Electric Circuits, 9th Edition P66 pot

16.7 The rms Value of a Periodic Function The rms value of a periodic function can be expressed in terms of the Fourier coefficients; by definition, ^ = ^ / fitrdt.. 16.8 The Exponentia

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

Objective 3—Be able to estimate the average power delivered to a resistor using a small number of Fourier

coefficients

16.7 The trapezoidal voltage function in Assessment

Problem 16.3 is applied to the circuit shown If

\2V m = 296.09 V and T = 2094.4ms, estimate

the average power delivered to the 2 fi resistor

Answer: 60.75 W

NOTE: Also try Chapter Problems 16.34 and 16.35

16.7 The rms Value of a Periodic Function

The rms value of a periodic function can be expressed in terms of the Fourier coefficients; by definition,

^ = ^ / fitrdt (16.79)

Representing /(f) by its Fourier series yields

tu+T

a v + 2 A i COS (iwat - 0„)

« = i

dt (16.80)

The integral of the squared time function simplifies because the only terms to survive integration over a period are the product of the dc term and the harmonic products of the same frequency All other products inte-grate to zero There fore Eq 16.80 reduces to

F - = w r ,r + „?,!""

'4 + 2 v

/ r = l z

4+2(^¼) (16-81)

Equation 16.81 states that the rms value of a periodic function is the square root of the sum obtained by adding the square of the rms value of each harmonic to the square of the dc value For example, let's assume that

a periodic voltage is represented by the finite series

v = 10 + 30cos(w(/ - $ t ) + 20cos(2w0/ - 02) + 5cos(3to0f - 03) + 2cos(5w(/ - 9 5 )

The rms value of this voltage is

V = V l O2 + (30/V2)2 + (20/V2-)2 + (5/V5)2 + (2/V2)2

= V7643 = 27.65 V

Usually, infinitely many terms are required to represent a periodic func-tion by a Fourier series, and therefore Eq 16.81 yields an estimate of the true rms value We illustrate this result in Example 16.5

16.8 The Exponential Form of the Fourier Series 627

Example 16.5 Estimating the rms Value of a Periodic Function

Use Eq 16.81 to estimate the rms value of the voltage Therefore,

in Example 16.4

V, = 27.01/V2 V, the rms value of the fundamental,

V 2 = 19.10/V2 V, the rms value of the second harmonic, From Example 16.4, the true rms value is 30 V We

nnn j m*r , , / - , , , , • approach this value by including more and more

harmon-V% = 9.00/V2 V, the rms value of the third harmonic, n , ,0 1 c , ^ t ,

V 5 = 5.40/V2 V, the rms value of the fifth harmonic through k = 9, the equation yields a value of 29.32 V

NOTE: Assess your understanding of this material by trying Chapter Problems 16.36 and 16.39

16.8 The Exponential Form

of the Fourier Series

The exponential form of the Fourier series is of interest because it

allows us to express the series concisely The exponential form of the

series is

oo

/(0 = 2 CjP**, (16.82)

ft =t-CX3

where

To derive Eqs 16.82 and 16.83, we return to Eq 16.2 and replace the

cosine and sine functions with their exponential equivalents:

(16.84)

Substituting Eqs 16.84 and 16.85 into Eq 16.2 gives

OO _ I

/ ( 0 = a v + ^—(e inco »' + e-'" bJ "') + —(e> n ^ - e ~ inmf )

«=i 2 2/'

COSrtW()f =

e //!w„r _|_ e-jnco{)i

2 ejnuytf _ g-jnwy/t

= «„ + 2 ( " 9 " r'"^ + ( " 2 k*W- (16-86)

Now we define C„ as

C„ = rfa, - jb ) = -f/-B,„ AZ = 1 , 2 , 3 , - - (16.87)

c„-\ 2 /*'«+7 ' 7f(t) cos na) () t dt — j — I / ( f ) sin nco r ' n+T () tdt

i r'u+T

= — / f(t) (cos nw^t — jsm no) { )t) dt

which completes the derivation of Eq 16.83 To complete the derivation of

Eq 16.82, we first observe from Eq 16.88 that

C0 = - / ,J{t)dt = a

Next we note that

./,,+7

(16.89)

(16.90)

c-« = ^ / " f{t)e^<dt = c;; = |(«„ + A )

Substituting Eqs 16.87,16.89, and 16.90 into Eq 16.86 yields

00

0 0 DO

/ 7 = 0 / / = 1

Note that the second summation on the right-hand side of Eq 16.91 is

equivalent to summing C n e ina} "' from - 1 to - c o ; that is,

« = i / / = - 1

Because the summation from - 1 to - c o is the same as the summation

from - c o to —1, we use Eq 16.92 to rewrite Eq 16.91:

00 —1

/(0 = ^cj*** + 2c>>"^

n=Q - c o

00

= 2C/^"W ( , ,> (16-93)

—oo

which completes the derivation of Eq 16.82

We may also express the rms value of a periodic function in terms of

the complex Fourier coefficients From Eqs 16.81,16.87, and 16.89,

^rms = \ K + 2 ^ V ^ ' (16-94)

, , Vfl?, + bl

Substituting Eqs 16.95 a n d 16.96 into E q 16.94 yields the desired

expression:

« = i

Example 16.6 illustrates the process of finding the exponential

Fourier series representation of a periodic function

16.8 The Exponential Form of the Fourier Series 629

Find the exponential Fourier series for the periodic

voltage shown in Fig 16.15

v(t)

-T/2 0 T/2 T-T/2 T T+T/2

Figure 16.15 A The periodic voltage for Example 16.6

Solution

Using - T / 2 as the starting point for the integration,

we have, from Eq 16.83,

C,

T

V,

T/2

V m e-' ),m ^dt

72

-jnioat

T \ -jnu>o

T/2

-r/2

IV

no}[)T

2V sin nco () T / 2

na)()T

Here, because v(t) has even symmetry, b„ = 0 for all n, and hence we expect C n to be real

Moreover, the amplitude of C n follows a (sin x)/x

distribution, as indicated when we rewrite

C„ = V m T sin (WQ>()T/2)

We say more about this subject in Section 16.9

The exponential series representation of v(t) is

v{t)= 2

T

V,

T

V m r\ sin Qiaw/2) ,

na)QT/2

sin (H(JO 0 T/2)

/ZO»()T/2

2

Jl'Ofil

,)nto (] t

• A S S E S S M E N T PROBLEMS

Objective 4—Be able to calculate the exponential form of the Fourier coefficients for a periodic waveform

16.8 Derive the expression for the Fourier

coeffi-cients C n for the periodic function shown

Hint: Take advantage of symmetry by using the

fact that C n = (a„ — jb n )/2

r(A)

- 2

- 8 - J

T—r 1 r i—r

4 8 12 ITJ-20 24 2T32 36 40 44

f(ms)

Answer: C„ = - / ^ ( 1 + 3 cos ,J f), n odd

16.9 a) Calculate the rms value of the periodic

cur-rent in Assessment Problem 16.8

b) Using C\—C n , estimate the rms value

c) What is the percentage of error in the value obtained in (b), based on the true value found in (a)?

d) For this periodic function, could fewer terms

be used to estimate the rms value and still insure the error is less than 1 %?

Answer: (a) V34 A;

(b) 5.777 A;

(c) -0.93 %;

(d) yes; if C\-C<) are used, the error is

-0.98 %

NOTE: Also try Chapter Problems 16.45 and 16.46

16.9 Amplitude and Phase Spectra

A periodic time function is defined by its Fourier coefficients and its period

In other words, when we know a v , a„, /?„, and T, we can construct / ( / ) , at least theoretically When we know a„ and b n , we also know the amplitude (A n ) and phase angle (-0,,) of each harmonic Again, we cannot, in general,

visualize what the periodic function looks like in the time domain from a description of the coefficients and phase angles; nevertheless, we recognize that these quantities characterize the periodic function completely Thus, with sufficient computing time, we can synthesize the time-domain wave-form from the amplitude and phase angle data Also, when a periodic driv-ing function is excitdriv-ing a circuit that is highly frequency selective, the Fourier series of the steady-state response is dominated by just a few terms Thus the description of the response in terms of amplitude and phase may provide an understanding of the output waveform

We can present graphically the description of a periodic function in terms of the amplitude and phase angle of each term in the Fourier series

of f(t) The plot of the amplitude of each term versus the frequency is

called the amplitude spectrum of f(t), and the plot of the phase angle

ver-sus the frequency is called the phase spectrum of / ( / ) Because the

ampli-tude and phase angle data occur at discrete values of the frequency (that

is, at <t>(), 2tt)(), 3 w0, ) , these plots also are referred to as line spectra

An Illustration of Amplitude and Phase Spectra

Amplitude and phase spectra plots are based on either Eq 16.38 (A„ and -0,,) or Eq 16.82 (C„) We focus on Eq 16.82 and leave the plots based on

Eq 16.38 to Problem 16.49 To illustrate the amplitude and phase spectra, which are based on the exponential form of the Fourier series, we use the periodic voltage of Example 16.6 To aid the discussion, we assume that

V m - 5 V and r = 7/5 From Example 16.6,

•r T T T S /

1.0-H 0.6 0'.4

0!2 _ L

10 - 8 - 6 - 4 - 2

- 0 4

K.rm

4 6 8 10

ft

Figure 16.16 • The pLot of C„ versus n when r - T/5,

for the periodic voltage for Example 16.6

Figure 16.17 A The plot of (sin x)/x versus x

C„ = V m T sin (rt«0r/2)

which for the assumed values of V,„ and T reduces to

C„= 1 sin (/277-/5)

Figure 16.16 illustrates the plot of the magnitude of C n from Eq 16.99 for

values of n ranging from - 1 0 to +10 The figure clearly shows that the

amplitude spectrum is bounded by the envelope of the |( sin -\:)/x| func-tion We used the order of the harmonic as the frequency scale because the

numerical value of 7 is not specified When we know T, we also know o>0

and the frequency corresponding to each harmonic

Figure 16.17 provides the plot of |(sinx)/x| versus x, where x is in radians It shows that the function goes through zero whenever x is an

integral multiple of TT From Eq 16.98,

tuo ()

/777 T

T

/77T

From Eq 16.100, we deduce that the amplitude spectrum goes through

zero whenever m/T is an integer For example, in the plot, T/T is 1/5, and therefore the envelope goes through zero at n = 5, 10,15, 10, 15, and so

16.9 Amplitude and Phase Spectra 6 3 1

on In other words, the fifth, tenth, fifteenth, harmonics are all zero As

the reciprocal of T/T becomes an increasingly larger integer, the number

of harmonics between every 77 radians increases If mr/T is not an integer,

the amplitude spectrum still follows the |( sin x)/x\ envelope However,

the envelope is not zero at an integral multiple of w0

Because C„ is real for all n, the phase angle associated with C„ is

either zero or 180°, depending on the algebraic sign of (sin /i7r/5)/(mr/5)

For example, the phase angle is zero for n = 0, ± 1 , ±2, ±3, and ±4 It is

not defined at n = ±5, because C±$ is zero The phase angle is 180° at

n - ±6, ±7, ± 8 , and ±9, and it is not defined at ±10 This pattern repeats

itself as n takes on larger integer values Figure 16.18 shows the phase

angle of C„ given by Eq 16.98

Now, what happens to the amplitude and phase spectra if f(t) is

shifted along the time axis? To find out, we shift the periodic voltage in

Example 16.6 t {) units to the right By hypothesis,

v(t) = 2 c^' jn&rf

/ 1 = - 0 0

(16.101)

Therefore

V(t - '0) = 2 C n e J,H0 ^-^ = 2 C n e- }na> ^e j,lw «', (16.102)

which indicates that shifting the origin has no effect on the amplitude

spec-trum, because

\C n \ = \C n e-''^% [16.103)

However, reference to Eq 16.87 reveals that the phase spectrum has

changed to -(0,, + na) 0 t () ) rads For example, let's shift the periodic voltage

in Example 16.1 r/2 units to the right As before, we assume that r = T/5:

then the new phase angle d' n is

B' n = -(0,, + nir/5) (16.104)

We have plotted Eq 16.104 in Fig 16.19 for n ranging from - 8 to +8 Note

that no phase angle is associated with a zero amplitude coefficient

You may wonder why we have devoted so much attention to the

amplitude spectrum of the periodic pulse in Example 16.6 The reason is

that this particular periodic waveform provides an excellent way to

illus-trate the transition from the Fourier series representation of a periodic

function to the Fourier transform representation of a nonperiodic

func-tion We discuss the Fourier transform in Chapter 17

180°

90°

++++

15 - 1 3 - 1 1 - 9 - 7 - 5 - 3 - 1

- 1 4 - 1 2 - 1 0 - 8 - 6 - 4 - 2

M M > M M

1 3 5

2 4

k k k k ]

9 11 13 15

8 10 12 14 16

0' fl

216°

144°

72°

_L_

-8 - 6 - 4 - 2

- 7 2 c

-144°

-216°

Figure 16.18 A The phase angle of C„ Figure 16.19 A The plot of B'„ versus n for Eq 16.104

• A S S E S S M E N T P R O B L E M

Objective 4—Be able to calculate the exponential form of the Fourier coefficients for a periodic waveform

Answer:

16.10 The function in Assessment Problem 16.8 is

shifted along the time axis 8 ms to the right

Write the exponential Fourier series for the

periodic current

NOTE: Also try Chapter Problems 16.49 and 16.50

A 60 -I

Ot) = " 2 - ( 1 + 3 cos ^ )e - 0 V / 2 ) ( « + i ) e ; ^ A >

7T „ _

w =-oo(odd)

Practical Perspective Active High-Q Fitters

Consider the narrowband op amp bandpass filter shown in Fig 16.20(a) The square wave voltage shown in Fig 16.20(b) is the input to the filter We know that the square wave is comprised of an infinite sum of sinusoids, one sinusoid at the same frequency as the square wave and all of the remaining sinusoids at integer multiples of that frequency What effect will the filter have on this input sum of sinusoids?

100 nF / ? 3 | l O k n

100 nF

«—•

+

(a)

v g (V)

15.65TT

•tips)

- 5 0 T T - 3 ' 5 T T - 2 5 T T - 1 5TT 0 12 5-7T 25TT 3 7 5 T T 50TT

15.6577 (b)

Figure 16.20 • (a) narrowband bandpass filter; (b) square wave input

The Fourier series representation of the square wave in Fig 16.20(b) is given by

vjt) = — V — sin —— cos ncont

77 «=fe„« 2

where A = 15.65TTV The first three terms of this Fourier series are given by

v g (t) = 62.6 cos oj 0 t - 20.87 cos 3«of + 12.52 cos 5co 0 t -

The period of the square wave is 507r /AS SO the frequency of the square wave is 40,000 rad/s

The transfer function for the bandpass filter in Fig 16.20(a) is

H(S) = -= r

s 2 + ps + oil

where K = 400/313, p = 2000 rad/s, a> 0 = 40,000 rad/s This filter has

a quality factor of 40,000/2000 = 20 Note that the center frequency of the bandpass filter equals the frequency of the input square wave

Multiply each term of the Fourier series representation of the square

wave, represented as a phasor, by the transfer function H(s) evaluated at

the frequency of the term in the Fourier series to get the representation of the output voltage of the filter as a Fourier series:

vM) = - 8 0 cos ujtf - 0.5 cos 3u) 0 t + 0.17 cos 5<o Q t -

Notice the selective nature of the bandpass filter, which effectively amplifies the fundamental frequency component of the input square wave and attenuates all of the harmonic components

Now make the following changes to the bandpass filter of Fig 16.20(a)

— let R x = 391.25 ft, R 2 = 74.4 O , R 3 = 1 k O , and C l = C 2 = 0.1 /xF

The transfer function for the filter, H(s), has the same form given above, but now K = 400/313, /3 = 20,000 rad/s, w0 = 40,000 rad/s The pass-band gain and center frequency are unchanged, but the pass-bandwidth has increased by a factor of 10 This makes the quality factor 2, and the result-ing bandpass filter is less selective than the original filter We can see this

by looking at the output voltage of the filter as a Fourier series:

vo(0 = — 80 cos a> 0 t — 5 cos 3w0r + 1.63 cos 5<o Q t —

The fundamental frequency of the input has the same amplification fac-tor, but the higher harmonic components have not been attenuated as

signif-icantly as they were when the filter with Q = 20 was used Figure 16.21 plots

the first three terms of the Fourier series representations of the input square wave and the resulting output waveforms for the two bandpass filters Note the nearly perfect replication of a sinusoid in Fig 16.21(b) and the distortion that results from the use of a less-selective filter in Fig 16.21(c)

t(txs)

Figure 16.21 A (a) The first three terms of the Fourier series of the square wave in Fig 16.20(b); (b) the first three terms of the Fourier

series of the output from the bandpass filter in Fig 16.20(a), where Q = 20; (c) the first three terms of the Fourier series of the output from the bandpass filter in Fig 16.20(a) with component values changed to give Q = 2

Summary

• A periodic function is a function that repeats itself

everv T seconds

Five types of symmetry are used to simplify the compu-tation of the Fourier coefficients:

A period is the smallest time interval (T ) that a

peri-odic function can be shifted to produce a function

iden-tical to itself (See page 604.)

The Fourier series is an infinite series used to represent

a periodic function The series consists of a constant

term and infinitely many harmonically related cosine

and sine terms (See page 607.)

The fundamental frequency is the frequency determined

by the fundamental period ( /0 = 1/7 or o>o = 2TJ-/ 0 )

(See page 607.)

The harmonic frequency is an integer multiple of the

fundamental frequency (See page 607.)

The Fourier coefficients are the constant term and the

coefficient of each cosine and sine term in the series

(See Eqs 16.3-16.5.) (See page 608.)

• even, in which all sine terms in the series are zero

• odd, in which all cosine terms and the constant term

are zero

• half-wave, in which all even harmonics are zero

• quarter-wave, half-wave, even, in which the series

contains only odd harmonic cosine terms

• quarter-wave, half-wave, odd, in which the series

con-tains only odd harmonic sine terms (See page 611.)

In the alternative form of the Fourier series, each har-monic represented by the sum of a cosine and sine

term is combined into a single term of the form

A n cos(nco {) t - 0,,) (See page 617.)

Problems 635

For steady-state response, the Fourier series of the

response signal is determined by first finding

the response to each component of the input signal The

individual responses are added (super-imposed) to

form the Fourier series of the response signal The

response to the individual terms in the input series is

found by either frequency domain or s-domain analysis

(See page 619.)

The waveform of the response signal is difficult to obtain

without the aid of a computer Sometimes the frequency

response (or filtering) characteristics of the circuit can

be used to ascertain how closely the output waveform

matches the input waveform (See page 620.)

Only harmonics of the same frequency interact to

pro-duce average power The total average power is the sum

of the average powers associated with each frequency (See page 623.)

The rms value of a periodic function can be estimated from the Fourier coefficients (See Eqs 16.81,16.94, and 16.97.) (See page 626.)

The Fourier series may also be written in exponential form by using Euler's identity to replace the cosine and sine terms with their exponential equivalents (See page 627.)

The Fourier series is used to predict the steady-state response of a system when the system is excited by a periodic signal The series assists in finding the steady-state response by transferring the analysis from the time domain to the frequency domain

Problems

Sections 16.1-16.2

16.1 Find the Fourier series expressions for the periodic

voltage functions shown in Fig P16.1 Note that

Fig P16.1(a) illustrates the square wave; Fig P16.1(b)

illustrates the full-wave rectified sine wave, where

v(t) = V m sm(<ir/T)t, 0 < t < T; and Fig P16.1(c)

illustrates the half-wave rectified sine wave, where

v(t) = Kmsin(27r/7)/,0 < t < T/2

Figure P16.1

v(t)

-T

V,n

0

-V It

(a)

16.2 Derive the Fourier series for the periodic voltage

shown in Fig P16.2, given that

2*7T

v(t) = 100 sin — / V ,

2TT

v(t) = 60 sin — t V,

0 < t <

t < T

T

Figure P16.2

v(t) V

T/4 T/2 37/4 T

16.3 For each of the periodic functions in Fig P16.3, specify

a) (o a in radians per second

b) f 0 in hertz

c) the value of a v

d) the equations for a k and b k

e) v(t) as a Fourier series