Partial Fractions - Lecture 7: The Partial Fraction Expansion
Partial FractionsMatthew M. PeetIllinois Institute of TechnologyLecture 7: The Partial Fraction Expansion
IntroductionIn this Lecture, you will learn: The Inverse Laplace Transform Simple FormsThe Partial Fraction Expansion How poles relate to dominant modes Expansion using single poles Repeated Poles Complex Pairs of PolesI Inverse LaplaceM. PeetLecture 7: Control Systems2 / 27
Recall: The Inverse Laplace Transform of a SignalTo go from a frequency domain signal, û(s), to the time-domain signal, u(t), weuse the Inverse Laplace Transform.Definition 1.The Inverse Laplace Transform of a signal û(s) is denoted u(t) Λ 1 û.Z u(t) Λ 1 û eıωt û(ıω)dω0 Like Λ, the inverse Laplace Transform Λ 1 is also a Linear system. Identity: Λ 1 Λu u. Calculating the Inverse Laplace Transform can be tricky. e.g.ŷ M. Peets4s3 s2 2s 1 3s3 2s2 s 1Lecture 7: Control Systems3 / 27
Poles and Rational FunctionsDefinition 2.A Rational Function is the ratio of two polynomials:û(s) n(s)d(s)Most transfer functions are rational.Definition 3.The point sp is a Pole of the rational function û(s) n(s)d(s)if d(sp ) 0. It is convenient to write a rational function using its polesn(s)n(s) d(s)(s p1 )(s p2 ) · · · (s pn ) The Inverse Laplace Transform of an isolated pole is easy:û(s) M. Peet1s pmeansu(t) e ptLecture 7: Control Systems4 / 27
Partial Fraction ExpansionDefinition 4.The Degree of a polynomial n(s), is the highest power of s with a nonzerocoefficient.Example: The degree of n(s) is 4n(s) s4 .5s2 1Definition 5.A rational function û(s) than the degree of d(s).n(s)d(s)is Strictly Proper if the degree of n(s) is less We assume that n(s) has lower degree than d(s) Otherwise, perform long division until we have a strictly proper remainders3 2s2 6s 72 s 1 2s2 s 5s s 5M. PeetLecture 7: Control Systems5 / 27
Poles and Inverse Laplace TransformsA Strictly Proper Rational Function is The Sum of Poles: We can usuallyfind coefficients ri such thatr1rnn(s) . (s p1 )(s p2 ) · · · (s pn )s p1s pn Except in the case of repeated poles.Poles Dominate the Motion: Because a signal is the sum of poles, Theinverse Laplace has the formu(t) r1 ep1 t . . . rn epn t pi may be complex.I If p are complex, r may be complex.ii Doesn’t hold for repeated poles.M. PeetLecture 7: Control Systems6 / 27
ExamplesSimple State-Space: Step Response21s 1 ŷ(s) (s 1)ss 1 s1y(t) e t 1(t)2Suspension System: Impulse Response s111112J qqŷ(s) J s2 M2JglJ M gl s M glM gls 2J2Js M gl 12J M gl ty(t) e 2J e 2J tJ M glSimple State-Space: Sinusoid Response s 1s1ŷ(s) (s 1)(s2 1)s2 1 s 111y(t) cos t e t22M. PeetLecture 7: Control Systems7 / 27
Partial Fraction ExpansionConclusion: If we can find coefficients ri such thatn(s)n(s) d(s)(s p1 )(s p2 ) · · · (s pn )rnr1 ··· , s p1s pnû(s) then this is the PARTIAL FRACTION EXPANSION of û(s).We will address several cases of increasing complexity:1. d(s) has all real, non-repeating roots.2. d(s) has all real roots, some repeating.3. d(s) has complex, repeated roots.M. PeetLecture 7: Control Systems8 / 27
Case 1: Real, Non-repeated RootsThis case is the easiest:ŷ(s) n(s)(s p1 ) · · · (s pn )where pn are all real and distinct.Theorem 6.For case 1, ŷ(s) can always be written asŷ(s) rnr1 ··· s p1s pnwhere the ri are all real constantsThe trick is to solve for the ri There are n unknowns - the ri . To solve for the ri , we evaluate the equation for values of s.IIIPotentially unlimited equations.We only need n equations.We will evaluate at the points s pi .M. PeetLecture 7: Control Systems9 / 27
Case 1: Real, Non-repeated RootsSolvingWe want to solveŷ(s) rnr1 ··· s p1s pnfor r1 , r2 , . . .For each ri , we Multiply by (s pi ).s pis pis pi · · · ri · · · rns p1s pis pns pis pi r1 · · · ri · · · rns p1s pnŷ(s)(s pi ) r1Evaluate the Right Hand Side at s pi : We getpi pipi pi · · · ri · · · rnr1pi p1pi pn00 r1 · · · ri · · · rn ripi p1pi pnSetting LHS RHS, we get a simple formula for ri :ri ŷ(s)(s pi ) s piM. PeetLecture 7: Control Systems10 / 27
Case 1: Real, Non-repeated RootsCalculating riSo we can find all the coefficients:ri ŷ(s)(s pi ) s pin(s)(s pi )(s p1 ) · · · (s pi 1 )(s pi )(s pi 1 ) · · · (s pn )n(s) (s p1 ) · · · (s pi 1 )(s pi 1 ) · · · (s pn )ŷ(s)(s pi ) This is ŷ with the pole s pi removed.Definition 7.The Residue of ŷ at s pi is the value of ŷ(pi ) with the pole s pi removed.Thus we haveri ŷ(s)(s pi ) s pi M. Peetn(pi )(pi p1 ) · · · (pi pi 1 )(pi pi 1 ) · · · (pi pn )Lecture 7: Control Systems11 / 27
Example: Case 1ŷ(s) 2s 6(s 1)(s 2)We first separate ŷ into parts:ŷ(s) r1r22s 6 (s 1)(s 2)s 1 s 2Residue 1: we calculate:2s 62s 6 2 64r1 (s 1) s 1 s 1 4(s 1)(s 2)s 2 1 21Residue 2: we calculate:2s 6 4 622s 6(s 2) s 2 s 2 2r2 (s 1)(s 2)s 1 2 1 1Thus4 2ŷ(s) s 1 s 221.81.61.4y(t)1.2Concluding,y(t) 4e t 2e 2t10.80.60.40.20M. Peet012Lecture 7: Control Systems3t45612 / 27
Case 2: Real, Repeated RootsSometimes ŷ has a repeated pole:ŷ(s) n(s)(s p1 )q (s p2 ) · · · (s pn )In this case, we CAN NOT use simple expansion (As in case 1). Instead wehaven(s)(s p1 p2 ) · · · (s pn ) r11r12r1qr2rn ··· ··· (s p1 ) (s p1 )2(s p1 )qs p2s pnŷ(s) )q (s The rij are still real-valued.Example: Find the rij(s 3)2r11r12r13r21r22 (s 2)3 (s 1)2s 2 (s 2)2(s 2)3s 1 (s 1)2M. PeetLecture 7: Control Systems13 / 27
Case 2: Real, Repeated RootsProblem: We have more coefficients to find. q coefficients for each repeated root.First Step: Solve for r2 , · · · , rn as before:If pi is not a repeated root, thenri M. Peetn(pi )(pi p1 )q · · · (pi pi 1 )(pi pi 1 ) · · · (pi pn )Lecture 7: Control Systems14 / 27
Case 2: Real, Repeated RootsNew Step: Multiply by (s p1 )q to get the coefficient r1q .(s p1 )q(s p1 )q(s p1 )q r12 · · · r1q2(s p1 )(s p1 )(s p1 )q(s p1 )q(s p1 )q · · · rn r2s p2s pn q 1 r11 (s p1 ) r12 (s p1 )q 2 · · · r1qŷ(s)(s p1 )q r11 r2(s p1 )q(s p1 )q · · · rns p2s pnand evaluate at the point s pi to get .r1q ŷ(s)(s p1 )q s p1 r1q is ŷ(p1 ) with the repeated pole removed.M. PeetLecture 7: Control Systems15 / 27
Case 2: Real, Repeated RootsTo find the remaining coefficients, we Differentiate:ŷ(s)(s p1 )q r11 (s p1 )q 1 r12 (s p1 )q 2 · · · r1q r2 (s p1 )q(s p1 )q · · · rns p2s pnto getd(ŷ(s)(s p1 )q )ds (q 1)r11 (s p1 )q 2 (q 2)r12 (s p1 )q 3 · · · r1,(q 1) r2rnd ··· (s p1 )qdss p2s pnEvaluating at the point s p1 .d(ŷ(s)(s p1 )q ) s p1 r1,(q 1)dsM. PeetLecture 7: Control Systems16 / 27
Case 2: Real, Repeated RootsIf we differentiate again, we getd2(ŷ(s)(s p1 )q )ds2 (q 1)(q 2)r11 (s p1 )q 3 · · · 2r1,(q 2) rnr2d2q ··· 2 (s p1 )dss p2s pnEvaluating at s p1 , we getr1,(q 2) 1 d2(ŷ(s)(s p1 )q ) s p12 ds2Extending this indefinitelyr1,j dq j 11(ŷ(s)(s p1 )q ) s p1(q j 1)! dsq j 1Of course, calculating this derivative is often difficult.M. PeetLecture 7: Control Systems17 / 27
Example: Real, Repeated RootsExpand a simple exampleŷ(s) r11r12s 3r2 22(s 2) (s 1)s 2 (s 2)s 1First Step: Calculate r2r2 s 3 1 32 s 1 2(s 2)2( 1 2)21Second Step: Calculate r12r12 s 3 2 31 s 2 1s 1 2 1 1The Difficult Step: Calculate r11 : d s 3 s 2r11 ds s 1 1s 31 2 311 s 2 2s 1 (s 1)2 2 1 ( 2 1)2 1 1M. PeetLecture 7: Control Systems18 / 27
Example: Real, Repeated RootsSo now we haveŷ(s) s 3 2 12 (s 2)2 (s 1)s 2 (s 2)2s 1Question: what is the Inverse Fourier Transform ofRecall the Power Exponential:1(s 2)2 ?0.351tm 1 e at (s a)m(m 1)!0.30.25y(t)0.20.15Finally, we have:0.1y(t) 2e 2t te 2t 2e t0.050M. Peet01234Lecture 7: Control Systems5t67891019 / 27
Case 3: Complex RootsMost signals have complex roots (More common than repeated roots).ŷ(s) 3s(s2 2s 5)has roots at s 0, s 1 2ı, and s 1 2ı.Note that: Complex roots come in pairs. Simple partial fractions will work, but NOT RECOMMENDEDI Coefficients will be complex.I Solutions will be complex exponentials.I Require conversion to real functions.Best to Separate out the Complex pairs as:ŷ(s) M. Peetn(s)(s2 as b)(s p1 ) · · · (s pn )Lecture 7: Control Systems20 / 27
Case 3: Complex RootsComplex pairs have the expansionk1 s k2r1rnn(s) 2 ··· (s2 as b)(s p1 ) · · · (s pn )s as b s p1s pnNote that there are two coefficients for each pair: k1 and k2 .NOTE: There are several different methods for finding k1 and k2 .First Step: Solve for r2 , · · · , rn as normal.Second Step: Clear the denominator. Multiply equation by all poles. k1 s k2r1rnn(s) (s2 as b)(s p1 ) · · · (s pn ) 2 ··· s as b s p1s pnThird Step: Solve for k1 and k2 by examining the coefficients of powers of s.Warning: May get complicated or impossible for multiple complex pairs.M. PeetLecture 7: Control Systems21 / 27
Case 3 ExampleTake the exampleŷ(s) 2(s 2)k1 s k2r1 2 (s 1)(s2 4)s 4s 1First Step: Find the simple coefficient r1 .r1 2(s 2)2 1 21 s 1 2 2 2 2s 4 1 455Next Step: Multiply through by (s2 4)(s 1).2(s 2) (s 1)(k1 s k2 ) r1 (s2 4)Expanding and using r1 2/5 gives:2s 4 (k1 2/5)s2 (k2 k1 )s k2 8/5M. PeetLecture 7: Control Systems22 / 27
Case 3 ExampleRecall the equation2s 4 (k1 2/5)s2 (k2 k1 )s k2 8/5Equating coefficients gives 3 equations: s2 term - 0 k1 2/5 s term - 2 k2 k1 s0 term - 4 k2 8/5An over-determined system of equations (but consistent) First term gives k1 2/5 Second term gives k2 2 k1 10/5 2/5 12/5 Double-Check with last term: 4 12/5 8/5 20/5M. PeetLecture 7: Control Systems23 / 27
Case 3 ExampleSo we have 2(s 2)21s 6ŷ(s) (s 1)(s2 4)5 s 1 s2 4 21s6 5 s 1 s2 4 s2 4 2 te cos(2t) 3 sin(2t)y(t) 51.51y(t)0.50 0.5 1 1.5M. Peet051015t2025Lecture 7: Control Systems3024 / 27
Case 3 ExampleNow consider the solution to A Different Numerical Example: (Nise)33/5 3s 2 s(s2 2s 5)s5 s2 2s 5What to do with term:s 2?s2 2s 5We can rewrite as the combination of a frequency shift and a sinusoid:s2M. Peets 2s 2(s 1) 1 2 2s 5s 2s 1 4(s 1)2 4(s 1)12 (s 1)2 4 2 (s 1)2 4Lecture 7: Control Systems25 / 27
Case 3 Example(s 1)1s 22 s2 2s 5(s 1)2 4 2 (s 1)2 4(s 1)(s 1)2 4is the sinusoid s2s 4 shifted by s s 1. s s 1 means multiplication by e t in the time-domain. (s 1)s 1 t 1Λ e Λ e t cos 2t(s 1)2 4s2 4 2 Likewise Λ 1 e t sin 2t(s 1)2 410.80.60.4y(t) 0.20 0.2 0.401 1ΛM. Peet 23s 2s2 2s 5 4t e5 t 678 1cos 2t sin 2t2Lecture 7: Control Systems26 / 27
SummaryWhat have we learned today?In this Lecture, you will learn: The Inverse Laplace Transform Simple FormsThe Partial Fraction Expansion How poles relate to dominant modes Expansion using single poles Repeated Poles Complex Pairs of PolesI Inverse LaplaceNext Lecture: Important Properties of the ResponseM. PeetLecture 7: Control Systems27 / 27
Partial Fraction Expansion De nition 4. The Degree of a polynomial n(s), is the highest power of s with a nonzero coe cient. Example: The degree of n(s) is 4 n(s) s4 :5s2 1 De nition 5. A rational function u(s) n(s) d(s) is Strictly Proper if the degree of n(s) is less than the degree of d(s). We assume that n (s)has lower degree than d
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