Steady-State
Sinusoidal Analysis
By
Dr. Tarek M. Said
tsaid@uark.edu
Assistant Professor
Electrical Engineering Department
College of Engineering Al-Majmaa University
Complex Numbers
Three forms
Rectangular form
A  a  jb
Polar form
A | A |  
Exponential form
A | A | e
j
Rectangular form
Im
A  a  jb




|A|
Real part a  R e [A ]

Imaginary part b  Im[ A ]
2
0
Magnitude a  b  A  A 
b
Angle   arctan
2
2
a

Conjugate

A
b
A  a  jb
Re
a
a b
2
2
Polar form
A | A |  
Im
|A|

0


A
b
Re
a
Conversion between Rectangular and polar
2
a  | A | cos 
b
  arctan
a
b  | A | sin 
A  a b
2
Exponential form
Im
Euler’s Identity
e
j
 cos   j sin 
|A|

0
A  A cos   j sin  
 Ae
j

A
b
Re
a
Arithmetic Operations
Given
 Adding
A  a  jb
B  c  jd
and subtracting
C  A  B  e1  je 2
 Product
C  A B  A e
 Ce
j C
C  AB
j A
 Be
e
j C
e1  a  c
e2  b  d
j B
e
j(  A   B )
C   A   B
•Arithmetic Operations
B  c  jd
j A
A
e
 Dividing
D  A B 
j B
Be
A j( A  B )
j D
D  De 
e
B
A
D  AB 
B
j D
j(  A   B )





D
A
B
e e
Given
A  a  jb
Sinusoidal Waveform
The sinusoidal function v(t) = VM sin  t is plotted (a) versus  t and (b) versus t.
Instantaneous value changes continuously with time
Average value = 0
Effective value is known as the root mean square, RMS
V
Peak Value
RMS value =
i.e. Vrms = m
2
2
Parameters of Sinusoidal Currents and Voltages
•Vm is the peak value, unit is volt
•ω is the angular frequency, unit is radians per second
•f is the frequency，unit is Hertz (Hz) or inverse second.
•θ is the phase angle, unit is radian or degree.
1
Frequency
f 
T
2
Angular frequency  
T
  2f
Resistor in an ac circuit
The voltage across the resistor is
changing with time, so the current
through the resistor must also be
time-varying.
I = V/R = Imax sin wt
Notice that as the voltage changes
sign,
the
current
changes
direction.
The voltage and the
current are in
phase. They peak
at the same time.
RMS Values
It is useful to compare ac circuits and dc circuits.
We cannot use, however, the average ac current
or voltage which are both zero. Instead we must
use the root mean square or rms values.
I rm s
I m ax

,
2
V rm s
V m ax

2
The rms value of current or voltage in an ac circuit
can be compared to the equivalent quantities in
a simple dc circuit.
Vrms = IrmsR
Pav = Irms2R = Vrms2/R
Root Mean Square Voltage and
Current
V (t )  Vmax sin(t )
I (t )  V (t ) / R  (Vmax / R) sin(t )
V2
Power  P 
 I 2R
R
2
Vmax
P
sin 2 (t )
R
2
Vmax
1
2
P 
 Vrms
/ R,
R 2
Vrms  VMax / 2
Vrms = Square root of the mean
(average) of V-squared.
Average AC Power consumed by a resistor
2
2
independent of form of V vs t curve
P  Vrms
/ R  I rms
R
Circuit Analysis
When a circuit has more than one element, a circuit analysis is required to
determine circuit parameters (v, i, power, etc.) in different parts of the
circuit. There are different methods for circuit analysis.
Time Domain Method
- applicable to both transient and
steady-state circuit analysis
- very useful for transient analysis
- difficult method (often requires
differentiation & integration of
sinusoidal functions)
Example:
v = Vm sin t
i = Im sin (t + )
Phasor Method
- only steady-state circuit analysis
- easy method
- sinusoidal functions represented by
magnitude (usually RMS value) and
phase angle
- Differentiation/integration replaced
by multiplication/division
V = Vm /00
2
I = Im /
2
Example: Time Domain Method
v1
Find the instantaneous value of v when
v1 = 0. Frequency, f = 60 Hz.
where,
v1 = 50 sin (377 t + 200) and
v2 = 10 sin (377 t + 100) volts.
+
v ~
v2
-
Sin (A+B) = Sin A·Cos B + Cos A·Sin B
v = v1 + v2
Vm sin(ωt+θ) = 50 sin (ωt + 200) + 10 sin (ωt + 100)
Vm (sin ωt·cos θ + cos ωt·sin θ ) = 50 (sin ωt·cos 200 + cos ωt·sin 200)
+ 10 (sin ωt·cos 100 + cos ωt·sin 100)
----1.
Vm cos θ = 56.83 ----2.
Vm sin θ = 18.84 ----3.
Example: Time Domain Method (continued)
Dividing Eqn 3 by Eqn 2 and taking the inverse tangent yields
θ = 18.340
Then from Eqn 3, Vm = 59.87 volts
From Eqn 1,
v = 59.87 sin (377t + 18.34) volts
v1 = 0  ωt = -20°
v = 59.87 sin (-20 + 18.34) V = -1.734 V when v1 is zero.
Example: Phasor Method
v1 = 50 sin (377 t + 200) volts and
v2 = 10 sin (377 t + 100) volts.
Phasors: V1 = 50 /200 volts, and V2 = 10 /100 volts
V = 50 /200 + 10 /100 volts
Complex Numbers in Polar form
V = 50 (cos 200 + j sin 200)
+ 10 (cos 100 + j sin 100)
Rectangular form
= 46.985 + j 17.101
+ 9.848 + j 1.736
Phasor Diagram
V1
= 56.833 + j 18.837
= 59.87
/18.340
V2
volts
v = 59.87 sin (377 t + 18.340) volts.
V
Example: Phasor Method
v1  t   20 cos t  45  V
v2  t   10 sin t  60
V
Find vs  v1  v2  ?
Solution: V1  20  45 V
V2  10  30 V
Vs  V1  V2
 20  45  10  30
 14.14  j14.14  8.660  j5
 23.06  j19.14
 29.97  39.7 V
 vs  t   29.97 cos t  39.7
V
Adding Sinusoids Using Phasors
Exercise
1.v1 (t )  10cos(t )  10sin(t )
2.i1 (t )  10cos(t  30 )  5sin(t  30 )
o
o
3.i2 (t )  20sin(t  90o )  15cos(t  60o )
Answers
1.v1 (t )  14.14 cos(t  45o )
2.i1 (t )  11.18cos(t  3.44o )
3.i2 (t )  30.4 cos(t  25.3o )
Circuit Elements:
 Active – supply energy, voltage or current source
e.g. battery, transistor, IC components
 Passive – absorb energy
e.g. resistor, inductor, capacitor
Linear Circuit
 obeys Ohm’s Law (i.e. v α i or v = Ri)
 if the i or v in any part of the circuit is sinusoidal, the i and v in every
other part of the circuit is sinusoidal and of the same frequency
Non-linear circuits do not obey Ohm’s Law.
Sinusoidal response of resistance
• Relationship between voltage and current
i  I m cos( t   )
i
v  Ri  RI m cos( t   )
u
R
 Vm cos( t   )
then：
U
Vrms  RI rms
Vrms Vm
R

I rms I m
u
I
i
0

t
Sinusoidal response of resistance
Substituting for current and voltage phasors
I
i
i  I m cos( t   )
I  I m 
Irms  I rms 
R
Vrms V

I rms I
v
R
R
V
Vrms  RI rms  RI rms  
V  RI  RI m  
Phasor diagram
I

V
0O
Sinusoidal response of Inductance
• Relationship between voltage and current
i  I m sin  t
di
v  L   LI m cos  t  Vm sin( t  90 0 )
dt
u、i have the same
frequency，u leads i
by 90o
i
u
0
π
2π
ωt
Sinusoidal response of Inductance
• Effective Value
i  I m sin  t
Vm   LI m
Definition
v  Vm sin( t  90 0 )
 Vrms   LI rms
Reactance
U rms U m
XL 

  L  2 fL
I rms
Im
Sinusoidal response of Inductance
• Phasor relations
i
Phasor diagram：


V  jX L0I
I  I 0 A
i  I m cos  t
L
v

I
v  Vm cos( t  90 0 )
V  Vm  90  jX L I m  jX L I m  0  jX L I
0

0
I
V

jωL
Z  j X L  j L 
V

I
Complex
impedance
Sinusoidal response of capacitance
•Relationship between voltage and current
i
v  Vm sin  t
dv
i  C  CU m sin(t  90 0 )
dt
 I m sin(t  90 0 )
u
I m   CVm  2 fCVm
Current leads voltage
by 90°
C
v
i
0

2
Sinusoidal response of capacitance
•Relationship between voltage and current
I m   CVm  2 fCVm
i
 I rms   CVrms  2 fCVrms
v
Reactance
Definition
Vrms U m
1
1
XC 



I rms I m  C 2 fC
C
Sinusoidal response of capacitance

i
v
I
• Phasor

C
U
-jωC
V  Vm  0 V
0
v  Vm sin  t

i  I m sin( t  90 ) A I  I m 90 0 A
0

Vm 0
Vm
Vm




j


j
X
C

0
I

90
j
I
I
m
m
m
I
U
0
diagram


U
I j
XC

U
•
It is given that the frequency of sinusoidal source is
50Hz, the rms or effective value is 10V, capacitor is 25μF,
determine the rms value of current. If the frequency is
5000Hz, then what is the rms value of current now？
Solution
when f =50Hz
i
1
1
XC 

 127.4
6
2 fC 2  3.14  50  (25  10 )
I rms
U rms
10


 0.078 A  78mA
X C 127.4
When f =5000Hz
X C 
1
1

 1.274
6
2 fC 2  3.14  5000  (25  10 )
I rms
U rms
10


 7 .8 A
XC
1.274
u
C
The higher frequency is under fixed rms value of voltage, the
bigger the rms value of current flowing through capacitor.
Complex Impedances of R L C
VR
R : ZR 
R
IR
| Z R | R
VL
 j L  j X L | Z L |  L
L : ZL 
IL
C : Z  VC  1   j X
C
C
I C j C
1
| Z C |
C
 0


2

 
2
Sinusoidal response of resistance
•
Power
Instantaneous power
p  vi
 2Vrms cos  t 2 I rms cos  t
 2Vrms I rms cos 2  t  Vrms I rms  Vrms I rms cos 2 t
No doubt, p≥0,Resistance always absorbs energy.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of resistance
•
Power
p  2V
rms I
2
cos
t  V
rms
rms
I rms (1  cos 2 t )
2VrmsIrms
p
VrmsIrms
t
0
i
v
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of resistance
•
Power
Average Power
——Average value of instantaneous power in a period
1
P
T

T
0
1
pdt 
T

T
0
Vrms I rms (1  cos 2 t ) dt
2
V
 Vrms I rms  I rms 2 R  rms
R
Average Power represents the consumed power, is also named as
Real power.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of Inductance
Power
•
u
i
di
i increases ，  0，
dt
p (t )  0, WL increases
t
0
i
i
i
i
u
u
u
u
UI
p

0
——magnitude of magnetic
field increases，inductance
absorbs energy
di
i decreases ，  0，
dt
p (t )  0, WL decreases



BASIC ELECTRONIC ENGINEERING
t
——magnitude of magnetic
field decreases，inductance
supplys energy
Department of Electronic Engineering
Sinusoidal response of Inductance
• Power
Average Power
1 T
P   pdt
T 0
1 T
  Vrms I rms sin 2 tdt  0
T 0
•Inductance is an energy-storage element rather
than an energy-consuming element
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of Inductance
• Power
v  Vm sin( t  90 0 )
i  I m sin  t
p  p L  vi  Vm I m sin  t  sin( t  90 )
 Vm I m sin  t  cos  t
 Vrms I rms sin 2 t
VrmsIrms
p
u
i
0
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
t
Sinusoidal response of Inductance
Reactive Power
The power flows back and forth to inductances and
capacitances is called reactive power Q, it is the peak
instantaneous power。
2
VLrms
2
Q  VLrms I Lrms 
 X L I Lrms
XL
Unit: Var
Reactive power is important because it causes power
dissipation in the lines and transformers of a power
distribution system. Specific Charge is executed by
electric-power companies for reactive power.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of capacitance
•Power
•Instantaneous power
p  vi  Vm sin  tI m sin( t  90 0 )
 Vrms I rms sin 2 t
UrmsIrms
p
u
i
0
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
t
Sinusoidal response of capacitance
•Power
u
i
t
0
i
i
i
i
u
u
u
u
UI
0


du
 0，p (t )  0
dt
——capacitance charges，it
absorbs energy
u decreases ，Wc decreases ，
p

u increases ，Wc increases ，

BASIC ELECTRONIC ENGINEERING
du
 0，p (t )  0
t dt
——capacitance discharges，
it supplys energy
Department of Electronic Engineering
Sinusoidal response of capacitance
•Power
Average Power
1
P
T

T
0
1
pdt 
T

T
0
Vrms I rms sin 2 tdt  0
•Capacitance is an energy-storage element rather
than an energy-consuming element.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Sinusoidal response of capacitance
•Power
Reactive Power
• Assuming the same current acts on the inductance and
capacitance respectively, the initial phase is 0，then
i  I m sin  t
vC  Vm sin( t  90 )
p  vi  Vm sin  tI m sin( t  90 0 )
 Vrms I rms sin 2 t
QC  VCrms I Crms   I
BASIC ELECTRONIC ENGINEERING
2
Crms
Department of Electronic Engineering
XC
Resonance In Electric Circuits

Any passive electric circuit will resonate if it has an inductor
and capacitor.

Resonance is characterized by the input voltage and current
being in phase. The driving point impedance (or admittance)
is completely real when this condition exists.

In this presentation we will consider (a) series resonance, and
(b) parallel resonance.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
Consider the series RLC circuit shown below.
V = VM 0
R
L
+
V _
C
I
The input impedance is given by:
Z  R  j ( wL 
The magnitude of the circuit current is;
BASIC ELECTRONIC ENGINEERING
I | I |
Department of Electronic Engineering
1
)
wC
Vm
R 2  ( wL 
1 2
)
wC
Series Resonance
Resonance occurs when,
1
wL 
wC
At resonance we designate w as w o and write;
1
wo 
LC
This is an important equation to remember. It applies to both series
And parallel resonant circuits.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
The magnitude of the current response for the series resonance circuit
is as shown below.
|I|
Vm
R
Vm
2R
Half power point
w1 wo w2
Bandwidth:
BASIC ELECTRONIC ENGINEERING
w
BW = wBW = w2 – w1
Department of Electronic Engineering
Series Resonance
• This situation is referred to as resonance
– the frequency at which is occurs is the resonant frequency
o 
1
LC
fo 
1
2 LC
– in the series resonant
circuit, the impedance is
at a minimum at resonance
– the current is at a maximum
at resonance
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
The peak power delivered to the circuit is;
2
V
P m
R
The so-called half-power is given when I 
V.m
2R
We find the frequencies, w1 and w2, at which this half-power
occurs by using;
1 2
2 R  R  ( wL 
)
wC
2
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
After some insightful algebra one will find two frequencies at which
the previous equation is satisfied, they are:
2
2
R
1
 R 
w2 
 


2L
2
L

 LC
R
1
 R 
w1  
 


2L
 2 L  LC
The two half-power frequencies are related to the resonant frequency by
wo  w1w2
The bandwidth of the series resonant circuit is given by;
BW  wb  w2  w1 
BASIC ELECTRONIC ENGINEERING
R
L
Department of Electronic Engineering
Series Resonance
• The series RLC circuit is an acceptor circuit
– the narrowness of bandwidth is determined
by the Q
Quality factor Q 
Resonant frequency o

Bandwidth
B
– combining this equation with the earlier
one gives
R
B
L
wo L
1
1 L
Q


 
R
wo RC R  C 
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
An Observation:
By using Q = woL/R in the equations for w1and w2 we have;
2
 1



1
w1  wo 
 
 1

 2Q

 2Q 


and
2
 1



1
w2  wo 
 
 1

2Q 
 2Q




BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
An Observation:
For high-Q circuits (Q > 10), the half-power frequencies are
symmetrical around the resonant frequency. one can safely use the
approximation;
w1  wo 
BW
2
and
w2  wo 
These are useful approximations.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
BW
2
Series Resonance
Example:
In the circuit shown, R = 2 Ω, L = 1 mH, and C = 0.4 μF.
(a) Find the resonant frequency and the half-power frequencies.
(b) Calculate the quality factor and bandwidth.
(c) Determine the amplitude of the current at ω0, ω1, and ω2.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Series Resonance
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Parallel Resonance
• The parallel arrangement is a rejecter circuit
– in the parallel resonant circuit, the impedance is at a
maximum at resonance
– the current is at a minimum at resonance
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Parallel Resonance
Background
Consider the circuits shown below:
V
I
R
L
1
1 
I  V   jwC 

R
jwL


C
L
R
V
C
I
BASIC ELECTRONIC ENGINEERING

1 
V  I  R  jwL 

jwC


Department of Electronic Engineering
Parallel Resonance
Duality
1
1 
I  V   jwC 

R
jwL



1 
V  I  R  jwL 

jwC


We notice the above equations are the same provided:
I
R
L
V
1
R
we make
makethe
theinner-change,
inner-change,
IfIf we
then one
oneequation
equationbecomes
becomes
then
the same
sameas
asthe
theother.
other.
the
For such
suchcase,
case,we
wesay
saythe
the
one
For
one
circuitisisthe
thedual
dual
other.
circuit
ofof
thethe
other.
C
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Parallel Resonance
Background
What this means is that for all the equations we have
derived for the parallel resonant circuit, we can use
for the series resonant circuit provided we make
the substitutions:
R
replaced be
L replaced by
C replaced by
BASIC ELECTRONIC ENGINEERING
1
R
C
L
Department of Electronic Engineering
Parallel Resonance
Example:
In the parallel RLC circuit in Fig. 14.27, let R = 8 k&, L = 0.2 mH,
andC = 8 μF.
(a) Calculate ω0, Q, and B.
(b) Find ω1 and ω2.
(c) Determine the power dissipated at ω0, ω1, and ω2.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Parallel Resonance
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
58
Resonance
Example :
Determine the resonant frequency for the circuit below.
1
jwL ( R 
)
( w LRC  jwL )
jwC
Z 

1
(
1

w
LC
)

jwRC
R  jwL 
jwC
2
IN
2
At resonance, the phase angle of Z must be equal to zero.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Resonance
( w LRC  jwL )
(1  w LC )  jwRC
2
2
For zero phase;
wL
wRC

( w LCR ) (1  w LC
2
2
This gives;
w LC  w R C 1
2
or
2
2
2
1
w
( LC  R C )
o
BASIC ELECTRONIC ENGINEERING
2
2
Department of Electronic Engineering
Transfer Function
The transfer function H(ω) of a circuit is the frequency-dependent ratio of a
phasor output Y(ω) (an element voltage or current) to a phasor input X(ω)
(source voltage or current).
A zero, as a root of the numerator polynomial, is a value that results in a
zero value of the function. A pole, as a root of the denominator polynomial,
is a value for which the function is infinite.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Transfer Function


We will start by considering very simple circuits
Consider the potential divider shown here

Clearly the transfer function is
vo
ZR
R
1



v i Z R  ZC R  j 1 1  j 1
C
CR

At high frequencies
–  is large, voltage gain  1

At low frequencies
–  is small, voltage gain  0
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Decibel scale
• All circuits either amplify or attenuate power.
A = Po/Pi = 100000
B = Po/Pi = 0.0002
• As the power ratios can be very large or very small, it is handy to use a
logarithmic scale, the decibel scale.
A(dB) = 10 log Po/Pi = 10 log 100000 = 50 dB
B(dB) = 10 log Po/Pi = 10 log 0.0002 = –37 dB
• Positive decibel readings indicate gain and negative values attenuation.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode Plots
Expressing in dB:
Given the tranfer function:
G ( jw) 
K ( jw / z 1)
( jw)( jw / p 1)
B
20 log | G( jw |  20 log K  20 log | ( jw / z  1) | 20 log | jw | 20 log | jw / p  1 |
B
We have 4 distinct terms to consider: 20logKB
20log|(jw/z +1)|
-20log|jw|
-20log|(jw/p + 1)|
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode Plots
The gain term, 20logKB, is just so many dB and this is a straight line on
Bode paper, independent of omega (radian frequency).
The term, - 20log|jw| = - 20logw, when plotted on semi-log paper is a
straight line sloping at -20dB/decade. It has a magnitude of 0 at w = 1.
20
-20db/dec
0
-20
=1
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode Plots
The term, - 20log|(jw/p + 1), is drawn with the following approximation: If w <
p we use the approximation that –20log|(jw/p + 1 )| = 0 dB, a flat line on the
Bode. If w > p we use the approximation of –20log(w/p), which slopes at
-20dB/dec starting at w = p.
It is easy to show that the plot has an error of -3dB at w = p and – 1 dB at w =
p/2 and w = 2p. One can easily make these corrections if it is appropriate.
2
0
0
-20db/dec
-20
-40
 =p
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode Plots
When we have a term of 20log|(jw/z + 1)| we approximate
it be a straight line of slop 0 dB/dec when w < z. We
approximate it as 20log(w/z) when w > z, which is a
straight line on Bode paper with a slope of + 20dB/dec.
20
+20db/dec
0
-20
-40
=z
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Example 1:
50, 000( jw  10)
G ( jw) 
( jw  1)( jw  500)
Given:
First: Always, get the poles and zeros in a form such that the
constants are
associated with the jw terms. In the above example
we do this by factoring out the 10 in the numerator and the
500 in the denominator.
G ( jw) 
Second:
50, 000 x10( jw /10  1)
100( jw /10  1)

500( jw  1)( jw / 500  1)
( jw  1)( jw / 500  1)
When you have neither poles nor zeros at 0, start the Bode
at 20log10K = 20log10100 = 40 dB in this case.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Example 1:
(continued)
Third: Observe the order in which the poles and zeros occur.
This is the secret of being able to quickly sketch the Bode.
In this example we first have a pole occurring at 1 which
causes the Bode to break at 1 and slope – 20 dB/dec.
Next, we see a zero occurs at 10 and this causes a
slope of +20 dB/dec which cancels out the – 20 dB/dec,
resulting in a flat line ( 0 db/dec). Finally, we have a
pole that occurs at w = 500 which causes the Bode
to slope down at – 20 dB/dec.
Before we draw the Bode we should observe the range
over which the transfer function has active poles and zeros.
This determines the scale we pick for the w (rad/sec)
at the bottom of the Bode.
The dB scale depends on the magnitude of the plot and
experience is the best teacher here.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode Plot Magnitude for 100(1 + jw/10)/(1 + jw/1)(1 + jw/500)
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Filters
• A filter is a circuit that is designed to pass
signals with desired frequencies and
reject or attenuate others.
• A low-pass filter passes low frequencies
and stops high frequencies.
• A high-pass filter passes high frequencies
and rejects low frequencies.
• A band-pass filter passes frequencies
within a frequency band and blocks or
attenuates frequencies outside the band.
• A band-stop filter passes frequencies
outside a frequency band and blocks or
attenuates frequencies within the band,.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
• Since the denominator has
real and imaginary parts, the
magnitude of the voltage gain is
Voltage gain 
1
 1 
1 

 CR 
2
• When 1/CR = 1
Voltage gain 
2
1
1

 0.707
1 1
2
• This is a halving of power, or a fall in gain of 3 dB
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
• The cutoff frequency is the frequency at which the transfer
function drops in magnitude to 70.71% of its maximum value.
It is also regarded as the frequency at which the power
dissipated in a circuit is half of its maximum value
• the angular frequency C at which this occurs is given by
1
1
cCR
1 1
c 
 rad/s
CR 
– where  is the time constant of the CR network. Also
c
1
fc 

Hz
2 2CR
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
• Substituting  =2f and CR = 1/ 2fC in the earlier equation
gives
vo
1


1
vi
1 j
1 j
CR
1
1
 1
( 2f )
 2f

c





1
1 j
fc
f
• This is the general form of the gain of the circuit
• It is clear that both the magnitude of the gain and the phase
angle vary with frequency
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
• Consider the behaviour of the circuit at different frequencies:
• When f >> fc
– fc/f << 1, the voltage gain  1
• When f = fc
vo
1
1
1 (1  j)
(1  j)




 0 .5  0 .5 j
v i 1  j fc 1  j 1  j  (1  j)
2
f
• When f << fc
vo
1
1
f


j
v i 1  j fc  j fc
fc
f
f
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• Transposing the C and R gives
1
vo
ZC
1
C 


v i Z R  ZC R  j 1 1  jCR
C
j
• At high frequencies
–  is large, voltage gain  0
• At low frequencies
–  is small, voltage gain  1
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• A similar analysis to before
gives
Voltage gain 
1
1  CR 2
• Therefore when, when CR = 1
Voltage gain 
1
1

 0.707
1 1
2
• Which is the cut-off frequency
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• Therefore
– the angular frequency C at which this occurs is given by
cCR  1
c 
1 1
 rad/s
CR 
– where  is the time constant of the CR network, and as before
c
1
fc 

Hz
2 2CR
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• Substituting  =2f and CR = 1/ 2fC in the earlier
equation gives
vo
1


v i 1  jCR
1
1 j

c

1
1 j
f
fc
• This is similar, but not the same, as the transfer
function for the high-pass network
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• Consider the behaviour of this circuit at different frequencies:
• When f << fc
– f/fc << 1, the voltage gain  1
• When f = fc
• When f
vo
1  j1  j  1  j  0.5  0.5 j
1


1  j
v i 1 j f
2
>> fc
fc
vo
f
1
1


 j c
f
v i 1 j f
f
j
fc
fc
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Low-Pass Filter
• Frequency response of
the low-pass network
– the gain response has
two asymptotes that
meet at the cut-off
frequency
– you might like to
compare this with
the Bode Diagram
for
a
high-pass
network
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
High-Pass Filter
• High-pass networks can also
be produced using RL circuits
– these behave similarly to the
corresponding CR circuit
– the voltage gain is
vo
ZL
jL
1
1




v i Z R  Z L R  jL 1  R 1  j R
jL
L
– the cut-off frequency is
c 
R 1
 rad/s
L 
BASIC ELECTRONIC ENGINEERING
fc 
c
R

Hz
2 2L
Department of Electronic Engineering
Low-Pass Filter
• Low-pass networks can also
be produced using RL circuits
– these behave similarly to the
corresponding CR circuit
– the voltage gain is
vo
ZR
R
1



v i Z R  Z L R  jL 1  j L
R
– the cut-off frequency is
R 1
c   rad/s
L 
BASIC ELECTRONIC ENGINEERING
fc 
c
R

Hz
2 2L
Department of Electronic Engineering
Bode diagrams
• Straight-line approximations
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode diagrams
• Creating more detailed Bode diagrams
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode diagrams
• The effects of several stages ‘add’ in bode diagrams
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Bode diagrams
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Example
For the circuit shown, obtain the transfer function Vo(ω)/Vi(ω).
Identify the type of filter the circuit represents and determine the
cut-off frequency. Take R1 = 100 & = R2, L = 2 mH
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Pass-band Filters
• Simple LC filters can be
produced using series or
parallel tuned circuits
– these produce
narrow-band filters
with a centre
frequency fo
fo 
1
2 LC
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Example
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Generalities:
The standard configuration of a two port:
I1
+
V1
_
Input
Port
I2
The Network
Output
Port
The network ?
The voltage and current convention ?
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
+
V2
_
Two Port Networks
Network Equations:
Impedance
Z parameters
Admittance
Y parameters
Transmission
A, B, C, D
parameters
V1 = z11I1 + z12I2
V2 = b11V1 - b12I1
V2 = z21I1 + z22I2
I2 = b21V1 – b22I1
I1 = y11V1 + y12V2
V1 = h11I1 + h12V2
I2 = y21V1 + y22V2
Hybrid
H parameters
I2 = h21I1 + h22V2
V1 = AV2 - BI2
I1 = g11V1 + g12I2
I1 = CV2 - DI2
V2 = g21V1 + g22I2
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Z parameters:
V
z  1
11 I
1
V
z  1
12 I
2
z
V
 2
21 I
1
V
z  2
22 I
2
I 0
2
I 0
1
I 0
2
I 0
1
z11 is the impedance seen looking into port 1
when port 2 is open.
z12 is a transfer impedance. It is the ratio of the
voltage at port 1 to the current at port 2 when
port 1 is open.
z21 is a transfer impedance. It is the ratio of the
voltage at port 2 to the current at port 1 when
port 2 is open.
z22 is the impedance seen looking into port 2
when port 1 is open.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Z parameters:
Example 1
Given the following circuit. Determine the Z parameters.
I1
8
I2
10 
+
V1
+
20 
_
20 
V2
_
Find the Z parameters for the above network.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Z parameters:
Example 1 (cont 1)
For z11:
For z22:
Z11 = 8 + 20||30 = 20 
Z22 = 20||30 = 12 
I1
For z12:
8
I2
10 
+
V
z  1
12 I
2
V1
I 0
1
20 xI2 x 20
V1 
 8 xI2
20  30
+
20 
20 
V2
_
Therefore:
BASIC ELECTRONIC ENGINEERING
_
8 xI2
z12 
8
I2
Department of Electronic Engineering

=
z 21
Two Port Networks
Z parameters:
Example 1 (cont 2)
The Z parameter equations can be expressed in
matrix form as follows.
V1   z11
V    z
 2   21
z12   I 1 



z 22   I 2 
V1  20 8   I 1 
V   
I 

 2   8 12  2 
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Z parameters:
Example 2 (problem 18.7 Alexander & Sadiku)
You are given the following circuit. Find the Z parameters.
I1
I2
4
1
+
+
+
V1
1
2
Vx
-
V2
2Vx
_
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
_
Two Port Networks
Z parameters:
V
z  1
11 I
1
Example 2 (continue p2)
I1
I 0
2
+
;
V1
2
Substituting gives;
V1
5
 z11  
or
I1
3
BASIC ELECTRONIC ENGINEERING
1
Vx
-
V2
2Vx
_
but V x  V1  I 1
3V1  I 1 
I1 
2
+
+
V
V  2V x
6V x  V x  2V x
I1  x  x

1
6
6
3V x
I1 
2
I2
4
1
Department of Electronic Engineering
Other Answers
Z21 = -0.667 
Z12 = 0.222 
Z22 = 1.111 
_
Two Port Networks
Modifying the two port network:
Earlier we found the z parameters of the following network.
I1
8
I2
10 
+
V1
+
20 
20 
_
V2
_
V1   20 8   I 1 
V    8 12  I 
  2
 2 
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Modifying the two port network:
We modify the network as shown be adding elements outside the two ports
6
+
10 v
_
I1
8
I2
10 
+
V1
+
20 
20 
_
We now have:
_
V1 = 10 - 6I1
V2 = - 4I2
BASIC ELECTRONIC ENGINEERING
V2
Department of Electronic Engineering
4
Two Port Networks
Modifying the two port network:
We take a look at the original equations and the equations describing
the new port conditions.
V1   20 8   I 1 
V    8 12  I 
  2
 2 
V1 = 10 - 6I1
V2 = - 4I2
So we have,
10 – 6I1 = 20I1 + 8I2
-4I2 = 8I1 + 12I2
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Modifying the two port network:
Rearranging the equations gives,
 I1  
I   8
 2 
26

16 

8
1
10 
 0 
 I1 
0.4545 
 I   -0.2273 
 
 2
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Y parameters:
I
y  1
11 V
1
I
y  1
12 V
2
y
I
 2
21 V
1
I
y  2
22 V
2
V 0
2
V 0
1
V 0
2
V 0
1
y11 is the admittance seen looking into port 1
when port 2 is shorted.
y12 is a transfer admittance. It is the ratio of the
current at port 1 to the voltage at port 2 when
port 1 is shorted.
y21 is a transfer impedance. It is the ratio of the
current at port 2 to the voltage at port 1 when
port 2 is shorted.
y22 is the admittance seen looking into port 2
when port 1 is shorted.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Going From Y to Z Parameters
For the Y parameters we have:
For the Z parameters we have:
V Z I
I Y V
From above;
1
V Y I  Z I
Therefore
Z  Y 1
z 
z
11
12
 
 
z
z
 21 22 
BASIC ELECTRONIC ENGINEERING
y 
 y
22
12




 Y
Y 
 y

y
11 
 21
 
 
 Y
Y 
Department of Electronic Engineering
where
Y  detY
Two Port Networks
Y Parameters and Beyond:
Given the following network.
I1
+
V1
I2
1
+
1
s
s
_
V2
_
1
(a) Find the Y parameters for the network.
(b) From the Y parameters find the z parameters
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Y Parameter Example
I
y  1
11 V
1
I1 = y11V1 + y12V2
I
y  1
12 V
2
V 0
2
V 0
1
I2 = y21V1 + y22V2
I1
+
V1
y
I2
1
I
 2
21 V
1
y
V 0
2
I
 2
22 V
2
+
1
s
s
_
V2
_
short
1
To find y11
2
s ) I  2 
V1  I 1 (
1
21 s
 2 s  1 
BASIC ELECTRONIC ENGINEERING
so
We use the above equations to
evaluate the parameters from the
network.
I
y  1
11 V
1
V 0
2
Department of Electronic Engineering
=
s + 0.5
V 0
1
Two Port Networks
Y Parameter Example
I1
y
I
 2
21 V
1
V 0
2
+
V1
I2
1
+
1
s
s
_
1
We see
V1   2I 2
BASIC ELECTRONIC ENGINEERING
I
y  2
21 V
1
= 0.5 S
Department of Electronic Engineering
V2
_
Two Port Networks
Y Parameter Example
I1
To find y12 and y21 we reverse
things and short V1
I
y  1
12 V
2
+
V1
+
1
s
s
_
1
V 0
1
We have
y
I
 2
22 V
2
V 0
1
We have
V2   2I1
I
y  1
12 V 2
short
I2
1
= 0.5 S
2s
V2  I 2
( s  2)
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
1
y22  0.5 
s
V2
_
Two Port Networks
Y Parameter Example
Summary:
Y
=
 y11
y
 21
y12   s  0.5
 0.5 


y22    0.5 0.5  1 s 
Now suppose you want the Z parameters for the same network.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Transmission parameters (A,B,C,D):
Example
Given the network below with assumed voltage polarities and
Current directions compatible with the A,B,C,D parameters.
I1
-I2
We can write the following equations.
V1 = (R1 + R2)I1 + R2I2
V2 = R2I1 + R2I2
+
V1
+
R1
R2
_
It is not always possible to write 2 equations in terms of the V’s and I’s
Of the parameter set.
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
V2
_
Two Port Networks
Transmission parameters (A,B,C,D):
Example (cont.)
V1 = (R1 + R2)I1 + R2I2
V2 = R2I1 + R2I2
From these equations we can directly evaluate the A,B,C,D parameters.
V1
A
V2
C
I1
V2
I2 = 0
I2 = 0
=
=
R1  R2
R2
1
R2
V1
B
 I 2 V2 = 0
=
I1
D
 I2
=
V2 = 0
R1
1
Electronic Engineering
BASIC
ELECTRONIC
ENGINEERING
Later we
will
see how to
interconnect Department
two of ofthese
networks together for a final answer
Two Port Networks
Transmission parameters (A,B,C,D):
The defining equations are:
V1   A B   V2 
I   
 I 

 1  C D  2 
V1
A
V2
I1
C
V2
I2 = 0
V1
B
 I2
V2 = 0
I2 = 0
I1
D
 I2
V2 = 0
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Hybrid Parameters:
The equations for the hybrid parameters are:
V1   h11
 I   h
 2   21
V1
h11 
I1
I2
h21 
I1
V2 = 0
V2 = 0
BASIC ELECTRONIC ENGINEERING
h12   I 1 
h22  V2 
V1
h12 
V2
I2
h22 
V2
I1 = 0
I1 = 0
Department of Electronic Engineering
Two Port Networks
Hybrid Parameters:
I1
I2
K1
+
V1
The following is a popular model used to represent
a particular variety of transistors.
+
+
K2V2
_
K3V1
K4
_
V2
_
We can write the following equations:
V1  AI 1  BV2
V2
I 2  CI 1 
D
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Two Port Networks
Hybrid Parameters:
V1  AI 1  BV2
I 2  CI 1 
V2
D
We want to evaluate the H parameters from the above set of equations.
V1
h11 
I1
I2
h21 
I1
=
K1
V2 = 0
=
K3
V2 = 0
BASIC ELECTRONIC ENGINEERING
V1
h12 
V2
I2
h22 
V2
Department of Electronic Engineering
=
K2
I1 = 0
=
I1 = 0
1
K
4
Two Port Networks
Another example with hybrid parameters.
Hybrid Parameters:
Given the circuit below.
I1
-I2
+
The equations for the circuit are:
+
R1
V1
R2
V2
_
_
V1 = (R1 + R2)I1 + R2I2
V2 = R2I1 + R2I2
The H parameters are as follows.
h11 
V1
I1
V2=0
=
R1
I2
=
-1
BASIC
ENGINEERING
I 1 ELECTRONIC
V =0
h21 
2
h12 
V1
V2
=
1
I2
h

=
22
Department of Electronic
VEngineering
2
I1=0
1
R2
I1=0
Two Port Parameter Conversions:
BASIC ELECTRONIC ENGINEERING
Department of Electronic Engineering
Interconnection Of Two Port Networks
Three ways that two ports are interconnected:
*
Y parameters
ya
Parallel
 y ya  
yb
*
Z parameters
z  za  
za
Series
yb 
zb
zb 
ABCD parameters
* Cascade
BASIC ELECTRONIC ENGINEERING
Ta
Tb
Department of Electronic Engineering
T  Ta  Tb 
Interconnection Of Two Port Networks
Consider the following network:
I1
Find
V2
V1
R1
I2
R1
+
+
V1
_
 R1  R2

V1   R
2
I   
 1  1
 R2
T1
R2
T2
 R  R
2
R1   1
  R2
 1
1 
  R2
BASIC ELECTRONIC ENGINEERING
R2

R1  V
 2 



I

1  2


Department of Electronic Engineering
V2
_
Interconnection Of Two Port Networks
 R1  R2

V1   R
2
I   
 1  1
 R2
 R  R
2
R1   1
  R2
 1
1 
  R2

R1  V


 2 
 I2 


1


Multiply out the first row:
  R  R  2 R 

 R  R 

2   1  V   1
2  R  R  (  I )
V1    1
1
2 
  R

 R
 1
 2
R




2
2
2


 

Set I2 = 0 ( as in the diagram)
Can be verified directly
by solving the circuit
BASIC ELECTRONIC ENGINEERING
V2
V1

R2 2
R1  3 R1 R2 R2 2
2
Department of Electronic Engineering