ECE2061

Review of Circuit Theory

© Monash University, Australia, 2008

Chapter 1 Jaeger and Blalock

Circuits and Systems

A system is a number of interconnected circuit blocks

The key to design is to be able to analyse the circuits individually, then predict how they will operate as a system.

Often the system model is at a more abstract level: each circuit is represented by its response (Vo/Vi), input impedance (Zin), and output impedance (Zout), which are usually all frequency dependent. Circuit analysis can find these values.

CL

RV(f)i V(f)o

FilterCircuitBlock

AmplifierCircuitBlock

AmplifierCircuitBlock

~v(t)input v(t)output

Input(s) Output(s)

Circuit

System

Zout(f)Zin(f)

Notational ConventionsIn Lecture 1 we showed that a nonlinear circuit could be approximated to a linear circuit by considering only signals close to a bias point (small signal analysis).

It is useful to develop a convention when we write variables to separate out the changing and bias parts of a signal. All variables are in italics.

For example, if we have a voltage on, say, the collector pin of a transistor, we call it vC (lower-case v, UPPER CASE C)

This can be divided into:• a changing signal part, vc (both lower-case)• a bias part (non changing or DC), VC (both UPPER CASE)

When we use the equation editor, these look like:

Sometimes we explicitly state time:C C cv V v= +

Section 1.3 ( ) ( )C C cv t V v t= +

Notational Conventions (cont.)

Iv

Section 1.3

ivIV

Ov

oV ov

I I iv V v= +

O O ov V v= +

This shows how the actual voltage on a circuit, vI, vO, is separated into the bias part (DC), VI, VO, plus a time-varying part, vi, vo, .

Amp

Iv Ov

Concepts from Circuit TheoryVoltage Division

1 1

2 2

1 2 1 2

1 2

1 21 2

1 2 1 2

..

Applying KVL to the loop.( )

and /( ) Combining gives the voltage division formulae

;

S

S

S S

S S

S S

v i Rv i R

v v v i R Ri v R R

R Rv v v vR R R R

==

= + = += +

= =+ +

Section 1.5

Sv 1v

2v

1R

2R

Si The currents through the two resistors must be equal

+-

Concepts from Circuit Theory

2i

Current Division

Section 1.5

Sv1R 2RSi

1iSi 1 2

1 1 2 2

1 2

1 2

1 2

1 2

21 2

1 2

Applying KVL to single node

/ ; / Solving for the voltage gives

1 .1 1

this is sometimes written.( )

combining the above gives

;

S

S S

S S S

S S

S S

i i ii v R i v R

R Rv i iR R

R R

v i R R

R Ri i i iR R

= += =

= =++

=

= =+

1

1 2

R R+

The voltages across the two resistors must be equal

Concepts from Circuit TheoryThévenin Equivalent (Example)

Section 1.5

ov

1R

SRSv1.iβ

+-

1i

150; 20 kΩ; =1 kΩ;SR Rβ = =

thv +-

1i thR

Thévenin equivalent

We need to find the values of vth and Rth that give the same voltage at the output of the circuit for any load upon the circuit.

First equate the open-circuit (no load) output voltages of the two circuitsUse KCL on the output node of the original circuit:

Replace the resistors by conductances

The current i1 is given by the voltage drop across R1

Combining the above gives

so(multiply top and bottom by R1Rs)

Thévenin Equivalent Voltage

s

osoRv

Rvv

i +−

=1

1.β

( ) osso vGvvGi +−= 11.β

( )os vvGi −= 11

( ) ( )[ ] oss vGGvG ++=+ 11 11 ββ

( )( )[ ]

( )( )[ ]11 ss

11

11

1RR

Rv

GGG

v sso ++

+=

+++

=β

ββ

β

Applying KCL:

These combine to give

which gives, using the definition of equivalent resistance:

xxsx vGivGiii 1111 and. −=+−−=

Thévenin Equivalent Resistance

β

( )[ ] xsx vGGi ++= 1.1β

( )[ ]

( )

( )( )1

1

1.

.11

1

1

1

1

+=

⎥⎦

⎤⎢⎣

⎡+

+

+=

++=

ββ

β

β

RR

RR

RR

GGR

s

s

s

sth

Concepts from Circuit TheoryNorton Equivalent (Example)

ov

1R

SRSv1.iβ

+-

1i

150; 20 kΩ; =1 kΩ;SR Rβ = =

ni

1i

thR

Norton equivalent

We need to find the values of in and Rn that give the same voltage at the output of the circuit for any load upon the circuit.

Norton Equivalent Circuit

( )

1 1 1 1

1

First equate the short-circuit output currents of the two networks. and Combining gives

1

n S

n S

i i i i G v

i G v

β

β

= + =

= +

Note that the Norton equivalent resistance is identical to the Théveninequivalent resistance.

Also note that: th n thv i R= Check it!

Acknowledgement: This note is extracted from Prof. AJ Lowery ©Monash University 2006

Frequency Spectrum of Signals

Section 1.6

Although we often measure and describe signals by how a voltage (or current) varies with time (a waveform) it is also useful to describe signal by what frequency components are within signals, because:

• We describe music by notes, which are discrete frequencies.• We divide the airways up into different bands for different purposes• We describe the spectral purity of a signal by how many unwanted frequency components there are within it• We quantify the fidelity of an amplifier by how little power lies at distortion frequencies compared with the test frequency we are amplifying• We are often interested in the frequency response of a circuit• etc, etc……….

Most importantly, for linear systems, we can analyse a system one frequency at a time (sinusoidal steady-state and phasor analysis), as we know that the frequencies don’t interact.

• A sinusoidal voltage source varies sinusoidally with time (or co-sinusoidally).

• It has a single frequency component at frequency f.

• The angular frequency is ω= 2πf (radians/second)

• Its peak value is Vm

• Its r.m.s. value is Vm/√2

• We choose a cosine function to represent it:

• Its phase angle defines its value at t=0

• We can similarly define a sinusoidal current source

• In the steady-state, a linear circuit subject to a sinusoidal input will have a sinusoidal output at the same frequency but generally with a different amplitude and different phase. A plot of this response versus frequency is called the frequency response of a circuit.

Sinusoidal Analysis

( ) cos( )mv t V tω θ= +

t

Phasor Representation & TransformA Phasor is a complex number that represents a sinusoidal signal’s phase and amplitude. It assumes we know the frequency that we are analysing at. Complex numbers are easily manipulated in mathematics.

Euler’s identity states:

Thus:

If we chose the cosine function as the basis of out analysis (cos(0) =1), then

The Phasor Transform, P, of a (co)sinusoidal function is written:

cos .sinjxe x j x± = ±

cos Real jxx e= sin Imag jxx e=

( ) cos( ) Real j t jm mv t V t V e eω θω θ= + =

P .cos( )jm mV e V tθ ω θ= = +V

( ). cos( ) .sin( )mV jθ θ= +V

Polar form

Rectangular form

Useful notationje θ θ≡ ∠

Time and Frequency Domains

Vm

Time, t

m( ) cos( ) Real V j t jmv t V t e eω θω θ= + =

Increasing θ

V

mReal V j t je eω θ

mImag V j t je eω θ

θVm

Polar form

V

mReal V j t je eω θ

mImag V j t je eω θ

.cos( )mV θ

sin( )mV θ

Rectangular form

Time Domain

PPhasor (Frequency) Domain

Inverse Phasor TransformThe time domain waveform can be extracted from the phasor representation by:

1. Multiply the phasor by

2. Extract the real part

This is known as the inverse phasor transform. Obviously it assumes you know the angular frequency, ω, that the analysis was done for.

Note that the power in the imaginary part is thrown away. This is OK if the notation is consistent.

j te ω

( ) Real j t jmv t V e eω θ=

Note that the textbook uses a sine to represent a zero phase shift phasor, rather than a cosine. This makes little difference, as most analysis looks at the phase shift across a circuit, rather than absolute phase.

Why Phasor Representation?• It is easy to differentiate exponentials

• Inductors and capacitors can be represented by complex impedances (reactances)

• It is much easier to add the effects of multiple sources using complex numbers than trigonometry: e.g. adding voltage sources

~

~

20cos( 30 )tω − °

40cos( 60 )tω + °

vo t)(

P

P

20cos( 30 )tω − °

40cos( 60 )tω + °

17.32 .10j−

20 .34.64j++

37.32 .24.64j+

P-1

44.72 33.43∠ °

( ) 44.72cos( 33.43 )ov t tω= + °

Rectangularform

Polar form

ECE2061 - AC Circuit Theory AC voltage or current has a time varying form of:

v (t)c

t

α

A

-A

0 T 2T

ω

which can be expressed as: ( ) ( )αω += tVtv m cos , or for current ( )βω += tIti m cos)( or ( ) ( )( ) ( )( )tjj

mtj

m eeVeVtv ωααω ReRe == +

A.1 Rev 1.0

ECE2061 AC Circuit Theory

The non time varying part is defined as a ‘phasor’, ie.

V ααα ∠=== VVeeV jjm

2 or I βββ ∠=== IIeeI jjm

2

where the bold font represents a phasor quantity. Note the use of rms quantities for the phasor magnitude.

PHASORS DEFINE THE NON-TIME VARYING INFORMATION OF TIME VARYING SINUSOIDS

Phasor Representation of AC sinusoid on complex plane

Ref

V

β

α

I

A.2 Rev 1.0

Passive Components in the Frequency DomainResistor ( ) . ( ) .v t R v t R= =V I

Inductor

Capacitor ( ) 1( ) . .dv ti t Cdt j Cω

= =V I

( )( ) . .di tv t L j Ldt

ω= =V I

. L LZ Z j Lω= =V I

1. C CZ Zj Cω

= =V I

or

or

Impedance of anInductor

Voltage leads Current

Impedance of a capacitor

Voltage lags CurrentReactance is the imaginary part of the impedance

ECE2061 AC Circuit Theory

Impedance of AC Circuit Elements Resistance: )()( tiRtv =

Assume sinusoidal ( )αω += tIti m cos)( gives ( )αω += tIRtv m cos)( or in phasor form V = RI

Inductance: dt

tdiLtv )()( =

Assume sinusoidal ( ) ( )( )αωαω +=+= tjmm eItIti Recos)( gives ( )( )αωω += tj

meILjtv Re)( or in phasor form V = jωLI

Capacitance: dttiC

tv ∫= )(1)(

Assume sinusoidal ( ) ( )( )αωαω +=+= tjmm eItIti Recos)( gives ( ) ⎟

⎠

⎞⎜⎝

⎛= +αω

ωtj

meICj

tv 1Re)(

or in phasor form V Cjω

1= I

Inductor impedance of Ljω causes a 900 advance of voltage phasor compared to current. Capacitor impedance of Cjω

1 causes a 900 retard of voltage phasor compared to current.

A.3 Rev 1.0

ECE2061 AC Circuit Theory

Example: phasor diagram for series R-L-C elements is:

Note that the sign convention for dc voltages and currents still exists for ac voltages and currents. The positive sign denotes the reference end of the phasor, ie. the end which is displaced by the phasor angle away from the complex plane reference angle. Ohm’s Law for AC circuits then becomes:

V = Z I where all quantities are complex numbers representing either the phasor voltage, phasor current or circuit element impedance. AC Networks are solved using standard circuit theory techniques, but with complex arithmetic.

Ref

V+ V LR

+ VL VIV RC+ VC

I

A.4 Rev 1.0

ECE2061 AC Circuit Theory

Standard Solution techniques are: Voltage and Current Divider Thevenin and Norton Equivalent Circuits Node and Mesh Analysis Impedance reduction But remember to define all reference directions. This is doubly important for AC networks because of the complication of the phasor angles. Lecture Example:

31.8μF

0.318H

100Ω 282sin(2π50t)

Question: Find the phasor currents flowing through each circuit element Solve by: Impedance reduction and voltage/current dividers Solution will be explained in lecture

A.5 Rev 1.0

Resonant Circuit in the Frequency Domain

C~vi(t)

L

R

1/jωC~Vi

jωL

RVo

1) Time Domain 2) Frequency Domain

vo(t)

2

/1/

1

L Cp

L C

Z ZZ

Z ZL C

j L j Cj LLC

ω ωω

ω

=+

=+

=− +

2

2

2

( )( )

1

1

(1 )

po

i p

ZV sV s Z R

j LLC

j L RLC

j Lj L R LC

ωωω

ωω

ω ω

=+

− +=+

− +

=+ −

3) Parallel L and C 4) Voltage Divider 5) Plot Magnitude( )( )

o

i

V sV s

2 1/( )LCω =

1

Using ‘s’ notation in the Frequency Domain

• For compactness, jω is often replaced by s• However, s can have special meanings:

– It can mean that we are in the Laplace Domain (we have transformed from the time domain using a Laplace transform)

– It can be a complex frequency equal to jω + σ

• See other units for more details

Decibels• A decibel is defined as the ratio of two powers, expressed as:

AdB = 10.log10 (P1/P2)

• If the two powers are developed into the same impedance (say, a resistor), then decibels can also be used to define a ratio of voltages:

AdB = 10.log10[(V1)2/(V2)2 ] = 20.log10 (V1/V2)

• Often a subscript is used to make a decibel into an absolute unit:– E.g. dBW is the power relative to one watt or simply 10log10(Power,watts)

• Decibels are often used to describe the combined performance of a chain of amplifiers, as adding dB is like multiplying the gains together.

• A factor of 2 in power is 3.01 dB. A factor of 2 in voltage is 6.02 dB.• A factor of 10 in power is 10 dB. A factor of 10 in voltage is 20 dB.

Section 10.2.4