circuit analysis ii - university of oxford · circuit analysis ii (ac circuits) syllabus complex...

Will Moore

MT 12

CIRCUIT ANALYSIS II

(AC Circuits)

Syllabus

Complex impedance, power factor, frequency response of AC networks

including Bode diagrams, second-order and resonant circuits, damping

and Q factors. Laplace transform methods for transient circuit analysis

with zero initial conditions. Impulse and step responses of second-order

networks and resonant circuits. Phasors, mutual inductance and ideal

transformers.

Learning Outcomes

At the end of this course students should:

1. Appreciate the significance and utility of Kirchhoff’s laws.

2. Be familiar with current/voltage relationships for resistors, capacitors

and inductors.

3. Appreciate the significance of phasor methods in the analysis of AC

circuits.

4. Be familiar with use of phasors in node-voltage and loop analysis of

circuits.

5. Be familiar with the use of phasors in deriving Thévenin and Norton

equivalent circuits

6. Be familiar with power dissipation and energy storage in circuit

elements.

2

7. Be familiar with methods of describing the frequency response of

AC circuits and in particular

8. Be familiar with the Argand diagram and Bode diagram methods

9. Be familiar with resonance phenomena in electrical circuits

10. Appreciate the significance of the Q factor and damping factor.

11. Appreciate the significance of the Q factor in terms of energy

storage and energy dissipation.

12. Appreciate the significance of magnetic coupling and mutual

inductance.

13. Appreciate the transformer as a means to transform voltage, current

and impedance.

14. Appreciate the importance of transient response of electrical

circuits.

15. Be familiar with first order systems

16. Be familiar with the use of Laplace transforms in the analysis of the

transient response of electrical networks.

17. Appreciate the similarity between the use of Laplace transform and

phasor techniques in circuit analysis.

Circuit Analysis II WRM MT12

3

AC Circuits

1. Basic Ideas

Our development of the principles of circuit analysis in Circuit Analysis I

was in terms of DC circuits in which the currents and voltages were

constant and so did not vary with time. Here in Circuit Analysis II we

extend our analysis to consider time varying currents and voltages and

thereby we are able to extend our analysis to include capacitors and

inductors. In our initial discussions we will limit ourselves to sinusoidal

functions. We choose this special case because, as you have now

learnt in P1, it allows us to make use of some very powerful and helpful

mathematical techniques. It is also a common waveform in nature and it

is easy to generate in the lab. However as you have also learnt in P1,

any waveform can be expressed as a weighted superposition of

sinusoids of different frequencies and hence if we analyse a linear circuit

for sinusoidal functions we can, by appropriate superposition, handle

any function of time.

Let's begin by considering a sinusoidal variation in voltage

tVv m cos

4

in which is the angular frequency and is measured in radians/second.

Since the angle t must change by 2 radians in the course of one

period, T, it follows that

2T

However the time period f

T1

where f is the frequency measured in

Hertz. Thus

fT

22

This is a simple and very important relationship. We naturally measure

frequency in Hz – the mains frequency in the UK is 50Hz – and it is easy

to measure the time period, fT 1 from an oscilloscope screen.

However as we will soon see, it is mathematically more convenient to

work in terms of the angular frequency . Mistakes may be easily made

because in practice the word frequency is commonly used to refer to


5

both and f. It is important in calculations to make sure that if

appears, then the correct value for f = 50 Hz, say, is = 100 rads/sec.

A simple point to labour I admit, but if I had a pound for every time

someone forgets and substitutes = 50 . . . . . . . . !!

In our example above, tVv m cos , it was convenient that

0 tVv m at . In general this will not be the case and the waveform will

have an arbitrary relationship to the origin t = 0 or, equivalently the origin

may have been chosen arbitrarily and the voltage, say, may be written in

terms of a phase angle, , as

tVv m cos

Alternatively, in terms of a different phase angle, , the same waveform

can be written

T

6

tVv m sin

where

2

The phase difference between two sinusoids is almost always measured

in angle rather than time and of course one cycle (i.e. one period)

corresponds to 2 or 360. Thus we might say that the waveform above

is out of phase with the earlier sinusoid by . When 2 we say

that the two sinusoids are said to be in quadrature. When the

sinusoids are in opposite phase or in antiphase.

2. RMS Values

We refer to the maximum value of the sinusoid, Vm, as the “peak” value.

On the other hand, if we are looking at the waveform on an oscilloscope,

it is usually easier to measure the “peak-to-peak” value 2Vm, i.e. from the

bottom to the top. However, you will notice that most meters are

calibrated to measure the root-mean-square or rms value. This is found,

as the name suggests, for a particular function, f, by squaring the

function, averaging over a period and taking the (positive) square root of

the average. Thus the rms value of any function f(x), over the interval x

to x+X, where X denotes the period is


7

dyyfX

fXx

xrms

21

For our sinusoidal function tVv m cos

The average of the square is given by

dttVT

m

T

22

0

cos1

where the time period 2T . At this point it's probably easiest to

change variables to t and to write 2cos12

1cos2 . Thus the

mean square value becomes

2

2cos122

1 22

0

2mm V

dV

The root mean square value, which is simply the positive square root of

this, may be written as

Vrms = Vm /√2 ≈ 0.7 Vm.

8

Since we nearly always use rms values in our AC analysis, we assume

rms quantities unless told otherwise so by convention we just call it V as

in:

V = Vm /√2.

So, for example, when we say that the UK mains voltage is 230V what

we are really saying is that the rms value 230V. Its peak or maximum

value is actually 230√2 ≈ 325 V.

To see the real importance of the rms value let's calculate the power

dissipated in a resistor.

Here the current is given by Rvi

tItR

Vi m

m coscos


9

where RVI mm and the power, vip , is given by

tR

Vp m 2

2

cos

If we want to calculate the average power dissipated over a cycle we

must integrate from 20 Ttt to . If we again introduce t ,

the average power dissipated, P, is given by

dR

VP m 2

2

0

2

cos.2

1

dR

VP m 2cos1

2

1.

2

1 2

0

2

RVP m 22

If we now introduce the rms value of the voltage 2mVV then the

average power dissipated may be written as

RVP 2

Indeed if the rms value of the current 2mII is also introduced then

RIRVP 22

10

which is exactly the same form of expression we derived for the DC

case.

Therefore if we use rms values we can use the same formula for the

average power dissipation irrespective of whether the signals are AC or

DC.

3. Circuit analysis with sinusoids

Let us begin by considering the following circuit and try to find an

expression for the current, i, after the switch is closed.


11

The Kirchhoff voltage law permits us to write

tVRidt

diL m cos

This is a linear differential equation, which you know how to solve.

We begin by finding the complementary function, from the homogeneous

equation:

0Ridt

diL

which yields the solution:

LRtAi exp

We now need to find the particular integral which, for the sinusoidal

"forcing function" tVm cos , will take the form tCtB sincos . Thus

the full solution is given by

tCtBLRtAti sincosexp

12

We see that the current consists of a "transient" term, LRtA exp ,

which eventually decays and becomes negligible in comparison with the

"steady state" response. The transient response arises because of the

sudden opening or closing of a switch but we will concentrate here on

the final sinusoidal steady state response. How long do we have to wait

for the steady state? If for example 100R and mHL 25 then

13 sec104 LR and so after only 1ms 018.04expexp LRt

and so any measurements we are likely to make on this circuit will be

truly 'steady state' measurements. Thus our solution of interest reduces

to

tCtBi sincos

In order to find B and C we need to substitute this expression back into

the governing differential equation to give

tVtCtBRtBtCL m cossincossincos


13

It is now a simple matter to compare coefficients of cost and sint to

obtain expressions for B and C which lead, after a little algebra, to

tLtR

LR

Vi m

sincos

22

If we now introduce the inductive reactance LXL we can write this

equation as

t

XR

Xt

XR

R

XR

Vi

L

L

LL

m sincos222222

The expression in curly brackets is of the form

ttt cossinsincoscos

and hence

i =Vm

R 2 +XL

2cos wt -j( )

where

j = tan-1 X L

R

æ

èç

ö

ø÷

Thus we see that the effect of the inductor has been to introduce a

phase lag between the current flowing in the circuit and the voltage

source. Similarly the ratio of the maximum voltage to the maximum

14

current is given by 22LXR which since it is a combination of

resistance and reactance is given the new name of impedance.

It is apparent that we could solve all networks containing combinations

of resistors, inductors and capacitors in this way. We would end up with

a series of simultaneous equations to solve – just as we did when

analysing DC circuits – the problem is that they would be simultaneous

differential equations which, given the effort we went through to solve

one equation in the simple example above, would be very tedious and

therefore rather error-prone. Fortunately there is an easier way.

We are saved because the differential equations we have to solve are

linear and hence the principle of superposition applies. This tells us

that if a forcing function v1(t) produces current i1(t) and a forcing function

tv2 provides current ti2 then tvtv 21 produces titi 21 . The

trick then is to choose a more general forcing function tvtvtv 21 in

which, say, tv1 corresponds to tVm cos and which made the

differential equation easy to solve. We achieve this with complex

algebra.

You should know that

exp j wt =coswt + j sinwt.

where j = (-1), [electrical engineers like to use i for current]

so let’s solve the differential equation with the general forcing function


15

v t( )=Vm

exp j wt =Vm

coswt + j Vm

sinwt

where

v1t( )=Re v t( ){ }=Re V

mexp j wt{ }=V

mcoswt

.

The solution will be of the form

i (t ) =I exp jwt

where will, in general, be a complex number. Then in

order to find that part of the full solution corresponding to the real part of

the forcing function, tjVm exp we merely need to find the real part of

ti . Thus

i1t( )=Re I exp jwt{ }=Re I exp j wt -j( ){ }

= I cos wt -j( )

Let's illustrate this by returning to our previous example where we tried

to solve:

tVRidt

diL m cos

Now, instead, we solve the more general case:

tjVRidt

diL m exp

16

and take the real part of the solution. As suggested above an

appropriate particular integral is tjIi exp which leads to

tjVtjRItjILj m expexpexp

The factor tj exp is common and hence

mVILjR

in which LjR may be regarded as a complex impedance. The

complex current I is now given by

j

LR

V

LjR

VI mm exp

22

with RL 1tan and hence

tLR

VtjIti m cosexpRe

22

which, thankfully, is the same solution as before but arrived at with

considerably greater ease.

Let us be clear about the approach. We have


17

(i) introduced a complex forcing function tjVm exp knowing that in

reality the voltage source must be real i.e. tjVm expRe .

(ii) We solved the equations working with complex voltages and

complex currents, tjV exp and tjI exp (or rather V and I since the

time dependence exp jt cancelled out).

(iii) Since the actual voltage is given by tjVm expRe the actual

current is given by tItjjItjI cosexpexpReexpRe .

(iv) Since the differential of is simply

and since we always take out as a

common factor, you may see now that our differential equations turn into

polynomial equations in jω (and you knew how to solve these at GCSE!)

This is a very powerful approach that will permit us to solve AC circuit

problems very easily.

18

4. AC Circuit theory -- Example

Let's now do an example to show, formally, how we can solve AC

problems. Let's imagine we want to find the steady state current, i2,

flowing through the capacitor in the following example

The two KVL loop equations may be written

Ri1+Ldi

1

dt+R i

1- i

2( )=Em

cos wt +a( )

and

R i2- i

1( )+1

Ci

2dt =0ò


19

Replacing ( ) +cos tEm by ( ) tjEtjEm exp=+exp 1 where

jEE m exp=1 and further introducing I1 and I2 via

tjIitjIi exp=exp= 2211 and we obtain

01

2

12

121

IRICj

R

EIRILjR

and, after a little algebra

NjM

jE

CLj

CR

LR

EI m

exp

2

12

where CRLRM and CLN 2 . We note that this may be

written, introducing MNtan

jNM

EI m exp

222

and hence the actual current tjIi expRe 22 may be written as

tNM

Eti m cos

222

20

5. Phasors

We have just introduced a very powerful method of circuit analysis. In

essence we have introduced the use of complex quantities to represent

sinusoidal functions of time. The complex number jAexp (often written

A ) when used in this context to represent tAcos is called a

phasor. Since the phase angle must be measured relative to some

reference we may call the phasor 0A the reference phasor.

Since the phasor, jAexp , is complex it may be represented in

Cartesian form jyx just like any other complex quantity and thus

22 +=

sin=

cos=

yxA

Ay

Ax

Further since the phasor is a complex quantity it is very easy to display it

on an Argand diagram (also in this context called a phasor diagram).

Thus the phasor jAexp is drawn as a line of length A at an angle to

the real axis.


21

We emphasise that this is a graphical representation of an actual

sinusoid ( ) +cos tA . The rules for addition, subtraction and

multiplication of phasors are identical to those for complex numbers.

Thus addition:

For multiplication it is easiest to multiply the magnitudes and add the

phases. Consider the effect of multiplying a phasor by j

( )2+exp=exp2exp=exp jAjAjjAj

which causes the phasor to be rotated by 90o.

22

Similarly, dividing by j leads to

2expexp2expexp1

jAjAjjAj

i.e. a rotation of –90o.

We finally note that it is usual to use rms values for the magnitude of

phasors.


23

6. Phasor relations in passive elements

Consider now a voltage tVm cos applied to a capacitor. As we have

indicated we elect to use the complex form tjVm exp and so omit the

"real part" as we calculate the current via

( ) tjVCjtjVdt

dC

dt

dvCi mm exp=exp==

If we now drop the tj exp notation and write the voltage phasor Vm as

V and the current phasor as I we have

VCjI = or ICj

V

1=

In terms of a phasor diagram, taking the voltage V as the reference

from which we confirm two things we already knew

24

(i) the ratio of the voltage to the current is C

1

- the reactance.

(ii) the current leads the voltage by 90. The pre-multiplying factor j

describes this.

For the inductance an analogous procedure leads to

ILjV =

Where the reactance is now jωL and, if we now take, say, the current as

the reference phasor we have

and here the current lags the voltage by 90.


25

[It is important to get these relationships the right way around and as a

check we may use the memory aid “CIVIL” – in a capacitor, the current

leads the voltage CIVIL and in an inductor, the current lags the voltage

CIVIL.]

Finally for a resistor we know that the current and voltage are in phase

and hence, in phasor terms

RIV =

26

7. Phasors in circuit analysis

We are now in a position to summarise the method of analysis of AC

circuits.

(i) We include all reactances as imaginary quantities ( )LXjLj = for

an inductor and cXjCj 1 for a capacitor.

(ii) All voltages and currents are represented by phasors, which usually

have rms magnitude, and one is chosen as a reference with zero phase

angle.

(iii) All calculations are carried out in complex notation.

(iv) The magnitude and phase of, say, the current is obtained as

jI exp . This can, if necessary, be converted back into a time varying

expression ( ) +cos2 tI .


27

Suppose we wish to find the current flowing through the inductor in the

circuit below

The reactances have been calculated and marked on the diagram. The

left hand voltage source has been chosen as reference and provides

10V rms. The right hand source produces 5V rms but at a phase angle

of 37 with respect to the 10V source. If we introduce phasor loop

currents I1 and I2 as shown then we may write KVL loop equations as

212

211

5103437exp5

10510

IjjIIjj

IIjI

o

where we have noted that 3+4=37sin5+37cos5=37exp5 jjj ooo. It

is routine to solve these simultaneous equations to give

5.12

71

jI

and

5.12

8+5.6=2

jI

and hence the current 21 III becomes

-

28

oo jj

I 8.86exp72.08.8672.05.12

95.0

Since rms values are involved, if we want to convert this into a function

of time we must multiply by 2 to obtain the peak value. Thus

otti 8.86cos02.1

In our example we do not know the value of but it was accounted for in

the value of the reactances. Since everything is linear and the sources

are independent it would be a good exercise for you to check this result

by using the principle of superposition.

We have used mesh or loop analysis in our examples so far. It is, of

course equally appropriate to use node-voltage analysis if that looks like

an easier way to solve the problem.

As an example let's suppose we would like to find the voltage V in the

circuit below where the reactances have been calculated corresponding

to the frequency, , of the source


29

It's probably as easy as anything to introduce two phasor node voltages

V1 and V. The two node voltage relationships may be written as

05

0

510

10 111

j

V

j

VVV

and

0105

0

105

0

51

j

V

j

V

j

VV

We note that in writing these equations no thought was given to whether

currents flowing into or out of the nodes were being considered. As in

the DC case it is merely necessary to be consistent. It is now

straightforward to solve these two simultaneous equations to yield

oV 6.71101 or, if the time domain result is required, remembering

that the voltage supply is 10V rms then ottv 6.71cos201 .

30

8. Combining impedances

As we have seen before the ratio of the voltage to the current phasors is

in general a complex quantity, Z, which generalises Ohms law, in terms

of phasors, to

IZV =

where Z in general takes the form

ee XjRZ +=

where the overall effect is equivalent to a resistance, Re, in series with a

reactance Xe. If Xe is positive the effective reactance is inductive

whereas negative values suggest that the effective reactance is

capacitative.

Consider the circuit below


31

ICj

LjRV

1

Thus the combined impedance Z is given by

XjRC

LjRZ

1

This may be visualised on an Argand or phasor diagram

We note that the reactance may be positive or negative according to the

relative values of L and C1 . Indeed at a frequency LC= we see

that 0=X and that the impedance is purely resistive. We will return to

this point later.

32

It is straightforward to show, and hopefully intuitive, that all the DC rules

for combining resistances in series and parallel carry over to

impedances. Thus if we have n elements in series, nZZZZ 321 ,,

Where

∑n

iineff ZZZZZZ

1321

and similarly for parallel elements

i

N

ieq ZZZZZ

1=+

1+

1+

1=

1

1=321

∑

We note that the inverse of impedance, Z, is known as admittance, Y.

Thus, as in the DC case it is sometimes more convenient to write

∑n

iieq YY

1=

=


33

and finally since Y is also a complex number it may be written

BjGY +=

where G is a conductance and B is known as the susceptance.

34

Example

Find the equivalent impedance of the circuit below

Using the usual combination rules gives

( )

CjLjR

CjLjR

ZZ

ZZZ

1++

1+

=+

=21

21

Which we can simplify to

RCjLC

LjRZ

21


35

9. Operations on phasors

We have just introduced a method of analysing AC circuits in terms of

complex currents and voltages. This method inevitably involves the

manipulation of complex phasor quantities and so we list below the

results for manipulating these quantities, which are, of course, simply the

standard rules for complex numbers. Sometimes it is easier to use the

a+jb notation and sometimes the rjr exp notation is easiest. We

summarise below the important relationships.

Addition and Subtraction

If

jdcIjbaI +=+= 21 and

then

( )dbjcaII ±+±=± 21

where the real and imaginary parts add/subtract

36

Multiplication

Here it is easiest by far to use the r rotation.

If 22211111 expexp jrIjrrI and

then 2121212121 exp rrjrrII

when we see the amplitudes multiply and the arguments add

For division we have

21

2

121

2

1

22

11

2

1 expexp

exp

r

rj

r

r

jr

jr

I

I

the amplitudes divide and the arguments subtract.


37

Complex conjugates also appear.

If

rjrbjaI exp

then the complex conjugate I*, is given by

rjrbjaI exp*

from which we see

IjIIIII Im2;Re2 **

where Re{ } denotes the real part the Im { } denotes the imaginary part.

38

Rationalising

We are often confronted with expressions of the form

djc

bja

+

+

And sometimes we wish to rationalise them. We do this in one of two

ways. The first is to multiply top and bottom by djc . This gives

2222 dc

adbcj

dc

bdca

djc

djc

djc

bja

djc

bja

Alternatively, we can write jba+ as 11 exp jr with 221 += bar and

ab=tan 1 . Similarly jdc+ may be written as 22 exp jr and hence

21

2

121

2

1

22

11 expexp

exp

r

rj

r

r

jr

jr

jdc

jba

where the amplitudes divide and the arguments subtract.

There is no golden rule as to which approach to take – it is determined

by the problem at hand. However, if you do not have a pressing need to

rationalise the expression, we will see some quite good reasons why we

may often prefer to stick with the factorised form and not rationalise at

all.


39

10. Phasor diagrams

Although direct calculation is easily carried out using phasors it is

sometimes useful to use a phasor (Argand) diagram to show the

relationships between say a voltage and current phasor graphically. In

this way it is easy to see their relative amplitudes and phases and hence

gain quick insight into how the circuit operates. As a simple example

consider the circuit below

Since the current, I, flows through both elements it is sensible to choose

this as the reference phasor. Having made this choice the voltage drop

across the resistor, IRVR = , whereas that across the capacitor,

ICjCjIVc . The sum of these voltages must equal V. The

phasor diagram is easily drawn as

40

from which it is clear that the voltage lags the current by an angle . The

angle may be obtained from the diagram as ( )CR 1tan= 1.

Let's consider another example

We could obtain the relationship between I and V by using the

equivalent impedance derived earlier. However we will use a phasor

diagram to show the various currents and voltages that appear across

the various components. Since the voltage, V, is the same across each

arm it is sensible to choose this as the reference phasor.

The relationships are

21112 IIIILjRIVVCjI and

The two phasor diagrams are


41

or, combining onto a single diagram

where denotes the phase angle between V and I. In the diagram

above I leads V by .

42

11. Thévenin and Norton equivalent circuits

The Thévenin and Norton theorems apply equally well in the AC case.

Here we replace any arbitrarily complicated circuit containing resistors,

capacitors, and inductors by a circuit whose behaviour, as far as the

outside world is concerned, is entirely equivalent.

The two choices are the Thévenin equivalent

and the Norton equivalent

The methods for determining V, I and Z are identical to those used in the

DC case. In general

(i) Calculate the open circuit voltage, Voc

(ii) Calculate the short circuit current, Isc


43

From which

sc

ocscoc I

VZIIVV ==,= and

We note, of course that in the absence of dependent sources it is often

easier to "set the sources to zero" and simply calculate the terminal

impedance Zab. We emphasise again that when a voltage source is "set

to zero" it is replaced by a short circuit whereas when a current source is

"set to zero", no current flows, and hence the source is replaced by an

open circuit.

Finally we note that the equivalent impedance Z is also frequently

referred to as either

(i) internal impedance

or

(ii) output impedance.

44

Example

Find the Norton and Thévenin equivalents of

Noting that there are no dependent sources it is easiest to calculate Zab

directly with the voltage source replaced by a short circuit. This is easy

since the circuit then reduces to an inductor in parallel with a resistor.

Thus

8+16=40+20

20.40== j

j

jZZ ab

We now need to calculate the open circuit voltage between a and b.

This is easy since the circuit is essentially a voltage divider

V

jj

j

jj

Voc

4.533.22

4.53exp3.224.63exp5

10exp50

10exp54020

20

Thus the Thévenin equivalent circuit takes the form

50


45

In order to find the Norton equivalent we need to find the current flowing

between the terminals a and b when they are shorted together. In this

case the circuit becomes

and

8025.180exp25.190exp40

10exp50

40

10exp50 j

j

j

j

jIsc

and hence the Norton equivalent becomes

46

We have elected to find the Norton equivalent directly. However it is

equally possible to transform between Thévenin and Norton equivalents

directly as we did in the DC case. It is left as an exercise to confirm that

Hence we could have worked out the Norton current source in our

example directly from the Thévenin equivalent as

8025.180exp25.16.26exp89.17

4.53exp36.22

816

04.53exp36.22

jj

j

j

j

Z

VI oc

which, of course, is the value we previously calculated.


47

12. Power in AC circuits

Let us consider the power flowing into a network as shown

We are usually interested in the average power P rather than the

instantaneous power p, especially for constant and sinusoidal currents

and voltages.

In general we can write the current and voltage, for the sinusoidal case,

as

tVv

tIi

m

m

cos

cos

where voltage is out of phase with the current by an angle . The phase

angle is, of course, determined by the actual network. The

instantaneous power p is given by

( ) +coscos== ttVIivp mm

48

which may be expanded noting that

BABABA coscos2coscos

to give

( ){ } +2cos+cos2

= tVI

pmm

or, introducing the rms values of current 2= mII and voltage

2= mVV

( ){ } +2cos+cos= tVIp

The first tem is constant whereas the second is periodic. Thus if we

average over a cycle the average power entering the network is given by

cos= VIP

The factor cos - the cosine of the phase angle between the current and

the voltage - is known as the power factor. For the networks we are

interested in, the power factor is determined by the elements in the

network.

The figure below shows the instantaneous power flow into the network

for three different values of .

(a) Shows the case where the network is purely resistive and there is

no phase angle between the current and the voltage and hence the

power factor, 1=cos (“unity power factor”). The power flow

fluctuates sinusoidally but is always into the network and is


49

therefore all dissipated by the resistive element(s) with an average

value of P =IV .

(c) Shows the other extreme where the voltage and current are in

quadrature ( = /2 or 90). In this case the energy flows into the

network for half the time but flows out again for the other half. The

power factor cosj is zero, P =0and there is no power dissipation.

The power flowing in must therefore be being stored in the reactive

component(s), e.g. an inductor, for half of the cycle before being

recovered again. As you found out at the end of P4 Electricity and

Magnetism we are sometimes interested in this reactive power or

reactive volt-amps (VAR), which is measured by its peak value

VAR =IV in this case.

(b) Shows the case for 0 < < /2. There is an average power

dissipation of cos= VIP in the resistive element(s) but also

energy flowing back and forth to the reactive element(s) and

VAR =IV sinf .

The combination of resistors, capacitors and inductors in the network

can be thought of as equivalent to impedance XjR+ .

50

In phasor notation we now have

( ) jZIXjRI

XIjRIV

exp=+=

+=

and

See here that V cosf = IR so that P =IV cosf = I 2R confirming that the

power is only dissipated in the resistor.

Also observe that V sinf = IX so that the reactive volt-amps is

VAR =IV sinf = I 2X .


51

We can illustrate this easily in a simple case: consider an inductor in

series with a resistor

We again take the actual current ( ) tIti cos2= where I represents the

rms value. The energy stored in the inductor is given by

tLIiLW 222 cos=2

1=

and hence the power flow due to the stored energy is given by

tLI

ttLIt

W

2sin

cossin2

2

2

∂

∂

and we recognise, for this circuit, that LX = and hence

tIV

tXIt

W

sinsin

sin2

∂

∂

52

where we have used the phasor diagram to observe that sin=VXI .

This quantity, which has a peak value of XI2 or, equivalently, sinVI , is

evidently the reactive volt-amps, VAR. The expression is equivalent to

those developed previously.

An alternative way to write these expressions is in terms of the complex

phasors ( )jVVI exp=and themselves. It is straightforward to write

average power, P as

{ }*Re=

cos=

IV

VIP

[If I is at an angle and V is at an angle ( + ), say, then I* is at an

angle (– ) and VI* is at an angle {( + ) + (– )} = and the projection

onto the real axis is VI cos .]

Similarly the reactive volt-amps, VAR, denoted by Q, may be written

{ }*Im=

sin=

IV

VIQ

Evidently these two quantities, P and Q, may be regarded as the real

and imaginary parts respectively of a complex quantity V I* whose

magnitude is VI and phase is . Thus


53

{ } { }

QjP

VIjVI

IVIVSIV

+=

sin+cos=

Im+Re== ***

The quantity S is simply called the volt-amps or VA and

22 +== QPVIS

54

13. Maximum Power Transfer

The internal or output impedance of a source restricts the amount of

power that we can extract from it. Let us imagine that the source is

represented by a Thévenin equivalent whose impedance may be written

ss XjR + . We further assume that the source is connected to a load

LLL XjRZ += .

Let us now assume that RL and XL may be chosen independently and

ask what their values should be in order to maximise the power

delivered to the load. We note that power can only be delivered to the

resistive part of LR since XL cannot consume power. Thus the average

power delivered to the load, PL, may be written

LL RIP 2=

and the current I is given by

( ) ( )LsLs

o

LLss

o

XXjRR

V

XjRXjR

VI

+++=

+++=


55

hence

( ) ( ){ }2

2

2

+++==

LsLs

LoLL

XXjRR

RVRIP

Since LX and LR are to be chosen we could simply differentiate with

respect to LX and LR to maximise P. However it is clear that P will be

maximised by setting XL = - Xs and hence

( )22

+=

Ls

Lo

LRR

RVP

If we now solve 0=L

L

dR

dP to find the maximum we find sL RR = .

Thus for maximum power transfer we require

ZL = RS - jXS = ZS*

where the asterisk denotes complex conjugate.

Under these conditions the power delivered to the load PL is given by

s

o

LL R

VRIP

4==

2

2

and the power dissipated in the source is given by

56

s

o

ss R

VRIP

4==

2

2

and hence the efficiency is given by

%50=+

==sL

L

PP

P

dissipatedpowerTotal

loadthetodeliveredPower

Thus half the power is dissipated in the source! This is clearly an

impractical proposition in the case of many generators and, indeed, is

undesirable in the case of the electricity supply companies!! Indeed it

would be spectacular to watch, albeit briefly, the effects of connecting a

matched load to a 500 MW power station. In this case, efficiency is

more important and hence Ps must be minimised in comparison to PL. In

the purely resistive case, for example, this requires that SL RR >> . On

the other hand in small signal electronic circuits handling small amounts

of power, matched operation is desirable because the maximum

available power in these applications is often rather small to begin with.

We emphasise that these results apply only if RL and XL can be chosen

separately. If, say, the load is restricted to being wholly resistive then it's

easy to show -- try it -- for maximum power transfer, we should choose

RL such that

sL ZR


57

Equally if the load consists of a fixed reactive element, XL, but an

arbitrary resistive element, RL, then for maximum power transfer we

should choose

( )22 ++= LssL XXRR

58

14. Power Factor correction

We have seen that AC power can be expressed as cosIVP where

cos denotes the power factor. The power factor has a significant

economic impact on the distribution and consumption of power since the

electricity utilities try to supply customers with a specified constant rms

voltage (240V in the UK). If the load has a small power factor then the

power company needs to provide a larger current to deliver the same

power to the load than would be the case if the power factor were high.

This is undesirable since the large current is supplied through

conductors of finite resistance, which leads to unnecessary losses. It is

particularly undesirable for the power companies since they have to

generate more power to supply the same average power to a customer

with low power factor equipment than would be required in the power

factor were high. The power companies have a very limited sense of

humour about this and tend to impose penalty charges to 'low power

factor' customers. So, what should we do if we assume that our load

has been designed to have as large a power factor as possible? For

example, most common AC motors have a lagging power factor of 0.9 at

best and often much worse.

Let us assume a load with a lagging power factor, cos , cosVIP L


59

Since the voltage, V, is fixed by the supply company we must find a way

to reduce the current supplied to I such that the supply "sees" a load with

an overall power factor of unity, IVP , without compromising the power

delivered to the load. One way to achieve this is to connect a reactive

(no power dissipation) element across the load such that the phasor sum

of the current passing through this element and IL add to I (which is in

phase with V). Since the load has a lagging power factor it is clear that

a capacitative element is a suitable candidate. Thus

and it is clear that

sinLc II

from which a suitable value of capacitor may be found.

60

Example

An electric motor consumes 2kW at a lagging power factor of 0.8 when

supplied at 240V and 50 Hz. Find the value of the capacitor needed to

correct the power factor to unity.

Since cosVIP L it follows that cosVPIL and hence

sin

cosV

PIc whence, since

c

IV c

we find

FfV

P

fV

IC c

83tan

22 2


61

15. Frequency Response

In our previous analysis we have considered networks of resistors,

capacitors and inductors and calculated voltages and currents in the

steady state due to sinusoidal forcing functions. In many practical cases

the frequency of this forcing or driving function may be variable or,

indeed, there may be several driving functions at different frequencies.

Since the reactance of an inductance, L , and a capacitor, -1 wC , vary

with frequency as shown below then, unless the network is purely

resistive, its response will inevitably depend on frequency.

The sensitivity of a network to frequency is sometimes undesirable, as in

an amplifier intended to amplify equally over a range of frequencies. On

62

the other hand it is sometimes desirable, as in an electrical filter

designed to separate out certain frequencies in a communication

system. In either case it is important to examine the response of a

network as a function of frequency.

As a simple example to illustrate frequency dependence consider the

circuit below in which a constant amplitude voltage source, say 1V, of

variable frequency is connected to a resistor and inductor in series.

Our previous analysis with te cos2 gives

tLR

ti cos2

22

with RL 1tan


63

At frequencies low enough that LR , we see that f » 0 and hence

i »2

Rcoswt . Thus the current is of reasonable amplitude and in phase

with the voltage.

At high frequencies on the other hand, RL and hence f » 90

giving i »2

wLcos wt -90( ) . In this case the current is relatively small,

due to the large impedance at high frequencies, and 90 out of phase

with the driving voltage. This behaviour, together with the phasor

diagrams is shown below

64

We note that the magnitude R2 + wL( )

2é

ëêù

ûú

-1 2

and the phase tan-1 wL R( )

vary smoothly with frequency as shown below.

1/R


65

16. Frequency Response Function

In the last section we assumed a specific input voltage 01 . It's more

convenient, however, to deal with the ratio of the phasor output to the

phasor input.

Further, the input phasor may represent a current or voltage and the

output phasor may also represent a voltage or current at some other part

of the circuit. This leads to the idea of a frequency response function.

As a trivial example, consider the voltage divider circuit below

If we assume that the output is not connected to anything (open circuit)

then

21

2

RR

R

V

VG

in

out

66

In this simple case the frequency response function, G, is constant. In

general, however, the frequency response function, G, will be a function

of frequency and hence, for an arbitrary network we can write

in

out

V

VjG

where, since we use the complex "" j notation, we have described the

frequency response function as jG .

There are two common ways of illustrating frequency response functions

graphically and these are described in the next two sections:


67

17. Polar Diagrams

For a given frequency response function the substitution of a particular

value of frequency, , will produce a particular complex number, jG ,

which can be represented as a point in the complex plane (Argand

diagram). A different value of will lead to a different complex value for

jG and is represented by a different point in the complex plane. Thus

we may represent the behaviour of the frequency response function

jG as the frequency varies as the locus of yjxjG with

as a parameter.

We'll now illustrate this by way of several examples. The most trivial is

the resistive potential divider for which we found

21

2

RR

RjG

68

which is frequency independent and hence represented by a single point

Consider now

and suppose we now define the frequency response function of interest

to be

LjRI

VjG


69

i.e. the complex impedance. The polar diagram now takes the form

A further example might be

but now the frequency response function of interest is

LjRV

IjG

1

70

i.e. the complex admittance. It is not quite so straightforward to draw the

polar diagram in this case. It's probably easiest to re-write this as

TjRRLjR

jG

1

11

1

11

Where we have introduced RLT which, since T must be

dimensionless, clearly T has the units of time. It is often called a time

constant and is something we will return to later. If we introduce the x

and y co-ordinates of jG via

21

11

T

Tj

RjyxjG

Hence

221

1;

1

11

T

T

Ry

TRx

from which it is straightforward to eliminate T to give

2

2

2

4

1

2

1

Ry

Rx

which is the equation of a circle. In order to find which part of the circle

is involved we note that TjGArg 1tan and hence the polar

diagram takes the form shown below where we also show the locus of

the phasor diagram as varies


71

As a final example we consider

iVCjR

R

CjR

CjVVV

11

1210

72

and hence the frequency response function jG , defined as 10 VV is

given by

Tj

TjjG

1

1

where we have again introduced a time constant RCT . In this case it

is clear that

TGArgG 1tan2;1

and hence the polar diagram is again semi-circular and takes the form

Since all frequencies are passed by this network with equal magnitude

this is an example of an "all pass" network.

There are several problems with this approach. This first is that these

diagrams are difficult to sketch without resort to computation for all but

the simplest frequency response function. The second is that the


73

frequency appears as a parameter and is lost in the representation

unless it is specifically marked on the diagram. The representation is

compact but an alternative approach (below) in which jG and

jGArg are drawn as a function of frequency, , on two separate

diagrams is more common.

74

18. Bode diagrams

An alternative to the polar (Argand) diagram method of representing the

frequency response of a function jG is to plot its magnitude, G , and

phase, GArg , as two separate functions of frequency, . As we will

see later, many frequency response functions can be factorised into

combinations of standard forms. Thus, in general, we might end up with

a frequency response function of the form

jGjG

GArgGArgGArgGArgGG

GG

jGjG

jGjGjG

4321

43

21

43

21

where jGi may be any of the forms we have come across before, i.e.

TjTj

TjTj

1

11

1

or, as we will see later, they may also be quadratic in form

221 TjTj .

It would, of course, be perfectly possible to plot

4321 GGGGjG directly as a function of frequency but it

would almost certainly require resorting to a calculator/computer in all

but the simplest cases. Fortunately this difficulty can be removed by


75

electing to plot the logarithm of jG rather than jG itself. This

approach turns the multiplication and division of the individual

magnitudes 21 , GG etc. into their addition and subtraction – a much

easier proposition. We also decide to use a logarithmic scale for

frequency so as to be able to represent a vastly wider range of

frequencies than would be possible with a linear scale. The ordinate of

the phase, GArg , frequency response is drawn linearly but with the

same logarithmic frequency scale. We elect to use logarithms to base

10 throughout and to measure the magnitude or amplitude in (somewhat

eccentric) logarithmic units called decibels (dB).

As an historical aside, this unit was first used to indicate the loudness of

sound. Thus a sound of intensity 2W was said to be louder than a

sound of intensity W1 by 1210log WW bels or 1210log10 WW decibels

(dB).

This was subsequently adopted by electrical engineers for the ratio of

electrical powers, 2W and 1W . When the powers happen to be

developed in the same or equivalent resistance then

1

2102

1

22

10

1

2102

1

22

10

1

210

log20log10

log20log10log10

I

I

I

I

V

V

V

V

W

W

Nowadays electrical engineers routinely use this interpretation of

decibels, dB, for voltage or current ratios without reference to the power.

i.e.

76

jGGdB 10log20

This approach to representing the frequency response of jG is due to

Hendrik Bode and yields two graphs

(i) The magnitude is plotted as dBG10log20 against frequency on a

logarithmic scale

(ii) The phase or argument, GArg is plotted against the same

logarithmic frequency scale.

We note that the usefulness of this approach is that if

4321 GGGGG then on a Bode plot we merely plot G10log20 as,

GdB, where

dBdBdBdB

dB

GGGG

GGGGG

4321

410310210110 log20log20lg20log20

Therefore we can build up the frequency response of a complicated

function merely by adding and subtracting the frequency responses of its

component parts.

The most complicated jGi that we will meet in this course will be the

ratio of two polynomials and since polynomials can always be factorized

there are very few basic elements that we need to look at. We will

consider these in detail before sketching more complicated looking

function.


77

Bode diagrams for standard forms

(i) KjG

In this case KjG and 0jGArg . If K=18, say,

dBKGdB 1.2518log20log20 1010

We note that K>1 corresponds to “gain” whereas K<1 describes

“attenuation”. If 25.0K , say, then dBGdB 1225.0log20 10 . The

negative sign arises because K<1.

(ii) TjjG

We see that

90log20 10 GArgTGdB and

78

and note that the magnitude Bode plot is a straight line with a slope of

20 dB/decade (i.e. a change of 20 dB for each ten-fold change in

frequency – note 2100log,110log 1010 etc.).

It is easy to generalise these results to ntjjG to give

GdB = 20.n.log10(ωT) and Arg{G} = n.90°

(iii) TjjG 1

We start by considering the magnitude plot

2

10

2

10 1log101log20 TTGdB

and look at the value of this expression for low frequencies 12T

and high frequencies 12T . For low frequencies

GdB

»10log10

1{ }=0dB


79

i.e. a low frequency asymptote which lies on the 0 dB axis.

At high frequencies, on the other hand,

TTGdB 10

2

10 log20log10

which gives a high frequency asymptote, of slope 20 dB/decade.

We note that these two asymptotes meet when 1T or T1 . This

frequency is sometimes known as a “corner frequency” and the point

at which the asymptotes meet is knows as a “break point”.

The asymptotes give a general idea as to what the frequency response

looks like but they are only asymptotes and so to find the actual value of

the function we have no alternative but to work out a few values. The

maximum difference between the asymptotic value and the actual values

occurs at the break point 1T . Here the actual value is

dBTGdB 32log101log10 10

2

10

or, strictly, 3.01 dB. Thus the frequency response takes the form below

where the actual response has been sketched alongside the asymptotes

80

The phase-frequency response is given by

TTjArgGArg 1tan1

which is a smooth curve varying between 0 and when T is very small

and 90 when 1T .

It is often convenient to use the straight-line approximation shown below

in which T1tan is approximated by a slope of 45/decade in the

region 101.0 T . (See HLT p167.)

We note that the actual curve and the approximation cross when 1T

and 45GArg in both cases and that the worst error is less than 6°.


81

Example

01.0125 jjG

The general approach is to regard this equation as containing two terms

25 and 01.01 j . We plot the Bode magnitude and phase frequency

responses for each constituent part separately and then merely add

them up to obtain the overall frequency response.

The magnitudes of each term in dBs are drawn initially in the left hand

diagram below. The 25K factor gives a constant gain of

dB2825log20 10 . The 01.01 j term exhibits a break point at T =

0.01 sec or a frequency of 100 rad/sec. The diagram on the left below

shows the actual response where the 3 dB correction has been made at

the breakpoint.

Straight line approximation

82

We now turn to the phase plot and note that the first factor (25) imparts

zero phase shift and hence the phase response is due entirely to

01.01 j and is shown below

We summarise below the asymptotes for magnitude and phase for the

four (two really) common factors.


83

Examples

We begin by considering the circuit below

Suppose we are interested in iVVjG 0 we have

84

Tj

Tj

CjR

R

V

V

i

11

0

where RCT . We now look at the solution in terms of the two factors

(i) The Tj term results in a linear gain of 20 dB/decade which passes

through 0dB when T1 .

(ii) The factor 11

Tj has a break point at T1 .

The overall response is shown below as the sum of the two components.

The phase response consists of two factors

(i) Tj leads to a phase (Argument) of + 90.


85

(ii) The term 11

Tj gives a phase lag increasing from zero at low

frequencies to 90 at high frequencies over the range T101 to

T10 .

The asymptotic and total phase shifts are shown

We note that this is an elementary example of a high pass filter. We

see from the Bode diagrams that frequencies lower than T1 are

attenuated whereas those higher are passed essentially unaltered –

note for these frequencies that 1.0 GdBGdB i.e and 0GArg .

We now move on to consider the voltage gain of an amplifier that, in a

particular case, might be given as

jj

jjG

510101.01

5.0

We look again at the solution in terms of the four individual factors

86

(1) The gain associated with the constant factor, 5.0 , is given by

dB65.0log205.0log20 1010 .

(2) The j term results in a linear gain of 20 dB/decade which passes

through 0 dB at =1 rad/sec.

(3) The factor 11.01

j has a breakpoint at w =10 rad/sec.

(4) The factor 15101 j has a breakpoint at

510 rad/sec.

The overall magnitude response is shown below as the sum of the

individual responses. The individual asymptotes are shown in the top

diagram together with the full response below which shows a 'mid-band'

gain of 34dB, corresponding to a numerical gain of 50

20

34

10


87

When dealing with the phase plot we have three basic factors:

(1) - 0.5 j leads to a phase (Argument) of -90.

(2) The term 101.01

j gives a phase lag increasing from zero (or

thereabouts) to -90 (or thereabouts) over the range 10 rad/sec to

1000 rad/sec as shown by the dotted line below.

88

(3) The term j5101 behaves in an analogous fashion over the

frequency range 104 to 106 rad/sec.

The asymptotic and total phase shifts are shown below.

Finally, for completeness, we include the quadratic factor that we will

meet later when we discuss resonance.

G jw( )=1+2z jwT + jwT( )2

This is an extremely common factor that occurs in many fields of applied

science. The parameter is called the damping factor, 0<<1, and T1

corresponds, as we will see later, to the resonant frequency. However,

for the time being, we can just regard it as a function whose frequency

response is to be sketched.


89

We begin with the magnitude response

222

10 21log20 TTGdB

and, as before, look for the high and low frequency asymptotes.

At low frequencies, 1T , and

dBGdB 01log20 10

which gives the low frequency asymptote as a straight line on the 0dB

axis.

At high frequencies, 1T , and only the highest powers of T are

retained. This gives

TGdB 10log40

The resulting high frequency asymptote is therefore a straight line with a

gradient of 40 dB/decade.

These two asymptotes meet in a breakpoint at T1 and, at this point

the actual value of the function is given by

2log20 10dBG

90

Since 0<<1 this value is negative for 0<<0.5 and positive 0.5<<1.

The asymptotic and actual magnitude frequency responses are plotted

below.

We finally note that when there is a negative peak, 0<<0.5, that it

occurs at a frequency a little less than T1 .

The phase response of this function is given by

2

1

1

2tan

T

TGArg

At low frequencies 0~GArg and at high frequencies

1802tan~ 1 TGArg whereas when 90;1 ArgT . This

suggests a straight-line asymptote as shown below. The actual form of

the response depends, of course, on the particular value of as

indicated on the right-hand side diagram.


91

Check out HLT page 168 to see accurate plots of the inverted form of

this second order function.

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

CLjRZ

1


93

Clearly Z is a minimum when L = 1/C when it has the real value Z =

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

frequency, , which is given by

LC

10

The voltages and current are related by

IC

jLjR

VVVV CLRS

At resonance, CL /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 0 , CIVC 0/ and IRVS . Thus

CRR

L

V

V

V

V

S

C

S

L

0

0 1

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

L

V

Vor

S

C

V

V

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

=w

0L

R=

1

w0CR

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, 0 .

[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 =w

0L

R 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

2

20 1

CLR

R

I

I

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

where the magnitude has fallen by 2/1 , i.e. 12 ,

We formally need to solve


97

2

1

12

2

CLR

R

or, equivalently,

RC

L

1

We take the negative sign to correspond to the lower frequency, 1.

Solving the resulting quadratic gives

LCL

R

L

R 1

22

2

1

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

1.

Similarly the higher frequency, 2, is obtained by taking the +R sign. In

this case we obtain

LCL

R

L

R 1

22

2

2

The frequency difference between the 3dB points is therefore

98

Dw = w2

- w1

=R

L

which together with our previous definition of R

LQ 0 permits us to

write

00

R

LQ

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

0Q as an alternative

definition.

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

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

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

equations for 2 and 1 together

210

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

2 and 1 will be symmetrical about 0 .


99

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

capacitor, VC. It is easy to write

CLjRCj

V

Cj

IV S

C1

and hence

LCjCRjV

V

S

C

21

1

Now, if we recall that 0 = 1/(LC) and CRRLQ 00 /1/ then we

may write

VC

VS

=1

1+ j1

Q

w

w0

æ

èçç

ö

ø÷÷+ j

w

w0

æ

èçç

ö

ø÷÷

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, ,

via

2

1Q

The Bode diagram corresponding to SC VVjG / are shown below

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

resonance peak.

100

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

168.]


101

Example

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

and the source VS = 2.0 V rms.

The resonant frequency, 0, is given by

rad/secLC

50 10

1

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

circuit is

100

R

LQ

and hence the bandwidth, Q/0 , or kHz.Qff 59.1/00

The current at resonance is

AR

VI S 2.00

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

inductor at resonance is

VQVV S 20

102

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 energy

stored energyQ 2

Parallel resonance

Another common circuit displaying resonance is the RLC parallel circuit

shown below.

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


103

LCj

RY

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

LC

10

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

resonant circuit.

By analogy, Q for the parallel circuit will become

CRL

RQ 0

0

but will still equate to

Q =w

0

Dw .

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

L

LR*Land

R

LRR

2

2222

*


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,

= (A/l) Ni and v = N d/dt

so that

v = L di/dt , where L = N2 (A/l) = N/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

= (A/l) N1i1 + (A/l) N2i2

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

V1 = N1 d/dt and V2 = N2 d/dt

And this is what we call mutual inductance.

Combining these equations we can write

v1=L

1

di1

dt+M

di2

dt

and

v2= M

di1

dt+L

2

di2

dt

where

L1 = N12 (A/l) and L2 = N2

2 (A/l)

are the self inductances of the two coils and

M = N1N2 (A/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 (A/l) and L2 < N2

2 (A/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=L

1

di1

dt+M

di2

dt

and

v2= M

di1

dt+L

2

di2

dt .

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=L

1

di1

dt+M

di2

dt

and

v2= M

di1

dt+L

2

di2

dt

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

ILjMIjV

MIjILjV

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

iii

v vv

.

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

dt

diMLLv

dt

diL

dt

diM

dt

diM

dt

diLv

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

iii

v vv

and hence

dt

diL

dt

diM

dt

diM

dt

diLv 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

ILjMIjV

MIjILjV

I2’

116

In an “ideal transformer” with no flux leakage, 22

12212 NNNLL and

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

alone as:

V1 = j L1 I1 + j NL1 I2

V2 = j NL1 I1 + j N2L1 I2

From which we deduce that

111

22 NVV

N

NV

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

''

21

12

11

11 NI

Lj

VI

L

M

Lj

VI


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

''

2212 ILjMIjZI L

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

gives

L

Lin

ZLj

LLMZLj

I

VZ

2

2122

1

1

1

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

large enough we approximate

Lin ZL

LZ

2

1

or

118

2N

ZZ 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

dt

dvCi

and so


121

vdt

dvRCE

or

RC

E

RC

v

dt

dv

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

T

E

T

v

dt

dv

We begin by considering the complementary function as the solution of

0T

v

dt

dv. A suitable candidate is mtAexp which leads to an auxiliary

equation

Tm

Tm

1;0

1

and hence the complementary function is T

tA 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 = ET

tA 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

T

tEv 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

0T

v

dt

dv

which gives a complementary function

T

tE

T

tBv 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

ERidt

diL

or

L

Ei

L

R

dt

di

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

L

E

T

i

dt

di

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

R

E

L

TE

and hence the full solution

R

E

T

tAi exp

The initial condition, 00 i leads to R

EA and hence

T

t

R

Ei 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 T

t

R

Ei 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

T

tAv 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

v = [v(0+) – vSS].exp(-t/T) + VSS

or

v = [initial value – steady state value].exp(-t/T) + steady state value

or, in words,

128

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

dt

diLv

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

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

130

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

dt

dvCi

and arguments analogous to those above require that voltage does not

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

initially uncharged 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

RCtR

Eti exp

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

large for a very short time. Although the 0R 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

cvdt

diLRiE

which together with dt

dvCi c gives a second order differential equation

for the voltage across the capacitor as

LCd 2v

c

dt 2+RC

dvc

dt+v

c=E

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

complementary function by setting 0E 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

LCL

R

L

Rm

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

R

LQ

LC

00

1 and

We also noted that Q was related to the damping factor

L

CR

Q 22

1 via . This permits us to write

10

2m

or

m

with 0 and 120 .

136

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

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

ttBtAEvc expexpexp

where the unknown constants are determined by the initial conditions,

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

to requiring 0dt

dvc . The full solution now takes the form

tttE

Evc

expexpexp2

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

imaginary. We consider these cases in turn

Case 1: z >1; R>2 L C

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

as

ttE

Evc

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

dt

dvCi c

or

ttEC

i

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: z =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

and hence the full solution becomes

tDtCEvc 0exp

which with initial conditions 0 ivc gives

ttEvc 00 exp11

which gives a response graphically similar to the previous case.


139

Case 3: z <1 ; R<2 L C

In this case becomes imaginary and so we write

njj 20

20 11

with

20 1 n

It is now straightforward to recast the general solution of page 136 by

replacing with nj as

tttEv n

n

nc expsincos1

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

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

rate determined by 0 and we speak of underdamped oscillation.

The frequency 20 1 n is the natural frequency. The same term

is sometimes applied to 0 but the two are strictly only equal when 1

(critical damping).

140

(See HLT p169)

Case 4: z =0;R=0

In this case there is no damping since 0R 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


141

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

frequency of the oscillatory transient 02

0 1 n , 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 (or Q) has an important role

in describing the transient behaviour:

If the damping is less than critical (<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 "j" 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 0t . Thus the terminal

voltage in the circuit below would be written tu10

144

The Laplace transform sU is given by

s

dtstdtsttusU1

exp1exp00

Hence

s

tu1

are Laplace transform pairs. Thus given s

sU1

the corresponding

function of time is the unit step, tu .

(ii) exponential decay

0;exp tttf

sdttssF

1exp

0


145

(iii) damped sinusoids

0;sin

cosexp

t

t

tttf

Here it's probably easiest to consider the composite function

tjtjtttg expsincosexp

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

discussed we can write the transform down directly as

222

1

sj

s

s

jssG

and hence by equating real and imaginary parts we find

146

2222

sinexp;cosexp

sttL

s

sttL

(iv) Laplace transform of a derivative

If the Laplace transform of tf is sF then

0fsFsdt

dfL

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

expressions for the second derivative is

002

2

2

ffssFsdt

fdL

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

sFsdt

dfL

and sFs

dt

fdL 2

2

2

(v) Laplace transform of an integral

s

sFdttfL

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 0t the governing

differential equation may be written very generally as

tuT

E

T

v

dt

dv

Where, as before, RCT and the unit step 00 ttu and 1, 0t .

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

sT

E

T

sVvsVs

1.0

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

00 v , gives

148

TssT

EsV

1

1.

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

1

11

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 0t , 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

1

1

2

1

L

R

LCL

Rs

LRs

sEsVc

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 sTexp .

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 0t together with a unit step down at Tt .

Thus

s

sT

s

sT

s

TtutuLtgL

exp1exp1

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

22.exp

s

ssT

we recognise this as the inverse transform of 22

s

s, i.e.

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

TtTt

s

ssTL

cosexpexp

22

1

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 nt n

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

dt

diLv

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

dt

dvCi

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

ICj

1VandILjVIRV

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

IsC

1VandIsLVIRV

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

the methods we developed in the AC theory ‘j’ 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

1

1

1

1

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

sTs

EsV

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

EsV

1)(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

/1

1

2

1

TsTsV

sV

where T1 = CR/2.

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

sTs

EsV exp11

and hence


161

sTTss

E

sTsTssT

EsV

exp1/1

11

2

exp1/1

1

2

1

11

2

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

TTtTtuTttuE

tv

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 please do consult one or more of the huge

number of textbooks on the subject.

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