ac circuit analysis - sharif university of technologyee.sharif.edu/~faez/ac_circuits.pdf · ac...

AC Circuit Analysis

Module: SE1EA5 Systems and Circuits

AC Circuit Analysis

Module: SE1EA5 Systems and Circuits

Lecturer: James GrimblebyyURL: http://www.personal.rdg.ac.uk/~stsgrimb/email: j.b.grimbleby reading.ac.ukj g y g

Number of Lectures: 10

Recommended text book:David Irwin and Mark NelmsBasic Engineering Circuit Analysis (8th edition)John Wiley and Sons (2005) ISBN: 0-471-66158-9

School of Systems Engineering - Electronic Engineering Slide 1James Grimbleby

AC Circuit AnalysisAC Circuit Analysis

Recommended text book:David Irwin and Mark NelmsBasic Engineering Circuit Analysis (8th edition)Paperback 816 pages J h Wil d S (2005)John Wiley and Sons (2005) ISBN: 0-471-66158-9

Price: £36


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 sinusoidsPhasorsComplex impedances of inductors and capacitorsDriving-point impedanceFrequency response of circuits – Bode plotsPower in ac circuitsE t i it d i d tEnergy storage in capacitors and inductorsThree-phase power


AC Circuit Analysis PrerequitiesAC Circuit Analysis Prerequities

You should be familiar with the following topics:

SE1EA5: Electronic CircuitsSE1EA5: Electronic CircuitsOhm’s LawSeries and parallel resistancespVoltage and current sourcesCircuit analysis using Kirchhoff’s Lawsy gThévenin and Norton's theoremsThe Superposition Theorem

SE1EC5: Engineering MathematicsComplex numbers



Lecture 1

AC Voltages and CurrentsReactive Components


AC WaveformsAC Waveforms

Sine waveform(sinusoid)

Square waveformSquare waveform

Sawtooth waveform

Audio waveformAudio waveform


FrequencyThe number of cycles per second of an ac waveform is known

Frequencyy p

as the frequency f, and is expressed in Hertz (Hz)

Voltage or Current 6 cycles → f = 6 Hzy

Time0 1s


FrequencyExamples:

Frequencyp

Electrocardiogram: 1 Hz

Mains power: 50 Hz

Aircraft power: 400 HzAircraft power: 400 Hz

Audio frequencies: 20 Hz to 20 kHz

AM radio broadcasting: 0.5 MHz – 1.5 MHz

FM di b d ti 80 MH 110 MHFM radio broadcasting: 80 MHz – 110 MHz

Television broadcasting: 500 MHz – 800 MHzg

Mobile telephones: 1.8 GHz


PeriodThe period T of an ac waveform is the time taken for a

Period

complete cycle:

frequencyperiod 1

=frequency

Voltage or Current T = 0.167 sT 0.167 s

Time0 1s


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?y p

present long enough for any transients to die away

Output V

VVin Timet

Vin

Vin=0 for t≤0


Why Sinusoidal?Why Sinusoidal?

A linear circuit will not change the waveform or frequency of a sinusoidal input (the amplitude and phase may be altered)

Power is generated as a sinusoid by rotating electrical machinery

Si id l i d l d iSinusoidal 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)fundamental pure sinusoid plus harmonics (Fourier Analysis)


Fourier AnalysisA 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

FundamentalFundamental

3 d h i3rd harmonic

5th harmonic


Representation of SinusoidsA sinusoidal voltage waveform v(t) of amplitude v0, and of

Representation of Sinusoidsg ( ) p 0

frequency f :tvftvtv ωsinπ2sin)( 00 ==

or: tvftvtv ωcosπ2cos)( 00 ==

where ω=2πf is known as the angular frequency

v fT /1=v fT /1=

tv0


Representation of SinusoidsThe sinusoid can have a phase term φ:

Representation of Sinusoids

A phase shift φ is equivalent to a time shift φ/ω

( )φω += tvtv sin)( 0

A phase shift φ is equivalent to a time shift -φ/ω

φτv

( )φω +tv sin0

ωτ =

t

( )φω +tv sin0

( )tv ωsin0

t

( )0

The phase is positive so the red trace leads the green traceSchool of Systems Engineering - Electronic Engineering Slide 15James Grimbleby

The phase is positive so the red trace leads the green trace

Resistors

Ceramic tube

Resistors

Ceramic tubecoated withConductive filmConductive film

Metal endi

v

Metal end cap

R i t RFil b

v

Resistance RRiv =

Film: carbonmetal

t l id(Ohm’s Law)

metal oxide


ResistorsResistors


Resistorsv

i

Resistors

Riv =Ohm’s Law:i

RSuppose that:

( )tvtv ωsin)(i, v

i v( )tvtv ωsin)( 0=

Then:

i v

Rtvti )()( =

Then: t

( )tRvR

ωsin0=R

Current in phase with voltage


Capacitors

i

Capacitors

Insulating

i

Conducting

dielectricv

Conducting electrodes

Di l t i i C it CdvCiCvq

Dielectrics: airpolymer

i

Capacitance C

dtCiCvq ==ceramic

Al203 (electrolytic)


CapacitorsCapacitors


Capacitorsv

i d

Capacitors

i

C dtdvCi =

Suppose that: ( )tvtv ωsin)( 0=

i, v

)(t)(ti( )tvtv ωsin)( 0

Then: )(tv)(ti

( )= sin)( 0 ωtvdtdCti t

( )⎞⎛

= cos0π

ωω tCv

⎟⎠⎞

⎜⎝⎛ +=

2sin0

πωω tCvCurrent leads voltage by π/2 (90°)


CapacitorsDoes a capacitor have a “resistance”?Capacitors

v, i ( )tvtv ωsin)( 0= ( )tCvti ωω cos)( 0=

t1t 2t1t 2t

∞==0)(

)( 01 vtitv 00

)()(

022 ==

ititv

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

0)( 1ti )( 02 iti


p

Capacitors

The reactance XC of a capacitor is defined:

Capacitors

The reactance XC of a capacitor is defined:

0ivXC =

where v0 is the amplitude of the voltage across the capacitor 0i

C

0 p g pand i0 is the amplitude of the current flowing through it

Thus:

fCCCvXC 2

110 ===fCCCvC πωω 20

The reactance of a capacitor is inversely proportional to its value and to frequency


InductorsInductors

Magnetisablecorecore

Copper wire

v

i

C i

v

I d t L

diL

Core: airferritei

Inductance L

dtdiLv =iron

silicon steel


InductorsInductors


Inductorsv

i di

Inductors

Suppose that:L dt

diLv =

Suppose that: ( )tvtv ωsin)( 0=

i, v

)(t )(ti

1Then:

)(tv )(ti

( )∫= sin1)( 0 ω

v

tvL

ti t

( )

⎞⎛

−= cos0 ωω

tL

v

⎟⎠⎞

⎜⎝⎛ −=

2sin0 πω

ωt

Lv

Current lags voltage by π/2 (90°)


InductorsThe reactance XC of an inductor is defined:

Inductors

C

0

0ivXC =

where v0 is the amplitude of the voltage across the inductor

0i

and i0 is the amplitude of the current flowing through it

Thus:fLL

LvvXc πωω

2/0 ===

The reactance of an ind ctor is directl proportional to its

Lv ω/0

The reactance of an inductor is directly proportional to its value and to frequency


Resistance and ReactanceResistance and Reactance

0ivX = 0→f ∞→f0i

0→f ∞→f

Resistance R R R R

Capacitance CCω1 short

circuitopencircuit

Inductance L Lω openi it

shorti itInductance L Lω circuitcircuit



L t 2Lecture 2

AC Analysis using Differential EquationsComplex NumbersComplex Numbers

Complex Exponential Voltages and Currents


AC Circuit AnalysisThe ac response of a circuit is determined by a differential

AC Circuit Analysisp y

equation:

R)()()( tvtRitv cin +=

)(

R )(ti

dttdvCti c )()( =)(tvin C )(tvc

)()()( tvdt

tdvRCtv cc

in +=dt

tvtvtdv i )()()(RC

tvRC

tvdt

tdv incc )()()(=+


AC Circuit AnalysisNow suppose that the input voltage vin is a sinusoid of angular

AC Circuit Analysis

frequency ω

Th t t lt ill b i id f th fThe output voltage vc will be a sinusoid of the same freqeuncy, but with different amplitude and phase:

( )( )φωω+=

=

tvtvtvtvin

cos)(cos)(

1

0

Expanding the expression for vc:

( )φω += tvtvc cos)( 1

c

tBtAtvtvtvc ωωφωφω sincossinsincoscos)( 11 +=−=

tBtAdt

tdvc ωωωω cossin)(+−=


dt

AC Circuit Analysis

The differential equation becomes:

AC Circuit Analysis

The differential equation becomes:

tvtBtAtBtA ωωωωωωω cossincoscossin 0=+++−

Comparing the coefficients of sinωt and cosωt on both sides of

tRC

tRC

tRC

tBtA ωωωωωωω cossincoscossin +++

p gthe equation:

0BRCA =+− ω

0vARCB =+ω

Solving these simultaneous linear equations in A and B:

2220

2220

11 CRRCvB

CRvA

ωω

ω +=

+=


AC Circuit Analysis

00 sincos RCvvBvvA ωφφ

AC Circuit Analysis

Thus:

2220

12220

11

sin1

cosCR

vBCR

vAω

φω

φ+

=−=+

==

Thus:RC

CRvv ωφ

ω−=

+= tan

11

22201

At an angular frequency ω=1/RC:CRω+1 222

−==42

01

πφvv

⎟⎠⎞

⎜⎝⎛ −=

4cos

2)(

420 πωtvtvc

The output voltage lags the input voltage by π/4 (45°)

⎠⎝ 42


p g g p g y ( )


1.0

0.7071

v 1/v

0e

gain

Vo

ltage

0 0

V

Angular frequency ω (rad/s)

0.01/RC 10/RC 100/RC0.1/RC0.01/RC


g q y ( )

Complex Numbers:Complex Numbers: Rectangular Form

Complex numbers can be represented in rectangular, polar or exponential form

Rectangular form:

h i h l i h i i ( d

jyxz +=

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

112 −=−= jj

Complex numbers are often the solutions of real problems, for example quadratic equations


p q q

Complex Numbers:Complex Numbers: Argand Diagram

Imaginary part

jyxz +=

Imaginaryaxis

yaxis

Real partpO x

Real axis


Complex Numbers: Polar FormPolar form:

θ∠rz

Complex Numbers: Polar Form

where r is the magnitude, and θ is the angle measured from th l i

θ∠= rz

the real axis:

I i

z

Imaginaryaxis r

θReal axis

O

θ


Complex Numbers:Complex Numbers: Exponential FormExponential form:

θjrez =

Euler’s identity:θθθ ijj θθθ sincos je j +=

1sin θ

θO

cos θ

The polar and exponential forms are therefore equivalent

cos θ


The polar and exponential forms are therefore equivalent

Complex Numbers: Conversion

z


z

r

θy

O

22

xO

Rectangular to polar:yz

yxzr

∠

+==

θθ tan

22

θP l t R t l

xyz =∠= θθ tan

θθ

sincos

ryrx

==Polar to Rectangular:


Complex Numbers: InversionIf the complex number is in rectangular form:

Complex Numbers: Inversionp g

1jyx

z+

=

))(( jyxjyxjyx

jyx

+−

=

+

))((jyx

jyxjyx−

=

−+

If th l b i i l ti l f

22 yx +

If the complex number is in polar or exponential form:

j11 φφ

jj e

AAez −==

11


Complex Numbers: ConversionWhen using the inverse tangent to obtain θ from x and y it is

Complex Numbers: Conversiong g y

necessary to resolve the ambiguity of π:

yy jyxz +=

y θ+−

yx

++

yx

xy

=θtanθ

+x−xx

−+

yx

−yx

jyxw −−=


Complex Numbers: ConversionWhen using the inverse tangent to obtain θ from x and y it is

Complex Numbers: Conversiong g y

necessary to resolve the ambiguity of π

1. Calculate θ using inverse tangent:

y1

/ θ /xy1tan−=θ

This should give a value in the range: -π/2 ≤ θ ≤ +π/2 (-90º ≤ θ ≤ +90º)

2. If the real part x is negative then add π (180º) :

xy1tan−+=πθ



C t t t l f2 π∠


Convert to rectangular form3

2 π∠=z

Real part: 1212

3cos2 =×=⎟

⎠⎞

⎜⎝⎛=πx

23 ⎠⎝

Imaginary part: 3232

3sin2 =×=⎟

⎠⎞

⎜⎝⎛=πy

Thus: 31 jz +=


Complex Numbers: ConversionComplex Numbers: Conversion

2

1 73231 jz +=

Imaginary

1.732

2=raxis

1

R l i)60(

3oπθ =

Real axisO 1 2



C t t l ti l f1


Convert to polar or exponential form:j

z+

=1

1

Magnitude:2

1

11

0122

22=

+== zr

211 22 +

Angle:

01t0t

)1(1

11 ππθ

⎟⎞

⎜⎛

⎟⎞

⎜⎛

+∠−∠=∠=

−−

jz

440

11tan

10tan 11 ππ

−=−=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛= −−

π

Thus: or 42

142

1π

π jezz−

=−∠=


242

Complex Numbers: ConversionComplex Numbers: Conversion

Real axisO π

0.5

O

Imaginary

)45(4

oπθ −=

Imaginaryaxis

0.5 11 j−2

11

1 jj

z =+

=


Complex Exponential Voltages

We shall be using complex exponential voltages and currents


We shall be using complex exponential voltages and currents to analyse ac circuits:

tj

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

tjVetv ω=)(

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:it unchanged, apart from a multiplying factor:

tjtjd tjtj VejVedtd ωω ω=



S th t l ti l lt i li d


Suppose that a complex exponential voltage is applied across a resistor:

tv )(tjVt ω)(

tjVRtvti =)()(

tjVetv ω=)()(ti

tjeRV ω=R

The current through the resistor is also a complexThe current through the resistor is also a complex exponential





Suppose that a complex exponential voltage is applied across a capacitor:

dttdvCti =)()(tjVetv ω=)(

tjVedtdC

dtω=

jVetv =)()(ti

tjCVejdt

ωω=C

The current through the capacitor is also a complexThe current through the capacitor is also a complex exponential





Suppose that a complex exponential voltage is applied across an inductor:

dttvL

ti )(1)( ∫=tjVetv ω=)(

tj dtVeL

ω1∫=

jVetv =)()(ti

tjVeLj

Lω

ω1

=L

The current through the inductor is also a complex

Ljω

The current through the inductor is also a complex exponential


Complex Exponential VoltagesComplex 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 exponentialsparts of the complex exponentials

Complex exponential: )sincos()( tjtetv tj ωωω +==Complex exponential:

Real voltage:

)sincos()( tjtetvc ωω +==

ttv ωcos)( =Real voltage: ttvr ωcos)( =



L t 3Lecture 3

PhasorsImpedancesImpedances

Gain and Phase ShiftFrequency Response


PhasorsIf the input voltage to a circuit is a complex exponential:

Phasors

then all other voltages and currents are also complex

tjcin evtv ω

0)( =

then all other voltages and currents are also complex exponentials:

tjtjjtj

tjtjjtjc

Iiiti

eVeevevtvωωφφω

ωωφφω

)(11

)(11

22

11

)(

)( ===+

+

tjtjjtjc eIeeieiti ωωφφω

22)(

22 22)( === +

where V1 and I2 are time-independent voltage and current phasors:

111φjevV =

222

11φjeiI

evV

=

=


PhasorsThe complex exponential voltages and currents can now be

Phasors

expressed:

tj

tjc eVtv

ω

ω11 )( =

Ph i d d t f ti b t i l f ti

tjc eIti ω

22 )( =

Phasors are independent of time, but in general are functions of jω and should be written:

( ) ( )ωω jIjV 21

However, when there is no risk of ambiguity the dependency will be not be shown explicitly

Note that upper-case letters are used for phasor symbols


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:

IVZ =

For a resistor:tRitv cc = )()(tjVetv ω)(

RIeVe

tRitvtjtj

cc

= ωω

)()(jc Vetv =)(

tjc Ieti ω=)(

RIV =R

So that: RIVZR ==


Impedance

F it

Impedance

For a capacitor:tdvCti c=)()(

VedCIe

dtCti

tjtj

c

ωω

=)(tj

c Vetv ω=)(tjIeti ω=)(

CVejIe

Vedt

CIe

tjtj

tjtj

ω ωω

ωω

=

=

C

c Ieti =)(

CVjICVejIe jj

ωω

==

So that: VZ 1CjI

VZC ω1

==


Impedance

F i d t

Impedance

For an inductor:tdiLtv c=)()(

IedLVe

dtLtv

tjtj

c

ωω

=)(tj

c Vetv ω=)(tjIeti ω=)(

LIejVe

Iedt

LVe

tjtj

tjtj

ω ωω

ωω

=

=c Ieti =)(

L

LIjVLIejVe jj

ωω

==

So that:V LjIVZL ω==


ImpedanceImpedance

IVZ = ∞→f0→f

Resistance R R R R

I

Resistance R R R R

1Capacitance C Cjω

1∞→Z 0→Z

Inductance L Ljω ∞→Z0→Z


ImpedanceAll the normal circuit theory rules apply to circuits containing

Impedancey pp y g

impedances

For example impedances in series:

4321 ZZZZZ +++=

and impedances in parallel:

4321 ZZZZZ +++=

and impedances in parallel:

11111+++=

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

4321 ZZZZZ+++

Other relevant circuit theory rules are: Kirchhoff’s laws, Thévenin and Norton's theorems, Superposition


ImpedanceImpedance

Potential divider:

Vin

21IZV

ZZVI in+

=I1Z

2

2ZV

IZV

in

out

=

=

VV

2

21ZV

ZZ

out

+=

outV2ZinV

212ZZ

ZVV

in

out+

=


AC Circuit AnalysisSuppose that a circuit has an input x(t) and an output y(t),

AC Circuit Analysispp p ( ) p y( )

where x and y can be voltages or currents

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

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

)()()cos()( tjtj Xereexretxtx ωωω

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

)()()cos()( 00tjtj Xereexretxtx ωωω ===

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

)()()cos()( 00tjtjj Yereeeyretyty ωωφφω ==+=


AC Circuit Analysis

Thus:

AC Circuit Analysis

Thus:

XY

xey j

=0

0φ

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

Xx0

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

Yy0Xx

yg ==0

0

and the phase shift is:

⎞⎛Y⎟⎠⎞

⎜⎝⎛∠=

XYφ



Using the potential divider formula:

ZV CcR

CjZZ

ZVV

RC

C

in

c

ω+

=

/1

R

RCjCj

ωω+

=/1

/1CinV cV

CRjω+=

11


AC Circuit AnalysisR

AC Circuit Analysis

Vc 1V VCRjV

Vin

cω+

=1

1CinV cV

1VVoltage gain:

2221

1

RCVVg

in

c

ω+==

Phase shift: CRVV

i

c ωφ 11 tan0tan −− −=⎟⎠

⎞⎜⎝

⎛∠=

CR

Vin

ω1tan−−=

⎠⎝


Frequency Response (RC = 1)Frequency Response (RC 1)

1.0

ain 0.7071

age

Ga

Volta

Angular Frequency (rad/s)

0.01 10 1000.10.01



Frequency Response (RC = 1)


Frequency Response (RC 1)


01 10 1000.1

shi

ft

π

Pha

se )45(4

°−−π

P

)90(2

°−−π


Frequency ResponseFrequency Response

CRωφ 1tan−−=2221

1

RCg

ω+=

0→ω 1→g )0(0 °→φ→g

1 1 π

)(φ

CR1

=ω2

1=g )45(

4°−−=

πφ

∞→ω 0→g )90(2

°−−→πφ

This is a low-pass response


p p


0→ω ∞→ω

0→CVinC VV →



ZVV R

C

RZZ

VVCR

RinR +

=

RinV RV

CRjCjR

RVin ω+=

/1

CRjCRjVV inR ω

ω+

=1

CRjCRj

VV

i

Rω

ω+

=1 CRjVin ω+1


AC Circuit AnalysisC

AC Circuit Analysis

CRjCRj

VVR ω

=1RinV RV CRjVin ω+1Rin R

V lt i CRVR ωVoltage gain:2221 RC

CRVVg

in

R

ω

ω

+==

⎞⎛Phase shift : CRVV

in

R ωφ 11 tantan −− −∞=⎟⎠

⎞⎜⎝

⎛∠=

CRωπ 1tan2

−−=

⎠⎝


2


1.0

ain 0.7071

age

gaVo

lta


0.01 10 1000.10.01




)90(2

°π

ift

)45(4

°π

ase

sh

4

Pha

100Angular Frequency (rad/s)

01 100.10.01




CRωπφ 1tan2

−−=2221 RC

CRgω

ω

+=

0→ω 0→g )90(2

°→πφ→g

1 1 π

)(2

φ

CR1

=ω2

1=g )45(

4°=

πφ

∞→ω )0(0 °→φ1→g

This is a high-pass response


g p p


0→ω ∞→ω0→ω ∞→ω

inR VV →0→RV



L t 4Lecture 4

Driving-Point Impedance


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:

)( ωjI

)()( ωω jVjZ =AC Circuit)( ωjV

)( ωjI

)()(

ωω

jIjZ =AC Circuit)( ωjV

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


Impedance

Z can be written in rectangular form:

Impedance

Z can be written in rectangular form:

)()()( ωωω jjXjRjZ +=

where R is the resistance and X is the reactance

Thus: XRZ 22 +=

RXZ 1tan−=∠

and:ZZR ∠= cos

ZZX ∠= sin


Symbolic and Numeric FormsSymbolic and Numeric Forms

CRjRZω+

=1

Symbolic Form

Substitute component values

Numeric Form31081

80−

=ωj

Z

Substitute frequency value

31081 ××+ ωj

Value at a given frequency

value

4024 jZ +=Value at a given frequency 4024 jZ +


Example 1Determine the driving-point impedance of the circuit at a

Example 1g p p

frequency of 40 kHz:

1ZZZ CR +=

1Cj

R +=ω

C = 200nF

1020010402125 93j ××××

+=−π

R = 25Ω

050270125

j

j

+=

Ω89.192505027.0

jj

−=


Example 1Example 1

89.1925 −= jZ Ω

89.1925 22 +=Z C = 200nF

93.3189.1925

=

+ZΩ

R = 25Ω

89.19tan 1−=∠ −Z

R = 25Ω

)5.38(6720.025

tan

°−−=

=∠Z

)(


Example 1What will be the voltage across the circuit when a current of

Example 1g

5 A, 40 kHz flows through it?

)89.1925(5 jIZV

−×==

f

V45.99125)89.1925(5

jj

−=×

In polar form:

)5.38(6720.0)93.315( °−−∠×== IZV

)5.38(6720.0159.7V)()(

°−−∠=



Example 2g p p

frequency of 20 Hz:

ZZZ CR+=

111

CjR

ZZZ CR

ω+=1 R = 80Ω C = 100μF

Z

jR

=1

μ

RCjR

Zω+/1

CRjω+=

1


Example 2Example 2

1 CRjRZ

+=

ωC = 100μF

801

6

CRj

=

+ ω

R = 80Ω8080101002021 6j ××××+ −π

)00511(80005.11j

j−

+=

Ω00407939005.11

)005.11(8022

j

j+

−

Ω00.4079.39 j−=


Example 2Example 2

00.4079.39 −= jZ Ω

00.4079.39 22 +=Z42.56= Ω R = 80Ω C = 100μF

00.40tan 1⎬⎫

⎨⎧−=∠ −Z

)245(78800

79.39tan

o−−=

⎭⎬

⎩⎨∠Z

)2.45(7880.0=


Example 2

What c rrent ill flo if an ac oltage of 24 V 20 H is applied

Example 2

What current will flow if an ac voltage of 24 V, 20 Hz is applied to the circuit?

ZVI ==

ZVI

78800425624−∠

=00407939

24−

=j

A3016030)2.45(7880.0A0.4254

7880.042.56

j+°∠=

∠)00.4079.39(24

00.4079.39

22+

=j

j

A3016.03.0 j+=

A3016.03.000.4079.39 22

+=+

j)2.45(7880.00.4254A °∠=


Phasor DiagramsWhere voltages or currents are summed the result can be

Phasor Diagramsg

represented by a phasor diagram: 321 VVVV ++=

Imaginary part

2V

Realpart

O

1V part1VV

3V


Example 2

A3.024==RI A3016.02410100202 6 jjIC =××××= −π

Example 2

80R jjC

0.3A

part

nary

p

0.2ACI

CR III +=

Imag

in

0 1A

RI R l

I 0.1A

0.2A

RI RealpartO

0.1A 0.3A



Example 3

frequency of 50 Hz:

ZZZZ CLR ++= R 1Cj

LjR ++=ω

ω24 Ω

L

101205021103650224 6

3

jj

j

+×××+=−

−πL36 mH

Ω53.2631.112410120502 6

jjj

−+=×××π

C120 μF

Ω22.1524 j−=120 μF


Example 3

221524 −= ΩjZ

Example 3

22.1524= ΩjZ

422822.1524 22

=

−=

ΩZ R

24 Ω42.28

⎫⎧

= Ω 24 Ω

L

2422.15tan 1

⎭⎬⎫

⎩⎨⎧−=∠ −Z 36 mH

)4.32(5652.0 °−−= C120 μF

)4.32(5652.042.28 °−−∠= ΩZ


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?

)221524(A10 jZIV

×==

ΩV2.152240

)22.1524(A10j

j−=

−×= Ω

In polar form:

ZIV)4.32(5652.0)42.2810( °−−∠×=

= ZIV

)4.32(5652.0V2.284 °−−∠=


Example 3

CLR VVVV ++=

Example 3

200VI i

CLR VVVV ++=

LV

Imaginary part

RVLV

Realt200V 400V

V

part

CVV

-200V


Example 4Determine the driving-point impedance of the circuit at a f f 400 H

Example 4

frequency of 400 Hz:111

+=

RC

ZZZZ CLR1

++

=

R2 Ω

CCj

LjR1

1 ωω

++

=

L1 mH

200 μFCjLjR

Z)/(1

1ωω ++

=

1 mH

LjRCjLjR

)(1 ωωω++

+=

LCCRjLjR

21 ωωω

+

+=


LCCRj1 ωω −+

Example 4LjRZ +

=ω

Example 4

1040022

13

2

j

LCCRjZ

××+

−+=

−π

ωω

1020010)4002(210200400211040022

6326

3

jj

××××−××××+

××+−−− ππ

π

263.1005.11513.22

jj

−++

=

005126330513.22j

jj

+−+

=

)005.12633.0()513.22(005.12633.0

22jj

j−−×+

=

+−

474.2852.1005.12633.0 22

jj

−=+


j

Example 4

47428521 jZ

Example 4

474.2852.1 −= jZ

R0913

474.2852.1 22 +=Z

2 ΩC

200 F

091.3=

L1 mH

200 μF

852.1474.2tan 1=∠ −Z

)2.53(9282.0852.1

°−−=

)2.53(9282.0091.3 °−−∠= ΩZ


Example 4What current will flow if an ac voltage of 120 V, 400 Hz is

Example 4

applied to the circuit?VI VI

120

=Z

I

120Z

I =

)47428521(120474.2852.1

120

+×−

=

jj

92820A82389282.0091.3

120

∠−∠

=

474.2852.1)474.2852.1(120

22 +

+×=

j08.3126.239282.0A82.38

j+=∠=

556.90.2974.222 +

=j

)(53.20.9282A82.3808.3126.23

°∠=+= j


)(

Example 4

Imaginary part

Example 4

III CRL +=Imaginary part

120IRL =

50ACI

23.2927.23 jZ

IRL

RL

−= I

120

j

Real120Z

IC

C =ea

partO 50AI32.60j= RLI


Admittance

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:

)( ωjI

1)()( ωω jIjY ==AC Circuit)( ωjV

)( ωjI

)()()(

ωωω

jZjVjY ==AC Circuit)( ωjV

Admittance Y is analogous to conductance in dc circuits and its unit is Siemens

)()()( ωωω jjBjGjY +=

where G is the conductance and B is the susceptance

)()()( ωωω jjBjGjY +


AdmittanceAdmittance

IY = ∞→f0→fV

Y ∞→f0→f

111Resistance R

R1

R1

R1

Capacitance C Cjω ∞→Y0→Y

Inductance L Lj1

∞→Y 0→YInductance L Ljω →Y 0→Y


Admittance

All the normal circ it theor r les appl to circ its containing

Admittance

All the normal circuit theory rules apply to circuits containing admittances

For example admittances in series:

4321

11111YYYYY

+++=

and admittances in parallel:

Oth l t i it th l Ki hh ff’ l

4321 YYYYY +++=

Other relevant circuit theory rules are: Kirchhoff’s laws, Thévenin and Norton's theorems, Superposition


Example 5

Determine the dri ing point admittance of the circ it at a

Example 5

Determine the driving-point admittance of the circuit at a frequency of 400 Hz:

YYYY C +=

/1/11 R

2 Ω

Cj

YY LRC

ω +=

+

1/1/1 2 Ω

L

C200 μF

LjRCj

ωω

++= L

1 mH

μ


Example 5Determine the driving-point admittance of the circuit at a f f 400 H

Example 5

frequency of 400 Hz:2104002102004002

36 jjY ×××+×××=

−− ππ

150270

1040022102004002 3

j

jjY

××++×××=

−ππ

51322513.22

15027.0

jj

j+

+=

R513.22513.225027.0 22

jj+

−+= R

2 ΩC

32.10513.225027.0 jj −

+= L

C200 μF

S2590.01939.02436.01939.05027.0

jjj

+=−+= 1 mH


S2590.01939.0 j+


L t 5Lecture 5

Resonant Circuits


Resonant CircuitsResonant Circuits

Passive resonant circuits must contain a resistor, capacitor and an inductor

The behaviour of resonant circuits changes rapidly around a particular frequency (the resonance frequency)

R i i b h i d bResonant circuits can be characterised by two parameters: the resonance frequency and the Q-factor

There are two basic resonant circuit configurations: series and parallelparallel


Resonant CircuitsResonant Circuits

Lg

dtd θω

−=ωθ=

dtd

θLdtdt

θ,ω ωθθ

t


Resonant CircuitsiL

Resonant Circuits

LC vC Ci

dtdv LC −=

Lv

dtdi CL =vC CdtLdt

i, vCv

iCLi

t


Parallel Resonant CircuitParallel Resonant Circuit

R LR LC

++1111 LRjZ ω

=ZZZZ LCR

++=

11 LjRLCRLj

Z 2

ωωω +−

=

LjCj

R ωω ++=

2

11LCRLj

Lj2 1/ ωω

ω+−

=

LRjRLCRLj

ωωω +−

=2

LCRLjLj

2/1 ωωω−+

=


j j


Ω= k5R μF1=C H1=L

Impedance is a maximum ( tΩ= k5R μF1=C H1=L (resonant frequency)

h

1

when:

2/1+=

ωωω

LCRLjLjZ

1=

LCω

624

/1

=

−+ωωωj

LCRLj610

1=

−624 101021 −− ×−××+ ωωj 310=



0→ω ∞→ω0→ω ∞→ω

0→Z0→Z



R LLjZ 2

ω=R L

CLCRLj 2/1 ωω −+

00 jLjZ ωω

1

01

0

RLjZ

jjZ =→→

ω

ω

R t f/

1

jLj

RRLj

LjZLC

===

ωωωωResonant frequency:

02 jCj

LCLjZ −=

−=

−→∞→

ωωωω


Parallel Resonant Circuit

5kΩ )90( °π


5kΩZ∠

)90(2

°

Z Z∠

0

Z

0 0

Z

)90( °−−π

Angular frequency (rad/s)

0.01000 10000100

)90(2



Quality Factor

The standard form for the denominator of a second-order

Quality Factor

The standard form for the denominator of a second-order system is:

20

20 //1 ωωωω −+ Qj

Compare this with the impedance Z:

00 //1 ωωωω+ Qj

Compare this with the impedance Z:

LjZ 2ω

=

So that:LCRLj

Z 2/1 ωω −+

LRQ

LC 00

1ω

ω ==

where Q is the quality-factor and ω0 is the resonant frequency

LLC 0ω


Quality FactorQuality Factor

Ω= k5R μF1=C H1=LΩ= k5R μF1=C H1=L

1010

11 360 ===

−LCω

5000

10

R 5101

500030=

×==

LRQω


Quality Factor

Z

Quality Factor

maxZ

2maxZ

Q0ωω =Δ

2 Q

Z

0 0


0.00ω



Quality Factor

2kΩ )90( °π

Quality Factor

2kΩ

Z∠

)90(2

°

Q=2

Z Z∠Z∠ Q

0

Z

0 0 )90( °−−π


0.01000 10000100

)(2



Quality Factor

10kΩ )90( °π

Quality Factor

10kΩZ∠

)90(2

°

Q=10

Z Z∠Q

0

Z

0 0

Z

)90( °−−π


0.01000 10000100

)(2




LICIRI

V1 Ωk5=R μF1=C H1=LV1

11 4R

IR ×=== −

1

1025000

11 4Resonance occurs in parallel resonant circuits

ωωω

jCjCj

IC ×=== −10/1

1 6parallel resonant circuits because the currents in the capacitor and

ωωωj

Lj

LjIL

−=

−==

1p

inductor cancel out


Parallel Resonant CircuitAt resonance:


:103=ω 1mAA102 4−×=IR

ImaginaryCI

10 6−×= jIC ω Real

Imaginarypart

A10 3−= j I partO1mARI

−=

jILLI

A10 3−−= j

L ω -1mA


j


Below resonance:R l

CIRI

:105.04

3×=ωI

Realpart

ImaginaryO 1mA

RI

10

A1026

4

−

−

×=

×=

jI

I

C

R

ω

IImaginarypart

A105.0

103−×=

×=

j

jIC ω-1mA

−=

jIL ωLI

A102 3−×−= jω

-2mA



Above resonance:2mA

:100.24

3×=ω CI

10

A1026

4

−

−

×=

×=

jI

I

C

R

ω I i

1mA

A100.2

103−×=

×=

j

jIC ωI

Imaginarypart

−=

jIL ωRealpartO 1mARI

A105.0 3−×−= jω

LI1mAR


Series Resonant CircuitSeries Resonant Circuit

ZZZZ LCR ++=

LjCj

R

LCR

ωω1

++=R

LCCRj

Cj

ωω

ω21−+

= L

LCCRj

Cj

ωω

ω21+

=

CjLCCRj

ωωω1 −+

= C



CjLCCRjZ

ωωω 21 −+

=Cjω R

∞−=−

=→→ jCj

CjZ

ωωω 10

L

=== RCj

CRjZLC

j

ωωω 1

∞==−

→∞→ jLjLCZ

CjLC

ωωω

ω2

C

∞==→∞→ jLjCj

Z ωω

ω



21 LCCRj 21 −+=

ωωω

CjLCCRjZ R = 200 Ω

6

626

1010102001

−

−− ×−××+=

ωωω

jj

L = 1H

624

6

101021

10−− ×−××+

×

ωω

ω

j

j L 1H

610−×=

ωj C = 1μF


Series Resonant Circuit

1kΩ )90( °π


1kΩ )90(2

°

ZZ

Z∠Z

0

200ΩZ∠

)90( °−−π

0Ω

200Ω

Angular frequency (rad/s)1000 10000100

)(20Ω




The standard form for the denominator of a second-order


The standard form for the denominator of a second order system is:

20

20 //1 ωωωω −+ Qj

Compare this with the admittance Y (= 1/Z):

00 //1 ωωωω+ Qj

p ( )

CjY 2ω

=

So that:LCCRj

Y 21 ωω −+

CRQ

LC 00

11ω

ω ==

where Q is the quality-factor and ω0 is the resonant frequency

CRLC 0ω


where Q is the quality factor and ω0 is the resonant frequency


1

1

10 =

LCω

1101

16 ××

=−

R = 200 Ω

rad/s103=L = 1H

10

=CR

Qω

C 1 F

200101101

63 ×××=

−

C = 1μF

520010110

=×××



Resonance occurs in series resonant circuits

R = 200 ΩRV

1A

because the voltages across the capacitor and

L = 1HLV

1Apinductor cancel out

L 1HLV

jj

RVR =×=

101

20016

C = 1μFCVωωωj

Cj

CjVC

−=

−==

101

ωω jLjVL =×= 1


Series Resonant CircuitAt resonance:


:103=ω 1kVV200VR =

LV

106jVC−

= RealImaginary

part

V103j

C

−=

ω partO RV 1kV

jVL = ωCV

V103j

jVL

=

= ω -1kV


Crystal ResonatorCrystal Resonator


Crystal Resonator

Equivalent circuit:

Crystal Resonator

f0 = 8.0 MHzR = 3 4 Ω

Equivalent circuit:

R

R = 3.4 ΩL1 = 0.086 mHC = 4 6 pFC1 = 4.6 pFC0 = 42 pF

L C011

0 == Qω

C1 127010035 70

0

=×=

CRQ

LC ωω

12701003.5 =×=



L t 6Lecture 6Frequency-Response FunctionFrequency Response Function

First-Order Circuits


Frequency-Response FunctionFrequency Response Function

Input OutputYX

)()( ωω jYjHFrequency-response function:)(

)(ω

ωjX

jH =Frequency-response function:

Voltage gain g: )()()( ω

ωω jHjXjYg ==

)( ωjX

)( jY ⎞⎛Phase shift φ: )()()( ω

ωωφ jHjXjY

∠=⎟⎠⎞

⎜⎝⎛∠=


Frequency-Response FunctionThe order of a frequency-response function is the highest

Frequency Response Function

power of jω in the denominator:

Fi t d 1First order:0/1

1)(ωω

ωj

jH+

=

Second order: 200 )/(/21

1)(ωωωω

ωjj

jH++

=

Third order:

00 )/(/21 ωωωω jj ++

1)( ωjH =Third order: 30

200 )/()/(/1

)(ωωωωωω

ωjjj

jH+++

=

The order is normally equal to (and cannot exceed) the number of reactive components


Example 1


Example 1


R

ZZZ

VV

RC

C

in

c+

=R

RCjCj

ZZV RCin

/1/1

+=

+

ωω CinV cV

jH

RCj1)(

/1

=

+

ω

ω

CRjjH

1:where11

)(+

ω

ωω

RCj:where

/1 00

=+

= ωωω


Example 1

R

Example 1

0/11)(ωω

ω+

=j

jHR

220

0

/1/1ωωωω

+

−=

jj

CinV cV

20

2 /1 ωω+

Gain: 20

2 /1

1)(ωω

ω+

== jHg0/1 ωω+

Ph hift /tan)( ωωφωφ ∠ jHPhase shift: 0/tan)( ωωφωφ −=∠= jH


Example 1Example 1

0→ω ∞→ω

0→g1→g


DecibelThe decibel is a measure of the ratio of two powers P1, P2 :

Decibelp 1 2

110log10dB P

=2

10log10dBP

=

It can also be used to measure the ratio of two voltages V1, V2:

2

21

102

21

10 log10//log10dB

VV

RVRV==

1

22

22

l20dB

/

V

VRV

2

110log20dB

V=


Decibel

DecibelsPower ratio

Decibel

DecibelsPower ratio60 dB20 dB

1000000100

10 dB20 dB100

106 dB3 dB

42

0 dB-3 dB

11/2 -3 dB

-6 dB20 dB

1/41/2

0 01 -20 dB-60 dB

0.010.000001


Decibel

DecibelsVoltage ratio

Decibel

DecibelsVoltage ratio60 dB20 dB

100010

10 dB20 dB10

√10 = 3.1626 dB3 dB

2√2 = 1.414

0 dB-3 dB

11/√2 = 0 7071 -3 dB

-6 dB20 dB

1/2 = 0.51/√2 = 0.7071

0 1 -20 dB-60 dB

0.10.001


Example 1Circuit is a first-order low-pass filter:

Example 1

1 /1

=g

p

01 /tan ωωφ −= −2

02 /1 ωω+

=g

0ωω << )dB0(1≈g )0(0 °≈φ

0ωω = )dB3(2

1−=g )45(

4°−−=

πφ2 4

0ωω >> )oct/dB6(0ωg )90( °−−≈πφ0ωω >> )oct/dB6(0 −≈

ωg )90(

2≈φ


Example 1

R = 1kΩ

Example 1

01

=RC

ωR 1kΩ

63 10101×

=−

C=1μF

3101010

=

Gain: 62 10/1

1)(ω

ω+

== jHg10/1 ω+

Phase shift: 310/tan)( ωφωφ ∠ jHPhase shift: 310/tan)( ωφωφ −=∠= jH


Bode PlotGain(dB) Phase(rad)

Bode Plot

0 dB 0-3 dB

-10 dBg

-20 dB 4π

−

gφ

30 dB

4

6 dB / octave-30 dB

π

-6 dB / octave

Frequency (rad/s)1000 10000 10000010010

-40 dB 2π

−


Frequency (rad/s)

Example 2


Example 2


L

ZZZ

VV

LR

R

in

R+

=V V

LjRR

LRin

+=

ω

RinV RV

RLjjH

LjR

=

+

/11)( ω

ω

RRLj

==

+

0:where1/1

ω

ω

Lj=

+= 0

0:where

/1ω

ωω


Example 2Example 2

0→ω ∞→ω0→ω ∞→ω

0→g1→g


Example 3


Example 3


C

ZZZ

VV

CR

R

in

c+

=

RCjRZZV CRin

/1 +=

+

ω

RinV cV

CRjjH

RCj

)(

/1

=

+ωω

ω

jCRj

jH

1:where/1

)(

0

+

ωωωω

ω

RCj:where

/1 00

0 =+

= ωωω


Example 3

/ωωj C

Example 3

00

/1/)(ωω

ωωω+

=j

jjH C

0 /11

ωω−=

j RinV RV

220

0

/1/1ωωωω

+

+=

jj

G i1)( ω == jHg

220 /1 ωω+

Gain: 220 /1

)(ωω

ω+

== jHg

Phase shift: ωωφωφ /tan)( 0=∠= jH


Example 3Example 3

0→ω ∞→ω0→ω ∞→ω

1→g0→g


Example 4


Example 4


R

ZZZ

VV

RL

L

in

L+

=

R

RLjLjZZV RLin

+=

+

ωω LinV LV

RLjjH

RLj

=

+/)( ωω

ω

RjRLj

jH+

0 :where//1

)(

ωωωω

ω

Lj=

+= 0

0

0 :where/1

ωωω


Example 4Example 4

0→ω ∞→ω0→ω ∞→ω

1→g0→g


gg

Example 4

Circuit is a first-order high-pass filter:

Example 4

11g

Circuit is a first order high pass filter:

ωωφ /tan 01−=22

0 /1 ωω+=g

0ωω << )oct/dB6(0ωω

≈g )90(2

°≈πφ

0ωω = )dB3(2

1−=g )45(

4°=

πφ

0

)dB0(1g

0 )(2

g )(4

φ

)0(0 °φ)dB0(1≈g0ωω >> )0(0 °≈φ



Bode Plot

0 dB2π

-3 dB

-10 dBg

-20 dB 4π

g

30 dB

4φ

-30 dB6 dB / octave

Frequency (rad/s)1000 10000 10000010010

-40 dB 0


Frequency (rad/s)

Example 5Using the potential divider formula:

Example 5g p

/1 CjRV

1R

122

/1/1

RCjRCjR

VV

in

c++

+=

ωω

iV tV2R

122

11CRjCRj

CRj+++=ωω

ω inV outV

C

212

12

)(11)(

RRCjCRjjH

jj

+++

=ω

ωω

212

2111:where/1

)(1j

RRCj

==+

=

++

ωωωωω

22

211

1 )(/1 CRRRCj ++ ωω


Example 5Example 5

1R

12

/1/1)(ωωωωω

jjjH

++

=R1j

inV outV2R

22

2 /1)( ωωω +== jHg

C

21

2 /1)(

ωωω

+jHg


Example 5Example 5

00→ω ∞→ω

1R2R2R

212RR

Rg+

→1→g


Example 5

Assuming that :21 ωω <<

Example 5

22

2 /1 ωω+=g

Assuming that :21 ωω <<

21

2 /1 ωω+=g

1ωω << )dB0(1≈g

21 ωωω <<<< )oct/dB6(1 −≈ωωg

2ωω >>

ω

21 Rg =≈ω2ωω >>

122 RRg

+ω


Example 5

/1 2+ j ωω

Example 5

1/1/1)(

12

++

=jjjH

ωωωωω

Ω9001 =R

)(1

211 +=

RRCω

1

)100900(101

6 +=

−Ω1002 =R

rad/s10

)100900(103=

+μF1=C

1001011

622

×==

−CRω

rad/s1042

=



Bode Plot

0 dB 0

-10 dB g

-20 dBφ

30 dB-30 dB

π

Frequency (rad/s)1000 10000 10000010010

-40 dB 2π

−


Frequency (rad/s)


L t 7Lecture 7

Second-Order Circuits


Example 1

R R

Example 1

V V

R R

inV cVC C

This circuit must be simplified before the frequency response f i b d i dfunction can be determined

A Thé i i l t i it i t d f th t tA Thévenin equivalent circuit is created of the components to the left of the red line


Example 1Thévenin equivalent circuit:

Example 1

R Z

CinV VCinV V

Cjω/1 CRjω+111

VRCj

CjVV in ωω+

=/1

/1

RR

CRjCjRZ

ωω +=+=

111

CRjVinω+

=1 CRj

RZω+

=1


Example 1

RCRjRω+1

Example 1

V

RCRjω+1

CRjVinω+1 C CV

RRCj

CjCRj

VV inC

ω

ωω ++

×+

=/1

/11

VCRj

RCjj

in

ωω

+++

11

/1

CRjCRjCRjCRj

Vin

ωωωω+

++×

+=

11

11


CRjω+1

Example 1Frequency-response function:

Example 1

1

11 CRjCRjCRj

VV inC ωω

×+

=

111

VCRj

CjCRjCRj

i

ωωωω+

+++

1)1()1( CRjCRjCRj

Vinωωω ++×+

=

222311)(

RCCRjjH

ωωω

−+=

R = 1 kΩ, C = 1μF:μ

623 1010311)(

−− ×−××+=

ωωω

jjH


101031 ×××+ ωωj

Example 1Example 1

0→ω ∞→ω

0→g1→g



Bode Plot

0 dB 0

-10 dB g

-20 dB φ

-30 dB-30 dB

π

Frequency (rad/s)1000 10000 10000010010

-40 dB π−


Frequency (rad/s)

Example 2

R C

Example 2

R

V V

C

inV RVRC

This circuit must be simplified before the frequency response s c cu us be s p ed be o e e eque cy espo sefunction can be determined

A Thévenin equivalent circuit is created of the components to the left of the red line


Example 2

CRjRω+1 C

Example 2

CRjω+1

Vin

C

CRjVinω+1 RVR

RCjR

RCRj

VV inR

ωω ++×

+=

/11

CRjVCRj

CjRj

in ωω

ω+

++1

/1

CRjCRjCRj

CRjCRj

Vin

ωωω

ωω

+++

×+

=

111


CRjω+1

Example 2Frequency-response function:

Example 2

11 CRjCRj

CRjCRj

VV inR ωω

ωω ++

×+

=

11

CRjVCRj

CRjj

in ωω

ω+

++

)1()1(CRj

CRjCRjCRjCRjVin

ωωωω

ω++×+

=

22231)(

RCCRjCRjjHωω

ωω−+

=

R = 1 kΩ, C = 1μF:3

623

3

10103110)(

−−

−

×−××+

×=

ωωωω

jjjH


101031+ ωωj

Example 2Example 2

0→ω ∞→ω

0→g0→g



Bode Plot

0 dB2π

-10 dB g

-20 dB φ

30 dB-30 dB

π

Frequency (rad/s)1000 10000 10000010010

-40 dB 2π

−


Frequency (rad/s)


Example 3

CjVC /1 ωRL

RLjCjCj

VV

in

C

1/1

/1++

=ωω

ω

CinV cV

LCCRjjH

11)( 2−+

=ωω

ω

Qj /)/(11

20

20 −+

=ωωωω

LQ

Qj11:and1:where

/)/(1

0

00

===

+

ω

ωωωω

CRCRQ

LC 00 ω


Example 3Example 3

0→ω ∞→ω

0→g1→g


Example 3Circuit is a second-order low-pass filter:

Example 3p

1)( jH )( jH20

20 /)/(1

)(ωωωω

ω−+

=Qj

jH )( ωjHg =

0ωω << 1)( ≈ωjH )dB0(1=g

0ωω = Qg =jQjH −=)( ω Qg

)t/dB12(20ω

jQj )(

20ω−0ωω >> )oct/dB12(2

0 −≈ω

g20)(

ωωω ≈jH


Example 3

Ω200RmH400L

Example 3

Ω200=RmH400=L

inV CVμF5.2=C

10111 30 ====ω

11040011

1010105.210400

3

6630

×

×××−

−−−

L

LCω

2106.1200

1105.210400

20011 5

6

3=×=

×

×==−C

LR

Q


Bode PlotGain(dB)

Bode Plot

20 dB2=Q 10=Q

0 dB0 dB

21

=Q

-20 dB12 dB / octave-12 dB / octave

Frequency (rad/s)1000 10000 10000010010

-40 dB


Frequency (rad/s)

Bode PlotPhase(rad)

Bode Plot

0 10=Q

2=Q1Q

2π

−2

1=Q

2

Frequency (rad/s)1000 10000 10000010010

π−


Frequency (rad/s)


Example 4g p

LjVL ωC R

LC

RLjCjj

Vin

L/1

2

++=

ωωinV LVL

LCCRjLCjH

1)( 2

2

−+

−=

ωωωω

L

Qj /)/(1/

22

20

2−=

ωω

LQ

Qj11:and1:where

/)/(1 20

20 −+

ω

ωωωω

CRCRQ

LC:and:where

00 ===

ωω


Example 4Example 4

0→ω ∞→ω

1→g0→g


Example 4Circuit is a second-order high-pass filter:

Example 4g p

)( jH20

2 /)( ωω−jH )( ωjHg =20

20

0/)/(1

/)(ωωωω

ωωω−+

=Qj

jH

2 2

0ωω << 20

2)(

ωωω −

≈jH )oct/dB12(20

2

ωω

≈g

0ωω = jQjH =)( ω

0

Qg =

0

jQj )( Qg

0ωω >> 1)( ≈ωjH )dB0(1=g


Bode PlotGain(dB)

Bode Plot

20 dB2=Q 10=Q

0 dB0 dB

21

=Q

-20 dB12 dB / octave

2

12 dB / octave

Frequency (rad/s)1000 10000 10000010010

-40 dB


Frequency (rad/s)

Bode PlotPhase(rad)

Bode Plot

π 10=Q

2=Q1Q

2π 2

1=Q

2

Frequency (rad/s)1000 10000 10000010010

0


Frequency (rad/s)


Example 5

RVC L

RLjCjR

VV

in

R/1 ++

=ωω inV RVR

LCCRjCRjjH

1)( 2−+=

ωωωω

QjQj

j

/)/(1)/(

220

+=

ωωωωωω

LQ

Qj11:and1:where

/)/(1

0

20

20

===

−+

ω

ωωωω

CRCRQ

LC:and:where

00 ===

ωω


Example 5Example 5

0→ω ∞→ω

0→g0→g


Example 5Circuit is a second-order band-pass filter:

Example 5

)( jH0 )/()( ωω QjjH )( ωjHg =20

20

0/)/(1

)()(ωωωω

ω−+

=Qj

QjjH

0ωω <<Q

jjH0

)(ωωω ≈ )oct/dB6(

0Qg

ωω

≈

0ωω = 1)( =ωjH

0

)dB0(1=g

0

jjH ω0)( −

1)( ωjH )dB0(1=g

)t/dB6(0ω0ωω >>Q

jjHωωω 0)( ≈ )oct/dB6(0 −≈

Qg

ωω


Bode PlotGain(dB)

Bode Plot

0 dB

2Q10=Q

2=Q

1Q

-20 dB2

=Q

Frequency (rad/s)1000 10000 10000010010

-40 dB


Frequency (rad/s)

Bode PlotPhase(rad)π

Bode Plot

2π

10=Q

2=Q1Q

02

1=Q

π

Frequency (rad/s)1000 10000 100000100102

π−


Frequency (rad/s)

Example 6Using the potential divider formula: R

Example 6

LjCjVC /1 + ωω

LC

RLjCjjj

Vin

C

1

/12

++=

ωω inV cVL

LCCRjLCjH

11)( 2

2

−+

−=

ωωωω C

QjLC

/)/(11

22

2−=

ω

LQ

Qj11:and1:where

/)/(1 20

20 −+

ω

ωωωω

CRCRQ

LC:and:where

00 ===

ωω


Example 6Example 6

0→ω ∞→ω

1→g1→g


Example 6Circuit is a second-order band-stop filter:

Example 6

)( jH20

2 /1)( ωω−jH )( ωjHg =20

20

0/)/(1

/1)(ωωωω

ωωω−+

=Qj

jH

0ωω << 1)( ≈ωjH )dB0(1≈g

0ωω = 0)( =ωjH )dB(0 −∞=g

1)( ωjH

)(g

0ωω >> 1)( ≈ωjH )dB0(1≈g


Bode PlotGain(dB)

Bode Plot

0 dB

2Q 2=Q 10=Q

-20 dB 21

=Q

Frequency (rad/s)1000 10000 10000010010

-40 dB


Frequency (rad/s)

Bode PlotPhase(rad)π

Bode Plot

2π

21

=Q

10=Q

2

0

10=Q

2=Q

π

Q

Frequency (rad/s)1000 10000 100000100102

π−


Frequency (rad/s)


L t 8Lecture 8

Power in AC Circuits


Power in AC CircuitsTo calculate the power in a circuit we shall need to make

Power in AC Circuits

use of some trigonometric identities:

BABABA ii)(BABABABABABA

sinsincoscos)cos(sinsincoscos)cos(

+=−−=+

Adding:

)cos()cos(1coscos

coscos2)cos()cos(

BABABA

BABABA

++

=−++

so that:

)cos()cos(2

coscos BABABA −++=

AAA 2cos21

210cos2cos

21cos2 +=+=


rms Voltages and CurrentsThe average power in a resistor is given by:

rms Voltages and Currents

∫T

)()(1∫=

T

dttitvT

P

20

)()(1)(ti

∫=T

dtR

tvT 0

2 )(1 )(tv R

∫=T

dttvTR

2

0

)(11

∫

∫

Trms dttVV

TR

22

0

)(1h ∫== rmsrms dttv

TV

R 0

2 )(:where



The root-mean-square voltage V determines the power


The root-mean-square voltage Vrms determines the power dissipated in a circuit:

2

RVP rms

2=

There is a similar expression for the power dissipated when e e s a s a e p ess o o t e po e d ss pated ea current Irms flows through a circuit:

22rmsRIP =

These expressions apply to any waveform


rms Voltages and CurrentsThe rms value of a sinusoid of amplitude (peak) value v0:


)(1 2 dttvVT∫

p (p ) 0

1

)(0

dttvT

V

T

rms ∫=

)(cos1

0

220 dttv

T

T∫= ω

)2cos(21

2112

0 dttT

vT∫ += ω

22

020

0

vv

TAverages to zero over

l t l2200 == a complete cycle:

T = 2π/ω



The UK mains power was until recently supplied at 240 V rms


The UK mains power was until recently supplied at 240 V rms and that in Europe 220 V rms

On 1 January 1995 the nominal voltage across Europe was harmonised at 230 V rms.harmonised at 230 V rms.

This corresponds to an amplitude of:s co espo ds to a a p tude o

2 V2302

20

×=

×= rmsVv

V325=


rms Voltages and CurrentsA mains power (230 V rms) electric fire has a resistance of


kW017123022VP rms

52 Ω:

kW017.152

230===

RVP rms

An audio amplifier which drives a 4 Ω loudspeaker at up to 150 W must supply a sinusoidal output voltage:pp y p g

6004150.2 =×==rms RPV

V5.24

6004150.

=

×

rms

rms

V

RPV

This corresponds to a sinusoid of peak value 34.6 V


rms Voltages and CurrentsSquare wave of amplitude ±v0:


v0

-v0

TT/2

22/ 22 )(11)(1 dtvdtvdttvVTTT∫+∫∫

T

2/0

00

0

1

)()( dtvT

dtvT

dttvT

V

TT

rms ∫ −+∫=∫=

020

0

20

1 vvdtT

vT

==∫=


Crest FactorThe ratio between the peak voltage and the rms voltage is k th t f t

Crest Factor

known as the crest factor:

peakVcf

rms

pV

cf =

For a sinusoid the crest factor is √2; for a square wave the crest factor is 1

For audio signals the crest factor depends on the source but is commonly 2 or higher

150 W f di i t 4 Ω l d k ld th f i150 W of audio into 4 Ω loudspeakers would therefore require peak voltages of 50 V or greater


Power in a Reactive LoadCapacitors and inductors store energy, but do not dissipate power

Power in a Reactive Load

power

ICIR

R C100 V rms

50 Hz

100

25Ω 200μF

100

A425

100

6

==RI

A6.28A10200502100100 6 =××××== −πC

C ZI

W40025

100 P2==


Instantaneous Power

)(tiFor sinusoidal voltages and currents:

Instantaneous Power

)(ti( )( )ω=

tititvtv

)(cos)( 0

)(tv Z( )φω += titi cos)( 0

Instantaneous power:

( ) ( )φωω coscos)()()(

00 +=×=

titvtitvtp( ) ( )( ) ( )φωω

φωωcoscoscoscos

00

00+=

+=

ttivtitv

φφω cos)2cos(21

00 ++= tiv


Average PowerAverage power:

T1

Average Power

∫=T

dttpT

P0

)(1

( ) ( )∫ +=T

dttitvT 0

00 coscos1 φωω

∫+∫ +=TT

dtivdttiv

T

0000

0

cos11)2cos(11 φφω ∫+∫ + dtT

ivdttT

iv0

000

00 cos2

)2cos(2

φφω

If T >> 1/ω: TIf T >> 1/ω:φcos1

21

000 dt

TivP

T∫=

φcos21

00

0

iv=


2

Average PowerAverage Power

φcos21

00ivP =)(ti

φ2 00

)(tv Zφcos21 2

0Zv

=

φcos212

20 Zi

Z

=2 0


Average Power

Average power:

Average Power

Average power:φcos

21

00ivP =

For a resistor:20

20

001110 RivivP ===→=φ

For a capacitor:

000 2220 Ri

RivP ===→=φ

o a capac to

02

=→= Pπφ

For an inductor:2

π 02

=→−= Pπφ


rms Voltages and CurrentsPower expressed in terms of rms voltages and currents:


φcos21

00ivP =

φ

φ

cos2212 00

IV rmsrms=

φ

φ

)W(cos

cos2

IV rmsrms

rmsrms

=

2V φcosZ

VP rms=

φcos2 ZIP rms=


Example 1Determine the average power dissipated in the circuit:

Example 1

Ω80R

F20C

Ω80=R230 V rms50 Hz

1

μF20=C

1

1+=

CjRZ

ω

1020502180 6×××

+=−j π

Ω)( 63 31 105178 1

Ω2.15980o∠=

−= j


Ω)(-63.31.105178.1 −∠=

Example 1Example 1

Ω80=R230 V rms50 H

μF20=C50 Hz

2cos

2= φ

ZVP rms

105.1cos1178

2302−=

W4.1331.178

=


Example 2Determine the average power dissipated in the circuit:Example 2

RR2 Ω

C80 V rms

L1 mH

C200 μF

400 Hz

1 mH

The driving-point impedance of this circuit at 400 Hz (calculated previously) is:(calculated previously) is:

9282.0091.3 −∠= ΩZ


Example 2

928200913 −∠Ω=Z

Example 2

9282.0091.3 −∠Ω=Z

R2 Ω

80 V

LC

200 μF

80 V rms400 Hz2V L

1 mH200 μF

cos

2

= φZ

VP rms

9283.0cos091.3

802−=

W1241=


Example 2Determine the average power dissipated in the circuit

Example 2

Since no power is dissipated in the capacitor we only need to calculate the power in the inductor resistor legto calculate the power in the inductor-resistor leg

+= LjRZLR ω R80 V rms400 Hz

1040022 3××+=

+−j

LjRZLR

π

ω2 Ω

C

400 Hz

8986.0212.3513.22∠=

+= j L1 mH

200 μF898603


Example 2Example 2

8986.0212.3 ∠=LRZ

R2 Ω2 Ω

C200 F

80 V rms400 Hzcos

2= φVP rms

L1 mH

200 μF

80

cos

2

φZ

P

W1241

8986.0cos212.3

80=

W1241=



L t 9Lecture 9

Power FactorTh Ph El t i PThree-Phase Electric Power


True and Apparent Power

The apparent power P in a circuit is:

True and Apparent Power

The apparent power Pa in a circuit is:

rmsrmsa IVP =

Apparent power is measured in VA

rmsrmsa IVP =

Apparent power is measured in VA

The true power P dissipated in a circuit is:

φcosrmsrms IVP =

True power is measured in W


Power FactorThe power factor is the ratio of the true power to the apparent

Power Factorp p pp

power:

IVP φφ coscos===

rmsrms

rmsrms

a IVIV

PPpf

where ø is the phase difference between voltage and currentwhere ø is the phase difference between voltage and current.

It does not matter whether ø is phase of the current withIt does not matter whether ø is phase of the current with respect to the voltage, or voltage with respect to the current, since:

φφ −= coscos


Example 1Determine the power factor, apparent power and true power

di i t d i th i it

Example 1

power dissipated in the circuit:

)(75 11 312Ω6051

Ω08.154o∠=

+= jZ

)(75.11.312Ω60.51 ∠=Ω4=R

Hz400,0Vrms8

2559.312.1cos ==pf mH6=L

Hz400,0Vrms8

VA3.4102

=== rmsrmsrmsa Z

VIVP

W0.105=×= aPpfPZ


Power Factor Correction

Most industrial loads have a poor (pf << 1) power factor


Most industrial loads have a poor (pf << 1) power factor

Examples are induction motors and inductor-ballast lightingExamples are induction motors and inductor ballast lighting

Power factor can be corrected by connecting a reactance inPower factor can be corrected by connecting a reactance in parallel with the load

This reduces the apparent power and the rms current without affecting the loadg

This is obviously desirable because it reduces the current yrating of the power wiring and supply


Power Factor CorrectionPower factor is normally corrected by connecting a reactive


element ZC in parallel with the load ZL :SI CI

Supply current: ISL d t I

LIVLoad current: IL

Correction current: IC LZ CZSV

A unity overall power factor will be obtained provided that VSu y o e a po e ac o be ob a ed p o ded a Sand IS are in phase:

)real(00 jGGVIS

S +=∠=



CLS jGIII 0


SI CI

IS

C

S

L

S

S jGVI

VI

VI

+=+=

11

0

LZ CZ

LISVCL

jGZZ

+=+ 011

LZ CZ

imagCimagL ZZ ⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡ 11

imagCimagL ⎦⎣⎦⎣

If IL leads VS then an inductor is used for correction

If IL lags VS then a capacitor is used for correction


Power Factor CorrectionCorrection of a lagging power factor load with a capacitor:


CLS III +=

CICLS III +=

SI Current (real part)

Current (imaginary

LI

(real part)part)

LI

Note that the magnitude of the supply current IS is less than that of the load IL


L

Example 2

Choose a suitable power factor correction component for the

Example 2

Choose a suitable power factor correction component for the circuit:

Ω4=RHz4000Vrms8

mH6=L

Hz400,0Vrms8

081541Ω08.154

jjZL +=

06195.001643.008.15408.1541

22 jjZL

−=+

−=

Thus: 06195.01 jZC

+=


Example 2Example 2

Ω4=RHz400,0Vrms8 μF465.2=C

mH6=L

Hz400,0Vrms8 μF465.2C

1 06195.01=+= ωCjj

ZC

μF465.24002

06195.0=

×=

πC


Example 280

=I

Example 2

)061950016430(80 −=

=

jZ

IL

L

II I

80956.4314.1

)06195.001643.0(80−=

=j

jLI

Ω4R

SI CI

80=

ZI

cC

Ω4=R

Hz4000Vrms8

F4652=C

956.406195.080

=×=

jj mH6=L

Hz400 μF465.2

956495643141

956.4+=

jjIII

j

CLS

314.1956.4956.4314.1

=+−= jj


Example 2

5A

Example 2

5A

ImaginaryCLS III +=

CIImaginary

part

5ASI Real

part5A part

LI

-5A


Example 3An electric motor operating from the 50 Hz mains supply has

Example 3p g pp y

a lagging current with a power factor of .80

The rated motor current is 6 A at 230 V so that the magnitude of 1/ZL is:

61 I 02609.0230

61===

S

L

L VI

Z

and the phase of 1/ZL is:

6435.08.0cos1 1 ±==⎭⎬⎫

⎩⎨⎧

∠ −Z

Since the current lags the voltage the negative phase is used

⎭⎩ LZ


g g g p

Example 3

015650020870643500260901−=−∠= j

Example 3

0156501

01565.002087.06435.002609.0 =∠=

Cjj

jZL

015650

01565.01=+= ωCjj

ZC

μF82.49502

01565.0=

×=

πC

Before correction: After correction:

13808013806230

××

=×=aPpfP

P1104

1104== aP

PP

110413808.0

=

×=×= aPpfP 8.4230

1104===

SS V

PI


Example 3

5A

Example 3

5A

Imaginary CLS III +=

5A

CIg ypart

5ASI Real

part

LI

p

-5A


Three-Phase Electric PowerMost ac power transmission systems use a three-phase

t

Three Phase Electric Power

system

Three phase is also used to power large motors and otherThree-phase is also used to power large motors and other heavy industrial loads

Three-phase consists of three sinusoids with phases 2π/3 (120º) apart(120 ) apart

This allows more power to be transmitted down a givenThis allows more power to be transmitted down a given number of conductors than single phase

A three-phase transmission system consists of conductors for the three phases and sometimes a conductor for neutral


p

Three-Phase Electric PowerThree Phase Electric Power

Three-phaseload

Three-phasegenerator


Three-Phase Electric Power

Phase to ne tral oltage

Three Phase Electric Power

π

Phase-to-neutral voltage v0Phase-to-phase voltage vp

v0v

)60(3

oπ

v0vp

3sin2 0vvp =

π

v0)120(

32 oπ

232

3

0v=

32

0

0

v=


Three-Phase Electric PowerUK domestic supply uses three -phase with a phase-to-

Three Phase Electric Powerpp y p p

neutral voltage v0 of 230 V rms (325 V peak)

This corresponds to a phase-to-phase voltage vp of 400 V rms (563 V peak)

Each property is supplied with one phase and neutral

If the phases are correctly balanced (similar load to neutral on h) th th ll t l t ieach) then the overall neutral current is zero

The UK electricit distrib tion net ork operates at 275 kV rmsThe UK electricity distribution network operates at 275 kV rms and 400 kV rms



L t 10Lecture 10

Energy Storage


Energy Storage

Reactive components (capacitors and inductors) do not

Energy Storage

Reactive components (capacitors and inductors) do not dissipate power when an ac voltage or current is applied

Power is dissipated only in resistors

Instead reactive components store energy

During an ac cycle reactive components alternately store energy and then release itgy

Over a complete ac cycle there is no net change in energy y g gystored, and therefore no power dissipation


Energy StorageThe voltage across a capacitor is increased from zero to V

Energy Storage

producing a stored energy E:

v0)()( dttitvE

T∫=v(t)

i0

)( dtdtdvCtv

T∫=

V

Cdv

0)(

dC

dtV∫

∫

dtdvCi =

01

dvvC ∫=

Tt

221CVE =

T


Energy StorageEnergy Storage

Example: calculate the energy storage in an electronic flash capacitor of 1000 μF charged to 400 V

21 2= CVE

40010100021 26 ×××= −

J802

=


Energy StorageThe current in an inductor is increase from zero to I

Energy Storage

producing a stored energy E:

)()( dttitvET∫=

vi(t)

0

)( dttidiLT∫

vi

LI

0)( dtti

dtL

I

∫= L

diL0

diiLI∫= dt

Lv =

Tt

221LIE =

T


Energy StorageEnergy Storage

Example: calculate the energy storage in a 2 mH inductor carrying a current of 10 A

21 2= LiE

10102212

23 ×××= −

J1.02

=


ac circuit analysis - sharif university of technologyee.sharif.edu/~faez/ac_circuits.pdf · ac...

Documents