Includes step-by-step procedures for designing analog and digital filters

Steven T. Karris

Includes step-by-step procedures for designing analog and

digital filters

TLAB® Applications

Orchard Publications

www.orchardpublications.com

with MA

Students and working professionals will find

Signals and Systems Signals and Systems with MATLAB® Applications,

with MATLAB® Applications Second Edition, to be a concise and easy-to-learn

text. It provides complete, clear, and detailed expla-

Second Edition

nations of the principal analog and digital signal Steven T. Karris

processing concepts and analog and digital filter design illustrated with numerous practical examples.

This text includes the following chapters and appendices: • Elementary Signals • The Laplace Transformation • The Inverse Laplace Transformation

• Circuit Analysis with Laplace Transforms • State Variables and State Equations • The Impulse Response and Convolution • Fourier Series • The Fourier Transform • Discrete Time Systems and the Z Transform • The DFT and The FFT Algorithm • Analog and Digital Filters • Introduction to MATLAB • Review of Complex Numbers • Review of Matrices and Determinants

Each chapter contains numerous practical applications supplemented with detailed instructions for using MATLAB to obtain quick solutions.

Steven T. Karris is the president and founder of Orchard Publications. He earned a bachelors degree in electrical engineering at Christian Brothers University, Memphis, Tennessee, a mas- ters degree in electrical engineering at Florida Institute of Technology, Melbourne, Florida, and has done post-master work at the latter. He is a registered professional engineer in California and Florida. He has over 30 years of professional engineering experience in industry. In addi- tion, he has over 25 years of teaching experience that he acquired at several educational insti- tutions as an adjunct professor. He is currently with UC Berkeley Extension .

Orchard Publications, Fremont, California Visit us on the Internet

www.orchardpublications.com or email us: info@orchardpublications.com

ISBN 0-9709511-8-3

$39.95 U.S.A.

Signals and Systems

with MATLAB® Applications

Second Edition

Steven T. Karris

Orchard Publications www.orchardpublications.com

publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher.

Direct all inquiries to Orchard Publications, 39510 Paseo Padre Parkway, Fremont, California 94538 Product and corporate names are trademarks or registered trademarks of the Microsoft™ Corporation and The

MathWorks™ Inc. They are used only for identification and explanation, without intent to infringe.

Library of Congress Cataloging-in-Publication Data

Library of Congress Control Number: 2003091595 ISBN 0-9709511-8-3

Preface

This text contains a comprehensive discussion on continuous and discrete time signals and systems with many MATLAB® examples. It is written for junior and senior electrical engineering students, and for self-study by working professionals. The prerequisites are a basic course in differential and integral calculus, and basic electric circuit theory.

This book can be used in a two-quarter, or one semester course. This author has taught the subject material for many years at San Jose State University, San Jose, California, and was able to cover all material in 16 weeks, with 2½ lecture hours per week.

To get the most out of this text, it is highly recommended that Appendix A is thoroughly reviewed. This appendix serves as an introduction to MATLAB, and is intended for those who are not familiar with it. The Student Edition of MATLAB is an inexpensive, and yet a very powerful software package; it can be found in many college bookstores, or can be obtained directly from

The MathWorks™ Inc., 3 Apple Hill Drive , Natick, MA 01760-2098 Phone: 508 647-7000, Fax: 508 647-7001 http://www.mathworks.com e-mail: info@mathwork.com

The elementary signals are reviewed in Chapter 1 and several examples are presented. The intent of this chapter is to enable the reader to express any waveform in terms of the unit step function, and subsequently the derivation of the Laplace transform of it. Chapters 2 through 4 are devoted to Laplace transformation and circuit analysis using this transform. Chapter 5 discusses the state variable method, and Chapter 6 the impulse response. Chapters 7 and 8 are devoted to Fourier series and transform respectively. Chapter 9 introduces discrete-time signals and the Z transform. Considerable time was spent on Chapter 10 to present the Discrete Fourier transform and FFT with the simplest possible explanations. Chapter 11 contains a thorough discussion to analog and digital filters analysis and design procedures. As mentioned above, Appendix A is an introduction to MATLAB. Appendix B contains a review of complex numbers, and Appendix C discusses matrices.

New to the Second Edition

This is an refined revision of the first edition. The most notable changes are chapter-end summaries, and detailed solutions to all exercises. The latter is in response to many students and working professionals who expressed a desire to obtain the author’s solutions for comparison with their own. The author has prepared more exercises and they are available with their solutions to those instructors who adopt this text for their class.

The chapter-end summaries will undoubtedly be a valuable aid to instructors for the preparation of presentation material.

The last major change is the improvement of the plots generated by the latest revisions of the MATLAB® Student Version, Release 13.

Orchard Publications Fremont, California www.orchardpublications.com info@orchardpublications.com

Chapter 1

Elementary Signals

T properties of the delta function are defined and derived. Several examples for expressing a vari-

his chapter begins with a discussion of elementary signals that may be applied to electric net- works. The unit step, unit ramp, and delta functions are introduced. The sampling and sifting

ety of waveforms in terms of these elementary signals are provided.

1.1 Signals Described in Math Form

Consider the network of Figure 1.1 where the switch is closed at time t = 0 .

v out open − terminals

− Figure 1.1. A switched network with open terminals.

We wish to describe v

out in a math form for the time interval – ∞ t +∞ << . To do this, it is conve-

nient to divide the time interval into two parts, – ∞t0 << , and 0 << t ∞ .

For the time interval – ∞t0 << , the switch is open and therefore, the output voltage v out is zero. In other words, v out = 0 for ∞ – << t 0 (1.1)

For the time interval 0 << t ∞ , the switch is closed. Then, the input voltage v S appears at the output, i.e.,

(1.2) Combining (1.1) and (1.2) into a single relationship, we get

We can express (1.3) by the waveform shown in Figure 1.2.

Signals and Systems with MATLAB Applications, Second Edition

1-1

Orchard Publications

Chapter 1 Elementary Signals

out

t Figure 1.2. Waveform for v out as defined in relation (1.3)

The waveform of Figure 1.2 is an example of a discontinuous function. A function is said to be discontinuous if it exhibits points of discontinuity, that is, the function jumps from one value to another without taking on any intermediate values.

1.2 The Unit Step Function u 0 () t

A well-known discontinuous function is the unit step function u 0 () t * that is defined as

u 0 () t = ⎨

It is also represented by the waveform of Figure 1.3.

u 0 () t 1

0 t Figure 1.3. Waveform for u 0 () t

In the waveform of Figure 1.3, the unit step function u 0 () t changes abruptly from to at 0 1 t = 0 . But if it changes at t = t 0 instead, it is denoted as u 0 ( t – t 0 ) . Its waveform and definition are as shown in Figure 1.4 and relation (1.5).

Figure 1.4. Waveform for u 0 ( t – t 0 )

* In some books, the unit step function is denoted as ut () , that is, without the subscript 0. In this text, however, we will reserve the ut () designation for any input when we discuss state variables in a later chapter.

1-2

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Unit Step Function

If the unit step function changes abruptly from to at 0 1 t = – t 0 , it is denoted as u 0 ( t + t 0 ) . Its waveform and definition are as shown in Figure 1.5 and relation (1.6).

−t

Figure 1.5. Waveform for u 0 ( t + t 0 ) ⎧ 0 t < – t 0

Example 1.1

Consider the network of Figure 1.6, where the switch is closed at time t = T .

v out open − terminals

− Figure 1.6. Network for Example 1.1

Express the output voltage v out as a function of the unit step function, and sketch the appropriate waveform.

Solution:

For this example, the output voltage v out = 0 for t < T , and v out = v S for t > T . Therefore, (1.7) v out = v S u 0 ( t – T ) and the waveform is shown in Figure 1.7.

Signals and Systems with MATLAB Applications, Second Edition

1-3

Orchard Publications

Chapter 1 Elementary Signals

v S u 0 ( t – T ) v out

Figure 1.7. Waveform for Example 1.1

Other forms of the unit step function are shown in Figure 1.8.

Figure 1.8. Other forms of the unit step function

Unit step functions can be used to represent other time-varying functions such as the rectangular pulse shown in Figure 1.9.

Figure 1.9. A rectangular pulse expressed as the sum of two unit step functions

Thus, the pulse of Figure 1.9(a) is the sum of the unit step functions of Figures 1.9(b) and 1.9(c) is

represented as u 0 ()u t – 0 ( t – 1 ) .

1-4

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Unit Step Function

The unit step function offers a convenient method of describing the sudden application of a voltage or current source. For example, a constant voltage source of 24 V applied at t = 0 , can be denoted as 24u 0 ()V t . Likewise, a sinusoidal voltage source vt () = V m cos ωt V that is applied to a circuit at

t = t 0 , can be described as vt () = ( V m cos ωt )u 0 ( t – t 0 )V . Also, if the excitation in a circuit is a rectangular, or triangular, or sawtooth, or any other recurring pulse, it can be represented as a sum (dif-

ference) of unit step functions.

Example 1.2

Express the square waveform of Figure 1.10 as a sum of unit step functions. The vertical dotted lines indicate the discontinuities at T 2T 3T , , and so on.

Figure 1.10. Square waveform for Example 1.2

Solution:

Line segment { has height , starts at A t = 0 , and terminates at t = T . Then, as in Example 1.1, this

segment is expressed as

(1.8) Line segment | has height – A , starts at t = T and terminates at t = 2T . This segment is expressed

v 1 () t = Au [ 0 ()u t – 0 ( t – T ) ]

(1.9) Line segment } has height A , starts at t = 2T and terminates at t = 3T . This segment is expressed as

v 2 () t = – A [ u 0 ( t – T )u – 0 ( t – 2T ) ]

(1.10) Line segment ~ has height – A , starts at t = 3T , and terminates at t = 4T . It is expressed as

v 3 () t = Au [ 0 ( t – 2T )u – 0 ( t – 3T ) ]

(1.11) Thus, the square waveform of Figure 1.10 can be expressed as the summation of (1.8) through (1.11),

v 4 () t = – A [ u 0 ( t – 3T )u – 0 ( t – 4T ) ]

that is,

Signals and Systems with MATLAB Applications, Second Edition

1-5

Orchard Publications

Chapter 1 Elementary Signals

Combining like terms, we get

Express the symmetric rectangular pulse of Figure 1.11 as a sum of unit step functions.

it () A

t – T ⁄ 2 0 T2 ⁄

Figure 1.11. Symmetric rectangular pulse for Example 1.3

Solution:

This pulse has height , starts at A t = – T2 ⁄ , and terminates at t = T2 ⁄ . Therefore, with reference to Figures 1.5 and 1.8 (b), we get

Express the symmetric triangular waveform of Figure 1.12 as a sum of unit step functions.

Figure 1.12. Symmetric triangular waveform for Example 1.4

Solution:

We first derive the equations for the linear segments { and | shown in Figure 1.13.

1-6

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Unit Step Function

Figure 1.13. Equations for the linear segments of Figure 1.12

For line segment {,

2 v () t = ⎛ --- t + 1 1 ⎞ u ⎛ 0 t + T ⎞ ⎝ – ⎠ --- u

(1.15) and for line segment |, v 2 () t = ⎛ – --- 2 t + ⎝ ⎞ 1 ⎠ u 0 ()u t – ⎛ t – --- T ⎞ 0 (1.16)

0 () t

Combining (1.15) and (1.16), we get

Express the waveform of Figure 1.14 as a sum of unit step functions.

vt () 3

Figure 1.14. Waveform for Example 1.5.

Solution:

As in the previous example, we first find the equations of the linear segments { and | shown in Fig- ure 1.15.

Signals and Systems with MATLAB Applications, Second Edition

1-7

Orchard Publications

Chapter 1 Elementary Signals

vt () 3

2 2t + 1

Figure 1.15. Equations for the linear segments of Figure 1.14

Following the same procedure as in the previous examples, we get

vt () = ( 2t + 1 )u [ 0 ()u t – 0 ( t – 1 ) ]3u + [ 0 ( t – 1 )u – 0 ( t – 2 ) ] + ( – t + 3 )u [ 0 ( t – 2 )u – 0 ( t – 3 ) ]

Multiplying the values in parentheses by the values in the brackets, we get

and combining terms inside the brackets, we get

(1.18) Two other functions of interest are the unit ramp function , and the unit impulse or delta function . We

vt () = ( 2t + 1 )u 0 ()2t1 t – ( – )u 0 ( t – 1 ) –u t 0 ( t – 2 ) + ( t – 3 )u 0 ( t – 3 )

will introduce them with the examples that follow.

Example 1.6

In the network of Figure 1.16 is a constant current source and the switch is closed at time i S t = 0 .

v C () t C−

Figure 1.16. Network for Example 1.6

1-8

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Unit Step Function

Express the capacitor voltage v C () t as a function of the unit step.

Solution:

The current through the capacitor is i C () t = i S = cons tan t , and the capacitor voltage v C () t is

1 t v C () t = ----

i C ()τ τ d *

where is a dummy variable. τ

Since the switch closes at t = 0 , we can express the current i C () t as

i C () t = i S u 0 () t

and assuming that v C () t = 0 for t < 0 , we can write (1.19) as

1 ----

i S u 0 ()τ τ d = C ∫ – ∞

u 0 ()τ τ d i t

v C () t = ----

+ S ---- u ()τ τ ∫ d

⎧⎪⎪⎨⎪⎪⎩ C ∫ 0 0

v C () t = ----- tu () t

Therefore, we see that when a capacitor is charged with a constant current, the voltage across it is a

linear function and forms a ramp with slope i S ⁄ C as shown in Figure 1.17.

v C () t slope = i S ⁄ C

Figure 1.17. Voltage across a capacitor when charged with a constant current source.

* Since the initial condition for the capacitor voltage was not specified, we express this integral with – ∞ at the lower limit of integration so that any non-zero value prior to t < 0 would be included in the integration.

Signals and Systems with MATLAB Applications, Second Edition

1-9

Orchard Publications

Chapter 1 Elementary Signals

1.3 The Unit Ramp Function u 1 () t The unit ramp function , denoted as u 1 () t , is defined as

u 1 () t =

u 0 ()τ τ ∫ d

where is a dummy variable. τ We can evaluate the integral of (1.23) by considering the area under the unit step function u 0 () t from

– ∞ to t as shown in Figure 1.18.

Area = 1 × τ = τ = t 1

Figure 1.18. Area under the unit step function from – ∞ to t

Therefore, we define u 1 () t as

u 1 () t = ⎨

Since u 1 () t is the integral of u 0 () t , then u 0 () t must be the derivative of u 1 () t , i.e.,

-----u d 1 () t = u 0 () t

Higher order functions of can be generated by repeated integration of the unit step function. For t

example, integrating u 0 () t twice and multiplying by 2 , we define u 2 () t as

u 2 () t = ⎨ 2 or

u 2 () t = 2 u 1 ()τ τ d (1.26)

Similarly,

u 3 () t = ⎨ 3 or

u 3 () t = 3 u 2 ()τ τ d (1.27)

and in general,

1-10

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Unit Ramp Function

u n () t = ⎨ n

u n () t = 3 u n – 1 ()τ τ d (1.28)

Also,

1 d u n – 1 () t = --- -----u n () t

In the network of Figure 1.19, the switch is closed at time t = 0 and i L () t = 0 for t < 0 .

i L () t v L () t `

L − Figure 1.19. Network for Example 1.7

Express the inductor current i L () t in terms of the unit step function.

Solution:

The voltage across the inductor is

v L () t = L -------

and since the switch closes at t = 0 ,

(1.31) Therefore, we can write (1.30) as

i L () t = i S u 0 () t

v L () t = Li S -----u t

But, as we know, u 0 () t is constant ( 0 or 1 ) for all time except at t = 0 where it is discontinuous. Since the derivative of any constant is zero, the derivative of the unit step u 0 () t has a non-zero value only at t = 0 . The derivative of the unit step function is defined in the next section.

Signals and Systems with MATLAB Applications, Second Edition

1-11

Orchard Publications

Chapter 1 Elementary Signals

1.4 The Delta Function δt () The unit impulse or delta function , denoted as δt () , is the derivative of the unit step u 0 () t . It is also

defined as

δτ ()τ d = u 0 () t

and δt () = 0 for all t ≠ 0 (1.34) To better understand the delta function δt () , let us represent the unit step u 0 () t as shown in Figure

1.20 (a).

1 Figure (a)

Area =1

Figure (b)

Figure 1.20. Representation of the unit step as a limit.

The function of Figure 1.20 (a) becomes the unit step as ε → 0 . Figure 1.20 (b) is the derivative of Figure 1.20 (a), where we see that as ε → 0 , 12 ⁄ε becomes unbounded, but the area of the rectangle remains . Therefore, in the limit, we can think of 1 δt () as approaching a very large spike or impulse at the origin, with unbounded amplitude, zero width, and area equal to . 1

Two useful properties of the delta function are the sampling property and the sifting property.

1.5 Sampling Property of the Delta Function δt () The sampling property of the delta function states that

(1.35) or, when a = 0 ,

1-12

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Sifting Property of the Delta Function

that is, multiplication of any function ft () by the delta function δt () results in sampling the function at the time instants where the delta function is not zero. The study of discrete-time systems is based on this property.

Proof:

Since δt () = 0 for t < 0 and t > 0 then,

ft ( )δ t () = 0 for t < 0 and t > 0 (1.37) We rewrite ft () as

(1.38) Integrating (1.37) over the interval – ∞ to t and using (1.38), we get

ft () = f0 () + [ ft ()f0 – () ]

[ f ()f0 τ – () ]δ τ ()τ d ∫ (1.39)

f ( )δ τ τ ()τ d =

f0 ( )δ τ ()τ d +

The first integral on the right side of (1.39) contains the constant term f0 () ; this can be written out- side the integral, that is,

δτ ()τ ∫ d

f0 ( )δ τ ()τ d = f0 ()

The second integral of the right side of (1.39) is always zero because

δt () = 0 for t < 0 and t > 0

and

[ f ()f0 τ – () ] τ = 0 = f0 ()f0 – () = 0

Therefore, (1.39) reduces to

δτ ()τ ∫ d

f ( )δ τ τ ()τ d = f0 ()

Differentiating both sides of (1.41), and replacing with , we get τ t

( )δ t ft () = f0 ( )δ t ()

Sampling Property of δ t ()

1.6 Sifting Property of the Delta Function δt () The sifting property of the delta function states that

Signals and Systems with MATLAB Applications, Second Edition

1-13

Orchard Publications

Chapter 1 Elementary Signals

ft ( )δ t α ( – )t d = f () α

that is, if we multiply any function ft () by δtα ( – ) , and integrate from – ∞ to +∞ , we will obtain the value of ft () evaluated at t = α .

Proof:

Let us consider the integral

ft ( )δ t α ( – ) t where a α b d <<

We will use integration by parts to evaluate this integral. We recall from the derivative of products that

(1.45) and integrating both sides we get

() d xy = xdy + ydx or xdy = ( ) ydx d xy –

xy d = xy – yx d ∫ (1.46) ∫

Now, we let x = ft () ; then, dx = ft ′ () . We also let dy = δtα ( – ) ; then, y = u 0 ( t – α ) . By substitution into (1.46), we get

ft ( )δ t α ( – )t d = ft ( )u 0 ( t – α ) –

u ( t – α )f t ′ () d t

We have assumed that a << αb ; therefore, u 0 ( t – α ) = 0 for αa < , and thus the first term of the right side of (1.47) reduces to fb () . Also, the integral on the right side is zero for αa < , and there-

fore, we can replace the lower limit of integration by . We can now rewrite (1.47) as a α

ft ( )δ t α ( – )t d = fb () – f ′ () t d t = fb ()fb – ()fα + ∫ ()

and letting a → – ∞ and b → ∞ for any α < ∞ , we get

ft ( )δ t α ( – )t d = f () ∫ α

Sifting Property of δ t ()

1-14

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Higher Order Delta Functions

1.7 Higher Order Delta Functions

An nth-order delta function is defined as the nth derivative of u 0 () t , that is,

δ δ () t = ----- u [ () ]

dt 0 t

The function δ' t () is called doublet , δ'' t () is called triplet , and so on. By a procedure similar to the derivation of the sampling property of the delta function, we can show that

(1.50) Also, the derivation of the sifting property of the delta function can be extended to show that

ft ( )δ' t a ( – ) = fa ( )δ' t a ( – )f – 'a ( )δ t a ( – )

ft ( )δ ( t – α )t d = () – 1 -------- f t [ ∫ ] n ()

Evaluate the following expressions:

a. 4 3t δt1 ( – )

b. t δt2 ( – )t ∫ d

c. 2 t δ' t 3 ( – )

Solution:

a. The sampling property states that 4 ft ( )δ t a ( – ) = fa ( )δ t a ( – ) For this example, ft () = 3t and

a = 1 . Then,

4 3t 4 δt1 ( – ) = { 3t

t = 1 }δ t 1 ( – ) = 3 δt1 ( – )

b. The sifting property states that

ft ( )δ t α ( – )t d = f () α . For this example, ft () = t and α = ∫ 2 .

Then,

t δt2 ( – )t d = f2 () = t t = 2 = ∫ 2

c. The given expression contains the doublet; therefore, we use the relation

Signals and Systems with MATLAB Applications, Second Edition

1-15

Orchard Publications

Chapter 1 Elementary Signals

ft ( )δ' t a ( – ) = fa ( )δ' t a ( – )f – 'a ( )δ t a ( – )

Then, for this example,

2 2 t 2 δ' t 3 ( – ) = t

t = 3 δ' t 3 ( – )d – -----t

a. Express the voltage waveform vt () shown in Figure 1.21 as a sum of unit step functions for the

time interval – 1 << t 7 s .

b. Using the result of part (a), compute the derivative of vt () and sketch its waveform.

−2 Figure 1.21. Waveform for Example 1.9

Solution:

a. We first derive the equations for the linear segments of the given waveform. These are shown in Figure 1.22.

Next, we express vt () in terms of the unit step function u 0 () t , and we get

Multiplying and collecting like terms in (1.52), we get

1-16

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Higher Order Delta Functions

2 Figure 1.22. Equations for the linear segments of Figure 1.21

b. The derivative of vt () is

dv ------ =

From the given waveform, we observe that discontinuities occur only at t = – 1 , t = 2 , and t = 7 . Therefore, δt1 ( – ) = 0 , δt4 ( – ) = 0 , and δt5 ( – ) = 0 , and the terms that contain these delta functions vanish. Also, by application of the sampling property,

2t δt1 ( + ) = { 2t t = – 1 }δ t 1 ( + ) = – 2 δt1 ( + ) ( – t + 3 )δ t 2 ( – ) = { ( – t + 3 ) t = 2 }δ t 2 ( – ) = δt2 ( – ) ( t – 6 )δ t 7 ( – ) = { ( t – 6 ) t = 7 }δ t 7 ( – ) = δt7 ( – )

and by substitution into (1.53), we get

Signals and Systems with MATLAB Applications, Second Edition

1-17

Orchard Publications

Chapter 1 Elementary Signals

dv ------ =

+ δt2 ( – )u + 0 ( t – 4 )u – 0 ( t – 5 ) + u 0 ( t – 7 )δt7 + ( – )

The plot of dv dt ⁄ is shown in Figure 1.23.

------ dv ( Vs ⁄ ) dt

Figure 1.23. Plot of the derivative of the waveform of Figure 1.21.

We observe that a negative spike of magnitude occurs at 2 t = – 1 , and two positive spikes of magnitude occur at 1 t = 2 , and t = 7 . These spikes occur because of the discontinuities at these points.

MATLAB * has built-in functions for the unit step, and the delta functions. These are denoted by the names of the mathematicians who used them in their work. The unit step function u 0 () t is referred

to as Heaviside(t), and the delta function δt () is referred to as Dirac(t) . Their use is illustrated with the examples below.

syms k a t;

% Define symbolic variables

u=k*sym('Heaviside(t-a)')

% Create unit step function at t = a

u= k*Heaviside(t-a)

d=diff(u) % Compute the derivative of the unit step function

d= k*Dirac(t-a)

* An introduction to MATLAB ® is given in Appendix A.

1-18

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Summary

int(d)

% Integrate the delta function

ans = Heaviside(t-a)*k

1.8 Summary

• The unit step function u 0 () t that is defined as

⎧ 0 t < 0 u 0 () t = ⎨ ⎩ 1 t > 0

• The unit step function offers a convenient method of describing the sudden application of a voltage or current source.

• The unit ramp function, denoted as u 1 () t , is defined as

u 1 () t =

u 0 ()τ τ ∫ d

• The unit impulse or delta function, denoted as δt () , is the derivative of the unit step u 0 () t . It is also defined as

δτ ()τ d = u 0 () ∫ t

and

δt () = 0 for all t ≠ 0

• The sampling property of the delta function states that

ft ( )δ t a ( – ) = fa ( )δ t ()

or, when a = 0 ,

ft ( )δ t () = f0 ( )δ t ()

• The sifting property of the delta function states that

ft ( )δ t α ( – )t d = f () ∫ α

• The sampling property of the doublet function δ' t () states that

ft ( )δ' t a ( – ) = fa ( )δ' t a ( – )f – 'a ( )δ t a ( – )

Signals and Systems with MATLAB Applications, Second Edition

1-19

Orchard Publications

Chapter 1 Elementary Signals

1.9 Exercises

1. Evaluate the following functions:

e. t e δt2 ( – )t ∫ d

f. sin t δ ⎛ t – ⎝ --- ⎞

a. Express the voltage waveform vt () shown in Figure 1.24, as a sum of unit step functions for

the time interval 0 << t 7 s .

b. Using the result of part (a), compute the derivative of vt () , and sketch its waveform.

Figure 1.24. Waveform for Exercise 2

1-20

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Solutions to Exercises

1.10 Solutions to Exercises

Dear Reader: The remaining pages on this chapter contain the solutions to the exercises. You must, for your benefit, make an honest effort to solve the problems without first looking at the

solutions that follow. It is recommended that first you go through and solve those you feel that you know. For the exercises that you are uncertain, review this chapter and try again. If your results do not agree with those provided, look over your procedures for inconsistencies and computational errors. Refer to the solutions as a last resort and rework those problems at a later date.

You should follow this practice with the exercises on all chapters of this book.

Signals and Systems with MATLAB Applications, Second Edition

1-21

Orchard Publications

Chapter 1 Elementary Signals

1. We apply the sampling property of the δt () function for all expressions except (e) where we apply the sifting property. For part (f) we apply the sampling property of the doublet.

We recall that the sampling property states that ft ( )δ t a ( – ) = fa ( )δ t a ( – ) . Thus,

We recall that the sampling property states that

ft ( )δ t α ( – )t d = f () ∫ α . Thus,

– e. 2 t e δt2 ( – )t d = t ∫ e

t = 2 = 4e = 0.54

We recall that the sampling property for the doublet states that

t = π2 ⁄ δ ⎝ t – --- – ⎠ ----- sin t

1-22

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Solutions to Exercises

– 2t

– vt 2t () = e u 0 ()e t – u 0 ( t – 2 ) 10tu + 0 ( t – 2 ) 30u – 0 ( t – 2 ) 10tu – 0 ( t – 3 ) 30u + 0 ( t – 3 ) – 10tu 0 ( t – 3 ) + 50u 0 ( t – 3 ) 10tu + 0 ( t – 5 ) 50u – 0 ( t – 5 ) 10tu + 0 ( t – 5 )

– 70u 0 ( t – 5 ) 10tu – 0 ( t – 7 ) 70u + 0 ( t – 7 )

– ------ 2t = – 2e u 0 ()e t + δt () + ( 2e + 10 )u 0 ( t – 2 ) + ( – e + 10t – 30 )δ t 2 ( – ) dt

Referring to the given waveform we observe that discontinuities occur only at t = 2 , t = 3 , and t = 5 . Therefore, δt () = 0 and δt7 ( – ) = 0 . Also, by the sampling property of the delta function

and with these simplifications (1) above reduces to

The waveform for dv dt ⁄ is shown below.

Signals and Systems with MATLAB Applications, Second Edition

1-23

Orchard Publications

Chapter 1 Elementary Signals NOTES

1-24

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Chapter 2

The Laplace Transformation

T and proved. It concludes with the derivation of the Laplace transform of common functions

his chapter begins with an introduction to the Laplace transformation, definitions, and properties of the Laplace transformation. The initial value and final value theorems are also discussed

of time, and the Laplace transforms of common waveforms.

2.1 Definition of the Laplace Transformation

The two-sided or bilateral Laplace Transform pair is defined as

– (2.1) st Lft { () }Fs = () = ft () e ∫ dt

– 1 σ jω L +

{ st Fs () }ft = = 1 () --------

Fs () e ∫ ds

2 πj σ jω –

– where 1 Lft { () } denotes the Laplace transform of the time function ft ()L , { Fs () } denotes the Inverse Laplace transform, and is a complex variable whose real part is , and imaginary part , s σ ω

that is, s = σ jω + . In most problems, we are concerned with values of time greater than some reference time, say t

t = t 0 = 0 , and since the initial conditions are generally known, the two-sided Laplace transform pair of (2.1) and (2.2) simplifies to the unilateral or one-sided Laplace transform defined as

2 πj ∫ σ jω –

L st { Fs () }f = () t = -------- Fs () e ds

The Laplace Transform of (2.3) has meaning only if the integral converges (reaches a limit), that is, if

– (2.5) st ft () e dt < ∫ ∞

To determine the conditions that will ensure us that the integral of (2.3) converges, we rewrite (2.5)

Signals and Systems with MATLAB Applications, Second Edition

2-1

Orchard Publications

Chapter 2 The Laplace Transformation

– σt – j ωt

ft ( )e e dt < ∞

– j The term ωt

e in the integral of (2.6) has magnitude of unity, i.e., e = 1 , and thus the condition for convergence becomes

– j ωt

ft ( )e dt < ∞

– σt

Fortunately, in most engineering applications the functions ft () exponential order are of * . Then, we can express (2.7) as,

and we see that the integral on the right side of the inequality sign in (2.8), converges if σσ > 0 . Therefore, we conclude that if ft () is of exponential order, Lft { () } exists if

Re s {} = σσ > 0 (2.9) where Re s {} denotes the real part of the complex variable . s Evaluation of the integral of (2.4) involves contour integration in the complex plane, and thus, it will

not be attempted in this chapter. We will see, in the next chapter, that many Laplace transforms can

be inverted with the use of a few standard pairs, and therefore, there is no need to use (2.4) to obtain the Inverse Laplace transform.

In our subsequent discussion, we will denote transformation from the time domain to the complex frequency domain, and vice versa, as

ft () ⇔ Fs ()

2.2 Properties of the Laplace Transform

1. Linearity Property

The linearity property states that if

f 1 ()f t , 2 ()…f t , , n () t

have Laplace transforms

* A function ft () is said to be of exponential order if ft () < ke

for all t ≥ 0 .

2-2

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Properties of the Laplace Transform

F 1 ()F s , 2 ()…F s , , n () s

respectively, and

c 1 ,, c 2 …c , n

are arbitrary constants, then,

c 1 f 1 ()c t + 2 f 2 ()…c t + + n f n () t ⇔ c 1 F 1 ()c s + 2 F 2 ()…c s + + n F n () s

Proof:

L { c 1 f 1 ()c t + 2 f 2 ()…c t + + n f n () t } =

[ c 1 f 1 ()c t + f ()…c + ∫ + 2 2 t n f n () t ] dt

1 f 1 () t e dt + c 2 f 2 () t e dt ∫ + ∫ …+c n f n () t e dt t 0 t 0 ∫ t 0

= c 1 F 1 ()c s + 2 F 2 ()…c s + + n F n () s

Note 1:

It is desirable to multiply ft () by u 0 () t to eliminate any unwanted non-zero values of ft () for t < 0 .

2. Time Shifting Property

The time shifting property states that a right shift in the time domain by units, corresponds to mul- a – tiplication by as e in the complex frequency domain. Thus,

ft ( – a ) e ∫ dt ∫

Now, we let t – a = τ ; then , t = τa + and dt = d τ . With these substitutions, the second integral on the right side of (2.13) becomes

– s ( τa + )

– as ∞

– f as () τ e d τ = e f () τ e d τ = e Fs ∫ ()

3. Frequency Shifting Property

The frequency shifting property states that if we multiply some time domain function ft () by an – exponential function at e where a is an arbitrary positive constant, this multiplication will produce a

shift of the s variable in the complex frequency domain by units. Thus, a

Signals and Systems with MATLAB Applications, Second Edition

2-3

Chapter 2 The Laplace Transformation

ft () e dt = Fs ( + a ∫ )

A change of scale is represented by multiplication of the time variable by a positive scaling factor t

a . Thus, the function ft () after scaling the time axis, becomes () f at .

4. Scaling Property

Let be an arbitrary positive constant; then, the a scaling property states that

() f at ⇔ ---F 1 ⎛⎞ --- s ⎝⎠

Proof:

– L f at st { () } = () f at e ∫ dt

and letting t = τa ⁄ , we get

f () τ e d --- = --- f () τ e d () τ = ---F ⎛⎞ ∫ ---

0 ⎝⎠ a a ∫ 0 a ⎝⎠ a

Note 3:

Generally, the initial value of − ft () is taken at t = 0 to include any discontinuity that may be present

at − t 0 . If it is known that no such discontinuity exists at t = 0 , we simply interpret f0 () as f0 () .

5. Differentiation in Time Domain

The differentiation in time domain property states that differentiation in the time domain corresponds = to multiplication by in the complex frequency domain, minus the initial value of − s ft () at t 0 . Thus,

f − 't () ----- f t () ⇔ sF s ()f0 ()

f 't () e ∫ dt

2-4

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Properties of the Laplace Transform

Using integration by parts where

vu d = uv – uv d ∫ (2.17) ∫

– st

– we let st du = f 't () and v = e . Then, u = ft () , dv = – se , and thus

lim − [ e fa ()f0 – () ] + sF s () = 0 – f0 () + sF s ()

The time differentiation property can be extended to show that

− – -------- f t −

-------- f t () ⇔

and in general

n – 1 -------- f t − n () ⇔ s Fs ()s –

To prove (2.18), we let

() = f 't () = ----- f t d

and as we found above,

Relations (2.19) and (2.20) can be proved by similar procedures.

We must remember that the terms f0 () f , − − '0 () f , '' 0 () , and so on, represent the initial conditions. Therefore, when all initial conditions are zero, and we differentiate a time function ft ()n times,

this corresponds to Fs () multiplied by to the s nth power.

Signals and Systems with MATLAB Applications, Second Edition

2-5

Chapter 2 The Laplace Transformation

6. Differentiation in Complex Frequency Domain

This property states that differentiation in complex frequency domain and multiplication by minus one, corresponds to multiplication of ft () by in the time domain. In other words, t

tf t () ⇔ – d ----- Fs ()

– L st { ft () } = Fs () = ft () e ∫ dt

Differentiating with respect to s, Leibnitz’s rule and applying * for differentiation under the integral, we get

-----F s ()

ds ∫ 0 ∫ 0 ∂ s

– st

– = st – [ tf t () ] e dt = – L tf t [ () ∫ ]

In general,

t ft () ⇔ () – 1 --------F s n ()

The proof for n ≥ 2 follows by taking the second and higher-order derivatives of Fs () with respect to . s

7. Integration in Time Domain

This property states that integration in time domain corresponds to Fs () divided by plus the initial s value of −

ft () at t = 0 , also divided by s . That is,

f () τ d τ ⇔ Fs () + () ----------- ------------ ∫ f0

* This rule states that if a function of a parameter is defined by the equation α F () α = ∫ fx ( , α )x d where f is some

known function of integration x and the parameter , a and b are constants independent of x and , and the par- α α

∫ a ∂ () α

dF tial derivative b ∂ f ⁄ ∂ α exists and it is continuous, then ------- = ----------------- x ∂ ( x , α ) d .

2-6

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Properties of the Laplace Transform Proof:

We express the integral of (2.23) as two integrals, that is,

f () τ d τ =

f () τ d τ + f () τ d τ

The first integral on the right side of (2.24), represents a constant value since neither the upper, nor the lower limits of integration are functions of time, and this constant is an initial condition denoted as −

f0 () . We will find the Laplace transform of this constant, the transform of the second integral on the right side of (2.24), and will prove (2.23) by the linearity property. Thus,

– st ∞

f0 () } =

f0 () e dt = f0 () e dt = f0 ( )e ∫ --------

f0 − ()0 × – ⎛ – f0 () ------------ ⎞ = f0 ⎝ ()

------------

This is the value of the first integral in (2.24). Next, we will show that

f () τ d τ ⇔ ∫ -----------

f () τ d ∫ τ

f () τ d τ = ∫ 0

Now,

g 't () } = Gs () = sL g t { () }g0 – () = Gs ()0 –

sL g t { () } = Gs () L { gt () } = Gs () -----------

⎫ L ⎨ f () τ d τ ⎬ = Fs () -----------

and the proof of (2.23) follows from (2.25) and (2.26).

Signals and Systems with MATLAB Applications, Second Edition

2-7

Chapter 2 The Laplace Transformation

8. Integration in Complex Frequency Domain

This property states that integration in complex frequency domain with respect to corresponds to s

division of a time function ft () by the variable , provided that the limit t lim -------- exists. Thus,

ft ()

-------- ⇔ Fs ()s d ∫

Proof:

– Fs st () = ft () e ∫ dt

Integrating both sides from to , we get s ∞

– Fs st ()s d = ft () e dt d ∫ s

Next, we interchange the order of integration, i.e.,

∞ – st

e d s ft ()t ∫ d

Fs ()s d =

and performing the inner integration on the right side integral with respect to , we get s

--------e d t = ∫ L ⎨ ∫ -------- ⎬

9. Time Periodicity

The time periodicity property states that a periodic function of time with period corresponds to T

– the integral sT ft () e dt divided by ( 1 – e ) in the complex frequency domain. Thus, if we let ft ∫ ()

T – st

be a periodic function with period , that is, T ft () = ft ( + nT ) , for n = 123 ,,, … we get the transform pair

– ft st () e ∫ dt

ft 0 ( + nT ) ⇔ -----------------------------

– sT

2-8

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Properties of the Laplace Transform Proof:

The Laplace transform of a periodic function can be expressed as

– { st ft () } = ft () e dt = ft () e dt + ft () e dt + ft () e dt + ∫ …

∫ 2T

In the first integral of the right side, we let t = τ , in the second t = τT + , in the third t = τ 2T + , and so on. The areas under each period of ft () are equal, and thus the upper and lower limits of integration are the same for each integral. Then,

– s ( τ 2T + { ) ft () } = f () τ e d τ + f ( τT + ) e d τ + f ( τ 2T + ) e d τ… +

– s ( τT + )

Since the function is periodic, i.e., f () τ = f ( τT + ) = f ( τ 2T + ) = … = f ( τ nT + ) , we can write (2.29) as

– sT

– 2sT

f () τ } = ( 1 + e + e + … )fτ () e d τ

By application of the binomial theorem, that is,

2 1 3 + a + a + a + … = 1 ------------

we find that expression (2.30) reduces to

f () τ e d ∫ τ

L { f () τ } = 0 ---------------------------------- – τe sT –

10. Initial Value Theorem

The − initial value theorem states that the initial value f0 () of the time function ft () can be found from its Laplace transform multiplied by and letting s s → ∞ .That is,

From the time domain differentiation property,

----- f t d () ⇔ sF s ()f0 – () − dt

Signals and Systems with MATLAB Applications, Second Edition

2-9

Chapter 2 The Laplace Transformation

∫ 0 dt

– ⎨ st ----- f t () ⎬ = sF s ()f0 – () − = ----- f t d ()

e dt

⎩ dt

Taking the limit of both sides by letting s → ∞ , we get

– lim st [ sF s ()f0 – − d () ]

= lim

T → ∞ ∫ ε dt

lim

----- f t () e dt

Interchanging the limiting process, we get

T → ∞ ∫ ε dt s → ∞

– lim st [ sF s ()f0 – () ] = lim ----- f t () lim e dt

and since

– lim st e = 0

the above expression reduces to

lim [ sF s ()f0 – () ] = 0

lim sF s () = f0 ()

11. Final Value Theorem

The final value theorem states that the final value f () ∞ of the time function ft () can be found from

its Laplace transform multiplied by s , then, letting s → 0 . That is,

From the time domain differentiation property,

d ----- f t − () ⇔

sF s ()f0 – ()

d – ⎨ st ----- f t () ⎬ =

sF s ()f0 – () =

∫ 0 dt

----- f t () e dt

⎩ dt

Taking the limit of both sides by letting s → 0 , we get

2-10

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

Properties of the Laplace Transform

s → 0 s → 0 T → ∞ ∫ ε dt

– lim st [ sF s ()f0 – () ] = lim lim ----- f t () e dt

and by interchanging the limiting process, we get

– lim st [ sF s ()f0 – () − ] = d

s → 0 T → ∞ ∫ ε dt s → 0

lim

----- f t () lim e dt

Also, since

– lim st e = 1

the above expression reduces to

lim [

sF s ()f0 () ] = lim

s → 0 T → ∞ ∫ ε dt

----- f t () dt = lim

ft ()

= lim [ fT ()fε –

() − ] = f ()f0 ∞ – ()

and therefore,

lim sF s () = f () ∞

12. Convolution in the Time Domain

Convolution * in the time domain corresponds to multiplication in the complex frequency domain, that is,

f 1 ( )f ∫ τ 2 ( t – τ ) d τ =

f 1 ( )f τ 2 ( t – τ ) d τ e dt

– = st f ∫

f 2 ( t – τ ) e dt d τ

We let t – τ = λ ; then, t = λτ + , and dt = d λ . By substitution into (2.35),

* Convolution is the process of overlapping two signals. The convolution of two time functions f 1 () t and f 2 () t is

denoted as f 1 ( )*f t 2 () t , and by definition, f 1 ( )*f t 2 () t = ∫ f 1 ( )f τ 2 ( t – τ ) d τ where is a dummy variable. We will τ

discuss it in detail in Chapter 6.

Signals and Systems with MATLAB Applications, Second Edition

2-11

Chapter 2 The Laplace Transformation

– L s λ { f 1 ( )*f t 2 () t } =

f 1 ( )e τ d τ f () λ e d ∫ λ

13. Convolution in the Complex Frequency Domain

Convolution in the complex frequency domain divided by 12 ⁄ πj , corresponds to multiplication in the time domain. That is,

f 1 ( )f t 2 () t ⇔ -------- F 1 ( )*F s 2 () s

f ( )f t () t e dt

and recalling that the Inverse Laplace transform from (2.2) is

σ jω +

f 1 ()

t = --------

2 πj ∫

by substitution into (2.37), we get

F 1 ( )e µ d µf () t e ∫ dt

--------

0 2 πj ∫

µt

σ jω 2

µ -------- )t

1 + σ jω

2 πj ∫ σ jω –

F 1 () µ

f 2 () t e dt d µ

We observe that the bracketed integral is F 2 ( s – µ ) ; therefore,

σ jω +

{ ( )f

Lf 1 t 2 () t } = --------

2 πj ∫

F 1 ( )F µ 2 ( s – µ )dµ = --------F ( )*F s

For easy reference, we have summarized the Laplace transform pairs and theorems in Table 2.1.

2.3 The Laplace Transform of Common Functions of Time

In this section, we will present several examples for finding the Laplace transform of common functions of time.

Example 2.1

Find L { u 0 () t }

2-12

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Laplace Transform of Common Functions of Time

TABLE 2.1 Summary of Laplace Transform Properties and Theorems

Property/Theorem

Time Domain

Complex Frequency Domain

2 Time Shifting

3 Frequency Shifting

4 Time Scaling

() f at

1 ---F ⎛⎞ --- s

a ⎝⎠ a

5 Time Differentiation

----- f t d ()

sF s ()f0 – () −

Chapter 2 The Laplace Transformation Solution:

We start with the definition of the Laplace transform, that is,

– L st { ft () } = Fs () = ft () e ∫ dt

For this example,

st ∞

u 0 () t } =

1 e dt = --------- = 0 – ⎛ – 1 ⎞ ∫ =

Thus, we have obtained the transform pair

Find L { u 1 () t }

Solution:

We apply the definition

– { st ft () } = Fs () = ft () e ∫ dt

and for this example,

We will perform integration by parts recalling that

uv d = uv – vu d ∫ (2.39) ∫

– – st du = 1 and v = e -----------

* This condition was established in (2.9).

2-14

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Laplace Transform of Common Functions of Time

By substitution into (2.39),

t = ------------- –

e = –e t ----------- e

------------- – --------

Since the upper limit of integration in (2.40) produces an indeterminate form, we apply L’ Hôpital’s rule * , that is,

lim ---------------- = lim -------- =

L {} t = Evaluating the second term of (2.40), we get 1 ----

Thus, we have obtained the transform pair

1 t ⇔ ---- 2

Find n L { t u

0 () t }

Solution:

To find the Laplace transform of this function, we must first review the gamma or generalized factorial function Γn () which is defined as

Γn () =

* Often, the ratio of two functions, such as ---------- fx () , for some value of x, say a, results in an indeterminate form. To

gx ()

work around this problem, we consider the limit lim ---------- fx () , and we wish to find this limit, if it exists. To find this

x → a gx ()

limit, we use L’Hôpital’s rule which states that if fa () = ga () = 0 , and if the limit ------f x d ()d ⁄ ------g x () as x dx

dx approaches a exists, then, d

lim ---------- fx () = lim ⎛ ⎝ ------f x dx ()d ⁄ ------g x () ⎞

x → a gx ()

x → a dx

Signals and Systems with MATLAB Applications, Second Edition

2-15

Chapter 2 The Laplace Transformation

The integral of (2.42) is an * improper integral but converges (approaches a limit) for all n > 0 . We will now derive the basic properties of the gamma function, and its relation to the well known

factorial function

n! = nn ( – 1 )n2 ( – ) ⋅⋅321 ⋅⋅

The integral of (2.42) can be evaluated by performing integration by parts. Thus, in (2.39) we let

– e dx and v = -----

and (2.42) is written as

Γn () = x e ------------

+ --- x e d x

With the condition that n > 0 , the first term on the right side of (2.43) vanishes at the lower limit x = 0 . It also vanishes at the upper limit as x → ∞ . This can be proved with L’ Hôpital’s rule by differentiating both numerator and denominator m times, where m ≥ n . Then,

m – 1 nx

lim x e ------------ = lim x = lim d -------- x ------------------- = lim d ------------------------------------ x = …

( n – 1 )n2 ( – )… n m ( – + 1 ) lim ------------------------------------------------------------------------------------ x

n – nn m ( –

1 )n2 ( – )… n m ( – + 1 )x

lim --------------------------------------------------------------------

Therefore, (2.43) reduces to

Γn () = 1 --- x e d x

and with (2.42), we have

* Improper integrals are two types and these are:

a. ∫ fx ()x d where the limits of integration a or b or both are infinite

b. ∫ fx ()x d where f(x) becomes infinite at a value x between the lower and upper limits of integration inclusive.

2-16

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Laplace Transform of Common Functions of Time

Γn () =

e d x = --- 1 x e d x

By comparing the integrals in (2.44), we observe that

Γn () = Γn1 ( + ) ---------------------

(2.46) It is convenient to use (2.45) for n < 0 , and (2.46) for n > 0 . From (2.45), we see that Γn () becomes

infinite as n → 0 . For n = 1 , (2.42) yields

and thus we have the important relation, Γ1 () = 1 (2.48)

From the recurring relation of (2.46), we obtain

and in general

(2.50) for n = 123 ,,, …

Γn1 ( + ) = n!

The formula of (2.50) is a noteworthy relation; it establishes the relationship between the Γn () function and the factorial n! We now return to the problem of finding the Laplace transform pair for n t u

0 t , that is,

n – { st t u

t e dt

To make this integral resemble the integral of the gamma function, we let st = y , or t = ys ⁄ , and

Signals and Systems with MATLAB Applications, Second Edition

2-17

Chapter 2 The Laplace Transformation

thus dt = dy s ⁄ . Now, we rewrite (2.51) as

1 = Γn ( + 1 ) = ∫ n!

----------- n + 1

Therefore, we have obtained the transform pair

t u 0 () t ⇔ ----------- n + 1 (2.52)

for positive integers of and n σ0 > .

Example 2.4

Find L { δt () }

Solution:

– { st δt () } = δt () e ∫ dt

and using the sifting property of the delta function, we get

– s0 { () δt () }

δt () e dt = e = ∫ 1

– st

Thus, we have the transform pair δt () ⇔ 1 (2.53) for all . σ

Example 2.5

Find L { δta ( – ) }

Solution:

– { st δta ( – ) } = δta ( – ) e ∫ dt

and again, using the sifting property of the delta function, we get

2-18

Signals and Systems with MATLAB Applications, Second Edition Orchard Publications

The Laplace Transform of Common Functions of Time

– { as δta ( – ) } = δta ( – ) e dt = ∫ e

– st

Thus, we have the transform pair – δta as ( – ) ⇔ e (2.54)

for . σ0 >

Example 2.6

– Find at L { e u

e ∫ dt

e e dt =

⎛ 1 ⎞ – ( s + a = )t – ----------- e =

----------- 1

Thus, we have the transform pair

– at

1 e u 0 () t ⇔ -----------

Find L ⎨ t e u 0 () t ⎬

Solution:

For this example, we will use the transform pair of (2.52), i.e.,

t u 0 () t ⇔ ----------- n + 1 (2.56)

and the frequency shifting property of (2.14), that is,

– e at ft () ⇔ Fs ( + a )

Signals and Systems with MATLAB Applications, Second Edition

2-19

Includes step-by-step procedures for designing analog and digital filters

Steven T. Karris