Chapter 9
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 9 Objectives
• Understand phasor concepts and be able to perform a phasor
transform and an inverse phasor transform.
• Be able to transform a circuit with a sinusoidal source into the
frequency domain using phasor concepts.
• Know how to use the following circuit analysis techniques to
solve a circuit in the frequency domain:
•
•
•
•
•
•
Kirchhoff’s laws;
Series and parallel simplifications;
Voltage and current division;
Thevenin and Norton equivalents;
Node-voltage method;
Mesh-current method.
Engr228 - Chapter 9, Nilsson 9E
1
Signal Definitions
• Time invariant signal
– Signal is an arbitrary constant when time progresses such as DC current
or DC voltage.
• Time-varying signal
– Signal varies with time such as AC current or AC voltage.
Periodic Signal
• A function f(t) is is said to be periodic if and only if there is
any T such that
f(t+T) = f(t) , for all t
where T is period and f is frequency
T=
1
f
In other words, f(t) is repeatable with a period of T.
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Properties of a Sine Wave
• The general form of sinusoidal wave is
v(t) = Vm sin(ωt + θ )
where
− Vm is the amplitude;
− ω is the angular frequency (radian/sec);
− θ is the phase shift (degrees or radians).
Frequency
period
volts
1
0.8
0.6
0.4
0.2
T=
0
1
f
-0.2
-0.4
-0.6
-0.8
-1
0
1
2
3
4
5
6
7
8
9
10
sec
Period ≈ 6.28 seconds, Frequency = 0.1592 Hz
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Amplitude
volts
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
1
2
3
4
5
6
7
8
9
10
sec
Peak: Blue 1 volt, Red 0.8 volts
Peak-to-Peak: Blue 2 volts, Red 1.6 volts
Average: 0
Phase Shift
Period=6.28
1
0.8
0.6
0.4
0.2
yblue = sin(t )
0
y red = 0.8 sin(t + 1)
-0.2
-0.4
-0.6
-0.8
-1
0
1
2
3
4
5
6
7
8
9
Red leads Blue by 57.3 degrees (1 radian)
Engr228 - Chapter 9, Nilsson 9E
10
φ=
1
× 360o = 57.3o
6.28
4
Sine Wave Phase
• The wave in red is said to lead the wave in green by θ.
• The wave in green sin(ωt) is said to lag the wave in red by θ.
• The units of θ and ωt must be consistent when computing the
sine function.
Basic AC Circuit Components
• AC Voltage and Current Sources (active elements)
• Resistors (R)
• Inductors(L)
passive elements
• Capacitors (C)
• Inductors and capacitors have limited energy storage
capability.
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AC Voltage and Current Sources
Voltage Sources
Current Sources
Amplitude = 10Vpeak
ω = 2πF so F = 1Hz
Phase shift = 45°
Sinusoidal Steady State (SSS) Analysis
• SSS is important for circuits containing capacitors and
inductors because these elements provide little value in
circuits with only DC sources.
• Sinusoidal means that source excitations have the form
VS cos(ωt + θ) or VS sin(ωt + θ).
• Since VS sin(ωt + θ) can be written as VS cos(ωt + θ - π/2),
we will use VS cos(ωt + θ) as the form of our general input
excitation.
• Steady state means that all transient behavior in the circuit
has died out, i.e. decayed to zero.
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Sinusoidal Steady State Response
• The SSS response of a circuit to a sinusoidal input is also a
sinusoidal signal with the same frequency but with possibly
different amplitude and phase shift.
v1(t) cos wave
v2(t) cos wave
cos wave
vL(t) cos wave
Forced Response to Sine Sources
When the source is sinusoidal, we often ignore the
transient/natural response and consider only the forced or
“steady-state” response.
The source is assumed to exist forever: −∞ < t < ∞
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Finding the Steady-State Response
1. Apply KVL:
L
di
+ Ri = Vm cos(ωt)
dt
2. Make a good guess:
3. Solve for the constants:
The Complex Forcing Function
Apply superposition and use:
Engr228 - Chapter 9, Nilsson 9E
e jθ = cos(θ ) + j sin(θ )
8
The SS Response via Complex Forcing Function
1. Apply KVL, assume vs=Vme jωt
L
di
+ Ri = v s
dt
2. Find the complex response
i(t) = Imejωt+θ
3. Find Im and θ, (discard the imaginary part).
Example: Sine Wave Analysis
Find the voltage on the capacitor.
Answer: vc(t)=298.5 cos(5t − 84.3 ) mV
◦
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Review of Complex Numbers
• Complex numbers can be viewed as vectors where the
X-axis represents the real part and the Y-axis represents
the imaginary part.
• There are two common ways to represent complex
numbers
– Rectangular form
3 + j4
– Polar form
5 ∠ 53o
Complex Number Forms
• Rectangular form: a + jb
• Polar form:
ρ∠θ
ρ = a2 + b2
b
a
θ = arctan 
a = ρ cos θ
b = ρ sin θ
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Complex Number Example
jω
3
σ
4
• Rectangular form: ρ = 4 + j3
• Polar form:
ρ = 5 ∠ 37°
Complex Math – Rectangular Form
p = a + jb
q = c + jd
• Addition and subtraction
x = p + q = (a + c) + j(b + d)
y = p – q = (a - c) + j(b - d)
• Example
p = 3 + j4
q = 1 - j2
x = p + q = (3 + 1) + j(4 - 2) = 4 + j2
y = p – q = (3 – 1) + j(4 – (-2)) = 2 + j6
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Complex Math – Rectangular Form
p = a + jb
q = c + jd
• Multiplication (easier in polar form)
x = p × q = ac + jad + jbc + j2bd = (ac – bd) + j(ad + bc)
• Example
p = 3 + j4
q = 1 - j2
x = p × q = [(3)(1) - (4)(-2)] + j[(3)(-2) + (4)(1)]
= 11 – j2
Complex Math – Rectangular Form
p = a + jb
q = c + jd
• Division (easier in polar form)
x=
p a + jb  (a + jb )(c − jd )   (ac + bd ) + j (bc − ad ) 
=
=
=

q c + jd  (c + jd )(c − jd )  
c2 + d 2

• Example
p = 3 + j4
x=
Engr228 - Chapter 9, Nilsson 9E
q = 1 - j2
p ((3)(1) + ( 4)(−2) ) + j ((4)(1) − (3)(−2) ) − 5 + j10
=
=
= −1 + j 2
q
12 + (−2) 2
5
12
Euler’s Identity
• Euler’s identity states that e jθ = cos(θ) + jsin(θ)
• A complex number can then be written as
r = a + jb = ρcos(θ) + jρsin(θ) = ρ[cos(θ) + jsin(θ)] = ρe jθ
• Using shorthand notation we write this as
ρe jθ ≡ ρ ∠ θ
Complex Math – Polar Form
x = a + jb = ρe jθ = ρ∠θ
p = m1e j (θ1 )
b
a
ρ = a2 + b2
θ = tan −1  
q = m2 e j (θ 2 )
• Addition and subtraction
– Too hard in polar – convert to rectangular coordinates.
• Multiplication
z = p × q = m1m2 e j (θ1 +θ 2 )
• Example
p = 6e
π 
j 
6
p = 6∠30°
Engr228 - Chapter 9, Nilsson 9E
q = 2e
π 
j 
2
q = 2∠90°
z = p × q = (6)(2 )e
π π 
j + 
6 2
= 12e
 2π 
j

 3 
z = p × q = 12∠120°
13
Complex Math – Polar Form
ρ = a2 + b2
x = a + jb = ρe jθ = ρ∠θ
p = m1e j (θ1 )
b
a
θ = tan −1  
q = m2 e j (θ 2 )
• Division
z = p÷q =
m1 j (θ1 −θ 2 )
e
m2
• Example
p = 6e
π 
j 
6
q = 2e
π 
j 
2
π π 
z = p ÷q =
 π
j − 
6 j  6 − 2 
e
= 3e  3  = 3∠ − 60o
2
More on Sinusoids
• Suppose you connect a function generator to any circuit
containing resistors, inductors, and capacitors. If the
function generator is set to produce a sinusoidal waveform,
then every voltage drop and every current in the circuit will
also be a sinusoid of the same frequency. Only the
amplitudes and phase angles will (may) change.
• The same thing is not necessarily true for waveforms of
other shapes like triangle or square waveforms.
• Fortunately, it turns out that sinusoids are not only the
easiest waveforms to work with, they're also the most
useful and occur quite frequently in real-world applications.
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Phasors
• A phasor is a vector that represents an AC electrical quantity
such as a voltage waveform or a current waveform.
• The phasor's length represents the peak value of the voltage
or current.
• The phasor's angle represents the phase angle of the voltage
or current.
• Phasors are used to represent the relationship between two or
more waveforms with the same frequency.
Phasors
• The diagram at the right shows two
phasors labeled v1 and v2.
• Phasor v1 is drawn at an angle of
0 and has a length of 10 units.
• Phasor v2 is drawn at an angle of
45 and is half as long as v1.
• In terms of the equations for
sinusoidal waveforms, this diagram
is a pictorial representation of the
equations
v1 = 10 cos(θ)
v2 = 5 cos(θ + 45 )
• The equations above and the
diagram convey the same
information.
°
°
°
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Phasors
• Phasors are complex numbers used to represent sinusoids of
a fixed frequency.
• Their primary purpose is to simplify the analysis of circuits
involving sinusoidal excitation by providing an algebraic
alternative to differential equations.
• A typical phasor current is represented as I = IM ∠φ
• For example, i(t) = 25cos(ωt + 45º) has the phasor
representation I = 25∠45º
• A phasor voltage is written as V = VM∠φ
• For example, v(t) = -15sin(ωt + 30º) = 15cos(ωt + 120º)
has the phasor representation V = 15∠120º
Example Problem 10.31 Hayt 7E
Useful trigonometric relationships:
sin(wt) = cos(wt - 90º)
-sin(wt) = cos(wt + 90º)
cos(wt) = sin(wt + 90º)
-cos(wt) = sin(wt - 90º)
Express each of the following currents as a phasor:
1. 12sin(400t + 110º)A
2. (-7sin800t – 3cos800t)A
3. 4cos(200t – 30º) – 5cos(200t + 20º)A
1. 12sin(400t + 110º)A = 12cos(400t + 20º)A = 12∠20ºA
2. (-7sin800t – 3cos800t)A = 7∠90º - 3∠0º =
(0 +7j) + (-3 + 0j) = -3 + 7j = 7.616∠113.2ºA
3. 4cos(200t – 30º) – 5cos(200t + 20º) = 4∠-30º -5∠20º =
(3.464 -2j) – (4.70 + 1.71j) = -1.235 – 3.71j = 3.91 ∠-108ºA
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Phasor Relationships for R, L, and C
• Now that we have defined phasor relationships for
sinusoidal forcing functions, we need to define phasor
relationships for the three basic circuit elements.
• The phasor relationship between voltage and current in a
circuit is still defined by Ohm’s law with resistance replaced
by impedance, a frequency sensitive form of resistance
denoted as Z(jω
ω).
• In terms of phasors, V = Z(jω
ω)I where V is a phasor
voltage, I is a phasor current, and Z(jω
ω) is the impedance of
the circuit element.
− Note that Z is a real number for resistance and a complex number
for capacitance and inductance.
• Since phasors are functions of frequency (ω
ω), we often refer
to them as being in the frequency domain.
• Circuits expressed as sinusoids are considered to be in the
time domain.
Phasors: The Resistor
In the frequency domain, Ohm’s Law takes the same form:
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Phasor Relationship for Resistors
• From Ohm’s law, and looking at the figure below,
vR (t ) = RiR (t ) = RI R e j (ωt +θ ) = VR e j (ωt +θ )
• In terms of phasors, IR = IRejθ and VR = VRejθ so the above
equation becomes
VR = RIR = ZR(jω)IR
• This result simply means that the impedance of an ideal
resistor is independent of frequency. In other words, resistors
act the same in AC circuits as they do in DC circuits.
Example
Find i(t) in the circuit
on the right.
v(t ) = i (t ) R
10 sin(2πt + π / 4) = i (t ) × 2
10 sin(2πt + π / 4)
2
i (t ) = 5 sin(2πt + π / 4)
i (t ) =
Note: In an ideal resistor, only the amplitude changes.
Frequency and phase remain the same.
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Phasor Relationship for Inductors
Finding the impedance (Z) of
an inductor:
v(t ) = L
di (t )
dt
1
1
v(t ) dt = ∫ A sin ωtdt
∫
L
L
A
A  − cos ωt 
= ∫ sin ωtdt = 

L
L ω 
π
A
A
=
(− cos ωt ) =
(sin ωt − )
ωL
ωL
2
i (t ) =
Phasors: The Inductor
Differentiation in time becomes multiplication in phasor form:
(calculus becomes algebra).
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Phasor Relationship for Inductors
• Looking at the figure below, if
iL(t) = ILej(ωt + θ)
and
vL(t) = VLej(ωt + φ)
then substituting these expressions into
vL = L
diL
dt
yields
VL e j (ωt +φ ) = L
d
I L e j (ωt +θ ) = jωLI L e j (ωt +θ )
dt
[
]
• In terms of phasors, IL = ILejθ and VL = VLejφ so the above
equation becomes
VL = jωLIL = ZL(jω)IL
• In other words, the inductor
impedance is jωL which is
dependent on frequency ω.
Phasor Relationship for Inductors
The relationship between voltage and current in the inductor
show that the current through an inductor lags the voltage
by 90º so it reaches its peak 1/4 cycle after the voltage
peaks.
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Comparison
i (t ) =
A
π
(sin ωt − )
ωL
2
Impedance of ωL
Phase shift of -90º
i (t ) =
A
(sin ωt )
R
Impedance of R
Phase shift of 0º
Phasors: The Capacitor
Differentiation in time becomes multiplication in phasor form:
(calculus becomes algebra).
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Phasor Relationship for Capacitors
• Looking at the figure below, if
iC(t) = ICej(ωt + θ) and
vC(t) = VCej(ωt + φ)
then substituting these expressions into
iC = C
dvC
dt
yields I C e j (ωt +θ ) = C
d
VC e j (ωt +φ ) = jωCVC e j (ωt +φ )
dt
[
]
• In terms of phasors, IC = ICejθ and VC = VCejφ so the above
equation becomes
IC = jωCVC = ZC(jω)VC
• In other words, the capacitor
impedance is 1/jωC which is
dependent on the frequency ω.
Phasor Relationship for Capacitors
Another way to find the
impedance of a capacitor
dv(t )
d ( A sin ωt )
=C
dt
dt
= AωC (cos ωt )
i (t ) = C
=
π
A
sin(ωt + )
2
 1 


 ωC 
Impedance of 1/jωC
Phase shift of +90º
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Phasor Relationship for Capacitors
The results from the previous page show that the current
through a capacitor leads the voltage by 90º.
Summary: Phasor Voltage/Current Relationships
Time Domain
Frequency Domain
Calculus (hard but real)
Algebra (easy but complex)
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Phasor Diagrams
The arrow for the phasor V on the phasor
diagram is a photograph, taken at ωt = 0, of
a rotating arrow whose projection on the
real axis is the instantaneous voltage v(t).
Example Phasor Diagram
If we assume I=1 ⁄ 0° A
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Phasor Diagram: Parallel RLC
Assume V = 1 /0 V
◦
Impedance
• We define impedance as Z = V/I or V = IZ
ZR=R
ZL=jωL
ZC=1/jωC
• Impedance is a complex number (with units of ohms)
– The real part of Z(jω
ω) is called the resistance.
– The imaginary part of Z(jω
ω) is called the reactance.
• Impedances in series or parallel can be combined using the
same “resistor rules” that you learned in Chapter 3.
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Impedance Relationships
• Some circuit applications deal with the inverse of
impedance which is called admittance. The admittance is
denoted as Y = 1/Z
YR=1/R
YL=1/jωL
YC=jωC
• If Z = R+jX; R is the resistance, X is the reactance: (units
of ohms Ω).
• If Y = G+jB; G is the conductance, B is the susceptance:
(units of siemens S).
Impedance Example
• Find the equivalent impedance, in polar form, for the circuit
below if ω = 0.333 rad/sec
1
Z EQ = R + jωL = 1 + j ⋅ 3 ⋅ = 1 + j = 2∠45o
3
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Example: Equivalent Impedance
Find the impedance of the network at ω = 5 rad/s
Answer: 4.255 + j4.929 Ω
Circuit Analysis Using Phasors
• Techniques that can be used in circuit analysis with phasors
– Ohm’s law;
– Kirchhoff’s voltage law (KVL);
– Kirchhoff’s current law (KCL);
– Nodal analysis;
– Mesh analysis;
– Thévenin's theorem;
– Norton’s theorem.
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Circuit Analysis Procedure Using Phasors
• Change the voltage/current sources into phasor form.
• Change R, L, and C values into phasor impedances.
• Use DC circuit analysis techniques normally, but the values
of voltage, current, and resistance can be complex numbers.
• Change back to the time-domain form if required.
Example Problem 10.46 Hayt 7E
A 20 mH inductor and a 30Ω resistor are in parallel. Find the
frequency ω at which: (a) |Zin| = 25Ω; (b) angle (Zin) = 25º;
(c) Re(Zin) = 25Ω; (d) Im(Zin) = 10Ω
(a)
(b)
(c)
(d)
ω = 2261 rad/s
ω = 3217 rad/s
ω = 3354 rad/s
ω = 572.9, 3927 rad/s
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Example Problem 9.25 Nilsson 9E
Find the frequency at which the impedance Zab is purely resistive.
Answer: Zab = 300k rad/second
Nodal Analysis Example
Find the phasor voltages V1 and V2.
Answer: V1=1 - j2 V and V2= -2 + j4 V
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Example Problem 9.55 Nilsson 9E
Use the node-voltage method to find VO.
Answer: VO = 138.078 – j128.22V = 188.43∠-42.88º V
Mesh Analysis Example
Find the currents i1(t) and i2(t).
i1(t) = 1.24 cos(103t + 29.7 ) A
i2(t) = 2.77 cos(103t + 56.3 ) A
◦
◦
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Example Problem 9.64 Nilsson 9E
Use the mesh current method to find the steady-state expression
for vo if vg = 130cos(10,000t)V.
Answer: vo = 56.57cos(10,000t – 45º)V
Example Problem 10.63 Hayt 7E
Find v2(t).
v2(t) = 34.36cos(ωt + 23.63º)V
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Example Problem 10.65 Hayt 7E
Find vx(t) in the circuit below if vs1 = 20cos1000t V and
vs2 = 20sin1000t V.
vx(t) = 70.71cos(1000t – 45º) V
Example Problem 10.68 Hayt 7E
Find vX(t).
vX(t) = 1.213cos(100t – 75.96º)V
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Example Problem 10.71 Hayt 7E
Compute the power dissipated by the 1Ω resistor.
P1Ω = 16.15mW
Superposition Example
The superposition principle applies to phasors; use it to find V1.
Answer: V1=V1L +V1R =(2 - j2)+(-1) = 1 - j2 V
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Thévenin Example
Thévenin’s theorem also applies to phasors; use it to find VOC
and ZTH in the circuit below.
Answer: Voc = 6 – j3 V
ZTH = 6 + j2 Ω
Example Problem 9.44 Nilsson 9E
Find the Thevenin equivalent circuit at terminals ab for
vg = 247.49cos(1000t + 45º ) V.
vTH = 350V = 350∠0º V
ZTH = 100 + j100Ω = 141.4∠45º Ω
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Example Problem 9.47 Nilsson 9E
Find the Thevenin equivalent circuit at terminals ab.
(Hint: Find Vth and In so Zth = Vth/In)
vTH = 10∠0º V
ZTH = 4.83∠0º Ω
Example Problem 10.81 Hayt 7E
Find the Thevenin equivalent circuit at terminals ab.
vTH = -50 + j150 = 158.11∠108.43º V
ZTH = j150Ω
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Example Problem 10.82 Hayt 7E
Find i(t).
i(t) = 51.07cos(1000t +42.23º) mA
Example Problem 10.87 Hayt 7E
(a) Find the Thevenin equivalent as seen by the capacitor.
(b) Derive an equation for the magnitude of the voltage ratio
Vout/Vs as a function of frequency.
(a) VTH = VS(405/505), RTH = 80.2Ω
1
(b) Vout =  405 


VS  505  1 + j 2.532×10−12ω
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Chapter 9 Summary
From the study of this chapter, you should:
• Understand phasor concepts and be able to perform a phasor
transform and an inverse phasor transform.
• Be able to transform a circuit with a sinusoidal source into the
frequency domain using phasor concepts.
• Know how to use the following circuit analysis techniques to
solve a circuit in the frequency domain:
•
•
•
•
•
•
Kirchhoff’s laws;
Series and parallel simplifications;
Voltage and current division;
Thevenin and Norton equivalents;
Node-voltage method;
Mesh-current method.
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