Higher Engineering Mathematics, Sixth Edition

146 Higher Engineering Mathematics

reaches 3A. Sketch one cycle of the waveform

showing relevant points.

⎡

⎢

⎢

⎢

⎢

⎢⎢

⎣

(a)5A,20ms,50Hz,

24 ◦ 45 ′lagging

(b)− 2 .093A

(c) 4 .363A

(d) 6 .375ms

(e) 3 .423ms

⎤

⎥

⎥

⎥

⎥

⎥⎥

⎦

14.6 Harmonic synthesis with

complex waveforms

A waveform that is not sinusoidal is called acomplex

wave.Harmonic analysisis the process of resolving a

complex periodic waveform into a series of sinusoidal

components of ascending order of frequency. Many of

the waveforms met in practice can be represented by the

following mathematical expression.

v=V 1 msin(ωt+α 1 )+V 2 msin( 2 ωt+α 2 )

+···+Vnmsin(nωt+αn)

and the magnitude of their harmonic components

together with their phase may be calculated using

Fourier series(see Chapters 66 to 69).Numerical

methodsareusedtoanalysewaveformsforwhich

simple mathematical expressions cannot be obtained.

A numerical method of harmonic analysis is explained

intheChapter70onpage637. Inalaboratory, waveform

analysis may be performed using awaveform analyser

whichproducesadirect readout ofthecomponent waves

present in a complex wave.

By adding the instantaneous values of the fundamen-

tal and progressive harmonics of a complex wave for

given instantsin time, the shape of a complex waveform

can be gradually built up. This graphical procedure is

known asharmonic synthesis(synthesis meaning ‘the

putting together of parts or elements so as to make up a

complex whole’).

Some examples of harmonic synthesis are con-

sidered in the following worked problems.

Problem 17. Use harmonic synthesis to construct

the complex voltage given by:

v 1 =100sinωt+30sin3ωtvolts.

The waveform is made up of a fundamental wave of

maximum value 100V and frequency,f=ω/ 2 πhertz

and a third harmonic component of maximum value

30V and frequency= 3 ω/ 2 π(= 3 f), the fundamental

and third harmonics being initially in phase witheach

other.

In Fig. 14.31, the fundamental waveform is shown

by the broken line plotted over one cycle, the periodic

timeTbeing 2π/ωseconds. On the same axis is plotted

30sin3ωt, shown by the dotted line, having a maximum

value of 30V and for which three cycles are completed

in timeTseconds. At zero time, 30sin3ωtis in phase

with 100sinωt.

The fundamental and third harmonic are combined by

adding ordinates at intervals to produce the waveform

forv 1 , as shown. For example, at timeT/12seconds,

the fundamental has a value of 50V and the third har-

monic a value of 30V. Adding gives a value of 80V for

waveformv 1 at timeT/12seconds. Similarly, at time

T/4seconds, the fundamental has a value of 100V and

the third harmonic a value of−30V. After addition,

the resultant waveformv 1 is 70V atT/4. The proce-

dure is continued betweent=0andt=Tto produce

the complex waveform forv 1. The negative half-cycle

of waveformv 1 is seen to be identical in shape to the

positive half-cycle.

If further odd harmonics of the appropriate amplitude

and phase were added tov 1 a good approximation to a

square wavewould result.

Problem 18. Construct the complex voltage

given by:

v 2 =100sinωt+30sin

(

3 ωt+

π

2

)

volts.

The peak value of the fundamental is 100volts and the

peak value of the third harmonic is 30V. However the

third harmonic has a phase displacement of

π

2

radian

leading (i.e. leading 30sin3ωtby

π

2

radian). Note that,

since the periodic time of the fundamental isTseconds,

the periodic time of the third harmonic isT/3seconds,

and a phase displacement of

π

2

radian or

1

4

cycle of the

third harmonic represents a time interval of(T/ 3 )÷4,

i.e.T/12seconds.

Figure 14.32 shows graphs of 100sinωt and

30sin

(

3 ωt+

π

2

)

over the time for one cycle of the fun-

damental. When ordinates of the two graphs are added

at intervals, the resultant waveformv 2 is as shown.

If the negative half-cycle in Fig. 14.32 is reversed it

can be seen that the shape of the positive and negative

half-cycles are identical.