Electric Circuits, 9th Edition P64 doc

In fact, the study and analysis of heat flow in a metal rod led the French mathe-matician Jean Baptiste Joseph Fourier 1768-1830 to the trigonometric series representation of a periodic

606 Fourier Series

4) + 2 7 /

Figure 16.1 A A periodic waveform

v{t)

V m

-v(t)

V,„

-T

(a)

(b)

Figure 16.2 • Output waveforms of a nonfiltered

sinu-soidal rectifier, (a) Full-wave rectification, (b) Half-wave

rectification

Another practical problem that stimulates interest in periodic func-tions is that power generators, although designed to produce a sinusoidal waveform, cannot in practice be made to produce a pure sine wave The distorted sinusoidal wave, however, is periodic Engineers naturally are interested in ascertaining the consequences of exciting power systems with a slightly distorted sinusoidal voltage

Interest in periodic functions also stems from the general observation that any nonlinearity in an otherwise linear circuit creates a nonsinusoidal periodic function The rectifier circuit alluded to earlier is one example of this phenomenon Magnetic saturation, which occurs in both machines and transformers, is another example of a nonlinearity that generates a nonsinusoidal periodic function An electronic clipping circuit, which uses transistor saturation, is yet another example

Moreover, nonsinusoidal periodic functions are important in the analysis of nonelectrical systems Problems involving mechanical vibra-tion, fluid flow, and heat flow all make use of periodic functions In fact, the study and analysis of heat flow in a metal rod led the French mathe-matician Jean Baptiste Joseph Fourier (1768-1830) to the trigonometric series representation of a periodic function This series bears his name and

is the starting point for finding the steady-state response to periodic exci-tations of electric circuits

V m

-Figure 16.3 A The triangular waveform of a cathode-ray

Vm

0

1/

v m n

7 27

(a)

V(

Vm

0

0

(c)

Figure 16.4 A Waveforms produced by function

generators used in laboratory testing, (a) Square wave, (b) Triangular wave, (c) Rectangular pulse

16.1 Fourier Series Analysis:

An Overview

What Fourier discovered in investigating heat-flow problems is that a

periodic function can be represented by an infinite sum of sine or cosine

functions that are harmonically related In other words, the period of any

trigonometric term in the infinite series is an integral multiple, or

har-monic, of the fundamental period T of the periodic function Thus for

peri-odic / ( / ) , Fourier showed that / ( / ) can be expressed as

oo

f(t) = a v + 5 X c o s n(a d + b n sin/iwo^ (16.2) < Fourier series representation of a periodic

«=i function

where n is the integer sequence 1,2,3,

In Eq 16.2, a v , a fn and b n are known as the Fourier coefficients and are

calculated from /(f) The term o>0 (which equals 2TT/T) represents the

fundamental frequency of the periodic function /(f) The integral

multi-ples of OJ() —that is, 2&>0, 3a>0, 4w0, and so on—are known as the harmonic

frequencies o f / ( / ) Thus 2w0 is the second harmonic, 3a>0 is the third

har-monic, and no) () is the «th harmonic of /(f)

We discuss the determination of the Fourier coefficients in

Section 16.2 Before pursuing the details of using a Fourier series in circuit

analysis, we first need to look at the process in general terms From an

applications point of view, we can express all the periodic functions of

interest in terms of a Fourier series Mathematically, the conditions on a

periodic function /(f) that ensure expressing /(f) as a convergent Fourier

series (known as Dirichlet's conditions) are that

1 /(f) be single-valued,

2 / ( / ) have a finite number of discontinuities in the periodic interval,

3 / ( / ) have a finite number of maxima and minima in the periodic

interval,

4 the integral

/ 1/(01*

exists

Any periodic function generated by a physically realizable source satisfies

Dirichlet's conditions These are sufficient conditions, not necessary

con-ditions Thus if /(f) meets these requirements, we know that we can

express it as a Fourier series However, if/(f) docs not meet these

require-ments, we still may be able to express it as a Fourier series The necessary

conditions on /(f) are not known

After we have determined /(f) and calculated the Fourier coefficients

(«,„ a,„ and b„), we resolve the periodic source into a dc source (a„) plus a

sum of sinusoidal sources (a„ and /?„) Because the periodic source is

driv-ing a linear circuit, we may use the principle of superposition to find the

steady-state response In particular, we first calculate the response to each

source generated by the Fourier series representation of/(f) and then add

the individual responses to obtain the total response The steady-state

response owing to a specific sinusoidal source is most easily found with

the phasor method of analysis

The procedure is straightforward and involves no new techniques of

circuit analysis It produces the Fourier series representation of the

steady-state response; consequently, the actual shape of the response is

608 Fourier Series

unknown Furthermore, the response waveform can be estimated only by adding a sufficient number of terms together Even though the Fourier series approach to finding the steady-state response does have some draw-backs, it introduces a way of thinking about a problem that is as important

as getting quantitative results In fact, the conceptual picture is even more important in some respects than the quantitative one

16.2 The Fourier Coefficients

After defining a periodic function over its fundamental period, we deter-mine the Fourier coefficients from the relationships

2 /

Fourier coefficients • ak = —I f(t) cos kco 0 t dt, (16.4)

Tjt(3

2 f'o+T

bk = r fit) sin ka)Qt dt (16.5)

1 J t0

In Eqs 16.4 and 16.5, the subscript k indicates the &th coefficient in the integer sequence 1,2,3, Note that a v is the average value of /(/), a k is

twice the average value of f{t) cos kcotf, and b k is twice the average value

of f(t) sin kcotf

We easily derive Eqs 16.3-16.5 from Eq 16.2 by recalling the

follow-ing integral relationships, which hold when m and n are integers:

sin mw0f dt = 0, for all m, (16.6) 'o

t 0 +T cos mcootdt = 0, for all m, (16-7)

I

Jtr,

t n +T

cos moitf sin nco^t dt = 0, for all m and n, (16.8)

t0+r

sin mo)()t sin no) {)t dt = 0, for all m # n,

= —, for m - n, (16.9)

2

to+T

cos mo) 0t cos ncoGt dt = 0, for all m # «,

T

= —, for m = n (16.10)

We leave you to verify Eqs 16.6-16.10 in Problem 16.5

To derive Eq 16.3, we simply integrate both sides of Eq 16.2 over

one period:

/ f(t)dt = / f a v 4- ^ a n cos n(x) () t + b lt sinna) {) t jdt

Jio Ji,) \ «=l /

/.?„+'/' oo rh+T

= / a v dt + 2 / (fl« c o s nwo* + ^« s m '""of) &

J/ tl 11= 1 Jfy

Equation 16.3 follows directly from Eq 16.11

To derive the expression for the A:th value of A„, we first multiply

Eq 16.2 by cos kio {) t and then integrate both sides over one period of /(f):

f{t) cos kco {) t dt = / a v cos k(o{) t dt

+ 2 ) / («/, cos «o> ( / cos &w () f + /}„ sin «6> ( / cos /ca> () f) d/

«=1 A,

'«

Solving Eq 16.12 for a k yields the expression in Eq 16.4

We obtain the expression for the kXh value of b n by first multiplying

both sides of Eq 16.2 by sin kcotf and then integrating each side over one

period of/(f) Example 16.1 shows how to use Eqs 16.3-16.5 to find the

Fourier coefficients for a specific periodic function

Find the Fourier series for the periodic voltage

shown in Fig 16.5

- 7 0 T IT

Figure 16.5 • The periodic voltage for Example 16.1

Solution

When using Eqs 16.3-16.5 to find a v , %, and b k , we

may choose the value of f0- FOT the periodic voltage

of Fig 16.5, the best choice for f() is zero Any other

choice makes the required integrations more

cum-bersome The expression for v(t) between 0 and 7 is

ly 1

The equation for a v is

This is clearly the average value of the waveform in Fig 16.5

The equation for the kth value of a n is

(lk = T I 1*7^ '

•Z—z COS k(Ont + ~ SmkiOnt

2V,

T 2

——z ( cos 2-rrk - 1) 0 for all k

610 Fourier Series

The equation for the A:th value of b„ is

2V "'f l i t i \

T

o

-vm

irk

y

m = —

-y

y m

2

The Fourier series for v(t) is

V °9 1

v in ^r\ l

X ~ smtt&w

K V V

sm (o {) t - -— sin 2oj () t - —— sin 3a>0? - • • •

1" 27T 37T

I / A S S E S S M E N T P R O B L E M S

Objective 1—Be able to calculate the trigonometric form of the Fourier coefficients for a periodic waveform

16.1 Derive the expressions for «„, a k , and b k for the

periodic voltage function shown if V m = 9ir V

v„

3

27/

3

4T

3

57/

3

IT

f s i n ^ V,

Answer: a v = 21.99 V,

NOTE: Also try Chapter Problems 16.1-16.3

c o s ^ ) V

16.2 Refer to Assessment Problem 16.1

a) What is the average value of the periodic voltage?

b) Compute the numerical values of ay — a 5

and b\ - b 5 c) If T = 125.66 ms, what is the fundamental

frequency in radians per second?

d) What is the frequency of the third harmonic

in hertz?

e) Write the Fourier series up to and including the fifth harmonic

Answer: (a) 21.99 V;

(b) -5.2 V , 2 6 V , 0 V , - 1 3 , and 1.04 V;

9 V, 4.5 V, 0 V, 2.25 V, and 1.8 V;

(c) 50 rad/s;

(d) 23.87 Hz;

(e) v(t) = 21.99 - 5.2 cos 50? + 9 sin 50? +

2.6 cos 100? + 4.5 sin 100* -1.3 cos 200? + 2.25 sin 200? + 1.04 cos 250? + 1.8 sin 250? V

Finding the Fourier coefficients, in general, is tedious Therefore any-thing that simplifies the task is beneficial Fortunately, a periodic function that possesses certain types of symmetry greatly reduces the amount of work involved in finding the coefficients In Section 16.3, we discuss how symmetry affects the coefficients in a Fourier series

16.3 The Effect of Symmetry

on the Fourier Coefficients

Four types of symmetry may be used to simplify the task of evaluating the

Fourier coefficients:

• even-function symmetry,

• odd-function symmetry,

• half-wave symmetry,

• quarter-wave symmetry

The effect of each type of symmetry on the Fourier coefficients is discussed

in the following sections

Even-Function Symmetry

A function is defined as even if

fit) = /(-0- (16.13) -< Even function

Functions that satisfy Eq 16.13 are said to be even because polynomial

functions with only even exponents possess this characteristic For even

periodic functions, the equations for the Fourier coefficients reduce to

2 fT/2

(16.14)

T/2

f{t) cos ktotfdt,

1 Jo

(16.15)

bk = 0, for all k (16.16)

Note that all the b coefficients are zero if the periodic function is even

Figure 16.6 illustrates an even periodic function The derivations of

Eqs 16.14-16.16 follow directly from Eqs 16.3-16.5 In each derivation,

we select t {) = -T/2 and then break the interval of integration into the

range from -T/2 to 0 and 0 to T/2, or

a r

-T/2 -T/2

0

/(0 dt

T/2

T/2

"fl f(t)dL (16.17)

-T

m

0 Figure 16.6 • An even periodic function,

/(0=/(-0-Now we change the variable of integration in the first integral on the

right-hand side of Eq 16.17 Specifically, we let t = -x and note that

/ ( 0 = f(~x) = /(•*) because the function is even We also observe that

x — T/2 when t = -T/2 and dt = —dx Then

/ ( 0 dt

-T/2 J'-T/2

0 /-7/2

f(x)(-dx) = / f(x) dx, (16.18)

612 Fourier Series

which shows that the integration from -T/2 to 0 is identical to that from

0 to 7/2; therefore Eq 16.17 is the same as Eq 16.14 The derivation of

Eq 16.15 proceeds along similar lines Here,

a k =

2 f°

/ ( / ) cos ka) {) t dt

T/2

V — / / ( / ) cos ko) {) t dt,

but

•4) ,.()

/ ( / ) cos ku>()tdt = / f(x) cos (-k00^)(-dx)

T/2 f(x) cos k(x) {) x dx (16.20)

As before, the integration from -T/2 to 0 is identical to that from 0 to T/2 Combining Eq 16.20 with Eq 16.19 yields Eq 16.15

All the b coefficients are zero when / ( / ) is an even periodic function, because the integration from —T/2 to 0 is the exact negative of the

inte-gration from 0 to T/2; that is,

0 ni)

/ ( / ) sin ka) {) t dt = / f(x)sm(-ko) {) x)(—dx) -T/2 JT/2

T/2 f(x) sin koi()X dx (16.21)

When we use Eqs 16.14 and 16.15 to find the Fourier coefficients, the

interval of integration must be between 0 and T/2

Odd-Function Symmetry

A function is defined as odd if Odd function •

Figure 16.7 • An odd periodic function

/(0 = -/(-0

Functions that satisfy Eq 16.22 are said to be odd because polynomial functions with only odd exponents have this characteristic The expres-sions for the Fourier coefficients are

a v = 0;

a k = 0,

6 * = f

for all k\

rT/2 / / ( / ) sin ka) {) t dt

/o

(16.23) (16.24)

(16.25)

Note that all the a coefficients are zero if the periodic function is odd

Figure 16.7 shows an odd periodic function

We u s e t h e same process t o derive Eqs 16.23-16.25 that we used t o

derive E q s 16.14-16.16 We leave t h e derivations to you in P r o b l e m 16.7

T h e evenness, or oddness, of a periodic function c a n b e destroyed by

shifting t h e function along t h e time axis In o t h e r words, t h e judicious

choice of w h e r e t - 0 m a y give a periodic function even o r o d d symmetry

For example, t h e triangular function shown in Fig 16.8(a) is neither even

n o r o d d H o w e v e r , w e c a n m a k e t h e function even, as shown in

Fig 16.8(b), or o d d , as shown in Fig 16.8(c)

Half-Wave Symmetry

A periodic function possesses half-wave symmetry if it satisfies the

constraint

E q u a t i o n 16.26 states that a periodic function has half-wave symmetry if,

after it is shifted one-half period a n d inverted, it is identical to t h e original

function For example, t h e functions shown in Figs 16.7 a n d 16.8 have

wave symmetry, w h e r e a s those in Figs 16.5 a n d 16.6 d o n o t N o t e that

half-wave symmetry is n o t a function of w h e r e t = 0

If a periodic function has half-wave symmetry, both a k a n d b k are zero

for even values of k M o r e o v e r , a v also is zero because t h e average value

of a function with half-wave symmetry is zero T h e expressions for t h e

Fourier coefficients a r e

a v = 0,

a k = 0,

(16.27)

7/2

1 -/()

h = y7

T/2

fit) cos ko) 0 t dt,

i

fit) sin k(i) {) t dt,

for k even;

for k odd;

for k even;

for k odd

(16.28)

(16.29)

(16.30)

(16.31)

(c)

Figure 16.8 A How the choice of where t = 0 can

make a periodic function even, odd, or neither, (a) A periodic triangular wave that is neither even nor odd (b) The triangular wave of (a) made even by shifting the

function along the t axis, (c) The triangular wave of (a) made odd by shifting the function along the t axis

We derive E q s 16.27-16.31 by starting with Eqs 16.3-16.5 a n d

choos-ing t h e interval of integration as —T/2 t o T/2 W e t h e n divide this r a n g e

into t h e intervals —T/2 to 0 a n d 0 t o T/2 For example, t h e derivation

for a k is

a k =

Jt,t

fit) cos kcjQt dt

T.L

T/2

fit) cos kco {) t dt

T L

T/2

.0

= — / fit) cos kcoot dt

T/2 T/2

+ ;=r / / ( f ) cos ka) {] t dt

614 Fourier Series

Now we change a variable in the first integral on the right-hand side of

Eq 16.32 Specifically, we let

Then

t = x - T/2

x = T/2, when t = 0;

x = 0, when t = —T/2;

dt = dx

We rewrite the first integral as

/ ( 0 cos kto () tdt = / f(x - T/2)coskto {) (x - T/2)dx (16.33)

Note that

cos ktO()(x — T/2) — cos (ko) {) x — kir) = cos kir cos kco () x

and that, by hypothesis,

f(x - T/2) = -f(x)

Therefore Eq 16.33 becomes

A

7/4

-A

7/2 37/4 7

(a)

/(0

/I

0

-i

7/4

-A

I

7/2 37/4 7

(b)

Figure 16.9 • (a) A function that has quarter-wave

symmetry, (b) A function that does not have

quarter-wave symmetry

./-7/2

f(t) cos kco () t dt

T/2

[-f(x)] cos k 77 cos ko) {) x dx (16.34)

t Incorporating Eq 16.34 into Eq 16.32 gives

7/2

a k = - ( 1 — cos kit) / f(t) cos &G>0f rf/ (16.35)

But cos kTT is 1 when k is even and - 1 when k is odd Therefore Eq 16.35

generates Eqs 16.28 and 16.29

We leave it to you to verify that this same process can be used to derive Eqs 16.30 and 16.31 (see Problem 16.8)

We summarize our observations by noting that the Fourier series rep-resentation of a periodic function with half-wave symmetry has zero aver-age, or dc, value and contains only odd harmonics

Quarter-Wave Symmetry

The term quarter-wave symmetry describes a periodic function that has

half-wave symmetry and, in addition, symmetry about the midpoint of the positive and negative half-cycles The function illustrated in Fig 16.9(a)

has quarter-wave symmetry about the midpoint of the positive and

nega-tive half-cycles The function in Fig 16.9(b) does not have quarter-wave

symmetry, although it does have half-wave symmetry

A periodic function that has quarter-wave symmetry can always be

made either even or odd by the proper choice of the point where t = 0

For example, the function shown in Fig 16.9(a) is odd and can be made

even by shifting the function 7/4 units either right or left along the t axis

However, the function in Fig 16.9(b) can never be made either even or

odd To take advantage of quarter-wave symmetry in the calculation of the

Fourier coefficients, you must choose the point where t = 0 to make the

function either even or odd

If the function is made even, then

a v = 0, because of half-wave symmetry;

a k = 0, for k even, because of half-wave symmetry;

8 fT/4

a k = — I f(t) cos kaitf dt, for k odd;

T Ju

b k = 0, for all /c, because the function is even (16.36)

Equations 16.36 result from the function's quarter-wave symmetry in

addition to its being even Recall that quarter-wave symmetry is

super-imposed on half-wave symmetry, so we can eliminate a r and a k for k even

Comparing the expression for a k , k odd, in Eqs 16.36 with Eq 16.29 shows

that combining quarter-wave symmetry with evenness allows the

shorten-ing of the range of integration from 0 to 7/2 to 0 to 7/4 We leave the

der-ivation of Eqs 16.36 to you in Problem 16.9

If the quarter-wave symmetric function is made odd,

a v = 0, because the function is odd;

a k = 0, for all k, because the function is odd;

b k = 0, for k even, because of half-wave symmetry;

8 f T/4

b k = ^ / f(t) sin kco {) tdt, for k odd (16.37)

Equations 16.37 are a direct consequence of quarter-wave symmetry and

oddness Again, quarter-wave symmetry allows the shortening of the

inter-val of integration from 0 to 7/2 to 0 to 7/4 We leave the derivation of

Eqs 16.37 to you in Problem 16.10

Example 16.2 shows how to use symmetry to simplify the task of

find-ing the Fourier coefficients