CHAPTER

32

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 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 voltage
across 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 be
arbitrary, but must be consistent with the solution of these differential equations. This means that
their impulse responses can only consist of exponentials and sinusoids. The Laplace transform
is a technique for analyzing these special systems when the signals are continuous. The ztransform is a similar technique used 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 Laplacetransform changes one signal into another according to some fixed set of rules
or 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 splane. 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 seldom
needed 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 real
numbers along the horizontal axis and imaginary numbers along the verticalaxis. The distance along the real axis is expressed by the variable, F, a lower

581

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The Scientist and Engineer's Guide to Digital Signal Processing

case 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: s ' F% jT. Just as
with the Fourier transform, signals in the s-domain are represented by capitalletters. For example, a time domain signal, x (t) , is transformed into an sdomain signal, X (s) , or alternatively, X (F,T) . The s-plane is continuous, andextends 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 complex
numbers, 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 how
the 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: e & F t . The left half of the s-plane
multiplies the time domain with exponentials that increase with time ( F < 0 ),while in the right half the exponentials decrease with time ( F > 0 ). Next, takethe complex Fourier transform of the exponentially weighted signal. The
resulting spectrum is placed along a vertical line in the s-plane, with the top
half of the s-plane containing the positive frequencies and the bottom halfcontaining the negative frequencies. Take special note that the values on the
y-axis of the s-plane ( F ' 0 ) are exactly equal to the Fourier transform of thetime domain signal.
As discussed in the last chapter, the complex Fourier Transform is given by:
4

X (T) '

mx(t ) e & jT t dt

&4

This can be expanded into the Laplace transform by first multiplying the time
domain signal by the exponential term:
4

X (F, T) 'm

[ x (t ) e & F t ] e & jTt d t&4

While this is not the simplest form of the Laplace transform, it is probably
the best description of the strategy and operation of the technique. To

Chapter 32- The Laplace Transform

583

2

x(t)
1

Amplitude

STEP 1
Start with the time domain signal
called x(t)

0
-1
-2

STEP 2Multiply the time domain signal by
an infinite number of exponential
curves, each with a different decay
constant, F. That is, calculate thesignal: x(t) e & Ft for each value of F
from negative to positive infinity.

-4

-3

-2

-1

0

1

2

3

4

Time

F = -3

F.T.

F = -2F = -1

F.T.

F= 0

F.T.

F= 1

F.T.

F= 2

F.T.

F= 3

F.T.

F.T.

STEP 3
Take the complex Fourier Transformof each exponentially weighted time
domain signal. That is, calculate:4m[ x (t ) e & Ft ] e & jTt d t&4

5

for each value of F from negative topositive infinity.

4

STEP 4
Arrange each spectrum along a
vertical line in the s-plane. The
positive frequencies are in theupper half of the s-plane while thenegative frequencies are in thelower half.

Imaginary axis ( jT)

3

X(s)Positive
Frequencies

2
1
0
-1
-2

Negative
Frequencies

-3
-4
-5
-5 -4 -3

-2 -1

0

1

Real axis (F)Increasing
Exponentials

2

3

4

5

spectrumfor F = 3

Decreasing
Exponentials

FIGURE 32-1
The Laplace transform. The Laplace transform converts a signal in the time domain, x(t) , into a signal in the s-domain,
X (s) or X(F, T) . The values along each vertical line in the s-domain can be found by multiplying the time domain signal
by an exponential curve with a decay constant F, and taking the complex Fourier transform. When the time domain is
entirely real, the upper half of the s-plane is a mirror image of the lower half.

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The Scientist and Engineer's Guide to Digital Signal Processing

place the equation in a shorter form, the two exponential terms can be
combined:4

X (F, T ) 'mx (t ) e &(F % jT )t d t&4

Finally, the location in the complex plane can be represented by the complex
variable, s, where s ' F%jT . This allows the equation to be reduced to an evenmore compact expression:
EQUATION 32-1
The Laplace transform. This equation
defines how a time domain signal, x (t) , isrelated to an s-domain signal, X (s) . The sdomain variables, s, and X( ) , are complex.
While the time domain may be complex, it isusually real.

4

X (s) '

mx(t ) e &s t dt

&4

This is the final form of the Laplace transform, one of the mostimportant equations in signal processing and electronics. Pay special
attention to the term: e & st , called a complex exponential. As shown by theabove 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 into
simpler components. The Laplace transform is a single equation relating x (t)and X (s) , not a step-by-step procedure. Equation 32-1 describes how to
calculate each point in the s-plane (identified by its values for F and T) basedon the values of F , T, and the time domain signal, x (t) . Using the Fouriertransform 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 ( F ' 0 ) are exactly
equal to the Fourier transform. As explained later in this chapter, this is a key
part of why the Laplace transform is useful.
To explore the nature of Eq. 32-1 further, let's look at several individual points
in 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 frequency
domain are related to the time domain signal. Each point in the frequency
domain, identified by a specific value of T, corresponds to two sinusoids,cos (Tt ) and sin (Tt ) . The real part is found by multiplying the time domain
signal by the cosine wave, and then integrating from -4 to 4. The imaginary
part is found in the same way, except the sine wave is used. If we are dealing
with the complex Fourier transform, the values at the corresponding negative
frequency, - T , will be the complex conjugate (same real part, negative
imaginary part) of the values at T. The Laplace transform is just an extensionof these same concepts.

Chapter 32- The Laplace Transform

s-Domain

585

Associated Waveforms
cos(40t)e-1.5t

60j

A+AN

A

20j
Time

0j

cos(40t)e0t
B+BN

-20j

CN

-40j

BN

Amplitude

Imaginary value ( jT )B

Amplitude

C

40j

AN

-60j
-3

-2

-1

0

1

2

Time

3

cos(40t)e1.5t

Real value (F)

C+CN
Amplitude

FIGURE 32-2
Waveforms 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 waveforms
associated with each location are sine and cosine waves offrequency T, with an exponentially changing amplitudecontrolled by F.

Time

Figure 32-2 shows three pairs of points in the s-plane: A&AN , B&B N , 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 domain
signal. In other words, treating these points in pairs bypasses the complex
math, 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 two
waveforms associated with each pair: cos (Tt ) e & Ft and sin (Tt ) e & Ft . For
instance, points A&AN are at a location of F '1.5 and T ' ±40 , and therefore
correspond to the waveforms: cos (40 t ) e & 1.5t and sin (40 t ) e & 1.5t . As shown in
Fig. 32-2, these are sinusoids that exponentially decreases in amplitude as time
progresses. In this same way, the sine and cosine waves associated with B&BN
have a constant amplitude, resulting from the value of F being zero. Likewise,
the sine and cosine waves that are associated with locations C&C N
exponentially increases in amplitude, since F is negative.

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The Scientist and Engineer's Guide to Digital Signal Processing

The value at each location in the s-plane consists of a real part and an
imaginary 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 weighted
sine wave is used instead. It looks like this in equation form, using the realpart of A&AN as an example:
4m

Re X (F'1.5, T'±40 ) 'x (t ) cos (40t ) e &1.5 t d t&4

Figure 32-3 shows an example of a time domain waveform, its frequency
spectrum, and its s-domain representation. The example time domain signal isa rectangular pulse of width two and height one. As shown, the complex
Fourier transform of this signal is a sinc function in the real part, and an
entirely zero signal in the imaginary part. The s-domain is an undulating twodimensional signal, displayed here as topographical surfaces of the real andimaginary parts. The mathematics works like this:4

X (s) '

m1

x(t ) e

&st

&4

dt '

m

1 e &st dt&1

In words, we start with the definition of the Laplace transform (Eq. 32-1), plugin the unity value for x (t) , and change the limits to match the length of thenonzero portion of the time domain signal. Evaluating this integral provides
the s-domain signal, expressed in terms of the complex location, s, and thecomplex value, X (s) :X (s) '

e s & e &ssWhile this is the most compact form of the answer, the use of complex
variables 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 F% jT , and then separate the real and imaginary parts:

Re X (F, T ) 'Im X (F, T ) 'F cos (T ) [e F &e &F ] % T sin(T ) [e F %e &F ]F2 % T2
F sin(T ) [e F %e &F ] & T cos (T ) [e F &e &F ]F2 % T2

Chapter 32- The Laplace Transform

587

Time Domain
Amplitude

1.5

FIGURE 32-3
Time, frequency and s-domains. A time
domain signal (the rectangular pulse) is
transformed into the frequency domainusing the Fourier transform, and into thes-domain using the Laplace transform.

1.0
0.5
0.0
-0.5
-4

-3

-2

-1

0

Time

1

2

3

4

Fourier
Transform

Laplace
Transform

Frequency Domain
Real Part

Real Part
15

Amplitude

2.0
1.5
1.0

0

0.5
0.0
-0.5
-1.0
-16

Amplitude

2.5

s-Domain

-15
-12

-8

-4

0

4

Frequency

8

12

16

16

-4
8

-2
0

0

Real axis (F)-82Imaginary axis (jT)

4 -16

Imaginary Part
Imaginary
Part

2.0
1.5

15

1.0
0.5

0

0.0
-0.5
-1.0
-16

-12

-8

-4

0

4

Frequency

8

12

Amplitude

Amplitude

2.5

-15

16

16

-4
8

-2
0

0

Real axis (F)-82Imaginary axis (jT)

4 -16

The 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

588

The Scientist and Engineer's Guide to Digital Signal Processingthat these equations reduce to the Fourier transform along the y-axis. This isdone by setting F to zero in the equations, and simplifying:

Re X (F, T) /
0

F' 0

'

2 sin(T)T

Im X (F, T) /
0

'

0

F' 0

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 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 is so dark you
cannot see any of the surrounding countryside. With nothing better to do, you
notice an altimeter on the wall of the passenger car and decide to keep track of
the 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 you
might be dangerous or demented, the conductor decides to respond: "Yes, I
guess that is true. Our destination is located at the base of a large mountain
range, 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 you
have directly measured the elevation along the way, you can rightly claim thatyou know everything about the relationship. In comparison, the conductor
knows this same complete information, but in a simpler and more intuitive
form: the location of the hills and valleys that cause the dips and humps along
the path. While your description of the signal might consist of thousands ofindividual measurements, the conductor's description of the signal will contain
only a few parameters.
To show how this is analogous to signal processing, imagine we are trying
to 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

589

However, this does not mean that you know the information in the simplest
way. In particular, you understand the frequency response as a set of values
that change with frequency. Just as in our train analogy, the frequency
response 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 does
the 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 are
extremely common in science and engineering. This is because sinusoids andexponentials are solutions to differential equations, the mathematics that
controls much of our physical world. For example, all of the following systems
are 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 differential
equations provides a mathematical way to find the impulse response.
Alternatively, we could measure the impulse response using suitable pulse
generators, oscilloscopes, data recorders, etc. Before we inspect the newly
found 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 impulse
response must have a value of zero until the input becomes nonzero at t ' 0 .
This 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 to
the differential equations that govern the system. Try as we might, we will
never find this type of system having an impulse response that is, forexample, a square pulse or triangular waveform. Third, the impulse
response will be infinite in length. That is, it has nonzero values that
extend from t ' 0 to t ' % 4. This is because sine and cosine waves have aconstant 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 t ' % 4. There is also the possibility that thesystem is unstable, for example, an amplifier that spontaneously oscillates
due to an excessive amount of feedback. In this case, the impulse response
will increase in amplitude as time increases, becoming infinitely large.
Even the smallest disturbance to this system will produce an unbounded
output.

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The Scientist and Engineer's Guide to Digital Signal Processing

The general mathematics of the Laplace transform is very similar to that of theFourier 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 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. An
example will show how this works.
The center column in Fig. 32-5 shows the impulse response of the RLC notch
filter discussed in Chapter 30. It contains an impulse at t ' 0 , followed by an
exponentially 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, that
determines the sinusoidal frequency, and F, that determines the decay rate. In
other 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 response
is probed by multiplying it with these waveforms, and then integrating theresult from t ' & 4 to % 4. This action is shown in the right column. Our goal
is to find combinations of F and T that exactly cancel the impulse response
being investigated. This cancellation can occur in two forms: the area under
the 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 produce
the "just barely infinite" type of cancellation are called poles. Poles and zeros
are 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 wheneverF > 0 (the right half of the s-plane). Since both the impulse response andthe probe becomes smaller with increasing time, the product of the two will
also have this same characteristic. When the product of the two waveforms
is integrated from negative to positive infinity, the result will be somenumber that is not especially interesting. In particular, a decreasing probe

s-plane diagram
FIGURE 32-4
Pole-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. (Figure30-5 is a continuation of this example).

e

d

jT
c

X

b

a

O

F
e!

X

d!

c!

O

b!

a!

Chapter 32- The Laplace Transform

Probing waveform, p(t)Impulse response, h(t)

3

2

2

4

1

1

0
-4

0
-1

-8

-2
-3
-4

-2

0

2

4

Time (µsec)

6

8

-3
-2

0

2

4

Time (µsec)

6

8

-4

3

2

2

4

1

1

0

0
-1

-8

-2

-12

-3
-4

-2

0

2

4

Time (µsec)

6

Amplitude

3

8

Amplitude
8

-2

0

2

4

Time (µsec)

6

8

1

1

0
-1
-2
-3

-4

-2

0

2

4

Time (µsec)

6

Amplitude

2

4

Amplitude

3

2

-8
8

0

2

4

Time (µsec)

6

8

1

1

0
-1

-3
-4

-2

0

2

4

Time (µsec)

6

Amplitude

2

4

Amplitude

3

-2
8

0

2

4

Time (µsec)

6

8

2

4

1

1

Amplitude

3

0
-1
-2
-3

-4

-2

0

2

4

Time (µsec)

6

8

2

4

Time (µsec)

6

8

6

8

area isinfinite

-4

2

-8

0

-3
-2

3

-12

-2

0

8

0

8

-1

12

-4

6

-2
-4

e. Too fast of increase

4

area isfinite

-4

2

-8

2

Time (µsec)

-3
-2

3

-12

0

0

8

0

-2

-1

12

-4

8

-2
-4

d. Exact cancellation (pole)

6

area isexactly zero

-4

3

-12

4

0

8

0

2

Time (µsec)

-1

12

-4

0

-3
-4

c. Too slow of increase-2-2

Amplitude

Amplitude

0
-1

12

-4

area isfinite-2-4b. Exact cancellation (zero)

Amplitude

Amplitude

3

8

-12

Amplitude

Multiply: p(t) × h(t)12Amplitude

Amplitude

a. Decreasing with time

Amplitude

591

-2

0

2

4

Time (µsec)

area isundefined

0
-1
-2
-3

-4

-2

0

2

4

Time (µsec)

6

8

-4

-2

0

2

4

Time (µsec)

68

FIGURE 32-5
Probing the impulse response. The Laplace transform can be viewed as probing the system's impulse response withvarious exponentially decaying sinusoids. Probing waveforms that produce a cancellation are called poles and zeros.
This illustration shows five probing waveforms (left column) being applied to the impulse response of a notch filter(center column). The locations in the s-plane that correspond to these five waveforms are shown in Fig. 32-4.

592

The Scientist and Engineer's Guide to Digital Signal Processing

cannot cancel a decreasing impulse response. This means that a stable system
will not have any poles with F > 0 . In other words, all of the poles in a stablesystem are confined to the left half of the s-plane. In fact, poles in the righthalf of the s-place show that the system is unstable (i.e., an impulse responsethat increases with time).
Figure (b) shows one of the special cases we have been looking for. When thiswaveform is multiplied by the impulse response, the resulting integral has avalue of zero. This occurs because the area above the x-axis (from the delta
function) is exactly equal to the area below (from the rectified sinusoid). The
values for F and T that produce this type of cancellation are called a zero of
the system. As shown in the s-plane diagram of Fig. 32-4, zeros are indicatedby small circles (•).
Figure (c) shows the next probe we can try. Here we are using a sinusoid that
exponentially increases with time, but at a rate slower than the impulse
response is decreasing with time. This results in the product of the two
waveforms also decreasing as time advances. As in (a), this makes the integralof the product some uninteresting real number. The important point being thatno type of exact cancellation occurs.
Jumping out of order, look at (e), a probing waveform that increases at afaster rate than the impulse response decays. When multiplied, the resulting
signal increases in amplitude as time advances. This means that the area under
the curve becomes larger with increasing time, and the total area fromt ' & 4 to % 4 is not defined. In mathematical jargon, the integral does notconverge. In other words, not all areas of the s-plane have a defined value.
The portion of the s-plane where the integral is defined is called the region-ofconvergence. In some mathematical techniques it is important to know what
portions of the s-plane are within the region-of-convergence. However, thisinformation is not needed for the applications in this book. Only the exact
cancellations are of interest for this discussion.
In (d), the probing waveform increases at exactly the same rate that the impulse
response decreases. This makes the product of the two waveforms have aconstant amplitude. In other words, this is the dividing line between (c) and(e), resulting in a total area that is just barely undefined (if the mathematicians
will forgive this loose description). In more exact terms, this point is on theborderline of the region of convergence. As mentioned, values for F and T that
produce this type of exact cancellation are called poles of the system. Poles
are indicated in the s-plane by crosses (×).

Analysis of Electric Circuits
We have introduced the Laplace transform in graphical terms, describing what
the waveforms look like and how they are manipulated. This is the mostintuitive way of understanding the approach, but is very different from how it
is actually used. The Laplace transform is inherently a mathematical technique;
it is used by writing and manipulating equations. The problem is, it is easy toChapter 32- The Laplace Transform

593

become lost in the abstract nature of the complex algebra and loose allconnection to the real world. Your task is to merge the two views together.
The Laplace transform is the primary method for analyzing electric circuits.
Keep in mind that any system governed by differential equations can behandled the same way; electric circuits are just an example we are using.
The brute force approach is to solve the differential equations controlling thesystem, providing the system's impulse response. The impulse response canthen be converted into the s-domain via Eq. 32-1. Fortunately, there is a better
way: transform each of the individual components into the s-domain, and thenaccount for how they interact. This is very similar to the phasor transformpresented in Chapter 30, where resistors, inductors and capacitors are
represented by R, jTL , and 1 /jTC , respectively. In the Laplace transform,
resistors, inductors and capacitors become the complex variables: R, s L , and1/sC . Notice that the phasor transform is a subset of the Laplace transform.
That is, when F is set to zero in s ' F% jT, R becomes R, s L becomes jTL ,and 1/sC becomes 1 /jTC .
Just as in Chapter 30, we will treat each of the three components as anindividual system, with the current waveform being the input signal, and thevoltage waveform being the output signal. When we say that resistors,
inductors and capacitors become R, s L , and 1/sC in the s-domain, this refers
to the output divided by the input. In other words, the Laplace transform of thevoltage waveform divided by the Laplace transform of the current waveform
is equal to these expressions.
As an example of this, imagine we force the current through an inductor to bea unity amplitude cosine wave with a frequency given by T0. The resulting
voltage waveform across the inductor can be found by solving the differential
equation that governs its operation:v (t ) ' Lddi (t ) ' L
cos (T0 t ) ' T0 L sin(T0 t )dt
dt

If we start the current waveform at t ' 0 , the voltage waveform will also startat this same time (i.e., i (t) ' 0 and v (t) ' 0 for t < 0 ). These voltage andcurrent waveforms are converted into the s-domain by Eq. 32-1:
4m

I (s) 'cos (T0 t ) e &st d t '04

V (s) 'm
0

T0 L sin(T0 t ) e &s t d t 'T02

T0 % s 2
T0 L s2

T0 % s 2

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To complete this example, we will divide the s-domain voltage by the s-domain
current, just as if we were using Ohm's law ( R ' V / I ):T0 L s
V(s)
'
I (s)

2

T0 % s 2
T0

'

sL

2

T0 % s 2
We find that the s-domain representation of the voltage across the inductor,
divided by the s-domain representation of the current through the inductor, isequal to sL. This is always the case, regardless of the current waveform westart with. In a similar way, the ratio of s-domain voltage to s-domain current
is always equal to R for resistors, and 1/sC for capacitors.
Figure 32-6 shows an example circuit we will analyze with the Laplacetransform, the RLC notch filter discussed in Chapter 30. Since this analysis
is the same for all electric circuits, we will outline it in steps.
Step 1. Transform each of the components into the s-domain. In other words,
replace the value of each resistor with R, each inductor with sL , and eachcapacitor with 1/sC . This is shown in Fig. 32-6.
Step 2: Find H (s) , the output divided by the input. As described in Chapter
30, this is done by treating each of the components as if they obey Ohm's law,with the "resistances" given by: R, s L , and 1/sC . This allows us to use thestandard equations for resistors in series, resistors in parallel, voltage dividers,etc. Treating the RLC circuit in this example as a voltage divider (just as in
Chapter 30), H (s) is found:H(s) '

Vout (s)Vin (s)'sL % 1/sC
sL % 1/sC'
R % sL % 1/sC
R % sL % 1/sCss'

L s 2 % 1/C
L s 2 % Rs % 1/C

As you recall from Fourier analysis, the frequency spectrum of the output
signal divided by the frequency spectrum of the input signal is equal to thesystem's frequency response, given the symbol, H (T) . The above equation isan extension of this into the s-domain. The signal, H (s) , is called the system'stransfer function, and is equal to the s-domain representation of the output
signal divided by the s-domain representation of the input signal. Further, H (s)is equal to the Laplace transform of the impulse response, just the same as
H (T) is equal to the Fourier transform of the impulse response.

Chapter 32- The Laplace Transform

595

Vin
R
FIGURE 32-6
Notch filter analysis in the s-domain. The
first step in this procedure is to replace theresistor, inductor & capacitor values with
their s-domain equivalents.

Vout
1/sC
sL

So far, this is identical to the techniques of the last chapter, except for usings instead of jT. The difference between the two methods is what happens fromthis point on. This is as far as we can go with jT. We might graph thefrequency response, or examining it in some other way; however, this is amathematical dead end. In comparison, the interesting aspects of the Laplacetransform have just begun. Finding H (s) is the key to Laplace analysis;
however, it must be expressed in a particular form to be useful. This requires
the algebraic manipulation of the next two steps.
Step 3: Arrange H (s) to be one polynomial over another. This makes thetransfer function written as:
EQUATION 32-2
Transfer function in polynomial form.

H (s) '

as 2 % b s % c
as 2 % bs % c

It is always possible to express the transfer function in this form if thesystem is controlled by differential equations. For example, the rectangularpulse shown in Fig. 32-3 is not the solution to a differential equation and its
Laplace transform cannot be written in this way. In comparison, any electric
circuit composed of resistors, capacitors, and inductors can be written in thisform. For the RLC notch filter used in this example, the algebra shown in step
2 has already placed the transfer function in the correct form, that is:

H (s) '

a s 2 % bs % c
a s 2 % bs % c'

L s 2 % 1/C
L s 2 % R s % 1/C

where: a ' L , b ' 0, c ' 1/C ; and a ' L, b ' R, c ' 1/C

Step 4: Factor the numerator and denominator polynomials. That is, break
the numerator and denominator polynomials into components that each contain

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a single s. When the components are multiplied together, they must equal theoriginal numerator and denominator. In other words, the equation is placed into
the form:

EQUATION 32-3
The factored s-domain. This form
allows the s-domain to be expressed aspoles and zeros.

H (s) '

(s & z 1) (s & z 2) (s & z 3)þ
(s & p 1) (s & p 2) (s & p 3)þ

The roots of the numerator, z1, z2, z3 þ, are the zeros of the equation, while theroots of the denominator, p1, p2, p3 þ, are the poles. These are the same zeros
and poles we encountered earlier in this chapter, and we will discuss how they
are used in the next section.
Factoring an s-domain expression is straightforward if the numerator anddenominator are second-order polynomials, or less. In other words, we caneasily handle the terms: s and s 2 , but not: s 3, s 4, s 5, þ. This is because theroots of a second-order polynomial, a x 2 % b x % c , can be found by using thequadratic equation: x ' & b ± b 2 & 4 a c / 2a . With this method, the transfer
function of the example notch filter is factored into:

H (s) '

(s & z 1) (s & z 2)(s & p 1) (s & p 2)

where:

z1 ' j / LC

p1 '& R % R 2 & 4L /C
2L

z2 ' & j / LCp2 '& R & R 2 & 4L /C
2L

As in this example, a second-order system has a maximum of two zeros andtwo poles. The number of poles in a system is equal to the number of
independent energy storing components. For instance, inductors and capacitors
store energy, while resistors do not. The number of zeros will be equal to, or
less than, the number of poles.
Polynomials greater than second order cannot generally be factored using
algebra, requiring more complicated numerical methods. As an alternative,
circuits can be constructed as a cascade of second-order stages. A good
example is the family of analog filters presented in Chapter 3. For instance,
an eight pole filter is designed by cascading four stages of two poles each. The
important point is that this multistage approach is used to overcome limitationsin the mathematics, not limitations in the electronics.

Chapter 32- The Laplace Transform597

pole-zero plot
frequency response

8

O

6

1.2

4

1.0

2
0

Amplitude

Imaginary axis ( T × 10 6 )

10

-2
-4
-6

O

-8

0.8
0.6
0.4
0.2

-10
-10 -8 -6 -4 -2 0

2

4

6

Real axis (F × 10 6 )

0.0

8 10

0

2

4

6

8

Frequency (T × 10 6 )

10

3
2
1

8

Amplitude

4

0
10

6

s-domain

4
2
0
-10

-2
-8

-6

-4
-4

-2

0

Imaginary axis
( T × 106 )-62Real axis ( F × 106 )

4

6

8

10

-8
-10

FIGURE 32-7
Poles and zeros in the s-domain. These illustrations show the relationship between the pole-zero plot, thes-domain, and the frequency response. The notch filter component values used in these graphs are:
R=220 S, C=470 DF, and L = 54 µH. These values place the center of the notch at T = 6.277 million, i.e.,a frequency of approximately 1 MHz.

The Importance of Poles and Zeros
To make this less abstract, we will use actual component values for the notch
filter we just analyzed: R ' 220 S, L ' 54 µH, C ' 470 DF . Plugging thesevalues into the above equations, places the poles and zeros at:

z1 ' 0 % j 6.277 × 106

p1 ' & 2.037 × 106 % j 5.937 × 106

z2 ' 0 & j 6.277 × 106

p2 ' & 2.037 × 106 & j 5.937 × 106

These pole and zero locations are shown in Fig. 32-7. Each zero isrepresented by a circle, while each pole is represented by a cross. This iscalled a pole-zero diagram, and is the most common way that s-domain data
are displayed. Figure 32-7 also shows a topographical display of the splane. For simplicity, only the magnitude is shown, but don't forget that

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The Scientist and Engineer's Guide to Digital Signal Processing

there is a corresponding phase. Just as mountains and valleys determine theshape of the surface of the earth, the poles and zeros determine the shape of thes-plane. Unlike mountains and valleys, every pole and zero is exactly the same
shape and size as every other pole and zero. The only unique characteristic apole or zero has is its location. Poles and zeros are important because they
provide a concise representation of the value at any point in the s-plane. Thatis, we can completely describe the characteristics of the system using only
a few parameters. In the case of the RLC notch filter, we only need to specify
four complex parameters to represent the system: z1, z2, p1, p2 (each consistingof a real and an imaginary part).
To better understand poles and zeros, imagine an ant crawling around the splane. At any particular location the ant happens to be (i.e., some value of s),
there is a corresponding value of the transfer function, H (s) . This value is acomplex number that can be expressed as the magnitude & phase, or as thereal & imaginary parts. Now, let the ant carry us to one of the zeros in the splane. The value we measure for the real and imaginary parts will be zero at
this location. This can be understood by examining the mathematicalequation for H (s) in Eq. 32-3. If the location, s, is equal to any of the zeros,
one of the terms in the numerator will be zero. This makes the entire expression
equal to zero, regardless of the other values.
Next, our ant journey takes us to one of the poles, where we again measure thevalue of the real and imaginary parts of H (s) . The measured value becomeslarger and larger as we come close to the exact location of the pole (hence thename). This can also be understood from Eq. 32-3. If the location, s, is equal
to any of the p's, the denominator will be equal to zero, and the division by
zero makes the entire expression infinity large.
Having explored the unique locations, our ant journey now moves randomly
throughout the s-plane. The value of H (s) at each location depends entirely on
the positioning of the poles and the zeros, because there are no other types offeatures allowed in this strange terrain. If we are near a pole, the value will
be large; if we are near a zero, the value will be small.
Equation 32-3 also describes how multiple poles and zeros interact to form thes-domain signal. Remember, subtracting two complex numbers provides the
distance between them in the complex plane. For example, (s & z1) is the
distance between the arbitrary location, s, and the zero located at z1 .
Therefore, Eq. 32-3 specifies that the value at each location, s, is equal to the
distance to all of the zeros multiplied, divided by the distance to all of thepoles multiplied.
This brings us to the heart of this chapter: how the location of the poles &zeros provides a deeper understanding of the system's frequency response.
The frequency response is equal to the values of H (s) along the imaginary
axis, signified by the dark line in the topographical plot of Fig. 32-7.
Imagine our ant starting at the origin and crawling along this path. Near
the origin, the distance to the zeros is approximately equal to the distanceto the poles. This makes the numerator and denominator in Eq. 32-3

Chapter 32- The Laplace Transform

599

Imaginary value

Pole-Zero Diagram

LaplacetransformooPhysical System
C
L

Frequency Response
Phasor transform

Amplitude

R

Evaluate
at F=0Real value

Frequency

FIGURE 32-8
Strategy for using the Laplace transform. The phasor transform presented in Chapter 30 (the method using
R, jTL , & & j /TC ) allows the frequency response to be directly calculated from the parameters of thephysical system. In comparison, the Laplace transform calculates an s-domain representation from thephysical system, usually displayed in the form of a pole-zero diagram. In turn, the frequency response canbe obtained from the s-domain by evaluating the transfer function along the imaginary axis. While bothmethods provide the same end result, the intermediate step of the s-domain provides insight into why thefrequency response behaves as it does.

cancel, providing a unity frequency response at low frequencies. The situation
doesn't change significantly until the ant moves near the pole and zero location.
When approaching the zero, the value of H (s) drops suddenly, becoming zero
when the ant is upon the zero. As the ant moves past the pole and zero pair,the value of H (s) again returns to unity. Using this type of visualization, it canbe seen that the width of the notch depends on the distance between the poleand zero.
Figure 32-8 summarizes how the Laplace transform is used. We start with aphysical system, such as an electric circuit. If we desire, the phasor transformcan directly provide the frequency response of the system, as described in
Chapter 30. An alternative is to take the Laplace transform using the four step
method previously outlined. This results in a mathematical expression for thetransfer function, H (s) , which can be represented in a pole-zero diagram.
The frequency response can then be found by evaluating the transfer
function along the imaginary axis, that is, by replacing each s with jT.
While both methods provide the same result, the intermediate pole-zerodiagram provides an understanding of why the system behaves as it does,
and how it can be changed.

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Filter Design in the s-Domain
The most powerful application of the Laplace transform is the design ofsystems directly in the s-domain. This involves two steps: First, the sdomain is designed by specifying the number and location of the poles andzeros. This is a pure mathematical problem, with the goal of obtaining thebest frequency response. In the second step, an electronic circuit is derived
that provides this s-domain representation. This is something of an art,
since there are many circuit configurations that have a given pole-zerodiagram.
As previously mentioned, step 4 of the Laplace transform method is very
difficult if the system contains more than two poles or two zeros. A
common solution is to implement multiple poles and zeros in successive
stages. For example, a 6 pole filter is implemented as three successive
stages, with each stage containing up to two poles and two zeros. Since
each of these stages can be represented in the s-domain by a quadratic
numerator divided by a quadratic denominator, this approach is called
designing with biquads.
Figure 32-9 shows a common biquad circuit, the one used in the filter
design method of Chapter 3. This is called the Sallen-Key circuit, after
R.P. Sallen and E.L. Key, authors of a paper that described this techniquein the mid 1950s. While there are several variations, the most common
circuit uses two resistors of equal value, two capacitors of equal value, andan amplifier with an amplification of between 1 and 3. Although notavailable to Sallen and Key, the amplifiers can now be made with low-cost
op amps with appropriate feedback resistors. Going through the four step
circuit analysis procedure, the location of this circuit's two poles can be
related to the component values:

EQUATION 32-4
Sallen-Key pole locations. Theseequations relate the pole position, T
and F, to the amplifier gain, A, theresistor, R, and capacitor, C.

F'

T'

A& 3
2RC

± & A 2 % 6A & 5
2RC

These equations show that the two poles always lie somewhere on a circle ofradius: 1/RC . The exact position along the circle depends on the gain of theamplifier. As shown in (a), an amplification of 1 places both of the poles on
the real axis. The frequency response of this configuration is a low-pass filter
with a relatively smooth transition between the passband and stopband. The-3dB (0.707) cutoff frequency of this circuit, denoted by T0 , is where the circle
intersects the imaginary axis, i.e., T0 ' 1/RC .

Chapter 32- The Laplace Transform

Amplitude

F

Vin

R

R
A

T0

jT

a. A = 1.0

C601

2 poles

Vout
CFrequency

T0

jT
Amplitudeb. A = 1.586
F

Frequency

T0
AmplitudejT

c. A = 2.5
FFrequency

T0

jT

d. A = 3.0
F

Amplitude

FIGURE 32-9
Sallen-Key characteristics. This circuit
produces two poles on a circle of radius
1/RC. As the gain of the amplifier isincreased, the poles move from the realaxis, as in (a), toward the imaginary axis,
as in (d).

Frequency

As the amplification is increased, the poles move along the circle, with acorresponding change in the frequency response. As shown in (b), an
amplification of 1.586 places the poles at 45 degree angles, resulting in thefrequency response having a sharper transition. Increasing the amplification
further moves the poles even closer to the imaginary axis, resulting in thefrequency response showing a peaked curve. This condition is illustrated in (c),where the amplification is set at 2.5. The amplitude of the peak continues togrow as the amplification is increased, until a gain of 3 is reached. As shownin (d), this is a special case that places the poles directly on the imaginary axis.
The corresponding frequency response now has an infinity large value at thepeak. In practical terms, this means the circuit has turned into an oscillator.
Increasing the gain further pushes the poles deeper into the right half of the splane. As mentioned before, this correspond to the system being unstable(spontaneous oscillation).
Using the Sallen-Key circuit as a building block, a wide variety of filter types
can be constructed. For example, a low-pass Butterworth filter is designedby placing a selected number of poles evenly around the left-half of the circle,
as shown in Fig. 32-10. Each two poles in this configuration requires one

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Sallen-Key stage. As described in Chapter 3, the Butterworth filter ismaximally flat, that is, it has the sharpest transition between the passband andstopband without peaking in the frequency response. The more poles used, thefaster the transition. Since all the poles in the Butterworth filter lie on thesame circle, all the cascaded stages use the same values for R and C. The only
thing different between the stages is the amplification. Why does this circularpattern of poles provide the optimally flat response? Don't look for an obvious
or intuitive answer to this question; it just falls out of the mathematics.
Figure 32-11 shows how the pole positions of the Butterworth filter can be
modified to produce the Chebyshev filter. As discussed in Chapter 3, theChebyshev filter achieves a sharper transition than the Butterworth at theexpense of ripple being allowed into the passband. In the s-domain, thiscorresponds to the circle of poles being flattened into an ellipse. The more
flattened the ellipse, the more ripple in the passband, and the sharper thetransition. When formed from a cascade of Sallen-Key stages, this requires
different values of resistors and capacitors in each stage.
Figure 32-11 also shows the next level of sophistication in filter design
strategy: the elliptic filter. The elliptic filter achieves the sharpest possible
transition by allowing ripple in both the passband and the stopband. In the sdomain, this corresponds to placing zeros directly on the real axis, with thefirst one near the cutoff frequency. Elliptic filters come in several varieties andare significantly more difficult to design than Butterworth and Chebyshev
configurations. This is because the poles and zeros of the elliptic filter do notlie in a simple geometric pattern, but in a mathematical arrangement involving
elliptic functions and integrals (hence the name).

1 polejT

2 polejT

F

F
FIGURE 32-10
The Butterworth s-plane. The low-pass
Butterworth filter is created by placing
poles equally around the left-half of acircle. The more poles used in the filter,the faster the roll-off.

3 poleFjT

6 poleF

jT

Chapter 32- The Laplace TransformjT

T0
Amplitude

Butterworth
F

Frequency

jT

Chebyshev

T0
Amplitude

FIGURE 32-11
Classic pole-zero patterns. These are thethree classic pole-zero patterns in filter
design. Butterworth filters have poles
equally spaced around a circle, resulting ina maximally flat response. Chebyshev
filters have poles placed on an ellipse,
providing a sharper transition, but at thecost of ripple in the passband. Elliptic
filters add zeros to the stopband. This
results in a faster transition, but withripple in the passband and stopband.

603

F

Frequency

o
o

jT

T0
Amplitude

EllipticFooFrequency

Since each biquad produces two poles, even order filters (2 pole, 4 pole, 6pole, etc.) can be constructed by cascading biquad stages. However, odd
order filters (1 pole, 3 pole, 5 pole, etc.) require something that the biquadcannot provide: a single pole on the imaginary axis. This turns out to benothing more than a simple RC circuit added to the cascade. For example, a
9 pole filter can be constructed from 5 stages: 4 Sallen-Key biquads, plus one
stage consisting of a single capacitor and resistor.
These classic pole-zero patterns are for low-pass filters; however, they can be
modified for other frequency responses. This is done by designing a low-passfilter, and then performing a mathematical transformation in the s-domain. We
start by calculating the low-pass filter pole locations, and then writing thetransfer function, H (s) , in the form of Eq. 32-3. The transfer function of thecorresponding high-pass filter is found by replacing each "s" with "1/s", andthen rearranging the expression to again be in the pole-zero form of Eq. 32-3.
This defines new pole and zero locations that implement the high-pass filter.
More complicated s-domain transforms can create band-pass and band-rejectfilters from an initial low-pass design. This type of mathematical manipulationin the s-domain is the central theme of filter design, and entire books are604

The Scientist and Engineer's Guide to Digital Signal Processing

devoted to the subject. Analog filter design is 90% mathematics, and only
10% electronics.
Fortunately, the design of high-pass filters using Sallen-Key stages doesn't
require this mathematical manipulation. The "1/s" for "s" replacement in thes-domain corresponds to swapping the resistors and capacitors in the circuit.
In the s-plane, this swap places the poles at a new position, and adds two zeros
directly at the origin. This results in the frequency response having a value ofzero at DC (zero frequency), just as you would expect for a high-pass filter.
This brings the Sallen-Key circuit to its full potential: the implementation of
two poles and two zeros.