The Laplace Transform

The two main techniques in signal processing, convolution and Fourier analysis, teach that a linear system can be completely understood from its impulse or frequency response. This is a very generalized approach, since the impulse and frequency responses

581CHAPTER32The Laplace TransformThe two main techniques in signal processing, convolution and Fourier analysis, teach that alinear system can be completely understood from its impulse or frequency response. This is avery generalized approach, since the impulse and frequency responses can be of nearly any shapeor form. In fact, it is too general for many applications in science and engineering. Many of theparameters in our universe interact through differential equations. For example, the voltageacross an inductor is proportional to the derivative of the current through the device. Likewise,the force applied to a mass is proportional to the derivative of its velocity. Physics is filled withthese kinds of relations. The frequency and impulse responses of these systems cannot bearbitrary, but must be consistent with the solution of these differential equations. This means thattheir impulse responses can only consist of exponentials and sinusoids. The Laplace transformis a technique for analyzing these special systems when the signals are continuous. The z-transform is a similar technique used in the discrete case.The Nature of the s-DomainThe Laplace transform is a well established mathematical technique for solvingdifferential equations. It is named in honor of the great French mathematician,Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplacetransform changes one signal into another according to some fixed set of rulesor equations. As illustrated in Fig. 32-1, the Laplace transform changes asignal in the time domain into a signal in the s-domain, also called the s-plane. The time domain signal is continuous, extends to both positive andnegative infinity, and may be either periodic or aperiodic. The Laplacetransform allows the time domain to be complex; however, this is seldomneeded in signal processing. In this discussion, and nearly all practicalapplications, the time domain signal is completely real.As shown in Fig. 32-1, the s-domain is a complex plane, i.e., there are realnumbers along the horizontal axis and imaginary numbers along the verticalaxis. The distance along the real axis is expressed by the variable, F, a lower The Scientist and Engineer's Guide to Digital Signal Processing582X(T) 'm4&4x(t) e&j T tdtX(F, T) 'm4&4[x(t) e&F t] e&j T tdtcase Greek sigma. Likewise, the imaginary axis uses the variable, T, thenatural frequency. This coordinate system allows the location of any point tobe specified by providing values for F and T. Using complex notation, eachlocation is represented by the complex variable, s, where: . Just ass ' F%j Twith the Fourier transform, signals in the s-domain are represented by capitalletters. For example, a time domain signal, , is transformed into an s-x(t)domain signal, , or alternatively, . The s-plane is continuous, andX(s) X(F, T)extends to infinity in all four directions. In addition to having a location defined by a complex number, each point in thes-domain has a value that is a complex number. In other words, each locationin the s-plane has a real part and an imaginary part. As with all complexnumbers, the real & imaginary parts can alternatively be expressed as themagnitude & phase.Just as the Fourier transform analyzes signals in terms of sinusoids, the Laplacetransform analyzes signals in terms of sinusoids and exponentials. From amathematical standpoint, this makes the Fourier transform a subset of the moreelaborate Laplace transform. Figure 32-1 shows a graphical description of howthe s-domain is related to the time domain. To find the values along a verticalline in the s-plane (the values at a particular F), the time domain signal is firstmultiplied by the exponential curve: . The left half of the s-planee& F tmultiplies the time domain with exponentials that increase with time ( ),F < 0while in the right half the exponentials decrease with time ( ). Next, takeF > 0the complex Fourier transform of the exponentially weighted signal. Theresulting spectrum is placed along a vertical line in the s-plane, with the tophalf of the s-plane containing the positive frequencies and the bottom halfcontaining the negative frequencies. Take special note that the values on they-axis of the s-plane ( ) are exactly equal to the Fourier transform of theF '0time domain signal. As discussed in the last chapter, the complex Fourier Transform is given by:This can be expanded into the Laplace transform by first multiplying the timedomain signal by the exponential term:While this is not the simplest form of the Laplace transform, it is probablythe best description of the strategy and operation of the technique. To Chapter 32- The Laplace Transform 583Real axis (F)-5 -4 -3 -2 -1 0 1 2 3 4 5-5-4-3-2-1012345F.T. F.T. F.T. F.T. F.T. F.T. F.T.Time-4 -3 -2 -1 0 1 2 3 4-2-1012PositiveFrequenciesNegativeFrequenciesDecreasingIncreasingExponentials Exponentialsx(t)X(s)F = -3 F = -2F = -1 F = 0 F = 1 F = 2 F = 3spectrumfor F = 3Imaginary axis (jT)AmplitudeSTEP 4Arrange each spectrum along avertical line in the s-plane. Thepositive frequencies are in theupper half of the s-plane while thenegative frequencies are in thelower half.m4&4[ x(t) e&Ft] e&j TtdtSTEP 2Multiply the time domain signal byan infinite number of exponentialcurves, each with a different decayconstant, F. That is, calculate thesignal: for each value of Fx(t) e&Ftfrom negative to positive infinity.STEP 1Start with the time domain signalcalled x(t)STEP 3Take the complex Fourier Transformof each exponentially weighted timedomain signal. That is, calculate:for each value of F from negative topositive infinity.FIGURE 32-1The Laplace transform. The Laplace transform converts a signal in the time domain, , into a signal in the s-domain,x(t). The values along each vertical line in the s-domain can be found by multiplying the time domain signalX(s) or X(F, T)by an exponential curve with a decay constant F, and taking the complex Fourier transform. When the time domain isentirely real, the upper half of the s-plane is a mirror image of the lower half. The Scientist and Engineer's Guide to Digital Signal Processing584X(F, T) 'm4&4x(t) e&(F %j T)tdtEQUATION 32-1The Laplace transform. This equationdefines how a time domain signal, , isx(t)related to an s-domain signal, . The s-X(s)domain variables, s, and , are complex.X( )While the time domain may be complex, it isusually real.X(s) 'm4&4x(t) e&stdtplace the equation in a shorter form, the two exponential terms can becombined:Finally, the location in the complex plane can be represented by the complexvariable, s, where . This allows the equation to be reduced to an evens ' F%jTmore compact expression:This is the final form of the Laplace transform, one of the mostimportant equations in signal processing and electronics. Pay specialattention to the term: , called a complex exponential. As shown by thee&stabove derivation, complex exponentials are a compact way of representing bothsinusoids and exponentials in a single expression.Although we have explained the Laplace transform as a two stage process(multiplication by an exponential curve followed by the Fourier transform),keep in mind that this is only a teaching aid, a way of breaking Eq. 32-1 intosimpler components. The Laplace transform is a single equation relating x(t)and , not a step-by-step procedure. Equation 32-1 describes how toX(s)calculate each point in the s-plane (identified by its values for F and T) basedon the values of , T, and the time domain signal, . Using the FourierF x(t)transform to simultaneously calculate all the points along a vertical line ismerely a convenience, not a requirement. However, it is very important toremember that the values in the s-plane along the y-axis ( ) are exactlyF ' 0equal to the Fourier transform. As explained later in this chapter, this is a keypart of why the Laplace transform is useful. To explore the nature of Eq. 32-1 further, let's look at several individual pointsin the s-domain and examine how the values at these locations are related to thetime domain signal. To start, recall how individual points in the frequencydomain are related to the time domain signal. Each point in the frequencydomain, identified by a specific value of T, corresponds to two sinusoids, and . The real part is found by multiplying the time domaincos(Tt) sin(Tt)signal by the cosine wave, and then integrating from -4 to 4. The imaginarypart is found in the same way, except the sine wave is used. If we are dealingwith the complex Fourier transform, the values at the corresponding negativefrequency, -T, will be the complex conjugate (same real part, negativeimaginary part) of the values at T. The Laplace transform is just an extensionof these same concepts. Chapter 32- The Laplace Transform 585Time12 30-1-2-3Real value (F)CCNBBNAANcos(40t)e-1.5tB+BNC+CNTimes-DomainTimeA+ANcos(40t)e0tcos(40t)e1.5tAssociated Waveforms60j40j20j0j-20j-40j-60jAmplitudeImaginary value ( jT)FIGURE 32-2Waveforms associated with the s-domain. Each locationin the s-domain is identified by two parameters: F and T.These parameters also define two waveforms associatedwith each location. If we only consider pairs of points(such as: A&AN, B&BN, and C&CN), the two waveformsassociated with each location are sine and cosine waves offrequency T, with an exponentially changing amplitudecontrolled by F.AmplitudeAmplitudeFigure 32-2 shows three pairs of points in the s-plane: A&AN, B&BN, andC&CN. Just as in the complex frequency spectrum, the points at A, B, & C (thepositive frequencies) are the complex conjugates of the points at AN, BN, & CN(the negative frequencies). The top half of the s-plane is a mirror image of thelower half, and both halves are needed to correspond with a real time domainsignal. In other words, treating these points in pairs bypasses the complexmath, allowing us to operate in the time domain with only real numbers.Since each of these pairs has specific values for F and ±T, there are twowaveforms associated with each pair: and . Forcos(Tt) e&Ftsin(Tt) e&Ftinstance, points A&AN are at a location of and , and thereforeF '1.5 T ' ±40correspond to the waveforms: and . As shown incos(40t) e&1.5tsin(40t) e&1.5tFig. 32-2, these are sinusoids that exponentially decreases in amplitude as timeprogresses. In this same way, the sine and cosine waves associated with B&BNhave a constant amplitude, resulting from the value of F being zero. Likewise,the sine and cosine waves that are associated with locations C&CNexponentially increases in amplitude, since F is negative. The Scientist and Engineer's Guide to Digital Signal Processing586ReX(F'1.5, T'±40) 'm4&4x(t) cos(40t) e&1.5tdtX(s) 'm4&4x(t) e&stdt 'm1&11 e&stdtX(s) 'es& e&ssReX (F, T) 'F cos(T) [eF&e&F] % T sin(T)[eF%e&F]F2% T2Im X (F, T) 'F sin(T)[eF%e&F] & T cos(T)[eF&e&F]F2% T2The value at each location in the s-plane consists of a real part and animaginary part. The real part is found by multiplying the time domain signalby the exponentially weighted cosine wave and then integrated from -4 to 4.The imaginary part is found in the same way, except the exponentially weightedsine wave is used instead. It looks like this in equation form, using the realpart of A&AN as an example:Figure 32-3 shows an example of a time domain waveform, its frequencyspectrum, and its s-domain representation. The example time domain signal isa rectangular pulse of width two and height one. As shown, the complexFourier transform of this signal is a sinc function in the real part, and anentirely zero signal in the imaginary part. The s-domain is an undulating two-dimensional signal, displayed here as topographical surfaces of the real andimaginary parts. The mathematics works like this:In words, we start with the definition of the Laplace transform (Eq. 32-1), plugin the unity value for , and change the limits to match the length of thex(t)nonzero portion of the time domain signal. Evaluating this integral providesthe s-domain signal, expressed in terms of the complex location, s, and thecomplex value, :X(s)While this is the most compact form of the answer, the use of complexvariables makes it difficult to understand, and impossible to generate a visualdisplay, such as Fig. 32-3. The solution is to replace the complex variable, s,with , and then separate the real and imaginary parts:F %jT Chapter 32- The Laplace Transform 587-4-2024-16-808160-15Real axis (F)Imaginary axis (jT)15-4-2024-16-80816150-15Real axis (F) Imaginary axis (jT)Real PartImaginaryPartFrequency-16 -12 -8 -4 0 4 8 12 16-1.0-0.50.00.51.01.52.02.5Frequency-16 -12 -8 -4 0 4 8 12 16-1.0-0.50.00.51.01.52.02.5Real PartImaginary Part Frequency Domain s-DomainTime-4 -3 -2 -1 0 1 2 3 4-0.50.00.51.01.5 Time DomainLaplaceTransformFourierTransformFIGURE 32-3Time, frequency and s-domains. A timedomain signal (the rectangular pulse) istransformed into the frequency domainusing the Fourier transform, and into thes-domain using the Laplace transform.AmplitudeAmplitudeAmplitudeAmplitude AmplitudeThe topographical surfaces in Fig. 32-3 are graphs of these equations. Theseequations are quite long and the mathematics to derive them is very tedious.This brings up a practical issue: with algebra of this complexity, how do weknow that we haven't made an error in the calculations? One check is to verify The Scientist and Engineer's Guide to Digital Signal Processing588ImX (F, T)/0F '0' 0ReX (F, T)/0F '0'2 sin(T)Tthat these equations reduce to the Fourier transform along the y-axis. This isdone by setting F to zero in the equations, and simplifying:As illustrated in Fig. 32-3, these are the correct frequency domain signals, thesame as found by directly taking the Fourier transform of the time domainwaveform. Strategy of the Laplace TransformAn analogy will help in explaining how the Laplace transform is used in signalprocessing. Imagine you are traveling by train at night between two cities.Your map indicates that the path is very straight, but the night is so dark youcannot see any of the surrounding countryside. With nothing better to do, younotice an altimeter on the wall of the passenger car and decide to keep track ofthe elevation changes along the route.Being bored after a few hours, you strike up a conversation with the conductor:"Interesting terrain," you say. "It seems we are generally increasing inelevation, but there are a few interesting irregularities that I have observed."Ignoring the conductor's obvious disinterest, you continue: "Near the start ofour journey, we passed through some sort of abrupt rise, followed by an equallyabrupt descent. Later we encountered a shallow depression." Thinking youmight be dangerous or demented, the conductor decides to respond: "Yes, Iguess that is true. Our destination is located at the base of a large mountainrange, accounting for the general increase in elevation. However, along theway we pass on the outskirts of a large mountain and through the center of avalley."Now, think about how you understand the relationship between elevation anddistance along the train route, compared to that of the conductor. Since youhave directly measured the elevation along the way, you can rightly claim thatyou know everything about the relationship. In comparison, the conductorknows this same complete information, but in a simpler and more intuitiveform: the location of the hills and valleys that cause the dips and humps alongthe path. While your description of the signal might consist of thousands ofindividual measurements, the conductor's description of the signal will containonly a few parameters. To show how this is analogous to signal processing, imagine we are tryingto understand the characteristics of some electric circuit. To aid in ourinvestigation, we carefully measure the impulse response and/or thefrequency response. As discussed in previous chapters, the impulse andfrequency responses contain complete information about this linear system. Chapter 32- The Laplace Transform 589However, this does not mean that you know the information in the simplestway. In particular, you understand the frequency response as a set of valuesthat change with frequency. Just as in our train analogy, the frequencyresponse can be more easily understood in terms of the terrain surrounding thefrequency response. That is, by the characteristics of the s-plane. With the train analogy in mind, look back at Fig. 32-3, and ask: how doesthe shape of this s-domain aid in understanding the frequency response?The answer is, it doesn't! The s-plane in this example makes a nice graph,but it provides no insight into why the frequency domain behaves as it does.This is because the Laplace transform is designed to analyze a specific classof time domain signals: impulse responses that consist of sinusoids andexponentials. If the Laplace transform is taken of some other waveform(such as the rectangular pulse in Fig. 32-3), the resulting s-domain ismeaningless.As mentioned in the introduction, systems that belong to this class areextremely common in science and engineering. This is because sinusoids andexponentials are solutions to differential equations, the mathematics thatcontrols much of our physical world. For example, all of the following systemsare governed by differential equations: electric circuits, wave propagation,linear and rotational motion, electric and magnetic fields, heat flow, etc. Imagine we are trying to understand some linear system that is controlled bydifferential equations, such as an electric circuit. Solving the differentialequations provides a mathematical way to find the impulse response.Alternatively, we could measure the impulse response using suitable pulsegenerators, oscilloscopes, data recorders, etc. Before we inspect the newlyfound impulse response, we ask ourselves what we expect to find. There areseveral characteristics of the waveform that we know without even looking.First, the impulse response must be causal. In other words, the impulseresponse must have a value of zero until the input becomes nonzero at .t ' 0This is the cause and effect that our universe is based upon.The second thing we know about the impulse response is that it will becomposed of sinusoids and exponentials, because these are the solutions tothe differential equations that govern the system. Try as we might, we willnever find this type of system having an impulse response that is, forexample, a square pulse or triangular waveform. Third, the impulseresponse will be infinite in length. That is, it has nonzero values thatextend from to . This is because sine and cosine waves have at ' 0 t ' %4constant amplitude, and exponentials decay toward zero without everactually reaching it. If the system we are investigating is stable, theamplitude of the impulse response will become smaller as time increases,reaching a value of zero at . There is also the possibility that thet ' %4system is unstable, for example, an amplifier that spontaneously oscillatesdue to an excessive amount of feedback. In this case, the impulse responsewill increase in amplitude as time increases, becoming infinitely large.Even the smallest disturbance to this system will produce an unboundedoutput. The Scientist and Engineer's Guide to Digital Signal Processing590FIGURE 32-4Pole-zero example. The notch filter has twopoles (represented by ×) and two zeros(represented by •). This s-plane diagramshows the five locations we will "probe" inthis example to analyze this system. (Figure32-5 is a continuation of this example).FjTabcdea!b!c!d!e!s-plane diagramOXOXThe general mathematics of the Laplace transform is very similar to that of theFourier transform. In both cases, predetermined waveforms are multiplied bythe time domain signal, and the result integrated. At first glance, it wouldappear that the strategy of the Laplace transform is the same as the Fouriertransform: correlate the time domain signal with a set of basis functions todecompose the waveform. Not true! Even though the mathematics is much thesame, the rationale behind the two techniques is very different. The Laplacetransform probes the time domain waveform to identify its key features: thefrequencies of the sinusoids, and the decay constants of the exponentials. Anexample will show how this works. The center column in Fig. 32-5 shows the impulse response of the RLC notchfilter discussed in Chapter 30. It contains an impulse at , followed by ant ' 0exponentially decaying sinusoid. As illustrated in (a) through (e), we willprobe this impulse response with various exponentially decaying sinusoids.Each of these probing waveforms is characterized by two parameters: T, thatdetermines the sinusoidal frequency, and F, that determines the decay rate. Inother words, each probing waveform corresponds to a different location in thes-plane, as shown by the s-plane diagram in Fig. 32-4. The impulse responseis probed by multiplying it with these waveforms, and then integrating theresult from . This action is shown in the right column. Our goalt ' &4 to %4is to find combinations of F and T that exactly cancel the impulse responsebeing investigated. This cancellation can occur in two forms: the area underthe curve can be either zero, or just barely infinite. All other results areuninteresting and can be ignored. Locations in the s-plane that produce a zerocancellation are called zeros of the system. Likewise, locations that producethe "just barely infinite" type of cancellation are called poles. Poles and zerosare analogous to the mountains and valleys in our train story, representing theterrain "around" the frequency response. To start, consider what happens when the probing waveform decreases inamplitude as time advances, as shown in (a). This will occur whenever (the right half of the s-plane). Since both the impulse response andF > 0the probe becomes smaller with increasing time, the product of the two willalso have this same characteristic. When the product of the two waveformsis integrated from negative to positive infinity, the result will be somenumber that is not especially interesting. In particular, a decreasing probe [...]... general mathematics of the Laplace transform is very similar to that of the Fourier transform. In both cases, predetermined waveforms are multiplied by the time domain signal, and the result integrated. At first glance, it would appear that the strategy of the Laplace transform is the same as the Fourier transform: correlate the time domain signal with a set of basis functions to decompose the waveform.... the input signal, and the voltage waveform being the output signal. When we say that resistors, inductors and capacitors become R, , and in the s-domain, this referssL 1/sC to the output divided by the input. In other words, the Laplace transform of the voltage waveform divided by the Laplace transform of the current waveform is equal to these expressions. As an example of this, imagine we force the. .. line in the s-plane, with the top half of the s-plane containing the positive frequencies and the bottom half containing the negative frequencies. Take special note that the values on the y-axis of the s-plane ( ) are exactly equal to the Fourier transform of theF '0 time domain signal. As discussed in the last chapter, the complex Fourier Transform is given by: This can be expanded into the Laplace. .. When the components are multiplied together, they must equal the original numerator and denominator. In other words, the equation is placed into the form: The roots of the numerator, , are the zeros of the equation, while thez 1 , z 2 , z 3 ỵ roots of the denominator, , are the poles. These are the same zerosp 1 , p 2 , p 3 ỵ and poles we encountered earlier in this chapter, and we will discuss how they are... in the discrete case. The Nature of the s-Domain The Laplace transform is a well established mathematical technique for solving differential equations. It is named in honor of the great French mathematician, Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. As illustrated in Fig. 32-1, the. .. waveform. Not true! Even though the mathematics is much the same, the rationale behind the two techniques is very different. The Laplace transform probes the time domain waveform to identify its key features: the frequencies of the sinusoids, and the decay constants of the exponentials. An example will show how this works. The center column in Fig. 32-5 shows the impulse response of the RLC notch filter discussed... words, we start with the definition of the Laplace transform (Eq. 32-1), plug in the unity value for , and change the limits to match the length of thex(t) nonzero portion of the time domain signal. Evaluating this integral provides the s-domain signal, expressed in terms of the complex location, s, and the complex value, :X(s) While this is the most compact form of the answer, the use of complex variables... relate the pole position, T and F, to the amplifier gain, A, the resistor, R, and capacitor, C. Filter Design in the s-Domain The most powerful application of the Laplace transform is the design of systems directly in the s-domain. This involves two steps: First, the s- domain is designed by specifying the number and location of the poles and zeros. This is a pure mathematical problem, with the goal... expressed as the magnitude & phase. Just as the Fourier transform analyzes signals in terms of sinusoids, the Laplace transform analyzes signals in terms of sinusoids and exponentials. From a mathematical standpoint, this makes the Fourier transform a subset of the more elaborate Laplace transform. Figure 32-1 shows a graphical description of how the s-domain is related to the time domain. To find the values... Fig. 32-3, these are the correct frequency domain signals, the same as found by directly taking the Fourier transform of the time domain waveform. Strategy of the Laplace Transform An analogy will help in explaining how the Laplace transform is used in signal processing. Imagine you are traveling by train at night between two cities. Your map indicates that the path is very straight, but the night . in the s-domain, this referssL 1/sCto the output divided by the input. In other words, the Laplace transform of thevoltage waveform divided by the Laplace. into the Laplace transform by first multiplying the timedomain signal by the exponential term:While this is not the simplest form of the Laplace transform,