AC circuits description with different signal

AC Circuit Analysis
Module: SE1EA5 Systems and Circuits
AC Circuit Analysis
Module: SE1EA5 Systems and Circuits
Lecturer: James Grimbleby
y
URL: http://www.personal.rdg.ac.uk/~stsgrimb/
email: j.b.grimbleby reading.ac.uk
j g y g
Number of Lectures: 10
Recommended text book:
David Irwin and Mark Nelms
Basic Engineering Circuit Analysis (8th edition)
John Wiley and Sons (2005)
ISBN: 0-471-66158-9
School of Systems Engineering - Electronic Engineering Slide 1
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
Recommended text book:
David Irwin and Mark Nelms
Basic 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
James Grimbleby

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 currents
Complex representation of sinusoids
Phasors
Complex impedances of inductors and capacitors
Driving-point impedance
Frequency response of circuits – Bode plots
Power in ac circuits
E t i it d i d t
Energy storage in capacitors and inductors
Three-phase power
James Grimbleby

AC Circuit Analysis Prerequities
AC Circuit Analysis Prerequities
You should be familiar with the following topics:
SE1EA5: Electronic Circuits
SE1EA5: Electronic Circuits
Ohm’s Law
Series and parallel resistances
p
Voltage and current sources
Circuit analysis using Kirchhoff’s Laws
y g
Thévenin and Norton's theorems
The Superposition Theorem
SE1EC5: Engineering Mathematics
Complex numbers
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
Lecture 1
AC Voltages and Currents
Reactive Components
James Grimbleby

AC Waveforms
AC Waveforms
Sine waveform
(sinusoid)
Square waveform
Square waveform
Sawtooth waveform
Audio waveform
Audio waveform
James Grimbleby

Frequency
The number of cycles per second of an ac waveform is known
Frequency
y p
as the frequency f, and is expressed in Hertz (Hz)
Voltage or
Current 6 cycles → f = 6 Hz
y
Time
0 1s
James Grimbleby

Frequency
Examples:
Frequency
p
Electrocardiogram: 1 Hz
Mains power: 50 Hz
Aircraft power: 400 Hz
Aircraft 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 MH
FM radio broadcasting: 80 MHz – 110 MHz
Television broadcasting: 500 MHz – 800 MHz
g
Mobile telephones: 1.8 GHz
James Grimbleby

Period
The period T of an ac waveform is the time taken for a
Period
complete cycle:
frequency
period
1
=
frequency
Voltage or
Current T = 0.167 s
T 0.167 s
Time
0 1s
James Grimbleby

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 inductors
p
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.
James Grimbleby

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
V
Vin
Time
t
V
in
Vin=0 for t≤0
James Grimbleby

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 i
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)
fundamental pure sinusoid plus harmonics (Fourier Analysis)
James Grimbleby

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
Fundamental
3 d h i
3rd harmonic
5th harmonic
James Grimbleby

Representation of Sinusoids
A sinusoidal voltage waveform v(t) of amplitude v0, and of
g ( ) p 0
frequency f :
t
v
ft
v
t
v ω
sin
π
2
sin
)
( 0
0 =
=
or: t
v
ft
v
t
v ω
cos
π
2
cos
)
( 0
0 =
=
where ω=2πf is known as the angular frequency
v f
T /
1
=
v f
T /
1
=
t
v0
James Grimbleby

The sinusoid can have a phase term φ:
A phase shift φ is equivalent to a time shift φ/ω
( )
φ
ω +
= t
v
t
v sin
)
( 0
A phase shift φ is equivalent to a time shift -φ/ω
φ
τ
v
( )
φ
ω +
t
v sin
0
ω
τ =
t
( )
φ
ω +
t
v sin
0
( )
t
v ω
sin
0
t
( )
0
The phase is positive so the red trace leads the green trace
James Grimbleby
The phase is positive so the red trace leads the green trace

Resistors
Ceramic tube
Resistors
Ceramic tube
coated with
Conductive film
Conductive film
Metal end
i
v
Metal end
cap
R i t R
Fil b
v
Resistance R
Ri
v =
Film: carbon
metal
t l id
(Ohm’s Law)
metal oxide
James Grimbleby

Resistors
Resistors
James Grimbleby

Resistors
v
i
Resistors
Ri
v =
Ohm’s Law:
i
R
Suppose that:
( )
t
v
t
v ω
sin
)
(
i, v
i v
( )
t
v
t
v ω
sin
)
( 0
=
Then:
i v
R
t
v
t
i
)
(
)
( =
Then:
t
( )
t
R
v
R
ω
sin
0
=
R
Current in phase with voltage
James Grimbleby

Capacitors
i
Capacitors
Insulating
i
Conducting
dielectric
v
Conducting
electrodes
Di l t i i C it C
dv
C
i
Cv
q
Dielectrics: air
polymer
i
Capacitance C
dt
C
i
Cv
q =
=
ceramic
Al203 (electrolytic)
James Grimbleby

Capacitors
Capacitors
James Grimbleby

Capacitors
v
i d
Capacitors
i
C dt
dv
C
i =
Suppose that:
( )
t
v
t
v ω
sin
)
( 0
=
i, v
)
(t
)
(t
i
( )
t
v
t
v ω
sin
)
( 0
Then:
)
(t
v
)
(t
i
( )
= sin
)
( 0 ωt
v
dt
d
C
t
i t
( )
⎞
⎛
= cos
0
π
ω
ω t
Cv
⎟
⎠
⎞
⎜
⎝
⎛
+
=
2
sin
0
π
ω
ω t
Cv
Current leads voltage by π/2 (90°)
James Grimbleby

Capacitors
Does a capacitor have a “resistance”?
Capacitors
v, i
( )
t
v
t
v ω
sin
)
( 0
= ( )
t
Cv
t
i ω
ω cos
)
( 0
=
t
1
t 2
t
1
t 2
t
∞
=
=
0
)
(
)
( 0
1 v
t
i
t
v 0
0
)
(
)
(
0
2
2 =
=
i
t
i
t
v
Thus “resistance” varies between ±∞: not a useful concept
0
)
( 1
t
i )
( 0
2 i
t
i
James Grimbleby
p

Capacitors
The reactance XC of a capacitor is defined:
Capacitors
The reactance XC of a capacitor is defined:
0
i
v
XC =
where v0 is the amplitude of the voltage across the capacitor
0
i
C
0 p g p
and i0 is the amplitude of the current flowing through it
Thus:
fC
C
C
v
XC
2
1
1
0 =
=
=
fC
C
Cv
C
π
ω
ω 2
0
The reactance of a capacitor is inversely proportional to its
value and to frequency
James Grimbleby

Inductors
Inductors
Magnetisable
core
core
Copper
wire
v
i
C i
v
I d t L
di
L
Core: air
ferrite
i
Inductance L
dt
di
L
v =
iron
silicon steel
James Grimbleby

Inductors
Inductors
James Grimbleby

Inductors
v
i di
Inductors
Suppose that:
L dt
di
L
v =
Suppose that:
( )
t
v
t
v ω
sin
)
( 0
=
i, v
)
(t )
(t
i
1
Then:
)
(t
v )
(t
i
( )
∫
= sin
1
)
( 0 ω
v
t
v
L
t
i t
( )
⎞
⎛
−
= cos
0 ω
ω
t
L
v
⎟
⎠
⎞
⎜
⎝
⎛
−
=
2
sin
0 π
ω
ω
t
L
v
Current lags voltage by π/2 (90°)
James Grimbleby

Inductors
The reactance XC of an inductor is defined:
Inductors
C
0
0
i
v
XC =
where v0 is the amplitude of the voltage across the inductor
0
i
and i0 is the amplitude of the current flowing through it
Thus:
fL
L
L
v
v
Xc π
ω
ω
2
/
0 =
=
=
The reactance of an ind ctor is directl proportional to its
L
v ω
/
0
The reactance of an inductor is directly proportional to its
value and to frequency
James Grimbleby

Resistance and Reactance
Resistance and Reactance
0
i
v
X = 0
→
f ∞
→
f
0
i 0
→
f ∞
→
f
Resistance R R R R
Capacitance C
C
ω
1 short
circuit
open
circuit
Inductance L L
ω
open
i it
short
i it
Inductance L L
ω
circuit
circuit
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 2
Lecture 2
AC Analysis using Differential Equations
Complex Numbers
Complex Numbers
Complex Exponential Voltages and Currents
James Grimbleby

AC Circuit Analysis
The ac response of a circuit is determined by a differential
AC Circuit Analysis
p y
equation:
R
)
(
)
(
)
( t
v
t
Ri
t
v c
in +
=
)
(
R )
(t
i
dt
t
dv
C
t
i c )
(
)
( =
)
(t
vin C )
(t
vc
)
(
)
(
)
( t
v
dt
t
dv
RC
t
v c
c
in +
=
dt
t
v
t
v
t
dv i )
(
)
(
)
(
RC
t
v
RC
t
v
dt
t
dv in
c
c )
(
)
(
)
(
=
+
James Grimbleby

AC Circuit Analysis
Now 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 f
The output voltage vc will be a sinusoid of the same freqeuncy,
but with different amplitude and phase:
( )
( )
φ
ω
ω
+
=
=
t
v
t
v
t
v
t
vin
cos
)
(
cos
)
(
1
0
Expanding the expression for vc:
( )
φ
ω +
= t
v
t
vc cos
)
( 1
c
t
B
t
A
t
v
t
v
t
vc ω
ω
φ
ω
φ
ω sin
cos
sin
sin
cos
cos
)
( 1
1 +
=
−
=
t
B
t
A
dt
t
dvc ω
ω
ω
ω cos
sin
)
(
+
−
=
James Grimbleby
dt

AC Circuit Analysis
The differential equation becomes:
AC Circuit Analysis
The differential equation becomes:
t
v
t
B
t
A
t
B
t
A ω
ω
ω
ω
ω
ω
ω cos
sin
cos
cos
sin 0
=
+
+
+
−
Comparing the coefficients of sinωt and cosωt on both sides of
t
RC
t
RC
t
RC
t
B
t
A ω
ω
ω
ω
ω
ω
ω cos
sin
cos
cos
sin +
+
+
p g
the equation:
0
B
RC
A =
+
− ω
0
v
A
RC
B =
+
ω
Solving these simultaneous linear equations in A and B:
2
2
2
0
2
2
2
0
1
1 C
R
RCv
B
C
R
v
A
ω
ω
ω +
=
+
=
James Grimbleby

AC Circuit Analysis
0
0 sin
cos
RCv
v
B
v
v
A
ω
φ
φ
AC Circuit Analysis
Thus:
2
2
2
0
1
2
2
2
0
1
1
sin
1
cos
C
R
v
B
C
R
v
A
ω
φ
ω
φ
+
=
−
=
+
=
=
Thus:
RC
C
R
v
v ω
φ
ω
−
=
+
= tan
1
1
2
2
2
0
1
At an angular frequency ω=1/RC:
C
R
ω
+
1 2
2
2
−
=
=
4
2
0
1
π
φ
v
v
⎟
⎠
⎞
⎜
⎝
⎛
−
=
4
cos
2
)
(
4
2
0 π
ωt
v
t
vc
The output voltage lags the input voltage by π/4 (45°)
⎠
⎝ 4
2
James Grimbleby
p g g p g y ( )

AC Circuit Analysis
AC Circuit Analysis
1.0
0.7071
v
1
/v
0
e
gain
Voltage
0 0
V
Angular frequency ω (rad/s)
0.0
1/RC 10/RC 100/RC
0.1/RC
0.01/RC
James Grimbleby
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
jy
x
z +
=
where x is the real part, y is the imaginary part (x and y are
both real numbers), and
1
1
2
−
=
−
= j
j
Complex numbers are often the solutions of real problems,
for example quadratic equations
James Grimbleby
p q q

Complex Numbers:
Complex Numbers:
Argand Diagram
Imaginary part
jy
x
z +
=
Imaginary
axis
y
axis
Real part
p
O x
Real axis
James Grimbleby

Complex Numbers: Polar Form
Polar form:
θ
∠
r
z
Complex Numbers: Polar Form
where r is the magnitude, and θ is the angle measured from
th l i
θ
∠
= r
z
the real axis:
I i
z
Imaginary
axis r
θ
Real axis
O
θ
James Grimbleby

Complex Numbers:
Complex Numbers:
Exponential Form
Exponential form:
θ
j
re
z =
Euler’s identity:
θ
θ
θ
i
j
j
θ
θ
θ
sin
cos j
e j
+
=
1
sin θ
θ
O
cos θ
The polar and exponential forms are therefore equivalent
cos θ
James Grimbleby
The polar and exponential forms are therefore equivalent

Complex Numbers: Conversion
z
z
r
θ
y
O
2
2
x
O
Rectangular to polar:
y
z
y
x
z
r
∠
+
=
=
θ
θ tan
2
2
θ
P l t R t l
x
y
z =
∠
= θ
θ tan
θ
θ
sin
cos
r
y
r
x
=
=
Polar to Rectangular:
James Grimbleby

Complex Numbers: Inversion
If the complex number is in rectangular form:
Complex Numbers: Inversion
p g
1
jy
x
z
+
=
)
)(
( jy
x
jy
x
jy
x
jy
x
+
−
=
+
)
)(
(
jy
x
jy
x
jy
x
−
=
−
+
If th l b i i l ti l f
2
2
y
x +
If the complex number is in polar or exponential form:
j
1
1 φ
φ
j
j
e
A
Ae
z −
=
=
1
1
James Grimbleby

When using the inverse tangent to obtain θ from x and y it is
g g y
necessary to resolve the ambiguity of π:
y
y jy
x
z +
=
y θ
+
−
y
x
+
+
y
x
x
y
=
θ
tan
θ
+
x
−
x
x
−
+
y
x
−
y
x
jy
x
w −
−
=
James Grimbleby

When using the inverse tangent to obtain θ from x and y it is
g g y
necessary to resolve the ambiguity of π
1. Calculate θ using inverse tangent:
y
1
/ θ /
x
y
1
tan−
=
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) :
x
y
1
tan−
+
= π
θ
James Grimbleby

C t t t l f
2
π
∠
Convert to rectangular form
3
2
π
∠
=
z
Real part: 1
2
1
2
3
cos
2 =
×
=
⎟
⎠
⎞
⎜
⎝
⎛
=
π
x
2
3 ⎠
⎝
Imaginary part: 3
2
3
2
3
sin
2 =
×
=
⎟
⎠
⎞
⎜
⎝
⎛
=
π
y
Thus: 3
1 j
z +
=
James Grimbleby

2
1 732
3
1 j
z +
=
Imaginary
1.732
2
=
r
axis
1
R l i
)
60
(
3
o
π
θ =
Real axis
O 1 2
James Grimbleby

C t t l ti l f
1
Convert to polar or exponential form:
j
z
+
=
1
1
Magnitude:
2
1
1
1
0
1
2
2
2
2
=
+
=
= z
r
2
1
1 2
2
+
Angle:
0
1
t
0
t
)
1
(
1
1
1 π
π
θ
⎟
⎞
⎜
⎛
⎟
⎞
⎜
⎛
+
∠
−
∠
=
∠
=
−
−
j
z
4
4
0
1
1
tan
1
0
tan 1
1 π
π
−
=
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
= −
−
π
Thus: or 4
2
1
4
2
1
π
π j
e
z
z
−
=
−
∠
=
James Grimbleby
2
4
2

Real axis
O π
0.5
O
Imaginary
)
45
(
4
o
π
θ −
=
Imaginary
axis
0.5 1
1 j
−
2
1
1
1 j
j
z =
+
=
James Grimbleby

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:
t
j
Thi i th ti l t i k f bt i i th
t
j
Ve
t
v 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:
t
j
t
j
d t
j
t
j
Ve
j
Ve
dt
d ω
ω
ω
=
James Grimbleby

S th t l ti l lt i li d
Suppose that a complex exponential voltage is applied
across a resistor:
t
v )
(
t
j
V
t ω
)
(
t
j
V
R
t
v
t
i =
)
(
)
(
t
j
Ve
t
v ω
=
)
(
)
(t
i
t
j
e
R
V ω
=
R
The current through the resistor is also a complex
The current through the resistor is also a complex
exponential
James Grimbleby

across a capacitor:
dt
t
dv
C
t
i =
)
(
)
(
t
j
Ve
t
v ω
=
)
(
t
j
Ve
dt
d
C
dt
ω
=
j
Ve
t
v =
)
(
)
(t
i
t
j
CVe
j
dt
ω
ω
=
C
The current through the capacitor is also a complex
The current through the capacitor is also a complex
exponential
James Grimbleby

across an inductor:
dt
t
v
L
t
i )
(
1
)
( ∫
=
t
j
Ve
t
v ω
=
)
(
t
j
dt
Ve
L
ω
1
∫
=
j
Ve
t
v =
)
(
)
(t
i
t
j
Ve
L
j
L
ω
ω
1
=
L
The current through the inductor is also a complex
L
jω
The current through the inductor is also a complex
exponential
James Grimbleby

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
parts of the complex exponentials
Complex exponential: )
sin
cos
(
)
( t
j
t
e
t
v t
j
ω
ω
ω
+
=
=
Complex exponential:
Real voltage:
)
sin
cos
(
)
( t
j
t
e
t
vc ω
ω +
=
=
t
t
v ω
cos
)
( =
Real voltage: t
t
vr ω
cos
)
( =
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 3
Lecture 3
Phasors
Impedances
Impedances
Gain and Phase Shift
Frequency Response
James Grimbleby

Phasors
If the input voltage to a circuit is a complex exponential:
Phasors
then all other voltages and currents are also complex
t
j
cin e
v
t
v ω
0
)
( =
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
ω
ω
φ
φ
ω
ω
ω
φ
φ
ω
)
(
1
1
)
(
1
1
2
2
1
1
)
(
)
( =
=
=
+
+
t
j
t
j
j
t
j
c e
I
e
e
i
e
i
t
i ω
ω
φ
φ
ω
2
2
)
(
2
2
2
2
)
( =
=
= +
where V1 and I2 are time-independent voltage and current
phasors: 1
1
1
φ
j
e
v
V =
2
2
2
1
1
φ
j
e
i
I
e
v
V
=
=
James Grimbleby

Phasors
The complex exponential voltages and currents can now be
Phasors
expressed:
t
j
t
j
c e
V
t
v
ω
ω
1
1 )
( =
Ph i d d t f ti b t i l f ti
t
j
c e
I
t
i ω
2
2 )
( =
Phasors are independent of time, but in general are functions
of jω and should be written:
( ) ( )
ω
ω j
I
j
V 2
1
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
James Grimbleby

Impedance
The impedance Z of a circuit or component is defined to be the
Impedance
ratio of the voltage and current phasors:
I
V
Z =
For a resistor:
t
Ri
t
v c
c = )
(
)
(
t
j
Ve
t
v ω
)
(
RIe
Ve
t
Ri
t
v
t
j
t
j
c
c
= ω
ω
)
(
)
(
j
c Ve
t
v =
)
(
t
j
c Ie
t
i ω
=
)
(
RI
V =
R
So that: R
I
V
ZR =
=
James Grimbleby

Impedance
F it
Impedance
For a capacitor:
t
dv
C
t
i c
=
)
(
)
(
Ve
d
C
Ie
dt
C
t
i
t
j
t
j
c
ω
ω
=
)
(
t
j
c Ve
t
v ω
=
)
(
t
j
Ie
t
i ω
=
)
(
CVe
j
Ie
Ve
dt
C
Ie
t
j
t
j
t
j
t
j
ω ω
ω
ω
ω
=
=
C
c Ie
t
i =
)
(
CV
j
I
CVe
j
Ie j
j
ω
ω
=
=
So that:
V
Z
1
C
j
I
V
ZC
ω
1
=
=
James Grimbleby

Impedance
F i d t
Impedance
For an inductor:
t
di
L
t
v c
=
)
(
)
(
Ie
d
L
Ve
dt
L
t
v
t
j
t
j
c
ω
ω
=
)
(
t
j
c Ve
t
v ω
=
)
(
t
j
Ie
t
i ω
=
)
(
LIe
j
Ve
Ie
dt
L
Ve
t
j
t
j
t
j
t
j
ω ω
ω
ω
ω
=
=
c Ie
t
i =
)
(
L
LI
j
V
LIe
j
Ve j
j
ω
ω
=
=
So that:
V
L
j
I
V
ZL ω
=
=
James Grimbleby

Impedance
Impedance
I
V
Z = ∞
→
f
0
→
f
Resistance R R R R
I
Resistance R R R R
1
Capacitance C
C
jω
1
∞
→
Z 0
→
Z
Inductance L L
jω ∞
→
Z
0
→
Z
James Grimbleby

Impedance
All the normal circuit theory rules apply to circuits containing
Impedance
y pp y g
impedances
For example impedances in series:
4
3
2
1 Z
Z
Z
Z
Z +
+
+
=
and impedances in parallel:
4
3
2
1 Z
Z
Z
Z
Z +
+
+
=
and impedances in parallel:
1
1
1
1
1
+
+
+
=
Other rele ant circ it theor r les are Kirchhoff’s la s
4
3
2
1 Z
Z
Z
Z
Z
+
+
+
Other relevant circuit theory rules are: Kirchhoff’s laws,
Thévenin and Norton's theorems, Superposition
James Grimbleby

Impedance
Impedance
Potential divider:
Vin
2
1
IZ
V
Z
Z
V
I in
+
=
I
1
Z
2
2
Z
V
IZ
V
in
out
=
=
V
V
2
2
1
Z
V
Z
Z
out
+
=
out
V
2
Z
in
V
2
1
2
Z
Z
Z
V
V
in
out
+
=
James Grimbleby

AC Circuit Analysis
Suppose that a circuit has an input x(t) and an output y(t),
AC Circuit Analysis
pp 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(
)
( t
j
t
j
Xe
re
e
x
re
t
x
t
x ω
ω
ω
and the real output voltage y(t) is the real part of the complex
)
(
)
(
)
cos(
)
( 0
0
t
j
t
j
Xe
re
e
x
re
t
x
t
x ω
ω
ω =
=
=
and the real output voltage y(t) is the real part of the complex
exponential output:
)
(
)
(
)
cos(
)
( 0
0
t
j
t
j
j
Ye
re
e
e
y
re
t
y
t
y ω
ω
φ
φ
ω =
=
+
=
James Grimbleby

AC Circuit Analysis
Thus:
AC Circuit Analysis
Thus:
X
Y
x
e
y j
=
0
0
φ
The voltage gain g is the ratio of the output amplitude to the
X
x0
The voltage gain g is the ratio of the output amplitude to the
input amplitude:
Y
y0
X
x
y
g =
=
0
0
and the phase shift is:
⎞
⎛Y
⎟
⎠
⎞
⎜
⎝
⎛
∠
=
X
Y
φ
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
Using the potential divider formula:
Z
V C
c
R
C
j
Z
Z
Z
V
V
R
C
C
in
c
ω
+
=
/
1
R
R
C
j
C
j
ω
ω
+
=
/
1
/
1
C
in
V c
V
CR
jω
+
=
1
1
James Grimbleby

AC Circuit Analysis
R
AC Circuit Analysis
Vc 1
V V
CR
j
V
V
in
c
ω
+
=
1
1
C
in
V c
V
1
V
Voltage gain:
2
2
2
1
1
R
C
V
V
g
in
c
ω
+
=
=
Phase shift: CR
V
V
i
c ω
φ 1
1
tan
0
tan −
−
−
=
⎟
⎠
⎞
⎜
⎝
⎛
∠
=
CR
Vin
ω
1
tan−
−
=
⎠
⎝
James Grimbleby

Frequency Response (RC = 1)
Frequency Response (RC 1)
1.0
ain
0.7071
age
Ga
Volta
Angular Frequency (rad/s)
0.0
1 10 100
0.1
0.01
James Grimbleby

0
1 10 100
0.1
shift
π
Phase
)
45
(
4
°
−
−
π
P
)
90
(
2
°
−
−
π
James Grimbleby

Frequency Response
Frequency Response
CR
ω
φ 1
tan−
−
=
2
2
2
1
1
R
C
g
ω
+
=
0
→
ω 1
→
g )
0
(
0 °
→
φ
→
g
1 1 π
)
(
φ
CR
1
=
ω
2
1
=
g )
45
(
4
°
−
−
=
π
φ
∞
→
ω 0
→
g )
90
(
2
°
−
−
→
π
φ
This is a low-pass response
James Grimbleby
p p

Frequency Response
Frequency Response
0
→
ω ∞
→
ω
0
→
C
V
in
C V
V →
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
Z
V
V R
C
R
Z
Z
V
V
C
R
R
in
R
+
=
R
in
V R
V
CR
j
C
j
R
R
Vin
ω
+
=
/
1
CR
j
CR
j
V
V in
R
ω
ω
+
=
1
CR
j
CR
j
V
V
i
R
ω
ω
+
=
1 CR
j
Vin ω
+
1
James Grimbleby

AC Circuit Analysis
C
AC Circuit Analysis
CR
j
CR
j
V
VR ω
=
1
R
in
V R
V
CR
j
Vin ω
+
1
R
in R
V lt i
CR
VR ω
Voltage gain:
2
2
2
1 R
C
CR
V
V
g
in
R
ω
ω
+
=
=
⎞
⎛
Phase shift : CR
V
V
in
R ω
φ 1
1
tan
tan −
−
−
∞
=
⎟
⎠
⎞
⎜
⎝
⎛
∠
=
CR
ω
π 1
tan
2
−
−
=
⎠
⎝
James Grimbleby
2

1.0
ain
0.7071
age
ga
Volta
0.0
1 10 100
0.1
0.01
James Grimbleby

)
90
(
2
°
π
ift
)
45
(
4
°
π
ase
sh
4
Pha
100
0
1 10
0.1
0.01
James Grimbleby

Frequency Response
Frequency Response
CR
ω
π
φ 1
tan
2
−
−
=
2
2
2
1 R
C
CR
g
ω
ω
+
=
0
→
ω 0
→
g )
90
(
2
°
→
π
φ
→
g
1 1 π
)
(
2
φ
CR
1
=
ω
2
1
=
g )
45
(
4
°
=
π
φ
∞
→
ω )
0
(
0 °
→
φ
1
→
g
This is a high-pass response
James Grimbleby
g p p

Frequency Response
Frequency Response
0
→
ω ∞
→
ω
0
→
ω ∞
→
ω
in
R V
V →
0
→
R
V
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 4
Lecture 4
Driving-Point Impedance
James Grimbleby

Impedance
Impedance
ratio of the voltage and current phasors:
)
( ω
j
I
)
(
)
(
ω
ω
j
V
j
Z =
AC Circuit
)
( ω
j
V
)
( ω
j
I
)
(
)
(
ω
ω
j
I
j
Z =
AC Circuit
)
( ω
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
James Grimbleby

Impedance
Z can be written in rectangular form:
Impedance
Z can be written in rectangular form:
)
(
)
(
)
( ω
ω
ω j
jX
j
R
j
Z +
=
where R is the resistance and X is the reactance
Thus: X
R
Z 2
2
+
=
R
X
Z 1
tan−
=
∠
and:
Z
Z
R ∠
= cos
Z
Z
X ∠
= sin
James Grimbleby

Symbolic and Numeric Forms
Symbolic and Numeric Forms
CR
j
R
Z
ω
+
=
1
Symbolic Form
Substitute component
values
Numeric Form
3
10
8
1
80
−
=
ω
j
Z
Substitute frequency
value
3
10
8
1 ×
×
+ ω
j
Value at a given frequency
value
40
24 j
Z +
=
Value at a given frequency 40
24 j
Z +
James Grimbleby

Example 1
Determine the driving-point impedance of the circuit at a
Example 1
g p p
frequency of 40 kHz:
1
Z
Z
Z C
R +
=
1
C
j
R +
=
ω
C = 200nF
10
200
10
40
2
1
25
9
3
j ×
×
×
×
+
=
−
π
R = 25Ω
05027
0
1
25
j
j
+
=
Ω
89
.
19
25
05027
.
0
j
j
−
=
James Grimbleby

Example 1
Example 1
89
.
19
25 −
= j
Z Ω
89
.
19
25 2
2
+
=
Z C = 200nF
93
.
31
89
.
19
25
=
+
Z
Ω
R = 25Ω
89
.
19
tan 1 −
=
∠ −
Z
R = 25Ω
)
5
.
38
(
6720
.
0
25
tan
°
−
−
=
=
∠Z
)
(
James Grimbleby

Example 1
What will be the voltage across the circuit when a current of
Example 1
g
5 A, 40 kHz flows through it?
)
89
.
19
25
(
5 j
IZ
V
−
×
=
=
f
V
45
.
99
125
)
89
.
19
25
(
5
j
j
−
=
×
In polar form:
)
5
.
38
(
6720
.
0
)
93
.
31
5
( °
−
−
∠
×
=
= IZ
V
)
5
.
38
(
6720
.
0
159.7V
)
(
)
(
°
−
−
∠
=
James Grimbleby

Example 2
Example 2
g p p
frequency of 20 Hz:
Z
Z
Z C
R
+
=
1
1
1
C
j
R
Z
Z
Z C
R
ω
+
=
1
R = 80Ω C = 100μF
Z
j
R
=
1
μ
R
C
j
R
Z
ω
+
/
1
CR
jω
+
=
1
James Grimbleby

Example 2
Example 2
1 CR
j
R
Z
+
=
ω
C = 100μF
80
1
6
CR
j
=
+ ω
R = 80Ω
80
80
10
100
20
2
1 6
j ×
×
×
×
+ −
π
)
005
1
1
(
80
005
.
1
1
j
j
−
+
=
Ω
00
40
79
39
005
.
1
1
)
005
.
1
1
(
80
2
2
j
j
+
−
Ω
00
.
40
79
.
39 j
−
=
James Grimbleby

Example 2
Example 2
00
.
40
79
.
39 −
= j
Z Ω
00
.
40
79
.
39 2
2
+
=
Z
42
.
56
= Ω R = 80Ω C = 100μF
00
.
40
tan 1
⎬
⎫
⎨
⎧−
=
∠ −
Z
)
2
45
(
7880
0
79
.
39
tan
o
−
−
=
⎭
⎬
⎩
⎨
∠Z
)
2
.
45
(
7880
.
0
=
James Grimbleby

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?
Z
V
I =
=
Z
V
I
7880
0
42
56
24
−
∠
=
00
40
79
39
24
−
=
j
A
3016
0
3
0
)
2
.
45
(
7880
.
0
A
0.4254
7880
.
0
42
.
56
j
+
°
∠
=
∠
)
00
.
40
79
.
39
(
24
00
.
40
79
.
39
2
2
+
=
j
j
A
3016
.
0
3
.
0 j
+
=
A
3016
.
0
3
.
0
00
.
40
79
.
39 2
2
+
=
+
j
)
2
.
45
(
7880
.
0
0.4254A °
∠
=
James Grimbleby

Phasor Diagrams
Where voltages or currents are summed the result can be
Phasor Diagrams
g
represented by a phasor diagram: 3
2
1 V
V
V
V +
+
=
Imaginary part
2
V
Real
part
O
1
V part
1
V
V
3
V
James Grimbleby

Example 2
A
3
.
0
24
=
=
R
I A
3016
.
0
24
10
100
20
2 6
j
j
IC =
×
×
×
×
= −
π
Example 2
80
R j
j
C
0.3A
part
nary
p
0.2A
C
I
C
R I
I
I +
=
Imagin
0 1A
R
I R l
I
0.1A
0.2A
R
I Real
part
O
0.1A 0.3A
James Grimbleby

Example 3
Example 3
frequency of 50 Hz:
Z
Z
Z
Z C
L
R +
+
= R
1
C
j
L
j
R +
+
=
ω
ω
24 Ω
L
10
120
50
2
1
10
36
50
2
24
6
3
j
j
j
+
×
×
×
+
=
−
−
π
L
36 mH
Ω
53
.
26
31
.
11
24
10
120
50
2 6
j
j
j
−
+
=
×
×
×
π
C
120 μF
Ω
22
.
15
24 j
−
=
120 μF
James Grimbleby

Example 3
22
15
24 −
= Ω
j
Z
Example 3
22
.
15
24
= Ω
j
Z
42
28
22
.
15
24 2
2
=
−
=
Ω
Z R
24 Ω
42
.
28
⎫
⎧
= Ω 24 Ω
L
24
22
.
15
tan 1
⎭
⎬
⎫
⎩
⎨
⎧−
=
∠ −
Z 36 mH
)
4
.
32
(
5652
.
0 °
−
−
= C
120 μF
)
4
.
32
(
5652
.
0
42
.
28 °
−
−
∠
= Ω
Z
James Grimbleby

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?
)
22
15
24
(
A
10 j
ZI
V
×
=
=
Ω
V
2
.
152
240
)
22
.
15
24
(
A
10
j
j
−
=
−
×
= Ω
In polar form:
ZI
V
)
4
.
32
(
5652
.
0
)
42
.
28
10
( °
−
−
∠
×
=
= ZI
V
)
4
.
32
(
5652
.
0
V
2
.
284 °
−
−
∠
=
James Grimbleby

Example 3
C
L
R V
V
V
V +
+
=
Example 3
200V
I i
C
L
R V
V
V
V +
+
=
L
V
Imaginary
part
R
V
L
V
Real
t
200V 400V
V
part
C
V
V
-200V
James Grimbleby

Example 4
f f 400 H
Example 4
frequency of 400 Hz:
1
1
1
+
=
R
C
Z
Z
Z
Z C
L
R
1
+
+
=
R
2 Ω
C
C
j
L
j
R
1
1
ω
ω
+
+
=
L
1 mH
200 μF
C
j
L
j
R
Z
)
/(
1
1
ω
ω +
+
=
1 mH
L
j
R
C
j
L
j
R
)
(
1 ω
ω
ω
+
+
+
=
LC
CR
j
L
j
R
2
1 ω
ω
ω
+
+
=
James Grimbleby
LC
CR
j
1 ω
ω −
+

Example 4
L
j
R
Z
+
=
ω
Example 4
10
400
2
2
1
3
2
j
LC
CR
j
Z
×
×
+
−
+
=
−
π
ω
ω
10
200
10
)
400
2
(
2
10
200
400
2
1
10
400
2
2
6
3
2
6
3
j
j
×
×
×
×
−
×
×
×
×
+
×
×
+
−
−
−
π
π
π
263
.
1
005
.
1
1
513
.
2
2
j
j
−
+
+
=
005
1
2633
0
513
.
2
2
j
j
j
+
−
+
=
)
005
.
1
2633
.
0
(
)
513
.
2
2
(
005
.
1
2633
.
0
2
2
j
j
j
−
−
×
+
=
+
−
474
.
2
852
.
1
005
.
1
2633
.
0 2
2
j
j
−
=
+
James Grimbleby
j

Example 4
474
2
852
1 j
Z
Example 4
474
.
2
852
.
1 −
= j
Z
R
091
3
474
.
2
852
.
1 2
2
+
=
Z
2 Ω
C
200 F
091
.
3
=
L
1 mH
200 μF
852
.
1
474
.
2
tan 1
=
∠ −
Z
)
2
.
53
(
9282
.
0
852
.
1
°
−
−
=
)
2
.
53
(
9282
.
0
091
.
3 °
−
−
∠
= Ω
Z
James Grimbleby

Example 4
What current will flow if an ac voltage of 120 V, 400 Hz is
Example 4
applied to the circuit?
V
I
V
I
120
=
Z
I
120
Z
I =
)
474
2
852
1
(
120
474
.
2
852
.
1
120
+
×
−
=
j
j
9282
0
A
82
38
9282
.
0
091
.
3
120
∠
−
∠
=
474
.
2
852
.
1
)
474
.
2
852
.
1
(
120
2
2
+
+
×
=
j
08
.
31
26
.
23
9282
.
0
A
82
.
38
j
+
=
∠
=
556
.
9
0
.
297
4
.
222 +
=
j
)
(53.2
0.9282
A
82
.
38
08
.
31
26
.
23
°
∠
=
+
= j
James Grimbleby
)
(

Example 4
Imaginary part
Example 4
I
I
I C
RL +
=
Imaginary part
120
IRL =
50A
C
I
23
.
29
27
.
23 j
Z
I
RL
RL
−
= I
120
j
Real
120
Z
I
C
C =
ea
part
O 50A
I
32
.
60
j
= RL
I
James Grimbleby

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:
)
( ω
j
I
1
)
(
)
(
ω
ω
j
I
j
Y =
=
AC Circuit
)
( ω
j
V
)
( ω
j
I
)
(
)
(
)
(
ω
ω
ω
j
Z
j
V
j
Y =
=
AC Circuit
)
( ω
j
V
Admittance Y is analogous to conductance in dc circuits and
its unit is Siemens
)
(
)
(
)
( ω
ω
ω j
jB
j
G
j
Y +
=
where G is the conductance and B is the susceptance
)
(
)
(
)
( ω
ω
ω j
jB
j
G
j
Y +
James Grimbleby

Admittance
Admittance
I
Y = ∞
→
f
0
→
f
V
Y ∞
→
f
0
→
f
1
1
1
Resistance R
R
1
R
1
R
1
Capacitance C C
jω ∞
→
Y
0
→
Y
Inductance L L
j
1
∞
→
Y 0
→
Y
Inductance L L
jω →
Y 0
→
Y
James Grimbleby

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:
4
3
2
1
1
1
1
1
1
Y
Y
Y
Y
Y
+
+
+
=
and admittances in parallel:
Oth l t i it th l Ki hh ff’ l
4
3
2
1 Y
Y
Y
Y
Y +
+
+
=
Other relevant circuit theory rules are: Kirchhoff’s laws,
Thévenin and Norton's theorems, Superposition
James Grimbleby

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
Y
Y
Y
Y C +
=
/
1
/
1
1 R
2 Ω
C
j
Y
Y L
R
C
ω +
=
+
1
/
1
/
1 2 Ω
L
C
200 μF
L
j
R
C
j
ω
ω
+
+
= L
1 mH
μ
James Grimbleby

Example 5
Determine the driving-point admittance of the circuit at a
f f 400 H
Example 5
2
10
400
2
10
200
400
2
3
6 j
j
Y
×
×
×
+
×
×
×
=
−
− π
π
1
5027
0
10
400
2
2
10
200
400
2
3
j
j
j
Y
×
×
+
+
×
×
×
=
−
π
π
513
2
2
513
.
2
2
1
5027
.
0
j
j
j
+
+
=
R
513
.
2
2
513
.
2
2
5027
.
0
2
2
j
j
+
−
+
=
R
2 Ω
C
32
.
10
513
.
2
2
5027
.
0
j
j
−
+
=
L
C
200 μF
S
2590
.
0
1939
.
0
2436
.
0
1939
.
0
5027
.
0
j
j
j
+
=
−
+
= 1 mH
James Grimbleby
S
2590
.
0
1939
.
0 j
+

AC Circuit Analysis
AC Circuit Analysis
L t 5
Lecture 5
Resonant Circuits
James Grimbleby

Resonant Circuits
Resonant 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 b
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
parallel
James Grimbleby

Resonant Circuits
Resonant Circuits
L
g
dt
d θ
ω
−
=
ω
θ
=
dt
d
θ
L
dt
dt
θ,ω ω
θ
θ
t
James Grimbleby

Resonant Circuits
iL
Resonant Circuits
L
C vC C
i
dt
dv L
C −
=
L
v
dt
di C
L =
vC C
dt
L
dt
i, v
C
v
i
C
L
i
t
James Grimbleby

Parallel Resonant Circuit
R L
R L
C
+
+
1
1
1
1 LR
j
Z
ω
=
Z
Z
Z
Z L
C
R
+
+
=
1
1 L
j
R
LCR
L
j
Z
2
ω
ω
ω +
−
=
L
j
C
j
R ω
ω +
+
=
2
1
1
LC
R
L
j
L
j
2
1
/ ω
ω
ω
+
−
=
LR
j
R
LCR
L
j
ω
ω
ω +
−
=
2
LC
R
L
j
L
j
2
/
1 ω
ω
ω
−
+
=
James Grimbleby
j j

Ω
= k
5
R μF
1
=
C H
1
=
L
Impedance is a
maximum
( t
Ω
= k
5
R μF
1
=
C H
1
=
L (resonant
frequency)
h
1
when:
2
/
1+
=
ω
ω
ω
LC
R
L
j
L
j
Z
1
=
LC
ω
6
2
4
/
1
=
−
+
ω
ω
ω
j
LC
R
L
j
6
10
1
=
−
6
2
4
10
10
2
1 −
−
×
−
×
×
+ ω
ω
j 3
10
=
James Grimbleby

0
→
ω ∞
→
ω
0
→
ω ∞
→
ω
0
→
Z
0
→
Z
James Grimbleby

R L
L
j
Z
2
ω
=
R L
C
LC
R
L
j 2
/
1 ω
ω −
+
0
0 j
L
j
Z
ω
ω
1
0
1
0
R
L
j
Z
j
j
Z =
→
→
ω
ω
R t f
/
1
j
L
j
R
R
L
j
L
j
Z
LC
=
=
=
ω
ω
ω
ω
Resonant frequency:
0
2
j
C
j
LC
L
j
Z −
=
−
=
−
→
∞
→
ω
ω
ω
ω
James Grimbleby

5kΩ )
90
( °
π
5kΩ
Z
∠
)
90
(
2
°
Z
Z
∠
0
Z
0 0
Z
)
90
( °
−
−
π
Angular frequency (rad/s)
0.0
1000 10000
100
)
90
(
2
James Grimbleby

Quality Factor
The standard form for the denominator of a second-order
Quality Factor
system is:
2
0
2
0 /
/
1 ω
ω
ω
ω −
+ Q
j
Compare this with the impedance Z:
0
0 /
/
1 ω
ω
ω
ω
+ Q
j
Compare this with the impedance Z:
L
j
Z
2
ω
=
So that:
LC
R
L
j
Z
2
/
1 ω
ω −
+
L
R
Q
LC 0
0
1
ω
ω =
=
where Q is the quality-factor and ω0 is the resonant frequency
L
LC 0
ω
James Grimbleby

Quality Factor
Quality Factor
Ω
= k
5
R μF
1
=
C H
1
=
L
Ω
= k
5
R μF
1
=
C H
1
=
L
10
10
1
1 3
6
0 =
=
=
−
LC
ω
5000
10
R
5
10
1
5000
3
0
=
×
=
=
L
R
Q
ω
James Grimbleby

Quality Factor
Z
Quality Factor
max
Z
2
max
Z
Q
0
ω
ω =
Δ
2 Q
Z
0 0
0.0
0
ω
James Grimbleby

Quality Factor
2kΩ )
90
( °
π
Quality Factor
2kΩ
Z
∠
)
90
(
2
°
Q=2
Z
Z
∠
Z
∠ Q
0
Z
0 0
)
90
( °
−
−
π
0.0
1000 10000
100
)
(
2
James Grimbleby

Quality Factor
10kΩ )
90
( °
π
Quality Factor
10kΩ
Z
∠
)
90
(
2
°
Q=10
Z
Z
∠
Q
0
Z
0 0
Z
)
90
( °
−
−
π
0.0
1000 10000
100
)
(
2
James Grimbleby

L
I
C
I
R
I
V
1 Ω
k
5
=
R μF
1
=
C H
1
=
L
V
1
1
1 4
R
IR ×
=
=
= −
1
10
2
5000
1
1 4
Resonance occurs in
parallel resonant circuits
ω
ω
ω
j
C
j
C
j
IC ×
=
=
= −
10
/
1
1 6
parallel resonant circuits
because the currents in
the capacitor and
ω
ω
ω
j
L
j
L
j
IL
−
=
−
=
=
1
p
inductor cancel out
James Grimbleby

At resonance:
:
103
=
ω
1mA
A
10
2 4
−
×
=
IR
Imaginary
C
I
10 6
−
×
= j
IC ω Real
Imaginary
part
A
10 3
−
= j I part
O
1mA
R
I
−
=
j
IL
L
I
A
10 3
−
−
= j
L
ω -1mA
James Grimbleby
j

Below resonance:
R l
C
I
R
I
:
10
5
.
0
4
3
×
=
ω
I
Real
part
Imaginary
O 1mA
R
I
10
A
10
2
6
4
−
−
×
=
×
=
j
I
I
C
R
ω
I
Imaginary
part
A
10
5
.
0
10
3
−
×
=
×
=
j
j
IC ω
-1mA
−
=
j
IL
ω
L
I
A
10
2 3
−
×
−
= j
ω
-2mA
James Grimbleby

Above resonance:
2mA
:
10
0
.
2
4
3
×
=
ω C
I
10
A
10
2
6
4
−
−
×
=
×
=
j
I
I
C
R
ω I i
1mA
A
10
0
.
2
10
3
−
×
=
×
=
j
j
IC ω
I
Imaginary
part
−
=
j
IL
ω
Real
part
O 1mA
R
I
A
10
5
.
0 3
−
×
−
= j
ω
L
I
1mA
R
James Grimbleby

Series Resonant Circuit
Z
Z
Z
Z L
C
R +
+
=
L
j
C
j
R
L
C
R
ω
ω
1
+
+
=
R
LC
CR
j
C
j
ω
ω
ω
2
1−
+
= L
LC
CR
j
C
j
ω
ω
ω
2
1+
=
C
j
LC
CR
j
ω
ω
ω
1 −
+
= C
James Grimbleby

C
j
LC
CR
j
Z
ω
ω
ω 2
1 −
+
=
C
jω
R
∞
−
=
−
=
→
→ j
C
j
C
j
Z
ω
ω
ω
1
0
L
=
=
= R
C
j
CR
j
Z
LC
j
ω
ω
ω
1
∞
=
=
−
→
∞
→ j
L
j
LC
Z
C
j
LC
ω
ω
ω
ω
2
C
∞
=
=
→
∞
→ j
L
j
C
j
Z ω
ω
ω
James Grimbleby

2
1 LC
CR
j 2
1 −
+
=
ω
ω
ω
C
j
LC
CR
j
Z R = 200 Ω
6
6
2
6
10
10
10
200
1
−
−
−
×
−
×
×
+
=
ω
ω
ω
j
j
L = 1H
6
2
4
6
10
10
2
1
10
−
−
×
−
×
×
+
×
ω
ω
ω
j
j L 1H
6
10−
×
=
ω
j C = 1μF
James Grimbleby

1kΩ )
90
( °
π
1kΩ )
90
(
2
°
Z
Z
Z
∠
Z
0
200Ω
Z
∠
)
90
( °
−
−
π
0Ω
200Ω
1000 10000
100
)
(
2
0Ω
James Grimbleby

The standard form for the denominator of a second order
system is:
2
0
2
0 /
/
1 ω
ω
ω
ω −
+ Q
j
Compare this with the admittance Y (= 1/Z):
0
0 /
/
1 ω
ω
ω
ω
+ Q
j
p ( )
C
j
Y
2
ω
=
So that:
LC
CR
j
Y
2
1 ω
ω −
+
CR
Q
LC 0
0
1
1
ω
ω =
=
where Q is the quality-factor and ω0 is the resonant frequency
CR
LC 0
ω
James Grimbleby
where Q is the quality factor and ω0 is the resonant frequency

1
1
1
0 =
LC
ω
1
10
1
1
6
×
×
=
−
R = 200 Ω
rad/s
103
=
L = 1H
1
0
=
CR
Q
ω
C 1 F
200
10
1
10
1
6
3
×
×
×
=
−
C = 1μF
5
200
10
1
10
=
×
×
×
James Grimbleby

Resonance occurs in
series resonant circuits
R = 200 Ω
R
V
1A
because the voltages
across the capacitor and
L = 1H
L
V
1A
p
inductor cancel out
L 1H
L
V
j
j
R
VR =
×
=
10
1
200
1
6
C = 1μF
C
V
ω
ω
ω
j
C
j
C
j
VC
−
=
−
=
=
10
1
ω
ω j
L
j
VL =
×
= 1
James Grimbleby

At resonance:
:
103
=
ω
1kV
V
200
VR =
L
V
106
j
VC
−
= Real
Imaginary
part
V
103
j
C
−
=
ω part
O R
V 1kV
j
VL = ω
C
V
V
103
j
j
VL
=
= ω
-1kV
James Grimbleby

Crystal Resonator
Crystal Resonator
James Grimbleby

Crystal Resonator
Equivalent circuit:
Crystal Resonator
f0 = 8.0 MHz
R = 3 4 Ω
Equivalent circuit:
R
R = 3.4 Ω
L1 = 0.086 mH
C = 4 6 pF
C1 = 4.6 pF
C0 = 42 pF
L C0
1
1
0 =
= Q
ω
C1
1270
10
03
5 7
0
0
=
×
=
CR
Q
LC ω
ω
1270
10
03
.
5 =
×
=
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 6
Lecture 6
Frequency-Response Function
Frequency Response Function
First-Order Circuits
James Grimbleby

Input Output
Y
X
)
(
)
(
ω
ω
j
Y
j
H
Frequency-response function:
)
(
)
(
ω
ω
j
X
j
H =
Voltage gain g: )
(
)
(
)
(
ω
ω
ω
j
H
j
X
j
Y
g =
=
)
( ω
j
X
)
( j
Y ⎞
⎛
Phase shift φ: )
(
)
(
)
(
ω
ω
ω
φ j
H
j
X
j
Y
∠
=
⎟
⎠
⎞
⎜
⎝
⎛
∠
=
James Grimbleby

The order of a frequency-response function is the highest
power of jω in the denominator:
Fi t d
1
First order:
0
/
1
1
)
(
ω
ω
ω
j
j
H
+
=
Second order:
2
0
0 )
/
(
/
2
1
1
)
(
ω
ω
ω
ω
ω
j
j
j
H
+
+
=
Third order:
0
0 )
/
(
/
2
1 ω
ω
ω
ω j
j +
+
1
)
( ω
j
H =
Third order: 3
0
2
0
0 )
/
(
)
/
(
/
1
)
(
ω
ω
ω
ω
ω
ω
ω
j
j
j
j
H
+
+
+
=
The order is normally equal to (and cannot exceed) the
number of reactive components
James Grimbleby

Example 1
Example 1
R
Z
Z
Z
V
V
R
C
C
in
c
+
=
R
R
C
j
C
j
Z
Z
V R
C
in
/
1
/
1
+
=
+
ω
ω C
in
V c
V
j
H
R
C
j
1
)
(
/
1
=
+
ω
ω
CR
j
j
H
1
:
where
1
1
)
(
+
ω
ω
ω
RC
j
:
where
/
1
0
0
=
+
= ω
ω
ω
James Grimbleby

Example 1
R
Example 1
0
/
1
1
)
(
ω
ω
ω
+
=
j
j
H
R
2
2
0
0
/
1
/
1
ω
ω
ω
ω
+
−
=
j
j
C
in
V c
V
2
0
2
/
1 ω
ω
+
Gain: 2
0
2
/
1
1
)
(
ω
ω
ω
+
=
= j
H
g
0
/
1 ω
ω
+
Ph hift /
tan
)
( ω
ω
φ
ω
φ ∠ j
H
Phase shift: 0
/
tan
)
( ω
ω
φ
ω
φ −
=
∠
= j
H
James Grimbleby

Example 1
Example 1
0
→
ω ∞
→
ω
0
→
g
1
→
g
James Grimbleby

Decibel
The decibel is a measure of the ratio of two powers P1, P2 :
Decibel
p 1 2
1
10
log
10
dB
P
=
2
10
log
10
dB
P
=
It can also be used to measure the ratio of two voltages V1, V2:
2
2
1
10
2
2
1
10 log
10
/
/
log
10
dB
V
V
R
V
R
V
=
=
1
2
2
2
2
l
20
dB
/
V
V
R
V
2
1
10
log
20
dB
V
=
James Grimbleby

Decibel
Decibels
Power ratio
Decibel
Decibels
Power ratio
60 dB
20 dB
1000000
100
10 dB
20 dB
100
10
6 dB
3 dB
4
2
0 dB
-3 dB
1
1/2 -3 dB
-6 dB
20 dB
1/4
1/2
0 01 -20 dB
-60 dB
0.01
0.000001
James Grimbleby

Decibel
Decibels
Voltage ratio
Decibel
Decibels
Voltage ratio
60 dB
20 dB
1000
10
10 dB
20 dB
10
√10 = 3.162
6 dB
3 dB
2
√2 = 1.414
0 dB
-3 dB
1
1/√2 = 0 7071 -3 dB
-6 dB
20 dB
1/2 = 0.5
1/√2 = 0.7071
0 1 -20 dB
-60 dB
0.1
0.001
James Grimbleby

Example 1
Circuit is a first-order low-pass filter:
Example 1
1
/
1
=
g
p
0
1
/
tan ω
ω
φ −
= −
2
0
2
/
1 ω
ω
+
=
g
0
ω
ω << )
dB
0
(
1
≈
g )
0
(
0 °
≈
φ
0
ω
ω = )
dB
3
(
2
1
−
=
g )
45
(
4
°
−
−
=
π
φ
2 4
0
ω
ω >> )
oct
/
dB
6
(
0
ω
g )
90
( °
−
−
≈
π
φ
0
ω
ω >> )
oct
/
dB
6
(
0 −
≈
ω
g )
90
(
2
≈
φ
James Grimbleby

Example 1
R = 1kΩ
Example 1
0
1
=
RC
ω
R 1kΩ
6
3
10
10
1
×
=
−
C=1μF
3
10
10
10
=
Gain: 6
2
10
/
1
1
)
(
ω
ω
+
=
= j
H
g
10
/
1 ω
+
Phase shift: 3
10
/
tan
)
( ω
φ
ω
φ ∠ j
H
Phase shift: 3
10
/
tan
)
( ω
φ
ω
φ −
=
∠
= j
H
James Grimbleby

Bode Plot
Gain(dB) Phase(rad)
Bode Plot
0 dB 0
-3 dB
-10 dB
g
-20 dB 4
π
−
g
φ
30 dB
4
6 dB / octave
-30 dB
π
-6 dB / octave
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB 2
π
−
James Grimbleby
Frequency (rad/s)

Example 2
Example 2
L
Z
Z
Z
V
V
L
R
R
in
R
+
=
V V
L
j
R
R
L
R
in
+
=
ω
R
in
V R
V
R
L
j
j
H
L
j
R
=
+
/
1
1
)
( ω
ω
R
R
L
j
=
=
+
0
:
where
1
/
1
ω
ω
L
j
=
+
= 0
0
:
where
/
1
ω
ω
ω
James Grimbleby

Example 2
Example 2
0
→
ω ∞
→
ω
0
→
ω ∞
→
ω
0
→
g
1
→
g
James Grimbleby

Example 3
Example 3
C
Z
Z
Z
V
V
C
R
R
in
c
+
=
R
C
j
R
Z
Z
V C
R
in
/
1 +
=
+
ω
R
in
V c
V
CR
j
j
H
R
C
j
)
(
/
1
=
+
ω
ω
ω
j
CR
j
j
H
1
:
where
/
1
)
(
0
+
ω
ω
ω
ω
ω
RC
j
:
where
/
1
0
0
0 =
+
= ω
ω
ω
James Grimbleby

Example 3
/ω
ω
j C
Example 3
0
0
/
1
/
)
(
ω
ω
ω
ω
ω
+
=
j
j
j
H
C
0 /
1
1
ω
ω
−
=
j
R
in
V R
V
2
2
0
0
/
1
/
1
ω
ω
ω
ω
+
+
=
j
j
G i
1
)
( ω =
= j
H
g
2
2
0 /
1 ω
ω
+
Gain: 2
2
0 /
1
)
(
ω
ω
ω
+
=
= j
H
g
Phase shift: ω
ω
φ
ω
φ /
tan
)
( 0
=
∠
= j
H
James Grimbleby

Example 3
Example 3
0
→
ω ∞
→
ω
0
→
ω ∞
→
ω
1
→
g
0
→
g
James Grimbleby

Example 4
Example 4
R
Z
Z
Z
V
V
R
L
L
in
L
+
=
R
R
L
j
L
j
Z
Z
V R
L
in
+
=
+
ω
ω
L
in
V L
V
R
L
j
j
H
R
L
j
=
+
/
)
(
ω
ω
ω
R
j
R
L
j
j
H
+
0 :
where
/
/
1
)
(
ω
ω
ω
ω
ω
L
j
=
+
= 0
0
0 :
where
/
1
ω
ω
ω
James Grimbleby

Example 4
Example 4
0
→
ω ∞
→
ω
0
→
ω ∞
→
ω
1
→
g
0
→
g
James Grimbleby
g
g

Example 4
Circuit is a first-order high-pass filter:
Example 4
1
1
g
Circuit is a first order high pass filter:
ω
ω
φ /
tan 0
1
−
=
2
2
0 /
1 ω
ω
+
=
g
0
ω
ω << )
oct
/
dB
6
(
0
ω
ω
≈
g )
90
(
2
°
≈
π
φ
0
ω
ω = )
dB
3
(
2
1
−
=
g )
45
(
4
°
=
π
φ
0
)
dB
0
(
1
g
0 )
(
2
g )
(
4
φ
)
0
(
0 °
φ
)
dB
0
(
1
≈
g
0
ω
ω >> )
0
(
0 °
≈
φ
James Grimbleby

Bode Plot
Gain(dB) Phase(rad)
Bode Plot
0 dB
2
π
-3 dB
-10 dB
g
-20 dB 4
π
g
30 dB
4
φ
-30 dB
6 dB / octave
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB 0
James Grimbleby
Frequency (rad/s)

Example 5
Example 5
g p
/
1 C
j
R
V
1
R
1
2
2
/
1
/
1
R
C
j
R
C
j
R
V
V
in
c
+
+
+
=
ω
ω
i
V t
V
2
R
1
2
2
1
1
CR
j
CR
j
CR
j
+
+
+
=
ω
ω
ω
in
V out
V
C
2
1
2
1
2
)
(
1
1
)
(
R
R
C
j
CR
j
j
H
j
j
+
+
+
=
ω
ω
ω
2
1
2
2
1
1
1
:
where
/
1
)
(
1
j
R
R
C
j
=
=
+
=
+
+
ω
ω
ω
ω
ω
2
2
2
1
1
1 )
(
/
1 CR
R
R
C
j +
+ ω
ω
James Grimbleby

Example 5
Example 5
1
R
1
2
/
1
/
1
)
(
ω
ω
ω
ω
ω
j
j
j
H
+
+
=
R
1
j
in
V out
V
2
R
2
2
2
/
1
)
(
ω
ω
ω
+
=
= j
H
g
C
2
1
2
/
1
)
(
ω
ω
ω
+
j
H
g
James Grimbleby

Example 5
Example 5
0
0
→
ω ∞
→
ω
1
R
2
R2
R
2
1
2
R
R
R
g
+
→
1
→
g
James Grimbleby

Example 5
Assuming that :
2
1 ω
ω <<
Example 5
2
2
2
/
1 ω
ω
+
=
g
Assuming that :
2
1 ω
ω <<
2
1
2
/
1 ω
ω
+
=
g
1
ω
ω << )
dB
0
(
1
≈
g
2
1 ω
ω
ω <<
<< )
oct
/
dB
6
(
1 −
≈
ω
ω
g
2
ω
ω >>
ω
2
1 R
g =
≈
ω
2
ω
ω >>
1
2
2 R
R
g
+
ω
James Grimbleby

Example 5
/
1 2
+ j ω
ω
Example 5
1
/
1
/
1
)
(
1
2
+
+
=
j
j
j
H
ω
ω
ω
ω
ω
Ω
900
1 =
R
)
(
1
2
1
1
+
=
R
R
C
ω
1
)
100
900
(
10
1
6
+
=
−
Ω
100
2 =
R
rad/s
10
)
100
900
(
10
3
=
+
μF
1
=
C
100
10
1
1
6
2
2
×
=
=
−
CR
ω
rad/s
104
2
=
James Grimbleby

Bode Plot
Gain(dB) Phase(rad)
Bode Plot
0 dB 0
-10 dB g
-20 dB
φ
30 dB
-30 dB
π
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB 2
π
−
James Grimbleby
Frequency (rad/s)

AC Circuit Analysis
AC Circuit Analysis
L t 7
Lecture 7
Second-Order Circuits
James Grimbleby

Example 1
R R
Example 1
V V
R R
in
V c
V
C C
This circuit must be simplified before the frequency response
f i b d i d
function can be determined
A Thé i i l t i it i t d f th t t
A Thévenin equivalent circuit is created of the components to
the left of the red line
James Grimbleby

Example 1
Thévenin equivalent circuit:
Example 1
R Z
C
in
V V
C
in
V V
C
jω
/
1 CR
jω
+
1
1
1
V
R
C
j
C
j
V
V in
ω
ω
+
=
/
1
/
1
R
R
CR
j
C
j
R
Z
ω
ω
+
=
+
=
1
1
1
CR
j
Vin
ω
+
=
1 CR
j
R
Z
ω
+
=
1
James Grimbleby

Example 1
R
CR
j
R
ω
+
1
Example 1
V
R
CR
jω
+
1
CR
j
Vin
ω
+
1 C C
V
R
R
C
j
C
j
CR
j
V
V in
C
ω
ω
ω +
+
×
+
=
/
1
/
1
1
V
CR
j
R
C
j
j
in
ω
ω
+
+
+
1
1
/
1
CR
j
CR
j
CR
j
CR
j
Vin
ω
ω
ω
ω
+
+
+
×
+
=
1
1
1
1
James Grimbleby
CR
jω
+
1

Example 1
Example 1
1
1
1 CR
j
CR
j
CR
j
V
V in
C ω
ω
×
+
=
1
1
1
V
CR
j
C
j
CR
j
CR
j
i
ω
ω
ω
ω
+
+
+
+
1
)
1
(
)
1
( CR
j
CR
j
CR
j
Vin
ω
ω
ω +
+
×
+
=
2
2
2
3
1
1
)
(
R
C
CR
j
j
H
ω
ω
ω
−
+
=
R = 1 kΩ, C = 1μF:
μ
6
2
3
10
10
3
1
1
)
(
−
−
×
−
×
×
+
=
ω
ω
ω
j
j
H
James Grimbleby
10
10
3
1 ×
×
×
+ ω
ω
j

Example 1
Example 1
0
→
ω ∞
→
ω
0
→
g
1
→
g
James Grimbleby

Bode Plot
Gain(dB) Phase(rad)
Bode Plot
0 dB 0
-10 dB g
-20 dB φ
-30 dB
-30 dB
π
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB π
−
James Grimbleby
Frequency (rad/s)

Example 2
R C
Example 2
R
V V
C
in
V R
V
R
C
This circuit must be simplified before the frequency response
s c cu us be s p ed be o e e eque cy espo se
function can be determined
A Thévenin equivalent circuit is created of the components to
the left of the red line
James Grimbleby

Example 2
CR
j
R
ω
+
1 C
Example 2
CR
jω
+
1
Vin
C
CR
j
Vin
ω
+
1 R
V
R
R
C
j
R
R
CR
j
V
V in
R
ω
ω +
+
×
+
=
/
1
1
CR
j
V
CR
j
C
j
R
j
in ω
ω
ω
+
+
+
1
/
1
CR
j
CR
j
CR
j
CR
j
CR
j
Vin
ω
ω
ω
ω
ω
+
+
+
×
+
=
1
1
1
James Grimbleby
CR
jω
+
1

Example 2
Example 2
1
1 CR
j
CR
j
CR
j
CR
j
V
V in
R ω
ω
ω
ω +
+
×
+
=
1
1
CR
j
V
CR
j
CR
j
j
in ω
ω
ω
+
+
+
)
1
(
)
1
(
CR
j
CR
j
CR
j
CR
j
CR
j
Vin
ω
ω
ω
ω
ω
+
+
×
+
=
2
2
2
3
1
)
(
R
C
CR
j
CR
j
j
H
ω
ω
ω
ω
−
+
=
R = 1 kΩ, C = 1μF:
3
6
2
3
3
10
10
3
1
10
)
(
−
−
−
×
−
×
×
+
×
=
ω
ω
ω
ω
j
j
j
H
James Grimbleby
10
10
3
1+ ω
ω
j

Example 2
Example 2
0
→
ω ∞
→
ω
0
→
g
0
→
g
James Grimbleby

Bode Plot
Gain(dB) Phase(rad)
Bode Plot
0 dB
2
π
-10 dB g
-20 dB φ
30 dB
-30 dB
π
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB
2
π
−
James Grimbleby
Frequency (rad/s)

Example 3
Example 3
C
j
VC /
1 ω
R
L
R
L
j
C
j
C
j
V
V
in
C
1
/
1
/
1
+
+
=
ω
ω
ω
C
in
V c
V
LC
CR
j
j
H
1
1
)
(
2
−
+
=
ω
ω
ω
Q
j /
)
/(
1
1
2
0
2
0 −
+
=
ω
ω
ω
ω
L
Q
Q
j
1
1
:
and
1
:
where
/
)
/(
1
0
0
0
=
=
=
+
ω
ω
ω
ω
ω
C
R
CR
Q
LC 0
0
ω
James Grimbleby

Example 3
Example 3
0
→
ω ∞
→
ω
0
→
g
1
→
g
James Grimbleby

Example 3
Circuit is a second-order low-pass filter:
Example 3
p
1
)
( j
H )
( j
H
2
0
2
0 /
)
/(
1
)
(
ω
ω
ω
ω
ω
−
+
=
Q
j
j
H )
( ω
j
H
g =
0
ω
ω << 1
)
( ≈
ω
j
H )
dB
0
(
1
=
g
0
ω
ω = Q
g =
jQ
j
H −
=
)
( ω Q
g
)
t
/
dB
12
(
2
0
ω
jQ
j )
(
2
0
ω
−
0
ω
ω >> )
oct
/
dB
12
(
2
0 −
≈
ω
g
2
0
)
(
ω
ω
ω ≈
j
H
James Grimbleby

Example 3
Ω
200
R
mH
400
L
Example 3
Ω
200
=
R
mH
400
=
L
in
V C
V
μF
5
.
2
=
C
10
1
1
1 3
0 =
=
=
=
ω
1
10
400
1
1
10
10
10
5
.
2
10
400
3
6
6
3
0
×
×
×
×
−
−
−
−
L
LC
ω
2
10
6
.
1
200
1
10
5
.
2
10
400
200
1
1 5
6
3
=
×
=
×
×
=
=
−
C
L
R
Q
James Grimbleby

Bode Plot
Gain(dB)
Bode Plot
20 dB
2
=
Q 10
=
Q
0 dB
0 dB
2
1
=
Q
-20 dB
12 dB / octave
-12 dB / octave
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB
James Grimbleby
Frequency (rad/s)

Bode Plot
Phase(rad)
Bode Plot
0 10
=
Q
2
=
Q
1
Q
2
π
−
2
1
=
Q
2
Frequency (rad/s)
1000 10000 100000
100
10
π
−
James Grimbleby
Frequency (rad/s)

Example 4
Example 4
g p
L
j
VL ω
C R
LC
R
L
j
C
j
j
Vin
L
/
1
2
+
+
=
ω
ω
in
V L
V
L
LC
CR
j
LC
j
H
1
)
(
2
2
−
+
−
=
ω
ω
ω
ω
L
Q
j /
)
/(
1
/
2
2
2
0
2
−
=
ω
ω
L
Q
Q
j
1
1
:
and
1
:
where
/
)
/(
1 2
0
2
0 −
+
ω
ω
ω
ω
ω
C
R
CR
Q
LC
:
and
:
where
0
0 =
=
=
ω
ω
James Grimbleby

Example 4
Example 4
0
→
ω ∞
→
ω
1
→
g
0
→
g
James Grimbleby

Example 4
Circuit is a second-order high-pass filter:
Example 4
g p
)
( j
H
2
0
2
/
)
(
ω
ω
−
j
H )
( ω
j
H
g =
2
0
2
0
0
/
)
/(
1
/
)
(
ω
ω
ω
ω
ω
ω
ω
−
+
=
Q
j
j
H
2 2
0
ω
ω << 2
0
2
)
(
ω
ω
ω
−
≈
j
H )
oct
/
dB
12
(
2
0
2
ω
ω
≈
g
0
ω
ω = jQ
j
H =
)
( ω
0
Q
g =
0
jQ
j )
( Q
g
0
ω
ω >> 1
)
( ≈
ω
j
H )
dB
0
(
1
=
g
James Grimbleby

Bode Plot
Gain(dB)
Bode Plot
20 dB
2
=
Q 10
=
Q
0 dB
0 dB
2
1
=
Q
-20 dB
12 dB / octave
2
12 dB / octave
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB
James Grimbleby
Frequency (rad/s)

Bode Plot
Phase(rad)
Bode Plot
π 10
=
Q
2
=
Q
1
Q
2
π 2
1
=
Q
2
Frequency (rad/s)
1000 10000 100000
100
10
0
James Grimbleby
Frequency (rad/s)

Example 5
Example 5
R
V
C L
R
L
j
C
j
R
V
V
in
R
/
1 +
+
=
ω
ω in
V R
V
R
LC
CR
j
CR
j
j
H
1
)
(
2
−
+
=
ω
ω
ω
ω
Q
j
Q
j
j
/
)
/(
1
)
/(
2
2
0
+
=
ω
ω
ω
ω
ω
ω
L
Q
Q
j
1
1
:
and
1
:
where
/
)
/(
1
0
2
0
2
0
=
=
=
−
+
ω
ω
ω
ω
ω
C
R
CR
Q
LC
:
and
:
where
0
0 =
=
=
ω
ω
James Grimbleby

Example 5
Example 5
0
→
ω ∞
→
ω
0
→
g
0
→
g
James Grimbleby

Example 5
Circuit is a second-order band-pass filter:
Example 5
)
( j
H
0 )
/(
)
(
ω
ω Q
j
j
H )
( ω
j
H
g =
2
0
2
0
0
/
)
/(
1
)
(
)
(
ω
ω
ω
ω
ω
−
+
=
Q
j
Q
j
j
H
0
ω
ω <<
Q
j
j
H
0
)
(
ω
ω
ω ≈ )
oct
/
dB
6
(
0Q
g
ω
ω
≈
0
ω
ω = 1
)
( =
ω
j
H
0
)
dB
0
(
1
=
g
0
j
j
H
ω0
)
(
−
1
)
( ω
j
H )
dB
0
(
1
=
g
)
t
/
dB
6
(
0
ω
0
ω
ω >>
Q
j
j
H
ω
ω
ω 0
)
( ≈ )
oct
/
dB
6
(
0 −
≈
Q
g
ω
ω
James Grimbleby

Bode Plot
Gain(dB)
Bode Plot
0 dB
2
Q
10
=
Q
2
=
Q
1
Q
-20 dB
2
=
Q
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB
James Grimbleby
Frequency (rad/s)

Bode Plot
Phase(rad)
π
Bode Plot
2
π
10
=
Q
2
=
Q
1
Q
0
2
1
=
Q
π
Frequency (rad/s)
1000 10000 100000
100
10
2
π
−
James Grimbleby
Frequency (rad/s)

Example 6
Using the potential divider formula: R
Example 6
L
j
C
j
VC /
1 + ω
ω
LC
R
L
j
C
j
j
j
Vin
C
1
/
1
2
+
+
=
ω
ω in
V c
V
L
LC
CR
j
LC
j
H
1
1
)
(
2
2
−
+
−
=
ω
ω
ω
ω C
Q
j
LC
/
)
/(
1
1
2
2
2
−
=
ω
L
Q
Q
j
1
1
:
and
1
:
where
/
)
/(
1 2
0
2
0 −
+
ω
ω
ω
ω
ω
C
R
CR
Q
LC
:
and
:
where
0
0 =
=
=
ω
ω
James Grimbleby

Example 6
Example 6
0
→
ω ∞
→
ω
1
→
g
1
→
g
James Grimbleby

Example 6
Circuit is a second-order band-stop filter:
Example 6
)
( j
H
2
0
2
/
1
)
(
ω
ω
−
j
H )
( ω
j
H
g =
2
0
2
0
0
/
)
/(
1
/
1
)
(
ω
ω
ω
ω
ω
ω
ω
−
+
=
Q
j
j
H
0
ω
ω << 1
)
( ≈
ω
j
H )
dB
0
(
1
≈
g
0
ω
ω = 0
)
( =
ω
j
H )
dB
(
0 −∞
=
g
1
)
( ω
j
H
)
(
g
0
ω
ω >> 1
)
( ≈
ω
j
H )
dB
0
(
1
≈
g
James Grimbleby

Bode Plot
Gain(dB)
Bode Plot
0 dB
2
Q 2
=
Q 10
=
Q
-20 dB 2
1
=
Q
Frequency (rad/s)
1000 10000 100000
100
10
-40 dB
James Grimbleby
Frequency (rad/s)

Bode Plot
Phase(rad)
π
Bode Plot
2
π
2
1
=
Q
10
=
Q
2
0
10
=
Q
2
=
Q
π
Q
Frequency (rad/s)
1000 10000 100000
100
10
2
π
−
James Grimbleby
Frequency (rad/s)

AC Circuit Analysis
AC Circuit Analysis
L t 8
Lecture 8
Power in AC Circuits
James Grimbleby

To calculate the power in a circuit we shall need to make
use of some trigonometric identities:
B
A
B
A
B
A i
i
)
(
B
A
B
A
B
A
B
A
B
A
B
A
sin
sin
cos
cos
)
cos(
sin
sin
cos
cos
)
cos(
+
=
−
−
=
+
Adding:
{ }
)
cos(
)
cos(
1
cos
cos
cos
cos
2
)
cos(
)
cos(
B
A
B
A
B
A
B
A
B
A
B
A
+
+
=
−
+
+
so that:
{ }
)
cos(
)
cos(
2
cos
cos B
A
B
A
B
A −
+
+
=
{ } A
A
A 2
cos
2
1
2
1
0
cos
2
cos
2
1
cos2
+
=
+
=
James Grimbleby

rms Voltages and Currents
The average power in a resistor is given by:
∫
T
)
(
)
(
1
∫
=
T
dt
t
i
t
v
T
P
2
0
)
(
)
(
1
)
(t
i
∫
=
T
dt
R
t
v
T 0
2
)
(
1 )
(t
v R
∫
=
T
dt
t
v
T
R
2
0
)
(
1
1
∫
∫
T
rms dt
t
V
V
T
R
2
2
0
)
(
1
h ∫
=
= rms
rms dt
t
v
T
V
R 0
2
)
(
:
where
James Grimbleby

The root-mean-square voltage V determines the power
The root-mean-square voltage Vrms determines the power
dissipated in a circuit:
2
R
V
P 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 e
a current Irms flows through a circuit:
2
2
rms
RI
P =
These expressions apply to any waveform
James Grimbleby

The rms value of a sinusoid of amplitude (peak) value v0:
)
(
1 2
dt
t
v
V
T
∫
p (p ) 0
1
)
(
0
dt
t
v
T
V
T
rms ∫
=
)
(
cos
1
0
2
2
0 dt
t
v
T
T
∫
= ω
)
2
cos(
2
1
2
1
1
2
0 dt
t
T
v
T
∫ +
= ω
2
2
0
2
0
0
v
v
T
Averages to zero over
l t l
2
2
0
0 =
= a complete cycle:
T = 2π/ω
James Grimbleby

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 V
230
2
2
0
×
=
×
= rms
V
v
V
325
=
James Grimbleby

A mains power (230 V rms) electric fire has a resistance of
kW
017
1
2302
2
V
P rms
52 Ω:
kW
017
.
1
52
230
=
=
=
R
V
P 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
600
4
150
.
2
=
×
=
=
rms R
P
V
V
5
.
24
600
4
150
.
=
×
rms
rms
V
R
P
V
This corresponds to a sinusoid of peak value 34.6 V
James Grimbleby

Square wave of amplitude ±v0:
v0
-v0
T
T/2
2
2
/
2
2
)
(
1
1
)
(
1
dt
v
dt
v
dt
t
v
V
T
T
T
∫
+
∫
∫
T
2
/
0
0
0
0
1
)
(
)
( dt
v
T
dt
v
T
dt
t
v
T
V
T
T
rms ∫ −
+
∫
=
∫
=
0
2
0
0
2
0
1
v
v
dt
T
v
T
=
=
∫
=
James Grimbleby

Crest Factor
The ratio between the peak voltage and the rms voltage is
k th t f t
Crest Factor
known as the crest factor:
peak
V
cf
rms
p
V
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 i
150 W of audio into 4 Ω loudspeakers would therefore require
peak voltages of 50 V or greater
James Grimbleby

Power in a Reactive Load
Capacitors and inductors store energy, but do not dissipate
power
Power in a Reactive Load
power
IC
IR
R C
100 V rms
50 Hz
100
25Ω 200μF
100
A
4
25
100
6
=
=
R
I
A
6.28
A
10
200
50
2
100
100 6
=
×
×
×
×
=
= −
π
C
C
Z
I
W
400
25
100
P
2
=
=
James Grimbleby

Instantaneous Power
)
(t
i
For sinusoidal voltages and currents:
Instantaneous Power
)
(t
i
( )
( )
ω
=
t
i
t
i
t
v
t
v
)
(
cos
)
( 0
)
(t
v Z
( )
φ
ω +
= t
i
t
i cos
)
( 0
Instantaneous power:
( ) ( )
φ
ω
ω cos
cos
)
(
)
(
)
(
0
0 +
=
×
=
t
i
t
v
t
i
t
v
t
p
( ) ( )
( ) ( )
φ
ω
ω
φ
ω
ω
cos
cos
cos
cos
0
0
0
0
+
=
+
=
t
t
i
v
t
i
t
v
{ }
φ
φ
ω cos
)
2
cos(
2
1
0
0 +
+
= t
i
v
James Grimbleby

Average Power
Average power:
T
1
Average Power
∫
=
T
dt
t
p
T
P
0
)
(
1
( ) ( )
∫ +
=
T
dt
t
i
t
v
T 0
0
0 cos
cos
1
φ
ω
ω
∫
+
∫ +
=
T
T
dt
i
v
dt
t
i
v
T
0
0
0
0
0
cos
1
1
)
2
cos(
1
1
φ
φ
ω ∫
+
∫ + dt
T
i
v
dt
t
T
i
v
0
0
0
0
0
0 cos
2
)
2
cos(
2
φ
φ
ω
If T >> 1/ω: T
If T >> 1/ω:
φ
cos
1
2
1
0
0
0 dt
T
i
v
P
T
∫
=
φ
cos
2
1
0
0
0
i
v
=
James Grimbleby
2

Average Power
Average Power
φ
cos
2
1
0
0i
v
P =
)
(t
i
φ
2
0
0
)
(t
v Z
φ
cos
2
1 2
0
Z
v
=
φ
cos
2
1
2
2
0 Z
i
Z
=
2
0
James Grimbleby

Average Power
Average power:
Average Power
Average power:
φ
cos
2
1
0
0i
v
P =
For a resistor:
2
0
2
0
0
0
1
1
1
0 Ri
v
i
v
P =
=
=
→
=
φ
For a capacitor:
0
0
0
2
2
2
0 Ri
R
i
v
P =
=
=
→
=
φ
o a capac to
0
2
=
→
= P
π
φ
For an inductor:
2
π
0
2
=
→
−
= P
π
φ
James Grimbleby

Power expressed in terms of rms voltages and currents:
φ
cos
2
1
0
0i
v
P =
φ
φ
cos
2
2
1
2
0
0
I
V rms
rms
=
φ
φ
)
W
(
cos
cos
2
I
V rms
rms
rms
rms
=
2
V
φ
cos
Z
V
P rms
=
φ
cos
2
Z
I
P rms
=
James Grimbleby

Example 1
Determine the average power dissipated in the circuit:
Example 1
Ω
80
R
F
20
C
Ω
80
=
R
230 V rms
50 Hz
1
μF
20
=
C
1
1
+
=
C
j
R
Z
ω
10
20
50
2
1
80
6
×
×
×
+
=
−
j π
Ω
)
( 63 3
1 105
178 1
Ω
2
.
159
80
o
∠
=
−
= j
James Grimbleby
Ω
)
(-63.3
1.105
178.1 −
∠
=

Example 1
Example 1
Ω
80
=
R
230 V rms
50 H
μF
20
=
C
50 Hz
2
cos
2
= φ
Z
V
P rms
105
.
1
cos
1
178
2302
−
=
W
4
.
133
1
.
178
=
James Grimbleby

Example 2
Determine the average power dissipated in the circuit:
Example 2
R
R
2 Ω
C
80 V rms
L
1 mH
C
200 μF
400 Hz
1 mH
The driving-point impedance of this circuit at 400 Hz
(calculated previously) is:
(calculated previously) is:
9282
.
0
091
.
3 −
∠
= Ω
Z
James Grimbleby

Example 2
9282
0
091
3 −
∠
Ω
=
Z
Example 2
9282
.
0
091
.
3 −
∠
Ω
=
Z
R
2 Ω
80 V
L
C
200 μF
80 V rms
400 Hz
2
V L
1 mH
200 μF
cos
2
= φ
Z
V
P rms
9283
.
0
cos
091
.
3
802
−
=
W
1241
=
James Grimbleby

Example 2
Determine 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 leg
to calculate the power in the inductor-resistor leg
+
= L
j
R
ZLR ω R
80 V rms
400 Hz
10
400
2
2 3
×
×
+
=
+
−
j
L
j
R
ZLR
π
ω
2 Ω
C
400 Hz
8986
.
0
212
.
3
513
.
2
2
∠
=
+
= j L
1 mH
200 μF
8986
0
3
James Grimbleby

Example 2
Example 2
8986
.
0
212
.
3 ∠
=
LR
Z
R
2 Ω
2 Ω
C
200 F
80 V rms
400 Hz
cos
2
= φ
V
P rms
L
1 mH
200 μF
80
cos
2
φ
Z
P
W
1241
8986
.
0
cos
212
.
3
80
=
W
1241
=
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 9
Lecture 9
Power Factor
Th Ph El t i P
Three-Phase Electric Power
James Grimbleby

True and Apparent Power
The apparent power P in a circuit is:
True and Apparent Power
The apparent power Pa in a circuit is:
rms
rms
a I
V
P =
Apparent power is measured in VA
rms
rms
a I
V
P =
Apparent power is measured in VA
The true power P dissipated in a circuit is:
φ
cos
rms
rms I
V
P =
True power is measured in W
James Grimbleby

Power Factor
The power factor is the ratio of the true power to the apparent
Power Factor
p p pp
power:
I
V
P
φ
φ
cos
cos
=
=
=
rms
rms
rms
rms
a I
V
I
V
P
P
pf
where ø is the phase difference between voltage and current
where ø is the phase difference between voltage and current.
It does not matter whether ø is phase of the current with
It does not matter whether ø is phase of the current with
respect to the voltage, or voltage with respect to the current,
since:
φ
φ −
= cos
cos
James Grimbleby

Example 1
Determine the power factor, apparent power and true power
di i t d i th i it
Example 1
power dissipated in the circuit:
)
(75 1
1 312
Ω
60
5
1
Ω
08
.
15
4
o
∠
=
+
= j
Z
)
(75.1
1.312
Ω
60
.
5
1 ∠
=
Ω
4
=
R
Hz
400
,
0Vrms
8
2559
.
312
.
1
cos =
=
pf mH
6
=
L
Hz
400
,
0Vrms
8
VA
3
.
410
2
=
=
= rms
rms
rms
a
Z
V
I
V
P
W
0
.
105
=
×
= a
P
pf
P
Z
James Grimbleby

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 lighting
Examples are induction motors and inductor ballast lighting
Power factor can be corrected by connecting a reactance in
Power 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 load
g
This is obviously desirable because it reduces the current
y
rating of the power wiring and supply
James Grimbleby

Power factor is normally corrected by connecting a reactive
element ZC in parallel with the load ZL :
S
I C
I
Supply current: IS
L d t I
L
I
V
Load current: IL
Correction current: IC
L
Z C
Z
S
V
A unity overall power factor will be obtained provided that VS
u y o e a po e ac o be ob a ed p o ded a S
and IS are in phase:
)
real
(
0
0 j
G
G
V
I
S
S +
=
∠
=
James Grimbleby

C
L
S j
G
I
I
I
0
S
I C
I
I
S
C
S
L
S
S j
G
V
I
V
I
V
I
+
=
+
=
1
1
0
L
Z C
Z
L
I
S
V
C
L
j
G
Z
Z
+
=
+ 0
1
1
L
Z C
Z
imag
C
imag
L Z
Z ⎥
⎦
⎤
⎢
⎣
⎡
−
=
⎥
⎦
⎤
⎢
⎣
⎡ 1
1
imag
C
imag
L ⎦
⎣
⎦
⎣
If IL leads VS then an inductor is used for correction
If IL lags VS then a capacitor is used for correction
James Grimbleby

Correction of a lagging power factor load with a capacitor:
C
L
S I
I
I +
=
C
I
C
L
S I
I
I +
=
S
I Current
(real part)
Current
(imaginary
L
I
(real part)
part)
L
I
Note that the magnitude of the supply current IS is less than
that of the load IL
James Grimbleby
L

Example 2
Choose a suitable power factor correction component for the
Example 2
Choose a suitable power factor correction component for the
circuit:
Ω
4
=
R
Hz
400
0Vrms
8
mH
6
=
L
Hz
400
,
0Vrms
8
08
15
4
1
Ω
08
.
15
4
j
j
ZL +
=
06195
.
0
01643
.
0
08
.
15
4
08
.
15
4
1
2
2
j
j
ZL
−
=
+
−
=
Thus: 06195
.
0
1
j
ZC
+
=
James Grimbleby

Example 2
Example 2
Ω
4
=
R
Hz
400
,
0Vrms
8 μF
465
.
2
=
C
mH
6
=
L
Hz
400
,
0Vrms
8 μF
465
.
2
C
1
06195
.
0
1
=
+
= ωC
j
j
ZC
μF
465
.
2
400
2
06195
.
0
=
×
=
π
C
James Grimbleby

Example 2
80
=
I
Example 2
)
06195
0
01643
0
(
80 −
=
=
j
Z
I
L
L
I
I I
80
956
.
4
314
.
1
)
06195
.
0
01643
.
0
(
80
−
=
=
j
j
L
I
Ω
4
R
S
I C
I
80
=
Z
I
c
C
Ω
4
=
R
Hz
400
0Vrms
8
F
465
2
=
C
956
.
4
06195
.
0
80
=
×
=
j
j mH
6
=
L
Hz
400 μF
465
.
2
956
4
956
4
314
1
956
.
4
+
=
j
j
I
I
I
j
C
L
S
314
.
1
956
.
4
956
.
4
314
.
1
=
+
−
= j
j
James Grimbleby

Example 2
5A
Example 2
5A
Imaginary
C
L
S I
I
I +
=
C
I
Imaginary
part
5A
S
I Real
part
5A part
L
I
-5A
James Grimbleby

Example 3
An electric motor operating from the 50 Hz mains supply has
Example 3
p 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:
6
1 I
02609
.
0
230
6
1
=
=
=
S
L
L V
I
Z
and the phase of 1/ZL is:
6435
.
0
8
.
0
cos
1 1
±
=
=
⎭
⎬
⎫
⎩
⎨
⎧
∠ −
Z
Since the current lags the voltage the negative phase is used
⎭
⎩ L
Z
James Grimbleby
g g g p

Example 3
01565
0
02087
0
6435
0
02609
0
1
−
=
−
∠
= j
Example 3
01565
0
1
01565
.
0
02087
.
0
6435
.
0
02609
.
0 =
∠
=
C
j
j
j
ZL
01565
0
01565
.
0
1
=
+
= ωC
j
j
ZC
μF
82
.
49
50
2
01565
.
0
=
×
=
π
C
Before correction: After correction:
1380
8
0
1380
6
230
×
×
=
×
=
a
P
pf
P
P
1104
1104
=
= a
P
P
P
1104
1380
8
.
0
=
×
=
×
= a
P
pf
P 8
.
4
230
1104
=
=
=
S
S
V
P
I
James Grimbleby

Example 3
5A
Example 3
5A
Imaginary
C
L
S I
I
I +
=
5A
C
I
g y
part
5A
S
I Real
part
L
I
p
-5A
James Grimbleby

Most ac power transmission systems use a three-phase
t
Three Phase Electric Power
system
Three phase is also used to power large motors and other
Three-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 given
This 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
James Grimbleby
p

Three-phase
load
Three-phase
generator
James Grimbleby

Phase to ne tral oltage
π
Phase-to-neutral voltage v0
Phase-to-phase voltage vp
v0
v
)
60
(
3
o
π
v0
vp
3
sin
2 0
v
vp =
π
v0
)
120
(
3
2 o
π
2
3
2
3
0
v
=
3
2
0
0
v
=
James Grimbleby

UK domestic supply uses three -phase with a phase-to-
pp 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 i
each) then the overall neutral current is zero
The UK electricit distrib tion net ork operates at 275 kV rms
The UK electricity distribution network operates at 275 kV rms
and 400 kV rms
James Grimbleby

AC Circuit Analysis
AC Circuit Analysis
L t 10
Lecture 10
Energy Storage
James Grimbleby

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 it
gy
Over a complete ac cycle there is no net change in energy
y g gy
stored, and therefore no power dissipation
James Grimbleby

Energy Storage
The voltage across a capacitor is increased from zero to V
Energy Storage
producing a stored energy E:
v
0
)
(
)
( dt
t
i
t
v
E
T
∫
=
v(t)
i
0
)
( dt
dt
dv
C
t
v
T
∫
=
V
C
dv
0
)
(
d
C
dt
V
∫
∫
dt
dv
C
i =
0
1
dv
v
C ∫
=
T
t
2
2
1
CV
E =
T
James Grimbleby

Energy Storage
Energy Storage
Example: calculate the energy storage in an electronic flash
capacitor of 1000 μF charged to 400 V
2
1 2
= CV
E
400
10
1000
2
1 2
6
×
×
×
= −
J
80
2
=
James Grimbleby

Energy Storage
The current in an inductor is increase from zero to I
Energy Storage
producing a stored energy E:
)
(
)
( dt
t
i
t
v
E
T
∫
=
v
i(t)
0
)
( dt
t
i
di
L
T
∫
v
i
L
I
0
)
( dt
t
i
dt
L
I
∫
= L
di
L
0
di
i
L
I
∫
= dt
L
v =
T
t
2
2
1
LI
E =
T
James Grimbleby

Energy Storage
Energy Storage
Example: calculate the energy storage in a 2 mH inductor
carrying a current of 10 A
2
1 2
= Li
E
10
10
2
2
1
2
2
3
×
×
×
= −
J
1
.
0
2
=
James Grimbleby

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