Laplace Transforms - Iowa State University

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Laplace transformsWhen we evaluate the performance of a circuit, there are many aspectsto consider. Two of the most important are:1. Step response. How does the output respond when the inputchanges abruptly, as in the case of a digital logic circuit? In otherwords, what is the transient response to large change in inputvoltage or current?2. Frequency response. What is the the response at the output whenthe input is a sinusoid? In particular, how does the sinusoidaloutput change when the frequency is varied?Both types of analysis were introduced in EE 201 — RC, RL, and RLCtransients and sinusoidal steady-state analysis. The mathematicalapproach in each case started with a consideration of the differentialequations that characterized the circuits, but the two approachesseemed to diverge. The transient analysis followed directly along thedifferential-equation route, but the AC analysis veered towards usingcomplex numbers, with the circuit being transformed into a new versionthat was analyzed using complex math.EE 230Laplace – 1

Recall from 201The step response of the capacitor voltage in a simple RC circuit. Theresult is a exponential transient with an RC time constant. The RLtransient case had similar behavior.RVfvi (t)Vi –Ct 0dvCdt vCRC vC (t)–at t 0, vC ViVfRCvC (t) Vf [Vf Vi] exp ( σt)11σ τRC(Go back the 201 notes, if you need to review.)EE 230Laplace – 2

Also from 201The sinusoidal response of the capacitor voltage in a simple RC circuit.In 201, we solved two ways: as a straight-forward differential equationand then by transforming the circuit and using impedances with complexanalysis. With the complex approach, transient effects were ignored.RdvCvCVA cos ωtdtRC RCvi (t) VAcosωt CvC (t)– σt(t)v Ae M cos (ωt θ)–CM ZRṽi VAeEE 230j0 –ZC ṽC–VA1 (ωRC)2ṽC θ arctan (ωRC)ZCZR ZCṽiṽC Me jθvC (t) M exp (jωt θ)Laplace – 3

We saw a similar relationships for second-order RLC circuits.Overdamped twodecaying exponentials.RVfvi (t)Vi –t 0CLA1, A2, σ1, and σ2 depend onR, L, C, and initial conditions.Rvi (t) VAsinωt –CL σ1t σ2t V V VAe AevC (t)]fi) [ 12( f– vC (t) B1e σ1t B2e σ2t M cos (ωt θ)–B1, B2, σ1, σ2, M, θ depend onVm, R, L, C, ω, and initial conditions.Again, using sinusoidal steady-stateanalysis, the sinusoidal part of thecapacitor voltage can be expressed as vC (t) Me j(ωt θ) vC Me jθEE 230Laplace – 4

Can these two approaches — the step-response using differentialequations and the AC method using a circuit transformation andcomplex analysis — be reconciled? In both cases, we are looking atetsame circuit — only the details of the source have changed. It seemsthat there might be a more unified approach to handling the twosituations. Solving differential equations is tedious but does works inevery case. The circuit transform approach offers simplicity, but can itbe made more general? The key the unification comes in consideringthe time dependence of the solutions in the two cases. In particular:For step-function problems, the solutions were exponentials,characterized by a decay rate (or rates for the 2nd-order case):vC (t) exp ( σt)σ decay rate, units of (seconds)–1.(Or nepers/s. Nepers are dimensionless.)For the steady-state sinusoidal problems, using the complex form, thesolutions are also exponentials, but this time complex:vC (t) exp (jωt)EE 230ω angular frequency, units of (seconds)–1.(Or rad/s. Radians are dimensionless.)Laplace – 5

Complex frequencyWe might consider combining these two exponentials into a singlequantity, which we could call complex frequency:s σ jω e st e σte jωts complex frequency, units of s–1.The complex frequency encompasses both transient and sinusoidalsituations. For pure step-function situations, ω 0 (DC) and s σ. Forsinusoidal steady-state situations, σ 0 and s jω.In fact, we witnessed this unified frequency notion in 201. In theunderdamped RLC transient, the capacitor voltage oscillated for a timebefore settling to the final voltage — both σ and ω were needed in thesolution.EE 230Laplace – 6

RecallRVfViVS(t) –From 201:CL vC(t)–vC (t) Vf (Vf Vi) eunderdamped a decayingoscillation σtσcos ωdt sin ωdtωd[]Using this new notion of complex frequency, we can re-write theunderdamped response as:vC (t) Vf EE 230(Vf Vi2σσ (σ jωd)t1 e 1 e (σ jωd)tωd )ωd )) ((Laplace – 7

The Laplace TransformThe idea of complex frequency leads inexorably to the Laplacetransform which is one of a number of integral transforms that allow foreasier solution of differential equations. The idea is to transform aproblem from one domain (or space) into a related domain, where,hopefully, the equations are easier to solve. Applying this method tocircuits, we will transform the differential equation from the timedomain to the frequency domain and find a solution in that form. Thenwe can transform back to the time domain to arrive at the final solution.You likely saw this method applied in your differential equations class.However, we will learn soon enough that transforming back from thefrequency domain is not really necessary. The frequency-domainrepresentation presents useful information without the need to transformback to the time domain.Working in the frequency domain is a key skill for EEs. Being able tosee how a system behaves in both the time domain and the frequencydomain leads to a much deeper understanding of a system and isessential for system design.EE 230Laplace – 8

The Laplace TransformGiven a function of time, f (t), we can transform it into a new, but related,function F(s).ℒ {f (t)} EE 230 0f (t) e st dt F (s) exp(–st) is the kernel of the transform, where s σ jω is the complexfrequency. By integrating from 0 to infinity, we “integrate out the time”, leaving afunction that depends only on s. The two variables s and t are complementary. If t is time, then s musthave units of inverse time, i.e. a frequency, and the product s·t is thendimensionless. This is the “one-sided” Laplace transform, since the integral starts att 0. There is a two-sided Laplace transform, but the extra integrationrange doesn’t really add to the utility of the transformation. In using theone-sided version, we assume that everything starts at t 0. The variable s is complex, and so F(s) must be complex function. Thishas implications when we attempt to use F(s) later.Laplace transform –9

Useful properties of Laplace TransformsGiven functions f(t) and f1(t), having L.T.s F(s) and F1(s)F (s) ℒ{f (t)}F1 (s) ℒ{f1 (t)}Here are a couple of obvious ones. These relations are easily provedusing the definition of at the Laplace transform. We will use these resultsconstantly when applying Laplace transforms to circuits.1. Multiply / divide by a constant m. (The number m could be complex.)ℒ{m f (t)} m ℒ{f (t)} m F (s)ℒ {f (t)} F (s)f (t)ℒ { m }mm2. Addition and subtraction.ℒ {f (t) f1 (t)} F (s) F1 (s)EE 230Laplace transform – 10

Here are the two key relationships for Laplace transforms. Withoutthese, the Laplace method would not be very useful.3. Differentiation. f (0) is the initial condition of the function at t 0.df (t)ℒ sF (s) f (0){ dt }4. Integration:ℒ { 0F (s)f (t) dt s}Higher order derivatives and integrals.d 2f (t)df2ℒ s F (s) sf (0) 2dt{ dt } t 0tF (s)ℒf (x) dx 2s{ 0 0}EE 230Laplace – 11

Some other interesting properties. (These will be used more extensivelyin EE 324.)1s5. Changing time scale: ℒ {f (at)} Fa (a)Expanding the time scale compresses the frequency scale.Compressing the time scale expands the frequency scale.6. Time shift: ℒ {f (t a)} e as F (s)7. Frequency shift: ℒ {e at f (t)} F (s a)Note the mathematical symmetry of the time and frequency shiftrelationships.There are many other important properties of Laplace transforms, but wewill leave the more advance details to EE 224, EE 324, and other generalsystems classes. Here we focus on the essentials needed to understandhow our basic electronic systems behave.EE 230Laplace transform – 12

ExampleFor a first example consider a linear ramp in time. The function is zerofor t 0, and then ramps up with a slope of 1: f (t) t.F (s) ℒ {f (t)} 0te stf (t)dtUse integration by parts: t st stte dt e 0s t st es 1F (s) 2s001 st e dts 01 st 2est 01 0 2smIf the slope is not 1, so that f (t) m t, then F (s) , bys2multiplicative property of LTs.EE 230Laplace – 13

Unit step functionWhen we studied transients in 201, we frequentlyused step-change sources, where the source valueabruptly jumped from one level to another. However,we didn’t develop a mathematical formalism — itwas just a jump between two values.u(t)01t 0Now we should be more formal. The basic step function, u (t) is definedby an abrupt change from 0 to 1 at t 0, making it a unit step function.Then an abrupt change in source voltage or current can be written as:Vfvs (t) Vf ·u(t)0t 0is (t) If ·u(t)0Ift 0The step could also go negative, in which case a the voltage would bevs (t) Vf u (t).EE 230Laplace transform – 14

In 201, we often used the situation where the source started at a non-zerolevel and stepped to another value.Vf VVivs(t) Vi V ·u(t)t 0 Vf ·u(t) [1 – u(t)] ViIf the step occurs at a time to 0, the step function would be shifted intime, u (t to).The unity-step function can also “turn on” other functions so that theyare zero for t 0. For example:vs (t) [VA cos ωt] u (t)is a function that is 0 for t 0, and then become a cosine for t 0.Since we will be using one-sided Laplace transforms, which are definedfrom t starting at zero and extending to infinity, we implicitly assumethat all source functions are multiplied by a unit step function, so thatthe functions are definitely zero for t 0.EE 230Laplace transform – 15

Transform of unit step, u(t)Apparently, we will need the Laplace transform for the unit step.F (s) ℒ{u (t)} stee st dt 0s 01 e st dt 01 0 ( s)1F (s) sThe transform for a typical step voltage source would be:Vaℒ {Va u (t)} sEE 230Laplace transform – 16

Decaying exponentialAnother function that we will use frequently is the simple exponential.We include a decay constant σ.f (t) e σtℒ {e σt} 0 0e σte st dte (s σ)t dt1 (s σ)t es σ1 s σ 0The transform for a growing exponential isℒ {eEE 2301} s σ σtLaplace transform – 17

sinusoids1 jωtRecall from Euler: cos ωt (e e jωt)21 jωt jωt stℒ {cos ωt} e eedt()2 01 (s jω)t1 (s jω)t edt edt2 02 0 12 (s jω)12 (s jω) e (s jω)t 012 (s jω)e (s jω)t 012 (s jω)s 2s ω2The derivation for sine is similar. The result is:ωℒ {sin ωt} 22s ωEE 230Laplace transform – 18

A few transformsf(t)F(s)impulseδ(t)1stepu(t)rampte σtexponentialsinesin ωtωs2 ω2cosinecos ωtss2 ω2phasore jωt1s jωωdamped sinedamped cosineEE 2301s1s21s σe σt sin ωte σtcos ωt(s σ)2 ω 2(s σ)(s σ)2 ω 2Laplace transform – 19

Analyzing a circuit in the frequency domainNow we are ready to apply the Laplace method to solve a problem. Wecould start with a generic differential equation, like was done in thedifferential equations class from the math department. However, wemay as well go directly to a circuit, since analyzing circuits in the timedomain leads to differential equations.The method is straight-forward:1. Using usual analysis techniques, find the differential equation for thequantity of interest in the circuit.2. Use Laplace methods to transform the entire equation into thefrequency domain. The differential equation in the time domainbecomes an algebra problem in the frequency domain.3. Use basic algebra to find a frequency-domain expression for theLaplace transform of the quantity of interest.4. Transform back from the frequency domain to the time domain.EE 230Later, we will see that step 4 is optional — it is not always necessary totransform back. We will not to do step 4 in the following examples.We will look the inverse transformation in the next set of notes.Laplace – 20

Example 1RFind the frequency domainexpression for the capacitorvoltage in the RC circuit at right.The source is a unit step with Vi 0for t 0 and Vf 10 V for t 0.Vfvi (t)Vi –Ct 0 vC (t)–The differential equation has been derived previously, but we will repeathere, just to be complete in this first example.iR iCvi (t) vC (t)RdvC (t)dt CdvC (t)dtvC (t)vi (t) RCRCTake the the Laplace transform of both sides of the equationℒEE 230{dvC (t)dtvi (t) ℒ{ RC }RC }vC (t)Laplace transform – 21

ℒ{dvC (t)dtvi (t) ℒ{ RC }RC }vC (t)Use the addition/multiplication properties of the LT to break it down a bit.ℒ{dvC (t)dt11 ℒ {vC (t)} ℒ {vi (t)}} RCRCThe three transforms are:ℒ {vC (t)} VC (s)ℒ{dvC (t)dt} sVC (s) vC (0)ℒ {vi (t)} ℒ {Vf u (t)} VfsNote that in this case, the initial condition is vc (0) 0, so the derivativeexpression is simplified. Putting it all back together,EE 230sVC (s) VC (s)RC VfsRCLaplace – 22

sVC (s) VC (s)RC VfsRCWith a bit of simple algebra, the frequency domain form of thecapacitor voltage isVC (s) 1s (s 1RC ) VfRCOf course, we don’t yet know the meaning of this function. In principlewe can transform it back to the time domain, and we will do that soonenough.More importantly, after a bit more practice, we will come realize thatthe frequency-domain form above tells us everything we need to knowabout the circuit’s behavior. And arriving at the frequency-domainexpression using the Laplace transform was quite easy.EE 230Laplace – 23

Example 2RSame circuit but with asinusoid source. Again, vi (t) VAcosωtassume that vc (0) 0. –C vC (t)–We won’t go through all the steps here — you should do that on yourown. The steps are very similar to example 1.dvCvCVA cos ωtdtRC RCVC (s)VAssVC (s) 2RCRC s ω 2VC (s) EE 2301(s2 ) (s ω21RC )V RCLaplace – 24

Example 3RVfRLC — a second-ordersystem — with a step input.vi (t)Vi –t 0Again, the analysis isabbreviated — fill in themissing steps for yourself.CL vC (t)–To keep it simple, use initial conditions, vC (0) 0[ dtdvCand iC (0) 0d 2vC (t)dt 2t 0 0 .]VfR dvC (t)1 vC (t) u (t)L dtLCLCVfR1s VC (s) sVC (s) VC (s) LLCsLC2VC (s) EE 2301s (s 2 RL s 1LC ) VfLCOnce the differential equation is inplace, the Laplace stuff is so easy!Laplace – 25

Example 4RAn RLC with a sine functionsource. Use the same initialconditions as in the previousRLC example.d 2vC (t)dt 2vi (t) VAsinωt –CL vC (t)–R dvC (t)1 vC (t) VA sin ωtL dtLCVAR1ωs VC (s) sVC (s) VC (s) 2LLCLC s ω 22VC (s) EE 230(s2ω RsL 12s(LC ) Vf ω 2) LCLaplace – 26

Example 5CHow about an op amp with astep input? Again, forsimplicity, use initial conditionof Vi 0, which translates toVfvi (t)Vit 0vC (0) 0 [ vo(0) 0 ].R2R1– vo (t)iR1 iR2 iCvi (t) Vf u (t) vo (t)dvo CR1R1R2dtVfVo (s) CsVo (s)sR1R2Vo (s) EE 2301s (s 1R2C ) VfR1CNote how similar this is to theresult of Example 1.Laplace – 27

Example 6COne more. Let’s do anop amp with a sinusoidalsource. Just for fun, use acomplex exponential forthe input sinusoid.R2vi (t) VAexp(jωt)R1– vo (t)vi (t) VA e jωt vo (t)dvo CR1R1R2dtVfVo (s) CsVo (s)R2R1 (s jω)Vo (s) EE 230(s 11sR2C ) ( Vf jω) R1CLaplace – 28

UnitsIn each of the examples, once the differential equation was transformed into thefrequency domain, the math was quite easy. That is the magic of Laplacetransforms — they turn messy differential equations is simpler algebra equations.Again, you may well have seen this your diff. eq. math class.However, there is an important distinction that we should emphasize. In atypical math class, the variables and equations have no units. When s wasintroduced as complementary variable in the LT process, it did not have anyspecific physical significance — it was just a means to an end. Transform afunction of t (or x or z or whatever) into a function of s, do some manipulations,and then transform back to the original variable. The “s” fades away.On the other hand, our circuits are physical, and the every quantity has units thatare tied to the physical meaning. So in transforming from time to frequency, theunits for s must be complementary — s is the complex frequency and it musthave units of inverse seconds (sec–1). (Note the potential confusion if we use sfor seconds as is typical in most situations.)Similarly, the transforms for voltage and current have defined units. In looking atthe definition of the LT, it is apparent that voltage in the time domain transformsto a frequency-domain quantity with units of volt-seconds (V·sec.) A frequencydomain current has units of amp-seconds (A·sec). These are subtle points, butthey are important.EE 230Laplace – 29

EE 230 Laplace transform – 9 The Laplace Transform Given a function of time, f (t), we can transform it into a new, but related, function F(s). exp(–st) is the kernel of the transform, where s σ jω is the complex frequency. By integrating fr

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