Dynamic System Response - PDF Free Download

Solutio of Liear, Costat-Coefficiet, Ordiary Differetial Equatios Classical Operator Method Laplace Trasform Method Laplace Trasform Properties 1 st -Order Dyamic System Time ad Frequecy Respose d -Order Dyamic System Time ad Frequecy Respose 1

Laplace Trasform Methods A basic mathematical model used i may areas of egieerig is the liear ordiary differetial equatio with costat coefficiets: a dq dt o 1 d q a a dq o o + 1 + + + aq o = 1 1 0 dt dt b d m q m 1 d q b b dq i i i m + bq m m 1 + + + m 1 1 0 dt dt dt q o is the output (respose) variable of the system q i is the iput (excitatio) variable of the system a ad b m are the physical parameters of the system i

Straightforward aalytical solutios are available o matter how high the order of the equatio. Review of the classical operator method for solvig liear differetial equatios with costat coefficiets will be useful. Whe the iput q i (t) is specified, the right had side of the equatio becomes a kow fuctio of time, f(t). The classical operator method of solutio is a three-step procedure: Fid the complimetary (homogeeous) solutio q oc for the equatio with f(t) = 0. Fid the particular solutio q op with f(t) preset. Get the complete solutio q o = q oc + q op ad evaluate the costats of itegratio by applyig kow iitial coditios. 3

Step 1 To fid q oc, rewrite the differetial equatio usig the differetial operator otatio D = d/dt, treat the equatio as if it were algebraic, ad write the system characteristic equatio as: Treat this equatio as a algebraic equatio i the ukow D ad solve for the roots (eigevalues) s 1, s,..., s. Sice root fidig is a rapid computerized operatio, we assume all the roots are available ad ow we state rules that allow oe to immediately write dow q oc : Real, urepeated root s 1 : 1 ad + a D + + ad+ a = 1 Real root s repeated m times: q oc st 1 = 1 1 0 0 c e st st st m q = c e + c te + c t e + + c t e oc 0 1 m st 4

Whe the a s are real umbers, the ay complex roots that might appear always come i pairs a ± ib: For repeated root pairs a ± ib, a ± ib, ad so forth, we have: at at q = c e si bt+ φ + c te si bt+ φ + oc at q = ce si bt+ φ oc a f a f 0 0 1 1 The c s ad φ s are costats of itegratio whose umerical values caot be foud util the last step. a f 5

Step The particular solutio q op takes ito accout the "forcig fuctio" f(t) ad methods for gettig the particular solutio deped o the form of f(t). The method of udetermied coefficiets provides a simple method of gettig particular solutios for most f(t)'s of practical iterest. To check whether this approach will work, differetiate f(t) over ad over. If repeated differetiatio ultimately leads to zeros, or else to repetitio of a fiite umber of differet time fuctios, the the method will work. The particular solutio will the be a sum of terms made up of each differet type of fuctio foud i f(t) ad all its derivatives, each term multiplied by a ukow costat (udetermied coefficiet). 6

If f(t) or oe of its derivatives cotais a term idetical to a term i q oc, the correspodig terms should be multiplied by t. This particular solutio is the substituted ito the differetial equatio makig it a idetity. Gather like terms o each side, equate their coefficiets, ad obtai a set of simultaeous algebraic equatios that ca be solved for all the udetermied coefficiets. Step 3 We ow have q oc (with ukow costats) ad q op (with o ukow costats). The complete solutio q o = q oc + q op. The iitial coditios are the applied to fid the ukow costats. 7

Certai advaced aalysis methods are most easily developed through the use of the Laplace Trasform. A trasformatio is a techique i which a fuctio is trasformed from depedece o oe variable to depedece o aother variable. Here we will trasform relatioships specified i the time domai ito a ew domai wherei the axioms of algebra ca be applied rather tha the axioms of differetial or differece equatios. The trasformatios used are the Laplace trasformatio (differetial equatios) ad the Z trasformatio (differece equatios). The Laplace trasformatio results i fuctios of the time variable t beig trasformed ito fuctios of the frequecy-related variable s. 8

The Z trasformatio is a direct outgrowth of the Laplace trasformatio ad the use of a modulated trai of impulses to represet a sampled fuctio mathematically. The Z trasformatio allows us to apply the frequecydomai aalysis ad desig techiques of cotiuous cotrol theory to discrete cotrol systems. Oe use of the Laplace Trasform is as a alterative method for solvig liear differetial equatios with costat coefficiets. Although this method will ot solve ay equatios that caot be solved also by the classical operator method, it presets certai advatages: Separate steps to fid the complemetary solutio, particular solutio, ad costats of itegratio are ot used. The complete solutio, icludig iitial coditios, is obtaied at oce. 9

There is ever ay questio about which iitial coditios are eeded. I the classical operator method, the iitial coditios are evaluated at t = 0 +, a time just after the iput is applied. For some kids of systems ad iputs, these iitial coditios are ot the same as those before the iput is applied, so extra work is required to fid them. The Laplace Trasform method uses the coditios before the iput is applied; these are geerally physically kow ad are ofte zero, simplifyig the work. For iputs that caot be described by a sigle formula for their etire course, but must be defied over segmets of time, the classical operator method requires a piecewise solutio with tedious matchig of fial coditios of oe piece with iitial coditios of the ext. The Laplace Trasform method hadles such discotiuous iputs very eatly. The Laplace Trasform method allows the use of graphical techiques for predictig system performace without actually solvig system differetial equatios. 10

All theorems ad techiques of the Laplace Trasform derive from the fudametal defiitio for the direct Laplace Trasform F(s) of the time fuctio f(t): st Lf() t = Fs = f() te dt s af 0 z t > 0 = a complex variable = σ+ iω This itegral caot be evaluated for all f(t)'s, but whe it ca, it establishes a uique pair of fuctios, f(t) i the time domai ad its compaio F(s) i the s domai. Comprehesive tables of Laplace Trasform pairs are available. Sigals we ca physically geerate always have correspodig Laplace trasforms. Whe we use the Laplace Trasform to solve differetial equatios, we must trasform etire equatios, ot just isolated f(t) fuctios, so several theorems are ecessary for this. 11

Liearity Theorem: L af t + af t = L af t + L af t = af s+ af s af af af af af af 1 1 1 1 1 1 Differetiatio Theorem: LdfO af L sf s f dt af 0 L L NM L NM L NM QP = O af af af df QP df sfs sf dt = 0 0 dt O 1 df QP df d f sfs s f s dt = 1 0 0 dt 0 1 dt af af af af f(0), (df/dt)(0), etc., are iitial values of f(t) ad its derivatives evaluated umerically at a time istat before the drivig iput is applied. 1

Itegratio Theorem: z af af a faf 1 Fs f 0 L ftdt= + s s af af Fs f a af kf L f a f 0 t = + k+ 1 s k= 1 s a af afa f af af f z z where f t = f t dt ad f a -0 f t = f t Agai, the iitial values of f(t) ad its itegrals are evaluated umerically at a time istat before the drivig iput is applied. 13

Delay Theorem: The Laplace Trasform provides a theorem useful for the dyamic system elemet called dead time (trasport lag) ad for dealig efficietly with discotiuous iputs. ut () = 10. t> 0 ut () = 0 t< 0 ut ( a) = 10. t> a ut ( a) = 0 t< a as Lf( t aut ) ( a) = e Fs ( ) 14

Fial Value Theorem: If we kow Q 0 (s), q 0 ( ) ca be foud quickly without doig the complete iverse trasform by use of the fial value theorem. lim f( t) lim sf( s) t = s 0 This is true if the system ad iput are such that the output approaches a costat value as t approaches. Iitial Value Theorem: This theorem is helpful for fidig the value of f(t) just after the iput has bee applied, i.e., at t = 0 +. I gettig the F(s) eeded to apply this theorem, our usual defiitio of iitial coditios as those before the iput is applied must be used. lim f( t) = lim sf( s) t 0 s 15

Impulse Fuctio p(t) 1/b Approximatig fuctio for the uit impulse fuctio b δ( t) t δ() t z+ ε ε δ() t = lim p() t b 0 = 0 t 0 δ() tdt L L δ() t = L NM du QP su() s s. dt = = 1 = 10 s The step fuctio is the itegral of the impulse fuctio, or coversely, the impulse fuctio is the derivative of the step fuctio. Whe we multiply the impulse fuctio by some umber, we icrease the stregth of the impulse, but stregth ow meas area, ot height as it does for ordiary fuctios. O = 1 ε> 0 16

A impulse that has a ifiite magitude ad zero duratio is mathematical fictio ad does ot occur i physical systems. If, however, the magitude of a pulse iput to a system is very large ad its duratio is very short compared to the system time costats, the we ca approximate the pulse iput by a impulse fuctio. The impulse iput supplies eergy to the system i a ifiitesimal time. 17

Iverse Laplace Trasformatio A coveiet method for obtaiig the iverse Laplace trasform is to use a table of Laplace trasforms. I this case, the Laplace trasform must be i a form immediately recogizable i such a table. If a particular trasform F(s) caot be foud i a table, the we may expad it ito partial fractios ad write F(s) i terms of simple fuctios of s for which iverse Laplace trasforms are already kow. These methods for fidig iverse Laplace trasforms are based o the fact that the uique correspodece of a time fuctio ad its iverse Laplace trasform holds for ay cotiuous time fuctio. 18

Time Respose & Frequecy Respose 1 st -Order Dyamic System Example: RC Low-Pass Filter i i i R out Dyamic System Ivestigatio e i C e out of the RC Low-Pass Filter 19

Measuremets, Calculatios, Maufacturer's Specificatios Model Parameter ID Which Parameters to Idetify? What Tests to Perform? Physical System Physical Model Math Model Experimetal Aalysis Assumptios ad Egieerig Judgemet Physical Laws Model Iadequate: Modify Equatio Solutio: Aalytical ad Numerical Solutio Actual Dyamic Behavior Compare Predicted Dyamic Behavior Modify or Augmet Make Desig Decisios Model Adequate, Performace Iadequate Model Adequate, Performace Adequate Desig Complete Dyamic System Ivestigatio 0

Zero-Order Dyamic System Model 1

Validatio of a Zero-Order Dyamic System Model

1 st -Order Dyamic System Model τ = time costat K = steady-state gai t = τ Slope at t = 0 q o Kq = τ is t e τ Kq is q o t = = 0 τ 3

How would you determie if a experimetallydetermied step respose of a system could be represeted by a first-order system step respose? qo() t = Kqis 1 e o () q t Kq q Kq is () t is t o 1 = e τ Kq is = e () t τ t τ qo t t t log10 1 = log10 e = 0.4343 Kqis τ τ Straight-Lie Plot: q ( t) o log10 1 vs. t Kqis Slope = -0.4343/τ 4

This approach gives a more accurate value of τ sice the best lie through all the data poits is used rather tha just two poits, as i the 63.% method. Furthermore, if the data poits fall early o a straight lie, we are assured that the istrumet is behavig as a first-order type. If the data deviate cosiderably from a straight lie, we kow the system is ot truly first order ad a τ value obtaied by the 63.% method would be quite misleadig. A eve stroger verificatio (or refutatio) of first-order dyamic characteristics is available from frequecyrespose testig. If the system is truly first-order, the amplitude ratio follows the typical low- ad highfrequecy asymptotes (slope 0 ad 0 db/decade) ad the phase agle approaches -90 asymptotically. 5

If these characteristics are preset, the umerical value of τ is foud by determiig ω (rad/sec) at the breakpoit ad usig τ = 1/ω break. Deviatios from the above amplitude ad/or phase characteristics idicate o-first-order behavior. 6

What is the relatioship betwee the uit-step respose ad the uit-ramp respose ad betwee the uit-impulse respose ad the uit-step respose? For a liear time-ivariat system, the respose to the derivative of a iput sigal ca be obtaied by differetiatig the respose of the system to the origial sigal. For a liear time-ivariat system, the respose to the itegral of a iput sigal ca be obtaied by itegratig the respose of the system to the origial sigal ad by determiig the itegratio costats from the zero-output iitial coditio. 7

Uit-Step Iput is the derivative of the Uit-Ramp Iput. Uit-Impulse Iput is the derivative of the Uit- Step Iput. Oce you kow the uit-step respose, take the derivative to get the uit-impulse respose ad itegrate to get the uit-ramp respose. 8

System Frequecy Respose 9

Bode Plottig of 1 st -Order Frequecy Respose db = 0 log 10 (amplitude ratio) decade = 10 to 1 frequecy chage octave = to 1 frequecy chage 30

Aalog Electroics: RC Low-Pass Filter Time Respose & Frequecy Respose i i e i R C i out e out e RCs + 1 R e i out i = i Cs 1 i out eout 1 1 = = whe iout = 0 e RCs + 1 τ s + 1 i 31

Time Respose to Uit Step Iput 1 0.9 0.8 0.7 Amplitude 0.6 0.5 0.4 0.3 R = 15 KΩ C = 0.01 µf 0. 0.1 0 0 1 3 4 Time (sec) 5 6 7 8 x 10-4 Time Costat τ = RC 3

Time Costat τ Time it takes the step respose to reach 63% of the steady-state value Rise Time T r =. τ Time it takes the step respose to go from 10% to 90% of the steady-state value Delay Time T d = 0.69 τ Time it takes the step respose to reach 50% of the steady-state value 33

Frequecy Respose R = 15 KΩ C = 0.01 µf 0 0-5 -0 Gai db -10-15 Phase (degrees) -40-60 -0-80 -5 10 10 3 10 4 10 5 Frequecy (rad/sec) Badwidth = 1/τ -100 10 10 3 ( ) ( ) Frequecy (ra e K K 0 K ωτ + ωτ + ωτ ωτ + out 1 ( iω ) = = = ta ωτ e 1 i i 1 1 ta 1 34

Badwidth The badwidth is the frequecy where the amplitude ratio drops by a factor of 0.707 = -3dB of its gai at zero or low-frequecy. For a 1 st -order system, the badwidth is equal to 1/ τ. The larger (smaller) the badwidth, the faster (slower) the step respose. Badwidth is a direct measure of system susceptibility to oise, as well as a idicator of the system speed of respose. 35

MatLab / Simulik Diagram Frequecy Respose for 1061 Hz Sie Iput τ = 1.5E-4 sec Sie Wave 1 tau.s+1 First-Order Plat output output iput iput Clock t time 36

Amplitude Ratio = 0.707 = -3 db Phase Agle = -45 Iput 1 0.8 Respose to Iput 1061 Hz Sie Wave 0.6 0.4 0. amplitude 0-0. -0.4 Output -0.6-0.8-1 0 0.5 1 1.5.5 3 3.5 4 time (sec) x 10-3 37

Time Respose & Frequecy Respose d -Order Dyamic System Example: -Pole, Low-Pass, Active Filter R 4 R 7 C 5 e i R 1 C R 3 - + R 6 - + e out Dyamic System Ivestigatio of the Two-Pole, Low-Pass, Active Filter 38

Physical Model Ideal Trasfer Fuctio R 7 1 R eout 6 R1R3CC 5 ( s) = ei 1 1 1 1 s + + + s+ RC RC RC RRCC 3 1 4 3 4 5 R 4 R 7 C 5 e i R 1 C R 3 - + R 6 - + e out 40

d -Order Dyamic System Model dq0 dq0 + 1 + 0 0 = 0 i a a a q b q dt dt 1 + + q0 = Kqi ω dq0 ζ dq0 dt ω dt a 0 ω = a 1 ζ = 0 0 K = steady-state gai 0 a a a b a udamped atural frequecy dampig ratio Step Respose of a d -Order System 41

dq0 ς dq0 dt ω dt 1 + ζ + q = Kq ω 0 i Step Respose of a d -Order System Uderdamped 1 ( ) ζω t q 1 o = Kqis 1 e si ω 1 ζ t + si 1 ζ ζ< 1 1 ζ Critically Damped ω ( ) t qo = Kq is 1 1+ωt e ζ = 1 Overdamped ( ζ+ ζ ) ζ+ ζ 1 1 ωt 1 e ζ 1 qo = Kq is ζ > 1 ζ ζ 1 ( ζ ζ 1) ωt + e ζ 1 4

Frequecy Respose of a d -Order System Laplace Trasfer Fuctio Q Q o i ( s) = s ω K ζs + + 1 ω Siusoidal Trasfer Fuctio Qo K 1 ζ ( iω ) = ta Q i ω ω ω 4ζ ω 1 + ω ω ω ω 43

Frequecy Respose of a d -Order System 44

-40 db per decade slope Frequecy Respose of a d -Order System 45

Some Observatios Whe a physical system exhibits a atural oscillatory behavior, a 1 st -order model (or eve a cascade of several 1 st -order models) caot provide the desired respose. The simplest model that does possess that possibility is the d -order dyamic system model. This system is very importat i cotrol desig. System specificatios are ofte give assumig that the system is d order. For higher-order systems, we ca ofte use domiat pole techiques to approximate the system with a d - order trasfer fuctio. 46

Dampig ratio ζ clearly cotrols oscillatio; ζ < 1 is required for oscillatory behavior. The udamped case (ζ = 0) is ot physically realizable (total absece of eergy loss effects) but gives us, mathematically, a sustaied oscillatio at frequecy ω. Natural oscillatios of damped systems are at the damped atural frequecy ω d, ad ot at ω. ω d =ω 1 ζ I hardware desig, a optimum value of ζ = 0.64 is ofte used to give maximum respose speed without excessive oscillatio. 47

Udamped atural frequecy ω is the major factor i respose speed. For a give ζ respose speed is directly proportioal to ω. Thus, whe d -order compoets are used i feedback system desig, large values of ω (small lags) are desirable sice they allow the use of larger loop gai before stability limits are ecoutered. For frequecy respose, a resoat peak occurs for ζ < 0.707. The peak frequecy is ω p ad the peak amplitude ratio depeds oly o ζ. ω p =ω 1 ζ peak amplitude ratio = ζ 1 ζ K 48

As ζ varies from 0 to 1, BW varies from 1.55ω to 0.64ω. For a value of ζ = 0.707, BW = ω. For most desig cosideratios, we assume that the badwidth of a d -order all pole system ca be approximated by ω. 50

G(s) Kω s s + ςω +ω s = ςω ± iω 1 ς s = 1, = σ± iω 1, d σt σ y() t = 1 e cosω dt+ siωdt ωd 1.8 t r rise time ω 4.6 t s settlig time ςω πς ( ) = ζ < 1 ς Mp e 0 1 overshoot ζ 1 0 ζ 0.6 0.6 ( ) Locatio of Poles Of Trasfer Fuctio Geeral All-Pole d -Order Step Respose 51

ω 1.8 t r σ 4.6 t s ( p ) ζ 0.6 1 M 0 ζ 0.6 Time-Respose Specificatios vs. Pole-Locatio Specificatios 5

Experimetal Determiatio of ζ ad ω ζ ad ω ca be obtaied i a umber of ways from step or frequecy-respose tests. For a uderdamped secod-order system, the values of ζ ad ω may be foud from the relatios: πζ π T = ω d M 1 ζ p e = ζ = π log ( ) e M p 1 + 1 ωd π ω d =ω 1 ζ ω = = 1 ζ T 1 ζ 53

Logarithmic Decremet δ is the atural logarithm of the ratio of two successive amplitudes. ( ) x() t ( + T) ζωt δ= l = l ( e ) =ζωt x t ζω π ζω π πζ ζ= = = = ω d ω 1 ζ 1 ζ δ= 1 l δ B B π +δ 1 + 1 Free Respose of a d -Order System π T = ω d ζωt ( ) = ( ω +φ) x t Be si t d 54

If several cycles of oscillatio appear i the record, it is more accurate to determie the period T as the average of as may distict cycles as are available rather tha from a sigle cycle. If a system is strictly liear ad secod-order, the value of is immaterial; the same value of ζ will be foud for ay umber of cycles. Thus if ζ is calculated for, say, = 1,, 4, ad 6 ad differet umerical values of ζ are obtaied, we kow that the system is ot followig the postulated mathematical model. For overdamped systems (ζ > 1.0), o oscillatios exist, ad the determiatio of ζ ad ω becomes more difficult. Usually it is easier to express the system respose i terms of two time costats. 55

q Kq For the overdamped step respose: t o τ 1 τ ( ζ+ ζ ) ζ+ ζ 1 1 ωt 1 e ζ 1 qo = Kq is ζ > 1 ζ ζ 1 ( ζ ζ 1) ωt + e ζ 1 τ1 τ e e 1 = + τ τ τ τ is 1 1 t where τ 1 1 τ ζ ζ ω ζ + ζ ω ( ) ( ) 1 1 1 56

To fid τ 1 ad τ from a step-fuctio respose curve, we may proceed as follows: Defie the percet icomplete respose R pi as: q o Rpi 1 100 Kqis Plot R pi o a logarithmic scale versus time t o a liear scale. This curve will approach a straight lie for large t if the system is secod-order. Exted this lie back to t = 0, ad ote the value P 1 where this lie itersects the R pi scale. Now, τ 1 is the time at which the straight-lie asymptote has the value 0.368P 1. Now plot o the same graph a ew curve which is the differece betwee the straight-lie asymptote ad R pi. If this ew curve is ot a straight lie, the system is ot secod-order. If it is a straight lie, the time at which this lie has the value 0.368(P 1-100) is umerically equal to τ. Frequecy-respose methods may also be used to fid τ 1 ad τ. 57

Step-Respose Test for Overdamped Secod-Order Systems 58

Frequecy-Respose Test of Secod-Order Systems 59