Theory of Electricity -Professor J R Lucas- Level 2-EE201_10_non_sinusoidal_part_2

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Theory of Electricity – Analysis of Non-sinusoidal Waveforms - Part 2 – J R Lucas – November 2001 1

Analysis of Non-Sinusoidal Waveforms – Part 2 – Laplace Transform

In the earlier section, we learnt that the Fourier Series may be written in complex form as

∑

∞

−∞= ⋅= n

t jn

n

oeC t f ω

)(

where the Fourier coefficient C n is given by

∫ +

− ⋅⋅=T t

t

t jn

n

o

o

o dt et f T

C ω

)(1

In the symmetrical form, the Fourier series is written with to = −T/2.

The Fourier series is written for a periodic function with period T, and discrete frequency

components are obtained for the waveform. We saw that the fundamental frequency ωo is

related to the period T by the expression ωo = 2π /T.

Now consider the following waveforms.

In figure 1(a), the period of repetition is quite small and in (b) somewhat larger. Waveform

(c) could be considered as one where the period of repetition has been increased up to infinity.

Thus any non-repetitive waveform may be considered as one which has a period T → ∞, and

the corresponding fundamental frequency 02

→∆== ω π

ω T

o.

It is also seen that the Fourier coefficient Cn in the symmetrical exponential series

0)(1 2

2

→∆=⋅⋅= ∫ −

− C dt et f T

C

T

T

t jn

noω

f(t)

t(a)

T

f(t)

t(b)

T

f(t)

t(c)

Figure 1 – Period of repetition gradually increased




The frequencies involved are no longer discrete but continuous, so that the general frequency

nωo corresponds to Σ ∆ω → ∫ dω = ω.

Thus for non-repetitive functions, the following can be written.

T → ∞

ω o → d ω

C n → dC

nω o → ω

π

ω

π

ω

22

1 d f

T

o →==

Thus the expression for the complex Fourier Coefficient Cn becomes

∫ ∞

∞−

−= dt et f d

dC t j .).(

2

ω

π

ω

dividing both sides by dω, this may be written as

∫ ∞

∞−

−== dt et f d

dC F

t j .).(2

1)( ω

π ω ω Definition of the Fourier Transform

The original function f(t) is now given as

∫ ∑∞

∞−

∞

−∞==⋅= t j

n

t jn

n edC eC t f o ω ω .)(

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

∫ ∞

∞−

= ω ω ω d eF t f

t j .).()( Fourier Inverse Transform

The Fourier Transform expression and the Fourier Inverse Transform expression together are

known as the 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.

If we examine the two transform expressions, we see that they look very similar except that

there is a difference of a negative sign in the exponent and a multiplying factor of 2π.

Thus we could define a symmetrical transform pair by using a factor of π 2 .In this case the Symmetric Fourier Transform is defined as

∫ ∞

∞−

−== dt et f d

dC F

t j

s .).(2

12)(

ω

π ω π ω

and the corresponding Symmetric Inverse Transform is defined as

∫ ∞

∞−

= ω ω π

ω d eF t f

t j

s .).(2

1)(

The Fourier Transform is useful in analysing transients in electrical circuits, especially wherethe elements are frequency dependant.




Fourier CosineTransform

A Fourier Cosine Transform F1(ω) may be defined when the non-repetitive waveform is even.

f(t) = f(-t) so that

∫

∞

⋅= 0

1 cos).(2

)( dt t t f F ω π ω , and

∫ ∞

⋅=0

1 cos).()( ω ω ω d t F t f

Fourier SineTransform

A Fourier Sine Transform F2(ω) may be defined when the non-repetitive waveform is odd.

f(t) = −f(-t) so that

∫

∞

⋅=0

2 sin).(2

)( dt t t f F ω π

ω , and

∫ ∞

⋅=0

2 sin).()( ω ω ω d t F t f

The Fourier Transform is sometimes expressed in terms of the sum of a sine and cosine

series, instead of the exponential series.

[ ]∫ ∞

⋅+=0

sin).(cos).()( ω ω ω ω ω d t Bt At f

where ∫

∞

∞− ⋅= dt t t f A ω π ω cos).(

1

)( , and

∫ ∞

∞−

⋅= dt t t f B ω π

ω sin).(1

)(

With functions which are non-repetitive, and does not decay till infinite time (such as the sine

waveform or the cosine waveform), the Fourier Integral Transform may not be obtained.

To avoid this problem, waveforms which do not decay may be artificially decayed by an

exponential factor to allow the integration. The integrated result is then exponentially

magnified to correct for the initial decay introduced. However, such exponential

magnification can also magnify numerical errors.

The Laplace Transform is defined based on this artificial decay.

Laplace Transform

In obtaining the Laplace Transform, any function f(t) is initially decayed artificially by an

exponential factor e-σt

, so that the new function always becomes integrable. However, the

decay would correspond to an exponential rise (rather than a decay) with negative time. The

Laplace transform is thus defined only for causal functions (functions that are caused and

hence are of zero value before time zero).

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

/ [f(t)] = F(s) = ∫ ∞

− ⋅⋅0

)( dt et f st

where s = σ + j ω is the Laplace operator




The Laplace operator s is also considered as a complex frequency.

If we compare with the Fourier Transform pair with a multiplier of 2π, then the Laplace

Inverse Transform takes the form

f(t) =

∫

∞+

∞−

⋅⋅ j

j

st dsesF j

σ

σ π )(

2

1

It is seen that the form of the transform has simplified from that of the Fourier Transform.

However, it is very rarely that the Inverse transform is calculated in this manner. It is

generally obtained from a knowledge of the transforms of common functions, generally

found in tabulated form.

The Laplace Transform is very useful in circuit transient analysis as it can convert

differential equations to linear algebraic equations.

Response of a linear Passive Bilateral Network

Consider a linear passive bilateral two-port network to which an excitation e(t) is given at one

port and which causes some response r(t) at the other port.

In general, the 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) = est → r(t) = F(p) . e

st= e

st. F(s)

Thus for an exponential excitation, the system has a transfer function r(t)/e(t) equal to F(s).

As stated earlier, any non-repetitive (or even repetitive) function may be broken up into a

series of exponentials. The coefficients of these exponentials are given by the Laplace

Transform.

Thus for any other excitation e(t), if the Laplace Transform E(s) is considered, it will be

related to the Laplace Transform R(s) of the response r(t) by the transfer function F(s).

Thus for any causal excitation e(t),

R(s) = F(s) . E(s)

One of the advantages of the Laplace Transform is that 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 time response.

Let us now consider the Laplace Transform of some special causal functions.

Laplace Transform of Special Causal Functions

(a) Unit impulse function δ (t)

The unit impulse has a value 0 at all values of t other than at t = 0 whereit has an infinite magnitude. Also the integral of the unit impulse

function over time is equal to 1.

Linear

Passive

bilateral

Network

e(t) r(t)

Figure 2 – Transfer Function

δ(t)

∞

t




/ [δ (t)] = ∫ ∞

− ⋅⋅0

)( dt et st δ = 1

If the unit impulse occurs at t = t i, rather than at t = 0, then the function is δ (t − t i).

/ [δ (t − t i)] =ist st edt et t −

∞− =⋅⋅∫

0i )-(δ

(b) Unit step function H(t)

The unit step has a value 0 for values of t < 0 and a value of 1 for t > 0.

/ [ H(t)] =s

dt edt et H st st 11)(

00

=⋅⋅=⋅⋅ ∫ ∫ ∞

−∞

−

If the unit step occurs at t = t i, rather than at t = 0, then the function is H(t − t i).

/ [ H(t − t i)] =s

edt edt et t H

i

i

st

t

st st −∞ −∞ − =⋅⋅=⋅⋅ ∫ ∫ 1)-(

0

i

(c) Causal exponential function eat

. H(t)

/ [eat. H(t)] =

asdt edt et H e

t asst at

−=⋅⋅=⋅⋅ ∫ ∫

∞−−

∞− 1

1)(.0

)(

0

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

/ [sin(ωt+φ). H(t)] = ∫ ∞

− ⋅⋅+0

)().sin( dt et H t st φ ω

)().sin(0

sF dt et st =⋅⋅+= ∫

∞−φ ω

F(s) = ∫ ∞ −∞−

⋅−

⋅+⋅−−

).+00

)cos(sin( dt s

et

s

et

st st

φ ω ω φ ω

= dt et ss

et

ss

st st

⋅⋅+⋅−−

⋅+⋅+ −

∞−

∫ )sin()cos(sin

2

2

0

φ ω ω

φ ω ω φ

= )(cossin

2

2

2sF

sss⋅−⋅+

ω φ

ω φ

(s2+ω2

). F(s) = s . sin φ + ω . cos φ

22

cos.sin.)(

ω

φ ω φ

++

=s

ssF with φ = 0

oand 90

othe following are obtained.

/ [sin ωt.H(t)] = 22

ω

ω

+s

, / [cos ωt.H(t)] = 22

ω +s

s

δ(t-ti)

1

t

H(t)

∞

t

ti

1t

H(t-ti)

ti

1

eat.H(t)

t

t




(e) Laplace Transform of the causal derivativedt

t f d )(

/ ∫ ∫ ∞

−∞

− ⋅=⋅⋅=00

)()()(

t f d edt et d

t f d

t d

t f d st st ∫ ∞

−∞− −−⋅=0

0.).).(()( dt est f t f e st st

= (−) f(0-) + s. ∫

∞−

0

.).( dt et f st

/ )0()(.)( −−=

f sF s

t d

t f d

It is worth noting that unlike in the case of the ordinary derivative, the transform of the

derivative also keeps information about the initial condition [i.e. f(0-)]

(f) An exponential multiplication of eat

in the time domain

An exponential multiplication of e

at

in the time domain corresponds to a shift of a in thes-domain.

/[eat

. f(t)] ∫ ∫ ∞

−−∞∞

− =⋅⋅=0

)(

00

.).()(.).( dt et f t f dt et f e t asst at

= F(s-a)

(g) A shift in the time domain

A shift in a the time domain f(t – a).H(t – a), corresponds to an exponential decay in the

s-domain.

/[ f(t-a).H(t-a)] ∫ ∞

−−−=0

.).().( dt eat H at f st

∫ ∞

−

−−− −−−=a

at ssa at d eat H at f e )(.).().()(

∫ ∞

−

−−=a

ssa d e f e τ τ τ .).(

∫ ∞

−−=0

.).( τ τ τ d e f e sas since f(τ) = 0 for τ < 0.

= e-as. F(s)

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

/[ f(t)] ∫ ∞

−=0

.).( t d et f st

∫ −=T

st t d et f 0

.).( + ∫ −T

T

st t d et f

2

.).( + ∫ −T

T

st t d et f

3

2

.).( +……

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

/[ f(t)] ∫ −=T

st t d et f 0

.).( + ∫ −+T

st t d eT t f 0

.).( + ∫ −+T

st t d eT t f 0

.).2( +……




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

/[ f(t)] ∫ −=T

st t d et f 0

.).( + ∫ −−T

st sT t d et f e0

.).( + ∫ −−T

st sT t d et f e0

2.).( +……

= [1+ e-sT

+ e-2sT

+ e-3sT

+ ……] ∫ −

T st

t d et f 0

.).(

/[ f(t)] ∫ −−−

=T

st

sT t d et f

e 0

.).(1

1

The transforms of other causal functions may be similarly obtained, and the table gives the

Laplace transforms for the common functions.

δ(t) unit impulse 1

H(t) unit steps

1

t ramp2

1

s

e-at

exponential decayas +

1

1- e-at

)( ass

a

+

t .e

-at

2)(

1

as +

e-at

- e-bt

double exponential))(( bsas

ab

++−

sin ω t sine wave 22 ω +s

s

0 t

0 t

0 t

0 t

0 t

0 t

0 t

0 t




sin (ω t +φ ) 22

sincos

ω

φ φ ω

++

s

s

cos ω t cosine wave22 ω +s

s

rectangular pulses

esT −−1

t n

nth order ramp 1

!+n

s

n

sinh at hyperbolic sine22

as

a

−

cosh at hyperbolic cosine22

as

s

−

a.f 1(t)+ b.f 2(t) addition a.F 1(s)+ b.F 2(s)

t d

t f d )(first derivative s F(s) – f(0

-)

n

n

t d

t f d )(n

thderivative s

nF(s) – ∑

=

−−

−n

jn

j jn

t d

f d s

1

1

)0(

∫ −

⋅t

dt t f 0

)( definite integral )(1

sF s

t.f(t)sd

sF d )(−

(-t)n.f(t)

n

n

sd

sF d )(

e-α t

.f(t) exponential multiplier F(s+α)

f(t-τ) shift e-sτ.F(s)

periodic function ∫ −

−−−

T st

sT dt et f

e 0)

.).(1(

1

0 t

0 t

0 t

(period T)

0 t(n > 0)




Transient Analysis of Circuits using the Laplace Transform

Electrical Circuits are usually governed by linear differential equations. Since derivatives and

integrals get converted to multiplications and divisions in the s-domain, the solution of circuit

equations can be converted to the solution of algebraic equations.

Let us first consider the representation of the three basic circuit components in LaplaceTransform analysis.

(a) Resistive Element R

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

)(

1

)( t v Rt i = )(

1

)( sV Rs I =

Thus the resistor may be represented by an impedance of value R even in the s-domain.

(b) Inductive Element L

v(t) = L .t d

t id )( V(s) = L s. I(s) – L.i(0

-)

)0(.)(1

)( −+= ∫ idt t v L

t i s

isV

Lss I

)0()(.

1)(

−

+=

Thus the inductor may be represented by an impedance of value L s 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. It is to be noted that the initial current i(0-) appears in both forms

of the equation and that one form can be obtained algebraically from the other, without

resorting to any additional information.

(c) Capacitive Element L

)0(.)(1

)(−+= ∫ vdt t i

C t v

s

vs I

CssV

)0()(.

1)(

−

+=

i(t) = C .t d

t vd )(I(s) = C s. I(s) – C.v(0

-)

Thus the capacitor may be represented by an impedance of value

Cs

1and either a series

voltage source or a parallel current source. As with the inductor these sources represent the

initial energy stored in the capacitor at time t = 0.

i(t)

v(t)

RI(s)

V(s)

RI(s)

V(s)

R

v(t)

i(t)L

I(s)

V(s)

L s L . i(0-)

− + I(s)

V(s)

Ls

1

s

i 0(−

i(t)

v(t)

CI(s)

V(s)

Cs

1

+ − s

v )0(−

C sI(s)

V(s)

C . v(0-)




In addition to Ohm’s Law and Kirchoff ’s laws, Superposition, Thevenin’s and Norton’s

theorems may also be applied to these transformed circuits in the s-domain.

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 then

substituting the initial and final conditions applicable.

Example 1

Find the Laplace transform of the following waveforms.

(a) (b)

Solution

(a)

using first principles

/ [f(t)] ∫ ∞

−=0

.).( t d et f st 0.).(...22

0

+−+= ∫ ∫ −−T

T

st T

st t d e E t d eT

t E

T

T

st T st T

st

s

e

E dt s

e

T E s

e

T

t

E

2

00...

1

.2..2 −−−−−=

−−−

∫ = ).()(.

2.2 2

0

2

sT sT

T st sT

ees

E

s

e

T

E

s

e E −−−−

−+−−−

= ).()(

)1(.

2.2 2

2

sT sT sT sT

ees

E

s

e

T

E

s

e E −−−−

−+−

−−

−

[ ] [ ]sT sT sT e

T s

E ee

s

E −−− −+−= 1.2

32

2

using properties from tables (this is not always possible)

≡

The part of the ramp from t = 0 to T can be considered as the addition of a positive ramp at

t=0, a negative ramp at t = T and a negative step of magnitude 2E at time t = T. The remaining

part of the waveform can be considered to be 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 will give the resultant waveform.

These will have Laplace transforms which will add up as follows.

t0

f(t)

T/2

E

t0

f(t)

T 2T

2E

E

t0

f(t)

T 2T

2E

Et0

f(t)

T 2T

2E

E

f(t) = E sin ωt for 0 < t < T/2

f(t) = 0 elsewhere




/ [f(t)] =

+−

−− −−−− sT sT sT sT e

s

E e

s

E e

s

E e

sT

E

sT

E 2

22...

2.

1.

21.

2

[ ] [ ]sT sT sT eT s

E ee

s

E −−− −+−= 1.2

32

2

which is the identical result that was obtained from the normal method.

(b)

We may also work this problem using tables.

The given waveform can also be considered as been built up of a causal sine wave startingat t = 0, and a negative of that waveform starting at T/2.

Thus the transform of the waveform is given by

/ [f(t)] =2

2222.

sT

es

s

s

s −

+−

+ ω ω =

−

+−

2

221

sT

es

s

ω

Example 2

Determine the transient voltage appearing

across the capacitor when the switch is closed at

time t = 0. Capacitor C is initially uncharged.

Solution

The transformed circuit is shown. The

capacitor has not been associated with a source

as there is no initial charge (or voltage) on the

capacitor.

Using potential divider action

RCsCs R

Cs

s A

sV out

+=

+=

+ 1

1

1

1

.

)(

22 ω ω

2222.

..

1)(

ω

ω

α

α

ω

ω

++=

++=∴

ss

A

s RCs

AsV out , where α =

RC

1

This can be split up as follows.

+−

−++

=2222

1..)(

ω

α

α α ω

α ω

s

s

s

AsV out

Using the tables, the inverse transform is then given as

+−

+= − t t e

Asv t

out ω ω

α ω

α ω

α ω α sincos..

)(22

t0

f(t)

T/2

E

t0

f(t)

T/2

E

R

vout

A sin ω t C

R

Cs

1

22

.

ω

ω

+s

A

V out(s)




Example 3

In a series LC circuit, initially the capacitor is charged to a voltage of V o and the inductor

does not carry any current. At time t = 0, a step voltage of magnitude E is applied to the

series combination. Determine the transient voltage across L.

Solution

The circuit is first transformed to the Laplace domain. The voltage source form is selected

for the capacitance because the circuit is a series circuit and that form makes calculationseasier. Since there is no initial current in the inductor, no source has been associated with the

inductor.

Cs Ls

s

V

s

E

s I 1

)(

0

+

−=

V(s) = L s.I(s)

Cs Ls

s

V

s

E

Ls

1

0

+

−⋅=

LC s

sV E

LCs

V E LCs

1

)(

1 202

0

+⋅−=

+

−⋅=

Let2

0

1ω =

LC ,

220 )()(

os

sV E sV

ω +⋅−=∴

t LC

V E t v1

cos)()( 0 ⋅−=∴

Example 4

Figure shows a circuit which has reached steady state with switch closed. If the switch S is opened at

time t=0, obtain an expression for the ensuing current through the inductor.

Solution

From potential divider action, under steady state conditions, half the supply voltage will dropacross R1 and half the voltage across R2. Therefore the voltage across the capacitor initially

will be 100/2 = 50 V, and the inductor current will be 100/20 = 5 A.

L s

I(s)

V(s)

sV 0

Cs1

s

E

i(t)

v(t) E.H(t)

C

+ −

V 0

L

S

R1 = 10 Ω

E = 100 V

C = 10 µF

R2 = 10 Ω

L = 10 mH




Transform the circuit to the Laplace domain.

Note the directions of the two sources. These correspond to the directions of the initial

voltage across the capacitor and the initial current through the inductor.

Note also that since switch S is now opened at t = 0, only the other two branches will become

part of the circuit.

Thus I(s)525

101001.0

05.050

101001.0

05.050

+++

=++

+=

ss

s

ss

s 72

101000

55000

+++

=ss

s

I(s)22

5.3122)500(

55000

+++

=s

s2222

5.3122)500(

5.31228006.0

5.3122)500(

)500(5

++×

+++

+=

ss

s

Finding the inverse transform from the standard expressions,

i(t) = t et et t

5.3122sin8006.05.3122cos5500500 −− + A

S

10

s

100

10

0.01 s

s10

106

I(s)

+

− s

50

+

− 0.01×5

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Theory of Electricity -Professor J R Lucas- Level 2-EE201_10_non_sinusoidal_part_2