Appendix 2.3 General Solutions for the Step Response of Third- and Fourth-Order Systems (with some unpleasant surprises!)

Similar documents
Y 0. Standing Wave Interference between the incident & reflected waves Standing wave. A string with one end fixed on a wall

Higher order derivatives

Linear-Phase FIR Transfer Functions. Functions. Functions. Functions. Functions. Functions. Let

First order differential equation Linear equation; Method of integrating factors

Thomas Whitham Sixth Form

Quasi-Classical States of the Simple Harmonic Oscillator

General Notes About 2007 AP Physics Scoring Guidelines

Multiple Short Term Infusion Homework # 5 PHA 5127

Thomas Whitham Sixth Form

A Propagating Wave Packet Group Velocity Dispersion

u x v x dx u x v x v x u x dx d u x v x u x v x dx u x v x dx Integration by Parts Formula

The second condition says that a node α of the tree has exactly n children if the arity of its label is n.

cycle that does not cross any edges (including its own), then it has at least

a 1and x is any real number.

Ch. 24 Molecular Reaction Dynamics 1. Collision Theory

First derivative analysis

Lecture 37 (Schrödinger Equation) Physics Spring 2018 Douglas Fields

Calculus concepts derivatives

SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Abstract Interpretation: concrete and abstract semantics

PHYS ,Fall 05, Term Exam #1, Oct., 12, 2005

Exam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.

The Matrix Exponential

DSP-First, 2/e. LECTURE # CH2-3 Complex Exponentials & Complex Numbers TLH MODIFIED. Aug , JH McClellan & RW Schafer

The Matrix Exponential

Problem Set 6 Solutions

surface of a dielectric-metal interface. It is commonly used today for discovering the ways in

Addition of angular momentum

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d

4. (5a + b) 7 & x 1 = (3x 1)log 10 4 = log (M1) [4] d = 3 [4] T 2 = 5 + = 16 or or 16.

Addition of angular momentum

Einstein Equations for Tetrad Fields

Function Spaces. a x 3. (Letting x = 1 =)) a(0) + b + c (1) = 0. Row reducing the matrix. b 1. e 4 3. e 9. >: (x = 1 =)) a(0) + b + c (1) = 0

1973 AP Calculus AB: Section I

Fourier Transforms and the Wave Equation. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation.

Hydrogen Atom and One Electron Ions

A. Limits and Horizontal Asymptotes ( ) f x f x. f x. x "±# ( ).

the output is Thus, the output lags in phase by θ( ωo) radians Rewriting the above equation we get

Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination

y cos x = cos xdx = sin x + c y = tan x + c sec x But, y = 1 when x = 0 giving c = 1. y = tan x + sec x (A1) (C4) OR y cos x = sin x + 1 [8]

The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function

That is, we start with a general matrix: And end with a simpler matrix:

Coupled Pendulums. Two normal modes.

22/ Breakdown of the Born-Oppenheimer approximation. Selection rules for rotational-vibrational transitions. P, R branches.

Finite Element Analysis

Differentiation of Exponential Functions

Slide 1. Slide 2. Slide 3 DIGITAL SIGNAL PROCESSING CLASSIFICATION OF SIGNALS

The graph of y = x (or y = ) consists of two branches, As x 0, y + ; as x 0, y +. x = 0 is the

(1) Then we could wave our hands over this and it would become:

Electrochemistry L E O

A RELATIVISTIC LAGRANGIAN FOR MULTIPLE CHARGED POINT-MASSES

6. The Interaction of Light and Matter

2. Finite Impulse Response Filters (FIR)

COMPUTER GENERATED HOLOGRAMS Optical Sciences 627 W.J. Dallas (Monday, April 04, 2005, 8:35 AM) PART I: CHAPTER TWO COMB MATH.

SPH4U Electric Charges and Electric Fields Mr. LoRusso

10. The Discrete-Time Fourier Transform (DTFT)

There is an arbitrary overall complex phase that could be added to A, but since this makes no difference we set it to zero and choose A real.

MSLC Math 151 WI09 Exam 2 Review Solutions

2F1120 Spektrala transformer för Media Solutions to Steiglitz, Chapter 1

ECE 2210 / 00 Phasor Examples

MATH 319, WEEK 15: The Fundamental Matrix, Non-Homogeneous Systems of Differential Equations

Sundials and Linear Algebra

EXST Regression Techniques Page 1

Sinusoidal Response Notes

2.3 Matrix Formulation

nd the particular orthogonal trajectory from the family of orthogonal trajectories passing through point (0; 1).

1 General boundary conditions in diffusion

Impedance Transformation and Parameter Relations

On the Hamiltonian of a Multi-Electron Atom

As the matrix of operator B is Hermitian so its eigenvalues must be real. It only remains to diagonalize the minor M 11 of matrix B.

Announce. ECE 2026 Summer LECTURE OBJECTIVES READING. LECTURE #3 Complex View of Sinusoids May 21, Complex Number Review

MCE503: Modeling and Simulation of Mechatronic Systems Discussion on Bond Graph Sign Conventions for Electrical Systems

Complex representation of continuous-time periodic signals

ANALYSIS IN THE FREQUENCY DOMAIN

UNTYPED LAMBDA CALCULUS (II)

BINOMIAL COEFFICIENTS INVOLVING INFINITE POWERS OF PRIMES

Alpha and beta decay equation practice

Types of Transfer Functions. Types of Transfer Functions. Types of Transfer Functions. Ideal Filters. Ideal Filters

4. Money cannot be neutral in the short-run the neutrality of money is exclusively a medium run phenomenon.

Math 34A. Final Review

u r du = ur+1 r + 1 du = ln u + C u sin u du = cos u + C cos u du = sin u + C sec u tan u du = sec u + C e u du = e u + C

Principles of Humidity Dalton s law

5. B To determine all the holes and asymptotes of the equation: y = bdc dced f gbd

Mathematics. Complex Number rectangular form. Quadratic equation. Quadratic equation. Complex number Functions: sinusoids. Differentiation Integration

Content Skills Assessments Lessons. Identify, classify, and apply properties of negative and positive angles.

Exercise 1. Sketch the graph of the following function. (x 2

INTRODUCTION TO AUTOMATIC CONTROLS INDEX LAPLACE TRANSFORMS

ph People Grade Level: basic Duration: minutes Setting: classroom or field site

Random Process Part 1

Sec 2.3 Modeling with First Order Equations

Section 11.6: Directional Derivatives and the Gradient Vector

Brief Introduction to Statistical Mechanics

Introduction to the Fourier transform. Computer Vision & Digital Image Processing. The Fourier transform (continued) The Fourier transform (continued)

ECE602 Exam 1 April 5, You must show ALL of your work for full credit.

Prod.C [A] t. rate = = =

Case Study 4 PHA 5127 Aminoglycosides Answers provided by Jeffrey Stark Graduate Student

COHORT MBA. Exponential function. MATH review (part2) by Lucian Mitroiu. The LOG and EXP functions. Properties: e e. lim.


y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)

Transcription:

P.Stariè, E.Margan Appnix 2. A2..1 A2..2 Contnts: Appnix 2. Gnral Solutions for th Stp Rspons of Thir- an Fourth-Orr Systms (with som unplasant surpriss!) Thr is no such thing as instant xprinc! ( Oppnhimr s Law ) For vry xprinc you pay with part of your lif. For inxprinc you pay with your whol lif. ( Yi-Ching ) Ar you xprinc? ( Jimi Hnrix ) Thir-orr systm with on pol ral an on complx conjugat pol pair... p.a2..2 Important not on )... p.a2..5 Intrprtation of th rsulting xprssion... p.a2..7 Thir-orr systm with all pols ral, on pol pair coincint... p.a2..8 A2.. Thir-orr systm with on ral pol, on complx conjugat pol pair an on ral ro... p.a2..10 A2..4 Thir-orr systm with on ral pol, on complx conjugat pol pair an on complx conjugat ro pair... p.a2..15 A2..5 Fourth-orr systm with two complx conjugat pol pairs... p.a2..19 Phas rror... p.a2..26 Intrprtation... p.a2..27 A2..6 Fourth-orr systm with two complx conjugat pol pairs an on ral ro... p.a2..28 A2..7 Fourth-orr systm with two pols ral, on complx conjugat pol-pair an on ral ro... p.a2..1 Intrprtation... p.a2..4 Fig. A2..1 Fig. A2..2 List of Figurs: Stp rspons an th phas rror in -pol MFA systm... p.a2..6 Stp rspons an th phas rror in 4-pol MFA systm... p.a2..26 Fig. A2.. Stp rspons an th phas rror in 4-pol MFED systm... p.a2..26 A2..1

P.Stariè, E.Margan Appnix 2. A2..1 Thir-orr systm with on pol ral an on complx-conjugat pol-pair A normali gnral form of a thir-orr all pol omain function: JÐ Ð Ð Ð Ð (A2..1.1) Th pol componnts ar: 5 4 5 4 5 (A2..1.2) Th stp rspons in th omain is: KÐ J Ð Ð Ð Ð (A2..1.) Th Invrs Laplac transform of KÐ is: Rsiu 0: gð KÐf rs ckð _! rs! lim (A2..1.4) Ä! Ð Ð Ð lim Ä! Ð Ð Ð Rsiu 1:! Ð! Ð! Ð! rs lim Ð (A2..1.5) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Ð Ð A2..2

P.Stariè, E.Margan Appnix 2. Rsiu 2: rs lim Ð (A2..1.6) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Rsiu : Ð Ð rs lim Ð (A2..1.7) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Ð Ð If w insrt th pol componnts th rsius ar: rs! (A2..1.8) Ð5 4 5 Ð5 4 rs (A2..1.9) Ð5 4 5 4 Ð5 4 5 Ð5 4 5 4 Ð5 5 4 5 4 Ð5 4 5 Ð5 5 4 5 4 4 cð5 5 5 4 5 c5 Ð5 5 4 Ð5 5 4 5 4 cð5 5 5 4 5 c5 Ð5 5 4 Ð 5 5 4 cð5 5 A2..

P.Stariè, E.Margan Appnix 2. Ð5 4 5 Ð5 4 rs (A2..1.10) Ð5 4 5 4 Ð5 4 5 Ð5 4 5 4 Ð5 5 4 5 4 Ð5 4 5 Ð5 5 4 5 4 4 cð5 5 5 4 5 c5 Ð5 5 4 Ð5 5 4 5 4 cð5 5 5 4 5 c5 Ð5 5 4 Ð 5 5 4 cð5 5 Ð5 4 Ð5 4 5 rs (A2..1.11) Ð5 5 4 Ð5 5 4 Ð 5 Ð5 5 5 W shall simplify th xprssions of rsius by using th following substitutions: A 5 Ð5 5 (A2..1.12) (A2..1.1) B 5 5 C Ð5 5 (A2..1.14) For th computation of gð w coul simply sum all rsius in th complx xponntial form, but for th human min it is always asir to intrprt th quations in trms of xprssions which rprsnt som part of th systm bhavior. Th complx conjugat pol pair trmins th systm rsonant frquncy an w woul lik to s its influnc on th stp rspons. rs rs 5 ÐA 4 B 5 4 5 ÐA 4 B 4 C 4 C 5 4 (A2..1.15) 5 5 A 4 B 4 A 4 B 4 C 4 4 5 5 A 4 A 4 4 B 4 4 B 4 C 4 4 4 4 5 A 5 B C 4 4 4 4 4 A2..4

P.Stariè, E.Margan Appnix 2. 5 5 A B C 4 4 4 4 4 Th two complx xponntials can b transform into a sin an cosin: 5 rs rs 5 A Ð B Ð (A2..1.16) C c sin cos Th trigonomtric sum in brackts can b furthr intrprt as a singl sin function, shift in phas an with moifi amplitu, in accoranc with th following trigonomtric transformation: U T sinð UcosÐ ÈT U sinš arctan (A2..1.17) T So: 5 rs rs 5 B A B C É sin Š arctan A (A2..1.18) Lt us rplac th arctangnt xprssion with th angl ): ) arctan B a 5 5 b arctan A 5 Ð5 5 (A2..1.19) Finally, w can writ th stp rspons function: 5 5 gð A B sinð ) C C Important not on ): É 5 5 (A2..1.20) r If w insrt th normali pol valus (of, say, a -orr Buttrworth) in Eq. A2..1.20 an chck th initial an final valus of gð for! an _, w obtain corrctly gð!! an gð_. Howvr, if w plot gð from! to & with?!þ!&, th rspons will b wrong, (s Fig. A2..1), sinc w know that for any function having th numbr of pols gratr than th numbr of ros by at last 2 th rivativ. gðî. must also b ro for p!. Whr is th rror? By trial an rror on might iscovr that th plot woul b corrct if th sign of th scon trm in gð is positiv. But by rchcking th calculation of rs rs w can vrify that th sign shoul b ngativ. So th rror coul b in th sign of ). But if w chck its valu w can s that )! (or at last vry clos to ro, pning on th prcision of th pol valus), thrfor th sign of ) os not mak any iffrnc. Now, what?? A2..5

P.Stariè, E.Margan Appnix 2. Th problm is hin in th tangnt function, which can not istinguish btwn th angls of! an 1 raians (apart from our inability to us a tru valu of 1, but insta only a finit numbr of its cimal placs). For th sam rason th function arctanatan 1b always rturns! an nvr 1Þ Howvr, whilst (for! ) sinð!!sinð! 1!, for Î th valu of sinð! is positiv an th valu of sinð 1 is ngativ! In short, it must b ( but not always!!!, s Fig. A2.. in Sc. A2..5): ) Ê ) 1 (A2..1.21) This can b vrifi if w plot gð by using Eq. A2..1.16 in plac of th mil trm of Eq. A2..1.20 th sin-cosin combination rtains th corrct phas information x Likwis th corrct rsult is obtain if w simply sum th rsius in thir complx xponntial form! 1.4 1.2 ERROR: θ 0 1.0 g() t 0.8 σ1 1.5000 0.6 Corrct: θ π ω 1 2.5981 σ.0000 0.4 0.2 θ arctan ω 1 ( 2 σ1 σ ) σ ( σ ) 2 1 1 σ ω 1 0.0 0 1 2 4 5 t/ T Fig. A2..1: To obtain th corrct stp rspons plot w must assign th valu 1 to th initial phas angl ) (an not ro, as suggst by th arctangnt opration). This is a nic illustration of why som mathmaticians rfr to trigonomtric transformations lik Eq. A2..1.17 as malvrsations! Whilst it is vry convnint to think of circuit action in trms of sin wavs, for numrical calculation w shall rathr stick to th complx xponntial form. A2..6

P.Stariè, E.Margan Appnix 2. Intrprtation of th rsulting xprssion In trms of circuit action th gnral thir-orr xprssion for intrprt as follows: gð can b a) Th valu of rs! (which in th cas of normali pols an normali gain is qual to 1) rprsnts th (normali) final valu to which th systm will stabili whn p_ ( or, sinc frquncy is th invrs of tim,! ). b) Th systm raction ow to th complx conjugat pol pair is rprsnt by th scon trm: th input stp xcits th systm into a sinusoial oscillation at its rsonant frquncy,, shift initially in phas by ). Th amplitu of this oscillation is normali by a ral corrction factor, th valu of which is trmin by th impanc valus at rsonanc. Th oscillation is amp by th ral an ngativ factor 5 in th xponnt, which nots nrgy issipation in rsistiv circuit lmnts an thus forcs th oscillations to cras with tim. c) Th final trm rprsnts th systm raction ow to th ral pol 5. Sinc th valu of 5 is also ngativ this is also a caying xponntial function, again corrct in amplitu by a ral factor, so that th sum of all trms at th tim! is ro. ) Thr rmains to b clar up th apparntly mystrious initial phas shift. ) Any circuit with a complx conjugat pol pair has a capacitanc an an inuctanc (in activ circuits th inuctiv bhavior can b simulat by a capacitanc with rgnrativ fback). Whn coupl togthr thy form a rsonant circuit in which th voltag an currnt bcom rivativs of ach othr, i.., a sin cosin rlationship. This maks sns of rlations such as Eq. A2..1.16. Th transint xcitation is a signal which changs in a vry short tim intrval, too short for th systm to follow, so all th nrgy gos into th xcitation of th systm s rsonanc an its consqunt rlaxation. At rsonanc thr is only on frquncy (for ach complx conjugat pol pair) by which th voltag an currnt vary, so thr is som sns in intrprting th phas as th ratio of th rlativ voltag an currnt amplitus, which in turn ar st by th impanc ratios. A2..7

P.Stariè, E.Margan Appnix 2. A2..2 Thir-orr systm with all pols ral, on pol pair coincint A normali gnral form of a thir-orr oubl pol omain function is: JÐ Ð Ð Ð (A2..2.1) Th pol componnts (all pols ral, on coincint pair) ar: 5 5 Th stp rspons in th omain is: KÐ Th invrs Laplac transform of KÐ is: J Ð Ð Ð (A2..2.2) (A2..2.) gð KÐf rs ckð _! Rsiu 0: rs! lim Ä! Ð Ð (A2..2.4) lim Ä! Ð Ð! Ð! Ð! Ð Owing to th oubl pol at rivativ:, th calculation of rs is on by limiting th. rs lim Ð (A2..2.5) Ä. Ð Ð lim Ä.. Ð Ð Ð lim Ä Ð lim Ä Ð A2..8

P.Stariè, E.Margan Appnix 2. Ð Ð Ð Rsiu 2 is calculat in th usual way: rs lim Ð (A2..2.6) Ä Ð Ð lim Ä Ð Ð Ð W insrt th pol componnts: rs! (A2..2.7) 5 5 Ð 5 5 5 rs 5 (A2..2.8) Ð5 5 5 5 rs (A2..2.9) Ð5 5 So th stp rspons is: Ð gð 5 5 5 5 5 5 5 5 (A2..2.10) Ð5 5 Ð5 5 With a littl rarrangmnt of th scon trm w can simplify th linar function of to finally obtain: 5 5 gð 5 5 5 5 5 5 Ð5 5 a b 5 5 Ð5 5 (A2..2.11) Sinc all thr pols ar ral th systm has no charactristic rsonant frquncy but only two caying xponntials. Again, th valu of th first rsiu (rs! ) is th normali final valu to which th systm will sttl. Also, sinc at! both xponntials ar qual to 1, th sum of th factors that multiply thm must b qual to in orr to mak gð!!. Th linar tim function within th scon trm slows own th rspons for small valus of. A2..9

P.Stariè, E.Margan Appnix 2. A2.. Thir-orr systm with on ral pol, on complx conjugat pol pair an on ral ro A normali gnral form of a -pol an 1-ro omain function is: Th pol an ro componnts ar: 5 4 5 4 5 5 Th stp rspons in th omain is: Ð Ð JÐ Ð Ð Ð Ð KÐ J Ð Ð Ð Ð (A2...1) (A2...2) (A2...) Th invrs Laplac transform of KÐ: gð KÐf rs ckð _! Rsiu 0: Ð rs! lim Ä! Ð Ð Ð (A2...4) lim Ä! Ð Ð Ð Ð Rsiu 1: Ð!! Ð! Ð! Ð! Ð rs lim Ð (A2...5) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Ð Ð Ð Ð Ð A2..10

P.Stariè, E.Margan Appnix 2. Rsiu 2: Ð rs lim Ð (A2...6) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Ð Ð Ð Ð Ð Rsiu : Ð rs lim Ð (A2...7) Ä Ð Ð Ð lim Ä Ð Ð Ð Ð Ð Ð Ð Ð Ð W now insrt th ral an imaginary componnts of th pols an th ro into th last thr rsius: Ð5 4 5 a5 4 b5 a5 4 b rs (A2...8) 5 Ð5 4 5 4 Ð5 4 5 4 5 c5 5a5 4 b 5 5 4 Ð5 5 4 4 5 c5 5a5 4 bð5 5 4 5 5 Ð5 5 4 Ð5 5 4 4 4 5 c5 5a5 4 bð5 5 4 5 5 a5 5 b 4 Th common ral factor, which will b us also in rs, is: K 5 5 a5 5 b (A2...9) A2..11

P.Stariè, E.Margan Appnix 2. Th rmaining factor in th numrator is sparat into its ral an imaginary part: c5 5 a5 4 bð5 5 4 ca5 b 5 5 a5 5 b 4 5 a5 5 b 4 ca5 b 55 5 a5 5 bca5 b 5 5 5 With th following substitutions: 4 ca5 b 5 a 5 5 b A a5 5 bc5 5 5 5 B c5 5 a 5 5 b (A2...10) w can writ: 4 5 rs K aa 4 Bb (A2...11) 4 Equally, for rs : Ð5 4 5 a5 4 b5 a5 4 b rs (A2...12) 5 Ð5 4 5 4 Ð5 4 5 5 4 5 c5 5a5 4 b 5 a 4 bð5 5 4 4 5 5 5 5 4 Ð5 5 4 c a b 5 5 Ð5 5 4 Ð5 5 4 4 4 5 c5 5a5 4 bð5 5 4 5 5 a5 5 b 4 W xtract th common ral factor factor into its ral an imaginary part: K as bfor an rarrang th rmaining c5 5 a5 4 bð5 5 4 c5 5 5 a5 5 b 5 4 a5 5 b 4 c5 5 5 5 c5 5 5 a5 5 b 5 4 c5 5 a 5 5 b A2..12

P.Stariè, E.Margan Appnix 2. By using th sam substitutions A an Bas bfor, w can writ: 4 5 rs K aa 4 Bb (A2...1) 4 For rs w hav: Ð5 5a5 4 ba5 4 b 5 rs (A2...14) 5 Ð5 5 4 Ð5 5 4 5 Ð 5 5 a5 b 5 cð5 5 5 K Th stp rspons is th sum of th rsius: 5 gð K A 4 B 4 4 5 K A 4 B 4 b 4 5 K (A2...15) Again, w transform th two complx conjugat rsius into a sin cosin pair: 4 4 aa 4 Bb aa 4 Bb 4 4 A B 4 4 4 4 4 A sin B cos casin Bcos This sin cosin pair can b furthr transform into a sin function with an appropriat phas an amplitu corrction: B Asin Bcos ÉA B sinœ arctan A Finally, w writ th stp rspons as: 5 gð K ÉA B 5 sina ) b K (A2...16) whr: ) arctan B A (A2...17) A2..1

P.Stariè, E.Margan Appnix 2. Howvr, as has alray bn shown in th first xampl, hr too w must incras th phas angl by 1: ) Ê ) 1 (A2...18) It is intrsting to not that th ro is a fr factor in th nominator of both K ( Eq. A2...9) an K ( Eq. A2...14). It will thrfor cras th group lay, an, if brought too clos to th complx plan origin, it woul also incras th ovrshoot. 5 A2..14

P.Stariè, E.Margan Appnix 2. A2..4 Thir-orr systm with on ral pol, on complx conjugat pol pair, an on complx conjugat ro pair A normali gnral form of a -pol an 2-ro omain function is: Ð Ð % Ð & JÐ Ð Ð Ð Ð Th pol an ro componnts ar: ß 5 4 5 %ß& 5 4 Th stp rspons in th omain is: % & Ð KÐ J Ð Ð Ð Ð (A2..4.1) (A2..4.2) (A2..4.) Th invrs Laplac transform of KÐ is: Rsiu 0: g Ð _ KÐf rs ckð! Ð % Ð & rs! lim (A2..4.4) Ä! Ð Ð Ð % & lim Ä! Ð % Ð & Ð Ð Ð % & Rsiu 1: Ð! % Ð! &! Ð! Ð! Ð! % & Ð % Ð & rs lim Ð (A2..4.5) Ä Ð Ð Ð % & lim Ä Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & A2..15

P.Stariè, E.Margan Appnix 2. Rsiu 2: Ð % Ð & rs lim Ð (A2..4.6) Ä Ð Ð Ð % & lim Ä Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & Rsiu : Ð % Ð & rs lim Ð (A2..4.7) Ä Ð Ð Ð % & lim Ä Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & Ð % Ð & Ð Ð % & W insrt th ral an imaginary componnts of th pols a ros: rs a5 4 b5 Ð5 4 5 4 Ð5 4 5 4 a5 4 ba5 4 bð5 4 5 4 Ð5 4 5 5 4 a5 4 b5 c5 5a5 4 b 5 a5 b4 Ð5 5 4 a5 4 b 4 a5 4 b5 c5 5 a5 4 b 5 Ð5 5 4 5 a5 bcð5 5 4 Now w can xtract th common ral factor: 5 K a5 bcð5 5 (A2..4.8) an multiply an rgroup th rmaining factors into ral an imaginary parts: a5 4 b 5 5 a5 4 b 5 Ð5 5 4 c5 a5 5 b 4 a 5 5 b c5 5 a5 4 b 5 A2..16

P.Stariè, E.Margan Appnix 2. c5 a5 5 b 4 a 5 5 bc5 5 5 a5 4 b c5 a5 5 b c5 5 5 5 4 a 5 5 bc5 5 5 5 5 4 c5 a5 5 b 5 a 5 5 b c5 a5 5 b c5 5 5 5 5 a 5 5 b 4 a 5 5 bc5 5 5 5 5 c5 a5 5 b f W substitut: A c5 a5 5 b c5 5 55 5 a 5 5 b (A2..4.9) B a 5 5 bc5 5 55 5c5 a5 5 b (A2..4.10) With this, rs can b writtn as: 4 5 rs K aa 4 Bb (A2..4.11) 4 Rpat th procur for rs : rs a5 4 b5 Ð5 4 5 4 Ð5 4 5 4 a5 4 ba5 4 bð5 4 5 4 Ð5 4 5 5 4 a5 4 b5 c5 5a5 4 b 5 a5 ba 4 bð5 5 4 a5 4 b 4 a5 4 b5 c5 5 a5 4 b 5 Ð5 5 4 5 a5 bcð5 5 4 W xtract K an rorr th rmaining numrator factors: a5 4 b 5 5 a5 4 b 5 Ð5 5 4 c5 a5 5 b 4 a 5 5 b c5 5 a5 4 b 5 c5 a5 5 b 4 a 5 5 bc5 5 5 a5 4 b A2..17

P.Stariè, E.Margan Appnix 2. c5 a5 5 b c5 5 5 5 4 a 5 5 bc5 5 5 5 5 4 c5 a5 5 b 5 a 5 5 b c5 a5 5 b c5 5 5 5 5 a 5 5 b 4 a 5 5 bc5 5 5 5 5 c5 a5 5 b f Using th sam substitutions as for rs, w hav: 4 rs K 5 aa 4 Bb (A2..4.12) 4 For rs w hav: rs a5 4 ba5 4 bð5 5 4 Ð5 5 4 a5 4 ba5 4 bð5 5 4 Ð5 5 4 5 a5 bc Ð 5 5 a5 bcð5 5 5 5 K (A2..4.1) Th sum of th rsius is th stp rspons sought: 5 gð K A 4 B 4 4 5 K A 4 B 4 b 4 5 K (A2..4.14) which can b rwrittn as: 4 4 4 4 5 5 gð K Œ A B K 4 an also as: gð K 5 A B K 5 a sin cos b (A2..4.15) an also as: 5 gð K ÉA B 5 sina ) b K (A2..4.16) whr th phas angl ) is: Incras by whr appropriat. Whn in ) B arctanœ 1 Œ 1 A in oubt, us ithr A2..4.14, or A2..4.15 A2..18

P.Stariè, E.Margan Appnix 2. A2..5 Fourth-orr systm with two complx conjugat pol pairs A four-pol normali gnral function is: JÐ Ð % % Ð Ð Ð Ð % (A2..5.1) In gnral w hav two complx conjugat pol pairs, with th following componnts: ß 5 4 (A2..5.2) 5 4 ß% Th stp rspons in th omain is a fifth-orr function: KÐ Th invrs Laplac transform of KÐ is: % J Ð Ð Ð Ð Ð % gð KÐf rs ckð _! % (A2..5.) Rsiu 0: % rs! lim (A2..5.4) Ä! Ð Ð Ð Ð % lim Ä! % Ð Ð Ð Ð % %! Ð! Ð! Ð! Ð! % Rsiu 1: % rs lim Ð (A2..5.5) Ä Ð Ð Ð Ð % lim Ä % Ð Ð Ð % % Ð Ð Ð % % Ð Ð Ð % Lt us insrt th pol componnts hr: rs a5 b 4 Ð5 4 Ð5 4 Ð5 4 Ð5 4 5 4 Ð5 4 5 4 Ð5 4 5 4 (A2..5.6) A2..19

P.Stariè, E.Margan Appnix 2. a5 4 b Ð5 4 Ð5 4 cð5 5 4Ð cð5 5 4Ð 4 5 Ð5 4 Ð5 4 Ð5 5 4Ð5 5 cð Ð Ð Ð f 5 4 Ð5 4 Ð5 4 Ð5 5 c 4 Ð5 5 Ð Ð5 Ð5 4 4 cð5 5 Ð 4 Ð5 5 5 4 W shall rationali th nominator (but only th part in th brackts th 4 factor 4 in front will b us latr with th trm to transform it into a sin an cosin, as was on in th thir-orr xampl) by multiplying both th numrator an th nominator by th complx conjugat of th nominator: Ð5 Ð5 4 Ð c 5 5 Ð 4 Ð5 5 5 Ð5 5 Ð c % Ð5 5 4 4 Bfor rgrouping th trms in th numrator into a ral an imaginary part w xtract th common ral factor: K Ð5 cð5 5 Ð % Ð5 5 (A2..5.7) Th rmaining xprssion in th numrator must b first multipli an thn rarrang to sparat th ral an imaginary part: Ð5 4 cð5 5 Ð 4 Ð5 5 5 cð5 5 Ð 4 cð5 5 Ð 45Ð5 5 Ð5 5 5 cð5 5 Ð Ð5 5 4 cð5 5 Ð 5 Ð5 5 In orr to simplify th xprssion w again introuc som substitutions. Ths sam substitutions will b us again for rs, whilst slightly iffrnt substitutions will b us for rs an rs %. A Ð5 5 Ð B Ð 5 5 (A2..5.8) (A2..5.9) A2..20

P.Stariè, E.Margan Appnix 2. So by taking into account th common factor K from abov w can writ: 4 rs K ˆ A B 4 A B 5 5 a 5 b (A2..5.10) 4 Rsiu 2: % rs lim Ð (A2..5.11) Ä Ð Ð Ð Ð % lim Ä % Ð Ð Ð % % Ð Ð Ð % % Ð Ð Ð % Entr th pol componnts: rs a5 4 Ð5 4 Ð5 4 Ð5 4 b Ð5 4 5 4 Ð5 4 5 4 Ð5 4 5 4 (A2..5.12) a5 Ð5 4 Ð5 4 b 4 cð5 5 4Ð cð5 5 4Ð 5 4 Ð5 4 Ð5 4 cð5 5 4 Ð5 5 Ð Ð 5 4 Ð5 4 Ð5 4 Ð5 5 c 4 Ð5 5 Ð Ð5 Ð5 4 4 cð5 5 Ð 4 Ð5 5 5 4 Again, w shall associat th factor 4 with th imaginary xponntial an thn rationali th xprssion in th brackts of th nominator by multiplying it (an, of cours, th numrator as wll) with its own complx conjugat: Ð5 Ð5 4 Ð5 5 Ð 4 Ð5 5 4 c 5 Ð5 5 Ð c % Ð5 5 4 W now xtract th sam common ral factor K as in rs (A2..7) an again rgroup th rmaining trms of th numrator into ral an imaginary parts: Ð5 4 cð5 5 Ð 4 Ð5 5 5 cð5 5 Ð Ð5 5 45Ð5 5 4 cð5 5 Ð A2..21

P.Stariè, E.Margan Appnix 2. 5 cð5 5 Ð Ð5 5 4 cð5 5 Ð 5 Ð5 5 By using th sam substitutions K, A, an Bas for rs, it is obvious that: 4 rs K ˆ A B 4 A B 5 5 a 5 b (A2..5.1) 4 Rsiu : % rs lim Ð Ä Ð Ð Ð Ð (A2..5.14) % lim Ä % Ð Ð Ð % % Ð Ð Ð % % Ð Ð Ð % Insrt th pol componnts: rs Ð5 4 Ð5 4 Ð5 4 Ð5 4 Ð5 4 5 4 Ð5 4 5 4 Ð5 4 5 4 Ð5 4 Ð5 4 Ð5 4 cð5 5 4Ð cð5 5 4Ð (A2..5.15) By taking out th sign of th Ð5 5 trm an using Ð5 5 insta, w can us similar substitution factors as with th prvious two rsius: Ð5 5 Ð5 4 4 4 c Ð5 5 4Ð c Ð5 5 4Ð 5 4 Ð5 4 Ð5 4 Ð5 5 c Ð 4 Ð5 5 4 Ð5 Ð5 4 5 Ð5 5 Ð 4 Ð5 5 4 Rationali th nominator: Ð5 Ð5 4 Ð c 5 5 Ð 4 Ð5 5 5 Ð5 5 c Ð % Ð5 5 4 W now xtract th common ral factor K, which for rs an rs % is slightly iffrnt from : K 4 A2..22

P.Stariè, E.Margan Appnix 2. K Ð5 cð5 5 Ð % Ð5 5 (A2..5.16) W rgroup th rmaining numrator trms into ral an imaginary part: Ð5 4 cð5 5 Ð 4 Ð5 5 5 cð5 5 Ð Ð5 5 45Ð5 5 4 cð5 5 Ð 5 cð5 5 Ð Ð5 5 4 cð5 5 Ð 5 Ð5 5 From this, th following substitutions ar us for both rs an rs : % C Ð5 5 Ð (A2..5.17) an again, as in Eq. A2..5.9: B Ð5 5 So: 4 rs K ˆ C B 4 C B 5 5 a 5 b (A2..5.18) 4 Rsiu 4: % rs % lim Ð % Ä % Ð Ð Ð Ð (A2..5.19) % lim Ä % % Ð Ð Ð % Ð Ð Ð % % % % % Ð Ð Ð % % % % Insrt th pol componnts: rs % 5 Ð 4 Ð5 4 Ð5 4 Ð5 4 Ð5 4 5 4 Ð5 4 5 4 Ð5 4 5 4 (A2..5.20) Ð5 Ð5 4 Ð5 4 4 c Ð5 5 4Ð c Ð5 5 4Ð A2..2

P.Stariè, E.Margan Appnix 2. 5 4 Ð5 4 Ð5 4 Ð5 5 c Ð 4 Ð5 5 Ð5 Ð5 4 4 5 Ð5 5 Ð 4 Ð5 5 4 Rationali th nominator: Ð5 Ð5 4 Ð5 5 Ð 4 Ð5 5 4 c 5 Ð5 5 c Ð % Ð5 5 4 Lt us again xtract th sam common ral factor K as for rs (A2..82) an rgroup th rmaining trms of th numrator into ral an imaginary parts: Ð5 4 cð5 5 Ð 4 Ð5 5 an, using th sam substitutions 5 cð5 5 Ð Ð5 5 45Ð5 5 4 cð5 5 Ð 5 cð5 5 Ð Ð5 5 4 cð5 5 Ð 5 Ð5 5 K, C, an B, as for rs, w obtain: 4 rs% K ˆ C B 4 C B 5 5 a 5 b (A2..5.21) 4 Now w can sum all fiv rsius an group th rlativ ral an imaginary trms: % rs K ˆ 5 A B 4 aa 5 Bb! 5 K ˆ 5 A B 4 aa 5 Bb 5 K ˆ 5C B 4 ac 5Bb 5 K ˆ 5C B 4 ac 5Bb 5 4 4 4 4 4 4 4 4 (A2..5.22) A2..24

P.Stariè, E.Margan Appnix 2. This w can writ as: % K 5 A B A 5 B 5 rs 4! 4 K A 5 B A 5 B 5 4 4 K C 5 B C 5 B 5 4 4 K C 5 B C 5 B 5 4 4 From th rsiu pairs rs ß an rs ß% w xtract th common factors an sum th xponntials with th imaginary xponnt: % rs! 4 K 4 4 4 5 ˆ 5 A B aa 5 Bb 4 K 4 4 4 4 ˆ 5 C B ac 5 Bb 4 5 Now w rplac th complx xponntial trms by thir quivalnt sin an cosin: gð K 5 ca5 A Bbsina b aa 5 Bbcosa b K 5 5 C B C 5 B ˆ sina b a bcosa b (A2..5.2) Finally, ach sin cosin pair can b transform in a singl sin function with th appropriat phas shift, as w i in th thr-pol cas ( Eq. A2..17; but, as xplain thr, w must chck if this opration will caus a wrong phas s th nxt pag!): K 5 gð Éa5 A Bb aa 5 B b sina ) b (A2..5.24) K 5 Éa5 C Bb ac 5 B b sina ) b whr: ) aa 5 B b an ) ac 5 B b arctan arctan 5 A B 5 C B (A2..5.25) A2..25

P.Stariè, E.Margan Appnix 2. Phas rror Th nxt two figurs show th pculiar problm with th fourth-orr systm phas if it is calculat by Eq. A2..5.24 an A.2..5.25. In th cas of an MFA systm, Fig. A2..2, both angls shoul b incras by 1 raians in orr to obtain th corrct rspons. But for a MFED systm, Fig. A2.., only ) ns to b corrct! 2.0 1.5 ERROR: θ 1 45 θ 45 g() t 1.0 0.5 Corrct: θ 1 15 θ 15 0.0 0 1 2 4 5 t T / th Fig. A2..2: A 4 -orr MFA systm stp rspons is mirror ovr th final valu if th phas is not corrct. 6 4 ERROR: θ 1 θ 27.87 88.18 2 g() t 0 ERROR: Corrct: θ 1 θ 1 +180 θ 0 0 θ +180 2 4 ERROR: θ 1 +180 θ +180 0 1 2 4 5 t T / Fig. A2..: Strangly, th 4 -orr MFED systm stp rspons is corrct if only th th scon phas angl is corrct. All th othr combinations ar wrong! A2..26

P.Stariè, E.Margan Appnix 2. Ths figurs clarly monstrat th unprictabl natur of th sin cosin to amplitu phas transformation ow to th half-circl prio of th arctangnt function. Whilst som insight into th circuit s bhavior can b gain by obsrving th initial amplitu of ach rsonanc mo, it is avisabl not to rly on this form of quation unlss you ar absolutly crtain of th rsult. To play it saf, th sum of th rsius in thir complx xponntial form is rcommn for th numrical computation. Intrprtation As for all stp rspons functions, th first rsiu rprsnts th normali final valu. Th fourth-orr systm has two istinct sinusoial rsonanc mos, an, ach ow to its own complx conjugat pol pair, ach with a iffrnt initial phas, ) an ), an with a iffrnt amplitu corrction factor. Also, ach mo 5 has its own amping function, 5 an. A2..27

P.Stariè, E.Margan Appnix 2. A2..6 Fourth-orr systm with two complx conjugat pol pairs an on ral ro A four-pol on-ro normali gnral function is: % Ð % JÐ Ð Ð Ð Ð % In gnral, th two complx conjugat pol pairs an th ro ar: ß 5 4 an 5 5 4 ß% Th stp rspons in th omain is: % KÐ J Ð Ð Ð Ð Ð % (A2..6.1) (A2..6.2) (A2..6.) Th invrs Laplac transform of KÐ is: gð KÐf rs ckð _! Sinc only th pols hav rsius th stp rspons calculation is similar to that in th prvious sction, with th iffrnc that ach rsiu is multipli by th normali iffrnc btwn th appropriat pol an th ro, as shown blow. Rsiu 0: % rs! lim (A2..6.4) Ä! Ð Ð Ð Ð % % % lim Ä! Ð Ð Ð Ð % %!! Ð! Ð! Ð! Ð! % Rsiu 1: % rs lim Ð (A2..6.5) Ä Ð Ð Ð Ð % % lim Ä Ð Ð Ð % % Ð Ð Ð % % Ð Ð Ð % Lt us insrt th pol an ro componnts hr: A2..28

P.Stariè, E.Margan Appnix 2. rs a5 b 4 Ð5 4 Ð5 4 Ð5 4 a5 4 5 b 5Ð5 4 5 4 Ð5 4 5 4 Ð5 4 5 4 (A2..6.6) W can now follow th sam path as in Sc. A2..5 for rs up to th rationaliation of th nominator, but th ning part will b slightly iffrnt. By using th trm K from Eq. A2..5.7, th common factor which w can xtract bcoms: K 5 Ð5 5 Ð5 5 š c Ð % Ð5 5 (A2..6.7) an th rmaining part of th numrator of Eq. A2..6.6 is rarrang as follows: a5 4 5 bð5 4 Ð c 5 5 Ð 4 Ð5 5 a5 5 b5 cð5 5 Ð Ð5 5 f 4 5 cð5 5 Ð Ð5 5 f 4 a5 5 bcð5 5 Ð 5 Ð5 5 cð5 5 Ð 5 Ð5 5 To follow th furthr vlopmnt mor asily, lt us us th sam substitutions A an B, as in Eq. A2..5.8 an A.2..5.9: A Ð5 5 Ð an B Ð5 5 an, by sparating th ral an imaginary parts, w can rwrit th rmaining numrator trms as: a5 5bc5 A B aa 5 Bb 4 5 A B a5 5 baa 5 Bb W shall us two nw substitutions for th abov ral an imaginary trms: M a5 5bc5 A B aa 5 Bb N c5 A B a5 5 baa 5 Bb (A2..6.8) (A2..6.9) So, w can writ: 4 K 5 rs am 4 Nb (A2..6.10) 5 4 By a similar procur w arriv at th following rlations for th rmaining rsius: A2..29

P.Stariè, E.Margan Appnix 2. 4 K 5 rs am 4 Nb (A2..6.11) 5 4 For rs an rs % w must us a slightly iffrnt substitution, rflcting th iffrnc btwn A, Eq. A2..5.8, an C, Eq. A2..5.17: P a5 5 bc5 C B ac 5 Bb (A2..6.12) Q c5 C B a5 5 bac 5 Bb (A2..6.1) So: 4 K 5 rs ap 4 Qb (A2..6.14) 5 4 4 K 5 rs% ap 4 Qb (A2..6.15) 5 4 whr K is th sam as in Eq. A2..5.16. Finally, w sum all th rsius: % K gð rs a M 4 N 5 b 5 4! 4 K M 4 N 5 a b 5 4 4 K P 4 Q 5 a b 5 4 4 K P 4 Q 5 a b 5 4 (A2..6.16) which can b rwrittn as: gð 4 K M 4 5 N 4 4 5 4 4 K P 4 4 4 5 Q 5 4 (A2..6.17) an, by rplacing th complx xponntials with thir quivalnt sin-cosin forms: gð K 5 M 5 sina b N cosa b K 5 P sina b Q cosa b 5 4 (A2..6.18) No, w shall not attmpt to xamin th possibility of running into yt anothr st of phas rrors by any furthr trigonomtric transformation! A2..0

P.Stariè, E.Margan Appnix 2. A2..7 Fourth-orr systm with two pols ral, on complx conjugat pol pair an on ral ro A four-pol on-ro normali gnral function is: % Ð % JÐ Ð Ð Ð Ð % (A2..7.1) In th cas of th two ral pols, on complx conjugat pol pair, an on ral ro, th componnts ar: ß 5 4 5 % 5 % 5 Th stp rspons in th omain is: (A2..7.2) % KÐ J Ð Ð Ð Ð Ð % (A2..7.) Th invrs Laplac transform of KÐ is: gð KÐf rs ckð _! Th stp rspons calculation is similar to that in th prvious sction, with th iffrnc that, insta of th scon complx conjugat pol pair, thr ar two simpl ral pols. As bfor, rs!. Th calculation of th rmaining four rsius follows. Ð % rs lim (A2..7.4) Ä Ð Ð Ð Ð % % % Ð Ð Ð % a b % Ð Ð Ð % a5 4 5 ba5 4 b5 5% 5 Ð5 4 5 4 Ð5 4 5 Ð5 4 5 % c5 5 a5 4 b5 5% 5 4 Ð5 5 4 Ð5 5 4 5 4 % a5 4 b 4 c5 5 a5 4 b5 5% 5 5 cð5 5 Ð5 5 4 Ð 5 5 5 4 % % A2..1

P.Stariè, E.Margan Appnix 2. W rationali th nominator by multiplying it with its own complx conjugat; to kp th ratio unchang w also multiply th numrator: 5 5% c5 5 a5 4 bcð5 5 Ð5 5 % 4 Ð 5 5 5% 5 š cð5 5 Ð5 5 Ð 5 5 5 % % From this w xtract th common ral factor (th ro will b ntr in th final xprssion): 5 5 % K ca5 5 ba5 5 % b a 5 5 5% b š (A2..7.5) Th rmaining two brackts in th numrator shoul b multipli an sparat into a ral an imaginary part: c5 5 5 4 5 ca5 5 ba5 5 b 4 a 5 5 5 b % % a5 5 5 bca5 5 ba5 5 b % 4 5 a5 5 ba5 5 b c % % % 4 a 5 5 5 ba5 5 5 b 5 a 5 5 5 b % % a5 5 5 bca5 5 ba5 5 b 5 a 5 5 5 b % % 4 5 ca5 5 ba5 5 b a 5 5 5 ba5 5 5 bf To simplify th xprssions w shall us th following substitutions: A a5 5 ba5 5 % b B 5 5 5% (A2..7.6) C 5 5 5 With ths substitutions w writ: 4 K rs ˆ CA 5 B 5 4 Š 5A CB (A2..7.7) 5 4 an rs is th complx conjugat of rs : 4 K rs ˆ CA 5 B 5 4 Š 5A CB (A2..7.8) 5 4 Bfor going on to rsius of ral pols, lt us rwrit th sum rs rs in th complx xponntial form, as wll as in th sin cosin form, as usually: A2..2

P.Stariè, E.Margan Appnix 2. rs rs 4 4 4 4 5 K ˆ CA 5 B Š 5A CB 5 4 K 5 CA 5 B A CB 5 5 ˆ sina b Š cosa b (A2..7.9) Th calculation of th two rsius of ral pols is lss maning: Ð % rs lim (A2..7.10) Ä Ð Ð Ð Ð % % Ð Ð Ð % a5 4 ba5 4 b5% 5 5 Ð5 5 4 Ð5 5 4 Ð5 5 5 % 5 a5 b5% 5 5 a5 5 b Ð5 5 5 % 5 Ð % % rs% lim (A2..7.11) Ä % Ð Ð Ð Ð % % Ð Ð Ð % % % % a5 4 ba5 4 b5 5% 5 Ð5 5 4 Ð5 5 4 Ð5 5 5 % % % 5 % a5 b5 5% 5 a5 5 b Ð5 5 5 % % 5 % W xtract a common ral factor from rs an rs : % a5 b K a5 5 b 5 5% Ð (A2..7.12) an w writ: K rs % (A2..7.1) 5 5 a 5 5 b 5 K rs % (A2..7.14) 5 5 a 5 5 b 5 % A2..

P.Stariè, E.Margan Appnix 2. Finally, w can writ th systm stp rspons: K gð 5 CA 5 B 5 A CB 5 ˆ sina b Š cosa b K 5 K 5% 5% a5 5b 5 a5% 5b (A2..7.15) 5 5 Intrprtation: Th valu of th first rsiu (rs! ) is th final valu to which th systm will sttl whn th tim variabl bcoms many tims gratr than th largst tim constant of th systm. Th transint part of th rspons is govrn by th systm rsonanc, which in turn is trmin by th imaginary part of th complx conjugat pol pair,. Th initial phas of th systm rsonanc is givn by th ratio of th cofficints of th cosin an sin trms. Ovrall, th systm s rsonanc amplitu is st by th ral factor K an th rspons is xponntially amp by th ral part 5 of th complx conjugat pol pair. Th transint is aitionally influnc by th two xponntial cays ow to both ral pols. It is intrsting to not that th ro appars as a fr factor in th nominator of all four rsius. This inicats that if th ro is plac too clos to th complx plan s origin th systm ovrshoot will incras. Concluing Rmarks: Th attntiv rar has probably not that th two rsius of ach complx conjugat pol pair also form a complx conjugat pair thmslvs. Sinc for th tim omain rspons w n th sum of all rsius, by summing ach complx conjugat pair of rsius will yil a oubl ral part of th pair (th imaginary parts cancl). Mathmatically this can b xprss as: 5 4 Ê rs A 4B,, rs rs rs f A This mans that w can spar ourslvs a grat al of work if w calculat only on rsiu for ach pol pair, tak only its ral part an oubl its valu. Any ral pol will also hav a singl ral rsiu an it is simply a to th rst (no oubling hr!). th Thrfor, for a 8 -orr systm w only n to fin th ral part of 8Î rsius if 8is vn, an a8 bî rsius if 8is o. A2..4