Let’s illustrate this by considering the series RLC ...

92

19. Resonance As we have seen previously, the reactance of an inductance increases

with frequency whereas the reactance of a capacitor decreases with

frequency. This suggests that in certain circuits it may be possible to

find a frequency at which the two reactances are equal and opposite.

When this occurs the impedance is purely resistive and a condition of

resonance occurs. Under these circumstances, depending on the

circuit, a large current may flow or a large voltage may develop across

part of the circuit.

Let’s illustrate this by considering the series RLC circuit below.

The impedance is given by

⎟⎠

⎞⎜⎝

⎛ω

−ω+=C

LjRZ 1

Circuit Analysis II WRM MT11

93

Clearly Z is a minimum when ωL = 1/ωC when it has the real value Z =

R. The frequency at which this occurs is called the “resonant”

frequency, ω0, which is given by

LC1

0 =ω

The voltages and current are related by

ICjLjR

VVVV CLRS

⎟⎠

⎞⎜⎝

⎛ω

−ω+=

++=

At resonance, CL ω=ω /1 , and the phasor diagram takes the form

And we see that, at resonance,

LC VV =

94

and that the current, I¸ and the voltage, VS, are in phase and are related

by

IRVS=

As the frequency changes from its values at resonance the current, I,

falls from the maximum value of VS/R. At resonance the impedance

(Z=R) is a minimum and the admittance (Y=1/Z) is a maximum. We plot

below the variation of current (and Y) with frequency. The graph is an

example of a resonance curve.

If R is small the current at resonance, VS/R, is very high and the

resonance curve is sharp. On the other hand if R is large then the

current is small and the resonance curve is much flatter.

At resonance, as we have seen, the magnitude of the voltages across

the capacitor and the inductor are equal, CL VV = , and may be larger

than the source voltage itself! In order to see this we note that


95

LIVL 0ω= , CIVC 0/ω= and IRVS = . Thus

CRRL

VV

VV

S

C

S

L

0

0 1ω

=ω

==

which can be made very high by making R small. The ratio, S

L

VV

or S

C

VV

represents the ‘voltage magnification’ and is known as the “quality factor”, Q, of the circuit. Thus

Q =magnitude of voltage across L or C at resonance

magnitude of voltage across whole circuit

=ω0LR

=1

ω0CR

A high Q also results in a sharp response curve. This suggests that the

circuit acts as a narrow band filter since it only passes significant current

near the resonant frequency, 0ω=ω .

[In the lab, when you build a tuned circuit for a radio receiver, the

inductor will always come with some resistance and this will limit the Q

that is achievable. We sometimes therefore refer to Q =ω0LR as the “Q

of the coil”.]

96

In order to have a quantitative measure of the sharpness of this peak we

could consider the frequency range where the magnitude has fallen to

some fraction of the maximum value. It is conventional to choose points

at which the magnitude has fallen by 3dB to 2/1 of the maximum.

These points are called ‘3 dB or half-power’ points. The later term arises

from the fact that -3 dB corresponds to a factor 21 in current or

voltage and hence to a factor of 1/2 in power.

If we write the maximum current at resonance as RVI S /0 = then the

current at any frequency is given by

220 1

⎟⎠

⎞⎜⎝

⎛ω

−ω+

=

CLR

RII

We now wish to find the spread of frequencies between the 3dB points

where the magnitude has fallen by 2/1 , i.e. ωΔ=ω−ω 12 ,

We formally need to solve


97

21

1 22

=

⎟⎠

⎞⎜⎝

⎛ω

−ω+C

LR

R

or, equivalently,

RC

L ±=⎟⎠

⎞⎜⎝

⎛ω

−ω1

We take the negative sign to correspond to the lower frequency, ω1.

Solving the resulting quadratic gives

LCLR

LR 1

22

2

1 +⎟⎠

⎞⎜⎝

⎛+−=ω

where the positive square root has been taken in order to give a positive

ω1.

Similarly the higher frequency, ω2, is obtained by taking the +R sign. In

this case we obtain

LCLR

LR 1

22

2

2 +⎟⎠

⎞⎜⎝

⎛+=ω

The frequency difference between the 3dB points is therefore

98

Δω = ω2 − ω1 =RL

which together with our previous definition of RL

Q 0ω= permits us to

write

ωΔ

ω=

ω= 00

RL

Q

and so the higher the Q the narrower the bandwidth, Δω, and the

sharper the resonance peak.

We note that although we first defined Q in terms of the ‘voltage

magnification’ we could just as easily used ωΔ

ω= 0Q as an alternative

definition.

We also note that the resonant frequency 0ω is not positioned midway

between the two 3dB frequencies 2ω and 1ω . It is, in fact, the

geometrical mean of the two as can be seen by multiplying the

equations for 2ω and 1ω together

210 ωω=ω

and hence, on a Bode plot where a logarithmic frequency scale is used,

2ω and 1ω will be symmetrical about 0ω .


99

Let us finally return to the circuit and consider the voltage across the

capacitor, VC. It is easy to write

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ω

−ω+ω

=ω

=

CLjRCj

VCjIV S

C 1

and hence

( ) LCjCRjVV

S

C21

1ω+ω+

=

Now, if we recall that LC=ω0 and CRRLQ 00 /1/ ω=ω= then we

may write

VCVS

=1

1+ j 1Q

ωω0

!

"##

$

%&&+ j ω

ω0

!

"##

$

%&&

2

We see that the denominator here is of the standard form we discussed

earlier where the Q factor is evidently related to the damping factor, ζ,

via

ς=21Q

100

The Bode diagram corresponding to ( ) SC VVjG /=ω are shown below

where it is again clear that a high Q – low damping -- results in a sharp

resonance peak.


101

[i.e. Exactly this frequency response function is plotted in HLT page

168.]

Example

In a particular series RLC circuit, R = 10 Ω , L = 1 mH and C = 0.1 µF

and the source VS = 2.0 V rms.

The resonant frequency, ω0, is given by

rad/secLC

50 101

==ω

which may be expressed as kHzf 9.152/00 =πω= . The Q of the

circuit is

100 =ω

=RL

Q

and hence the bandwidth, Q/0ω=ωΔ , or kHz.Qff 59.1/00 ==Δ

The current at resonance is

ARV

I S 2.00 ==

and the magnitude of the voltage across either the capacitor or the

inductor at resonance is

102

VQVV S 20==

General remark on Q

We have defined Q in a way that is convenient for our ‘electrical’

purposes. However, since all resonant systems – mechanical and

electrical – have a common basis in energy, it is also possible to show

that

cycle per dissipated energystored energyQ π= 2

Parallel resonance

Another common circuit displaying resonance is the RLC parallel circuit

shown below.


103

The admittance of the circuit (Y = 1/Z) is given by

⎟⎠

⎞⎜⎝

⎛ω

−ω+=L

CjR

Y 11

This is analogous to the equation for Z of the series resonant circuit but

with impedance/resistance/reactance replaced by

admittance/conductance/susceptance. The current will be a maximum at

the resonant frequency

LC1

0 =ω

and will have the same properties as the voltage has in a series

resonant circuit.

By analogy, Q for the parallel circuit will become

CRLRQ 00

ω=ω

=

but will still equate to

Q =ω0

Δω .

104

A practical resonant circuit

Because inductors always have some resistance, a more practical

parallel resonant circuit is given below.

Unfortunately the analysis is more complicated but with a little maths we

can show that this practical circuit is equivalent to

with

( ) ( )LLR*Land

RLRR 2

2222

*ω

ω+=

ω+=


105

For high enough Q, we can use the simple approximations in HLT (page

167).

106

20. Mutual Inductance

The action of the inductor relies on the presence of a varying current to

give rise to a varying magnetic field, which then induces a voltage in the

coil that produced it. Let’s wind a conductor around a high permeability

core and pass a current i through it:

You learnt in P4 that when there is a closed magnetic path (i.e. the

complete path is not shown in this figure) and no flux leakage,

φ = (µA/l) Ni and v = N dφ/dt

so that

v = L di/dt , where L = N2 (µA/l) = φ/i

We call this self inductance


107

Now let’s take two windings following around core and once more

assume no flux leakage. For good measure, suppose the first wire has

N1 turns and the second wire has N2 turns.

A flux is produced in the core from both currents and totals

φ = (µA/l) N1i1 + (µA/l) N2i2

Equally, if this flux changes, it induces a voltage in both windings

V1 = N1 dφ/dt and V2 = N2 dφ/dt

And this is what we call mutual inductance.

Combining these equations we can write

v1=L1

di1dt+M

di2dt

and

v2 =M

di1dt

+L2di2dt

where

L1 = N12 (µA/l) and L2 = N2

2 (µA/l)

are the self inductances of the two coils and

M = N1N2 (µA/l)

108

is the mutual inductance between the two coils.

Perfect Coupling

Note that in this ideal world of perfect flux coupling, M = √(L1L2).

Magnetic Coupling Coefficient

In practice we always get some flux leakage, i.e. not all the flux stays in

the core and therefore not all the flux links all the coils. We should

therefore expect that

L1 < N12 (µA/l) and L2 < N2

2 (µA/l)

and

M < √(L1L2)

We can then write

M = k . √(L1L2)

where k (0 < k < 1) is known as the “coupling coefficient”.


109

The dot notation

We assumed above that both coils were wound in the same direction

around the core. Of course if we reverse the direction of one winding, its

flux and its induced voltage will change sign so it is important that we

know the relative directions of our two windings. The dot notation is

used to specify the ends of the coils that have the same polarity.

In order to see how to use the dot notation, let's consider the coupled

circuit below.

110

When, as here, we define i1 and i2 as both “into” the dotted end and also

v1 and v2 as both “to” the dotted ends, the terms in our mutual coupling

equations will all have the positive signs that we saw above, i.e.

v1=L1

di1dt+M

di2dt

and

v2 =M

di1dt

+L2di2dt .

All your analysis can be performed using these standard equations to

represent mutual coupling, but some people like to redraw the circuit

with the mutual coupling replaced by dependent sources, thus:

Strictly speaking the dots are now superfluous but have been retained to

help avoid errors when marking the polarity of the induced e.m.f.. We

emphasise that we have taken account of the coupling by introducing

v1 v2


111

the voltage sources. We may now analyse each circuit independently,

giving of course:

v1=L1di1dt+M

di2dt

and

v2 =Mdi1dt

+L2di2dt

Should you choose to define a current or a voltage in the opposite

direction, you will have to change the sign of that variable in these

equations. Therefore I strongly recommend committing these two

equations to memory along with notation that the voltages are measured

to the dotted ends and the currents go into the dotted ends so that all the

terms appear positive. Otherwise it is very easy to finish up with the

wrong sign. If you decide that you actually want to work with the current

coming “out of” the right hand side say, it is a simple matter to define it

as a new variable, e.g. i3 = –i2.

Finally we note that if you are working in the frequency domain with

phasors, the equations become

2212

2111

ILjMIjVMIjILjV

ωω

ωω

+=

+=

112

21. Mutually coupled circuits in series

Using the definitions of v1, v2, i1 & i2 above and our two mutual coupling

equations, the series connections here can be expressed by the

following three further equations:

21

21

iiiv vv

==

+=

.

From these five equations, we can eliminate v1, v2, i1 & i2 to obtain:

( )dtdiMLLv

dtdiL

dtdiM

dtdiM

dtdiLv

221

21

++=

+++=


113

and so we see that effective inductance of the circuit is MLL 221 ++ . In

this case the coils have been connected so that the mutual effect is to

increase the effective inductance. If the coils were re-arranged and

connected so that the dots appear as

the connections can now be expressed by

21

21

iiiv vv−==

−=

and hence

dtdiL

dtdiM

dtdiM

dtdiLv 21 +−−=

and in this case the effective inductance is MLL 221 −+ .

We note that when the coupling is perfect, 21LLM = , the effective

inductance in the first configuration (series aiding) may be written as

( )221 LLLeff +=

whereas in the second configuration (series opposing),

v

114

( )221 LLLeff −=

For the special case of LLL == 21 gives LLeff 4= in the first case and

zero in the second case! (Both answers have simple practical

consequences – can you think what these might be?)

It is left as an exercise to confirm that you can use the dot notation

confidently to show that the circuit below

has an effective inductance of 12mH. It's easy but take care!!


115

22. The transformer

In essence a transformer is nothing more than a pair of coupled coils

with the expectation that we will supply power to one side (which we call

the “primary” winding) with N1 turns and take power out of the other side

(the “secondary” winding) with N2 turns. We call N1 / N2 the “turns ratio”

N.

Our standard equations are

and, because we like to se a current coming out of the secondary, we

need the supplementary equation

I2’ = – I2

2212

2111

ILjMIjVMIjILjV

ωω

ωω

+=

+=

I2’

116

In an “ideal transformer” with no flux leakage, 221

2212 NNNLL == and

21LLM = so we can write the equations for V1 and V2 in terms of L1

alone as:

V1 = jω L1 I1 + jω NL1 I2

V2 = jω NL1 I1 + jω N2L1 I2

From which we deduce that

111

22 NVV

NNV ==

Thus we can use a transformer to “transform” power from one voltage to

another by selecting an appropriate turns ratio. For example, to "step

down" the mains 230V & 50 Hz to something more useful for electronic

circuits, say 12V. Alternatively a transformer can be used to "step up"

voltages from a generator for transmission over large distances.

Since our transformer does not dissipate any power, it is tempting to

think that V.I is the same at the primary and secondary and that when

the voltage is 'transformed' by a factor N the current will be 'transformed'

by the inverse factor 1/N. However, this would to be to ignore the phase

shift in the output current. By eliminating I2 we obtain

''2

1

12

11

11 NI

LjVI

LM

LjVI +=+=

ωω


117

from which we see that the ideal current relationship will be approached

by making 1L large. That’s why we like lots of windings!

We conclude this section by working out the input impedance of an

ideal transformer when it is connected to a load ZL. When the load is in

place we have LZIV '22 = and the equations become

'2111 IMjILjV ωω −=

''2212 ILjMIjZI L ωω −=

The input impedance is defined as 11 IVZin = which, after a little algebra

gives

( )L

Lin ZLj

LLMZLjIVZ

+ω

−ω+ω==

2

2122

1

1

1

For our ideal ideal transformer with 212 LLM = when the inductances are

large enough we approximate

Lin ZLLZ2

1=

or

118

2NZZ L

in =

Therefore, as well as being useful for transforming currents and voltages

transformers are also used to transform impedances. In this case the

source "sees" an effective load of 2NZL . This can be useful when

coupling signals to very low impedance loads. Indeed if you elect to use

earphones in the radio you will build in the DBT exercise you may have

occasion to use a transformer in this way.


119

23. Transient response

In our previous analysis of circuits we were concerned with the 'steady

state' solutions for current and voltage, which occur long after the source

or sources have been connected to the circuit. Since the current/voltage

relationships for capacitors and inductors involve time derivatives they

clearly do not respond instantly to abrupt changes in current and

voltage. It takes time for the effects of an abrupt change to die away

and for the final steady state conditions to be reached.

When we began discussing AC circuit theory we set up the governing

differential equation and proceeded to solve it using standard

mathematical techniques. The full solution consisted of two terms. The

first, the complementary function, corresponded to the transient

response which eventually decayed away whereas the second, the

particular integral, led to the final steady state response. In the RL

circuit you built in the laboratory the steady state was achieved in a

matter of milliseconds and so the transient component was not evident

in the measurements you took. However, a millisecond, or even a

microsecond, can be a long time in electrical engineering and hence

transients deserve our attention.

We will now discuss in more detail the effects of abruptly connecting or

disconnecting sources to electrical circuits. Previously we have only

dealt with sinusoidally varying sources but we will now remove this

restriction. In order to introduce the topic logically we will begin by

considering first order systems; so called because they contain only

resistive and one reactive element and lead to first order differential

equations before moving on to second order systems containing

120

resistors and two reactive elements. At first we will solve the differential

equations using standard 'classical' mathematics before introducing a

powerful (and simple) technique based on the Laplace transform.

First order transients

We begin by asking how the voltage across the capacitor increases with

time as it is charged by the battery after the switch is closed. We

assume the capacitor is initially uncharged.

When the switch is closed

vRiE +=

The current/voltage relationship for the capacitor gives

dtdvCi =

and so


121

vdtdvRCE +=

or

RCE

RCv

dtdv

=+

where, evidently, as we have seen before, the product RC must have

the dimensions of time. Thus we set RC=T and we formally now need to

solve

TE

Tv

dtdv

=+

We begin by considering the complementary function as the solution of

0=+Tv

dtdv . A suitable candidate is ( )mtAexp which leads to an auxiliary

equation

Tm

Tm 1;01

−==+

and hence the complementary function is TtA −exp . The particular

integral is determined by the specific forcing function applied. In this

case it is constant (a battery) and hence a suitable particular integral is

Ev = . The full solution is provided by

122

v = Complementary function + Particular integral

v = ETtA +−exp

The unknown constant, A, is determined by the initial conditions, i.e. the

value of v just after the switch is closed. Since the capacitor was initially

uncharged

( ) 00 == +tv

where the notation, 0+, is used to indicate time just after closure of the

switch. This condition leads to EA −= and hence the full solution is

given by

⎟⎠

⎞⎜⎝

⎛ −−=TtEv exp1 .


123

The factor RCT = is known as the time constant and indicates how

quickly the capacitor charges.

We can now make some general remarks about the form of this solution

(i) The complementary function, TReA − , determines the transient

response which decays away at a rate determined by the time

constant, T. The unknown constant, A is determined by the initial

conditions.

(ii) The particular integral leads to the final steady state solution. As

such it is determined by the specific forcing function – constant in

this case, (sinusoidal in our previous discussion) – and not by the

initial conditions. It is, of course, independent of the initial

conditions.

Let us now look at what happens after the capacitor has charged up to a

voltage E and is discharged through a resistor in the manner shown

below.

124

Again we want to determine the variation of v with time. The governing

differential equation is as before but with zero forcing function

0=+Tv

dtdv

which gives a complementary function

TtE

TtBv −=−= expexp

where the unknown constant, B has been determined from the initial

condition that the capacitor was charged to in voltage E

( ) Ev =+0

Formally we note that the particular integral is zero indicating, as it must,

that no voltage appears across a discharged capacitor.


125

Let us now consider the following RL series circuit where we assume no

current to be flowing in the inductor before the switch is closed.

Kirchhoff leads to

ERidtdiL =+

or

LEi

LR

dtdi

=+

As before we note that TRL = must have the units of time and hence

LE

Ti

dtdi

=+

126

which is mathematically, formally, equivalent to the differential equation

we previously encountered. The complementary function is again given

by

TtA −exp

with time constant RLT = and unknown constant A. The particular

integral is

RE

LTE=

and hence the full solution

RE

TtAi +−= exp

The initial condition, ( ) 00 =+i leads to REA −= and hence

⎭⎬⎫

⎩⎨⎧ −−=

Tt

REi exp1

which again, has the standard form with a time constant, RLT = .

It is left as an exercise to show that if we discharge the inductor through

the resistor that the current decays as Tt

REi −= exp .


127

General remark

In the case of both the RC and RL circuits we have essentially solved a

first order differential equation with a time constant T. This led to a

complementary function for, say voltage, v, but it could equally be

current – it depends on the specific case – of the form

TtAv −= exp

together with a particular integral giving the final steady state value, VSS

SSVv =

The unknown constant, A, is determined by the initial condition ( )+= 0vv .

Thus

( ) SSVAv +=+0

or

( )[ ] SSSS VTtvVv +−−= + exp0

or,

[ ] valuestatesteadyvalueinitialvaluestatesteady +−−=Ttv exp

128

or, in words,

the variable rises (falls) from its initial value to its final steady

state value exponentially with a time constant T


129

This observation permits us to sketch the transient response of many

simple circuits by inspection.

Initial Conditions

In order to sketch the transient response directly we need to be able to

find the voltage and current in a resistor, inductor and capacitor

immediately after a switch has been open or closed.

Resistor

Since the current and voltage are related by IRV = it is clear that any

instantaneous change in, say, voltage will be accompanied by an equally

instantaneous change in current.

Inductor

Since the current and voltage are related by

dtdiLv =

a sudden change in current would result in an infinite voltage. Since this

is implausible we can conclude that closing a switch to connect an

inductor to a source will not cause current to flow at the initial instant,

+= 0t , i.e.

( ) 00 =+i

130

i.e. it will act as an open circuit.

If, of course, a current 0I was flowing before the switch was closed it will

continue to flow and

( ) 00 Ii =+

These observations permit us to introduce the idea of an equivalent

circuit at += 0t

Capacitor

In this case

dtdvCi =

and arguments analogous to those above require that voltage does not

change instantly since the current cannot be infinite. If the capacitor is

initially unchanged then

( ) 00 =+v


131

i.e. the capacitor acts as a short circuit. If the capacitor is initially

changed to 0V

( ) 00 Vv =+

24. Example

If the capacitor is initially uncharged it acts like a short circuit at += 0t

and hence

132

( ) REi =+0

The final steady state current is zero and hence the transient response

may be written as

( ) RCtREti −= exp

We note that if the resistance were small (zero) that ( )ti would be very

large for a very short time. Although the 0=R case could not happen in

practice it is often a useful "special case" and the discontinuity in current

is described by an impulse function, which has infinite magnitude for a

vanishingly short time.

As a further example we investigate the current flowing in the circuit

below after the switch is closed.


133

When the switch has been open long enough for the inductor to act as a

short circuit to DC a current ( )21 RREi += and a time constant 1RLT =

permits us to sketch

134

Second order transient

Consider

Kirchhoff gives

cvdtdiLRiE ++=

which together with dtdv

Ci c= gives a second order differential equation

for the voltage across the capacitor as

LCd 2vcdt 2

+RCdvcdt

+vc =E

We now solve the equation by our standard method of first obtaining the

complementary function by setting 0=E on the right hand side and

looking for solutions of the form mtAexp . The auxiliary equation takes

the form


135

012 =++ mRCmLC

which has solutions

LCLR

LRm 1

42 2

2

−±−=

In order to make some general comments we now recast this into non

circuit-specific notation following our discussion of resonance. In that

context we introduced

RLQ

LC0

01 ω

==ω and

We also noted that Q was related to the damping factor

LCR

Q 221==ζζ via . This permits us to write

⎟⎠⎞⎜

⎝⎛ −ζ±ζ−ω= 10

2m

or

β±α−=m

with ζω=α 0 and 120 −ζω=β .

136

We note that the particular integral, - corresponding to the steady value

of vc – is simply Evc = . The full solution is therefore

{ } ttBtAEvc α−β−+β+= expexpexp

where the unknown constants are determined by the initial conditions,

0=cv and 0=i . We note that the latter condition is formally equivalent

to requiring 0=dtdvc . The full solution now takes the form

( ){ ( ) } tttEEvc α−β−α−β+βα+ββ

−= expexpexp2

It is clear that according to the value of ζ the factor β may be real or

imaginary. We consider these cases in turn

Case 1: ζ >1; R>2 L C

In this case β is real and α>β. It is convenient to write the solution

as

( ) ( ){ ( ) ( ) }ttEEvc β+αβ−α−β−α−β+αβ

−= expexp2

where we see that the transient solution of the response is the difference

between two exponentials and is negative. The voltage cv follows the


137

usual gradual rise to its steady state value typical over overdamped

systems.

We note that the current is given by

dtdv

Ci c=

or

( ) ( ){ }ttEC

i β+α−−β−α−β

ω= expexp

2

20

Thus the current is proportional to the difference in two exponentials of

equal amplitude but different decay rates and hence rises from zero to a

maximum before finally falling to its steady state value of zero

138

Case 2: ζ =1 ; R=2 L C

For this special case of critical damping the general solution is no longer

valid since the auxiliary equation has two equal roots. The

complementary function in this case must take the form

( ) tDtC 0exp ω−+

and hence the full solution becomes

( ) tDtCEvc 0exp ω−++=

which with initial conditions 0== iv c gives

( ){ }ttEvc 00 exp11 ω−ω+−=

which gives a response graphically similar to the previous case.


139

Case 3: ζ <1 ; R<2 L C

In this case β becomes imaginary and so we write

njj ω=ζ−ω=−ζω=β 20

20 11

with

2

0 1 ζ−ω=ωn

It is now straightforward to recast the general solution of page 131 by

replacing β with nj ω as

⎭⎬⎫

⎩⎨⎧

α−⎟⎟⎠

⎞⎜⎜⎝

⎛ω

ω

α+ω−= tttEv n

nnc expsincos1

In this case the solution is very different from before. The transient

solution is oscillatory, at a frequency nω , but its amplitude decays at a

rate determined by ( )0ωζ=α and we speak of underdamped oscillation.

The frequency 20 1 ζ−ω=ωn is the natural frequency. The same term

is sometimes applied to 0ω but since the two are strictly only equal when

1=ζ (critical damping)

140

Case 4: ζ =0;R=0

In this case there is no damping since 0=R means there is no element

in which energy can be dissipated. The voltage across the capacitor in

this undamped case is given by

{ }tEvc 0cos1 ω−=


141

We observe that for small damping, representing minimal loss, that the

frequency of the oscillatory transient 02

0 1 ω≈ζ−ω=ωn , the resonance

frequency in the steady state. This is not surprising since this is the

frequency at which energy naturally oscillates between the reactive

elements and this is actually what is happening during transient

oscillation. Naturally with increased damping the effect is reduced.

We therefore see that the damping factor ζ (or Q) has an important role

in describing the transient behaviour:

If the damping is less than critical (ζ<1) the transient

behaviour is oscillatory otherwise it is not.

142

25. The Laplace Transform

The classical approach we have just used to find the transient response

and indeed the final steady state response for simple circuits may be

extended to ever more involved circuits but at the expense of ridiculously

increased complexity. An alternative approach is required. Since we

are concerned with turning forcing functions, often constants, on and off

the mathematical technique of choice to solve the differential equations

is based on the use of the Laplace transform. This is particularly useful

for our purposes because

(i) it gives rise to the ability to write circuit equations in a very general

way just as the "jω" method does for sinusoidal forcing functions.

(ii) the initial conditions are dealt with automatically. The solution does

not contain any arbitrary constants.

(iii) Transients are dealt with automatically.

(iv) and, very importantly, the answers can be looked up easily in

H.L.T.!!

We recall that the Laplace transform of a function of time ( )tf is

defined as

( ){ } ( ) ( ) dtsttfsFtfL −== ∫∞

exp0


143

where negative values of t are excluded. In mathematics since only

positive values of the variable t are permitted this is often called 'one-

sided' transform.

Arguably one of the most useful properties, from our point of view, of the

Laplace transform, is that for a given ( )tf there is one and only one

( )sF and vice versa ( )sF and ( )tf are called transform pairs.

( ) ( )tfsF ↔

Hence if we know the value of ( )sF and we want to know ( )tf , or vice

versa, we simply look it up in the tables!! What could be easier?

Although you will have worked out Laplace transforms in maths we'll

introduce a few 'electrically useful' transform pairs below.

(i) the unit step, ( )tu

This function, which is useful to describe the closing of a switch say is a

function that switches from 0 abruptly to 1 at 0=t . Thus the terminal

voltage in the circuit below would be written ( )tu10

144

The Laplace transform ( )sU is given by

( ) ( )s

dtstdtsttusU 1exp1exp00

=−=−= ∫∫∞∞

Hence

( )s

tu 1↔

are Laplace transform pairs. Thus given ( )s

sU 1= the corresponding

function of time is the unit step, ( )tu .

(ii) exponential decay

( ) 0;exp >α−= tttf

( ) ( )α+

=α+−= ∫∞

sdttssF 1exp

0


145

(iii) damped sinusoids

( ) 0;sincos

exp >⎩⎨⎧

ω

ωα−= t

tt

ttf

Here it's probably easiest to consider the composite function

( ) { } ( )tjtjtttg ω−α−=ω+ωα−= expsincosexp

since this is formally equivalent to the exponential function we have just

discussed we can write the transform down directly as

( )( ) ( ) ( ) ω+α+

ω+

ω+α+

α+=

ω−α+= 222

1s

jss

jssG

and hence by equating real and imaginary parts we find

146

{ }( )

{ }( ) 2222 sinexp;cosexp

ω+α+

ω=ωα−

ω+α+

α+=ωα−

sttL

ssttL

(iv) Laplace transform of a derivative

If the Laplace transform of ( )tf is ( )sF then

( ) ( )+−=⎭⎬⎫

⎩⎨⎧ 0fsFsdtdfL

Where ( )+0f represents the initial value of ( )tf . The corresponding

expressions for the second derivative is

( ) ( ) ( )++ ʹ′−−=⎭⎬⎫

⎩⎨⎧

0022

2

ffssFsdtfdL

where ( )⋅f denotes the first derivative.

Fortunately for you in the P2 course (though not in the P1 course), we

will only use Laplace in the case of zero initial conditions when these

expressions simplify to

( )sFsdtdfL =

⎭⎬⎫

⎩⎨⎧

and ( )sFs

dtfdL 22

2

=⎭⎬⎫

⎩⎨⎧

(v) Laplace transform of an integral

( ) ( )ssFdttfL

t=⎭⎬⎫

⎩⎨⎧∫0


147

Having listed a number of Laplace transforms let's now illustrate their

use formally by returning to the RC circuit we considered previously but

redrawn below

Since the switch is assumed to be closed at 0=t the governing

differential equation may be written very generally as

( )tuTE

Tv

dtdv

=+

Where, as before, RCT = and the unit step ( ) 00 <= ttu and 1, 0>t .

We now take Laplace transforms of all terms in the equation to give

( ) ( ) ( )sT

ETsVvsVs 1.0 =+− +

If we assume, as before, that the capacitor is initially uncharged

( )00 =+v , gives

148

( )( )TssT

EsV11.+

=

The form of ( )sV is almost familiar, but not quite. In order to be able to

find the inverse transform we need to rearrange it as

( )⎭⎬⎫

⎩⎨⎧

+−=

TssEsV

111

which are functions we have met before and permits us to transform

back to the time domain to give

( ) ( ){ }TttuEtv −−= exp

which, if we restrict ourselves to 0>t , may be written in a more familiar

way as

( ) { }TtEtv −−= exp1

which is, of course, the solution we obtained previously but now with

considerably greater ease since we had no arbitrary constants to find.

We emphasise that the procedure involves

(i) transform the differential equation and include the initial conditions


149

(ii) manipulate the equation to obtain ( )sV is an easily 'invertible' form.

This, sadly, may involve the use of partial fractions in more involved

cases.

(iii) transform back to the time domain to obtain ( )tv .

It would be a useful exercise to return now to the second order RLC

circuit, transform the governing differential equation to obtain the

Laplace transform ( )sVc of ( )tvc with zero initial conditions as,

( ) ( )

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛ +

+−= 22

11

2

1

LR

LCLRs

LRss

EsVc

It is now routine – try it – to transform back to the time domain to

reproduce the damped and undamped cases we discussed previously.

We end this section by listing a further few useful properties of the

Laplace transform.

150

The Shifting theorem

We show below a function ( )tf along with a "shifted" version. The

function ( )Ttu − is a unit step that is switched on at Tt = .

If ( )sF denotes the Laplace transform of ( )tf then

( ) ( ){ } ( )sFsTTtuTtfL −=−− exp

i.e. when the time function is shifted along by T the Laplace transform is

multiplied by sT−exp .

This property is very useful when building up the transforms of

complicated functions. Suppose we want to consider a pulse that might

correspond to the opening and subsequent closing of a switch. The time

function, ( )tg , might look like


151

that is a unit step up at 0=t together with a unit step down at Tt = .

Thus

( ){ } ( ) ( ){ }

ssT

ssT

s

TtutuLtgL−−

=−

−=

−−=

exp1exp1

In reverse if we are required to find the inverse transform of, say,

( ) 22.expω+α+

α+−

sssT

we recognise this as the inverse transform of ( ) 22 ω+α+

α+

ss , i.e.

tt ωα− cosexp but "shifted" by T. Thus the inverse transform is given by

( )( ) ( )TtTt

sssTL −ω−α−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

ω+α+

α+−− cosexpexp

221

152

Thus we look up the transform in tables as usual and then introduce the

shifting function.

The tables which are available to you in the examination are to be found

in H.L.T. and are reproduced below for convenience.

Unit step at ( )tut ,0= s1

( )!11 −− ntn ns1

ate− ( )as +1

( )( )atea −−11 ( ){ }ass +1

atcos ( )22 ass +

atcosh ( )22 ass −

atsin ( )22 asa +

atsinh ( )22 asa −

( )( )ata cos11 2 = ( ){ }221 ass +

( )( )atata sin1 3 − ( ){ }2221 ass +

atte − ( )21 as +

( )ate at −− 1 ( )2ass +

( )( )atatata cossin21 3 − ( )2221 as +

( ) atat sin2 ( )222 ass +

att cos ( ) ( )22222 asas +−

bte at cos− ( ) ( ){ }22 basas +++

bte at sin− ( ){ }22 basb ++


153

26. The Laplace transform in circuit analysis. If we wanted to find how the voltage across a certain component varied

with time due to some arbitrary forcing function it would be perfectly

possible, using appropriate combinations of node-voltage and loop

equations, to set up the required (simultaneous) differential equations.

We could then take Laplace transforms of both sides of all the equations

or, to put it more technically, we would transform from the time domain

to the s-domain. The resulting simultaneous algebraic equations could

then be solved to find an expression for the Laplace transform of the

desired ‘output’ in terms of the Laplace transform of the ‘input’ source.

Once this frequency response function has been obtained it becomes a

routine matter to introduce the actual transform of the source and then

transform back to the time domain by looking up the inverse transforms

in tables. However it is possible to miss out the initial steps of writing the

differential equations and transforming them by working directly with

transformed voltages and currents V(s) and I(s) together with the

appropriate s-domain version of ‘Ohm’s Law’ for resistors, capacitors

and inductors.

Representation of circuit elements in the s-domain.

Resistor

The v-i equation for a resistor in the time domain is

iRv =

154

The Laplace transform of this equation is

)()( sIRsV =

The transfer function RsZsIsV == )()(/)( is the s-domain impedance

of the element.

Inductor

In this case the time domain v-i relationship is

dtdiLv =

which, when transformed into the s-domain assuming zero initial

conditions becomes

V(s) = sL I(s)


155

Capacitor

Finally we consider the v-i characteristics of a capacitor as

dtdvCi =

which becomes in the s-domain, assuming zero initial conditions,

I(s) = sC V(s)

or

V(s) = (1/sC) I(s)

I hope that, by now, although this approach is new, it is beginning to feel

a little familiar. When we were interested in sinusoidal sources (forcing

functions) we found that the mathematical technique which enabled us

to solve the differential equations most easily required us to write ‘Ohm’s

Law’ for the resistor, inductor and capacitor, in phasor notation as

156

ICj1VandILjVIRVω

=ω==

respectively. In the more general s-domain case the relationships are

IsC1VandIsLVIRV ===

which are seen to be identical if we merely replace jω by s. Thus all

the methods we developed in the AC theory ‘jω’ case carry over to the s-

domain by simple substitution.

Example

Let’s revisit the RC circuit we have already analysed twice before

already i.e.

Where )()(1 tuEtv = since the switch is closed at t=0. Since the

capacitor is assumed to be uncharged the circuit may be redrawn s-

domain notation below. We have also re-arranged the circuit to make it

clearly of the form of a voltage divider.


157

We can immediately write

( ) ( )

( )sVsCR

sV

sCR

sCsV

1

12

11

1

1

+=

+=

which, introducing T = RC and sEsV /)(1 = gives

( )sTsEsV+

=1

)(2

which is the same function we obtained previously but we have found it

much more easily and directly here.

As a further example consider the circuit below together with its s-

domain representation.

158

In this case

( )sV

sCR

RsV 12 1)(

+=

which with sEsV /)(1 = and T = RC gives

Ts

EsV1

)(2+

=

The inverse is readily looked up in tables to give

TtEtv /exp)(2 −=

We know, from physical reasoning, that when the switch opens,

Ev =+ )0(2 , whereas when it has been open for a long time the capacitor

acts as an open circuit and hence 0)(2 =∞v . These observations are

confirmed, of course, by the expression we have just obtained. However

it can also be checked from the form of V(s) by using the initial and final

value theorems. We know


159

{ } EsT

ELim

Ts

sELimssVLimvalue initial sss =⎭⎬⎫

⎩⎨⎧

+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+== ∞→∞→∞→ 11

)(2

whereas the final value is given by

{ } 01

)( 020 =⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+== →→

Ts

sELimssVLimvalue final ss

We note the procedure is

(i) Introduce ‘generalised’ s-domain impedances, sL and 1/sC.

(ii) Find the appropriate transfer function. In more complicated cases

this may/will involve writing node-voltage and/or loop equations.

(iii) Substitute the Laplace transform of the input signal (forcing

function)

(iv) Convert back to the time domain by looking up the inverse

transforms in tables.

As a final example let’s consider the following circuit where the input

voltage is a pulse of duration T and we are required to find the voltage

)(2 tv .

160

The generalised impedance of the RC parallel combination is given by

( )CRsR +1/ and hence, after a little trivial algebra

( )( ) 111

2

/11

21

TsTsVsV

+=

where CRT =1 .

We have already shown how the shift theorem may be used to obtain

the Laplace transform of a pulse by regarding it as the superposition of

two time-delayed step functions. The details were given previously

which permit us to write

( ) ( )sTsEsV −−= exp11


161

and hence

( )( )

( )

( )sTTss

E

sTsTssT

EsV

−−⎭⎬⎫

⎩⎨⎧

+−=

−−+

=

exp1/111

2

exp1/11

2

1

112

The inverse of the expression in curly brackets is readily obtained as

1/exp)( Tttu −− . The second bracket merely indicates, as we have

seen before that the full solution consists of the difference between two

identical, but time shifted, expressions. Thus

( ) [ ]{ }112 /)(exp)(/exp)(2

TTtTtuTttuEtv −−−−−−−=

which, depending on the relationship between T and T1, might look

something like

162

27. Disclaimer!!

The notes are reasonably self-contained but they certainly don’t pretend

to tell the whole story so do consult one or more of the huge number of

textbooks on the subject.

If you spot any errors, do please tell me: [email protected].

Let’s illustrate this by considering the series RLC ...

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