waveform analysis

8/13/2019 Waveform Analysis

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1

Analysis of Non-Sinusoidal WaveformsJ . R. Lucas

DC and AC Waveforms

Up to the present, analysis has been carried out with

direct waveforms and sinusoidal alternating waveforms

Figure 1(a)direct waveform Figure 1(b)sinusoidal waveform

a t a ta t = A

a(t) = Amsin(t+)


2/116

Non-sinusoidal Waveforms J R Lucas May 20112

Other Waveformsa t a t a t

a t

t

a t a t

t

T

T T

a b c

d e (f)

t t

a t

t

(h) (i)

Fi ure 2


3/116


(a), (b), (c) and (d) are uni-directional, although not purely direct.

(e) and (f) are repetitive with zero mean value,

(c), (h) and (i) are repetitive waveforms with finite mean values.(g) is alternating but non-repetitive and mean value is non-zero.

Basically two groups of waveforms

repetitivenon-repetitive (will be dealt with later)Repetitive waveform

only one period T needs to be definedcan be broken up to a fundamental component

(corresponding to period T) and its harmonics.

uni-directional term (direct component) may exist.


4/116


Fourier SeriesNamed after the French mathematician who first

presented the series in 1822.o= 2/Tf(t) = Fo+ F1cos ( t+1) + F2cos (2 t+2)

+ F3cos (3 t+3) + F4cos (4 t+4)+ F5cos (5 t+5) +

f(t) = Ao/2 + A1cos t + A2cos 2 t + A3cos 3 t+ A4cos 4 t + + B1sin t + B2sin 2 t + B3sin 3 t +

)sincos(2

)(1

tnBtnAAtf on

non

o


5/116


Symmetry in Waveforms

Many periodic waveforms exhibit symmetry.

Use of symmetry helps reduce tedious calculations

(i) Even symmetry

(ii) Odd symmetry

(iii) Half-wave symmetry

Even Symmetry

Region beforey-axisis mirror image of region aftery-axis.

f(t) = f(-t)


6/116


a(t)

t

(a)

a(t)

t

(b)

a(t)

t

(c)

a t

t

(d)

a t

t

(e)

t

(f)

a t

Figure 3Waveforms with Even symmetry


7/116


Simplest waveforms with even symmetry

cosinewaveformfigure 3 (a)

directwaveformfigure 3 (b).Even symmetry can exist in both periodic and non-periodic waveforms.

Both direct terms as well as varying terms can exist in

such waveforms.

If waveform is defined for only t0,the remainder is automatically known by symmetry.


8/116


Odd SymmetryRegion before y-axis is negative of mirror image of region

aftery-axis.

f(t) = () f(t)a t

t

(a)

a t

t

(b)

a t

t

(c)

a t

t

(d)

a t

t

(e)

t

(f)

a t

Figure 4Waveforms with Odd symmetry


9/116


Simplest waveforms with odd symmetry

sinewaveformfigure 4 (a)

rampwaveformfigure 4 (b).Odd symmetry can exist in both periodic and non-periodic waveforms.

However, only varying terms can exist in such

waveforms.

Direct terms cannot exist in odd waveforms.

If the waveform is defined for only t0,the remainder is automatically known by symmetry.


10/116


Decomposition into odd and even functions

Waveforms in general need not be either odd or even.

They can be split into a combination of odd and even.Let f(t) = fodd(t) + feven(t)

Then f(-t) = fodd(-t) + feven(-t) = fodd(t) + feven(t)Addition and subtraction thus gives

)]()()(21 tftftfodd

and

)]()()(2

1 tftftfeven

verifying the assumption.


11/116


Half-wave Symmetry

A function f(t)exhibits half-wave symmetry, when one

half of waveform is exactly equal to the negation of

previous or next half of waveform.

i.e. )()()()()( 22TT tftftf

t

(a)

a(t)

t

(b)

a(t)

t

(c)

Figure 5Waveforms with Half-wave symmetry

a(t)


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The simplest form ofHalf-wave Symmetryissinusoidal

waveformin figure 5(a).

Half-wave symmetry can only exist in periodic

waveforms.

Only varying terms can exist in such waveforms.

Direct terms cannot exist in half-wave symmetrical

waveforms.If the waveform is defined for only one half cycle, not

necessarily starting from t=0, the remaining half of the

waveform is automatically known by symmetry.


13/116


Some useful Trigonometric Properties

0.sin Tt

to

o

o

dtt

0.cos Tt

to

o

o

dttwith T = 2

0.sin Tt

t

o

o

o

dttn

0.cos Tt

t

o

o

o

dttn

0.cos.sin

Tt

too

o

odttmtn for all values of mand n

mnwhen

mnwhen0.sin.sin

2T

Tt

too

o

o

dttmtn

mnwhen

mnwhen0.cos.cos

2T

Tt

t

oo

o

o

dttmtn


14/116


Evaluation of Coefficients Anand Bn

)sincos(

2

)(

1

tnBtnAAtf on

n

ono

First term of Fourier Series is written as Ao/2 rather than Ao.

Zero belongs equally to positive half and negative half.

gives only Aofor positive half

evaluated with same general expression for An, n = 0.Ao/2also corresponds to direct component of waveform.

3o 2o o 0 o 2o 3o Negative half Positive hal


15/116


Tt

t

onn

on

Tt

t

oTt

t

dttnBtnAdtA

dttf0

0

0

0

0

0

)sincos(2

)(1

using properties of trigonometric functions,only first term can give a non zero integral.

i.e.

TA

dtA

dttf oTt

t

o

Tt

t

2

0

2

)(0

0

0

0

Tt

to

o

o

dttfT

A )(2

or mean value= Tt

t

oo

o

dttfT

A )(1

2 gives same result forAo.


16/116


To determine remaining values of An,

multiply each term by cos mot and integrate.

Tt

t

oon

n

on

Tt

t

oo

Tt

t

o dttmtnBtnAdttm

A

dttmtf

0

0

0

0

0

0

cos)sincos(cos2cos)( 1

using properties of trigonometric functions,

only cos notterm on right hand side of equation will give

a non zero integral, and that too only when m=n.

Tt

t

on

o

o

dttntfT

A cos)(2

Similarly

Tt

t

on

o

o

dttntfT

B sin)(2


17/116


Analysis of Symmetrical Waveforms

Even Symmetry

0 to T/2 corresponds to mirror image fromT/2 to 0.

Consider period from T/2 to T/2 for integration.

Figure 6Analysis of even waveform

f(t)

t

T

T/2T/2


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T

T

on dttntfT

A21

21

cos)(2

2

0

0

2

cos)(2

cos)(2 T

oT

on dttntfTdttntf

TA

In first part of expression, t is replaced by t

2

0

0

2

cos)(2

)()(cos)(2

T

oT

on dttntfT

dttntfT

A

Since function is even,

f(t) = f(t), and cos(nt) = cos(nt).Thus equation simplifies to


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2

0

0

2

cos)(2

)(cos)(2

)(

T

oT

on dttntfTdttntf

TA

Interchanging upper and lower limits of first integralremoves negative sign in front of first integral.

Thus

2

0

cos)(22T

on dttntfT

A

In a similar manner

2

0

0

2

sin)(2

)()(sin)(2

T

oT

on dttntfT

dttntfT

B

f(t) = f(t), and sin(nt) = sin(nt).In this case the two integrals are equal in magnitude but

have opposite signs so that they cancel out on addition.


20/116


For waveforms witheven symmetry

Anis calculated as twice integral over half cycle from zero.

Bn= 0 for all values of n

Can only have cosineterms and a direct term.

1

cos

2

)(

n

ono tnAAtf

where

2

0

cos)(4T

on dttntfT

A


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Odd Symmetry

From 0 to T/2 corresponds to negated mirror image of

waveform fromT/2 to 0.Consider period from T/2 to T/2 for integration.

f(t) = f(t)

Figure 7Analysis of odd waveform

f(t)

t

T

T/2T/2


22/116


T

T

on dttntfT

A21

21

cos)(2

2

0

0

2

cos)(2

cos)(2 T

oT

on dttntfTdttntf

TA

In first part of expression, t is replaced by t.

2

0

0

2

cos)(2

)()(cos)(2

T

o

T

on dttntfTtdtntf

TA

0cos)(2

cos)(2

)(2

0

2

0

T

o

T

on dttntfTdttntf

TA

for all nfor odd waveform


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In a similar way,

2

0

sin)(22T

on dttntf

T

B

For waveforms withodd symmetry

An= 0 for all values of n

Bnis calculated as twice integral over half cycle from zero.

Can only have sineterms and no direct term

1

sin)(n

on tnBtf

where 2

0

sin)(4T

on dttntfT

B


24/116


Half-wave Symmetry

Half-cycle from

(to+T/2) to (to+T)

corresponds to

negated value of

previous half cycle

from to to (to+T/2).

Tt

t

on

o

o

dttntfT

A cos)(2

Tt

Tt

o

Tt

t

on

o

o

o

o

dttntfTdttntf

TA

2

2

cos)(2cos)(2

In second part of expression t is replaced by tT/2.

T

f t

t

Figure 8Analysis of waveform with

half wave symmetry


25/116


22

)2

()2

(cos)2

(2

cos)(2

Tt

t

o

Tt

t

on

o

o

o

o

TtdTtnTtfTdttntf

TA

f(t T/2) = f(t) for half-wave symmetry,and since oT = 2 cos no(tT/2) = cos (notn)has a value of ()cos not when n is odd, andhas a value of cos not when n is even.

Thus

2

cos)(22Tt

t

on

o

o

dttntfT

A when n is odd

and An= 0 when n is even


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Similarly

2

sin)(22Tt

t

on

o

o

dttntfT

B when n is odd

and Bn= 0 when n is even

With half-wave symmetry,

even harmonics do not exist for odd harmonics both coefficients An and Bn can be

obtained by taking double the integral over any half cycle.

Most practical waveforms have half-wave symmetry

due to natural causes.


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Summary for waveforms with symmetr ical properties

1. With even symmetry,

Bnis 0 for all n,

Anis twice integral over half a cycle from zero.

2. With odd symmetry,

Anis 0 for all n,

Bnis twice integral over half a cycle from zero.

3. With half-wave symmetry,

Anand Bnare 0 for even n,

Anand Bnare twice the integral over any half cycle

for odd n.


28/116


4. With half-wave symmetryand either evensymmetry or odd symmetrypresent,

Anand Bnare 0 for even n,

four times integral over quarter cycle from time

zero, for odd n for Anor Bn

zero for the remaining coefficient Bnor An.

5.Note that in any waveform,

Ao/2corresponds to mean value of waveform;

a symmetrical property may sometimes be obtained

by subtracting mean value from waveform.


29/116


Piecewise Continuous waveforms

Most practical waveforms arecontinuous and single valued

(i.e. having a single value atany particular instant).When sudden changes occur(such as in switching), or in

square waveforms, near vertical lines could occur givingmulti-value instants.As long as these multi-values occur over finite bounds,the waveform is single-valued and continuous in pieces,or said to be Piecewise continuous.Analysis can be carried out using the Fourier Series for

both continuous or piecewise continuous waveforms.

Figure 9

Piecewise continuous waveform


30/116


31/116


Example 1

Find the Fourier Series of

the piecewise continuous

rectangular waveformshown in figure 10.

Solution

Period of waveform = 2TMean value of waveform = 0. Ao/2 = 0Waveform has even symmetry. Bn= 0 for all n

Waveform has half-wave symmetry.An, Bn= 0 for even n

a(t)

t

E

-E

0 T/2 3T/2-T/2-3T/2

Figure 10Rectangular waveform


32/116


Ancan be obtained, for odd values of n,as 4 times integral over quarter cycle as follows.

42

0cos)(2

24T

on dttntaTA for odd n

2sin

4

2sin

4sin

4cos

4

0

22

0

n

n

ETn

Tn

EtnE

TndttnE

T

o

o

T

o

o

T

o

i.e. A1= 4E/,A3=4E/3A5=4E/5

A7= 4E/7

a(t) =

.......7

7cos

5

5cos

3

3coscos

4 tttt

E oooo


33/116


Figure 11 shows synthesis of the rectangular waveform

using the Fourier components obtained.

Figure 11Fourier Synthesis of Rectangular Waveform


34/116


The waveforms shown correspond to

(i) original waveform,

(ii) fundamental component only,

(iii) fundamental + third harmonic,

(iv) fundamental + third harmonic + fifth harmonic,

(v) fundamental, third, fifth and seventh harmonics.

It can be seen that with addition of each component, the

waveform approaches original waveform more closely,

however without an infinite number of components itwill never become exactly equal to original.

F f f i h i fi 12


35/116


Frequency spectrum of waveform is shown in figure 12.

Now consider the same rectangular waveform but with

a few changes.

Figure 12Line Spectrum

0 o 2o 3o 4o 5o 6o 7o t

Amplitude

E l 2b(t)


36/116


Example 2

Find the Fourier series

of the waveform shown

in figure 13.

Solution

Waveform does not

have any symmetrical properties, although it is virtuallysame waveform as in example 1.

Period = 2T,

mean value = E/2If E/2 is subtracted from the waveform b(t), half wave

symmetry is observed.

b(t)

t

2E

-E0 2T/3 5T/3-T/3-5T/3


If f i hif d b T/6 l f i


37/116


If waveform is shifted by T/6 to left, even symmetry is

observed.

Consider waveform

b1(t) = b(t T/6) +E/2,shown in figure 14.

b1(t) differs from

waveform a(t) infigure 10 in magnitude only (1.5 times).

Thus b1(t) can be obtained directly from earlier analysis.

b1(t) = 1.5a(t) = .......

77cos

55cos

33coscos6 ttttE oooo

where o2T = 2

b1(t)

t

3E/2

-3E/2

0 T/2 3T/2-T/2-3T/2

Figure 14Modified waveform

b( ) b ( T/6) E/2 l T/6 /6


38/116


b(t) = b1(t +T/6) + E/2, also oT/6 = /6

=

.......

7

)6

77cos(

5

)6

55cos(

3

)2

3cos()

6cos(

6

2

tttt

EE oooo

If problem was worked from first principles the resultant would be

above answer.

Figure 15 shows the corresponding line spectrum.

Only differences from earlier are that amplitudes are 1.5 times

higher and a d.c. term is present.

Figure 15Line Spectrum0 o 2o 3o 4o 5o 6o 7o t

Amplitude

E l 3


39/116


Example 3

Find Fourier Series of

triangular waveform

shown in figure 16.

Solution

Period of waveform = 2T, o.2T = 2Mean value = 0. Ao/2= 0Has odd symmetry. An = 0 for all nHas half-wave symmetry. An, Bn = 0 for even n

4

2

0

sin2

2

24T

on dttntT

E

TB

=dt

n

tn

T

E

n

tntT

ET

o

o

T

o

o 2

02

2

0

2

cos8cos8

for odd n

y(t)

t

E

-E

0 T/2 2T-T-2T

Figure 16Triangular waveform

T

TT


40/116


= 222 )(

02

(sin82cos

2

8 )

o

o

o

o

n

n

T

E

n

nT

T

E TT

= 2)(2

sin82

cos4

n

nE

n

nE

Substituting values

B1= 8E/2, B3 = 8E/(32, B5 = 8E/(52, B7 = 8E/(72, .

y(t) =

...........

7

7sin

5

5sin

3

3sinsin

82222

tttt

E oooo

Consider the derivative of theoriginal waveform y(t),

has waveform shown in figure

17. This corresponds to same

type of rectangular waveform inexample 1, except that the

amplitude is 2/T times higher.

a(t)

t

2E/T

-2E/T

0 T/2 3T/2-T/2-3T/2



41/116

E l 4 a(t)


42/116


Example 4

Find Fourier series of

the waveform shownin figure 18.

Solution

Period = T, T = 2 Mean value = 0, Ao/2 = 0Has half-wave symmetry. even harmonics absent.

2

4

4

0

2

0

cos)4

(4

cos4

cos)(4

T

To

T

o

T

on dttntT

EE

TdttnE

Tdttntf

TA

=

2

44

24

0

sin)

4(

4sin)

4(

4sin4T

T o

o

T

T

o

o

T

o

o dtn

tn

T

E

Tn

tntT

EETn

tnET

a(t)

t

E

-E0 T/4 3T/4-T/4-3T/4

Figure 18Periodic waveform

)(ii TTT


43/116


= 4

2

22

24

)(

)cos(16sin)(

4sin4T

T

o

o

o

To

o

To

n

tn

T

E

n

nE

Tn

nET

since T = 2 T/4 = and T/2 =

2)2(

)2

coscos(

162

sin4

2

2sin

4

n

nn

En

nE

n

n

EAn for odd n

Substituting different values of n, we have

A1= 24

02

EE

= 1.0419E

A3= -0.1672E, A5= 0.1435E, A7= -0.08267ESimilarly, the Bnterms for odd n are given as follows.

TTT


44/116


2

4

4

0

2

0

sin)4

(4

sin4

sin)(4

T

To

T

o

T

on dttntT

EE

TdttnE

Tdttntf

TB

=

2

44

2

4

0 )(

cos)

4(

4

)(

cos)

4(

4

)(

cos4T

T o

oT

T

o

o

T

o

o dtn

tn

T

E

Tn

tntT

EETn

tnET

= 4

2

22

24

)(

sin)(16cos4)cos1(4T

T

o

o

o

To

o

To

n

tn

T

E

n

nETn

nET

since T = 2 T/4 = and T/2 =

2

)2(

)2

sinsin(

16

2

cos4

2

)2

cos1(

4

n

nn

E

n

nE

n

n

EBn

forodd nSubstituting different values of n,

B1= 0.4053E, B3= -0.04503E, B5= 0.0162E, B7= -0.00827

d c term = 0


45/116


d.c. term = 0

amplitude 1 =22 4053.00419.1 = 1.1180, amplitude 3 = 0.1731,

amplitude 5 = 0.1444, amplitude 7 = 0.08309, .

Figure 19 shows synthesised waveform (red) and its components up to

the 29th harmonic (odd harmonics only) along with the originalwaveform (black).

Figure 20 shows line spectrum of waveform of first 7 harmonics.

Figure 19Synthesised waveform Figure 20Line Spectrum

0 o 2o 3o 4o 5o 6o 7o t

Amplitude

Example 5


46/116


Example 5

Figure 21 shows a

waveform from a

power electroniccircuit.

Determine its Fourier

Series.

f(t) is defined as

follows for one cycle.

f(t) = 100 cos 314.16 t for0.333 < t < 2.5 ms

f(t) = 86.6 cos (314.16 t0.5236) for 2.5 < t < 3.0 ms

t (ms)

Figure 21Power electronics waveform

-0.333 0 2.5 3.0 5.833

f(t)

Solution


47/116


SolutionWaveform does not have any symmetrical properties.

T = 0.003333 s, o= 2/T = 1885 rad/s

003.0

0025.0

0025.0

000333.0

cos)(2

cos)(2

dttntfTdttntf

TA oon

003.0

0025.0

0025.0

000333.0

cos)5236.016.314cos(6.862

cos16.314cos1002

dttnt

T

dttnt

T oo

003.0

0025.0

0025.0

000333.0

)5236.016.314cos()5236.016.314cos(

6.86

)16.314cos()16.314(cos(100

dttnttntT

dttnttntT

oo

oo

0025.0

)16314sin()16314sin(100 tnttnt


48/116


003.0

0025.0

000333.0

16.314

)5236.016.314sin(

16.314

)5236.016.314sin(6.86

16.314

)16.314sin(

16.314

)16.314sin(100

o

o

o

o

o

o

o

o

n

tnt

n

tnt

T

n

tnt

n

tnt

T

003.0

0025.0

0025.0

000333.0

188516.314

)18855236.016.314sin(

188516.314

)18855236.016.314sin(

003333.0

6.86

188516.314

)188516.314sin(

188516.314

)188516.314sin(

003333.0

100

n

tnt

n

tnt

n

tnt

n

tnt

188516.314

)7125.45236.07854.0sin(

188516.314

)7125.45236.07854.0sin(

188516.314

)655.55236.09422.0sin(

188516.314

)655.55236.09422.0sin(

1098.25

188516.314

)6277.01046.0sin(

188516.314

)6277.01046.0sin(

188516.314

)712.47854.0sin(

188516.314

)712.47854.0sin(1030

3

3

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

Thus A1, A2, A3, A4, A5, A6, . can be determined.Similarly B1, B2, B3, B4, B5, B6, . can be determined.

Fourier Series of waveform can then be determined.

Remaining calculations are left to reader as an exercise.

Effective Value of a Periodic Waveform


49/116


Effective Value of a Periodic Waveform

is defined in terms of power dissipation and is hence the

same as the r.m.s. value of the waveform.

Tto

to

effective dttaT

A )(1 2

Since the periodic waveform may be defined as

11

sincos2

)(n

onn

ono tnBtnAAta

Tto

to non

non

oeff dttnBtnAAT

A

2

11sincos

21

2


50/116


Tto

to non

non

oeff dttermsproducttnBtnA

A

TA

1

2

1

2

2

sincos2

1

Using trigonometric properties, only square terms will give non-zero integrals.

Product termswill all give zero integrals.

1

2

1

22

1

2

1

2

2

222222

1

n

n

n

no

nn

nn

o

eff

BAATB

TAT

A

TA

2

oAis d.c. term,

2

)( 22 nn BA

is r.m.s. value of nth

harmonic

Effective value or r.m.s. value of a periodic waveform is

square root of sum of squares of r.m.s. components.


51/116

Using trigonometric properties only similar terms from


52/116


Using trigonometric properties, only similar terms from

v(t)and i(t)can give rise to non-zero integrals.

P =Vdc.Idc+ (1/2)VnIncos(n -n)

= Vdc.Idc+ Vrms,nIrms,ncosnTotal power is given as sum of the powers of the

individual harmonics including the fundamental and thedirect term.

Example 6


53/116


Example 6

Determine effective values of voltage and current, total power

consumed, overall power factor and fundamental displacement

factor, if Fourier series of voltage (V) and current (A) arev(t) = 5 + 8 sin( t + /6) + 2sin3 ti(t)=3+5sin(t+/2)+1sin(2t- /3)+1.414cos(3t+/4)

Solution22

2

2

2

2

85

rmsV = 7.681 V

222

2

2

414.1

2

1

2

53

rms

I

= 4.796 AP =53+(8/2).(5/2).cos(/3)+0 + (2/2).(1.414/2).cos(/2+/4)

= 15 + 101 = 24 W

Overall power factor


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Overall power factor

Overall power factor of a periodic waveform is defined

as ratio of active power to apparent power.

Overall power factor = powerapparent

poweractive

Overall power factor = 24/(7.681 4.796) = 0.651With non-sinusoidal waveforms, power factor is not

associated with lead or lag as these no longer have any

meaning.

Fundamental displacement factor


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Fundamental displacement factor

Fundamental displacement factor corresponds to power

factor of the fundamental.

It tells by how much the fundamental component of

current is displaced from the fundamental component of

voltage, and hence is also associated with the terms lead

and lag.Fundamental displacement factor (FDF)

= cos 1= cos (/2 - /6) = cos/3 = 0.5 lead

Note that the term lead is used as the original current isleading the voltage by an angle /3.

Analysis of Circuits in presence of Harmonics in Source


56/116


Analysis of Circuits in presence of Harmonics in Source

Due to presence of non-linear devices in the system,

voltages and currents get distorted from sinusoidal.

Thus it becomes necessary to analyse circuits in thepresence of distortion in source.

This can be done by using the Fourier Series of supply

voltage and the principle of superposition.For each frequency component, circuit is analysed, as

for pure sinusoidal quantities, using normal complex

number analysis, and the results are summed up to give

the resultant waveform.

Example 7


57/116


Example 7

Determine voltage across load R for supply voltage e(t)

applied to circuit shown in figure 22.

e(t) = 100 + 30 sin(300t + /6) + 20 sin 900t

+ 15 sin (1500t - /6) + 10 sin 2100t

e(t)

L = 50mH

r = 10

C = 100 F R = 100

Figure 22Circuit with distorted source

Solution


58/116


Solution

For the d.c. term,

10100

100

100 dcV

. Vdc= 90.91 VFor any a.c. term,

if Vnmis peak value of nth

harmonic of output voltage,

)1)(()1(

)1(//

//

2

2CRjrLjR

R

CRjRrLjCRj

R

RCrLRC

E

V

nm

nm

)100101003001)(10050.0300(100

1006

njnj

EV nmnm

)31)(1015(100

100

njnjEV nmnm

100


59/116


njnEV nmnm

4545110

1002

for the fundamental

V1m=30100/(65+j45)=3000/79.0634.7o=37.95-34.7o

V3m=20100/(-295+j135)=2000/324.42155.4o=6.16-155.4o

V5m=15100/(-1015+j225)=1500/1039.64167.5o=1.44-167.5o

V7m=10100/(-2095+j315) =1000/2118.5171.4o=0.47-171.4ov(t) = 90.91+37.95sin(300t+)+6.16sin(900t155.4o)

+1.44 sin(1500t)+0.47sin (2100t 171.4o)

v(t) = 90.91 + 37.95 sin(300t

) + 6.16 sin (900t 155.4o

)+ 1.44 sin(1500t) + 0.47 sin (2100t171.4o)

Harmonic Analysis of Graphical Waveforms


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Harmonic Analysis of Graphical Waveforms

Many waveforms, when obtained experimentally, are

known at discrete points and not as a function of time.

Fourier coefficients need to be obtained graphically.

When discrete data is available, approximations are made.

Integrals used are replaced by finite summations.

If period T = 2/is divided into N equal intervals t,

NN

Tt

2

N

pp

N

pp

T N

ppo tf

NNTtf

Tttf

Tdttf

TA

110 1

)(2).(2).(2).(2

similarly


61/116


similarly

N

pppn tntf

NA

1

cos)(2

N

pppn tntf

NB

1

sin)(2

Values of cos ntpand sin ntpcalculated at mid interval.

pN

tp 2

Example 8 i(t)


62/116


Example 8Figure shows waveform

obtained from a graph

plotter which has been readoff at 24 discrete intervals

of time in one period.

Find fundamental and third-

harmonic of half-wave

symmetric waveform.One half-cycle of waveform

is defined as discrete

ordinates as follows.

2 360o

24 divisions, 1 div= 15o

= p 0 1 2 3 4 5 6 7 8 9 10 11 12

i(tp) 0 6.5 10.4 12.5 14.1 15.9 18.2 21.8 27.8 19.2 10 5 0

Figure 23discrete waveform

i(t)

t

Solution


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SolutionWaveform has half-wave symmetry no even harmonicsIntegration (or summation) is done only over half a cycle.

It is seen that mean value of waveform is zero so that Ao/2 = 0.

Consider numerical evaluation of individual coefficients.

12

1

12

11

cos)(6

1cos)(

24

22

p pp pp

ptittiA

= (1/6).[6.5 cos 15o+ 10.4 cos 30o + 12.5 cos 45o+ 14.1 cos 60o+15.9 cos 75o+18.2 cos 90o + 21.8 cos 105o+ 27.8 cos 120o

+ 19.2 cos 135o+ 10 cos 150o+ 5 cos 165o+0 cos 180o]

= (1/6).(-11.32)= -1.89

12

i)(1tiB


64/116


1

1 sin)(6 p

p ptiB

= (1/6).[ 6.5 sin 15o + 10.4 sin 30o + 12.5 sin 45o + 14.1 sin 60o

+ 15.9 sin 75o +18.2 sin 90o + 21.8 sin 105o + 27.8 sin 120o+ 19.2 sin 135o+ 10 sin 150o+ 5 sin 165o+0 sin 180o]

= (1/6).(126.5) = 21.1

A3= (1/6).[6.5 cos45o+10.4 cos 90o+ 12.5 cos 135o+ 14.1 cos 180o

+15.9 cos 225o+18.2 cos 270o+ 21.8 cos 315o+ 27.8 cos 360o+ 19.2 cos 45o+ 10 cos 90o+ 5 cos 135o+0 cos 180o]

= (1/6).(23.67) = 3.94

B3 = (1/6).[6.5 sin 45o+ 10.4 sin 90o+ 12.5 sin 135o+ 14.1 sin 180o

+15.9 sin 225o+18.2 sin 270o+ 21.8 sin 315o+ 27.8 sin 360o

+ 19.2 sin 45o+ 10 sin 90o+ 5 sin 135o+0 sin 180o]

= (1/6).(6.09) = 1.015

Determination of higher harmonics would not be accurate


65/116


g

unless more points are defined on waveform.

fundamental=1.89cost+21.1sint=21.2 sin(t5.1o)

thirdharmonic=3.94cos3t+1.01sin3t=4.07sin(3t+75.6o)In practical waveforms higher harmonics are generally smaller

in amplitude.

If the fifth and higher harmonics are neglected,

i(t) = 21.2 sin(t5.1o) + 4.07 sin(3t +75.6o)

Effective value = 207.4

2

2.21 22

= 15.26 A

Error caused would generally be only in the second decimal

place in the above example.

Complex form of the Fourier Series


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Complex form of the Fourier Series

It would have been noted that the only frequency terms

that were considered were positive frequency terms

going up to infinity but that time was not limited topositive values.

Mathematically speaking, frequency can have negative

values, but as will be obvious, negative frequency termswould have a corresponding positive frequency term

giving the same Fourier component.

In the complex form, negative frequency terms are alsodefined.

Using the trigonometric expressions


67/116


Us g e go o e c e p ess o s

ej

= cos + j sin and ej= cos j sin rewrite Fourier series.

)sincos(2

)(1

tnBtnAAtf on

non

o

jee

Bee

AA

tftjntjn

nn

tjntjn

n

ooooo

222)(

1

This can be re-written in the following form

222

0)(

1

nntjn

n

nntjno jBAejBA

ejA

tf oo

It is to be noted that Bois always 0, so that thej0with

Aomay be written as jBo. Also ej0

= 1

nn jBAC


68/116


Thus defining 2nn

n

jC

,

we have 2

000

jAC

and 2

nn

n

jBAC

the term on right hand side outside summation can be

written as Co ej0

and first term inside summation

becomes Cn e

jn

t

.

Since

Tt

t

on

o

o

dttntfT

A cos)(2

,

Tt

t

on

o

o

dttntfT

A )(cos)(2

=An

and

Tt

t

on

o

o

dttntf

T

B sin)(2

,

Tt

t

on

o

o

dttntf

T

B )(sin)(2

=

Bn

22

nnnnn

jBAjBAC

i.e. Second term inside summation becomes C n e- jn

t.


69/116


i.e. Second term inside summation becomes C-n e .

The three sets of terms in equation correspond to zero

term, positive terms and negative terms of frequency.

Thus Fourier Series may be written in complex form as

n

tjn

noeCtf

)(

and Fourier coefficient Cncan be calculated as follows.

2

nnn

jBAC

Tt

t

oo

o

o

dttnjtntfT

]sin[cos)(2

2

1

Tt

t

tjnn

o

o

o dtetfT

C

)(1

The Fourier Series may be written in complex form as


70/116


y p

n

tjn

noeCtf

)(

where the Fourier coefficients Cnare given by

Tt

t

tjn

n

o

o

o dtetfT

C

)(1

In symmetrical form,Fourier series is written with to= T/2.Fourier series is written for a periodic function with period T,

and discrete frequency components are obtained for the

waveform.The fundamental frequency ois related to the period T by theexpression o= 2/T.

Consider the following waveforms.f


71/116


gf t

t(a) T

f(t)

t(b)T

f(t)

t(c)

Figure 1Period of repetition gradually increased

1(a) period of repetition is quite small


72/116


( ) p p q

1(b) period of repetition somewhat larger

1(c) period of repetition has been increased up to .

Any non-repetitive waveform may be considered as one

which has a period T , and the corresponding

fundamental frequency0

2

T

o .

Fourier coefficients Cnin symmetrical exponential series

0)(1 2

2

Cdtetf

TC

T

T

tjn

no

Frequencies involved are no longer discrete but continuous.

General frequency nocorresponds to d= .

For non-repetitive functions, following can be written.


73/116


p g

T o d

Cn dCno

22

1 dfT

o Expression for complex Fourier Coefficient Cnbecomes

dtetfd

dC tj .).(2

dividing both sides by d,

dtetfd

dCF tj .).(

2

1)(

Fourier Transform

The original function f(t) is now given as


74/116


tj

n

tjn

n edCeCtf o .)(

from the definition, dC = F().d, so that

deFtf tj .).()( Fourier Inverse Transform

Fourier Transform expression and Fourier Inverse Transformexpression together are known as Fourier Transform Pair.

If we multiply the Fourier Transform by a constant and divide

the Inverse Transform also by the same constant, we would

again get a modified transform pair.We could define a symmetrical transform pair by using a factor

of 2 .

Symmetric Fourier Transform


75/116


y

dtetfd

dCF tjs .).(

2

12)(

and correspondingSymmetric Inverse Transformis

deFtf tjs .).(2

1)(

Fourier Transform is useful in analysing transients in electrical

circuits, especially where the elements are frequencydependant.


76/116

Laplace Transform


77/116


Laplace Transform

In obtaining the Laplace Transform, any function f(t)

is initially decayed artificially by an exponential factore

-t, so that new function always becomes integrable.

However, the decay would correspond to an exponential

rise (rather than a decay) with negative time.Laplace transform is thus defined only for causal

functions (functions that are caused and hence are of

zero value before time zero).

Laplace Transform of a time function f(t) is thus


78/116


defined as

L [f(t)] = F(s) =

0

)( dtetf st

where s = + j is the Laplace operatorLaplace operatorsis also a complex form of frequency.

Laplace Inverse Transform takes the form

f(t) =

j

j

st dsesFj

)(

2

1

It is seen that the form of the transform has simplified

from that of the Fourier Transform, but not the inverse.

It is very rarely that the Inverse transform is calculated


79/116


in this manner.

It is generally obtained from a knowledge of transforms

of common functions, generally found in tabulatedform.

The Laplace Transform is very useful in circuit

transient analysis as it can convert differential equationsto linear algebraic equations.

Response of a linear Passive Bilateral Network


80/116


Response r(t) would be related to the input e(t) by an

ordinary linear differential equation.r(t) = F(p) . e(t)

where p = d/dt = differential operator

Consider an exponential excitation function est

.i.e. e(t) = e

str(t) = F(p) . est = est. F(s)

Linear

Passive

bilateral

e(t) r(t)

Figure 2Transfer Function


81/116

One of the advantages of the Laplace Transform is that


82/116


it converts ordinary differential equations in to algebraic

equations, so that the solution is fairly simple.

The inverse transform is then obtained to get the timeresponse.

Let us now consider the Laplace Transform of some

special causal functions.

Laplace Transform of Special Causal Functions


83/116


(a) Unit impulse function (t)Unit impulse has a value 0 at all values

of t other than at t = 0 where it has aninfinite magnitude. Integral of unit

impulse function over time is equal to 1.

L [(t)] =

0

)( dtet st = 1

If unit impulse occurs at t = ti, rather than at t = 0, then

the function is (t ti).L [(t ti)] = istst edtett

0i )-(

(t)

t

t-ti t

ti

(b) Unit step function H(t) H(t)


84/116


Unit step has a value 0 for values of

t < 0 and a value of 1 for t > 0.

L [H(t)] = sdtedtetH stst 1

1)(00

If the unit step occurs at t = ti,

rather than at t = 0,

then the function isH(t ti).

L [H(t ti)] = s

edtedtettH

i

i

st

t

stst

1)-(

0 i

1

t

1

t

H(t-ti)

ti


85/116

(d) Causal Sinusoidal function sin(t+) .H(t)


86/116


L [sin(t+).H(t)]

= 0

)().sin( dtetHt st

=)().sin(

0

sFdtet st

F(s) =

00

)cos(sin( dts

et

s

et

stst

=dtet

ss

et

ss

stst

)sin()cos(sin

2

2

0

t

)(cossin 2

sF


87/116


=)(cos

22 sFsss

(s2+2). F(s) = s . sin + . cos

22

cos.sin.)(

s

ssF

with = 0o

and 90

o

the following are obtained.

L [sin t.H(t)] = 22

s ,

L[cos t.H(t)] = 22 ss

( ) L l T f f h l d i itfd )(


88/116


(e) Laplace Transform of the causal derivative dt

L

00

)()()(

tfdedtetd

tfd

td

tfd stst

=

0

0 .).).(()( dtestftfe stst

= (f(0-) + s.

0

.).( dtetf st

L )0()(.)(

fsFs

td

tfd

Note: Unlike in case of ordinary derivative, transform of

derivative also keeps information about initial condition

[i.e.f(0-)]

(f) E i l l i li i fat

i h i d i


89/116


(f)Exponential multiplication of eat

in the time domain

An exponential multiplication of eat in the time domain

corresponds to a shift of a in the s-domain.

L [eat. f(t)]

0

)(

0

0

.).()(.).( dtetftfdtetfe tasstat

= F(s-a)

(g) A shift in the time domain


90/116


A shift of ain time domain f(ta).H(ta), corresponds

to an exponential decay in the s-domain.

L [f(t-a).H(t-a)] 0

.).().( dteatHatf st

a

atssa atdeatHatfe )(.).().( )(

a

ssa defe .).(

0

.).( defe sassince f() = 0 for

= e-as. F(s)

(h) F d f f( ) h d T


91/116


(h) For a periodic waveform f(t) with period T

L [f(t)]

0

.).( tdetf st

Tsttdetf

0

.).(+

T

T

st tdetf2

.).(+

T

T

st tdetf3

2

.).(+

using a change of variables, may be re-written as

L [f(t)] T

st tdetf0

.).( +

Tst tdeTtf

0

.).( +

Tst tdeTtf

0

.).2( +

Since the function is periodic,f(t) = f(t+T) = f(t+2T) =..

L [f(t)]

T

st tdetf .).( +

TstsT tdetfe .).( +

T

stsT tdetfe 2 .).( +


92/116


L [f(t)] 0

+0

+0

+

= [1+ e-sT+ e-2sT+ e-3sT+ ]

T

st

tdetf0 .).(

L [f(t)]

T

st

sT tdetf

e 0.).(

1

1

The transforms of other causal functions may be

similarly obtained.

Table of Laplace transforms for common causal functions


93/116


(t) unit impulse 1

H(t) unit step s

1

t ramp 21

s

0 t

0 t

0 t

e-at

exponential decay

1


94/116


e exponential decay as

1- e-at

)( ass

a

t .e-at

2)(1as

e-a-e-btdouble exponential ))(( bsasab

0 t

0 t

0 t

0 t


95/116


sin t sine wave 22

s

sin (t +) 22sincos

s

s

cos t cosine wave 22 ss

0 t

0 t

0 t

sT1


96/116


rectangular pulse s

e sT1

tn

nthorder ramp 1!ns

n

sinh at hyperbolic sine 22 as

a

cosh at hyperbolic cosine 22 ass

0 t

0 t(n > 0)

a.f1(t)+ b.f2(t) addition a.F1(s)+ b.F2(s)

fd )(


97/116


td

tfd )(

first derivative s F(s)f(0-)

n

n

td

tfd )(

nth

derivative

n

jn

jjnn

td

fdssFs

1

1

)0()(

t

dttf0

)( definite integral

)(1 sFs

t.f(t) sd

sFd )(


98/116

Transient Analysis of Circuits using Laplace Transform


99/116


Electrical Circuits are usually governed by linear

differential equations.

Since derivatives and integrals get converted to

multiplications and divisions in the s-domain, solution

of circuit equations can be converted to solution of

algebraic equations.First consider representation of the three basic circuit

components in Laplace Transform analysis.

(a) Resistive Element R1


100/116


v(t) = R . i(t) V(s) = R . I(s)

)(

1

)( tvRti )(1

)( sVRsI

Thus the resistor may be represented by an impedance

of value R even in the s-domain.

i(t)

v(t)

R I(s)

V(s)

R I(s)

V(s)

1/R

(b) I nductive Element L 1


101/116


v(t) = L . tdtid )(

V(s)= Ls. I(s)L.i(0-)

)0(.)(1

)( idttvLti s

isVLs

sI )0()(.1)(

Thus inductor may be represented by an impedance of valueLs

and either a series voltage source or a parallel current source.

These sources represent the initial energy stored in the inductor

at time t = 0. Thus the initial current i(0-)appears.

V(s)v t

i t L I(s) L sL.i(0-)

+ I sV(s)

Ls

1

s

i )0(

(c) CapacitiveElement C

V( ))0(


102/116


)0(.)(1)( vdttiCtv svsICssV )0()(.1)(

i(t) = C . tdtvd )(

I(s) = Cs.V(s)C.v(0-)

Capacitor may be represented by an impedance of value Cs

1

and either a series voltage source or a parallel current source.

V(s)

i(t)

v t

C I(s)

V(s)

Cs

1

+s

v )0(

C sI s

C.v(0-)

Transient Analysis


103/116


Using these circuits, and the transforms of source

voltages and/or currents, the system transients could be

obtained.

You would by now have realised that this method is

much less tedious than the solution of the differential

equations to find the transient solutions and thensubstituting the initial and final conditions applicable.

Example 1

Fi d th L l t f f th f ll i f


104/116


Find the Laplace transform of the following waveforms.

(a)

(b)

f(t) = E sin t for 0 < t < T/2f(t) = 0 elsewhere

t0

f(t)

T/2

E

t0

f(t)

T 2T

2E

E

Solution

( )


105/116


(a)

using first principles

L[f(t)]

0

.).( tdetf st 0.).(...22

0

T

T

stT

st tdeEtdeT

tE

T

T

stT stTst

s

eEdt

s

e

T

E

s

e

T

tE

2

00

...1

.2..2

=

).(

)(

.2.2 2

0

2

sTsT

TstsT

ee

s

E

s

e

T

E

s

eE

).()(

)1(.

2.2 22

sTsTsTsT

ees

E

s

e

T

E

s

eE

sTsTsT eTs EeesE 1.23 22

Alternate solution

using splitting up into known forms (ramp step etc)


106/116


using splitting up into known forms (ramp, step, etc)

(this is not always possible)

Part of ramp from t = 0 to T can be considered as theaddition of a positive ramp at t=0, a negative ramp at

t = T and a negative step of magnitude 2E at time t = T.

t0

f(t)

T 2T

2E

E t0

f(t)

T 2T

2E

E

Remaining part of waveform can be considered to be

made up of a negative step waveform of magnitude E at


107/116


made up of a negative step waveform of magnitude E at

t = T, and a positive step also of magnitude E at t = 2T.

Superposition of these waveforms gives the resultantwaveform.

These have Laplace transforms which will add up as

follows.

L[f(t)]

sTsTsTsT es

Ees

Ees

EesT

E

sT

E 222

...2

.1

.21

.2

sTsTsT

eTs

E

ees

E

1.2

3 22

Identical to result obtained from the normal method.

(b) f(t)f(t)E


108/116


Consider working this problem by splitting waveform.

Can be considered as been built up of a causal sine wavestarting at t = 0, and a negative of sine wave starting at T/2.

Thus the transform of the waveform is given by

L[f(t)] =

2

2222 .

sT

es

s

s

s

=

2

22 1

sT

es

s

t0 T/2

E

t0 T/2


109/116

Using potential divider action1


110/116


RCsCs

R

Cs

sA

sVout

1

1

1

1

.

)(

22

2222 .

..

1)(

ss

A

sRCs

AsVout , where = RC

1

This can be split into partial fractions as follows.

2222

1..)(

s

s

s

AsVout

Using the tables, the inverse transform is then given as

tteA

sv tout

sincos..

)(22

Example 3

In the series LC circuit 1


111/116


In the series LC circuit

shown, initially capacitor

is charged to a voltage Voand inductor does not

carry any current.

At time t = 0, a step voltageof magnitude E is applied to

the series combination.

Determine the transient voltage across L.

I(s)

V(s)

s

V0Cs

1

s

E


112/116

s

V

s

E

I )(

0


113/116


Cs

Ls

sssI1

)(

Cs

Ls

s

V

s

E

LssILssV1

)(.)(

0

LC

s

sVE

LCs

VELCs

1)(

1 202

0

Let2

0

1

LC ,220

)()(

os

sVEsV

tLC

VEtv 1cos)()( 0

Example 4

Figure shows a circuit which has reached steady state


114/116


Figure shows a circuit which has reached steady state

with switch closed.

If the switch S is opened at time t=0, obtain anexpression for the ensuing current through the inductor.

S

R1= 10

E = 100 V

C = 10 F

R2= 10

L = 10 mH


115/116


116/116

waveform analysis

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