ECE 1311: Electric Circuits
Chapter 9 & 10: Phasor and Steady-state analysis
Overview of Sinusoids and Phasor
 Sinusoidal steady-state analysis
 Phasors
 Phasor relationship for circuit elements
 Impedance and Admittance
 Phasor Analysis
 Practice!!!
What if… vs is a sinusoidal signal?
Determine i(t) and v(t)?
Note: vs = 10 V? Is it DC?
Introduction: Sinusoids (1)
 A sinusoid is a signal that has the form of the sine or cosine function.
T =
2

where
Vm = the amplitude of the sinusoid
ω = the angular frequency in radians/s
Ф = the phase
How do we know which sinusoid is leading?
In the above, V2 leads V1 by ᶲ
or , V1 lags V2 by
ᶲ
The starting point of V2
occurs first in time before V1
Introduction: Sinusoids (2)
• A periodic function is one that satisfies v(t)
= v(t + nT), for all t and for all integers n.
T=
2
,

f =
1
Hz
T
Therefore,  = 2f
• Only two sinusoidal values with the same
frequency can be compared by their
amplitude and phase difference.
• If phase difference is zero, they are in
phase; if phase difference is not zero, they
are out of phase.
Note: Ф is the phase
Example 1:
o
Given a sinusoid, 5 sin( 4t − 60 ), calculate its
amplitude, phase, angular frequency, period, and
frequency.
Note: in general,
Solution:
Amplitude = 5, phase = –60o, angular frequency
= 4 rad/s, Period = 0.5 s, frequency = 2 Hz.
Trigonometric Identities
Transforming a sinusoid
Note: When comparing two sinusoids, we need
to express both as either sine or cosine with
positive amplitudes
Relating Cosine & Sine in Graphical form
Adding two sinusoids using Graphical Form
Example 2:
Find the phase angle between i1 = −4 sin( 377t + 25 o )
and i2 = 5 cos(377t − 40 o ) , does i1 lead or lag i2?
Solution:
One method is to change i2 into sin
Since sin(ωt+90o) = cos ωt, then
i2 = 5 sin( 377t − 40 o + 90 o ) = 5 sin( 377t + 50 o )
i1 = −4 sin( 377t + 25o ) = 4 sin( 377t + 180 o + 25o ) = 4 sin( 377t + 205o )
therefore, i1 leads i2 155o.
Note: we need to change from –ve sin
to +ve sin.
Phasor (1)
 Phasor: a complex number that represents the amplitude and
phase of a sinusoid
 It can be represented:
a. Rectangular z = x + jy = r (cos  + j sin  )
z = r 
j
z
=
re
c. Exponential
b. Polar
where
r = x2 + y2
 = tan −1
y
x
Phasor (2)
Phasor (3)
Mathematic operation of complex number:
1.
Addition
z1 + z 2 = ( x1 + x 2 ) + j ( y1 + y 2 )
2.
Subtraction
z1 − z 2 = ( x1 − x2 ) + j ( y1 − y2 )
3.
Multiplication
z1 z 2 = r1r2  1 + 2
4.
Division
z1 r1
= 1 −  2
z 2 r2
5.
Reciprocal
6.
Square root
7.
Complex conjugate
8.
Euler’s identity
1 1
=  −
z
r
z = r 
z = r  2
z  = x − jy = r  −  = re − j
e  j = cos   j sin 
Example 3
 Evaluate the following complex numbers:
a.
[(5 + j2)( −1 + j4) − 5 60 o ] 1/2
b.
10 + j5 + 340o
+ 10 30o
− 3 + j4
Solution:
a. 15.5 + j13.67,
b. 8.293 + j2.2
Phasor Transform (1)
• Phasor will be defined from the cosine function in all our
proceeding study. If a voltage or current expression is in
the form of a sine, it will be changed to a cosine by
subtracting from the phase.
Phasor Transform (2)
Example 4
Example 4
Transform the following sinusoids to phasors:
i = 6cos(50t – 40o) A
v = –4sin(30t + 50o) V
Solution:
a. I = 6 − 40 A
b. Since –sin(A) = cos(A+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor => V = 4140 V
Example 5
Find the sinusoids corresponding to phasors:
a. V = − 1030 V
b.
I = j(5 − j12) A
Solution:
a) v(t) = 10cos(t + 210o) V
5
) = 13 22.62
12
b) Since I = 12 + j5 = 12 2 + 52  tan −1 (
i(t) = 13cos(t + 22.62o) A
Phasor Transform (3)
 The differences between v(t) and V:
 v(t) is instantaneous or time-domain representation
V is the frequency or phasor-domain representation.
 v(t) is time dependent, V is not.
 v(t) is always real with no complex term, V is generally
complex.
 Note: Phasor analysis applies only when frequency is
constant; when it is applied to two or more sinusoid signals
only if they have the same frequency.
Phasor Transform (4)
 Relationship between differential, integral operation in
phasor listed as follow:
v(t )
V = V
dv
dt
jV
 vdt
V
j
Example 6a
Use phasor approach, determine the current i(t) in a circuit
described by the integro-differential equation.
di
4i + 8 idt − 3 = 50 cos(2t + 75)
dt
Answer: i(t) = 4.642cos(2t + 143.2o) A
Example 6b
Use phasor approach, determine the current
v(t) in a circuit described by the integrodifferential equation.
Answer: v(t) 5.3 cos(5t – 88o) V
Phasor Relationship (1)
 Resistor:
Time domain
V = RI
Phasor/frequency domain
Bold to indicate phasor
Phasor Relationship (2)
 Capacitor:
Time domain
Phasor/frequency domain
Assume no initial condition, steady state condition
Phasor Relationship (3)
 Inductor:
Time domain
Phasor/frequency domain
Assume no initial condition, steady state condition
Phasor Transform (8)
Resistor:
Inductor:
Capacitor:
Phasor Transform (9)
Example 6
If voltage v(t) = 6cos(100t – 30o) is applied to a
50 μF capacitor, calculate the current, i(t),
through the capacitor.
Answer: i(t) = 30 cos(100t + 60o) mA
Impedance and Admittance (1)
 Generalization of Ohm’s law
 The impedance Z of a circuit is the ratio of the phasor
voltage V to the phasor current I, measured in ohms Ω.
 The admittance Y is the reciprocal of impedance,
measured in siemens (S).
Impedance and Admittance (2)
Impedances and admittances of passive elements
Element
R
Impedance
Z=R
L
Z = jL
C
1
Z =
jC
Admittance
1
Y=
R
Y=
1
jL
Y = jC
Impedance and Admittance (3)
 = 0; Z = 0
Z = jL
 → ; Z → 
 = 0; Z → 
Z=
1
jC
 → ; Z = 0
Example 7
Refer to Figure below, determine v(t) and i(t).
vs = 5 cos(10t )
Answers: i(t) = 1.118cos(10t – 26.56o) A; v(t) = 2.236cos(10t + 63.43o)V
Example 8
Determine the input impedance of the circuit in figure below
at ω =10 rad/s.
Answer: Zin = 32.38 – j73.76
Can we apply KVL?
How about KCL?
Phasor Analysis
 Four main steps:
Transform all independent sources to phasors
2. Calculate the impedance of all passive elements
3. Apply analysis method (Ohm’s, KCL, KVL, volatge div,
current div and etc)
4. Apply inverse transform to obtain time-domain expression
for currents and voltages of interest
1.
Impedance and Admittance (2)
Impedances and admittances of passive elements
Element
R
Impedance
Z=R
L
Z = jL
C
1
Z =
jC
Admittance
1
Y=
R
Y=
1
jL
Y = jC
Practice 1
Ans: Z= 16-j12
Y = 0.04+j0.03
Practice 2
Ans: 30-j40
Practice 3: Voltage divider
Ans: 32<90
Practice 4: Current divider
Ans: 28cos(500t-116.57)mA
Practice 5: Kirchoff’s law
Ans:Ia = 3<-45A, Ic = 4.24<0A, Vg = 127.2V
Practice 6b: Nodal Analysis
Find v1(t) and v2(t) using Nodal analysis.
Answer:
v1(t) = 11.32 sin(2t + 60.01) V
v2(t) = 33.02 sin(2t + 57.12) V
Practice 6a: Nodal Analysis
(additional QUIZ)
Find ix using nodal analysis
Practice 7: Mesh Analysis*
Find I0 using mesh analysis.
Answer: Io = 1.194<65.44 A
Mesh analysis
Find Io
I0 = 6.12<144.78
Practice 8: Superposition
Calculate vo in the circuit of figure shown below using the
superposition theorem.
Vo = 4.631 sin(5t – 81.12) + 1.051 cos(10t – 86.24) V
Practice 9: Source transformation
Find Io in the circuit of figure below using the concept of source
transformation.
Io = 3.288<99.46 A
Practice 10: Thevenin and Norton
Find the Thevenin and Norton equivalent at terminals a–b of
the circuit below.
VTH = 18.97<-51.57 V
Zth =12.4 – j3.2
What about maximum power transfer?
Extra practice
Extra practice
Extra practice
Extra practice
Extra practice
Extra practice