Electronics

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ELECTRONICS

DENNIS DUNNDENNIS DUNNDENNIS DUNNDENNIS DUNN

UNIT 2/PH/SKSummer Term 2000

Lectures ~16 hoursWorkshops ~8 hours

Laboratories ~6 hours

Reference Text Book:"Electronics" by D I Crecroft, D A Gorham & J J Sparkes (Published by Chapman & Hall)

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SYLLABUSAnalogue ElectronicsCircuit Laws:

Kirchhoff's current and voltage laws

Linear Circuits:

Linearity; time-invariance

Special properties of linear, time-invariant systems

Concepts of Fourier analysis

Complex Representation of Sinusoidal Signals

Review of complex numbers and functions

Complex form of sinusoidal functions

Complex impedances

Sinusoidal Properties of Linear Circuits

Thevenin and Norton equivalent circuits for sinusoidal signals

Input and output impedances and their significance

Frequency response function; relation to gain and phase

Bode plots for simple systems

Simple RC and LCR Circuits

First-order RC circuits

Second-order LCR circuits; resonance, decay, Q-factor

Operational Amplifiers

The ideal operational amplifier

Non-ideal properties: input impedance, output impedance, input bias currents, input offset voltage,

output current and voltage limits, slew-rate limit

Operational Amplifier Circuits

Feedback

Infinite-gain approximation

Inverting and non-inverting mode circuits

Filters

Sum and difference amplifiers

Beyond the infinite-gain approximation; gain-bandwidth product

Digital ElectronicsDigital Signals:

Digital devices; truth-tables; combinational circuits

Noise immunity; advantages (and disadvantages) of digital systems

Boolean Algebra

Rules of Boolean algebra derived from the device properties

Correspondence between algebraic expressions and digital circuits

Use of algebra to produce simpler circuit designs

Design Procedures

Design procedures for combinational digital circuits using NAND or NOR gates

Bistable Systems

NOR and NAND bistables; property of memory

Clocked Digital Devices

Properties of D and JK flip-flops; memory systems; counters

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Practical ElectronicsTwo design projects:

� Operational amplifier band-pass filter design.

� Combinational digital circuit design.

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AimsTo introduce students to principles and methods of analogue and digital electronics; and to reinforce the priciplesby means of practical work.

Learning OutcomesAfter the analogue part unit each student should be able to:

� state Kirchhoff's circuit laws and apply these to simple circuits;

� describe the concepts (but not the techniques) of Fourier Analysis;

� perform a sinusoidal analysis of simple linear systems using the complex representation and determine thefrequency response function, gain and phase;

� sketch a Bode plot of the gain of first and second-order systems;

� analyse first and second-order RC and LCR circuits;

� state the ideal properties of an operational amplifier and describe how real operational amplifiers deviate fromthe ideal;

� design simple filter circuits using operational amplifiers.

After the digital part unit each student should be able to:

� state the advantages in using digital electronics;

� specify the truth tables for NOT, AND, OR, NAND and NOR gates;

� state the rules of Boolean algebra and how these rules relate to the properties of the basic gates;

� use the rules to simplify algebraic expressions and decsribe the significance of the simplification;

� use design procedures for systems of NAND or NOR gates to design combinational circuits;

� describe the properties of NOR and NAND bistables;

� describe the properties of clocked D- and JK-flip-flops;

� show how D- and JK-flip-flops can be used in simple memory and counter systems.

Student should be able to apply the above skills to the practical design and construction of simple analogue anddigital circuits.

Unit PrerequisitesBasic skills in calculus as in 1/PH/H Mathematical Physics.

Use of complex numbers and the complex representation of waves as in 1/PH/L1 Waves and Optics

Term(s): Summer Size of unit: 1

Unit Code: 2/PH/SK Modules: PH312

Department: Physics

Pre-requisites: A-level Mathematics and/or Physics (or near equivalents)

Co-requisites: None

Excluded units: None

Required for: Some Part 2 Laboratory units

Convener: Dr D Dunn

Teaching and learning methods:

Lectures, workshops, and practical laboratory sessions.

Contact hours: Lectures 16

Workshops 8

Practical Laboratories 6

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Assessment Weight

Continuous assessment:

Assigned problems 10%

Design projects 20%

Formal University Examination (One hour, April) 70%

Requirement for pass: An average of at least 40%

Re-assessment: 1-hour examination in September

28 February 2000

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Mathematical BackgroundThe mathematics needed in this unit can be revised from the following FLAP modules:

� M1.1Arithmetic and algebra

� M1.2 Numbers, units etc

� M1.3Functions & graphs

� M1.5 Exponential and logarithm functions

� M1.6 Trigonometric functions

� M1.7 Series expansions

� M4.1 Introducing differentiation

� M4.2 Basic differentiation

� M4.3 Further differentiation

� M4.4 Stationary points & graph sketching

� M4.5 Taylor expansions

� M3.1 Introducing complex numbers

� M3.2 Polar representation of complex numbers

� M3.3 Demoivre’s theorem and complex algebra

There may be other topics but these will be introduced as they are needed.

Summaries of the FLAP mathematics units can be obtained directly from our Web page:

http://www.physics.reading.ac.uk/flap/maths.html

Preliminary Exercises

1. The quantity A is given by

��

���

��

�

ba

A21

1. Which of the following 3 expressions is also equal to A ?

baiii

abab

ii

bai

21)(

2)(

2)(

�

�

�

2. Provide numerical examples to demonstrate the inequality:baba �

��111

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3. Provide numerical examples to show that (in most cases)

� � baba ���22

Give an example when the two sides are equal.

4. Differentiate the following functions with respect to t:

i. )sin()(1 attf �� �

ii. )exp()( 22 btattf ��

iii. 23 1

2)(

tt

tf�

�

In each case show that df/dt is different from f/t

4. Evaluate each of the following expressions (no calculators !)

i. 2,1;11

1����

�

ba

ba

ii. 92

,31

; �� baba

iii. 34

,41

;1

1��

�

ba

ba

5. Solve the following equation for x:

xx

y�

��

112

6. Use the Taylor series ...)0(2

)0()0()(2

������� fx

fxfxf to determine the first 3 terms in the

power series expansion of each of the following:

i. )sin()( ttg ��

ii. )sin()( �� �� tth

7. {An} represents a sequence of numbers A0, A1, A2, … etc. Are the following quantities the same or different:

����

10

1

10

1

;s

sk

k AA

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8. f(x) is a real function defined by

22

1)(

axxf

�

�

Are the following quantities the same or different:

i. ��2

11 )( dxxfI

ii. ��2

12 )( dzzfI

iii. � ��

2

13 )1( dqqfI

iv. � ��

3

24 )1( dzzfI

9. Using the same function as in 8, write out the expression for the following integral (do not try to evaluate it)

� ����

t

tdttftI0

)()(

10. In a particular class

10 students have black shoes;

10 students have black trousers; and

5 students have black shoes and black trousers.

How many students have black shoes or black trousers ?

11. Solve the following equations for x and y ( in terms of a, b, c, d, e and f):

feydx

cbyax

��

��

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ANALOGUE ELECTRONICS

CHAPTER 1: CIRCUIT LAWS

1.1 THE BASIC CONSTITUENTS

Electric Current i(t)This is the rate of flow of electrical charge (usually electrons) across a surface. The "t" is included to emphasize thatthe current is a function of time. The "surface" is the cross-sectional area of the medium which is conducting thecharge. The unit of current is the Amp (A) and 1 Amp is the flow of 6.24 �1018 electrons per second across asurface, that is 1 Coulomb per second.

If the total charge which has crossed a surface from time 0 to time t is q(t) then the current at time t is

i(t) = dq(t)

dtand not q(t)/t !

Voltage and Voltage Difference v(t)The voltage difference between two points A and B is the work done by the applied electric field in moving oneCoulomb of charge from A to B. This is

(t)(t)-vv = t),(v BAa drrE ��B

A

AB = (t)

Ea(r,t) is the applied electric field.

I distinguish between voltage difference v(t) and potential difference �(t): The potential difference is

ABA

B

A B(t) = (t) - (t)� � �� �E r dr( ,t) =

where E is the total electric field in the conducting medium This is the sum of the applied electric field and anyelectric field which exists within the medium when no current is flowing.

The applied electric field, and hence voltage difference, is the cause of the current.

Note that these expressions are merely to show the relationship to electromagnetism: In electronics we workdirectly with the voltage differences: There is no need, in this unit, to evaluate line integrals !

The direction of the current is from the high voltage end to the low voltage end.

The unit of voltage difference is the Volt (V).

The voltage at a particular point in a circuit can be defined by choosing a reference point O as the voltage origin.This origin is assigned a voltage 0. The voltage at any other point A is the voltage difference between A and O.

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Voltage SourceThis is the energy source for an electrical system. If the voltage source is vs(t) then this is the energy gained by oneCoulomb of charge as it passes through the source. The primary source of the (static or time-dependent) voltage isusually some chemical process: Electrolytic process in a battery; burning coal or oil at a power station.

Power p(t)The instantaneous power in a conductor is

p(t) i(t)v(t)�

where i(t) is the current flowing in the conductor and v(t) is the voltage difference across the conductor. This isthe amount of work done per second in forcing the current through the conductor (and is usually emitted as heat).The unit of power is the Watt (W).

ResistorThis is a component in which the current i(t) passing through it isproportional to the voltage difference v(t) across it.

Gv(t)i(t);Ri(t)v(t) ��

R is the resistance and its unit is Ohms (�); G is the conductance andits unit is Siemens (S=�-1).

The direction of the current within the resistor is from the highvoltage end towards the low voltage end.

The resistance R is usually temperature dependent: In a metal theresistance increases with increasing temperature; in a semiconductorthe opposite is usually (but not always) the case.

In circuits resistors usually have resistance in the range 1� - 10M�.

CapacitorThis is a component which stores a charge q(t) proportional to thevoltage v(t) across it.

q(t) Cv(t)�

This relation can be expressed in terms of the current flowing i(t):

i(t) = Cdv(t)

dtThis is the more useful relation because it is much easier to measure current than it is to measure charge.

A capacitor essentially consists of two parallel metal plates separated by a dielectric material and, strictly speaking,it stores a charge q(t) on one plate and a charge -q(t) on the other.

The unit of capacitance is the Farad (F) and in circuits capacitors typically have capacitances in the range 100pF -10�F.

In normal commercial capacitors the metal plates are rolled in the form of a "swiss roll".

InductorThis is a component in which the voltage difference across it is proportional to the rate of change of the current:

Li(t)

i(t)

i(t)

v(t)

v(t)

C

v(t)

R

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v(t) = Ldi(t)

dtL is the inductance and its unit is the Henry (H). An inductor consists of wire wound onto a cylindrical coremade of magnetic material (usually ferrite). Inductors used in circuits usually have inductances in the range 1�H -100mH. However real inductors are far from ideal -- they always have some resistance -- and it is better to avoidtheir use.

Voltage DifferencesYou should notice that in all the above component formulas it is the voltage difference which occurs; not thevoltage at one end of the device.

1.2 NODES AND LOOPS

A node is a point in a circuit at which two or more circuitelements are connected: For example points N1, ... N8 in thediagram.

A loop is any closed path in a circuit: For example N2-N3-N6-N7-N2 in the diagram.

Often nodes which only connect two circuit components,such as N1, N4, N5, and N8, need not be considered sincethese can be dealt with trivially.

The circle denotes a voltage source that may be static or time-varying.

1.3 KIRCHHOFF'S CIRCUIT LAWS

Kirchhoff's Current Law

The algebraic sum of the currents entering any node is zero.

i (t) 0kk� �

Kirchhoff's Voltage Law

The algebraic sum of voltage differences and voltage sources around any circuit loop is zero.

kkv (t) = 0�

Notes

Nodes & Loops

N8 N7

N1 N2

N6 N5

N3 N4

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(i) It is very important to take proper account of the directions of the currents and the sense of the voltages in theapplication of these laws.

(ii) These laws apply at all times and are not restricted to static (i.e. time-independent) currents and voltages.

(iii) Strictly Kirchhoff's voltage law should include any induced voltage due to time-dependent magnetic fields(Faraday-Lenz law of electromagnetism). However the effect of this can be reduced by careful circuit layout.The induced voltage in any circuit loop is

dSrB �� t),(dtd

(t) = -vind

where the integral is over the surface area of the loop and B(r,t) is the external time-varying magnetic field. Suchmagnetic fields can originate from nearby mains cables or from local radio signals.

The effect of such induced voltages can be reduced by ensuring that the surface area enclosed by such loops is as small aspossible.

Strictly for Kirchhoff's current law to apply the current i(t) has to include Maxwell's displacement current:

i(t) i(t)+ddt

( ,t) d� ��D r S

where D(r,t) is the electric displacement field (usually proportional to the electric field) and the integral is over thecross-sectional area of the conductor through which the current is flowing.

In normal copper conductors D is very small and may be neglected.

The complications of notes (iii) & (iv) will be ignored for the rest of this unit: They will re-appear if you study the physics of semiconductor devices in Part III.

1.4 METHOD OF WORKING

Here are some guidelines for a method of analysing problems in electronics (and the much of the rest of physics).

(i) Assign algebraic symbols to all the physical quantities in the problem, even to those quantities whosevalues are known.

(ii) Choose easily identifiable symbols. For example, R1, R2, ... for resistors, C1, C2, ... for capacitors, L1,L2, ... for inductors.

(iii) In the algebraic process make sure that you add or subtract only those quantities which have the sameunits. It makes no physical sense to add quantities that have different units.

(iv) Insert any known values only at the end of the analysis.

Using algebraic symbols makes it very easy to change the component values: only the very last step of thecalculation needs to be changed.

Error Detection

Checking that quantities being added have the same units is a very simple but powerful method of detecting errorsin a calculation. For example if the term (R1 + R2R3) appears anywhere in your analysis then you have made anerror: R1 has unit � whereas R2R3 has unit �2. If the numerical values of R1, R2 and R3 had been used, ratherthan algebraic symbols, it would have been impossible to detect such an error.

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The units of resistors, capacitors and inductors are related:

H = �s; F = s�-1; H = s2F-1; F = s2H-1.

Particularly useful combinations to remember are:

[RC] = s = [L/R],

where [RC] means the unit of RC. That is, the unit of resistance � capacitance is the second as is the unit ofinductance divided by resistance.

Exercise 1.1The following terms exist as part of a calculation, where R's, L's and C's indicate resistors, inductors and capacitorsand t is time. Indicate those which must be wrong:

(i) RR RR R12 3

2 3�

�

; (ii) RR R

R R12 3

2 3�

�

;

(iii) R C R L1 2 2 1� ; (iv) R CLR1 2

1

2�

(v) exp( )�R C t1 2 ; (vi) exp( )�R t1 ;

(vii) sin( )LR

t2

1; (viii) cos( )

RC

t2

1.

Example 1.1Applying Kirchhoff's current at node N1:

i (t) = i (t) - i (t) 3 1 2 .

In each case the arrow denotes thedirection which has been designated as'positive'. So, for example, when i1(t) ispositive the current is moving to the rightand when i1(t) is negative it is moving tothe left.

i1(t)

i3(t)

i2(t)

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Example 1.2

v (t) = (v (t) + v (t) + v (t)s 1 2 3 )

In each case the voltage difference is taken asthat at the arrow head minus that at the arrowtail.

This means, for example, that if v1(t) ispositive then the current flowing through thedevice is from left to right.

1.5 COMBINATIONS OF RESISTORS AND CAPACITORS

Applying Kirchhoff's voltage and current laws we get the equations:

i (t) = i (t) = i (t);

v (t) = v (t) + v (t).s 1 2

s 1 2

From the definitions of resistance we have:

v (t) = R i (t)

v (t) = R i (t)1 1 1

2 2 2

Hence the relationship between is(t) and vs(t) is:

v (t) = (R + R ) i (t)s 1 2 s .

This means that if we were to replace the two resistors by one then to have the same current-voltage relationshipthis single resistor would be

R = R + R1 2

That is two resistors R1 and R2 in series are equivalent to a single resistance (R1 + R2).

Exercise 1.2Using Kirchhoff's current and voltage laws derive

(i) the equivalent resistance to two resistors in parallel;

(ii) the equivalent capacitance for two capacitors in series and

(iii) the equivalent capacitance for two capacitors in parallel.

vs(t)

v3(t)

v1(t)

v2(t)

is(t)vs(t)

i1(t)

v1(t)

R1

v2(t)

R2 i2(t)

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1.6 INTERNAL RESISTANCE

All practical voltage sources have internalresistance. This means that a practical voltagesource can be considered as an ideal voltagesource vs in series with a resistor Rs.

This resistor is not an added component butis an integral part of the voltage source (e.g.battery)

Exercise 1.3A 1.5V battery has an internal resistance of 0.75�. Calculate what is the output voltage when the current takenfrom the battery is

(i) 0

(ii) 200 mA.

What is the load resistor in case (ii) ?

Note: ‘Load’ is electronics jargon for something attached to the output terminals of a circuit.

1.7 VOLTAGE DIVIDER

A voltage source vs(t) can be divided down by using the circuitshown.

Exercise 1.4Prove, using the circuit laws, the relation shown in the figure.

This voltage divider will be used many times inmany different guises throughout the course: makesure that you understand it.

Exercise 1.5This problem should show that care is needed in the use of thevoltage divider. Suppose a 12V static voltage source has beendivided down to 6V using two 100 � resistors. This is then applied to a 100 � load resistor. Find the voltageacross the load resistor.

Internal Resistance

vs(t)

Rs

R1 vo(t) = ----------------- vs(t) R1 + R2

vs(t)

vo(t)R1

R2

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1.8 NODAL ANALYSIS

A general method of circuit analysis is :

(i) Choose one of the circuit nodes as the reference node -- usually one of the terminals of a voltage source.This is assigned a voltage 0 and denoted by an inverted triangle.

(ii) Give each node voltage (relative to the reference node) a symbolic name v1, v2, ... et cetera.

(iii) Some nodes may have their voltages determined directly by voltage sources -- label these accordingly.

(iv) If there are n unknown voltages then apply Kirchhoff's current law at the n nodes corresponding to theseunknowns.

Example:Note that this is quite a difficult problem!

Consider the circuit in the diagram. The output vo istaken across the terminals labelled v2 and v2+vo. Thevoltage at the one node is labelled v2+vo rather than,say v3 because the required output voltage is thedifference between this voltage and v2. v1 is equal tothe input voltage vs and so there are two unknownvoltages v2, which we do not want, and vo the outputvoltage. We apply Kirchhoff's current law at thenodes with voltages v2 and (v2+vo):

2v : sv - 2v

2R +

ov

3R -

2v

5R = 0

( 2v + ov ): ( 1v - 2v - ov )

1R -

( 2v + 0v )

4R

- ov

3R = 0

Each of the terms on the left-hand-side of these equations is the current through one of the resistors.

These equations are easier to manipulate if we use the conductances (G = 1/R) instead of the resistances. If we usethe first of these to determine v2 and then substitute this into the second we obtain:

o1 5 2 4

2 5 1 4 3 3 1 4sv =

G G - G G( G + G )(G + G + G ) + G ( G + G )

v

This is called a bridge circuit and a typical use is with a strain gauge.

A strain is the change in length divided by the original length for any object. A strain gauge is essentially a metalresistor whose resistance varies with the strain (i.e. squashing or stretching) applied to it. In a typical application, ifR1 is the variable strain gauge resistor, R2, R4 and R5 would be chosen to be equal to the unstrained value of R1

and hence the output voltage would be zero for zero strain. For small strains and hence small changes in R1 theoutput voltage would be proportional to the strain.

Exercise 1.6In the above example suppose that R1 = R + �R where �R is proportional to the strain and that the other fourresistors are equal to R. Determine the output voltage.

vs(t)

R4

R1

v0+v2

R3

R5

v1

R2

v2

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CHAPTER 2: LINEAR, TIME-INVARIANTSYSTEMS

We are now going to consider a particular class of systems – linear, time-invariant systems – which are importantin physics (of which electronics is but a small part). We first have to say what we mean by these terms.

Consider a physical system with one input atwhich we apply some time-dependent signalx(t) and one output at which is produced theresulting signal y(t).

In electrical circuits x(t) and y(t) are usuallyinput and output voltages or currents.However in other areas of physics x(t) and y(t)could be the intensities of incident andreflected light beams; the stress and strain ona solid body; the force applied to and thedisplacement of some object and so on.

Basically x(t) is the 'cause' of some effect andy(t) is the resulting 'effect'.

LinearitySuppose that if a signal x1(t) is applied at the input this gives rise to an output y1(t) and that if a different signalx2(t) were to be applied to the input this would give rise to an output y2(t).

The system is said to be linear if, and only if:

(i) Linear Scaling -- input [c x1(t)] gives rise to output [c y1(t)] for any number c (including negativenumbers) and for any input function x1(t);

and (ii) Linear Superposition -- input (x1(t)+x2(t)) gives rise to output (y1(t)+y2(t)) for any inputfunctions x1(t) and x2(t).

These two conditions are equivalent; that is, either one implies the other. (This is not obvious and I shall notprove it ! )

In order to prove that a system is linear it is sufficient to prove either that (i) Linear Scaling or (ii) LinearSuperposition applies.

Time-InvarianceSuppose, as before, that input x1(t) gives rise to output y1(t).

A system is said to be time-invariant if, and only if, input x1(t�t0) gives rise to output y1(t�t0) for any t0 and x1(t).

General System

x(t)

Input

y(t)

Output

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Note that the function x1(t�t0) has the same shape as function x1(t) except that is shifted in time by amount t0.So in a time-invariant system shifting the input in time by t0 simply produces a time-shift of the output by thesame amount.

The concept of linearity can be extended to systems with more than one input and with more than one output. Insuch a system linearity implies that we can deal with the inputs one at a time. That is we first put all but one ofthe inputs to zero and analyse the system; then move on to the next input and so on and finally add all the variouscontributions to the output(s).

It is important to realise the linearity does NOT simply mean that y(t) = a x(t) !!

Example: Linear, Time-Invariant SystemsThe following input-output relations all represent linear, time-invariant (LTI) systems.

( i ) y ( t ) = a x ( t )

( i i ) y ( t ) = d x ( t )

d t

( i i i ) y ( t ) = c x ( t ) d t-

t

�

� � �

Example: Linear but Non-Time-Invariant systems(i) y(t) = acos(wt)x(t)

(ii) y(t) = bsin(wt)dx(t)

dt

(iii) y(t) = sin(wt )x(t )dt-

t

�

� � � �

Example: Non-Linear but Time-Invariant systems

(i) y(t) = a x (t)

(ii) y(t) = bx(t)dx(t)

dt

(iii) y(t) = c x (t )dt

2

-

t3

�

� � �

Example: Non-Linear and Non-Time-Invariant Systems (i) y(t) = acos(wt) x (t)

(ii) y(t) = bsin(wt)x(t)dx(t)

dt

(iii) y(t) = c sin(wt ) x (t )d t

2

-

t

2

�

� � � �

You should ensure that you can demonstrate that each of the above examples has the stated properties.

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Worked exampleConsider one of the example systems quoted above. In particular consider a system in which the input x(t) andoutput y(t) are related by:

y(t) = b sin(wt)dx(t)

dtLinearity

How do we test for linearity ?

First consider an input x1(t). (Note: x1(t) is just a name we choose for this input.) Using the above formula thecorresponding output, let’s call this y1(t), is

dt

(t)dxt)(w(t)=b y 1sin1 .

Now consider another input x2(t) which is just cx1(t) where c is a constant. Again using the above formula thecorresponding output, let’s call this y2(t), is

dt

(t))d(cxt)(w=b

dt

(t)dxt)(w(t)=by 1sin2sin2 .

Now we ask the question : Is y2(t) equal to cy1(t) ?

Since c is a constant it can be brought outside the derivative in the above equation. We then clearly see that y2(t)is indeed equal to cy1(t) and hence the system is linear.

Time-Invariance

How do we test for time-invariance ? We start with input and corresponding output, x1(t) and y1(t), where theoutput is given as before by

dt

(t)1dxt)sin(wb=(t)1 y

Next we consider another input x3(t) which is a time-shifted version of x1(t): x t x t t3 1 0( ) ( )� � . The output

corresponding to x3(t), which we call y3(t), is obtained by inserting the function x3(t) into the general formula

dt

))0t-(t1d(xt)sin(wb=

dt

(t)3dxt)sin(wb=(t)3 y .

Now we ask the question: Is y3(t) equal to y1(t�t0) ? y1(t�t0) can be obtained by taking the above formula for y1(t)

and replacing t by t�t0. y1 (t - t0

) = b sin(w t - t0

))dx1 (t - t

0)

dt( . This is Not the same as y3(t) because in one

case we have sin(�t) and in the other sin(�(t-t0)). Hence the system is not time-invariant.

Exercise 2.1Determine whether or not the following relations between input x(t) and output y(t) represent linear, time-invariant systems:

(i) y(t) = ax(t) + b

(ii) y(t) = f(t)x(t)

(iii) y(t) = a|x(t)|

(iv) y(t) = sin( w (t - t ))x(t )d t-

t

0

�

� � � �

where a, b and �0 are constants and f(t) is an arbitrary function of time.

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2.1 LINEAR, TIME-INVARIANT SYSTEMS AND SINUSOIDAL INPUTSIt is important to realise that, in general, even for a linear time-invariant system, the output function y(t) does nothave the same shape as the input function x(t).

In order to emphasise that the input andoutput, in general have different shapes, weconsider the simple RC circuit shown in thefigure. This circuit is typically used where wedo want to change the shape in order toremove 'noise' from electrical signals: Theoutput is a 'smoothed' version of the input.

Exercise 2.2Using the circuit laws, derive the relationbetween the input voltage vs(t) and the outputvoltage vo(t) and show that the system is linearand time-invariant.

In this case if the input voltage is a step function (that is, the voltage is switched on at time 0 )

s sv (t) = V , t 0

= 0 , t < 0

�

then the output voltage is

0 sv (t) = V (1 - exp(-t

RC)), t 0

= 0, t < 0

�

This clearly does not have the same shape as the input.

This has demonstrated that, in general, there is no simple relationship between the shapes of the input and outputsignals.

Sinusoidal InputsA sinusoidal function is a sine or cosine function or a linear combination of sine and cosine functions:

sin(wt)

cos(wt)

asin(wt)+ bcos(wt)

Sinusoidal inputs have very special properties for LTI systems: If the input x(t) is a sinusoidal function withangular frequency � then so is the output. That is the input and output have the same shape. The onlymodifications are a change in amplitude and a shift along the time axis.

If the input x(t) to a LTI system is

R-C Circuit

vs(t) C

R

vo(t)

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x(t) = X0 cos(�t)

then the output function y(t) has to be in the form

y(t) = X0 G(�) cos(�t + �(�)) .

� is the angular frequency of the sinusoid; the normal frequency is f = �/2�. G(�) is called the gain (even if it lessthan one) and (�) is called the phase shift. These are, in general, functions of angular frequency.

Note that the gain G(�) is a positive number and that the phase shift (�) is in radians and is restricted to therange (-�.,�.).

If we consider the input to be a combination of sinusoids at different frequencies �n then, because of the linearproperty, the output is the same combination of the corresponding output sinusoids.

input: x(t) = A cos( t)

output : y(t) = A G( )cos( t + ( ))n

n n

nn n n

�

�

�

� � � �n

The individual terms in the summations are called the Fourier (or frequency ) components of x(t) or y(t).

Non-Linear SystemsThere are many types of non-linear systems. In the type which occurs most often if a sinusoidal signal X0 cos(�t) isapplied to the input then the output contains combinations of sinusoids of frequency �, 2�, 3�, 4�, ...etc.

In other non-linear systems, called chaotic, an input X0 cos(�t) can give rise, if X0 is large enough, to an outputwhich contains a continuum of frequencies.

2.2 FOURIER ANALYSIS

The above property of sinusoidal inputs to LTI systems would be interesting but not very important if it were notfor the concepts involved in Fourier Analysis.

The essential concept is that (almost) any function can be constructed using sinusoids as the building blocks.

There are two forms of construction: A periodic function x(t) can be written in the form

x(t) = X cos( w t + (w ))n

n n n� �

and if this is applied to the input of a LTI system the corresponding output function y(t) occurs in the form.

y(t) = X G( w )cos( w t + ( w ) + ( w ) )n

n n n n n� � �

For a general, non-periodic, input the sums over �n are replaced by integrals over �.

x(t) = X( )cos( t-

� � � � �

�

�

� � ( ))d

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y(t) = X( ) cos( t-

� � � � � � � �G d( ) ( ) ( ))�

�

� � �

The details of Fourier Analysis are not important at this stage in your course but the concepts are.

These concepts provide a method of analysing any LTI system.

What we want to know about such a system is: If we apply a particular input x(t) what will be the output y(t) ?

If we know how the system responds to sinusoidal inputs then we can answer this question however complicatedthe required input x(t) may be.

The method which enables us to do this is:

(i) Determine, either by calculation or experiment, the response of your LTI system to sinusoidal inputs.That is determine the gain G(�) and phase �(�) as a function of angular frequency � (or frequency f = �/2�).

(ii) Obtain a Fourier Representation of the required input x(t). That is, express x(t) as a sum (or integral) ofsinusoidal terms. You will learn how to do this in detail later in the Physics course.

(iii) The output function y(t) can then be constructed: Each Fourier Component of x(t) is multiplied byG(�n) and is phase shifted by �(�n).

The existence of this method explains why we put so much emphasis on determining the response of LTI systemsto a sinusoidal input even if such an input would not normally be applied to the system.

Consider for example an amplifier designed to be used with a microphone. This would normally be used toamplify the electrical signals produced from speech or music. Such signals are much more complicated thansimple sine waves and it would be very unusual if a simple sine wave were ever to be applied to the input in itsnormal usage. Yet in designing and testing the amplifier we would concentrate on its response to sinusoidalsignals.

You should make sure that you understand this apparent paradox.

Example: An RC CircuitConsider the RC circuit used in Exercise 2.2 and suppose that the input vs(t) is

vs(t) = Vs cos(�t).

According to the theory the output vo(t) must be in the form

vo(t) = Vs G(�)cos(� t + � (�)).

If we insert these forms into the equation which you should have derived in Exercise 2.2 we get, after dividingboth sides by Vs and multiplying by RC,

-RCwG(w)sin(wt + (w)) + G(w)cos(wt + (w)) = cos(wt)� �

In order to identify G(�) and �(�) we have to expand the sin and cos functions on the left-hand-side and thenidentify the coefficients of cos(�t) and sin(�t).

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Exercise 2.3Perform these expansions, determine the resulting equations and solve them for G(�) and �(�) .

The solution you should have obtained is :

G(w) = 1

1 + (wRC )

(w) = - arctan(wRC)

2

�

This method of determining the sinusoidal response is extremely unwieldy and we next look at a much bettermethod.

Exercise 2.4(i) Explain why, in the design or analysis of a LTI system, we concentrate on the way the system responds to

sinusoidal inputs even if in its ‘normal life’ it may never see such an input.

(ii) Invent a function f(t) and plot this over a range 0 < t < 10. On the same graph plot f(t-1) and f(t+1). Commenton the results.

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CHAPTER 3: COMPLEX REPRESENTATIONOF SINUSOIDS

Suppose that the input x(t) = X0 cos(�t) to a LTI system produces an output

y(t) = X0 G(�) cos(� t + � (�))

then the input x(t) = X0 sin(�t) would produce the output

y(t) = X0 G(�) sin(� t + � (�))

with the same G and (�) because the sine function is simply a shifted version of the cosine function.

Exercise 3.1Prove that the above statement is true. Hint: express sin in terms of cos

As a mathematical trick, we can consider the possibility of using complex input functions. Consider the inputfunction

x(t) = A cos(�t) + j A sin(�t)

where j = (-1). (Note: j rather than i because i is used for current.)

Because of linearity we can immediately write down the output which corresponds to this complex input:

y(t) = A G(�) cos(� t + � (�)) + j A G(�) sin(� t + � (�)) .

These can be expressed more concisely using exponential functions: The complex input

x(t) = A exp(j� t)produces the complex output

y(t) = A G(�) exp[j(� t + � (�))].

We can still extract the real physical responses of the system: The real part of the output isthat produced by the real part of the input and the imaginary part of the output is thatproduced by the imaginary part of the input.

We can make the above relations appear even simpler. We define a complex quantity

H(�) = G(�) exp(j� (�))

and call this the frequency response function of the LTI system. In terms of this function we have

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input x(t) = A exp(j� t) gives output y(t) = A H(�) exp(j� t) .

The significance of using H(�) is that it is very easy to determine: It is much easier to find the complex quantityH(�) than to find the real quantities G(�) and �(�). The normal method is to first find H(�) and then to workout G(�) and �(�) from H(�).

The reason for this is as follows: If the input has time-dependence exp(j�t) then so does every current and voltagein the system. This effectively means that time can be eliminated from the problem. It also means that capacitors andinductors can be treated just as easily as resistors.

Consider first a circuit consisting of a single capacitor C. The relationship between the voltage vc(t) across thecapacitor and the current ic(t) flowing to and from it is

cc

i (t) = Cdv (t)

dtIf these two quantities both have complex exponential time-dependence

c c c ci (t) = I (w)exp(jwt) , v (t) = V (w)exp(jwt)

then the capacitor equation becomes

c cI (w)exp(jwt) = jwC V (w)exp(jwt)

The exponential time factor occurs on both sides and can be cancelled out.

This means that that there is a simple relationship between the complex amplitudes of the voltage and currentwhich is independent of time:

c cV (w ) = 1

jwCI (w)

Compare this with what we would have had if we had used real sinusoidal signals: one side would have had asin(�t) term and the other would have had a cos(�t) term which, of course, could not be cancelled.

Because of the cancellation of the complex exponential time factors the capacitor equation can be written entirelyin terms of the complex amplitudes.

If we do this, in exactly the same way, for a simple resistor we get VR (�) = R IR (�): These relations have the sameform except that the capacitor has a complex resistance 1/(j�C). We call such a complex resistance animpedance.

So if we use a complex representation for the sinusoidal currents and voltages we do not need to explicitly involvetime: We can simply deal with the complex amplitudes and components such as capacitors and inductors can betreated in the same way as resistors.

Exercise 3.2Show that the impedance of an inductor L is j�L

3.1 METHOD OF ANALYSISThe response of LTI systems to sinusoidal voltages (or currents) can be analysed as follows:

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(i) Work with the complex amplitudes V1(�), V2(�),.., I1(�), I2(�).. of the voltages and currents. Do notinclude the complex exponential time-dependent factors exp(j����t) at this stage .

(ii) Use the appropriate complex impedances for the circuit elements: R for a resistor, j�L for an inductorand 1/(j�C) for a capacitor.

(iii) The analysis then proceeds exactly as for static voltages and currents, using Kirchhoff's current andvoltage laws.

(iv) The ratio of the complex output voltage amplitude to the complex input voltage amplitude is H(�) thefrequency response function.

(v) The gain G(�) and the phase shift �(�) can then be determined from H(�) by

G(�) = �H(�)�, �(�) = ph(H(�)).

ph(Z) denotes the phase angle of the complex number Z.

(vi) At the end of the calculation multiply the input and output voltage (or current) amplitudes by exp(j�t)and take the real part of each to get the real sinusoidal responses.

Example: An RC CircuitConsider (yet again !) the RC circuit as we have looked at before.

If we use complex exponential voltages and currents then we only need to deal with the complex amplitudes: Thetime-dependent factor need not be included. In this case the circuit consists of two impedances, R and (1/j�C), inseries and the analysis is exactly the same as for the voltage divider treated earlier. The frequency responsefunction is simply the complex output voltage amplitude over the complex input voltage amplitude:

H( ) = 1

(1 + j RC)�

�

The ease with which this result was obtained should be compared with the previous analysis using real sinusoidalfunctions.

The complex amplitude of the output voltage is

oi

V (w) = V (w)

(1 + jwRC)

where Vi(�) is the complex amplitude of the input voltage. The gain and phase are simply

|V ( )| = |V ( )|

1 + ( RC)ph(V ( )) = ph(V ( )) - arctan( RC)

oi

2

o i

�

�

�

� � �

and hence the final real input and output voltages are

v (t) = V (w) cos(wt +i i i� )

oi

2 iv (t) = |V (w)|

1 + (wRC )cos(wt + - arctan(wRC))�

where �i is ph(Vi(�)).

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3.2 MAGNITUDE AND PHASE OF A COMPLEX NUMBER

It is common to obtain the frequency response function H(�) as a product of complex terms. We are then facedwith the problem of determining the magnitude and phase of this complex product.

Suppose we have a complex number Z given by

Z = ( Z Z .... )( Z Z ... )

1 2

a b

The conventional method of dealing with this as taught, for example, in A-Level mathematics is: Explicitlymultiply the terms in numerator; multiply the terms in denominator; rationalize the resulting complex fractionand when the result is finally in the form (X + jY), determine the magnitude as (X2+Y2) and the phase asph(X+jY).

Do not use the conventional method: It is very prone to errors and almost always leads to very complicatedexpressions which are much harder to analyse than the original expression.

A much better method is to determine the magnitude and phase as:

|Z| = |Z ||Z |...|Z ||Z |...

ph(Z) = (ph( Z ) + ph( Z ) + ... ) - (ph( Z ) + ph( Z ) + ... )

1 2

a b

1 2 a b

In the case of the phase, enough (2�)'s should be added or subtracted to put the result into the range (-�,�).

You should be aware that determining the phase of (X + jY) as arctan(Y/X) sometimes gives the wrong answers.For example the phase angle of (-1 - j) is clearly �3�/4 (draw it !) yet arctan(Y/X) gives �/4.

The arctan method gives the correct result only if X > 0: If X < 0 and Y < 0 use arctan and then subtract �; if X < 0and Y > 0 use arctan then add �.

Worked example

Consider a complex number Z given by Zj jj j

�

� �

� �

( ) ( )( ) ( )5 7 6 27 5 2 6

3 2

3 2 . How do we work out the magnitude &

phase ?

We can work on each term separately.

(5 + 7j): � � � � � �

5 7 5 7 74 5 7 7 5

5 7 5 7 74 5 7 3 7 5

2 2

3 2 23 3 3

� � � � � �

� � � � � �

j ph j

j ph j

; ( ) arctan( / )

; ( ) arctan( / )

(7+5j) : � � � � )7/5arctan(3)57(;745757

)7/5arctan()57(;745757

333223

22

������

��� ���

������

jphj

jphj

(6+2j) : � � � � � �

6 2 6 2 40 6 2 2 6

6 2 6 2 40 6 2 2 2 6

2 2

2 2 22 2 2

� � � � � �

� � � � � �

j ph j

j ph j

; ( ) arctan( / )

; ( ) arctan( / )

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(2+6j) : � � � � � �

2 6 2 6 40 2 6 6 2

2 6 2 6 40 2 6 2 6 2

2 2

2 2 22 2 2

� � � � � �

� � � � � �

j ph j

j ph j

; ( ) arctan( / )

; ( ) arctan( / )

Putting these together we get:

� � � �ph Z( ) arctan( / ) arctan( / ) arctan( / ) arctan( / ). . . .

.

� � � �

� � � � � � � �

� �

3 7 5 2 2 6 3 5 7 2 6 23 0 951 2 0 322 3 0620 2 1249

3429This phase, although correct, is outside the normal �� to +� range. To get into this range we add or subtract anappropriate number of 2�. The resulting phase is -0.287 (in degrees this is -16.45�)If you do not believe that this method is easier, repeat this Exercise using the conventional method!

Exercise 3.3Determine the magnitude and phase of the complex numbers:

(i) Z = (1 + j )

(1 - j )

(ii) Z = 1

(1 + j)(1 + 2j)

3

3

(iii) Z(2 j)(3 4j)(1 2j)(4 3j)

�

� �

� �

Exercise 3.4Determine the complex impedance of aresistor R and a capacitor C in parallel. Hencedetermine the frequency response functionH(�) for the circuit in the diagram.

Hint: Regard this as a voltage divider.

R1

R2 C2

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3.3 THEVENIN AND NORTON EQUIVALENT CIRCUITSSuppose we have an electronic instrument with two output terminals. This may involve a complicated circuit withmany hundreds of components. It is perhaps surprising that no matter how complicated this instrument we candescribe its output properties in terms of justtwo functions.

The methods that enable us to do this areknown as equivalent circuits.

Imagine a circuit which is subject to a complexsinusoidal input and hence in which all thevoltages and currents have the same complexexponential time-dependent factor. If weconsider any pair of terminals, then the circuitlooked at from these terminals is equivalent to:

(i) a voltage source VT(�) inseries with an impedance Z(�),

or (ii) a current source IN(�) inparallel with an impedance Z(�).

The first of these is called the Thevenin equivalent and the second the Norton equivalent.

VT(�) is the complex amplitude of the open-circuit voltage across these pair of terminals. That is the voltagemeasured with no external components connected across the terminals.

IN(�) is the complex amplitude of the short-circuit current measured at these terminals. That is the current whichflows when the terminals are connected by zero resistance --that is shorted-circuited.

Z(�) is the impedance seen at those terminals, defined by Z(�) = VT(�)/IN(�)

Exercise 3.5Determine the Thevenin equivalent at the output terminals of the circuit in Exercise 3.4 .

These equivalent circuits mean that at a given pair of terminals the properties of a circuit, howevercomplicated, can be described by just two functions – either VT(����) and Z(����) or IN(����) and Z(����)

Using this equivalent circuit it is very easy to determine the changes that occur when if we connect some extraimpedance across the particular pair of terminals.

If a load impedance ZL is connected across the pair of terminals we are considering the complex amplitudes of thecurrent through and the voltage across R are

L TL

L

L NL

V ( ) = V ( )Z ( )

Z ( ) + Z( )

I ( ) = I ( )Z( )

Z ( ) + Z( )

� �

�

� �

� �

�

� �

These results can be derived by applying the voltage divider result.

These equations can be used to measure the equivalent output impedance Z(�) of a pair of terminals of a circuit.

A

B

A

B

A

B

VT

Z

Z

IN

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A method is to use a load resistance (ZL = R) and to measure the amplitude of the output voltage (or current) as afunction of R and from these results deduce Z(�). This is not particularly easy ! In order to avoid damage to thecircuit it is better to start with a large resistance, say of magnitude 1 M�: A small impedance may draw too muchcurrent from the circuit. If the impedance is written as Z(�) = X(�) + jY(�), where X and Y are real, then for largevalues of R

|V ( )||V ( )|

1 - XR

, R L

T

�

�

� � � and so X(�) can be identified (at the particular frequency we are using).

As the resistance is reduced

|V ( )||V ( )|

R

X + Y , R 0L

T2 2

�

�

� �

which can be used to identify Y2 and hence the magnitude (but not the sign) of Y.

In simple cases where the output impedance is real then this can be measured very easily: Start from a large loadresistor and gradually reduce this until the voltage across the load is just a half of the open circuit voltage. At thispoint the load resistor is equal to the (real) output impedance.

The equivalent output impedance of a pair of terminals can be calculated from the circuit diagram by determiningthe impedance between the two terminals when all independent voltage and current sources have been replaced bytheir internal impedances. For an ideal voltage source the internal impedance is zero and for an ideal currentsource the internal impedance is infinite. So we replace an ideal voltage source by a short-circuit and an idealcurrent source by an open-circuit.

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3.4 INPUT AND OUTPUT EQUIVALENTSWhat do we need to know about a circuit ?

There are three essential pieces of information:

The relationship between the input voltage and (open-circuit) output voltage;

The way the output voltage changes when we connect some load impedance across the output terminals;

The current that is required at the input to the circuit

We can use the Thevenin equivalent circuits to provide all this information.

For a simple LTI system with one pair of input terminals and one pair of output terminals we can construct thefollowing equivalent circuit:

Vs

Zs

Zi

Zo

Zload

Vout

This shows a LTI system (denoted by the double-lined box) to which is applied a voltage source with complexamplitude Vs(�) and source impedance Zs(�) and which has a load impedance ZL(�) connected across its outputs.

Zin(�) is the input impedance: This is the ratio of the complex input voltage amplitude to the complex inputcurrent amplitude. It is the Thevenin equivalent impedance at the input terminals (in the absence of any voltagesource).

At the output terminals the Thevenin equivalent voltage source is H(�)Vi(�) where Vi(�) is the complex voltageamplitude at the input terminals and H(�) is the frequency response function. Zo(�) is the output impedance andis the Thevenin equivalent impedance.

Notice, in general, that Vi(�) � Vs(�) and that Vo(�) � H(�)Vi(�) : These quantities have to be determined usingthe impedance values and the general expressions are

)(V)H()(Z + )(Z

)(Z = )(V

)(V)(Z + )(Z

)(Z = )(V

ioL

Lo

ssi

ii

��

��

�

�

�

��

�

�

Ideal CircuitsFor most purposes the ideal LTI circuit has infinite input impedance and has zero output impedance. So in theideal LTI circuit no current is taken at the input, which means that the input voltage is independent of the sourceimpedance, and the output voltage does not change when we attach a load impedance.

For such an ideal system we have

)(V)H( = )(V

)(V = )(V

io

si

���

��

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In practice we try to achieve these ideals by making the input impedance as large as possible and the outputimpedance as low as possible.

Exercise 3.6Consider the linear systems shown in the figure.

R1

R2

C2

C1

Suppose that a voltage source with complex voltage amplitude Vs(�) and with internal impedance Zs(�) is appliedto the input of the RC circuit. Determine the open-circuit output voltage and the output impedance.

Suppose next that the input of the second circuit (R2-C2) is connected to the output of the first. Find the outputvoltage and output impedance of the combined circuit.

3.5 FREQUENCY RESPONSE FUNCTIONS

In simple electronic systems the frequency response function must occur in the form

H(w) = b + b jw . . . + b (jw )

a + a jw . . . + a (jw )0 1 m

m

0 1 nn

That is, H(�) is one polynomial in (j�) divided by another polynomial in (j�). The order n of the denominatorpolynomial is called the order of the system. This is a useful classification and it is also a guide to the likelycomplexity of the resulting circuit: The larger n the more complicated is the circuit.

The order of numerator is usually less or equal to the order of the denominator: m � n.

Note that frequency response functions for general physical systems can be much more complicated !

First-Order SystemsThere are two simple first order systems:

(i) H(w) = H w

w + jw

(ii) H(w) = jwH

w + jw

0 0

0

0

0

The first is a low-pass filter and the second is a high-pass filter. The term filter is applied to a circuit that can beused to remove unwanted components from an electrical signal.

Exercise 3.7Determine the gain and phase for these two frequency response functions. What is the significance of the names?

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Bode PlotsA useful way of presenting the gain of a system is via a Bode plot. This is simply a log-log plot of the gain G(�)versus angular frequency � or the gain versus the frequency f (=�/2�). This is usually done by plotting gain versusf on log-log paper rather than plotting log(gain) versus log(f) on linear paper. These two ways of plotting areequivalent but the first method is much easier and this is the method I shall use.

A Bode plot of a first-order systemwith H0=1000 and f0(=�0/2�)=1000Hz is shown in the figure:

The two straight lines are a goodapproximation to the actual curve.One of the straight lines -- theconstant -- gives the low frequencygain; the other, which gives the highfrequency gain 'slope' on the log-logplot of �1. A 'slope' of �1 on the log-log plot means that if f increases by afactor the gain decreases by the samefactor.

Exercise 3.8Draw the Bode plot for a high-passfilter with H0=100 and f0 ( = �0/2�) =100Hz.

FIRST-ORDER LOW-PASS FILTER: PHASE

-1.6E+00

-1.4E+00

-1.2E+00

-1.0E+00

-8.0E-01

-6.0E-01

-4.0E-01

-2.0E-01

0.0E+001.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05

f Hz

Phas

e

In the case of phase it is conventional to plot the actual phase (not the logarithm because the phase may benegative) against the logarithm of the frequency or, equivalently, to plot the phase verus frequency on linear-logpaper.

The figure above shows the variation of phase with frequency for the first-order low-pass filter considered above.Notice that the phase starts at zero for low frequencies, tends to ��/2 for high frequencies and is equal to ��/4 at�=�0.

FIRST-ORDER LOW-PASS FILTER: GAINFIRST-ORDER LOW-PASS FILTER: GAINFIRST-ORDER LOW-PASS FILTER: GAINFIRST-ORDER LOW-PASS FILTER: GAIN

1.00E+00

1.00E+01

1.00E+02

1.00E+03

1.00E+04

1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05

f Hz

Gai

n

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Exercise 3.9Sketch the variation of the phase versus log-frequency for a first-order high-pass filter with H0=100 andf0=�0/2�=100Hz.

Second-Order SystemsSecond-order systems have frequency response function with denominators that are quadratic in (j�). It isconventional to write this denominator in the form

(j ) + j

Q + 2 0

02

�

��

�

�0 is called the characteristic angular frequency and Q is called the Q-factor.

The most common second systems are:

(i) Low-Pass Filter

H( ) = H

+ jQ

-

0 02

02 0 2

�

�

�

��

�

The significance of the various parameters can be seen by considering the high and low frequency limits, and thevalue at �=�0 :

QH = )H(

» , H- )H(

« , H )H(

00

02

200

00

�

��

�

��

���

�

�

Notice that the high frequency components are reduced. The shape of the function H(�) for ���0 is controlledby the value of Q. For small values of Q, Q<1/2, the gain varies smoothly with �; for large values of Q, Q>1/2,the gain has a peak at

� � = 1 - 1

2Q0 2

and the height of this peak is

0

2

H Q

1 - 1

4 Q

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SECOND-ORDER LOW-PASS FILTER: GAIN

1.0E+00

1.0E+01

1.0E+02

1.0E+03

1.0E+04

1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05f Hz

Gai

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The figure shows examples of this second-order low-pass filter with H0=1000, f0=�0/2�=1kHz and for four valuesof Q: Q = 1/2, 1/2 , 2.and 5 The case Q=1/2 is special: This is known as a Butterworth filter and it has theflattest low frequency region. Notice that on this Bode plot the high frequency regions have a 'slope' of -2.

The phase function for this filter is as shown below

S E C O N D - O R D E R L O W - P A S S F IL T E R : P H A S E

- 3 .5 E + 0 0

- 3 .0 E + 0 0

- 2 .5 E + 0 0

- 2 .0 E + 0 0

- 1 .5 E + 0 0

- 1 .0 E + 0 0

- 5 .0 E - 0 1

0 .0 E + 0 01 .0 E + 0 1 1 .0 E + 0 2 1 .0 E + 0 3 1 .0 E + 0 4 1 .0 E + 0 5

f H z

Phas

e

(ii) Band-Pass Filter

H( ) = H (j )

- + jQ

0

02 2 0

�

��

� �

��

0

This is a filter which allows a band of frequencies around �=�0 to pass through.

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The gain, in this case, has a maximum at � = �0 and its maximum value is H0Q. The width of the peak iscontrolled by the value of Q and is approximately �0/Q. The phenomenon where a particular quantity has asharp peak in the magnitude of its frequency response is known as resonance and occurs in a wide range ofphysical properties (as well as in electrical circuits).

The limiting values of H(�) at high and low frequencies are:

H( ) - Hj

, «

H( ) Hj

, »

00

0

00

0

�

�

�

� �

�

�

�

� �

�

�

SECOND-ORDER BAND-PASS FILTER: GAINSECOND-ORDER BAND-PASS FILTER: GAINSECOND-ORDER BAND-PASS FILTER: GAINSECOND-ORDER BAND-PASS FILTER: GAIN

1.0E+00

1.0E+01

1.0E+02

1.0E+03

1.0E+04

1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06f Hz

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The figure shows the gain as a function of frequency (f = �/2�) for a band-pass filter with H0= 1000,f0=�0/2�=1000 Hz and for four values of Q: Q = 1/2, 1/2, 2 and 5. The larger Q the sharper is the resonancepeak.

(iii) High-Pass Filter

H( ) = -H

- + jQ

02

02 2 0

�

�

� �

� �

At �=�0 the frequency response function becomes jH0Q and the high and low frequency limits are:

H( ) H , «

H( ) - H , »

0 0

002

2 0

� � �

�

�

�

� �

�

�

(iv) Notch Filter

H( ) = H ( - )

- + jQ

0 02 2

02 0

�

� �

� �

��2

This is a filter which removes frequencies in a narrow band around �=�0. The width of this band is approximately�0/Q. At both high and low frequencies H(�) tends to H0.

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SECOND-ORDER NOTCH FILTER: GAINSECOND-ORDER NOTCH FILTER: GAINSECOND-ORDER NOTCH FILTER: GAINSECOND-ORDER NOTCH FILTER: GAIN

1.0E+01

1.0E+02

1.0E+03

1.0E+02 1.0E+03 1.0E+04f Hz

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n

The figure shows the gain as a function of frequency (f = �/2�) for a notch filter with H0= 1000, f0=�0/2�=1000Hz and for four values of Q: Q = 1/2, 1/2, 2 and 5. The larger Q the sharper is the ‘notch’.

LCR CircuitsThese second-order filters can all be made using a resistor R, an inductor L and a capacitor C in series.

In each case the characteristic angular frequency is �0=1/(LC) and the Q-factor is Q = 1R

LC

.

In the low-pass filter the output is taken across the capacitor; in the band-pass filter the output is taken across theresistor; in the high-pass across the inductor; and in the notch filter it is taken across the capacitor and theinductor.

Exercise 3.10Derive the frequency response functions for these four circuits.

Exercise 3.11Design a notch filter with f0=�0/2�=16kHz and Q=1/2.

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CHAPTER 4: THE OPERATIONALAMPLIFIER

4.1 THE IDEAL OPERATIONAL AMPLIFIER

V p s+

V p s- 4

7 2

3

6

V -

V +

V o

An operational amplifier is an active electronic device. This means that it requires a separate power supply inorder to operate. In fact it requires a two-sided power supply +Vps-0-(-Vps) as shown in the diagram. The positiveterminal of the power supply +Vps should be connected to pin 7 and the negative terminal -Vps should beconnected to pin 4. The centre terminal of the two-sided power supply is used as the common reference node ofany operational amplifier circuit and must be connected to the circuit board.

The numbers shown on the diagram are the pin numbers for an 8-pin operational amplifier. Pins 1, 5 and 8 arenot shown.

There are two signal inputs labelled V- (pin 2) and V+ (pin 3) and these are called the inverting and non-invertinginput terminals respectively. The signal output V0 is on pin 6.

In an ideal operational amplifier the input-output relation, using complex amplitudes for the input and outputvoltages, is

0 + -

0 0

0

V ( ) = A( )(V ( ) - V ( ))

A( ) = A + j

� � � �

�

�

� �

In this input-output relation all the voltages are measured relative to the reference node (i.e. the centre terminal ofthe power supply).

A(�) is called the open-loop frequency response function of the operational amplifier. A(�) has a first-order low-pass form: A0 is the low-frequency gain and is very large, typically 200,000; �0 is the lower cut-off angularfrequency and is typically 5�2� rads/sec.

The products A0�0 and A0f0 (=A0�0/2�) have special significance: They are the angular frequency and normalfrequency respectively at which the gain becomes (approximately) 1.

The product A0f0 is called the gain-bandwidth product.

For all but very low frequencies the open-loop frequency response can be approximated by

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A( ) A

jA fj f

0 0 0 0�

�

�

� �

The corresponding gain is, on a Bode plot, a straight line with slope �1 that goes through the point (�0, 1).

The ideal operational amplifier has infinite input impedance at its two signal input terminals and zero outputimpedance at its signal output terminal. This means that no current enters the input terminals and that thevoltage at the output terminal is not affected when some load impedance is connected.

It is very rare for an operational amplifier to be used directly as an amplifier with frequency response A(�).

One reason for this is that the parameters A0 and �0 of the open-loop frequency response are not specifiedprecisely in that they can vary by as much as a factor 2 one device to another (of the same type). The normal wayto use an operational amplifier is to employ negative feedback. This means that some connection is made fromthe output of the operational amplifier back to the inverting signal input. The frequency response of the completecircuit is then largely determined by the external components -- resistors and capacitors -- rather than by the valuesof A0 and �0 .

There are two distinct forms of operational amplifier circuits: The inverting mode and the non-inverting mode.

4.1 INVERTING MODECIRCUIT

The operational amplifier circuitshown in the diagram is an invertingmode circuit. The negative feedbackis provided by the impedance Z2(�)and goes from the operationalamplifier output back to the invertinginput.

Exercise 4.1Using Kirchhoff's current law at theinverting input of the operationalamplifier and the above input-outputrelation, show that the frequencyresponse function for the invertingmode circuit is

H( ) = -ZZ

1

(1 + (Z + ZZ A( )

))

2

1 2 1

1

�

�

For those frequencies at which

| A( )| « |(1 + Z ( )Z ( )

)|2

1�

�

�

the expression for the frequency response function simplifies to

H( ) -Z ( )Z ( )

2

1�

�

�

�

and so depends only on the external components Z2(�) and Z1(�).

Z1 Z2 -

+

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Remember that at low frequencies the magnitude of A(�) is very large so that the above inequality is likely to bevalid over a wide frequency range.

Exercise 4.2Show that the input impedance of the inverting mode circuit is

)A1

+(1

)ZA

Z+Z + (1

Z = Z 1

12

1in

For those frequencies at which the magnitude of A(�) is large compared with the magnitude of (Z1+Z2)/Z1, theinput impedance is approximately Z1.

4.2 NON-INVERTING MODECIRCUIT

In the non-inverting mode the circuitinput is connected directly to the non-inverting input of the operationalamplifier.

In this case the expression for the

��

���

����

�

�

���

�

�

)A(Z

Z + Z+ 1

Z

Z

= )H(

1

12

�

�1

21

I will leave this as an exercise to derivethis.

The input impedance to the circuit isjust that of the operational amplifieritself ie almost infinite.

For those frequencies at which

| A( )| « |(1 + Z ( )Z ( )

)|2

1�

�

�

the expression for the frequency response function simplifies to

)(Z

)(Z1 )H(1

2

�

�

� ��

and so depends only on the external components Z2(�) and Z1(�).

The 1 in this expression for H(�) is a nuisance: We often want the expression to get very small for either low orhigh (or sometimes both) frequencies. The 1 prevents this.

A more versatile version of the non-inverting mode circuit, which involves two more impedances is shown in thenext figure.

Z1 Z2

-

+

input output

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The frequency response function in thiscase is

� ���

���

����

�

�

���

�

�

)A(ZZ + Z+ 1

Z

Z

ZZ

Z= )H(

1

12

�

�1

2

43

41

and for those frequencies at which

| A( )| « |(1 + Z ( )Z ( )

)|2

1�

�

�

the expression for the frequency responsefunction simplifies to

� � ���

����

��

� 1

2

43

4 1Z

Z

ZZ

Z= )H(�

4.3 INFINITE-GAIN APPROXIMATIONThe approximation made above

0)(

1�

�A

is called the infinite-gain approximation. Applying this approximation to

))(V-)(V)(A(=)(V -+0 ����

yields the main result of the infinite gain approximation:

)(V)(V -+ �� �

That is the large gain forces the voltages at the inverting and non-inverting inputs to be zero.

This is the usual first step in the design of operational amplifier circuits. The reason for this is that A(�) is notusually specified precisely so that we would like to make the design depend on A(�) as little as possible.

4.4 IMPEDANCESSuitable values for the impedances are

COMPONENT IMPEDANCE ZResistance R

Capacitance 1/(j�C)

Resistance & Capacitance in Series R + 1/(j�C)

Resistance & Capacitance in Parallel R/(1+ j�RC)

4.5 AN INVERTING BAND-PASS CIRCUITA very useful circuit is the inverting band-pass circuit: This removes low and high frequencies and amplifiesfrequencies in the mid-band region.

Z1 Z2

-

+

input output

Z3

Z4

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In this case the mid-band gain isR2/R1; the lower cutoff frequency (inHz) is 1/(2�R1C1) ; and the uppercutoff frequency (in Hz) is1/(2�R2C2).

The lower cutoff frequency is thefrequency below which the circuitstarts to remove the low frequencies;

The upper cutoff frequency is thefrequency above which the circuitstarts to remove the high frequencies;

4.6 A NON-INVERTING BAND-PASS CIRCUITA similar circuit with the same usage is the non-inverting band-pass circuit: This removes low and highfrequencies and amplifies frequencies in the mid-band region.

In this example Z1=R1; Z2 =R2/(1+j�R2C2) that isR2 in parallel with C2 ;

Z3= 1/ j�C3 Z4 = R4 .

For this circuit the mid-band gain is (R1+R2)/R1;the lower cutoff frequency (in Hz) is 1/(2�R3C4); and the upper cutoff frequency (in Hz) is1/(2�R2C2).

This circuit is not as efficient as thecorresponding inverting mode circuit at removingvery high frequencies. Above a frequency

2

21

2221

R

RR

CR

�

�

the reduction in the high frequency componentsstops

The Bode plot for this circuit is of the form:

R1 R2

-

+

input

output

C1 C2

R1 R2 -

+

input output

C3

R4

C2

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f

gain 1+R2/R1

1/2piR1C1 1/2piR2C2

1

The frequency at which the gain becomes 1 is 222

21

21

CRR

RR

�

�

EFFECT OF THE FINITE GAINAt the end of the design procedure, for the inverting or non-inverting circuits, using the infinite-gainapproximation we must check that the approximation is valid over the required frequency range. This is done bychecking

| A( )| « |(1 + Z ( )Z ( )

)|2

1�

�

�

If we use the fact that, except at very low frequency,

A( ) A

jA fj f

0 0 0 0�

�

�

� �

then the infinite-gain approximation is valid if

00max1

2max 1 fA

Z

Zf ��

The product A0f0 is known as the gain-bandwidth product. So the required criterion is that the product of themaximum frequency and the maximum gain must be less than the gain-bandwidth product.

A typical value for this gain-bandwidth product is 1MHz.

The remedy if this condition is not satisfied is either

use an operational amplifier with a larger gain-bandwidth product; or

split the circuit into two parts with each part having one opamp and a smaller maximum gain.

This checking can be done graphically.

The method is:

Draw the Bode plot of the gain worked out using IGA.

On the same plot include the opamp gain. Except at very low frequencies this is simply a straight line of slope�1 passing through the point (1, A0f0).

The opamp line effectively cuts off the gain of the circuit. So where the opamp gain is to the right of thecalculated circuit gain the IGA calculation is valid.

Consider two examples shown below. In circuit 1 the components are:

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Circuit 1: R1=1 k�; R1=10 k�; C2=1nF; C3=1�F; R4=10 k�; A0f0=3 MHz

Using the IGA the mid-bandband gain is 11; the lower and upper cut-off frequencies are 16Hz and 16kHz; andthe frequency at which the gain becomes one is 176kHz. The figure below shows this. The figure includes IGAgain, the ‘straight-line approximation’ to the IGA gain; the opamp gain and the exact gain of the circuit. In thiscase the IGA is valid up to about 3 MHz

Non-Inverting Bandpass filter1

1.0E-01

1.0E+00

1.0E+01

1.0E+02

1.0E+00 1.0E+01 1.0E+02 1.0E+03 1.0E+04 1 .0E+05 1.0E+06 1.0E+07f

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Circuit 1: R1=100�; R1=100 k�; C2=100 pF; C3=1�F; R4=10 k�; A0f0=3 MHz

Using the IGA the mid-bandband gain is 1000; the lower and upper cut-off frequencies are 16Hz and 16kHz; andthe frequency at which the gain becomes one is 176kHz. The figure below shows this. The figure includes IGAgain, the ‘straight-line approximation’ to the IGA gain; the opamp gain and the exact gain of the circuit. In thiscase the IGA is only valid up to about 3 kHz

Non-Inverting BandPass Filter 2

1.0E-01

1.0E+00

1.0E+01

1.0E+02

1.0E+03

1.0E+04

1.0E+05

1.0E+06

1.0E+00 1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07

f

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4.7 UNITY GAIN BUFFERA very useful circuit, or more correctlysub-circuit, is the unity gain buffer:

In this case the output voltage is(within the infinite-gainapproximation) equal to the inputvoltage.

The special properties are that theinput impedance is (almost) infiniteand the output impedance is (almost)zero. This means that it can be usedto join two sub-circuits withoutaffecting their properties. So if it isused to join a circuits with frequencyresponse functions H1 and H2 thenthe combined circuit has frequencyresponse function H1 H2.

4.8 NON-IDEAL PROPERTIESI have been assuming ideal operational amplifiers up to now. How do real operational amplifers differ from theideal.

Output impedance

A typical output impedance for an operational amplifier is of magnitude 100�. Thus we can still assume the idealoutput properties if all impedances connected to the opamp output are much larger than this value .

Input impedance

There are two types of operational amplifier input, one based on bipolar transistors (for example 741 opamp) andone based on field-effect transistors (for example 071 opamps).

For the bipolar input op-amps a typical input impedance is 1M� in parallel with 5pF.

For the bipolar input op-amps a typical input impedance is 1012 � in parallel with 5pF.

Output Voltage Limits

The output voltage is bounded by maxoutV and min

outV . These two values are almost the same as the upper and

lower values of the power supply.

This has the effect of truncating the output waveform.

Output Current Limit

The output current is limited in magnitude by IL. A typical value for this maximum is 20mA. So opamps cannotdeliver much current!

Slew-Rate Limit (SRL)This rather strange name refers to the maximum rate of change of the output voltage:

SRLdt

tdvo�

)(

A typical value for SRL is 106 V s�1 (or 1V per microsecond).

In practice for each design each of these limitations needs to be checked to see if it presents a problem.

-

+

input output