Masters G.M. Renewable and Efficient Electric Power Systems

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88 FUNDAMENTALS OF ELECTRIC POWER

where ω = 2πf = 2π/T . The coefﬁcients can be found from



f(t)cos nωt dt, n = 0, 1, 2 ... (2.81)

and



f(t)sin nωt dt, n = 1, 2, 3 ... (2.82)

Under special circumstances, the series in (2.80) simpliﬁes. For example, when

there is no dc component to the waveform (average value = 0), the ﬁrst term,

, drops out:

= 0 : when average value, dc = 0 (2.83)

For functions with symmetry about the y-axis, the series contains only cosine

terms. That is,

cosines only: when f(t) = f(−t) (2.84)

For the series to contain only sine terms, it must satisfy the relation

sines only: when f(t) =−f(−t) (2.85)

Finally, when a function has what is called half-wave-symmetry it contains no

even harmonics. That is,

no even harmonics: when f



t +



=−f(t) (2.86)

Examples of these properties are illustrated in Fig. 2.31. Notice the waveform

in Fig. 2.31c is very much like the current waveform for a full-wave rectiﬁer,

which says that its Fourier spectrum will have only cosine terms with only

odd harmonics.

(a)

= 0

sines only

no even harmonics

≠ 0

sines & cosines

even & odd harmonics

= 0

cosines only

no even harmonics

(b) (c)

Figure 2.31 Examples of periodic functions, with indications of special properties of

their Fourier series representations.

POWER QUALITY 89

Example 2.11 Harmonic Analysis of a Square Wave. Find the Fourier series

equivalent of the square wave in Fig. 2.31a, assuming that it has a peak value

of 1 V.

Solution We know by inspection, using (2.83) to (2.86), that the series will

have only sines, with no even harmonics. Therefore, all we need are the even

coefﬁcients b

from (2.82):



f(t)sin nωt dt =





T/2

1 · sin nωt dt +



T/2

(−1) · sin nωt dt



Recall that the integral of a sine is the cosine with a sign change, so



−1

nω

cos nωt



t=T/2

t=0

nω

cos nωt



t=T

t=T/2



nωT



(−1)



cos nω

− cos nω · 0



+ cos nωT − cos nω



Substituting ω = 2πf = 2π/T gives

2π



−cos



2π



+ 1 + cos



2π

· nT



− cos



2π



nπ

[−2cosnπ + 1 +cos 2nπ ]

Since this is half-wave symmetric, there are no even harmonics; that is, n is an

odd number. For odd values of n,

cos nπ = cos π =−1and cos2nπ = cos 0 = 1

That makes for a nice, simple solution:

nπ

That is,

= 1.273,b

3π

= 0.424,b

5π

= 0.255 ...

So the series is

(Square wave with amplitude of 1) =



sin ωt +

sin 3ωt +

sin 5ωt +···



90 FUNDAMENTALS OF ELECTRIC POWER

t t

(a) (b)

Figure 2.32 Showing the sum of the ﬁrst two terms (a) and ﬁrst three terms (b) of the

Fourier series for a square wave along with the square wave that it is approximating.

To show how quickly the Fourier series for the square wave begins to

approximate reality, Fig. 2.32 shows the sum of the ﬁrst two terms of the

series, and the sum of the ﬁrst three terms, along with the square wave that

it is approximating. Adding more terms, of course, will make the approximation

more and more accurate.

The ﬁrst few terms in the harmonic analysis of the square wave derived

in Example 2.11 are presented in tabular form in Table 2.2. Two columns are

shown for the harmonics: one that lists the amplitudes of each harmonic, and the

other listing the harmonics as a percentage of the fundamental amplitude. Both

presentations are commonly used.

Another convenient way to describe the harmonic analysis that the Fourier

series provides is with bar graphs such as those shown in Fig. 2.33 for a

square wave.

An example harmonic spectrum for the current drawn by an electronically

ballasted compact ﬂuorescent lamp (CFL) is shown in Fig. 2.34. Notice only odd

harmonics are present, which is the case for all periodic half-wave symmetric

TABLE 2.2 Harmonic Analysis of a Square Wave

with Amplitude 1

Harmonic Amplitude Percentage of fundamental

1 1.273 100.0%

3 0.424 33.3%

5 0.255 20.0%

7 0.182 14.3%

9 0.141 11.1%

11 0.116 9.1%

13 0.098 7.7%

15 0.085 6.7%

Only harmonics through the 15th are shown.

POWER QUALITY 91

13579111315 13579111315

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

Harmonic

100

110

120

Harmonic

(a) (b)

Percentage

Amplitude

Figure 2.33 First few harmonics for a square wave with amplitude 1: (a) By amplitude

of harmonics. (b) By percentage of fundamental.

waveforms. Also notice that the harmonics are still sizable as far out as they are

shown, which is all the way out to the 49th harmonic!

We can reconstruct the waveform for the CFL whose current is shown in

Fig. 2.34. The bulb is run on a 60-Hz (2π · 60 = 377 rad/s) voltage source so,

taking a few values from Fig. 2.34, we can write the ﬁrst few terms of the

160

140

120

100

Harmonic

Current (mA)

Figure 2.34 Harmonic current spectrum for an 18-W, electronic-ballast compact ﬂuo-

rescent lamp. Currents are rms values.

92 FUNDAMENTALS OF ELECTRIC POWER

current as

i(A)=

√

2(0.145 cos ωt + 0.140 cos 3ωt + 0.132 cos 5ωt

+ 0.115 cos 7ωt +···) (2.87)

where ω = 377 rad/s. Notice the

√

2 converts the rms values of current given in

the ﬁgure to their full magnitude. A plot of (2.87) is shown in Fig. 2.35.

Even with just a few of the harmonics included, the “gulps” of current similar

to those shown in the full-wave rectiﬁer circuit of Fig. 2.23 are clearly evident.

2.8.2 Total Harmonic Distortion

While the Fourier series description of the harmonics in a periodic waveform

contains all of the original information in the waveform, it is usually awkward to

work with. There are several simpler quantitative measures that can be developed

and more easily used. For example, suppose we start with a series representation

of a current waveform that is symmetric about the y axis:

i =

√

2(I

cos ωt + I

cos 2ωt + I

cos 3ωt +···)(2.88)

where I

is the rms value of the current in the nth harmonic. The rms value of

current is therefore

rms



)

avg





√

2(I

cos ωt + I

cos 2ωt + I

cos 3ωt +···)



avg

(2.89)

While algebraically squaring that inﬁnite series in the parentheses looks menac-

ing, it falls apart when we fool around a bit with the algebra. Starting with the

−1

Current (A)

34567

Figure 2.35 The sum of the ﬁrst through seventh harmonics of current for the CFL of

Fig. 2.34.

POWER QUALITY 93

square of a sum of terms

(a + b + c +···)

= a

+ b

+ c

···+2ab + 2ac + 2bc +··· (2.90)

we see that each term is squared, and then in addition there are combinations of

all possible cross products between terms. Therefore,

rms



2[(I

cos ωt)

+ (I

cos 2ωt)

+···+(2I

cos ωt · cos 2ωt)

+ (2I

cos ωt · cos 3ωt) +···]

avg

(2.91)

Fortunately, the average value of the product of two sinusoids of differing fre-

quency is zero; that is,

[cos nωt · cos mωt]

avg

= 0forn = m(2.92)

Also, we note that

cos

nωt)

avg

(2.93)

That leaves

rms





+···





+ I

··· (2.94)

So, the rms value of current when there are harmonics is just the square root of

the sum of the squares of the individual rms values for each frequency. While this

was derived for a Fourier series involving just a sum of cosines, it holds for the

general case in which the sum involves cosines and sines.

In the United States, the most commonly used measure of distortion is called

the total harmonic distortion (THD), which is deﬁned as

THD =



+ I

+···

(2.95)

Notice that since THD is a ratio, it doesn’t matter whether the currents in (2.95)

are expressed as peak values or rms values. When they are rms values, we can

recognize (2.95) to be the ratio of the rms current in all frequencies except the

fundamental divided by the rms current in the fundamental. When no harmonics

are present, the THD is zero. When there are harmonics, there is no particular

limit to THD and it is often above 100%.

Example 2.12 Harmonic Distortion for a CFL. A harmonic analysis of the

current drawn by a CFL yields the following data. Find the rms value of current

and the total harmonic distortion (THD).

94 FUNDAMENTALS OF ELECTRIC POWER

Harmonics rms Current (A)

10.15

30.12

50.08

70.03

90.02

Solution From (2.94) the rms value of current is

rms

= [(0.15)

+ (0.12)

+ (0.08)

+ (0.03)

+ (0.02)

]

1/2

= 0.211 A

From (2.95) the total harmonic distortion is

THD =



(0.12)

+ (0.08)

+ (0.03)

+ (0.02)

0.15

= 0.99 = 99%

so the total rms current in the harmonics is almost exactly the same as the rms

current in the fundamental.

2.8.3 Harmonics and Voltage Notching

As we have seen, diode-rectiﬁers draw currents in bursts and those bursts can be

described in terms of their THD. One effect of these current surges is that there

will be a voltage drop in the connecting lines during those surges, which results in

momentary drops in the voltage that reaches the load. To illustrate this problem,

consider the circuit shown in Fig. 2.36 in which a sinusoidal voltage source sends

current through a line to a load. To keep everything as simple as possible, the

line impedance is modeled as just having resistance with no reactive component.

Figure 2.37 shows what we might expect if current surges are modeled as

simple, rectangular pulses. During current pulses, there will be voltage drop in

the line equal to line current times line resistance. The voltage reaching the load

Load

load

Source

voltage

Line

resistance

load

Figure 2.36 A simple circuit to illustrate voltage notching.

POWER QUALITY 95

load

∆

line

i R

Figure 2.37 Source voltage, line current, voltage drop across the line, and resulting

notched voltage to the load.

will correspondingly be decreased by that line voltage drop. In between pulses,

the full source voltage reaches the load. The result is a “notch” in the load voltage

as shown. Also, there will be harmonic distortion, this time on the voltage signal

rather than the current, that can be analyzed and measured in the same ways we

just dealt with current harmonic distortion.

It is important to note that the nonlinear load causing the notch may also be

affected by the notch. Moreover, other loads that may be on the same circuit

will also be affected. Those current surges could even cause notches in the volt-

age delivered to neighboring facilities. When notching is a potential problem,

capacitors may be added to help smooth the waveform.

2.8.4 Harmonics and Overloaded Neutrals

One of the most important areas of concern associated with harmonics is the

potential for the neutral wire in four-wire, three-phase, wye-connected systems

to overheat, potentially to a degree sufﬁcient to initiate combustion (Fig. 2.38).

The current in the neutral wire is always given by

= i

+ i

(2.96)

For balanced three-phase currents, with no harmonics, we know from (2.56) and

(2.60) that the current in the neutral wire is zero:

√

phase

[cos ωt + cos(ωt + 120

◦

) + cos(ωt − 120

◦

)] = 0 (2.97)

96 FUNDAMENTALS OF ELECTRIC POWER

Load A

Load B

Load C

−

Figure 2.38 Showing the neutral line current in a four-wire, three-phase, wye-

connected circuit.

In setting up a building’s electrical system, every effort is made to distribute the

loads so that approximate balance is maintained, with the intention of minimizing

current ﬂow in the neutral. Under the assumption that the neutral carries little

or no current, early building codes allowed smaller neutral wires than the phase

conductors. It has only been since the mid-1980s that the building code has

required the neutral wire to be a full-size conductor.

The question now is, What happens when there are harmonics in the phase

currents? Even if the loads are balanced, harmonic currents cause special prob-

lems for the neutral line. Suppose that each phase carries exactly the same current

(shifted by 120

◦

of course), but now there are third harmonics involved. That is,

√

2[I

cos ωt + I

cos 3ωt]

√

2{I

cos(ωt + 120

◦

) + I

cos[3(ωt + 120

◦

)]} (2.98)

√

2{I

cos(ωt − 120

◦

) + I

cos[3(ωt − 120

◦

)]}

Now, notice what happens to the third harmonics in phases B and C due to

the following:

cos[3(ωt ± 120

◦

)] = cos(3ωt ± 360

◦

) = cos 3ωt (2.99)

This means that the current in the neutral wire, which is the sum of each of the

three phase currents, is

√

2[I

cos ωt + I

cos(ωt + 120

◦

) + I

cos(ωt − 120

◦

) + I

cos 3ωt

+ I

cos 3ωt + I

cos 3ωt] (2.100)

Since the sum of the three cosines of equal magnitude—but each out of phase

with the other by 120

◦

—is zero, the fundamental currents drop out and we are

left with

= 3

√

cos 3ωt (2.101)

POWER QUALITY 97

In terms of rms values,

= 3I

(2.102)

That is, the rms current in the neutral line is three times the rms current in each

line’s third harmonic. There is considerable likelihood, therefore, that harmonic-

generating loads will cause the neutral to carry even more current than the phase

conductors, not less!

The same argument about harmonics adding in the neutral applies to all of

the harmonic numbers that are multiples of 3 (since 3 × n × 120

◦

= n360

◦

= 0

◦

That is, the third, sixth, ninth, twelfth, ...harmonics all add to the neutral current

in an amount equal to three times their phase-current harmonics. Notice, by the

way, that harmonics not divisible by 3 cancel out in the same way that the

fundamental cancels—for example, the second harmonic:

n−2nd harmonic

√

{cos(2ωt) + cos[2(ωt + 120

◦

)] + cos[2(ωt − 120

◦

)]}

(2.103)

The terms in the bracket are

cos(2ωt) + cos(2ωt + 240

◦

) + cos(2ωt − 240

◦

) = cos 2ωt + cos(2ωt − 120

◦

)

+ cos(2ωt + 120

◦

) = 0

so the second harmonic currents cancel just as the fundamental did. This will be

the case for all harmonics not divisible by 3.

For currents that show half-wave symmetry, there are no even harmonics, so

the only harmonics that appear on the neutral line for balanced loads of this sort

will be third, 9th, 15th, 21st, ... etc. These odd harmonics, divisible by three,

are called triplen harmonics.

Example 2.13 Neutral-Line Current. A four-wire, wye-connected balanced

load has phase currents described by the following harmonics:

Harmonic rms Current (A)

1 100

350

520

710

11 4

13 2

Find the rms current ﬂowing in the neutral wire and compare it to the rms

phase current.