Electric Circuits, 9th Edition P65 pptx

616 Fourier Series Example 16.2 Finding the Fourier Series of an Odd Function with Symmetry Find the Fourier series representation for the cur-rent waveform shown in Fig, 16.10.. 16.5 A

616 Fourier Series

Example 16.2 Finding the Fourier Series of an Odd Function with Symmetry

Find the Fourier series representation for the

cur-rent waveform shown in Fig, 16.10

-T/2

Figure 16.10 • The periodic waveform for Example 16.2

In the interval O s t < 7/4, the expression for /(f) is

m-f,

Thus

8 f T/44I

bk = — —risinkaitfdt

1 Jo *

321m I sin ktoot t cos kco()t

2 2

T l \ kW 0 k(0()

TfA

Solution

We begin by looking for degrees of symmetry in the

waveform We find that the function is odd and, in

addition, has half-wave and quarter-wave

symme-try Because the function is odd, all the a

coeffi-cients are zero; that is, a v = 0 and a k = 0 for all k

Because the function has half-wave symmetry,

bk = 0 for even values of k Because the function

has quarter-wave symmetry, the expression for b k

for odd values of k is

8j,/« kir

2 7 sin — {k is odd)

Kr £

TT

The Fourier series representation of /(f) is

i(t) =—r* 2J ~ sin —- sin «<W 0 /

TT" « = 1,3.5, « 2

—r sin w Qt - - sin 3<oGt

r/4

b k =-= f /(f) sin ko) {] t dt

/ A S S E S S M E N T PROBLEM

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

16.3 Derive the Fourier series for the periodic

volt-age shown

\7Vm « sin (mr/3)

Answer: v g(t) = —z— 2J z smnco0t

TT «=1,3,5, n

MO

vm

-vm

/ \ 1 \

0 T/6 T/3 r/s

—

1 1

2 7 / 3 57/6

/ / T

NOTE: Also try Chapter Problems 16.11 and 16.12

16.4 An Alternative Trigonometric Form of the Fourier Series 617

16.4 An Alternative Trigonometric Form

of the Fourier Series

In circuit applications of the Fourier series, we combine the cosine and

sine terms in the series into a single term for convenience Doing so allows

the representation of each harmonic of v(t) or i(t) as a single phasor

quan-tity The cosine and sine terms may be merged in either a cosine

expres-sion or a sine expresexpres-sion Because we chose the cosine format in the

phasor method of analysis (see Chapter 9), we choose the cosine

expres-sion here for the alternative form of the series Thus we write the Fourier

series in Eq 16.2 as

fit) = av + ^Ancos(no){)t - 6,X (16.38)

«=i

where A n and d„ are defined by the complex quantity

an - jbn = Vflg + bl/-6 n = A a/-9„ (16.39)

We derive Eqs 16.38 and 16.39 using the phasor method to add the cosine

and sine terms in Eq 16.2 We begin by expressing the sine functions as

cosine functions; that is, we rewrite Eq 16.2 as

00

/ ( / ) = a v + 2X,cosncotf + b n cos(nco 0 t - 90°) (16.40)

Adding the terms under the summation sign by using phasors gives

and

®{b„ cos(nco () t - 90°)} = b n / - 9 0 ° = -jb n (16.42)

Then

2P{«„ cos(«a>o* + b n cos(ti(OQt — 90°)} = a n - jb n

= Val + %/-$„

= A„/-0„ (16.43)

When we inverse-transform Eq 16.43, we get

ancosno)()t + bncos(nw0t - 90°) = ^jA,,/-6,,}

= Ancos(no)0t - 6n) (16.44)

Substituting Eq 16.44 into Eq 16.40 yields Eq 16.38 Equation 16.43

corresponds to Eq 16.39 If the periodic function is either even or odd,

A n reduces to either a n (even) or b„ (odd), and 6 n is either 0° (even) or

90° (odd)

The derivation of the alternative form of the Fourier series for a given

periodic function is illustrated in Example 16.3

618 Fourier Series

Example 16.3 Calculating Forms of the Trigonometric Fourier Series for Periodic Voltage

a) D e r i v e the expressions for a k and b k for the

peri-odic function shown in Fig 16.11

b) Write t h e first four terms of the Fourier series

r e p r e s e n t a t i o n of v(t) using the format of

E q 16.38

J L

T_ T_ yr T 5T 3T IT 2T

4 2 4 4 2 4

Figure 16.11 • The periodic function for Example 16.3

Solution

a) T h e voltage v(t) is neither even n o r odd, n o r

does it have half-wave symmetry T h e r e f o r e we

use Eqs 16.4 a n d 16.5 to find a k and b k Choosing

to as zero, we obtain

a k =

T

V m cos ko) {) t dt + (0) cos ka> {) t dt

JTjA

2V m smkcotf

kcor

7/4

0

Vjn kTJ^

kir 2

and

2 [ T/A

b k = — I V m sin ktntf dt

1 Jo 2V m ( — cos koot

k(o {)

b) Tlie average value of v{t) is

a„ = -— = - v

T h e values of a k — jb k for k = 1,2, and 3 are

«i - jb\ V,

IT

V,

TT

V2V,

/ - 4 5 ° ,

V V (h — lb~> = 0 - / = / —90 ,

3 " J IT TT L

«3 _ jb 3 = -y,>

3TT

Vzv,

3ir / - 1 3 5 '

Thus the first four t e r m s in the Fourier series

representation of v(t) are

V V2V V v(t) =^r + -—^-cos^t - 45°) + -f-cos(2co7T 77 () t - 90°)

V2V + ——^cos(3o>o* - 135°) +

/ A S S E S S M E N T PROBLEM

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

1 6 4 a) C o m p u t e A\—A 5 a n d 0 i ~ 0 5 for t h e periodic

function shown if V m — 9TT V

b) Using the format of E q 16.38, write the

Fourier series for v(t) up t o a n d including

the fifth h a r m o n i c assuming T = 125.66 m s

Answer: (a) 10.4,5.2,0,2.6,2.1 V, a n d - 1 2 0 ° , - 6 0 ° ,

n o t defined, - 1 2 0 ° , - 6 0 ° ;

(b) v(t) = 21.99 + 10.4cos(50/ - 120°) +

5.2cos(100r - 60°) + 2.6 cos(200f - 120°) + 2.1 cos(250/ - 60°) V

3

IT

3

4 J

3

57'

3

IT

NOTE: Also try Chapter Problem 16.22

16.5 An Application

Now we illustrate how to use a Fourier series representation of a periodic

excitation function to find the steady-state response of a linear circuit The

RC circuit shown in Fig 16.12(a) will provide our example The circuit is

energized with the periodic square-wave voltage shown in Fig 16.12(b)

The voltage across the capacitor is the desired response, or output, signal

The first step in finding the steady-state response is to represent the

peri-odic excitation source with its Fourier series After noting that the source has

odd, half-wave, and quarter-wave symmetry, we know that the Fourier

coefficients reduce to b k, with k restricted to odd integer values:

b

T

4V rrk

774

V m sin kaj {] t dt

(k is odd)

Then the Fourier series representation of v„ is

4V

IT

CO -t

- j? — sin nco {) t

(16.45)

(16.46)

Writing the series in expanded form, we have

v., = sin cunt + —— sin 3oW

16.5 An Application

(a)

v„

-V,

(b)

Figure 16.12 A An RC circuit excited by a periodic

voltage, (a) The RC series circuit, (b) The square-wave

voltage

4V 4V

in • c ^ v in • *7 ,

—— sin 5(ti5 7T {) t + —— sin 7ct>777 0/ + (16.47)

Tlie voltage source expressed by Eq 16.47 is the equivalent of

infi-nitely many series-connected sinusoidal sources, each source having its

own amplitude and frequency To find the contribution of each source to

the output voltage, we use the principle of superposition

For any one of the sinusoidal sources, the phasor-domain expression

for the output voltage is

V r

v„ = 1 + jo)RC (16.48)

All the voltage sources are expressed as sine functions, so we interpret a

phasor in terms of the sine instead of the cosine In other words, when we

go from the phasor domain back to the time domain, we simply write the

time-domain expressions as sin(atf + 6) instead of cos(wf + 6)

The phasor output voltage owing to the fundamental frequency of the

sinusoidal source is

V,„ = {4V

m /ir)/Qf

Writing V„i in polar form gives

cii

where

1 + j(ti {) RC '

( 4 V J / - / 3 ,

TrVl + (tilR 2Cr

0i = tan~ X(ti{)RC

(16.49)

(16.50)

(16.51)

From Eq 16.50, the time-domain expression for the fundamental

fre-quency component of v (, is

4V

sin(ttiof - ft) (16.52)

T T V I + (4RC 2

We derive the third-harmonic component of the output voltage in a

simi-lar manner The third-harmonic phasor voltage is

(4V„;/377)/cy

Y " 3 J3(OQRC

4V

f = Z z & > (16.53)

3 T T V I + 9<4R 2 C

where

The time-domain expression for the third-harmonic output voltage is

4V V03 = , '" = = =sin(3<o0f - j83) (16.55) 3TT V I + 9wg^2C2

Hence the expression for the &th-harmonic component of the output

voltage is

v ok = '" = sin(/cw,/ - j8jt) (& is odd), (16.56)

where

We now write down the Fourier series representation of the output

voltage:

ff /, = u l t t V I + (HW0i?C)2

The derivation of Eq 16.58 was not difficult But, although we have an

ana-lytic expression for the steady-state output, what v 0(t) looks like is not

imme-diately apparent from Eq 16.58 As we mentioned earlier, this shortcoming is

a problem with the Fourier series approach Equation 16.58 is not useless,

however, because it gives some feel for the steady-state waveform of v (> (t), if

we focus on the frequency response of the circuit For example, if C is large,

1/ncooC is small for the higher order harmonics Thus the capacitor short

cir-cuits the high-frequency components of the input waveform, and the higher

order harmonics in Eq 16.58 are negligible compared to the lower order

har-monics Equation 16.58 reflects this condition in that, for large C,

4Vm °° 1 v<> ~ S ^ 2 -^sin(Aio)0r - 90°)

a ~ 2 -jcosnwof (16.59)

7T(D()RC ,, = 1¾ It

Equation 16.59 shows that the amplitude of the harmonic in the output

is decreasing by 1/n 2, compared with 1/n for the input harmonics If C is

so large that only the fundamental component is significant, then to a

first approximation

~4V V»{t) « ; ^ c o s w( )f , (16.60)

16.5 An Application 621

and Fourier analysis tells us that the square-wave input is deformed into a

sinusoidal output

Now let's see what happens as C —>0 The circuit shows that v () and v g

are the same when C = 0, because the capacitive branch looks like an

open circuit at all frequencies Equation 16.58 predicts the same result

because, as C —> 0,

Wm * 1 ,

But Eq 16.61 is identical to Eq 16.46, and therefore v 0 —* v g as C —*• 0

Thus Eq 16.58 has proven useful because it enabled us to predict that

the output will be a highly distorted replica of the input waveform if C is

large, and a reasonable replica if C is small In Chapter 13, we looked at

the distortion between the input and output in terms of how much

mem-ory the system weighting function had In the frequency domain, we look

at the distortion between the steady-state input and output in terms of

how the amplitude and phase of the harmonics are altered as they are

transmitted through the circuit When the network significantly alters the

amplitude and phase relationships among the harmonics at the output

rel-ative to that at the input, the output is a distorted version of the input

Thus, in the frequency domain, we speak of amplitude distortion and

phase distortion

For the circuit here, amplitude distortion is present because the

ampli-tudes of the input harmonics decrease as 1/rc, whereas the ampliampli-tudes of

the output harmonics decrease as

1 1

n V l + (na> QRC)2'

This circuit also exhibits phase distortion because the phase angle of each

input harmonic is zero, whereas that of the nth harmonic in the output

sig-nal is - tan"1 ri(o0RC

An Application of the Direct Approach

to the Steady-State Response

For the simple RC circuit shown in Fig 16.12(a), we can derive the

expres-sion for the steady-state response without resorting to the Fourier series

representation of the excitation function Doing this extra analysis here

adds to our understanding of the Fourier series approach

To find the steady-state expression for v 0 by straightforward circuit

analysis, we reason as follows The square-wave excitation function

alter-nates between charging the capacitor toward +V„, and —V m After the

circuit reaches steady-state operation, this alternate charging becomes

periodic We know from the analysis of the single time-constant RC circuit

(Chapter 7) that the response to abrupt changes in the driving voltage is

exponential Thus the steady-state waveform of the voltage across the

capacitor in the circuit shown in Fig 16.12(a) is as shown in Fig 16.13

The analytic expressions for v„{t) in the time intervals 0 < t < T/2

and T/2<t<T are

Vo = Vm + (V, - VJe^RC, 0 < t < T/2; (16.62)

Vo = ~V m + (V 2 + V m )e-^™ RC , T/2 < t < T (16.63)

We derive Eqs 16.62 and 16.63 by using the methods of Chapter 7, as

sum-marized by Eq 7.60 We obtain the values of V\ and V 2 by noting from

Eq 16.62 that

Vl = Vm + (1/, - Vm)e-TI2RC\ (16.64)

Toward + V.„ Toward + V

\ \

Toward —V m Toward —V

Figure 16.13 • The steady-state waveform of v 0 for the circuit in Fig 16.12(a)

622 Fourier Series

Small C

Figure 16.14 • The effect of capacitor size on the

steady-state response

and from Eq 16.63 that

V 1 = -V ln + (V 2 + V m )e- T ? 2RC Solving Eqs 16.64 and 16.65 for V\ and V 2 yields

y, = -y = —™i L

Substituting Eq 16.66 into Eqs 16.62 and 16.63 gives

2V,

V ° " j " j |_ e -f/2RC ->/RC , 0 < t < T/2

(16.65)

(16.66)

(16.67)

and

1 + e-r/ac

ff-(y/2)]/RC F/2 S / =S 7 (16.68) Equations 16.67 and 16.68 indicate that vw(0 has half-wave symmetry

and that therefore the average value of v 0 is zero This result agrees with the Fourier series solution for the steady-state response —namely, that because the excitation function has no zero frequency component, the response can have no such component Equations 16.67 and 16.68 also show the effect of changing the size of the capacitor If C is small, the

exponential functions quickly vanish, v a = Vm between 0 and T/2, and

v a = —V m between T/2 and T In other words, v a —* v% as C —> 0 If C is

large, the output waveform becomes triangular in shape, as Fig 16.14 shows Note that for large C, we may approximate the exponential

terms e~' /RC and C ,-['-(772)1/KC b y t h e H n e a r t e r m s j _ ( t/RC) and

1 - {[t - (T/2)]/RC}i respectively Equation 16.59 gives the Fourier

series of this triangular waveform

Figure 16.14 summarizes the results The dashed line in Fig 16.14 is the input voltage, the solid colored line depicts the output voltage when

C is small, and the solid black line depicts the output voltage when C is large

Finally, we verify that the steady-state response of Eqs 16.67 and 16.68 is equivalent to the Fourier series solution in Eq 16.58 To do so we simply derive the Fourier series representation of the periodic function described by Eqs 16.67 and 16.68 We have already noted that the periodic voltage response has half-wave symmetry Therefore the Fourier series

contains only odd harmonics For k odd,

«* =

T!1 ( 2V e' t/IiC (y _ Z K"'C ;—

-8RCV,,,

cos kco {)t dt

T[\ + (kco{)RC): (k is odd), (16.69)

bk = - J \Vm - - + e_T/2RC )

4V, $kco(ymR2C2 kir T[\ + (kiOuRC)2] (k is odd) (16.70)

To show that the results obtained from Eqs 16.69 and 16.70 are consistent with Eq 16.58, we must prove that

4V„, 1

Vol + b 2k =

k7T V l + (ka> {) RC) 2' and that

— = ~ko)[)RC

(16.71)

(16.72)

16.6 Average-Power Calculations with Periodic Functions 623

We leave you to verify Eqs 16.69-16.72 in Problems 16.23 and 16.24

Equations 16.71 and 16.72 are used with Eqs 16.38 and 16.39 to derive the

Fourier series expression in Eq 16.58; we leave the details to you in

Problem 16.25

With this illustrative circuit, we showed how to use the Fourier series

in conjunction with the principle of superposition to obtain the

steady-state response to a periodic driving function Again, the principal

short-coming of the Fourier series approach is the difficulty of ascertaining the

waveform of the response However, by thinking in terms of a circuit's

fre-quency response, we can deduce a reasonable approximation of the

steady-state response by using a finite number of appropriate terms in the

Fourier series representation (See Problems 16.27 and 16.29.)

S S E S S M E N T P R O B L E M

Objective 2—Know how to analyze a circuit's response to a periodic waveform

16.5 The periodic triangular-wave voltage seen on

the left is applied to the circuit shown on the

right Derive the first three nonzero terms in

the Fourier series that represents the

steady-state voltage v 0 if V m = 281.25ir2 mV and the

period of the input voltage is 200-7T ms

Answer: 2238.83 cos(10; 5.71°) + 239.46 cos(30/

-16.70°) + 80.50 cos(50f - 26.57°) + mV

16.6 The periodic square-wave shown on the left is

applied to the circuit shown on the right

a) Derive the first four nonzero terms in the

Fourier series that represents the

steady-state voltage v 0 if V„, = 210-77 V and the

period of the input voltage is 0.277 ms

b) Which harmonic dominates the output

voltage? Explain why

1

i

y,n

0

~vm

1

T/2

1

T

100 kft -^vw—

+

100 nF o a

Answer: (a) 17.5 cos(10,000r + 88.81°) +

26.14cos(30,000^ - 95.36°) + 168cos(50,0000 +

17.32 cos(70,000/ + 98.30°) + V;

(b) The fifth harmonic, at 10,000 rad/s, because the circuit is a bandpass filter with a center frequency of 50,000 rad/s and a quality factor of 10

10 kH

!20nF

+

20 mH v„

NOTE: Also try Chapter Problems 16.27 and 16.28

16.6 Average-Power Calculations

with Periodic Functions

If we have the Fourier series representation of the voltage and current at

a pair of terminals in a linear lumped-parameter circuit, we can easily

express the average power at the terminals as a function of the harmonic

voltages and currents Using the trigonometric form of the Fourier series

expressed in Eq 16.38, we write the periodic voltage and current at the

terminals of a network as

00

v = V dc + 2Xcos(/io>of - Q m )< (16.73)

DO

' = ' d c + ^,I a COS(na) {r t - B tn ) (16.74)

/ ( = 1

The notation used in Eqs 16.73 and 16.74 is defined as follows:

V dc = the amplitude of the dc voltage component,

Vn = the amplitude of the nth-harmonic voltage,

Qvn - the phase angle of the nth-harmonic voltage,

/d c = the amplitude of the dc current component,

/n = the amplitude of the nth-harmonic current,

d in = the phase angle of the nth-harmonic current

We assume that the current reference is in the direction of the

refer-ence voltage drop across the terminals (using the passive sign

conven-tion), so that the instantaneous power at the terminals is w'.The average

power is

j rh+T j ft tt +T

P = ~ / P dt = T Vt dL ( 1 6-7 5 )

To find the expression for the average power, we substitute Eqs 16.73 and

16.74 into Eq 16.75 and integrate At first glance, this appears to be a

for-midable task, because the product vi requires multiplying two infinite

series However, the only terms to survive integration are the products of

voltage and current at the same frequency A review of Eqs 16.8-16.10

should convince you of the validity of this observation Therefore

Eq 16.75 reduces to

y ^ d c ^ d c f

t 0 +T oo_ i fh+T

v„i„co$(na){)t - em)

n=\ l Jt a

Now, using the trigonometric identity

1 1

cos a cos (3 = - c o s ( a - / 3 ) + — cos(a + /3),

we simplify Eq 16.76 to

1 °° V I f' 0+T

p = vdc/dc + 7 S - ^ r1 / I c o s( 0 - - *bd

1 ;i=i z A,

The second term under the integral sign integrates to zero, so

P = ^d c/d c + 2 - ^ c o s ( 0 , „ - 0 in) (16.78)

Equation 16.78 is particularly important because it states that in the case

of an interaction between a periodic voltage and the corresponding periodic

current, the total average power is the sum of the average powers obtained

from the interaction of currents and voltages of the same frequency Currents

and voltages of different frequencies do not interact to produce average

16.6 Average-Power Calculations with Periodic Functions 625

power Therefore, in average-power calculations involving periodic

func-tions, the total average power is the superposition of the average powers

associated with each harmonic voltage and current Example 16.4 illustrates

the computation of average power involving a periodic voltage

A s s u m e that the periodic square-wave voltage in

E x a m p l e 16.3 is applied across the terminals of a

15 H resistor The value of V m is 60 V, and that of T

is 5 ms

a) Write the first five nonzero terms of the Fourier

series representation of v(t) U s e the

trigono-metric form given in Eq 16.38

b) Calculate the average power associated with

each term in (a)

c) Calculate the total average power delivered to

the 15 O resistor

d) W h a t percentage of the total power is delivered

by the first five terms of the Fourier series?

Solution

a) The dc component of v(t) is

(60)(7/4)

T = 15 V

From Example 16.3 we have

A ] = V2 6O/77 = 27.01 V,

0i = 45°,

A 2 = 60/TT = 19.10 V,

e2 = 90°,

A 3 = 20 V2/TT = 9.00 V,

03 = 135°,

A4 = 0,

04 = 0 ° ,

A5 = 5.40 V,

05 = 45°,

2TT 277(1000)

w () = 40077 r a d / s

Thus, using the first five nonzero terms of the Fourier series,

17(f) = 15 + 27.01 cos(40077/ - 45°)

+ 19.10COS(800T7/ - 90°)

+ 9 0 0 C O S ( 1 2 0 0 T 7 / - 135°)

+ 5 4 0 C O S ( 2 0 0 0 T 7 / - 45°) + - - - V

b) T h e voltage is applied to the terminals of a resis-tor, so we can find t h e power associated with each term as follows:

152

P d c = 15" = 1 5 W'

1 92

P3 = _ _ = 2.7 0 W,

1 5.42

, , = - _ _ = 0.97 W

c) To obtain t h e total average power delivered to the 15 0 resistor, we first calculate the r m s value

of v(t):

V =

r rms

/(60) 2 (774)

T = V 9 0 0 = 30 V

The total average p o w e r delivered to t h e 15 (1 resistor is

302 ,

PT = — = 60 W

d) T h e total power delivered by the first five nonzero terms is

P = Pd c + P { + P 2 + P 3 + P 5 = 55.15 W

This is (55.15/60)(100), o r 91.92% of the total