Sinusoid Steady-State Analysis
Objectives:
1)  Be able to perform a phasor transform
and its inverse;
2)  Be able to phasor-transform a circuit;
3)  Solve arbitrarily complex circuits with
sinusoidal sources using phasor
method.
Sinusoid Steady-State Analysis
Review the terminology of sinusoids
v(t) = Vmcos(ωt + φ) V
• Vm : Amplitude or magnitude
• ω : angular frequency [rad/s]
• t : time [s]
• φ : phase angle [deg]
• f = ω /2π : frequency [Hz]
• T = 1/f : period [s]
What is the frequency of the sinusoid
given below?
i(t) = 36 cos(4πt + 45°) mA
A. 4πt
B. 4π Hz
C. 4π rad/s
What is the phase angle of the
sinusoid given below?
v(t) = 50 cos(3000t - 60°) V
A.  -60 deg
B.  -60 rad
C. 60 deg
Suppose we want to find the value of the
sinusoidal current described below at
t = 5 ms. When we substitute this value
for t into the sinusoid, the argument for
the cosine function has what units?
i(t) = 0.2 cos(50πt + 45°) A
A.  degrees
B.  radians
C.  A mix of degrees and
radians
Sinusoid Steady-State Analysis
Evaluating a sinusoid at a specified time
i(0.005) = 0.2 cos(50πt + 45°) A
50π has the units rad/s
50πt has the units radians
45 has the units degrees
Need to convert 50πt to degrees, or 45 to
radians, and set your calculator
appropriately!
2π/360° = 50π(0.005)/x° → x = 45°
Thus, i(0.005) = 0.2 cos(45° + 45°) A
= 0.2 cos(90°) = 0 A
Sinusoid Steady-State Analysis
Suppose v(t) = Vmcos(ωt + φ) V in the circuit
below:
The equation for i(t) is
L
di (t )
+ Ri (t ) = Vm cos(ωt + φ )
dt
The solution of this equation for i(t) is
i (t ) =
where
− Vm
2
R + (ωL)
2
cos(φ − θ )e −( R L ) t +
θ = tan −1 (ωL R)
− Vm
2
R + (ωL)
2
cos(ωt + φ − θ )
Consider the equation for the current
in the RL circuit with the sinusoidal
source:
i (t ) =
− Vm
2
R + (ωL)
2
cos(φ − θ )e
−( R L ) t
+
− Vm
2
R + (ωL)
2
cos(ωt + φ − θ )
Which term(s) go to 0 as t → ∞?
A.  The first term
B.  The second term
C.  Both terms
D.  Neither term
Sinusoid Steady-State Analysis
i (t ) =
− Vm
2
R + (ωL)
2
cos(φ − θ )e −( R L ) t +
− Vm
2
R + (ωL)
2
cos(ωt + φ − θ )
• The first term is the transient term – it
decays to 0 as t goes to infinity
• The second term is the steady-state term –
it persists for all time greater than 0. This
term
•  Is sinusoidal
•  Has the same frequency as the input voltage
•  Has a different magnitude and phase angle
compared to the input voltage
Sinusoid Steady-State Analysis
The circuit analysis techniques we are
studying will allow us to calculate the
sinusoidal steady-state response of the
circuit – that is, the circuit’s response
to a sinusoidal input once the transient
response has effectively decayed to 0.
This is also called the AC Steady-State
(ACSS) response.
Sinusoid Steady-State Analysis
Phasor
•  A complex number in polar form, with a
magnitude and phase angle
•  Derived from a sinusoid using the phasor
transform
Euler' s identity :
⇒
Now
e ± jθ = cosθ ± j sin θ
cosθ = Re {e jθ }
v (t ) = Vm cos(ωt + φ )
= Vm Re {e j (ωt +φ ) }
= Vm Re {e jωt e jφ }
= Re {Vm e jωt e jφ }
Sinusoid Steady-State Analysis
Phasor transform
•  Extracts a sinusoid’s magnitude and phase
angle
•  Transforms a function of time into a function of
frequency
V = P {Vm cos(ωt + φ )} = Vm e jφ = Vm∠φ
Inverse phasor transform
•  Turns a phasor back into a sinusoid, if
someone tells you the frequency
v (t ) = P -1{V} = P -1{Vm∠φ } = Vm cos(ωt + φ )
Suppose
i(t) = 36 cos(4πt + 45°) mA
Then the phasor transform of i(t) is
A.  36 mA
B.  36∠45° mA
C.  36∠4πt mA
Sinusoid Steady-State Analysis
Adding or subtracting sinusoids in the time
domain is hard (you need trig identities!) But
if you phasor-transform the sinusoids, it is
easy to combine them. For example,
i (t ) = [5 cos(300t + 36.87°) + 10 cos(300t − 53.13°)] A
I = 5∠36.87° + 10∠ − 53.13° = 11.18∠ − 26.57° A
∴
i (t ) = P −1{I } = 11.18 cos(300t − 26.57°) A
Sinusoid Steady-State Analysis
Consider the equations that relate
voltage and current in a resistor,
inductor, and capacitor in the time
domain, and in the phasor domain.
Sinusoid Steady-State Analysis
Resistors:
Time domain
Phasor (frequency)
domain
P
→
i (t ) = I m cos(ωt + θ )
∴
v(t ) = Ri (t ) = RI m cos(ωt + θ )
I = I m∠θ
∴
V = RI = RI m∠θ
Note that the voltage and current phasors for a resistor
have the same phase angle – therefore we say that the
voltage and current in a resistor are “in phase”.
Sinusoid Steady-State Analysis
Inductors:
Time domain
Phasor (frequency)
domain
P
→
i (t ) = I m cos(ωt + θ )
∴
v (t ) = L di (t ) dt
I = I m∠θ
∴
V = −ωLI m e j (θ −90° )
= −ωLI m e jθ e − j 90°
= −ωLI m sin(ωt + θ )
= −ωLI m e jθ ( − j )
= −ωLI m cos(ωt + θ − 90°)
= jωLI m e jθ = jωLI
In an inductor, the voltage leads the current by 90°
Sinusoid Steady-State Analysis
Capacitors:
Time domain
Phasor (frequency)
domain
P
→
v (t ) = Vm cos(ωt + θ )
∴
i (t ) = C dv (t ) dt
V = Vm∠θ
∴
I = −ωCVm e j (θ −90° )
= −ωCVm e jθ e − j 90°
= −ωCVm sin(ωt + θ )
= −ωCVm e jθ ( − j )
= −ωCVm cos(ωt + θ − 90°)
= jωCVm e jθ = jωCV
In a capacitor, the voltage lags the current by 90°
The impedance of a resistor is
A.  Always a real number
B.  Always an imaginary number
C.  A complex number with real
and imaginary parts
The impedance of an inductor is
A.  A real number
B.  A positive imaginary
number
C.  A negative imaginary
number
The impedance of a capacitor has the
units
A.  Farads [F]
B.  Henries [H]
C.  Ohms [Ω]
Impedance is a phasor.
A.  True
B.  False
Sinusoid Steady-State Analysis
Summary:
•  In the time domain
•  Resistor:
•  inductor:
•  Capacitor:
v(t) = Ri(t)
v(t) = Ldi(t)/dt
i(t) = Cdv(t)/dt
•  In the phasor domain
V = ZI
•  Z is impedance, defined as the ratio of V to I
•  Z has the units Ohms [Ω]
•  Resistor: ZR = R
•  Inductor: ZL = jωL
•  Capacitor: ZC = 1/jωC = -j/ωC
Sinusoid Steady-State Analysis
Summary:
•  In the time domain
•  Ohm’s law (only for resistors): v = Ri
•  KVL (around a loop):
v1 + v2 + … + vn = 0
•  KCL (at a node):
i1 + i2 + … + i n = 0
•  These three laws lead to all other time-domain
circuit analysis techniques
•  In the phasor domain
•  Ohm’s law (for R, L, C): V = ZI
•  KVL (around a loop):
V1 + V2 + … + Vn = 0
•  KCL (at a node):
I1 + I2 + … + In = 0
•  These three laws mean we can use all time-domain
circuit analysis techniques in the phasor domain!
Sinusoid Steady-State Analysis
Circuit in the
time domain
P
Hard –
calculus!
Solution in the
time domain
Circuit in the
phasor domain
Easy –
algebra!
P -1
Solution in the
phasor domain
Sinusoid Steady-State Analysis
Steps in ACSS Analysis:
1.  Redraw the circuit (the phasor transform does not
change the components or their connections).
2.  Phasor transform all known v(t) and i(t).
3.  Represent unknown voltages and currents with V
and I.
4.  Replace component values with impedance (Z)
values.
5.  Use any circuit analysis method(s) to write
equations and solve them with a calculator.
6.  Inverse-transform the result, which is a phasor,
back to the time domain.
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
1.  Redraw the circuit.
2.  Phasor transform all
known voltages and
currents.
3.  Represent unknown
voltages and
currents with phasor
symbols
The next step is to calculate the
impedance of all resistors, inductors,
and capacitors. Actually, the
impedance of the resistors doesn’t
require calculation.
A.  True
B.  False
The impedance of inductors and
capacitors is calculated using
A.  The source frequency
B.  The component value
C.  Both A and B
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
4.  Calculate the
impedances of all
resistors, inductors,
and capacitors.
Z L = jωL = j ( 200,000)(40 µ ) = j8Ω
−j
−j
ZC =
=
= − j5Ω
ωC (200,000)(1µ )
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
5.  Use DC circuit analysis techniques to find the
phasor V.
•
In this problem, we’ll simplify the circuit by replacing
all impedances by one equivalent impedance, and
use Ohm’s law for phasors.
Z eq = 10 || (6 + j8) || − j5 = (4 − j 3)Ω
V = ZI = (4 − j 3)(8∠0°) = (32 − j 25)V
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
6.  Inverse phasor-transform the result to get back
to the time domain.
V = (32 − j 24)V = 40∠ − 36.87° V
v(t ) = P −1{40∠ − 36.87°} = 40 cos( 200,000t − 36.87°) V
The most direct way to calculate the
phasor I1 is
A.  Voltage division
B.  Current division
C.  Source transform
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
5.  Use DC circuit analysis techniques to find the
phasor I1.
•
Use current division.
I1 =
Z eq
Z1
Is =
(4 − j 3)
(8∠0°)
10
= (3.2 − j 2.4) A
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
6.  Inverse phasor-transform the result to get back
to the time domain.
I1 = (3.2 − j 2.4) A = 4∠ − 36.87° A
i1 (t ) = P −1{4∠ − 36.87°} = 4 cos( 200,000t − 36.87°) A
The most direct way to calculate the
phasor V2 is
A.  Voltage division
B.  Current division
C.  Node voltage method
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
5.  Use DC circuit analysis techniques to find the
phasor V2.
•
Use voltage division.
ZL
j8
V2 =
V =
(32 − j 24 )
ZR + ZL
6 + j8
= (32 + j 0) V
Sinusoid Steady-State Analysis
Example 9.7
is(t) = 8cos(200,000t) A.
Find v(t), i1(t), and v2(t)
in the steady-state.
6.  Inverse phasor-transform the result to get back
to the time domain.
V2 = (32 + j 0) V = 32∠0° V
v2 (t ) = P −1{32∠0°} = 32 cos 200,000t V
Sinusoid Steady-State Analysis
Example 9.7
Suppose we can vary the frequency of the current
source. What frequency will cause is(t) and v(t) to
be in-phase in the steady-state?
If is(t) and v(t) are in-phase in the
steady-state, this means
A.  Their phase angles are the same.
B.  The equivalent impedance seen by the
current source has a phase angle of 0°.
C.  The equivalent impedance seen by the
current source is purely resistive.
D.  All of the above.
Sinusoid Steady-State Analysis
1
1
1
= +
+ jωC
Z eq 10 6 + jωL
=
(6 + jωL)(6 − jωL)
10(6 − jωL)
10( jωC )(6 + jωL)(6 − jωL)
+
+
10(6 + jωL)(6 − jωL) 10(6 + jωL)(6 − jωL)
10(6 + jωL)(6 − jωL)
36 + ω 2 L2 + 60 − 10 jωL + 10 jωC (36 + ω 2 L2 )
=
10(36 + ω 2 L2 )
⎧⎪ 1 ⎫⎪ − 10ωL + 10ωC (36 + ω 2 L2 )
Im ⎨ ⎬ =
=0
2 2
10(36 + ω L )
⎪⎩ Z eq ⎪⎭
L C − 36 40 1 − 36
2
∴
ω =
=
∴
L2
( 40 µ ) 2
∴
− 10ωL + 10ωC (36 + ω 2 L2 ) = 0
ω = 50,000 rad/s
Sinusoid Steady-State Analysis
Example 9.7
Suppose we can vary the frequency of the current
source. What frequency will cause is(t) and v(t) to
be in-phase in the steady-state?
For ω = 50,000 rad/s
Z L = jωL = j (50,000)(40 µ ) = j 2Ω
P
−j
−j
ZC =
=
= − j 20Ω
ωC (50,000)(1µ )
∴
Z eq = 10 || (6 + j 2) || − j 20 = 4Ω
Sinusoid Steady-State Analysis
Steps in ACSS Analysis:
1.  Redraw the circuit (the phasor transform does not
change the components or their connections).
2.  Phasor transform all known v(t) and i(t).
3.  Represent unknown voltages and currents with V
and I.
4.  Replace component values with impedance (Z)
values.
5.  Use any circuit analysis method(s) to write
equations and solve them with a calculator.
6.  Inverse-transform the result, which is a phasor,
back to the time domain.
Sinusoid Steady-State Analysis
AP 9.12
is(t) = 10cos(50,000t) A;
vs(t) = 100sin(50,000t) V.
Find v(t) in the steadystate.
1.  Redraw the circuit.
2.  Phasor transform all
known voltages and
currents.
3.  Represent unknown
voltages and currents
with phasor symbols
vs (t ) = 100 sin 50,000t V
= 100 cos(50,000t − 90°) V
Sinusoid Steady-State Analysis
AP 9.12
is(t) = 10cos(50,000t) A;
vs(t) = 100sin(50,000t) V.
Find v(t) in the steadystate.
4.  Calculate the
impedances of all
resistors, inductors, and
capacitors.
Z L = jωL = j (50,000)(100µ ) = j5Ω
ZC =
−j
−j
=
= − j 2.22Ω
ωC (50,000)(9 µ )
Which of the following circuit analysis
techniques provides a one-equation
method to find the phasor voltage?
A.  Mesh current method
B.  Source transformation
C.  Node voltage method
Sinusoid Steady-State Analysis
AP 9.12
is(t) = 10cos(50,000t) A;
vs(t) = 100sin(50,000t) V.
Find v(t) in the steadystate.
5.  Use DC circuit
analysis techniques to
find the phasor V.
•  Use the node
V
voltage method.
− 10∠0° + +
5
V
V V - j100
+ +
=0
− j 2.22 j5
20
⎛ 1
1
1
1 ⎞
V⎜⎜ +
+ + ⎟⎟ = 10 + j5
⎝ 5 − j 2.22 j5 20 ⎠
∴
V = 10 − j 30 V
In writing the equations previously,
we replaced the complex number
100∠-90° with -j100. Your calculator
can do this for you by putting its
representation of complex numbers in
A.  Rectangular form
B.  Polar form
C.  Approximate form
Convert the complex number -15 to
polar form.
A.  -15∠0°
B.  15∠-90°
C.  15∠180°
Sinusoid Steady-State Analysis
AP 9.12
is(t) = 10cos(50,000t) A;
vs(t) = 100sin(50,000t) V.
Find v(t) in the steadystate.
6.  Inverse phasor-transform the result to get back to the
time domain.
V = 10 − j 30 V = 31.62∠ − 71.57°V
P
−1
{31.62∠ − 71.57°}= 31.62 cos(50,000t − 71.57°) V
Sinusoid Steady-State Analysis
AP 9.13
Find the value of the
phasor current I in the
phasor-transformed circuit
shown.
5.  Use DC circuit analysis techniques to find the
phasor I.
Which of the circuit analysis methods
below will generate equations that, when
solved, directly result in the value of the
phasor current?
A.  Node voltage method
B.  Mesh current method
C.  Source transformation
Sinusoid Steady-State Analysis
AP 9.13
Find the value of the
phasor current I in the
phasor-transformed circuit
shown.
− 33.8 + (1 + j 2)I + (3 − j5)(I − I1 ) = 0
Mesh current equations: (3 − j5)(I − I) + 2(I + 0.75V ) = 0
1
1
x
Vx = − j5(I − I1 )
Standard form:
I(1 + j 2 + 3 − j5) + I1 ( −3 + j5) + Vx (0) = 33.8
I( −3 + j5) + I1 (3 − j5 + 2) + Vx [2(0.75)] = 0
I( − j5) + I1 ( j5) + Vx ( −1) = 0
Sinusoid Steady-State Analysis
AP 9.13
Find the value of the
phasor current I in the
phasor-transformed circuit
shown.
Solving:
I = 29 + j 2 A
I1 = 19.4 + j 6 A
Vx = −20 + j 48 V
In preparation for the inverse phasor
transform, we need to convert the
phasor current to
A.  Rectangular form
B.  Polar form
C.  Standard form
If we convert 29 + j2 to polar form,
its phase angle will be
A.  Positive
B.  Negative
C.  I need my calculator!
Sinusoid Steady-State Analysis
AP 9.13
Find the value of the
phasor current I in the
phasor-transformed circuit
shown.
6.  Inverse phasor-transform the result to get back to the
time domain. We don’t have to complete this step for
this problem!
Sinusoid Steady-State Analysis
AP 9.10
v1(t) = 240 cos(4000t + 53.13°) V; v2(t) = 96 sin(4000t) V. Find
vo(t) in the steady-state, using a series of source
transformations.
1.  Redraw the circuit.
2.  Phasor transform all
known voltages and
currents.
3.  Represent unknown
voltages and currents
with phasor symbols
v2 (t ) = 96 sin 4000t V
= 96 cos( 4000t − 90°) V
Sinusoid Steady-State Analysis
AP 9.10
v1(t) = 240 cos(4000t + 53.13°) V; v2(t) = 96 sin(4000t) V. Find
vo(t) in the steady-state, using a series of source
transformations.
4.  Calculate the
impedances of all
resistors, inductors,
and capacitors.
Z L = jωL = j (4000)(0.015) = j 60Ω
ZC = − j ωC = − j (4000)(25 6 µ ) = − j 60Ω
We can check our impedance values
to be sure that the resistor
impedances are ______ and the
capacitor impedances are _______.
A.  Positive real numbers;
negative real numbers
B.  Positive real numbers;
negative imaginary numbers
C.  Positive imaginary numbers;
negative imaginary numbers
Sinusoid Steady-State Analysis
AP 9.10
v1(t) = 240 cos(4000t + 53.13°) V; v2(t) = 96 sin(4000t) V. Find
vo(t) in the steady-state, using a series of source
transformations.
5.  Source transform
240∠53.13°
= 3 .2 − j 2 .4 A
j 60
96∠ − 90°
= − j 4 .8 A
− j 60
Sinusoid Steady-State Analysis
AP 9.10
v1(t) = 240 cos(4000t + 53.13°) V; v2(t) = 96 sin(4000t) V. Find
vo(t) in the steady-state, using a series of source
transformations.
5.  Source transform
3.2 − j 2.4 − ( − j 4.8) = 3.2 − j 2.4 A
j 60 || 30 || − j 60 || 20 = 12Ω
Vo = (3.2 + j 2.4)(12) = 38.4 + j 28.8 V
Sinusoid Steady-State Analysis
AP 9.10
v1(t) = 240 cos(4000t + 53.13°) V; v2(t) = 96 sin(4000t) V. Find
vo(t) in the steady-state, using a series of source
transformations.
6.  Inverse phasor-transform the result to get back to the
time domain.
Vo = 38.4 + j 28.8 = 48∠36.87° V
vo (t ) = P
−1
{Vo }= 48 cos(4000t + 36.87°) V