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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+)
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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
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(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.
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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
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Non-sinusoidal Waveforms J R Lucas May 20115
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)
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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
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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.
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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
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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.
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Non-sinusoidal Waveforms J R Lucas May 201110
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.
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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.
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Non-sinusoidal Waveforms J R Lucas May 201113
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
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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
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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.
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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
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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.
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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
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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
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
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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
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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.
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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.
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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
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Non-sinusoidal Waveforms J R Lucas May 201131
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
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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
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Figure 11 shows synthesis of the rectangular waveform
using the Fourier components obtained.
Figure 11Fourier Synthesis of Rectangular Waveform
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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
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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)
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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
Figure 13Rectangular waveform
If f i hif d b T/6 l f i
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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
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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
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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
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= 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
Figure 17Rectangular waveform
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E l 4 a(t)
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Non-sinusoidal Waveforms J R Lucas May 201142
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
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= 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
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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
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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
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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
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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
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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
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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
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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.
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Using trigonometric properties only similar terms from
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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.
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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
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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
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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
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( ) 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.
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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
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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
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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.
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Laplace Transform
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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
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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
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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
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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
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One of the advantages of the Laplace Transform is that
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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
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(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)
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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
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(d) Causal Sinusoidal function sin(t+) .H(t)
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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
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=)(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 )(
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(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
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(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
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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
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(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 .).( +
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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
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(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
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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
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sin t sine wave 22
s
sin (t +) 22sincos
s
s
cos t cosine wave 22 ss
0 t
0 t
0 t
sT1
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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 )(
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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 )(
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Transient Analysis of Circuits using Laplace Transform
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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
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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
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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(
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)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
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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
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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
( )
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(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)
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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
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Non-sinusoidal Waveforms J R Lucas May 2011107
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
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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
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Using potential divider action1
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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
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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
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s
V
s
E
I )(
0
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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
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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
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