AC circuit analysis

AC Circuit AnalysisModule: SE1EA5 Systems and Circuits AC Circuit Analysis Module: SE1EA5 Systems and Circuits Lecturer: James Grimblebyy URL: http://www.personal.rdg.ac.uk/~stsgrimb/ em

AC Circuit Analysis

Module: SE1EA5 Systems and Circuits

AC Circuit Analysis

Module: SE1EA5 Systems and Circuits

Lecturer: James Grimblebyy

URL: http://www.personal.rdg.ac.uk/~stsgrimb/

email: j.b.grimbleby reading.ac.ukj g y g

Recommended text book:

David Irwin and Mark Nelms

Basic Engineering Circuit Analysis (8th edition)

John Wiley and Sons (2005)

ISBN: 0-471-66158-9

AC Circuit Analysis

Recommended text book:

David Irwin and Mark Nelms

Basic Engineering Circuit

AC Circuit Analysis Syllabus

This course of lectures will extend dc circuit analysis to deal

AC Circuit Analysis Syllabus

This course of lectures will extend dc circuit analysis to deal with ac circuits

The topics that will be covered include:

AC voltages and currentsComplex representation of sinusoidsPhasors

Complex impedances of inductors and capacitorsDriving-point impedance

Frequency response of circuits – Bode plotsPower in ac circuits

Energy storage in capacitors and inductorsThree-phase power

AC Circuit Analysis Prerequities

AC Circuit Analysis Prerequities

You should be familiar with the following topics:

SE1EA5: Electronic Circuits

Ohm’s LawSeries and parallel resistancespVoltage and current sourcesCircuit analysis using Kirchhoff’s Lawsy gThévenin and Norton's theorems

The Superposition Theorem

SE1EC5: Engineering Mathematics

Complex numbers

AC Circuit Analysis

Lecture 1

AC Voltages and Currents Reactive Components

Examples:

Frequency

pElectrocardiogram: 1 HzMains power: 50 Hz

Aircraft power: 400 HzAudio frequencies: 20 Hz to 20 kHz

AM radio broadcasting: 0.5 MHz – 1.5 MHz

FM di b d ti 80 MH 110 MH

FM radio broadcasting: 80 MHz – 110 MHzTelevision broadcasting: 500 MHz – 800 MHzgMobile telephones: 1.8 GHz

Why Linear?

We shall consider the steady-state response of linear ac

Why Linear?

We shall consider the steady state response of linear ac

circuits to sinusoidal inputs

Linear circuits contain linear components such as resistors, capacitors and inductorsp

A linear component has the property that doubling the voltage across it doubles the current through it

Most circuits for processing signals are linear

Analysis of non-linear circuits is difficult and normally requires the use of a computer

Why Steady-State?

Steady-state means that the input waveform has been

Why Steady State?

Sinusoidal carrier waves are modulated to transmit

information (radio broadcasts)

Any periodic waveform can be considered to be the sum of a fundamental pure sinusoid plus harmonics (Fourier Analysis)

Fourier Analysis

A square waveform can be considered to consist of a

f d t l i id t th ith dd h i i id

Fourier Analysis

fundamental sinusoid together with odd harmonic sinusoids

Square wave Sum

Fundamental

3 d h i

3rd harmonic

5th harmonic

ft v

t

or: v(t) = v0 cos2 ft = v0 cos t

where =2 f is known as the angular frequency

t

Ceramic tube

Resistors

Ceramic tubecoated withConductive film

Metal end

i v

Metal end cap

Resistors

Then:

t

( )t R

dv C i

Cv q

Cv

ceramic

Al 0 (electrolytic)

Capacitors

Suppose that:

( )t v

dt

d C t

)

( 1 v0t

i

t v

0

0)

(

)(

0 2

2 = =

i t

i

t v

Thus “resistance” varies between ±∞: not a useful concept

0)

(t1

p

fC C

Magnetisablecore

Copper wire

Core: air

ferritei

Inductance L

dt

di L

ironsilicon steel

Inductors

Suppose that:

( )t v

)(t

( )

∫

= 1 sin)

v

t

v L

Current lags voltage by /2 (90°)

The reactance of an inductor is directly proportional to its

value and to frequency

Resistance and Reactance

(t Ri t v t

)(

dt

t

dv C t

)

()

dt

t

dv RC t

dt

t v

t v t

RC

t v

RC

t v dt

t

=+

=

t v

t v

t v

t

cos)

(

cos)

(

1 0

Expanding the expression for v c:

( + )

= v t t

c

t B

t A

t v

t v

t

t B

t A

t

cossin

)(

+

−

=

B t

A t

B t

−

Comparing the coefficients of sin t and cos t on both sides of

t RC

t RC

t RC

t B

t

p gthe equation:

0

B RC

−

0

v A

RC

Solving these simultaneous linear equations in A and B:

2 2 2

0 2

2 2

0

1

RCv B

C R

v A

+

=+

=

AC Circuit Analysis

Thus:

2 2 2

0 1

2 2 2

0 1

1

sin1

cos

C R

v

B C

=

=

Thus:

RC C

1

At an angular frequency =1/RC:

C R

0 1

v v

)(

42

v t

⎠

⎝ 42

0.01/RC

where x is the real part, y is the imaginary part (x and y are

both real numbers), and

11

2 = − j = −

j

Complex numbers are often the solutions of real problems,

Complex Numbers: Polar Form

Polar form:

∠

r z

Complex Numbers: Polar Form

where r is the magnitude, and is the angle measured from

Complex Numbers: Conversion

x

O

Rectangular to polar:

y z

y x

z r

Complex Numbers: Inversion

If the complex number is in rectangular form:

Complex Numbers: Inversion

1

jy x

z

+

=

))(

(x jy x jy

jy x

jy x

(

jy x

jy x

Complex Numbers: Conversion

When using the inverse tangent to obtain from x and y it is

Complex Numbers: Conversion

x

y

=tan

jy x

Complex Numbers: Conversion

When using the inverse tangent to obtain from x and y it is

Complex Numbers: Conversion

necessary to resolve the ambiguity of

1 Calculate using inverse tangent:

Complex Numbers: Conversion

Complex Numbers: Conversion

Convert to rectangular form

Complex Numbers: Conversion

2

1 732 z = 1+ j 3Imaginary

(3

o

=

Real axis

Complex Numbers: Conversion

Complex Numbers: Conversion

Convert to polar or exponential form:

j

z

+

=11

Magnitude:

2

11

1

01

2 2

12 + 2Angle:

0

1t

0t

)1

(1

44

01

1tan

1

0tan 1 1 ⎟ = − = −

Complex Numbers: Conversion

Real axisO

0.5O

Imaginary

)45

(4

Complex Exponential Voltages

We shall be using complex exponential voltages and currents

Complex Exponential Voltages

We shall be using complex exponential voltages and currents

to analyse ac circuits:

t j

Thi i th ti l t i k f bt i i th

t j

Ve t

This is a mathematical trick for obtaining the ac response

without explicitly solving the differential equations

It works because differentiating a complex exponential leaves

it unchanged apart from a multiplying factor:

t j t

j

Ve j

Ve dt

Complex Exponential Voltages

S th t l ti l lt i li d

Complex Exponential Voltages

Suppose that a complex exponential voltage is applied

across a resistor:

t

t j

V

t )

(

t j

V R

t

v t

t j

Ve t

e R

V

=

R

The current through the resistor is also a complex

The current through the resistor is also a complex

exponential

Complex Exponential Voltages

S th t l ti l lt i li d

Complex Exponential Voltages

Suppose that a complex exponential voltage is applied

across a capacitor:

dt

t

dv C t

t j

Ve t

t j

Ve dt

d C

dt

=

j

Ve t

CVe j

dt

=

C

The current through the capacitor is also a complex

The current through the capacitor is also a complex

exponential

Complex Exponential Voltages

S th t l ti l lt i li d

Complex Exponential Voltages

Suppose that a complex exponential voltage is applied

across an inductor:

dt t

v L

t

t j

Ve t

t

j dt

Ve L

Ve L

The current through the inductor is also a complex

exponential

Complex Exponential Voltages

A complex exponential input to a linear ac circuits results in all voltages and currents being complex exponentials

Of course real voltages are not complex

The real voltages and currents in the circuit are simply the real parts of the complex exponentials

Complex exponential: v (t) = e j t (= cos t + j sin t)

Complex exponential:

Real voltage:

)sin

cos(

)

t t

Real voltage: v r (t) = cos t

AC Circuit Analysis

Lecture 3

Phasors Impedances Gain and Phase Shift Frequency Response

then all other voltages and currents are also complex

exponentials:

t j t

j j

t j

t j t

j j

t

j c

I i

i t

i

e V e

e v e

v t

v

) (

1 1

)

( 1 1

2 2

1 1

)(

j j

1 v e j

j

e v V

=

=

Phasors are independent of time, but in general are functions

of j and should be written:

The impedance Z of a circuit or component is defined to be the

Impedance

The impedance Z of a circuit or component is defined to be the

ratio of the voltage and current phasors:

t j

Ve t

RIe Ve

t Ri t

v

t j t

j

c c

=

)()

d C Ie

dt

C t

i

t j t

Ie t

CVe j

Ie

Ve dt

C Ie

t j t

j

t j t

I

CVe j

I V

d L Ve

dt

L t

v

t j t

Ie t

LIe j

Ve

Ie dt

L Ve

t j t

j

t j t

V

LIe j

V

1

++

+

=Other rele ant circ it theor r les are Kirchhoff’s la s

4 3

2

Z

Other relevant circuit theory rules are: Kirchhoff’s laws,

Thévenin and Norton's theorems, Superposition

Potential divider:

2 1

IZ V

Z Z

IZ V

2

2 1

Z V

Z Z

2

Z Z

Z V

AC Circuit Analysis

Suppose that a circuit has an input x(t) and an output y(t),

AC Circuit Analysis

where x and y can be voltages or currents

The corresponding phasors are X(j ) and Y(j )

The real input voltage x(t) is a sinusoid of amplitude x0:

)(

)(

)cos(

)(t x t re x e j t re Xe j t x

and the real output voltage y(t) is the real part of the complex

)(

)(

)cos(

)(t x0 t re x0e j t re Xe j t

and the real output voltage y(t) is the real part of the complex

exponential output:

)(

)(

)cos(

)(t y0 t re y0e j e j t re Ye j t

=

0 0

The voltage gain g is the ratio of the output amplitude to the

y

0 0

and the phase shift is:

Z Z

Z V

V

R C

C in

c

+

=

/1

R

R C

/

1C

in

CR j

+

=1

1

1

1

R C V

V g

2 − °

−

1

R C

AC Circuit Analysis

Z V

C

R

Z Z

V

V

C R

R in

C j

CR j

CR

j V

CR j

AC Circuit Analysis

C

AC Circuit Analysis

CR j

CR j

1 C R

CR V

V g

Frequency Response (RC = 1)

)90

0

0.10.01

Angular Frequency (rad/s)

Frequency Response

CR

1

tan2

−

−

=

2 2 2

1 C R

CR g

AC Circuit Analysis

Lecture 4

Driving-Point Impedance

The impedance Z of a circuit or component is defined to be the

Impedance

The impedance Z of a circuit or component is defined to be the

ratio of the voltage and current phasors:

)( j I

)

()

( j V j

AC Circuit)

( j

V

)( j I

)(

)

(

j I

j

AC Circuit)

( j

V

Impedance Z is analogous to resistance in dc circuits and its

units are ohms

When Z applies to a 2-terminal circuit (rather than simple

component) it is known as the driving-point impedance

)( j R j jX j

Z Z

Symbolic and Numeric Forms

CR j

R Z

+

=1Symbolic Form

Example 1

Determine the driving-point impedance of the circuit at a

Example 1

g p pfrequency of 40 kHz:

1

Z Z

1

C j

10200

1040

2

125

9 3

+

050270

125

05027

Example 1

89.19

25 −

= j Z

89.19

252 + 2

=

93.31

89.19

19tan 1 −

=

∠Z −

)5.38(

6720

(5

6720

0)

93.315

6720

0159.7V

)(

Example 2

Determine the driving-point impedance of the circuit at a

Example 2

g p pfrequency of 20 Hz:

Z Z

C

j R

Z Z

= 1

R

C j

R

Z

+/

1

CR j

+

=1

Example 2

1 j CR

R Z

+

=

C = 100 F80

1

6

CR j

10020

2

1+ j × × × −6 ×

)0051

1(80

005

11

7939

005

11

)005

11

(

80

2 2

39 − j

=

Example 2

00.4079

00

40tan 1⎨⎧− ⎬⎫

=

∠Z −

)245(

78800

79.39tan

7880

0

=

4256

24

−

∠

=00

4079

30

)2.45(7880

0A

0.4254

7880

042

.56

00.4079

.39(24

00.4079

.39

2 2

+

j

A3016

03

0 + j

=A

3016

03

.0

00.4079

00.4254A ∠ °

=

Phasor Diagrams

Where voltages or currents are summed the result can be

Phasor Diagrams

grepresented by a phasor diagram: V =V1 +V2 +V3

Imaginary part

2

V

Realpart

O

1

V V

3

V

Example 2

A3.0

1

C j

L j

L

10120

502

110

3650

L

36 mH

53.2631

.1124

10120

50

j j

j

−+

.15

24 − j

Example 3

2215

24 −

= j Z

Example 3

22.1524

= j Z

4228

22.15

.28

15tan 1

5652

5652

042

=

Z

Example 3

What voltage will be generated across the circuit if an ac

Example 3

What voltage will be generated across the circuit if an ac

current of 10 A, 50 Hz flows though it?

)2215

24(A

)22.1524

(A10

)4.32(

5652

0)

42.2810

5652

0V

2

=

Example 3

C L

R V V V

R V V V

R C

Z Z

Z

1

++

j R

1

1 ++

L j

R

Z

)/(

1

1

++

=

1 mH

L j

R C j

L j

R

)(

1+ +

+

=

L j

Example 4

L j

R

Example 4

10400

22

j Z

×

×+

−+

=

−

10200

10)

4002

(210

200400

21

10400

2

2

6 3

2 6

1005

.11

513

22

j

j

−+

+

=

0051

26330

513

22

j j

12633

.0()513

22

(

005

12633

.0

2 2

j j

12633

0 2 + j 2

Example 4

4742

2852

474

2852

1

474

2tan 1

=

∠Z −

)2.53(

9282

0

852

9282

0091

=

Z

(120

474

2852

.1

A8238

9282

0091

.3

2852

.1

)474

2852

.1(

.23

9282

0A

82.38

9

0.2974

222 +

08.3126

23 +

Example 4

Imaginary part

Example 4

I I

j

The admittance Y of a circuit or component is defined to be

Admittance

The admittance Y of a circuit or component is defined to be

the ratio of the current and voltage phasors:

)

( j

I

1)

()

( j I j

AC Circuit)

)(

)

(

j Z j

V

j

AC Circuit)

( j

V

Admittance Y is analogous to conductance in dc circuits and

its unit is Siemens

)(

)(

)( j G j jB j

2 1

11

11

1

Y Y

Y Y

and admittances in parallel:

Oth l t i it th l Ki hh ff’ l

4 3

2

Y

Other relevant circuit theory rules are: Kirchhoff’s laws,

Thévenin and Norton's theorems, Superposition

/1/

1

2

C j

/1/

R

C j

+

+

1 mH

210

200400

2

3

6 j j

15027

0

10400

22

10200

2

513

22

15027

.0

22

513

2

25027

0

2 2

j j

+

−+

2

C

32.10

513

2

25027

01939

.05027

AC Circuit Analysis

Lecture 5

Resonant Circuits

Resonant circuits can be characterised by two parameters:

the resonance frequency and the Q-factor

There are two basic resonant circuit configurations: series and parallel

Resonant Circuits

L

g dt

dt

,

t

Parallel Resonant Circuit

C

++ 1 1

1

Z Z

L j

LC R

L j

R LCR

L

LC R

L j

Parallel Resonant Circuit

Ω

= k5

Impedance is a maximum

1when:

L j

L

j Z

1

=

LC

6 2

4

/1

L j

2

102

1+ j × × − − × − 3

10

=

Parallel Resonant Circuit

Parallel Resonant Circuit

L

j Z

0

R

L

j Z

j L

j

R R

L j

L

j Z

=Resonant frequency:

0

C

j L

Parallel Resonant Circuit

−Angular frequency (rad/s)

0.0

100

)90

(2

Angular frequency (rad/s)

=

So that:

LC R

L j

5000

3 0

Angular frequency (rad/s)

100

)(

2

Parallel Resonant Circuit

j C

j

/1

parallel resonant circuits

because the currents in

the capacitor and

j L

j L

j

pinductor cancel out

Parallel Resonant Circuit

Parallel Resonant Circuit

.0

R

I

10

A10

5.0

2× −3

−

Parallel Resonant Circuit

Above resonance: 2mA

:100

.2

0.2

5

0 × −3

−

R

Series Resonant Circuit

Z Z

Z

L

j C

j R

L C

R

1 ++

LC CR

j

C j

j

C j

2

1+

=

C j

LC CR

j

1+ −

Series Resonant Circuit

C j

LC CR

j Z

CR

j Z

Z

Series Resonant Circuit

LC CR

j

6

6 2

6

10

1010

4

6

1010

21

Series Resonant Circuit

−0

Series Resonant Circuit

The standard form for the denominator of a second-order

Series Resonant Circuit

The standard form for the denominator of a second order

system is:

2 0

2

=

So that:

LC CR

Series Resonant Circuit

110

Series Resonant Circuit

because the voltages

across the capacitor and

L

V

j j

R

101

2001

j C

j

j L

j

Series Resonant Circuit

Crystal Resonator

Crystal Resonator

0 0

127010

03

=

Frequency-Response Function

Input X Y Output

)

()

( j Y j H

Frequency-response function:

)(

)

(

j X

)

(

j

H j

)

(

j

H j

X

j Y

Frequency-Response Function

The order of a frequency-response function is the highest

Frequency Response Function

power of j in the denominator:

First order:

0

/1

1)

(

j

j H

+

=

Second order:

2 0

0 ( / )/

21

1)

(

j j

j

H

++

=Third order:

0

0 ( / )/

2

1+ j + j

1)

( j

Third order:

3 0

2 0

0 ( / ) ( / )/

1

)

(

j j

j

j

H

++

+

=The order is normally equal to (and cannot exceed) the

Z V

V

R C

C in

j

C j

Z Z

/1

R C

j

1)

(

/1

=

+

CR j

j H

1:

where1

1

)(

R

2 2

0 0

/1

/

1+

1)

(

+

=

= H j g

0

/

1+

Ph hift ∠H( j ) tan /Phase shift: = ∠H( j ) tan = − / 0

dB = P

2 10

log10

2 1

10 10log

/

/log

10

dB

V

V R

2 2

l20dB

/

V

V R

V

1 10

log20

dB

V

=

DecibelsPower ratio

Decibel

DecibelsPower ratio

60 dB

20 dB

1000000100

10 dB

20 dB100

10

6 dB

3 dB

42

0 dB-3 dB

1

-6 dB

20 dB1/4

1/2

0 01 -20 dB

-60 dB0.01

0.000001

DecibelsVoltage ratio

Decibel

DecibelsVoltage ratio

60 dB

20 dB

100010

11/√2 = 0 7071 -3 dB

-6 dB

20 dB1/2 = 0.5

1/√2 = 0.7071

0 1 -20 dB0.1

= −

2 0

1+Phase shift: ∠H( j ) tan /103

Phase shift: = ∠H( j ) tan = − /103

1000 10000 100000100

10

Z V

V

L R

R in

R

R

L R

j

j H

L j

R

=

+

/1

1)

(

R

R L

/1

1

Z V

V

C R

R in

c

+

=

R C

j R

Z Z

H

R C

j

)(

/1

=

+

j

CR j

j H

1:

where/

1

)(