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Steady State
Analysis
Ch. 9 – Steady State Analysis
2
Sinusoidal Source
This may be voltage or current
i  Im cos t   

Vm is the amplitude

φ is the phase angle (expressed in radians)
ω is the angular frequency (expressed in radians/s)
T is the period (expressed in seconds)
f is the frequency expressed is Hz or s-1




v  Vm cos t   
Root Mean Square (RMS)
Proof: VRMS 
Vm
1
2T
1 T  t0 2
1 T t0 1  cos  2 t    
2
V
cos

t


dt

V
dt 


m
m
T t0
T t0
2

T  t0
t0
dt  
T  t0
t0

cos  2 t     dt  Vm
VRMS 
Vm
2
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V
1
T  0  m
2T
2
Ch. 9 – Steady State Analysis
3
Angular frequency and frequency
Frequency is the number of cycles per second
Angular frequency corresponds to angular displacement 2π multiply by
the frequency.


Notes on sinusoidal

All sinusoidal expressions involve the angular frequency and time
and NOT the frequency f

Sinusoidal are periodic with period 2π.

The phase φ is the amount of shift a
sinusoidal exhibits as compared to a
reference and it is expressed in radians
vRef  Vm cos t 
v  Vm cos t   
iRef  Im cos t 
i  Imcos t   
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Ch. 9 – Steady State Analysis
4
 Example: Response of RL circuit to a sinusoidal source
Applying KVL :
di
L  Ri  Vm cos t   
dt

The solution of this non-homogeneous differential equation
comprises the solution of the homogeneous plus a particular
solution of the non-homogeneous i  t   in  i f
i t  
Vm
R  L
2
2 2
cos     e R / Lt 
tan  
Vm
R  L
2
L
R
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2 2
cos t     
Ch. 9 – Steady State Analysis
5
 Example: Response of RC circuit to a sinusoidal source
R
Applying KVL:
dvC
RC
 vc  Vm cos t   
dt

v +-
C
The solution of this non-homogeneous differential equation
comprises the solution of the homogeneous plus a particular
solution of the non-homogeneous v  t   vn  v f
vC  t  
Vm
1 R C
2
2
cos     e

t
RC

Vm
1 R C
tan    RC
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2
2
cos t     
Ch. 9 – Steady State Analysis
6
 Phasor Concept
Given a sinusoidal signal

v  t   Vm cos t   
e j  cos   j sin 
Euler’s identity
•

We can use Euler’s equality to rewrite it as


v  t   Vm cos t     Vm e jt     Vme jt e j    Vme j e jt 

The complex quantity Vm e j carries the amplitude and phase
angle of a given sinusoidal signal
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Ch. 9 – Steady State Analysis
7
Phasor abbreviation
 This complex quantity is defined as the phasor representation
or the phasor transform of v(t)
V
m cos t     
 Vme j
Vme  Vm
j

We can use the phasor transform to simplify finding the steady state behavior of our
circuits under sinusoidal excitation.

The assumption is that under sinusoidal excitation, the steady state response is also
a sinusoidal with the same angular frequency, but a different phase.
ELEC 250 – Summer 2015
Ch. 9 – Steady State Analysis
8
VI relationships
General Assumptions

We consider sinusoidal signals at steady state
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Ch. 9 – Steady State Analysis
9
Resistor
IV R
 Note that the resistor itself doesn’t change the phase of the circuit
Proof:
v  Vm cos t     Ri
V  Vme j
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i  Vme j R  I= V R
Ch. 9 – Steady State Analysis
10
 Inductance
 2
V  j LI   Le j 90  I m e j    LI m e j e j 90   LI me j  90 
 Note that the inductor changes the phase of the circuit
 Note that voltage leads the current
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V  jLI
 ( L ) I m    90 
Ch. 9 – Steady State Analysis
11
 Capacitance
I  jCV
 2
or
  jC  Vme j   CVme

j  90
 CVm  90
I
1
  90 I mi  m  i  90 
C
C
 Note that the capacitor changes the phase of the circuit
 Note that voltage lag the current
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j

C
Ch. 9 – Steady State Analysis
12
Phasor Transform in use to solve a circuit

This can be rewritten as
jLI  RI  V
ss


Note that
Vm
R  L
2
2 2
I m e j (  )
Vm e
j
e
j t   
where tan  
L
R
ELEC 250 – Summer 2015
.
Ch. 9 – Steady State Analysis
13
Leading and Lagging visuals
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Ch. 9 – Steady State Analysis
14
 Impedance and Reactance
All equations relating the phasor of the voltage to that of the current are of the form
V = ZI
The complex quantity Z is called the impedance while the imaginary part of it is
called the reactance
Circuit Element
Impedance
Resistor
R
Inductor
Capacitor
j L
j
C
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Reactance
L
1
C
Ch. 9 – Steady State Analysis
15
Impedance and Reactance cont’d

Impedance is a complex number, but not a phasor.
 That is, it does not implicitly involve time through ωt

Reactance is analogous to resistance in that it is a property of circuit
elements that opposes (resists) the passage of current.
 However, this is due to inductive (magnetic) or capacitive
(electrostatic) phenomena rather than thermodynamic ones
(resistance).

A circuit element may exhibit both reactive and resistive behavior. This is
expressed as the impedance which is a complex number, its real part
corresponds to its resistivity, the imaginary part corresponds to its
reactance
ELEC 250 – Summer 2015
Ch. 9 – Steady State Analysis
16
 Kirchhoff’s laws

Kirchhoff’s Voltage Law (KVL)

2


k
0
 I1 
2


k
0
1
Proof:
Kirchhoff’s Current Law (KCL)
Proof:
 i  t   0   I cos t    
   I e  e    0     I e   e    0 
 I e  0 I 0
k k
k
j
k
j
j t
mk
k
j
k
k
mk
mk
k
k
j t
mk
k
k
k
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Ch. 9 – Steady State Analysis
17
Impedances in series
and in parallel
Wye<->Delta

Same rules apply!
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Ch. 9 – Steady State Analysis
18
 Impedances in series
ab


1
Proof using KVL:
ab


2



1
1


n
2


n
2


n


1

ab
2



ab

n
1

2


n
 Impedances in parallel
1
1
1



Zab Z1 Z 2

1
Zn
Proof using KCL:

1

2


n

Zab

Z1

Z2


Zn
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
1
1
1



Zab Z1 Z 2

1
Zn
Ch. 9 – Steady State Analysis
19
 Admittance

Admittance is defined as the inverse of impedance
1
Y = = G + jB
Z

G is called conductance
B is called susceptance
Not that we can write the rule for
combining impedances in parallel,
in terms of the admittances as
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Yeq  Y1  Y2 
Yn
Ch. 9 – Steady State Analysis
20
Delta<->Wye

The same rules apply as in the time domain
DELTA =>WYE
Zb Zc
Z1 =
Z a + Z b + Zc
Z a Zc
Z2 =
Za + Zb + Zc
Zb Za
Z3 =
Z a + Z b + Zc
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WYE=>DELTA
Za =
Z1Z 2 + Z 2 Z 3 + Z 3 Z1
Z1
Z1Z 2 + Z 2 Z3 + Z 3 Z1
Z2
Z Z + Z 2 Z3 + Z3 Z1
Zc = 1 2
Z3
Zb =
Ch. 9 – Steady State Analysis
21
Source Transformations
Thevenin and Norton Equivalent


The same techniques used for resistive circuits apply as well.
Resistances are replaced by inductances and the time domain
currents and voltages are replaced by their frequency domain
phasors
VTH = Z N I N
IN =
ELEC 250 – Summer 2015
VTH
ZTH
and ZTH  Z N
Ch. 9 – Steady State Analysis
22
VTH = Z N I N
ZTH = Z N
VTH
IN =
ZTH
ELEC 250 – Summer 2015
Ch. 9 – Steady State Analysis
23
Node-Voltage and Mesh-Current analysis methods

Exactly the same techniques are used here.

We replace voltages and currents in the time domain
(steady state) with their phasor representations in the
frequency domain.

We use impedance and admittances in place of resistance
and conductance.
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