6.302 Feedback Systems Recitation 4: Complex Variables and the s-plane Prof. Joel L. Dawson

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1 Number 1 quesion: Why deal wih imaginary and complex numbers a all? One answer is ha, as an analyical echnique, hey make our lives easier. Consider passing a cosine hrough an LTI filer wih impulse response h(): x() = cosω h() y() Because his is a linear and ime-invarian sysem, we can apply he convoluion inegral o figure ou y(): y() = - h(τ)cosω(-τ)dτ Evaluaing his can be easy or hard, depending on he form of h(). Bu, if we ake advanage of Euler s Ideniy, cosω = ½ (e jω + e -jω ) and so y() = ½ - h(τ)[e jω(-τ) + e -jω(-τ) ]dτ = ½ [ h(τ)e -jωτ dτ]e jω + ½ [ - - h(τ)e -j(-ω)τ dτ]e -jω H(jω) H(-jω) LAPLACE TRANSFORM = ½ H(jω)e jω + ½ H(-jω)e -jω Exponenials, even complex exponenials, are eigenfuncions of LTI sysems.

2 So complex numbers appear o be convenien. Bu in our sysems, we re always dealing wih real signals in he sense ha here is no oher or imaginary par. Wha is i abou our LTI mahemaical machinery ha les us plunge ino he complex s-plane and re-emerge safely in he world of real signals? Le s look a how we ge ino he s-plane in he firs place. Saring wih a real ime-domain signal h(), we use he Laplace Transform evaluaed on he jω axis: H(jω) = - h()e -jω d = h()cosω d - j - - h()sinωd We can say ha H(jω) is conjugae symmeric: Re {H(jω)} = Re {H(-jω)} Im {H(jω)} = -Im {H(-jω)} So he real par of H(jω) is an even funcion of ω, while he imaginary par is an odd funcion of ω. Jus he fac ha h() is real guaraneed his. Grea. We re in he world of he s-plane now, and all of our sysem funcions, frequency responses, ec., have his propery of being conjugae symmeric. Convince yourself ha any produc of H(jω)G(jω) will be conjugae symmeric if H(jω) and G(jω) have his propery separaely. 2

3 Time passes. We do our manipulaions in he s-plane, passing hrough filers, amplifiers, and moors o our hears conen. We have our oupu Y(jω), and i s ime o go back o he ime domain. We pull ou our rusy inverse Laplace Transform, and choose as our conour of inegraion he jω axis: y() = 1 2π - Y(jω)e jω dω Le Y R (jω) = Re{Y(jω)} Y Im (jω) = Im {Y(jω)} y() = 2π 1 - (Y R +jy im )(cosω + jsinω)dω = 1 (Y R cosω - Y Im sinω)dω + j 2π - (Y R sinω - Y Im cosω)dω - 2π real par imaginary par Our hope is ha he imaginary par is zero. Le s examine he erms: Y R sinω Y Im cosω ƒ ODD (ω) even odd odd even produc is ODD produc is ODD ω sum is an ODD funcion inegraing over an odd funcion gives ZERO! We re lef wih he purely real funcion of ime: y() = 1 2π - (Y R cosω - Y Im sinω)dω 3

4 Our LTI machinery, hen, gives us he convenience of complex variables and he guaranee ha our resuls will be real funcions in ime. Properies of he Laplace Transform DIFFERENTIATION: L dƒ d = sf(s) - ƒ(0) INTEGRATION: [ ƒ(τ)dτ] = L 0 F(s) s INITIAL VALUE THEOREM: > 0+ ƒ() = s > sf(s) FINAL VALUE THEOREM: > ƒ() = s >0 sf(s) We can use hese properies o make our calculaions easier. 1 Example 1: H(s) = ; skech h() s 2 We know ha or a ramp: L -1 {1/s} is u(). Using he inegraion propery, we know ha h() mus be u()d, h() 0 1 4

5 Example 2: H(s) = s+2 s+1 skech he sep response. 1 s+2 Iniial value: s > s = 1 1 s+2 Final value: s >0 s = 2 How i ges from sar o finish: 1 s+2 = A B + A sep u() ~e - u() ƒ() 2 1 τ=1 Class Exercise: Skech sep response of H(s) = 1-s s+1

6 { { } } Feedback Sysems Complex numbers and vecor geomery Recall ha for ordinary vecors, A = a 1 x ˆ + a 2 ŷ. And B = b 1 x ˆ + b ˆ 2 y, he vecor A - B had a very definie meaning: A - B = (a 1 - b 1 ) x ˆ + (a 2 - b 2 ) ŷ ŷ A A - B B xˆ For complex numbers, we do a similar hing: u = a 1 + ja 2 v = b 1 + jb 2 } u - v = (a 1 - b 1 ) + j(a 2 - b 2 ) We rea complex numbers like vecors. jω So wih poles and zeros, we appeal o a geomerical picure. s F(s) = vecor vecor (s-z 1 ) (s-z 2 ) (s-p 1 ) (s-p 2 ) s-p 1 p 1+ s-z 1 s-z 2 z 1 z 2 σ vecor vecor p 2 + s-p 2 The vecor s - z, has a magniude (R Z1 ) and an angle, or phase φ Z1. We can wrie F(s) as F(s) = r z1 e jφ z1 r z2 e jφ z2 r p1 e jφ p1 r p2 e jφ p2 r z1 r = z2 r p1 r e j(φ + φ - φ - φ ) z1 z2 p1 p2 p2 6