Linear Circuits 2
Assoc. Prof., Dr. Sc. Trần Hoài Linh
School of Electrical and Electronics Engineering
linh.tranhoai@hust.edu.vn, thlinh2000@yahoo.com
Subject contents
• Slides + video: Videos from Dr. Nguyen Cong Phuong
• Midterm test and Final test: 3 questions (9 points) + 1 point for presentation
• Text book(s): Linear Circuit 2 (Tran Hoai Linh, HUST, 2023)
Printed version: https://nxbbachkhoa.vn/linear-circuit-2-b11141.html
Online version: 3 volumes
https://nxbbachkhoa.vn/linear-circuit-2-vol-1-b11205.html
https://nxbbachkhoa.vn/linear-circuit-2-vol-2-b11206.html
https://nxbbachkhoa.vn/linear-circuit-2-vol-3-b11207.html
1
Subject contents
1. Sinusoids and Phasors
2. Sinusoidal Steady-State Analysis
3. AC Power Analysis
4. Three-phase Circuits
5. Magnetically Coupled Circuits
6. The Laplace Transform
7. Frequency Response
8. Two-port Networks
1. Sinusoids and Phasors
1.1. Sinusoids
1.2. Complex Numbers
1.3. Phasors
1.4. Phasor Relationships for Circuit Elements
2
1.1. Sinusoids
A sinusoidal function (a sinusoid)
g (t )  Gm sin t   
where:
Gm – the amplitude
ω – the angular frequency (in radians/s)
Φ(t) = ωt + θ – the phase
Φ(0) = θ – the initial phase
The frequency:
The period:

(Hz)
2
2 1
T
 (s)

f
f 
1.1. Sinusoids
v1 and v2 are in phase
v1 leads v2
v1 lags v2
Note: American text books use cosine as the preferred default, Vietnamese text books use sine.
3
1.1. Sinusoids
• RMS (or Effective) value of a sinusoid:
g (t )  Gm sin t     Grms 
Gm
2
• Linear algebraic operations on sinusoids:

A
A sin t   B cos t   A2  B 2 
sin t  
2
2
 A B

cos t  
A2  B 2

B
 A2  B 2 sin t   
where: cos =
or: tan =
A
A B
2
2
, sin 
B
A  B2
2
B
A
1.1. Sinusoids
• Linear algebraic operations on sinusoids:
A sin t     B sin t     A sin t  cos   cos t  sin    B sin t  cos   cos t  sin  
  A cos   B cos   sin  t    A sin   B sin   cos t 
 C sin t   D cos t 
 ...
Note:
- Similar operations can be performed for cosine functions.
- Don’t consider operations for sinusoids of different frequencies.
- Graphic operations using vectors are good at visualization but not at accuracy.
Remark: Using complex numbers can greatly help with algebraic operations on sinusoid.
4
1.2. Complex numbers
• Imaginary number:
j  1
Note: Official mathematical symbol is “i” but in Circuit Theory “I” is preferred for currents
• Rectangular form of a complex number:
z  x j y
x  Re  z  ; y  Im  z 
• Polar form of a complex number:
z  r  e j  r
r  z  magnitude  z  ;   z  phase  z 
Graphical representation of
a complex number
1.2. Complex numbers
• Euler’s form:
e j  cos   j  sin 
• Operation on complex numbers using calculator:
 Addition, Subtraction, Multiplication, Division,…
 Conversion between two forms
 Not directly supported by the device: Solve a system of equations
for complex variables
Note:
- Later versions of Casio are also good.
- Other calculators available for supporting complex number operations.
Casio fx-570ES/
fx-570ES Plus
5
1.3. Phasors
• Phasor is an operator on a sinusoid using complex number:
P  V  P v (t )  Vm  e j  Vm   V 
v  t   Vm sin t    
 
rms
2
2
• Sinor is the full representation of a sinusoid using complex number :
S S v(t )  Vm  e j t  
v  t   Vm sin t    
 
2
Note:
- Circuit Theory mostly use Phasors!
- Phasors don’t contain information on ω (since we considers each frequency at a time.
- American standard uses magnitude, not RMS value for phasor.
- Different notations of phasor (from Dr. N.C.Phuong: P(v(t)) = V)
1.3. Phasors
• Phasor is a linear operator:
 n

n
n
 i 1

i 1
i 1
P   ai vi (t )    ai P  vi (t )    aiVi
• Phasor of a derivation:
 dv 
if P  v(t )   V then P     j   V
 dt 
• Phasor of an integration:


1 
if P  v (t )   V then P  v  t  dt 
V
j
6
1.3. Phasors
• Using phasor to perform algebraic operation on sinusoids:
i4 (t )  i1 (t )  i2 (t )  i3 (t )  2sin  5t  20   3sin  5t  60   5sin  5t  15 


2
3
5
8.446
I4  I1  I2  I3 
20 
60 
  15 
13.60
2
2
2
2
i4 (t )  8.446sin  5t  13.60 
1.4. Phasors Relationships for Circuit Elements
• General idea:
Convert the time-domain relationship between voltage and current signals to
the relationship between the phasors of voltage and current signals


f  u (t ), i (t )   0  F U , I  0
Example:
u (t )  A  i (t )  B
di
 C  i  dt  U  ...
dt
7
1.4. Phasors Relationships for Circuit Elements
• Independent voltage source:

i (t ) :
I
uba (t )  e(t )
U  E
u (t ) :
U
iab (t )  j (t )
I  J
ba
• Independent current source:

ab
1.4. Phasors Relationships for Circuit Elements
• Dependent voltage sources:
8
1.4. Phasors Relationships for Circuit Elements
• Dependent current sources:
1.4. Phasors Relationships for Circuit Elements
• Resistor:

i  t   I m sin t     u  t   R  i  t   RI m sin t   
U RI m

 R or U  R  I
I 
I
m
• Capacitor:
du
 CU m sin t    90 
dt
U m 
U
1
1
1 

I  CU   90  C  90 = jC or U  jC  I
m
u  t   U m sin t     i  t   C

9
1.4. Phasors Relationships for Circuit Elements
• Inductor:
i  t   I m sin  t     u  t   L

di
  LI m sin t    90 
dt
U  LI m  90

  L90  j L
I m
I
or
U  j L  I
• Inductors with mutual coupling effect: (will get back in Chapter 5)
1.4. Phasors Relationships for Circuit Elements
10
1. Sinusoids and Phasors
1.1. Sinusoids
1.2. Complex Numbers
1.3. Phasors
1.4. Phasor Relationships for Circuit Elements
2. Sinusoidal Steady-State Analysis
2.1. Introduction
2.2. Branch currents method
2.3. Nodal voltages method
2.4. Loop currents analysis
2.5. Circuit analysis using superposition theorem
2.6. Thévenin and Norton equivalent circuits
2.7. Sources transformation
2.8. Op-Amp AC circuits
11
2.1. Introduction
• Circuit Analysis:
 Input: A circuit schematic and the values of all elements.
 Output: Find all the signals (voltage, current and power)
• Circuit Synthesis: propose a circuit schematic (or circuit elements’ values) to achieve
expected outputs or properties.
• Known from Linear Circuit 1: Basic circuit analysis methods are based on Kirchhoff’s
system of equation. To analyze a circuit:
 Identify the state variables
 From the system of equations (based on Kirchhoff equations)
 Solve the system to find the state variables (using math tools)
 Find circuit’s signals from the state variables (using Kirchhoff equations and elements’
characteristic equations)
2.1. Introduction
• Example of an AC circuit: How many Kirchhoff equations? How many KCL, how many
KVL equations?
12
2.1. Introduction
• Example of an AC circuit: 4 Kirchhoff equations
i1 (t )  i2 (t )  i3 (t )

i3 (t )  j4 (t )  i5 (t )

u R1 (t )  uL 2 (t )  e1 (t )  0
uC 3 (t )  u R5 (t )  u L 2 (t )  0
2.1. Introduction
• Example of an AC circuit: Converting K’s equations to equation for branch currents:
i1 (t )  i2 (t )  i3 (t )

i3 (t )  j4 (t )  i5 (t )

di
 R1  i1 (t )  L2 2  e1 (t )  0
dt


di
  i3 (t )  dt  R5  i5 (t )  L2 2  0
dt

13
2.1. Introduction
• Example of an AC circuit: Transforming to phasors: Phasor of circuit ↔ phasor of
system of equations
i1 (t )  i2 (t )  i3 (t )

i3 (t )  j4 (t )  i5 (t )

u R1 (t )  uL 2 (t )  e1 (t )  0
uC 3 (t )  u R 5 (t )  u L 2 (t )  0
 I1  I2  I3

 I3  J 4  I5

U R1  U L 2  E 1  0



U C 3  U R 5  U L 2  0
2.1. Introduction
• Very important: The similarity in structure and equations’ formula will lead to the
similarity in circuit analysis methods!
 I1  I 2  I 3

 I3  J 4  I5

U R1  U R 2  E1  0
U R 3  U R 5  U R 2  0
 I1  I2  I3

 I3  J 4  I5

U R1  U L 2  E 1  0
U  U  U  0
R5
L2
 C3
14
2.2. Branch currents method
• What the main idea of the method?
 I1  I 2  I 3

 I3  J 4  I5

U R1  U R 2  E1  0
U R 3  U R 5  U R 2  0
 I1  I2  I3

 I3  J 4  I5

U R1  U L 2  E 1  0
U  U  U  0
R5
L2
 C3
2.2. Branch currents method
• What the main idea of the method?
 I1  I 2  I 3

 I3  J 4  I5

U R1  U R 2  E1  0
U R 3  U R 5  U R 2  0

 I1  I 2  I 3  0

 I3  I5   J 4

 R1  I1  R2  I 2  E1
  R2  I 2  R3  I 3  R5  I 5  0
15
2.2. Branch currents method
• What the main idea of the method?
 I1  I2  I3

 I3  J 4  I5

U R1  U L 2  E 1  0



U C 3  U R 5  U L 2  0

 I1  I2  I3  0

 I3  I5   J 4

 R1  I1   j L2   I2  E 1

  j L   I  1 I  R  I  0
2
2
3
5
5

jC3

• Solve the system to achieve all branch currents
→ Find all the voltages
→ Find all the powers (to be discussed later)
2.3. Nodal voltage method
• What the main idea of the method?
 KCL equations:
 I1  I2  I3

 I 3  J 4  I5
 Transform branch currents as functions
of nodal voltages:
V  E 1 
V
V  Vb  Vb
I1  a
; I 2  a ; I3  a
; I5  .
R1
Z L2
ZC 3
R5
 Substitute back to KCL equations:

 1
E
1
1  
1 

 Vb  1
 
  Va 
ZC 3
R1
 R1 Z L 2 Z C 3 

1  
 1   1

  Z  Va   Z  R   Vb  J 4
C
3

C
3
5


16
2.3. Nodal voltage method
Example: Values of the elements are:
E 1  12 20 V; J 4  1  15 A; R1  5 ;
Z L 2  j 3 ; Z C 3   j 5 ; R5  8 .
• Substitute into the equation:
 1
E
1
1  
1 

 Vb  1
 
  Va 
ZC 3
R1
 R1 Z L 2 ZC 3 

1  
 1   1

 Z  Va   Z  R   Vb  J 4
C
3
C
3
5




Va  8.096 62.93

Vb  2.860 76.88
VR1  Va  E 1  8.205  22.25

VL 2  Va  8.096 62.93



VC 3  Va  Vb  5.36555.55
VR 5  Vb  2.860 76.88

 I1  1.641  22.25

 I 2  2.699  27.07

 I 3  1.073145.55
 I5  0.358 76.88

2.3. Nodal voltage method
Example 2.3 from Text book: (Circuit with dependent source(s))
E 1  100 V; R1  10 ; J 2  1.5 20 A;
Z C 3   j 5 ; Z L 5  j 7 ; k  0.5.
17
2.3. Nodal voltage method
Example 2.4 from Text book: (Circuit with nodes with known voltages)
E 1  12 20 V; R2  5 ; Z L3  j 3 ; Z C 4   j 5 ;
J  1  15 A; Z   j8 ; Z  j 4 .
5
C6
L6
2.4. Loop currents method
• What the main idea of the method?
 KVL equations → branch currents equatipns:
 R1  I1  Z L 2  I2  E 1

 Z C 3  I3  R5  I5  Z L 2  I2  0
 Transform branch currents as functions of loop
currents (unknown and discharging):
I  I ; I  I  I ; I  I ; I  I  J .
1
a
2
a
b
3
b
5
b
4
 Substitute back to KVL-based equations:

 R1  Z L 2   Ia  Z L 2  Ib  E 1

  Z L 2  Ia   ZC 3  R5  Z L 2   Ib   R5  J 4
18
2.4. Loop currents method
Example: Values of the elements are:
E 1  12 20 V; J 4  1  15 A; R1  5 ;
Z L 2  j 3 ; Z C 3   j 5 ; R5  8 .
• Substitute into the equation:
 5  j 3  Ia  j 3  Ib  12 20

 j 3  Ia   8  j 2   Ib  8  15
 Ia  1.641  22.25

 I b  1.073145.55

 I1  Ia  ...

 I 2  Ia  Ib  ...


 I 3  I b  ...
 I5  Ib  J 4  ...

VR1  R1  I1  ...

VL 2  Z L 2  I2  ...


VC 3  ZC 3  I 3  ...
VR 5  R5  I5  ...

2.5. Circuit analysis using superposition theorem
Superposition theorem: In linear circuits having multiple independent sources, the
response of an element equals to the algebraic sum of the responses of that element by
considering one independent source at a time (other independent sources are set to 0 at
that time).
i3 (t )  i31 (t ) e only  i32 (t ) j only  i35 (t ) e only
1
P

 I3  I31
e1 only
2
 I32
j2 only
 I35
5
e5 only
Let circuit’s elements are:
E 1  120 V; J 2  1.530 A; E 5  15  30 V;
Z1  5  j 4 ; Z 3  10  j 2 ; Z 4   j 5 ; Z5  2  j1 ;
19
2.5. Circuit analysis using superposition theorem
Example: E 1  120 V; J 2  1.530 A; E 5  15  30 V;
Z1  5  j 4 ; Z 3  10  j 2 ; Z 4   j 5 ; Z5  2  j1 ;
Component I31:
Z eq1   Z 4  Z 5   Z3  Z1  17.5  j 6

E
I31  1  0.649  18.92
Z eq1
2.5. Circuit analysis using superposition theorem
Example: E 1  120 V; J 2  1.530 A; E 5  15  30 V;
Z1  5  j 4 ; Z 3  10  j 2 ; Z 4   j 5 ; Z5  2  j1 ;
Component I32:
Z eq 2   Z 4  Z5   Z 3  12.5  j 2

I32 
Z1
J 2  0.519 49.74
Z1  Z eq 2
20
2.5. Circuit analysis using superposition theorem
Example: E 1  120 V; J 2  1.530 A; E 5  15  30 V;
Z1  5  j 4 ; Z 3  10  j 2 ; Z 4   j 5 ; Z5  2  j1 ;
Component I35:
Z eq 5   Z1  Z 3   Z 4   Z5  3.659  j 4.111

E
I55  5  2.72618.32
Z eq 5

I35 
Z4
I55  0.907104.51
Z 4   Z1  Z 3 
Sum up all components:
I3  I31  I32  I35  1.28655.83

i3 (t )  1.286 2 sin t  55.83  (A)
2.5. Circuit analysis using superposition theorem
Example: (With dependent source(s))
E 1  120 V; J 2  1.530 A; J 4  0.7 I1 A;
Z1  5  j 4 ; Z3  10  j 2 ; Z 5  2  j1 ;
21
2.6. Thevenin and Norton equivalent circuits
Thevenin theorem: Any linear circuit with two terminals can be replaced by a single voltage
source and a single series impedance Zab.
Norton theorem: Any linear circuit with two terminals can be replaced by a single current source
and a single parallel impedance Zab.
E Th  U ab-open ;
J N  Iab-short ;
E Th  Z ab  J N
2.6. Thevenin and Norton equivalent circuits
Example: Find the equivalent Thevenin and Norton circuits seen on two nodes a and b
E1  12 20 V; J 4  1  15 A;
R1  5 ; Z L 2  j 3 ; ZC 3   j 5 ; R5  8.
E Th  U ab -open ;
Z ab 
E Th
J
J N  Iab-short ;
N
or Z ab all independent sources are set to 0
or
E ext
Iext all independent sources are set to 0
22
2.6. Thevenin and Norton equivalent circuits
Example: Find the equivalent Thevenin and Norton circuits seen on two nodes a and b
E 1  120 V; J 2  1.530 A; J 4  0.7 I1 A;
R1  5 ; Z C 3   j 2 .
E Th  U ab -open ;
Z ab 
E Th
J N
J N  Iab -short ;
or Z ab all independent sources are set to 0
or
E ext
I
ext all independent sources are set to 0
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
23
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V
J1  1
R1
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V
J12  J1  J 2  1  J 2
R1
24
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V4  Z C 4  J 3
 V

V12 = R1  J12  R1   1  J 2   V1  R1  J 2
 R1

2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
Z14  R1  ZC 4
V124 = V12 +V4  V1  R1  J 2  ZC 4  J 3
25
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V  R1  J 2  Z C 4  J 3
V
J124 = 124  1
Z14
R1  ZC 4
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V  R1  J 2  Z C 4  J 3
V
J124 = 124  1
Z14
R1  ZC 4
26
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V1246  Z14  J1246
J1246 = J124  J 6
2.7. Sources transformation
The equivalence of Thevenin and Norton circuits can be used to transform, simplify circuits.
Example: Find I5 from the given circuit.
V  V1246
I5  5
R5  Z 14
V5  V1  R1  J 2  Z C 4  J 3   R1  Z C 4   J 6

R5  R1  ZC 4


27
2.8. OP_AMP AC Circuit
• The phasor model of the OP-AMP:
i  (t )  0; i  (t )  0  I   0; I   0




va (t )  vb (t )
Va  Vb
• Primary method for analyzing OPAMP circuit is ……
2.8. OP_AMP AC Circuit
Example: Find the transfer function of the given circuit.
• From KCL for node (b):
V  Vb Vb
R3
IR 4  IR 3  out

 Vb 
Vout  Va
R4
R3
R3  R4


• From KCL for node (a):
V  Va Va
R  ZC 2 
IC 2  IR1  d

 Vd  1
Va
ZC 2
R1
R1
• From KCL for node (d):

 1
V  Vs Vd  Vout
V  0
V
1
1  Vout
IC1  IR 2  IC 2  d

 d
 s  Vd 



ZC 2
R2
R1  Z C 2
ZC 2
 R2 R1  Z C 2 Z C 2  R2
Vout
 ...
V
s
28
2.8. OP_AMP AC Circuit
Example: Find the transfer function of the given circuit.
2.8. OP_AMP AC Circuit
Example: Find the transfer function of the given circuit.
29
2. Sinusoidal Steady-State Analysis
2.1. Introduction
2.2. Branch currents method
2.3. Nodal voltages method
2.4. Loop currents analysis
2.5. Circuit analysis using superposition theorem
2.6. Thévenin and Norton equivalent circuits
2.7. Sources transformation
2.8. Op-Amp AC circuits
3. AC Power Analysis
3.1. Instantaneous and average power
3.2. Apparent power and power factor
3.3. Complex power
3.4. Conservation of AC powers
3.5. Power factor correction
3.6. Maximum power transfer
30
3.1. Instantaneous and average power
• Instantaneous power for sinusoid:
p (t )  u (t )  i (t ) (W)
Let the sinusoids are: u  t   U m sin  t    ;
then:
i  t   I m sin  t   
p  t   u  t   i  t   U m sin t     I m sin t   
1
1
 U m I m cos      U m I m cos  2t     
2
2

Constant + AC with double frequency
In Linear circuit , we are interested mostly in average value of instantaneous power:
1
p  t   Paverage  U m I m cos      U rms I rms cos     (W)
2
3.1. Instantaneous and average power
• Average power for sinusoid:
P  U rms I rms cos  U  I  (W)
Remark: Pay attention for the directions of the voltage and the current and the power
absorption and/or generation.
Example for basic load:
 Resistors:

U2 
U  I  0  P  U rms I rms   R 2  I rms  rms 

R 

 Capacitors: U  I  90  P  0
 Inductors:
U  I  90  P  0
31
3.2. Apparent power and power factor
• Average power for sinusoid:
P  U rms I rms cos  U  I 
  Apparent Power    Power Factor 
New terms:
 Apparent power:
S  U rms I rms (VA)
 Power factor:
pf  cos  U  I 
 Power factor angle:   U  I  arccos  pf 


Note: For a load, the power factor angle is also the angle of the load’s impedance!
3.3. Complex power
• To calculate P from the phasors of voltage and current:
U  U rms  ; I  I rms   P  U  I  cos    
• Another approach to calculate P:
U  U  ; I  I 

*
rms
rms
U  I  U rms   I rms      U rms  I rms      
 U rms  I rms   e 
j   
 U rms  I rms   cos      j  U rms  I rms   sin    

S
 P  j  U rms  I rms   sin    
 P  j Q
32
3.3. Complex power
• Another approach to calculate P :

U  I*  P  j  U rms  I rms   sin    
S
 P  j Q
New terms:
 Complex power: S  U  I*
 Reactive power: Q  U rms  I rms   sin    
  (VAR)
 Im S
→ New dependencies:
S  U rms  I rms = S
S  P2  Q2
3.3. Complex power
• Reactive power of basic loads:
Q  U rms I rms sin  U  I  (VAR)
 Resistors:
U  I  0  Q  0
 Capacitors:
2
I
2
U  I  90  QC  U rms I rms  C  U rms
  rms
C
 Inductors:
2
U
2
U  I  90  QL  U rms I rms  rms   L  I rms
L
33
3.4. Conservation of AC powers
• Conservation for P :

 pgenerated (t )   pabsorb (t )
 Pgenerated   Pabsorb
Can be extended for reactive powers, complex powers but NOT for apparent powers:
 Qgenerated   Qabsorb
 Sgenerated   Sabsorb
 S generated   Sabsorb
3.4. Conservation of AC powers
Example: (See chapter 2) Values of the elements:
E 1  12 20 V; J 4  1  15 A; R1  5 ;
Z L 2  j 3 ; Z C 3   j 5 ; R5  8 .
 I1  1.641  22.25

 I 2  2.699  27.07

 I 3  1.073145.55
 I5  0.358 76.88

VR1  8.205  22.25

VL 2  8.09762.93

VC 3  5.36555.55
VR 5  2.864 76.88

34
3.5. Power factor correction
Average power:
P  U rms I rms  pf
In practice, Urms is fixed:

I rms 
P
U rms  pf
→ When P and Urms are fixed, a big pf means smaller currents!


pf  cos U  I  1

to reduce the phase difference between voltage and current!
3.5. Power factor correction
• Compensating elements are connected in parallel with the loads
• Compensating elements should have resistance = 0 → capacitive or inductive
elements are used.
• Usually the compensating element has opposite power factor angle comparing with
the load.
35
3.5. Power factor correction
• Same solution presented using P, Q and
the complex power.
Q  Q1  Q2  P (tan 1  tan  2 )
3.5. Power factor correction
• Same solution presented using P, Q and the complex power.
Q  Q1  Q2  P (tan 1  tan  2 )
Note: Formula for reactive power of capacitors and inductors!
C:
2
Q  C  U rms

Ccomp 
Q
2
  U rms
L:
Q
2
U rms
L

Lcomp 
2
U rms
  Q
36
3.6. Maximum power transfer
Standard problem: For a given circuit and a selected Zload. Find the value of Zload to
receive maximum power from the circuit.
Only 1 load is considered → the rest of the circuit can be replaced with Thevenin
equivalent circuit (similar results can be achieved with Norton circuit)
3.6. Maximum power transfer
Let’s the impedances of the circuit and the load are: ZTh  RTh  jX Th ; Zload  RL  jX L
The current through the load:
E Th
E Th
E Th
I 


Z ab  Z load ( Rab  jX ab )  ( RL  jX L ) ( Rab  RL )  j ( X ab  X L )
The average power absorbed by the load:
2
RL  E Th
2
P  RL  I 
( Rab  RL ) 2  ( X ab  X L ) 2
37
3.6. Maximum power transfer
The average power absorbed by the load:
P  RL  I 
2
2
RL  E Th
( Rab  RL )2  ( X ab  X L ) 2
The general case:
2
2
2
( Rab  RL )  ( X ab  X L )  ( Rab  RL )  4  Rab  RL  P 
2
E Th
4  Rab
→ Pmax can be achieved when Rab = RL and Xab + XL = 0 → ZL = Zab*
3.6. Maximum power transfer
The average power absorbed by the load:
2
P  RL  I 
2
RL  E Th
( Rab  RL )2  ( X ab  X L ) 2
Homework: We have only resistive load, i.e. XL ≡ 0. What value of RL gives us
maximum power?
38
3. AC Power Analysis
3.1. Instantaneous and average power
3.2. Apparent power and power factor
3.3. Complex power
3.4. Conservation of AC powers
3.5. Power factor correction
3.6. Maximum power transfer
4. Three-phase circuits
4.1. Three-phase sources and three-phase loads
4.2. Three-phase circuit analysis
4.3. Balanced three-phase circuits
4.4. Unbalanced three-phase system and the dynamic loads
39
4.1. Three-phase sources and three-phase loads
The three-phase sources: Wye-connected (star) or Delta-connected (triangle)
Ideal sources
Non-ideal sources
Note: For ideal sources connected in delta, KVL requires that
t :
e AB (t )  eBC (t )  eCA (t )  0
4.1. Three-phase sources and three-phase loads
The three-phase loads: Wye-connected (star) or Delta-connected (triangle)
Three-phase loads
40
4.2. Three-phase circuits analysis
Combinations of three-phase sources and loads give different configurations.
YY circuit
Y-Y circuit
YΔ circuit
ΔΔ circuit
4.2. Three-phase circuits analysis
Example of Y-Y circuit:
• Define the phase impedances:
Z pA  Z wA  Z la
Z pB  Z wB  Zlb
Z pC  Z wC  Zlc
• Nodal voltage method:
 1
E
E
E
1
1
1 




 Vn  A  B  C
 Z pA Z pB Z pC Z N 
Z pA Z pB Z pC


Y  Y  Y  Y V  Y  E  Y  E  Y

 pA

Vn 
pB
pC
N
 n
pA
A
pB
B
Y-Y circuit

pC  EC
YpA  E A  YpB  E B  YpC  E C
YpA  YpB  YpC  YN
41
4.2. Three-phase circuits analysis
Example of Y-Y circuit:
• The voltage of the neutral node:
Vn 
YpA  E A  YpB  E B  YpC  E C
YpA  YpB  Y pC  YN
• The phase currents:



 I A  YpA  E A  Vn

 IB  YpB  E B  Vn

 IC  YpC  E C  Vn

 I N  YN  Vn



Y-Y circuit
4.2. Three-phase circuits analysis
Example of YΔ circuit:
• Convert the Δ load to Y load: the circuit
becomes YY.
• Solve the equivalent YY circuit.
• Solve the original circuit using known
signals from YY circuit.
YΔ circuit
42
4.3. Balanced three-phase circuits
Balanced (symmetrical) three-phase:
• Loads are identical.
• 3 transmission lines are identical
(the neutral wire can be different.
• The sources are symmetrical:
 The sources are sinusoid of the same frequency
(i.e. the same angular frequency).
 The amplitudes of the three sources are the
same.
 The phases of the three sources are shifted by
the same angle.
YΔ circuit
eA (t )  2 Erms sin  t   A   E A  Erms  A

eB (t )  2 Erms sin  t   B   E B  Erms  B


eC (t )  2 Erms sin t  C   EC  Erms C
 B   A   C   B   A   C  
4.3. Balanced three-phase circuits
Symmetrical sources:
eA (t )  2 Erms sin  t   A   E A  Erms  A

eB (t )  2 Erms sin  t   B   E B  Erms  B


eC (t )  2 Erms sin  t  C   EC  Erms C
 B   A   C   B   A   C  

YΔ circuit
a  1  e j
 E B  a  E A

 EC  a  E B


 E A  a  EC
→ There are 3 possible cases:
zero sequence
a0  10

a  1  a1  1  120 positive sequence (default)
a  1120
negative sequence
 2
3
43
4.3. Balanced three-phase circuits
Symmetrical sources:
 E B  a  E A

 EC  a  E B


 E A  a  EC
zero sequence
a0  1 0

a1  1  120 positive sequence (default)
a  1120
negative sequence
 2

i :
ai3  1

2
i  1, 2 : 1  ai  ai  0
 2
2
a1  a2 and a2  a1
(a)
(b)
(c)
Relative phasors of the sources in zero sequence (a),
positive sequence (b) and negative sequence (c)
4.3. Balanced three-phase circuits
Example of Y-Y balanced circuit:
• Nodal voltage method:
Vn 

YpA  E A  YpB  E B  YpC  E C
YpA  Y pB  YpC  YN

YpA  E A + E B  E C
 0
YpA  Y pB  YpC  YN
 I A 
E A
;
Z pA
E
IB  B  a1  I A ;
Z pB
Y-Y circuit
E
IC  C  a12  I A ;
Z pC
I N  0
44
4.3. Balanced three-phase circuits
Y-Δ conversion for balanced sources:
 E AB  E A  E B  1  1  120   E A  330  E A

 E BC  E B  E C  1  1  120   E B  a1  E AB

2 



 ECA  EC  E A  1  1  120   EC  a1  E AB
Power in balanced circuits:


E AB
EA 
330

 E  a  E
1
A
 B
 E  a 2  E
1
A
 C


PB  VB  IB  cos  VB  IB 
 V A  I A  cos   V A  120    I A  120  
 V A  I A  cos  V A  I A 
PA  V A  I A  cos V A  I A
 PA
 PA  PB  PC
Analogically:
PE A  PEB  PEC
4.4. Unbalanced three-phase system and the dynamic loads
In the case of un-balanced sources:
E A , E B , E C , E A0 , E A1 , E A2 :
E A 
E A0

E A1
E

E

a  E
B
E C
A0

E A0

Zero sequence

1
A1
2 
a1  E A1



Positive sequence



E A 2
a2  E A2
a22  E A2





E A0  E A1  E A2
E  E  E
B0
B1
B2
 E C 0  E C1  E C 2
Negative sequence
45
4.4. Unbalanced three-phase system and the dynamic loads
In the case of un-balanced sources:
E A , E B , E C , E A0 , E A1 , E A2 :
E A 
E A0

E A1
E

E

a  E
B
A0
E C
E A0



Zero sequence

 E A  1 1
  
 E B   1 a1
 E C  1 a 2
  
1
Since
1 1

1 a1
1 a 2
1

1

a12 
a1 
1
1
A1
2 
a1  E A1






Positive sequence
E A 2
a2  E A2
a22  E A2



E A0  E A1  E A2
E B 0  E B1  E B 2
 E C 0  E C1  E C 2


Negative sequence
1   E A0 
 

a12    E A1 
a1   E A 2 
1 1
1
 1 a12
3
1 a1
1

a1 
a12 
4.4. Unbalanced three-phase system and the dynamic loads
In the case of un-balanced sources:

1 1
 E A0 
   1
2
 E A1   3 1 a1
1 a
 E A2 



1

 E A0
1   E A  
   
a1    E B    E A1

a12   E C  

 E A2

1 
E A  E B  E C
3
1 

E A  a12  E B  a1  E C
3
1 

E A  a1  E B  a12  E C
3







Example: The 3-phase source has E A  22010 V; E B  210  100 V; E C  215110 V.

 E A0 


 E A1 


 E A2 

1 
E A  E B  E C
3
1 
E A  a 2  E B  a  E C
3
1 
E A  a  E B  a 2  E C
3


 37.25817.40 V

  209.9786.63 V

  27.616174.12 V
46
4.4. Unbalanced three-phase system and the dynamic loads
Dynamic loads: The load that impedance depends on the characteristic of the source.
Example: A three-phase Y-connected dynamic load has its impedance (for each phase)
when connected to a positive balanced source is Z1=40+j30, when connected to a
negative balanced source is Z2=20+j20; and when connected to a zero sequence source
is Z0=10+j5. Find the phase currents when this dynamic load is connected to the Yconnected source: E A  22010 V; E B  210  100 V; E C  215110 V.

 E A0  37.25817.40 V


 E A1  209.978 6.63 V


 E A2  27.616174.12 V

Vn  E A0
 0  IB 0  IC 0  0
Z0
E
209.978 6.63
 I A1  A1 
 4.200  30.24
Z1
40  j 30
E
27.616174.12
 I A2  A2 
 0.976129.12
Z1
20  j 20
 I A0 
4.4. Unbalanced three-phase system and the dynamic loads
Dynamic loads: The load that impedance depends on the characteristic of the source.
Example: A three-phase Y-connected dynamic load has its impedance (for each phase)
when connected to a positive balanced source is Z1=40+j30, when connected to a
negative balanced source is Z2=20+j20; and when connected to a zero sequence source
is Z0=10+j5. Find the phase currents when this dynamic load is connected to the Yconnected source: E A  22010 V; E B  210  100 V; E C  215110 V.

 I A0  0
 IB 0  IC 0  0

2
 I A1  4.200  30.24  IB1  4.200  150.24; IC1  a1  I A1  4.20089.76

2
 I A2  0.976129.12  IB 2  a2  I A 2  0.976  110.88; IC 2  a2  I A2  0.9769.12
 I A  I A0  I A1  I A2  3.305  24.26

 IB  IB 0  IB1  IB 2  4.993  143.12





 I C  I C 0  I C1  I C 2  4.46477.30
47
4. Three-phase circuits
4.1. Three-phase sources and three-phase loads
4.2. Three-phase circuit analysis
4.3. Balanced three-phase circuits
4.4. Unbalanced three-phase system and the dynamic loads
5. Magnetically coupled circuits
5.1. Mutual Inductance and Self Inductance
5.2. The Dot Conversion for Mutual Coupled Inductors
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
5.4. Power Transferred in Coupled Circuits
5.5. The Thévenin – Norton Equivalent Sources for Circuits with Magnetically Coupling
48
5.1. Mutual Inductance and Self Inductance
The total flux passing through an inductor when there is a
field from another inductor:
 1 (t )   11 (t )   21 (t )  L1  i1 (t )   21 (t )

 2 (t )   22 (t )   12 (t )  L2  i2 (t )   12 (t )
 1 (t )  L1  i1 (t )  M 21  i2 (t )


 2 (t )  L2  i2 (t )  M12 (t )  i1 (t )
 1 (t )  L1  i1 (t )  M  i2 (t )
M12  M 21  M


 2 (t )  L2  i2 (t )  M  i1 (t )
The magnetic field of inductor
L1 and the partial penetration
into inductor L2
M  mutual inductance

k
M ij
L1 L2
  1  coupling coefficient
5.1. Mutual Inductance and Self Inductance
The voltage – current relationship for inductors:
d 1
di1
di2

v1 (t )  dt  L1 dt  M dt

v (t )  d 2  L di2  M di1
2
 2
dt
dt
dt
v  0
v2  0
For DC circuits: i1 , i2  const   1
For AC circuits: Transform to phasors
di1
di2

v1 (t )  L1 dt  M dt

v (t )  L di2  M di1
2
 2
dt
dt



 V1  L1 j I1  M j I2
 V2  L2


Z L  j L



V1  Z L1  I1  Z M  I 2
Z M  j M


V2  Z L2  I2  Z M  I1
j I2  M j I1



49
5.2. The dot conversion
The voltage – current relationship for inductors:
d 1
di1
di2

v1 (t )  dt  L1 dt  M dt

v (t )  d 2  L di2  M di1
2
 2
dt
dt
dt
To facilitate the signs in the equation, each inductor having
mutual coupling effect will have one terminal marked with a dot (●) or a star (*) and:
• If the branch currents are selected both entering the dotted terminals or both leaving the
dotted terminals then the sign of mutual inductance is positive.
• If one inductor has its own current entering the dotted terminal and the other inductor has its
own current leaving the dotted terminal then the sign of mutual inductance is negative.
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
Due to the changes in voltage-current relationship, the nodal voltage method and the
equivalent impedance method are NOT preferred!
The Branch current method: Changes when converting KVL-based equations into branch
current equations.


VZ 1  VL1  V1  0


VZ 2  VL 2  0
 Z1  I1   Z L1  I1  Z M  I2   V1  0


 Z 2  I2  Z L 2  I2  Z M  I1  0
 Z1  Z L1   I1  Z M  I2  V1



 Z M  I1   Z 2  Z L 2   I 2  0


50
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
The Branch current method:
 Z1  Z L1   I1  Z M  I2  V1

  Z M  I1   Z 2  Z L 2   I2  0
Substitute the values:
V1  120 V ; Z1  5  j 3 ; Z L1  j8 ;
Z L 2  j 6 ; Z M  j 6 ; Z 2  5  j 2 .

 I1  0,999  38,57 ( A)


 I 2  0,93612,77 ( A)
VZ 1  Z1  I1  5,826  7, 61 (V )

VZ 2  Z 2  I2  5,041  34,57 (V )



VL1  Z L1  I1  Z M  I 2  6, 2737,06 (V )
V  Z  I  Z  I  5,042170,97 (V )
L2
2
M
1
 L2
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
The Branch current method:
 I1  I2  I3

VZ 1  VL1  VL 2  V1  0
V  V  0
L2
 Z3


 I  I  I
2
3
 1


 Z1  I1  Z L1  I1  Z M  I2  Z L 2  I2  Z M  I1  V1  0

 Z 3  I3  Z L 2  I2  Z M  I1  0
 I1  I2  I3  0

 Z1  Z L1  Z M   I1   Z L 2  Z M   I2  V1
  Z  I  Z  I  Z  I  0
L2
2
3 3
 M 1


 


51
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
The Branch current method:
Substitute the values: V1  120 V ; Z1  5  j 3 ; Z L1  j8 ;
Z L 2  j 6 ; Z M  j 6 ; Z3  5  j 2 .
 I1  I2  I3  0

 Z1  Z L1  Z M   I1   Z L 2  Z M   I2  V1
 Z  I  Z  I  Z  I  0
L2
2
3 3
 M 1

 I1  0, 443  33,53 ( A)

 I 2  0,653  130,19 ( A)


 I 3  0,83117,81 ( A)

VZ 1  Z1  I1  5,826  7,61 (V )

VL1  Z L1  I1  Z M  I2  6, 2737, 06 (V )



VL 2  Z L 2  I 2  Z M  I1  5,042170,97 (V )
V  Z  I  5,041  34,57 (V )
3 3
 Z3
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
The Loop current method: Only changes when converting KVL-based equations to
branch currents equations.
 Z1  Z L1  Z M   I1   Z L 2  Z M   I2  V1

 Z M  I1  Z L 2  I2  Z3  I3  0
I1  Ia ; I2  Ia  Ib ; I3  Ib



 Z1  Z L1  Z M   Ia   Z L 2  Z M    Ia  Ib   V1


 0
  Z M  Ia  Z L 2   Ia  Ib   Z 3  Ib
 Z1  Z L1  Z L 2  2Z M   Ia   Z L 2  Z M   Ib  V1

 0
   Z M  Z L 2   Ia   Z L 2  Z 3   Ib
 Ia  ...

 I b  ...
52
5.4. Power transfer in circuit with coupled inductors
The voltage – current relationship for inductors causes that their phase difference is not
90o anymore!
VL1 
j L1  I1  j M  I2



perpendicular to I1

NOT perpendicular to I1
I1  0,999  38,57 A; I2  0,93612, 77 A and
VL1  Z L1  I1  Z M  I2  6, 2737,06 V

VL 2  Z L 2  I2  Z M  I1  5,042170,97 V


PL1  Re VL1  IL*1  6, 273  0,999  cos  7.06  38.57 
 4.382 (W)


PL 2  Re VL 2  IL* 2  5.042  0.936  cos 170.97  12.77 
 4.382 (W)
5.4. Power transfer in circuit with coupled inductors
PL1  4.382 W; PL2  4.382 W
PV1  9.373 W; PZ1  4,99 W; PZ 2  PL1   PL2
 PL1  PL2  0

 PV1  PZ1  PZ 2
4.382W
4.382W
4.382W
" electrically "
" magnetically "
" electrically "
Circuit 
 L1 
 L2 
 Circuit
One inductor takes the power from the circuit, and “transfers” it to the 2nd inductor. The 2nd
inductor gives back the whole received power to the circuit so it looks like the second inductor
“generates” power to the circuit. The absolute value of power of each inductor is called the
power transferred between the inductors.
The inductor taking the power from the circuit is called the primary inductor, the inductor
giving back the power to the circuit is called the secondary inductor.
53
5.5. Thévenin – Norton Equivalent Sources for Circuits with
Magnetically Coupling
E Th  U ab -open ;
E Th
J N
E
or ext
I
J N  Iab -short ;
Z ab 
ext all independent sources are set to 0
(not recommended) or Z ab all independent sources are set to 0
5.5. Thévenin – Norton Equivalent Sources for Circuits with
Magnetically Coupling
Example: Find the equivalent Thevenin and Norton circuits seen on two nodes a and b
V1  120 V ; Z1  5  j 3 ; Z L1  j8 ;
Z L 2  j 6 ; Z M  j 6 .
The open voltage: Branch current method
VZ1  VL1  V1  0



Z1  I1  Z L1  I1  Z M  0  V1  0

I1 

Vab  open  VL2  Z L2  0  Z M  I1 
V1
 0.993  65.56
Z1  Z L1
ZM
V1
Z1  Z L1
54
5.5. Thévenin – Norton Equivalent Sources for Circuits with
Magnetically Coupling
Example: Find the equivalent Thevenin and Norton circuits seen on two nodes a and b
The short current: Branch current method



 Z1  I1  Z L1  I1  Z M  I2  V1  0

 Z L 2  I2  Z M  I1  0
ZM
Iab  short  I2 
 V1
2
Z

Z

Z

Z
 1 L1  L 2 M
The sources’ impedance:
Vab open
Z ab =

I
ab  short
ZM
V1
Z1  Z L1
ZM
2
2
Z1  Z L   Z L  Z M

ZM

Z 
1
 V1
2
 Z1  Z L   Z L  Z M
1
2
Z1  Z L1
L2
Z1  Z L1
2
5.5. Thévenin – Norton Equivalent Sources for Circuits with
Magnetically Coupling
Example: Find the equivalent Thevenin and Norton circuits seen on two nodes a and b
2nd approach to find the impedance of the sources: Using
external (current) source and turning off all independent
sources.

VZ 1  VL1  0
Z1I1   Z L1I1  Z M Iext   0
Z M 

I1 
I ext
Z1  Z L1
and Vext  VL 2   Z L 2  I2  Z M  I1

 Z L 2  Iext 

2
ZM
Iext
Z1  Z L1
The sources’ impedance:

2
Vext  Z1  Z L1  Z L 2  Z M
Z ab =

Z1  Z L1
Iext
 ZL2 
2
ZM
Z1  Z L1
 Z1  Z L1  Z L 2  Z M2 
Z1  Z L1
I ext
55
5. Magnetically coupled circuits
5.1. Mutual Inductance and Self Inductance
5.2. The Dot Conversion for Mutual Coupled Inductors
5.3. Analysis Methods For AC Circuits With Mutual Coupling Effect
5.4. Power Transferred in Coupled Circuits
5.5. The Thévenin – Norton Equivalent Sources for Circuits with Magnetically Coupling
6. The Laplace Transform
6.1. The Laplace Transform
6.2. Properties of The Laplace Transform
6.3. The Inverse Laplace Transform
6.4. Laplace Transformation of Circuit Elements
6.5. Laplace Transformation of a Circuit
6.6. Transient Analysis Using Laplace Transform
6.7. Transfer function in Laplace domain
56
6.1. Laplace Transform
The Laplace Transform:

L  f (t )  F ( s )   f (t )e st dt
0
Example:


L[V0 ]   V0e st dt   V0 e st
V
V
  0  0  1  0
0
s
s
s
0

L  E0  e at   E0   e at e st dt   E0 e( s a )t
sa
0

0

E0
sa
6.1. Laplace Transform
Other selected Laplace Transform:
f (t )  E0  sin t   F ( s )  E0

;
s  2
s
f (t )  E0  cos t   F ( s )  E0 2
s  2
  cos   s  sin 
f (t )  E0  sin t     F ( s )  E0
s2   2
s  cos     sin 
f (t )  E0  cos t     F ( s )  E0
s2   2
...
2
57
6.2. Properties of The Laplace Transform
• Linearity:
L  f1 (t )   F1 ( s); L  f 2 (t )   F2 ( s) 
L  a1  f1 (t )  a2  f 2 (t )   a1  F1 (s )  a2  F2 (s)
2
s
4  5  cos 60  s  sin 60 
3

s 1
s 2  25
L  2  3sin(t )  4sin(5t  60)    2
• Transform of a derivation:
L  f (t )   F ( s )  L 
df 

  s  F ( s )  f (0 )
 dt 
• Transform of a time-shift:
L  f (t )   F ( s )  L  f (t  T0 )   F ( s)  e sT0
…
6.3. Inverse Laplace Transform
a) Definition:
F ( s )  L1  F ( s)   f (t ) 
  jT
1
lim 
F ( s)  e st  dt

j  2 T   jT
•
The integrating formula is not convenient for computation.
•
In Circuit Theory problems, most of the cases F(s) is the division of two polynomials of s.
 We can use the Heaviside method.
(If the F(s) function is simple enough, we can directly use the fundamental inverse
transform and superimpose the responses).
58
6.3. Inverse Laplace Transform
b) Heaviside method:
Let F(s) be the division of two polynomials F1(s)/F2(s), perform 3 following steps:
1.
Find the roots si of the denominator F2(s): si = ? to have F2(si) = 0.
2.
Find the corresponding coefficient Ai for single root (or Ai,k) for multiple root :
•
For single roots si: one coefficient Ai
Ai 
F1( s )
F2( s) s s
i
•
For multiple root of N-order: N coefficients Ai,k for k=N,…,1:
Ai ,k 
1 dk 
N F (s) 
s  si  1 
k 
k ! ds 
F2 ( s )  s s
i
3.
Form the final response:
f (t ) 

Ai  e sit 
single root si


multiple root s j 1k  root's order
A j ,k
t k 1
st
 e j for t  t0
(k  1)!
6.3. Inverse Laplace Transform
Example: (Single, real roots)
U (s) 
F1 ( s ) 12( s  1)( s  4)

F2 ( s ) s ( s  2)( s  3)
1. Find the roots si of the denominator F2(s): s1  0; s2  2; s3  3.
2. Find the corresponding coefficient Ai:
Ai 
F1 ( s )
12( s  1)( s  4)

F2 ( s) s  s
3s 2  10s  6 s  si
i

A1  8; A2  12; A3  8
3. Form the final response:
3
u (t )   Ai  e si t  8  12e 2t  8e 3t (for t  0)
i 1
Note: The real roots should be NOT POSITIVE!
59
6.3. Inverse Laplace Transform
U (s) 
Example: (Single, complex roots)
F1 ( s ) 12( s  1)( s  4)

F2 ( s ) s ( s 2  4 s  5)
s1  0; s2  2  j; s3  2  j.
1. Find the roots si of the denominator F2(s):
2. Find the corresponding coefficient Ai:
 A1  9,6

F ( s)
12( s  1)( s  4)
Ai  1


 A2  1, 2  j8,4  8, 485  81,87
F2 ( s) s s
3s 2  8s  5 s si

*
i
 A3  A2  1,2  j8,4  8,48581, 87
3. Form the final response:
3
u (t )   Ai  e si t  9,6  2  8, 485  e 2t  cos(t  81,87)
i 1
 9,6  16, 97  e 2t  sin(t  8,13) (for t  0)
6.3. Inverse Laplace Transform
Note: If there are complex roots:
1. When the coefficient of F2(s) are real, the complex roots form conjugated pair!
si   is a complex root  s j  si* is also a root.
2. The corresponding coefficients Ai and Aj are also conjugated:
A j  Ai*
3. The corresponding components in final response are also conjugated:
s t

A j  e j  Ai  e si t
→ Their sum is a real function:


*

s t
Re s
Ai  e si t  A j  e j  2  Re Ai  e si t  2 Ai e  i  cos  Im  si   Ai 
4.
The complex roots should have the real part NOT POSITIVE!
60
6.3. Inverse Laplace Transform
Example: (Double, real roots)
U (s) 
F1 ( s)
10s 2  4

F2 ( s) s ( s  1)( s  2) 2
s1  0; s2  1; s3,4  2.
1. Find the roots si of the denominator F2(s):
2. Find the corresponding coefficient Ai:
For i  1,2 : Ai 
F1 ( s )
10s 2  4
 3
 A1  1; A2  14
F2 ( s ) s s 4 s  15s 2  16 s  4 s s
i
i
For i  3,4 : A3,2  ( s  2) 2  U ( s)
A3,1 
s 2

2
10 s  4
 22
s ( s  1) s 2
d 10 s 2  4  10 s 2  8s  4


ds  s ( s  1) 
 s(s  1)2
 13
s 2
3. Form the final response:
u (t )  u (t )  1  14  e t  13  e 2t  22  t  e 2t (for t  0)
6.4. Laplace Transformation of Circuit Elements
Main idea:
L
Basic circuit elements:
a. Independent voltage source:
b. Independent current source:
L
L
uba (t )  e(t ) 
U ba ( s )  E ( s )
L
L
iab (t )  j (t ) 
 I ab ( s )  J ( s)
61
6.4. Laplace Transformation of Circuit Elements
L
c. Resistors:
L
u (t )  R  i (t ) 
U ( s)  L  u (t )   L  R  i (t )   R  L  i (t )   R  I ( s )
d. Capacitors:
L
i (t )  C 
du L
du
du

 I ( s)  L  C    C  L    C   s  L  u (t )   u (0) 
dt
dt 

 dt 
 U ( s) 
1
u (0)
I (s) 
sC
s
Question: What is the Norton equivalent circuit for capacitors?
6.4. Laplace Transformation of Circuit Elements
e. Inductors:
L
u (t )  L 
di L
di
di

 U ( s )  L  L    L  L    L   s  L  i (t )   i (0) 
dt
dt
dt


 
 sL  I ( s)  L  i (0)
Question: What is the Norton equivalent circuit for inductors?
62
6.4. Laplace Transformation of Circuit Elements
f. Inductors with coupling:
L


di1
di2

 u1 (t )  L1  dt  M  dt
U ( s )   sL1  I1 ( s)  L1i1 (0)    sM  I 2 ( s )  Mi2 (0) 
 1

u (t )  L  di2  M  di1 U 2 ( s )   sL2  I 2 ( s)  L2i2 (0)    sM  I1( s )  Mi1 (0) 
2
 2
dt
dt
U1 ( s )   sL1  I1 ( s)  sM  I 2 ( s)    L1i1 (0)  Mi2 (0) 

U 2 ( s )   sL2  I 2 ( s)  sM  I1 ( s )    L2i2 (0)  Mi1 (0) 
U1 ( s ) 

U 2 ( s ) 
 sL1  I1 (s)  sM  I 2 ( s)   E ps1 ( s)
 sL2  I 2 ( s)  sM  I1 (s)   E ps 2 ( s)
Note: Above figure and formulas were for POSITIVE mutual inductance. Readers are asked to form the
formulas for NEGATIVE cases!
6.4. Laplace Transformation of Circuit Elements
g. OP-AMP:
For ideal OP-AMP (A= ∞ ):
i  (t )  0; i  (t )  0  I  ( s )  0; I  ( s )  0


a (t )  b (t )
 a ( s )  b (s )
63
6.4. Laplace Transformation of Circuit Elements
h. Dependent sources:
a) VCVS
e(t )  k  ucd (t )  E ( s)  k  U cd ( s )
b) CCVS
e(t )  k  icd (t )  E ( s )  k  I cd ( s )
6.4. Laplace Transformation of Circuit Elements
h. Dependent sources:
c) VCCS
d) CCCS
j (t )  k  ucd (t )  J ( s )  k  U cd ( s )
j (t )  k  icd (t )  J ( s )  k  I cd ( s )
64
6.4. Laplace Transformation of a Circuit
Note: 1. The transformation is performed for circuit after transient time.
2. There could be new sources due to initial conditions of capacitors and inductors
3. Be careful about the directions of the new sources!
6.4. Laplace Transformation of a Circuit
Note: Due to the linearity of Laplace transformation → the forms of Laplace transformed Kirchhoff’s equations
are very similar to the phasor forms.
→ We can use all known circuit analysis methods in a very similar way for Laplace circuits.
65
6.5. Transient Analysis using Laplace Transformation
Let’s consider the transient time is at t=0.
4 steps in analyzing Laplace transformed circuits:
Step 1: Find the initial conditions for all capacitors and inductors from the circuit before
transient time and in steady state.
Step 2: Draw the Laplace image of the circuit after the transient time.
Step 3: Analyze the Laplace image of the circuit using known methods as for phasor circuits to
find all the Laplace transformed signals U(s) and I(s) (Power is non considered by now)
Step 4: Find the required signals in time-domain using the inverse Laplace transform
u(t)=L-1(U(s)),… (for example using Heaviside method):
4.1: Find the roots of the denominator.
4.2: Find the corresponding coefficients.
4.3: Form the final response.
6.5. Transient Analysis using Laplace Transformation
Note:
1. If the transient time is at t0 ≠ 0: Shift the time axis by using t’ = t - t0 to analyze the circuit for
transient time at t’ = 0.
2. Steps 1 and 3 use the known circuit analysis methods.
3. The Laplace circuit may contains new sources → the impedance equivalence method and
the superposition method are not recommended.
66
6.5. Transient Analysis using Laplace Transformation
Example:
1. R-C circuit with DC source
Analyze the circuit in transient state to find uC(t), where
E  12V ; R  5; C  0,1F .
Step 1: Before t=0: uC(0-)=0.
Step 2: Draw the Laplace image of the circuit after t=0.
6.5. Transient Analysis using Laplace Transformation
Step 3: Analyze the Laplace-transformed circuit
U C (s) 
1
sC
E
E
12


1 s s  sRC  1 s (0.5s  1)
R
sC
Step 4: Find inverse Laplace transform u(t)=L-1(U(s)),… (using Heaviside method)
4.1: Denominator’s roots: s1  0; s2  2.
4.2: Find the corresponding coefficients:
Ai 
F1 ( s )
12

 A1  12; A2  12
F2 ( s ) s s s  1 s si
i
4.3: Form the final response (for t≥0)
uC (t )  A1e s1t  A2e s2t  12  12e2t
Question: Write the function for the signal for entire time axis (t=-∞ → ∞).
67
6.5. Transient Analysis using Laplace Transformation
Example:
2. R-C circuit with AC source
Analyze the circuit in transient state to find uC(t), where
e(t )  E0 sin t   12sin(5t ) V; R  5 ; C  0,1 F.
Step 1: Before t=0: uC(0-)=0.
Step 2: Draw the Laplace image of the circuit after t=0.
6.5. Transient Analysis using Laplace Transformation
Step 3: Analyze the Laplace-transformed circuit
U C ( s) 
E
1
1
1
60
E (s) 
 2 0 2 2
1
sC R 
 sRC  1 s  
s  25 (0.5s  1)
sC


Step 4: Find inverse Laplace transform u(t)=L-1(U(s)),… (using Heaviside method)
4.1: Denominator’s roots:
 
s1  2; s2  j5; s3   j 5  s1* .
4.2: Find the corresponding coefficients:
F ( s)
60
Ai  1

2
F2( s) s  s 1.5s  2 s  12.5 s  si
i
 A1  4.138; A2  2.069  j 0.828  2.228  158.20  A3  A2*  2.228158.20
4.3: Form the final response (for t≥0)
uC (t )  A1e s1t  A2e s2t  A2e s3t  4.138  e 2t  2  2.228  cos  5t  158.20 
 4.138  e 2t  4.456  sin  5t  68.20 
Question: Check the new steady state in the circuit!
68
6.5. Transient Analysis using Laplace Transformation
Example:
3. R-L-C circuit with DC source
Analyze the circuit in transient state to find uC(t) and iL(t), where
E  12 V; R  9 ; L  1 H; C  0,05 F.
Step 1: Before t=0: uC(0-)=0; iL(0-)=0.
Step 2: Draw the Laplace image of the circuit after t=0.
6.5. Transient Analysis using Laplace Transformation
Step 3: Analyze the Laplace-transformed circuit
I L ( s) 
E ( s)
1
R  sL 
sC
U C ( s )  ...

E

1
s L  sR 
C
2
12
s 2  9s  20
Step 4: Find inverse Laplace transform u(t)=L-1(U(s)),… (using Heaviside method)
4.1: Denominator’s roots:
s1  4; s2  5.
4.2: Find the corresponding coefficients:
Ai 
F1 ( s )
12

 A1  12; A2  12.
F2 ( s ) s s 2s  9 s si
i
4.3: Form the final response (for t≥0)
uC (t )  A1e s1t  A2e s2t  12  e4t  12  e 5t
Question: 1. Find the voltage on the capacitor.
2. Select the values of R, L and C to have complex roots and find the signals for that cases.
3. Compare with the results from classical method (in LC 1)
69
6.5. Transient Analysis using Laplace Transformation
Example:
4. Circuit with dependent source(s)
Analyze the circuit in transient state to find uC6(t), where
V1  10 V; R1  8 ; I 2  1 A; V3  VR1  0.75VR1;
R4  4 ; R5  4 ; V5  12 V; C6  0.2 F.
Step 1: Before t=0:
uC 6 (0 )  U R 4 
R4
V5  6
R4  R5
6.5. Transient Analysis using Laplace Transformation
Step 2: Draw the Laplace image of the circuit
Step 3: Analyze the circuit, for example with the nodal voltage method
 1 1 1 1 
 1  
V5 ( s ) VC 60 ( s )
 





 
Va ( s)   
V1 ( s)  I 2 ( s) 
R4
R5
ZC 6 
R5
ZC 6
 R1
 R1 R4 R5 ZC 6 
3 s  5
10s  6 s  30
 Va (s ) 
s 1
s ( s  5)

20 4

 Vb (s )  Va ( s )   V1 ( s )  Va ( s )  
3  2 s  15 
s ( s  5)
70
6.5. Transient Analysis using Laplace Transformation
Step 4: Find inverse Laplace transform u(t)=L-1(U(s)),… (using Heaviside method) Vb ( s ) 
4.1: Denominator’s roots:
s1  0; s2  5.
3  2 s  15 
s( s  5)
4.2: Find the corresponding coefficients:
Ai 
3  2 s  15 
F1 ( s )

 A1  9; A2  3.
F2 ( s) s s
2 s  5 ss
i
i
4.3: Form the final response (for t≥0)
uC6 (t )  A1e s1t  A2e s2t  9  3  e5t
Question: 1. Find the voltage on the capacitor C6 when the switch K is opened at t=0.
6.6. Transfer function
The transfer function is defined as the ration of an output response Y(s) to an input (excitation) signal X(s).
H (s) 
Y (s)
X (s)
When X(s) and Y(s) are electrical signals,
71
6.6. Transfer function
Example: Find the two transfer functions in the given circuit.
H1 ( s ) 
Vo ( s )
V ( s)
and H 2 ( s )  o
Vi ( s )
I o (s)
6.6. Transfer function
Example: a) Find the transfer function in the given circuit.
H1 ( s ) 
Vo ( s )
Vi ( s )
b) Find the output response when vi(t) = 1(t) V
c) Find the output response when vi(t) = 8cos(2t) V
72
6. The Laplace Transform
6.1. The Laplace Transform
6.2. Properties of The Laplace Transform
6.3. The Inverse Laplace Transform
6.4. Laplace Transformation of Circuit Elements
6.5. Laplace Transformation of a Circuit
6.6. Transient Analysis Using Laplace Transform
6.7. Transfer function in Laplace domain
7. Frequency Response
7.1. Trigonometric Fourier series and the frequency spectra of a signal
7.2. Circuit analysis for periodical signal
7.3. The phasor transfer function and its frequency spectra.
7.4. The Bode plots
7.5. Passive filters
7.6. Active filters
73
7.1. Trigonometric Fourier series and the frequency spectra of a signal
n   :
A periodic function g(t) with period T:
1
T
g (t )  g (t  nT )
The frequency:
f0 
The angular frequency:
0  2 f 0 
The Fourier theorem:
f (t )  a0  a1 sin 0t   a2 sin  20t   a3 sin  30t   
(Hz)
2
T
(rad/s)
 b1 cos 0t   b2 cos  20t   b3 cos  30t   ...

 a0    an sin  n0t   bn cos  n0t  
k 1

or
 A0   An
k 1
sin
 n0t  n 
cos
7.1. Trigonometric Fourier series and the frequency spectra of a signal

The Fourier theorem: f (t )  a0    an sin  n0t   bn cos  n0t  
k 1

or
 A0   An sin  n0t  n 
k 1
T
with:
a0 
1
f (t )dt
T 0
an 
2
f (t )sin  n0t  dt
T 0
bn 
2
f (t )cos  n0t  dt
T 0
T
for n  1
T
for n  1
74
7.1. Trigonometric Fourier series and the frequency spectra of a signal
Example:
1,
f (t )  
0,
kT  t  kT  1
kT  1  t  kT  2
Computation:
a0 
T
1
2
  1 1 1
1
1
f
(
t
)

dt

1

dt

0

dt

    2 t  0  2
T 0
2  0
1

an 
2
2
1
 1

f (t )sin  n0t   dt    1sin  n t  dt   
cos n t   
(cos n  1)

T0
2 0
n
 n
0
T
 2

  n
0
n is odd
n is even
T
bn 
1
1
1
1
2
2
 1

f (t ) cos  n0t  dt    cos  n t  dt   sin  n t    0

T0
2 0
 n
0
7.1. Trigonometric Fourier series and the frequency spectra of a signal
Example:
 2
1

a0  ; An  an   n
2
0
1,
f (t )  
0,
kT  t  kT  1
kT  1  t  kT  2
n is odd
n is even
The amplitude spectrum
The phase spectrum
(for cosine function)
75
7.2. Circuit analysis for periodical signal
Periodical sources:
e(t )  A0  A1 sin 0t  1   A2 sin  20t   2     Ak sin  k0 t   k   ...
 E0  e1 (t )  e2 (t )  ...  ek (t )  ...
j (t )  B0  B1 sin 0t  1   B2 sin  20t   2     Bk sin  k0t   k   ...
 J 0  j1 (t )  j2 (t )  ...  jk (t )  ...
7.2. Circuit analysis for periodical signal
Periodical sources → Have to use Superposition theorem to analyze each frequency
and them sum up the results (in time domain)
Example: The circuit has R = 0.4 Ω, C = 50 mF. The source is:
1 2
2
2
 cos  t  90  
cos  3 t  90  
cos  5 t  90   
2 
3
5
 0.5  0.637 cos  t  90   0.212cos  3 t  90   0.127 cos  5 t  90   
vs (t ) 
Solution:
• DC component:
• 1st component:
• 2nd component:
VC  DC  V0  0.5 V
ZC 
1
    ZC   j
  j 6.366  VC  AC1 
V1  0.636  93.60 V
C
R  ZC
ZC 
1
  3  Z C   j
  j 2.122  VC  AC2 
V2  0.208  100.68 V
C
R  ZC
76
7.2. Circuit analysis for periodical signal
• 3rd component:
• 4th component:
ZC 
1
  j1.273  VC  AC3 
V3  0.121  107.44 V
C
R  ZC
ZC 
1
  7  Z C   j
  j 0.909  VC  AC4 
V4  0.0832  113.75 V
C
R  ZC
  5  Z C   j
Superimpose 5 components in time domain:
vC (t )  0.5  0.636cos  t  93.60   0.208cos  3 t  100.68   0.121cos  5 t  107.44 
0.0832cos  7 t  113.75  
Remark:
• When the frequency increases, the impedance of the capacitor is decreased and it tends
to zero at very high frequency.
• The amplitudes of the output components decrease with the frequency → this type of circuits
tends to let the lower frequencies to pass and tends to reduce the higher frequencies → this
type of circuits is a low-pass filter.
7.3. The phasor transfer function and its frequency spectra
The phasor transfer function H(jω) is a frequency-dependent
ratio between an output and an input signals
Y ( j )
H ( j ) 
X ( j )
Remark:
• jω that make H(jω) = 0 are called “zeros”,
• jω that make H(jω) = ∞ are called “poles”.
Example:
H ( j ) 
j  ( j  200)
( j  10)( j  40)( j  60)
• Zeros: jω = 0; jω = -200.
• Poles: jω = -10; jω = -40; jω = -60.
77
7.3. The phasor transfer function and its frequency spectra
Y ( j )
The phasor transfer function: H ( j ) 
X ( j )
New terms:
• |H(jω)| is the amplitude spectrum of the transfer function,
• /_H(jω) is the phase spectrum of the transfer function.
1
1  j RC
Example:
H ( j ) 

H ( j ) 

H ( j )  
1
1

2
1  j RC
1    RC 
1
  arctan  RC 
1  j RC
7.3. The phasor transfer function and its frequency spectra
1
1  j RC
Example:
H ( j ) 

H ( j ) 
1
1

;
2
1  j RC
1   RC 
Amplitude spectrum
H ( j )  
1
  arctan  RC 
1  j RC
Phase spectrum
78
7.4. Passive filters
Electrical filters are frequency specific: the signals in some ranges of frequencies are more
attenuated, in other frequencies are less attenuated or even enhanced.
Basic types of filters:
Ideal
low-pass filter
Ideal
high-pass filter
1 cut-off frequency ω0
Ideal
band-pass filter
Ideal
band-stop filter
2 cut-off frequencies ω1 < ω2
7.4. Passive filters
Low-pass filters:
Ideal
low-pass filter
Actual
low-pass filter
Simple R-C
low-pass filter
Question: Where is the cut-off frequency?
79
7.4. Passive filters
Low-pass filters: The cut-off frequency is set where the output power is 50% of the input power,
i.e. the output voltage (or current) is 1/ 2 of the input signal.
Ideal
low-pass filter
Actual
low-pass filter
H ( j0 ) 
Identifying the cut-off
frequency for the filter
1
1  0 RC 
Note: 1/ 2 can be approximated by 0.7. Also in decibel scale, 20 log
2

1
1
 0 RC  1  0 
RC
2
≈ −3 𝑑𝐵 .
7.4. Passive filters
High-pass filters:
Ideal
high-pass filter
Actual
high-pass filter
Simple R-C
high-pass filter
80
7.4. Passive filters
High-pass filters:
Ideal
high-pass filter
Actual
high-pass filter
Cut-off frequency:
H ( j0 ) 
Simple R-C
high-pass filter
0 RC
1  0 RC 
2

1
2
 0 RC  1  0 
1
RC
7.4. Passive filters
Band-pass filters:
Ideal
high-pass filter
Actual
high-pass filter
Simple R-L-C
band-pass filter
81
7.4. Passive filters
Band-pass filters:
Ideal
band-pass filter
Actual
band-pass filter
Cut-off frequency:
H ( j0 ) 
Identifying cut-off frequencies
for band-pass filter
0 RC
1  0 RC 
2

1
2
 0 RC  1  0 
1
RC
7.5. Active filters
Passive filters: The maximum of amplitude transfer is  1.
Active filters can generate output signal with amplitude transfer >1.
Low-pass filter:
H ( j )  

H ( j ) 

Cut-off frequency:

Rf
1
Ri 1  jC f R f
Rf
1
Ri
1 Rf C f
max H ( j ) 

Rf
Ri

2
at max  0
1
1 Rf
H ( j0 ) 
max H ( j ) 
2
2 Ri
2
1
1  0 R f C f  2  0 
Rf C f

Example of low-pass active
filter

82
7.5. Active filters
High-pass filter:
H ( j )  
Rf
Ri 
H ( j ) 

max H ( j ) 

j R f Ci
1  j Ri Ci
 R f Ci

Cut-off frequency:
1
jCi

1   Ri Ci 
Rf
2
at max  
Ri
Example of high-pass active
filter
1
1 Rf
H ( j0 ) 
max H ( j ) 
2
2 Ri
1  0 Ri Ci   2 0 Ri Ci 
2
2
 0 
1
Ri Ci
7.5. Active filters
Band-pass filter:
Example of band-pass active filter
Rf


j RC2   R f 
j RC2
1
1
H ( j )   

 
 
Ri 1  j RC1 1  j RC2
 1  j RC1  1  j RC2  Ri 
Cut-off frequencies:
1
RC2


1 
from high- pass section

1
RC1


2 

0  12 ; B  2  1 ; Q 
0
B
from low- pass section
83
7.5. Active filters
Band-stop filter:
Example of band-pass active filter
Rf 
V
j RC2 
1
H ( j )  o  




Vi
Ri  1  j RC1 1  j RC2 
1
RC1


1 
Cut-off frequencies:
from low- pass section

1
RC2

2 

0  12 ; B  2  1 ; Q 
0
B
from high- pass section
7.6. Bode plots
Ideas of Bode plots:
Amplitude spectrum in linear scales
Amplitude spectrum with log(x) scale
Amplitude spectrum with
log(x)-log(y) scales
1
 100
RC
log10 0   2
Cut-off frequency: 0 
84
7.6. Bode plots
Ideas of Bode plots:
Amplitude spectrum with log(x)-log(y) scales
0 
Amplitude spectrum in decibel scales (log(x) – 20*log(y))
1
1
 H  j0  
 log10 H  j0   0.15  20log10 H  j0   3( dB )
RC
2
7.6. Bode plots
Ideas of Bode plots:
y0
y  20 x  b
Amplitude spectrum with log(x)-log(y) scales
0 
Amplitude spectrum in decibel scales (log(x) – 20*log(y))
  0
0
2
 
1

1

  RC 
 20log10
 20log10 1     

RC
0
1  j RC
 0 
 20log10   20log10   20log10 0   0

0
Note: The two approximating lines intersect at ω = ω0!
85
7.6. Bode plots
• When the transfer function is the ratio of two polynomials, each polynomial can be presented as the
product of 4 basic elements: constant, (jω)n, (jω + a)n, ((jω)2 + b(jω) + c)n.
H ( j ) 
j  500
j ( j  40)
2
 j   150  j   10000
2
• Convert basic elements to standardized format: constant, (jω)n, (1+ j(ω/a))n, (1 + A(jω/a) + (jω/a)2)n.
 

500  1  j

500 


2
2
2

j ( j  40)  j   150  j   10000
 j  2  150  j   1

 j   402  1  j  10000 

40 
10000
10000



 
5 
3.125 10  1  j

500 


2 
2




 j   1  j     j   1.5  j   1
40    100 

 100  
H ( j ) 

j  500

7.6. Bode plots
• Convert the standardized format to decibel scale:
H ( j ) 

 

3.125 105  1  j

500 

2 
2




 j   1  j     j   1.5  j   1
40    100 

 100  


     
20 log10 H ( j )  20log10 3.125 105  20log10 1  j
 20log10 j  2  20log10 1  j
 20log10 1  1.5  j
 j

500
40
 100   100 


2
• Draw the Bode plot for each of the basic component:
 Bode plot (amplitude) for the constant term: y  20log10 K  const
This term’s plot is a straight line with constant height = 20log10K
(the slope of the line is 0)
86
7.6. Bode plots
• Draw the Bode plot for each of the basic component :
 Bode plot (amplitude) for the (jω)N term:
y  20log10 j
N
 20 N  log10 
This term’s plot is a line intersecting frequency axis
at ω = 1 (or log10ω = 0).
The slope of the line is 20N/decades
Amplitude spectrum of (jω) (N=1)
7.6. Bode plots
• Draw the Bode plot for each of the basic component :
 Bode plot (amplitude) for the (1+ j(ω/ω0))N term:
20log10 1  j

0
N
  0
0



20 N  log10   20 N  log10   20 N  log10 0   0
0

− This term’s plot can be approximated with 2 lines intersecting on
the frequency axis at ω = ω0.
− On log scale, we can approximate “ω << ω0” by “ω  0.1ω0”;
“ω >> ω0” by “ω ≥ 10ω0”
  0
Amplitude spectrum of
(1+ j(ω/ω0)) (N=1)
− The slope of the 2nd line is 20N/decades.
87
7.6. Bode plots
• Draw the Bode plot for each of the basic component :
 Bode plot (amplitude) for the (1 + A(jω/ω0) + (jω/ω0)2)N term:
20log10 1  A
 j 


0  0 
j
2 N
  0
0

2N

 
20log10    40 N  log10   40 N  log10 0   0
 0 

− This term’s plot can be approximated with 2 lines intersecting on
the frequency axis at ω = ω0.
− On log scale, we can approximate “ω << ω0” by “ω  0.1ω0”;
“ω >> ω0” by “ω ≥ 10ω0”
  0
Amplitude spectrum of
(1 + A(jω/ω0) + (jω/ω0)2) (N=1)
− The slope of the 2nd line is 40N/decades.
7.6. Bode plots
• Sum up all the components’ plots: For example using method of “tracking slope”:
 Each component has 1 or 2 linear segments,
 (From math) Sum of 2 segments has the slope equal the sum of 2 slopes.
2


     
20 log10 H ( j )  20log10 3.125 105  20log10 1  j
20log10 j 2  20log10 1  j
20log10 1  1.5  j
 j




500
40
 
 100   100 






Slope_3=-20

Slope_1=0 for all ω
Slope_2 increased by
Slope_4 decreased by


20dB/decade at ω=500
40
for all ω
Slope_5 decreased by
40dB/decade at ω=100
40dB/decade at ω=40
ω
-∞
100
500
∞
Slope_2
‖
0
‖
0
‖
0
‖
+20
‖
Slope_3
‖
-20
‖
-20
‖
-20
‖
-20
‖
Slope_4
‖
0
‖
-40
‖
-40
‖
-40
‖
Slope_5
‖
0
‖
0
‖
-40
‖
-40
‖
Slope_Σ
‖
-20
‖
-60
‖
-100
‖
-80
‖
88
7.6. Bode plots
• Sum up all the components’ plots:
2


     
20 log10 H ( j )  20log10 3.125 105  20log10 1  j
20log10 j 2  20log10 1  j
20log10 1  1.5  j
 j




500
40
 
 100   100 






Slope_3=-20

Slope_1=0 for all ω
Slope_2 increased by
Slope_4 decreased by


20dB/decade at ω=500
ω
-∞
Slope_Σ
‖
for all ω
40
-20
‖
100
-60

Slope_5 decreased by
40dB/decade at ω=100
40dB/decade at ω=40
‖
500
-100
‖
∞
-80
‖

At   1: value of the 20log10 H comes from 20log10 3.125×10-5 and 20log10 j  4.5  0  4.5
Before   1: slope  20dB / decade  at   0.1  range  log10 1 / 0.1 =1(decade)  value is:  4.5  20  1  24.5
From   1  40 : slope  20dB / decade; range  log10  40 / 1 =1.6(decade)  final value is:  4.5  20  1.6  36.5
From   40  100 : slope  60 dB / decade; range  log10 100 / 40  =0.4(decade)  final value is:  36.5  60  0.4  60.5
From   100  500 : slope  100dB / decade; range  log10  500 / 100  =0.7(decade)  final value is:  60.5  100  0.7  130.5
From   500   : slope  80dB / decade 
at   1000  range  log10 1000 / 500  =0.3(decade)  final value is:  130.5  80  0.3  154.5
7. Frequency Response
7.1. Trigonometric Fourier series and the frequency spectra of a signal
7.2. Circuit analysis for periodical signal
7.3. The phasor transfer function and its frequency spectra
7.4. Passive filters
7.5. Active filters
7.6. The Bode plots
89
8. Two-port networks
8.1. Introduction
8.2. Parameters of two-port networks
8.3. Relationships between parameters
8.4. Analysis of circuits with two-ports
8.5. The Thévenin – Norton theorems for circuits with two-ports
8.6. Interconnection of networks
8.1. Introduction
The idea of 1-port networks and 2-port networks:
A (DC) circuit seen as
an 1-port network
Two characteristic signals: Uin, Iin
A (DC) circuit seen as a
2-port network
Four characteristic signals: U1, I1, U2, I2.
Requirement: For 1-port, the entering current and the leaving current have to be the same!
90
8.1. Introduction
Characteristic equations for 2-port networks:
• Characteristic equations describe dependencies
between input/output signals.
• In LC2, we use equations to present 02 selected
signals as functions of the other 02 →
U1 = f1(U2, I2)
I1 = f2(U2, I2)
Four characteristic signals: U1, I1, U2, I2.
→ How may possible ways?
Note: We limit the consideration for 2-port not containing active (independent) sources!
8.2. Parameters of two-port networks
1. Transmission parameters (T)
U1  f1 U 2 , I 2   a  U 2  b  I 2  c

 I1  f 2 U 2 , I 2   d  U 2  e  I 2  f
We limit for 2-port not containing independent sources
→c=f=0
t  U 
U  t  U  t  I
U   t
U 
  1 11 2 12 2 or  1    11 12    2   T   2 
 I1  t21  U 2  t22  I 2
 I1  t21 t22   I 2 
 I2 
with tij  
For AC circuits:
U 1   t11 t12  U 2 
U 2 



T

  
 
  
 I1  t21 t22   I 2 
 I2 
with tij  
91
8.2. Parameters of two-port networks
1. Transmission parameters (T)
U1   t11 t12  U 2 
 I   t
 
 1   21 t22   I 2 
How to determine the parameters tij?
• Method 1: directly from the Kirchhoff equations when
the circuit’s structure is known.
• Method 2: Using two special cases (U2 = 0 or I2 = 0)
• ...
8.2. Parameters of two-port networks
1. Transmission parameters (T)
Example:
• Method 1: directly from the Kirchhoff equations when
the circuit’s structure is known.
 I1  I 2  I 3

U1  R1  I1  R2  I 2 +U 2
U   R  I  R  I
2
2
3
3
 2
 R 
1
U 2  1  2  I 2
R3
R3 


U 2   R2  I 2  R3   I1  I 2   I1 

1
 R  


R 
RR 
U1  R1   U 2  1  2  I 2   R2  I 2 +U 2   1  1 U 2   R1  R2  1 2  I 2
R3  
R3 
 R3

 R3 

92
8.2. Parameters of two-port networks
1. Transmission parameters (T)
Example:



R1 
R1 R2 
U1  1 
U 2   R1  R2 
 I2
R3 

 R3 


 R2 
1

 I1  R U 2   1  R  I 2
3
3 



 R1
1  R
3
T
 1

 R3
R1R2 
R3 


R2
1

R3

R1  R2 
8.2. Parameters of two-port networks
1. Transmission parameters (T)
Example: Method 2: Using two special cases (U2 = 0 or
I2 = 0)
U1   t11
 I   t
 1   21

U1
,
t11 
U 2 I 0
t12  U 2  
2


t 22   I 2  
I1
,
t21  U
2
I

0

2
t12 
U1
I 2 U 0
2
t22 
I1
I 2 U 0
2
93
8.2. Parameters of two-port networks
1. Transmission parameters (T)
Example: Method 2: Using two special cases (U2 = 0 or
I2 = 0)

R  R3
U1
R
 1
1 1
t11 
U 2 I 0
R3
R3

2
I2  0  
I
1
t  I1
 1 
 21 U
R3  I1 R3
2 I2 0



R R   R 
U1
R R
  R1  2 3    1  2   R1  R2  1 2
t12 
I
R

R
R
R3
2 U 2 0
2
3 
3 



U2  0  
I
I1
R
t22  1

1 2
R3
I 2 U 0
R3

I1
2

R2  R3
8.2. Parameters of two-port networks
2. Inverse transmission parameters (B)
U 2   b11 b12  U1 
 I   b
 
 2   21 b22   I1 
b11 
U2
,
U1 I 0
b12 
I2
,
U1 I 0
b22 
1
b21 
2
also
BT
U2
I1 U 0
1
I2
I1 U 0
1
1
94
8.2. Parameters of two-port networks
3. Impedance parameters (Z)
U1   z11
U    z
 2   21
z11 
z12   I1 

z22   I 2 
U1
,
I1 I 0
z12 
U2
,
I1 I 0
z22 
2
z21 
U1
I 2 I 0
1
2
U2
I 2 I 0
1
8.2. Parameters of two-port networks
4. Admittance parameters (Y)
 I1   y11
I    y
 2   21
y11 
y12  U1 
U 
 Y 1 



y22  U 2 
U 2 
I1
,
U1 U 0
y12 
I2
,
U1 U  0
y22 
2
y21 
2
also
YZ
I1
U 2 U 0
1
I2
U 2 U 0
1
1
95
8.2. Parameters of two-port networks
5. Hybrid parameters (H)
U1   h11
   h
 I 2   21
h11 
h12   I1 
 I1 
 H 



h22  U 2 
U 2 
U1
,
I1 U  0
h12 
I2
,
I1 U 0
h22 
2
h21 
U1
U 2 I 0
1
2
I2
U 2 I 0
1
8.2. Parameters of two-port networks
6. Inverse hybrid parameters (G)
 I1   g11
U    g
 2   21
g11 
g12  U1 
U 
 G   1



g 22   I 2 
 I2 
I1
,
U1 I  0
g12 
U2
,
U1 I  0
g 22 
2
g 21 
2
also
GH
I1
I 2 U 0
1
U2
I 2 U 0
1
1
96
8.3. Relationships between parameters of two-port networks
6. Inverse hybrid parameters (G)
Inverse pairs:
B  T1; Y  Z 1 ;
G  H 1
Other pairs: can deduce directly from math equations:
Example from Z to H as follow:
U  z I  z I (1) ? U1  h11I1  h12U 2
Z  H :  1 11 1 12 2


U 2  z21I1  z22 I 2 (2)
 I 2  h21 I1  h22U 2
z
1
(2)  I 2   21 I1 
U2
z22
z22


h21
h22
det  Z 
z z  z 12 z21
z
z
(1)  U1  11 22
I1  12 U 2 
I1  12 U 2
z22
z22
z22
z22



h11
h12
8.4. Analysis of circuits with two-ports
Reviews of known methods:
• The modified branch currents method,
• The loop current method (with matrix Z)
• The nodal voltage method (with matrix Y)
• The equivalent resistance method (with matrix T).
The power absorbed by the two-port:
P  Pin  Pout
Note: Relative directions of voltage and current on a port decide if the power is
“in” or “out” of the two-port.
97
8.4. Analysis of circuits with two-ports
The modified branch currents method:
• Unknowns are: branch currents + input/output voltages of two-port(s)
• Equations: Kirchhoff’s equations and characteristic equations of two-port(s).
Examples:
• How many unknown?
• How many Kirchhoff’s equations and how many
characteristic equations of two-port?
8.4. Analysis of circuits with two-ports
Examples: The modified branch currents method
 I3  I1  I4
 I3  I1  I4


 I2  I4  I5
 I2  I4  I5




 Z 3  I3  U 1  V3
U Z 3  U 1  V3  0
7 unkowns:


I1 , I2 , I3 , I4 , I5 ,U 1 ,U 2
U Z 4  U 2  U 1  0 
 Z 4  I4 +U 2  U 1  0
7 equations





U Z 5  U 2  0
 Z 5  I5  U 2  0




U 1  z11 I1  z12 I2
U 1  z11I1  z12 I2






U 2  z21 I1  z22 I 2
U 2  z21I1  z22 I 2
98
8.4. Analysis of circuits with two-ports
The loop currents method (with Z matrix):
• Unknowns are: loop currents
• Equations: KVL equations → Equations for branch currents → Equations for loop
currents .
Examples:
• How many unknown?
• How many unknown loop currents?
8.4. Analysis of circuits with two-ports
Examples: Loop currents method (with Z matrix)

 Z3  I3  U 1  V3

 Z 4  I4  U 1  U 2  0



 Z5  I 5  U 2  0
 Z3  I3  z11I1  z12 I2  V3


 Z 4  I4  z11 I1  z12 I2  z21I1  z22 I2  0

 Z5  I5  z21 I1  z22 I2  0




 


Branch currents as function of loop currents:
I1  Ia  Ic ; I2  Ib  Ic ; I3  Ia ; I4  Ic ; I5  Ib
99
8.4. Analysis of circuits with two-ports
Examples: Loop currents method





 

 Z 3  I3  z11I1  z12 I2  V3


 Z 4  I4  z11I1  z12 I2  z21 I1  z22 I2  0

 Z 5  I5  z21 I1  z22 I2  0
 Z3  z11  Ia  z12 Ib   z11  z12  Ic  V3

 z21  z11  Ia   z22  z12  Ib   Z 4  z11  z12  z21  z22  Ic  0




 z21I a   Z 5  z22  I b   z21  z22  I c  0

8.4. Analysis of circuits with two-ports
The nodal voltages method (with Y matrix):
• Unknowns are: nodal voltages
• Equations: KCL equations → Equations for nodal voltages
Examples:
• How many KCL equations?
• How many unknown nodal voltages?
100
8.4. Analysis of circuits with two-ports
Examples: Nodal voltages method (with Y matrix)
 I3  I1  I4

 I 4  I2  I5
Branch currents as function of nodal voltages:
V  V3  Va  Vb  Vb
I3  a
; I4 =
; I5 
;
Z3
Z4
Z5
I  y  V  y  V ; I  y  V  y  V
1

 11
a
12
b

2
 21
a
22
b

 1

 1

V

1
Va  V3 Va  Vb



 y11   Va   
 y12   Vb  3

 0 
 y11  Va  y12  Vb 
Z3
Z3
Z4

 Z 3 Z 4

 Z4



1   
1
1  
 y  V  y  V  Va  Vb  Vb  0
 
22
b
 21 a
  y21  Z   Va    y22  Z  Z   Vb  0
Z
Z
4
5

4 
4
5

 




8.4. Analysis of circuits with two-ports
The equivalent impedance method (with T matrix):


U 1 t11  U 2  t12  I2 U 2  Z 2 I 2 t11  Z 2  t12
Z in 


I1 t21  U 2  t22  I2
t21  Z 2  t22
Two special cases:
t
Z 2  0  Z in  12
t22
t
Z 2    Z in  11
t21
The equivalent impedance of a two-port with
a load at the output port
101
8.4. Analysis of circuits with two-ports
Examples: equivalent impedance method (with T matrix)
• Simplify the circuit using equivalent impedance
formula
• Analyze the simplified circuit (can find U1, I1)
• Go back to original circuit to find U2, I2 (if needed).
8.5. Thévenin – Norton theorems for circuits with two-ports
Note: Circuits with two-ports are still linear → can still used known formulas for
Thevenin and Norton theorems:
ETh  U ab open (or ETh  U ab open )
J N  I ab  short (or J N  Iab  short )
ETh
= equivalent resistance on a-b when all independent sources are set to 0
JN
E
(or Z ab = Th = equivalent impedance...)
J
Rab =
N
102
8.5. Thévenin – Norton theorems for circuits with two-ports
Examples: Find R2 to receive maximum
power from the circuit. Let:
Vs1  50 V, R1  10 ;
20  
 4
T= 
.
2 
0.1 S
Find the Thevenin voltage:
t
4
Rin _ open  11 
 40
t21 0.1
Vs1
50

1
R1  Rin _ open 10  40

I1_ open 

U 2 _ open 
I1_ open
t21

I2 0
1
 10
0.1
8.5. Thévenin – Norton theorems for circuits with two-ports
Find the Norton current:
t
20
Rin _ short  12 
 10
t22
2

I1_ short 

I 2 _ short 
Vs1
50

 2.5
R1  Rin _ short 10  10
I1_ short
t22

V2  0
2.5
 1.25
2
Find the resistance of the equivalent sources:
Rab 
ETh
10

8
J N 1.25
103
8.6. Interconnection of networks
• Two-ports have more connectors, then there are different ways to connect them.
• Each type of connection may be easier presented using a specific set of
parameters.

U 1  U 3  U 5



U 2  U 4  U 6
Z  Z1  Z 2
Series connection of two two-port networks
8.6. Interconnection of networks
• Two-ports have more connectors, then there are different ways to connect them.
• Each type of connection may be easier presented using a specific set of
parameters.

 I1  I3  I5

 I 2  I4  I6
Y  Y1  Y2
Parallel connection of two two-port networks
104
8.6. Interconnection of networks
• Two-ports have more connectors, then there are different ways to connect them.
• Each type of connection may be easier presented using a specific set of
parameters.

U 1 
U 2 
   T1    ;
 I1 
 I2 
T  T1  T2
U 3 
U 2 

T
.

 
2 
 I2 
 I3 
Cascade connection of two two-port networks
8. Two-port networks
8.1. Introduction
8.2. Parameters of two-port networks
8.3. Relationships between parameters
8.4. Analysis of circuits with two-ports
8.5. The Thévenin – Norton theorems for circuits with two-ports
8.6. Interconnection of networks
105