Chapter 9
Sinusoidal Steady–State
Analysis
9.1-9.2
9.3
9.4
9.5-9.9
The Sinusoidal Source and Response
The Phasor
Impedances of Passive Elements
Circuit Analysis Techniques in the
Frequency Domain
9.10-9.11 The Transformer
9.12
Phasor Diagrams
1
Overview

We will generalize circuit analysis from
constant to time-varying sources (Ch7-14).

Sinusoidal sources are particularly important
because: (1) Generation, transmission,
consumption of electric energy occur under
sinusoidal conditions. (2) It can be used to
predict the behaviors of circuits with nonsinusoidal sources.

Need to work in the realm of complex numbers.
2
Key points

What is the phase of a sinusoidal function?

What is the phasor of a sinusoidal function?

What is the phase of an impedance? What are
in-phase and quadrature?

How to solve the sinusoidal steady-state
response by using phasor and impedance?

What is the reflected impedance of a circuit with
transformer?
3
Section 9.1, 9.2
The Sinusoidal Source and
Response
1.
2.
Definitions
Characteristics of sinusoidal
response
4
Definition

A source producing a voltage varying
sinusoidally with time: v(t) = Vm cos(t + ).
: Phase angle,
Vm: Amplitude.
determines the
value at t = 0.
: Angular frequency, related to period T via  = 2/T.
The argument t changes 2 radians (360) in one period.
5
More on phase angle

Change of phase angle shifts the curve along
the time axis without changing the shape
(amplitude, angular frequency).

Positive phase (>0),  the curve is shifted
to the left by   in time, and vice versa.
Vm cos(t+)
Vm cos(t)
6
Example: RL circuit (1)


Consider an RL circuit with zero initial current
i (t  0  )  0 and driven by a sinusoidal voltage
source vs (t )  Vm cos(t   ) :
d
By KVL: L i  Ri  Vm cos(t   ).
dt
7
Example: RL circuit (2)

The complete solution to the ODE and initial
condition is (verified by substitution):
i (t )  itr (t )  iss (t ),
itr (t )  
iss (t ) 
Vm cos(   )
R  L
2
Vm
R 2   2 L2
2 2
e
( R L ) t
Transient response,
vanishes as t  .
cos(t     ) Steady-state response,
lasts even t  .
  tan 1 L R .
8
Characteristics of steady-state response
of this example exhibits the following
characteristics of steady-state response:
 iss(t)
iss (t ) 
Vm
R  L
2
2 2
cos(t     )
1.
It remains sinusoidal of the same frequency as
the driving source if the circuit is linear (with
constant R, L, C values).
2.
The amplitude differs from that of the source.
3.
The phase angle differs from that of the source.
9
Purpose of Chapter 9

Directly finding the steady-state response
without solving the differential equation.

According to the characteristics of steady-state
response, the task is reduced to finding two real
numbers, i.e. amplitude and phase angle, of the
response. The waveform and frequency of the
response are already known.

Transient response matters in switching. It will
be dealt with in Chapters 7, 8, 12, 13.
10
Section 9.3
The Phasor
1.
2.
Definitions
Solve steady-state response by
phasor
11
Definition

The phasor is a constant complex number
that carries the amplitude and phase angle
information of a sinusoidal function.

The concept of phasor is rooted in Euler’s
identity, which relates the (complex)
exponential function to the trigonometric
functions: e  j  cos   j sin  .
 
 
 cos   Re e j , sin   Im e j .
12
Phasor representation

A sinusoidal function can be represented by the
real part of a phasor times the “complex carrier”.


e  Re V  e 
Vm cos(t   )  Vm Re e j (t  )

 Re Vm e

1.
j
j t
j t
phasor carrier
A phasor can be represented in two forms:
Polar form (good for , ):
Vm
j
V  Vm e  Vm ,
2.
Rectangular form (good for +, -):
V  Vm cos   jVm sin  .

Imag.
real
13
Phasor transformation

A phasor can be regarded as the “phasor
transform” of a sinusoidal function from the time
domain to the frequency domain:
j


V  P Vm cos(t   )  Vm e .
time domain

freq. domain
The “inverse phasor transform” of a phasor is a
sinusoidal function in the time domain:

P -1V  Re Ve jt
 V
m
cos(t   ).
14
Time derivative  Multiplication of constant
Time
domain:
d
Vm cos(t   )  Vm sin(t   )
dt
 Vm cos(t    90 ),
d2
2
V
t
Vm cos(t   ).
cos(



)



2 m
dt
d

j (  90 )
P  Vm cos(t   )   Vm e
Frequency
 dt

domain:
  Vm e
j
e
j 90
 jV,
 d2

P  2 Vm cos(t   )   ( j ) 2 V   2 V.
 dt

15
How to calculate steady-state solution by phasor?

Step 1: Assume that the solution is of the form:

Re Ae

j
e 
j t
Step 2: Substitute the proposed solution into the
differential equation. The common time-varying
factor e jt of all terms will cancel out, resulting in
two algebraic equations to solve for the two
unknown constants {A, }.
16
Example: RL circuit (1)

Q: Given vs (t )  Vm cos(t   ), calculate iss(t).
Assume
iss (t )  I m cos(t   ).
d
 L iss (t )  Riss (t )
dt
 Vm cos(t   ),
d
 L I m cos(t   )  RI m cos(t   )  Vm cos(t   ),
dt
 LI m sin(t   )  RI m cos(t   )  Vm cos(t   ),
17
Example: RL circuit (2)

By cosine convention:
 LI m cos(t    90 )  RI m cos(t   )  Vm cos(t   ),

 Re LI m e
j (   90 )
 
 
,

e jt  Re RI m e j e jt  Re Vm e j e jt ,

   
 Re jL  R Ie   ReVe  .
 Re jLIe jt  Re RIe jt  Re Ve jt
j t

j t
A necessary condition is:
 j L  R  I e
j t
 Ve
j t
,   jL  R  I  V.
18
Example: RL circuit (3)

A more convenient way is directly transforming
the ODE from time to frequency domain:
d
L iss (t )  Riss (t )  Vm cos(t   ),
dt
 L j I  RI  V ,  jL  R I  V.

The solution can be obtained by one complex (i.e.
two real) algebraic equation:
j
V
e
V
j
, i.e. I m e  m
.
I
jL  R
jL  R
19
Section 9.4
Impedances of The Passive
Circuit Elements
1.
2.
3.
Generalize resistance to impedance
Impedances of R, L, C
In phase & quadrature
20
What is the impedance?

For a resistor, the ratio of voltage v(t) to the
current i(t) is a real constant R (Ohm’s law):
v (t )
R
. …resistance
i (t )

For two terminals of a linear circuit driven by
sinusoidal sources, the ratio of voltage phasor V
to the current phasor I is a complex constant Z:
V
Z  . …impedance
I
21
The i-v relation and impedance of a resistor

i(t) and v(t) reach the peaks simultaneously (in
phase),  impedance Z = R is real.
22
The i-v relation and impedance of an inductor (1)

Assume i (t )  I m cos(t  i )
d
 v (t )  L i (t )
dt
 L I m sin(t  i )
 LI m cos(t  i  90 ).

By phasor transformation:
V  L  jI
V
 Z   jL.
I
23
The i-v relation and impedance of an inductor (2)

v(t) leads i(t) by T/4 (+90 phase, i.e. quadrature)
 impedance Z = jL is purely positive imaginary.
d
v(t )  L i (t )
dt
24
The i-v relation and impedance of an capacitor (1)
d
i (t )  C v(t ).
dt
V
1
I  C  j V ,  Z  
.
I
jC
25
The i-v relation and impedance of a capacitor (2)

v(t) lags i(t) by T/4 (-90 phase, i.e. quadrature)
 impedance Z  1  jC  is purely negative
imaginary.
26
More on impedance

Impedance Z is a complex number in units of
Ohms.

Impedance of a “mutual” inductance M is jM.

ReZ   R, ImZ   X are called resistance and
reactance, respectively.

Although impedance is complex, it’s not a
phasor. In other words, it cannot be transformed
into a sinusoidal function in the time domain.
27
Section 9.5-9.9
Circuit Analysis Techniques
in the Frequency Domain
28
Summary

All the DC circuit analysis techniques:
1.
KVL, KCL;
2.
Series, parallel, -Y simplifications;
3.
Source transformations;
4.
Thévenin, Norton equivalent circuits;
5.
NVM, MCM;
are still applicable to sinusoidal steady-state
analysis if the voltages, currents, and passive
elements are replaced by the corresponding
phasors and impedances.
29
KVL, KCL
n

KVL: v1 (t )  v2 (t )    vn (t )   vq (t )  0,
q 1
n
n
q 1
q 1

 Vmq cos(t   q )   Re Vmq e

n
j
  Re Vmq e q
 q 1
j ( t  q )


 jt
e  0,

V1  V2  ...  Vn  0.

KCL:
i1(t) + i2(t) +… + in(t) = 0, 
I1  I 2  ...  I n  0.
30
Equivalent impedance formulas

Impedances in series
Z ab   Z j
j

Impedances in parallel
1
1

Z ab
j Z j
31
Example 9.6: Series RLC circuit (1)

Q: Given vs(t)=750 cos(5000t+30),  i(t)=?
Z L  jL  j (5000)(32 10 3 )  j160 ,

1
1

j
  j 40 ,
Z C 
6
(5000)(5 10 )
j C

V  75030 V,
 s
32
Example 9.6: Series RLC circuit (2)
Z ab  90  j160  j 40  90  j120
 90 2  120 2  tan 1 (120 90)  15053.13 ,
Vs
75030 V

I


5


23
.
13
A,

Z ab 15053.13 
 i (t )  5 cos(5000t  23.13 ) A.
33
Thévenin equivalent circuit

Terminal voltage phasor and current phasor are
the same by using either configuration.
34
Example 9.10 (1)

Q: Find the Thévenin circuit for terminals a, b.
a
b

Apply source transformation to {120V, 12, 60}
twice to get a simplified circuit.
35
Example 9.10 (2)
100  (10  j 40  120)I  10Vx ,  (130  j 40)I  10Vx  100 (1)
Vx  100  10I  ( 2)
36
Example 9.10 (3)
I
 900
 18  126.87 A,
I
30  j 40
VTh  10(100  10I)  120I  835.22  20.17 V.
37
Example 9.10 (4)
VT
Ia 
 j 40  (12 // 60)
VT

,
10  j 40
Vx  (12 // 60)I a  10I a ,
VT  10Vx
,
Ib 
120
VT  100I a I a VT 1 VT
VT
 


,
IT  I a  I b  I a 
120
6 120 6 10  j 40 120
Z Th  VT IT  91.2  j 38.4 .
38
Section 9.10, 9.11
The Transformer
1.
2.
Linear transformer, reflected
impedance
Ideal transformer
39
Summary

A device based on magnetic coupling.

Linear transformer is used in communication
circuits to (1) match impedances, and (2)
eliminate dc signals.

Ideal transformer is used in power circuits to
establish ac voltage levels.

MCM is used in transformer analysis, for the
currents in various coils cannot be written by
inspection as functions of the node voltages.
40
Analysis of linear transformer (1)

Consider two coils wound around a single core
(magnetic coupling):

+
+

Z11
Mesh current
equations:
 Vs  ( Z s  R1  jL1 )I1  jMI 2 ,

0   jMI1  ( R2  jL2  Z L )I 2 .
Z22
41
Analysis of linear transformer (2)
jM
jM
Z 22
 I1 
Vs , I 2 
I1 
Vs .
2
2
2
2
Z11Z 22   M
Z 22
Z11Z 22   M
Zint
 Z int
Z11
Z22
Vs Z11Z 22   2 M 2
2M 2


 Z11 
.
I1
Z 22
Z 22
42
Input impedance of the primary coil
Zint
Zab
Z ab  Z int  Z s  R1  jL1   Z r , Z r 
2M 2
Z 22
2M 2

.
R2  jL2  Z L

Zr is the equivalent impedance of the secondary
coil and load due to the mutual inductance.

Zab = ZS is needed to prevent power reflection.
43
Reflected impedance
Zr 
 M
2
Z 22
2
 M

Z


*
Z
Z 22 Z 22
 22
 M
2
2
*
22
2
 *
 Z 22 .


Z22

Linear transformer reflects (Z22)* into the
primary coil by a scalar multiplier (M/|Z22|)2.
44
Example 9.13 (1)

Q: Find the Thévenin circuit for terminals c, d.
I2
c
d
VTh = Vcd. Since I2 = 0,  Vcd = I1jM, where
Vs
3000
I1 

 79.67  79.29 A.
Z11 (500  j100)  ( 200  j 3600)
 VTh  (79.67  79.29 )  ( j1200)  95.610.71 V.
45
Example 9.13 (2)
c
Short

ZTh
Z11
d
ZTh = (100+j1600) + Zr, where Zr is the reflected
impedance of Z11 due to the transformer:
Z11  (500  j100)  ( 200  j 3600)  700  j 3700  .
2
2
 M  * 

1200
 Z11  
 (700  j 3700),
Z r  

 700  j 3700 
Z
11




Z Th  (100  j1600)  Z r  171.09  j1224.26  .
46
Characteristics of ideal transformer

An ideal transformer consists of two
magnetically coupled coils with N1 and N2 turns,
respectively. It exhibits three properties:
1.
Magnetic field is perfectly confined within the
magnetic core,  magnetic coupling coefficient
is k = 1,  M  L1 L2 .
2.
3.
The self-inductance of each coil Li  N i2  is
large, i.e. L1 = L2  .
The coil loss is negligible: R1 = R2  .
47
Current ratio
L1L2

By solving the two mesh equations of a general
linear transformer:
if L2 >> |ZL|
I1
Z 22
jL2  Z L
 


I 2 jM
j L1L2
L2 N 2

.
L1 N1
48
Voltage ratio
L1L2
+
+
V1
V2


 V1  jL1I1  jMI 2 ,
jM
I1 into 
 Substitute I 2 
Z 22
 V2  Z L I 2 .
jL2 + ZL
V1 Z 22  jL1    2 M 2 L1
L1 N1





.
V2
jMZ L
M
L2 N 2
49
Input impedance
L1L2
Zin

+
+
V1
V2


By the current and voltage ratios,
in-phase
2
2
 N1 
Z ab V1 I1 V1 I 2  N1 
 Z L .
 ,  Z in  Z ab  


 
Z L V2 I 2 V2 I1  N 2 
 N2 
2
 N1 
 R2  Z L .
 For lossy transformer,  Z ab  R1  
 N2 
50
Polarity of the voltage and current ratios
51
Example 9.14 (1)

Q: Find v1, i1, v2, i2.
Zs
ZL
2500
cos(400t)
52
Example 9.14 (2)
25000  (0.25  j 2)I1  V1  (1)
V2  (0.2375  j 0.05)I 2 ,

  V1  (23.75  j 5)I1  (2)
V1  10V2 , I 2  10I1 , 
25000
( 2)  (1) : I1 
 100  16.26 , i1  100 cos( 400t  16.26 ).
24  j 7
By (2) : V1  ( 23.75  j5)(100  16.26 )  2427  4.37 , v1  
53
Section 9.12
Phasor Diagrams
54
Definition

Graphical representation of -7-j3 = 7.62-156.8
on the complex-number plane.

Without calculation, we can anticipate a
magnitude >7, and a phase in the 3rd quadrant.
55
Example 9.15 (1)

Q: Use a phasor diagram to find the value of R
that will cause iR to lag the source current is by
45° when  = 5 krad/s.
j1
-j0.25
Vm
Vm
Vm


 Vm  90 , IC 
 4Vm90 , I R 
 Vm0.
IL 
j1
 j 0.25
R
56
Example 9.15 (2)

By KCL, Is = IL + IC + IR. Addition of the 3 current
phasors can be visualized by vector summation
on a phase diagram:
To make Is = 45,
j3Vm IR = 3Vm,
 R = 1/3 .
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Key points

What is the phase of a sinusoidal function?

What is the phasor of a sinusoidal function?

What is the phase of an impedance? What are
in-phase and quadrature?

How to solve the sinusoidal steady-state
response by using phasor and impedance?

What is the reflected impedance of a circuit with
transformer?
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