2) y(t) = D x(t)
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ECE 350 D. van Alphen 6 Total Response, Example continued From Lecture 3, pp. 6 – 7, we found that a system with diff. eq: In Chapter 4, we will learn how to find the impulse response, which is: h(t) = ( 2e -2t - e -t ) u(t) So, if the input is x(t) = 10 e -3t u(t), the zero-state response is: (D 2 +3D+2) y(t) = D x(t), and IC’s: y(0) = 0, y(0) = -5 has zero-input response: y 0 (t) = -5 e -1t + 5 e -2t x(t) * h(t) = [10 e -3t u(t)] * [( 2e -2t - e -t ) u(t)] = 20 [ (e -3t u(t) * e -2t u(t)] – 10 [ (e -3t u(t) * e -t u(t)]
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ECE 350 D. van Alphen 7 Total Response, Example continued Repeating: ) t ( u ) 2 ( 3 e e t 2 t 3 x(t) * h(t) = 20 [ (e -3t u(t) * e -2t u(t)] – 10 [ (e -3t u(t) * e -t u(t)] = 20 e -2t u(t) - 20 e -3t u(t) – 5 e -t u(t) + 5 e -3t u(t) = [20 e -2t -15e -3t –5e -t ] u(t) y(t) = { (-5 e -1t + 5 e -2t )+ [20 e -2t - 15 e -3t -t ] } u(t) = { -10 e -1t + 25 e -2t -3t } u(t) ) t ( u ) 1 ( 3 e e t t 3 ( ) ) t ( u e e t 3 t 2 = ( ) ) t ( u e e ) 5 (. t 3 t = Conv. Table Entry #4
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ECE 350 D. van Alphen 8 Review: Classical Solution to Differential Equations General solution is sum of two components: y(t) = y n (t) + y φ (t) •y c (t): the complementary solution, or the homogeneous solution, or the “natural response” of the system – A linear combination (weighted sum) of characteristic modes – Similar to zero-input response, but with different coefficients φ (t): the particular solution, or the “forced response” of the system – Determined by “form matching” the input to known categories of input given in a table (Lathi, Table 2.2, p. 199) that shows the corresponding forced response
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ece_350_lecture4_sp10_v5 - ECE 350 Lecture 4 Lecture...

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