AC STEADY-STATE ANALYSIS
LEARNING GOALS
SINUSOIDS
Review basic facts about sinusoidal signals
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
Behavior of circuits with sinusoidal independent sources
and modeling of sinusoids in terms of complex exponentials
PHASORS
Representation of complex exponentials as vectors. It facilitates
steady-state analysis of circuits.
IMPEDANCE AND ADMITANCE
Generalization of the familiar concepts of resistance and
conductance to describe AC steady state circuit operation
PHASOR DIAGRAMS
Representation of AC voltages and currents as complex vectors
BASIC AC ANALYSIS USING KIRCHHOFF LAWS
ANALYSIS TECHNIQUES
Extension of node, loop, Thevenin and other techniques
SINUSOIDS
x ( t ) = X M sin ω t
Adimensional plot
As function of time
X M = amplitude or maximum value
ω = angular frequency (rads/sec)
ω t = argument (radians)
T=
2π
ω
" leads by θ "
= Period ⇒ x (t ) = x (t + T ), ∀t
1 ω
=
= frequency in Hertz (cycle/sec )
T 2π
ω = 2π f
f =
" lags by θ "
BASIC TRIGONOMETRY
ESSENTIAL IDENTITIES
sin(α + β ) = sin α cos β + cosα sin β
cos(α + β ) = cosα cos β − sin α sin β
sin( −α ) = − sin α
cos(−α ) = cosα
SOME DERIVED IDENTITIES
sin(α − β ) = sin α cos β − cosα sin β
cos(α − β ) = cosα cos β + sin α sin β
1
1
sin α cos β = sin(α + β ) + sin(α − β )
2
2
1
1
cosα cos β = cos(α + β ) + cos(α − β )
2
2
APPLICATIONS
π
cosω t = sin(ω t + )
2
π
sin ω t = cos(ω t − )
2
cos ω t = − cos(ω t ± π )
sin ω t = − sin(ω t ± π )
RADIANS AND DEGREES
2π radians = 360 degrees
180
θ (rads) =
θ (degrees)
π
ACCEPTED EE CONVENTION
π
sin(ω t + ) = sin(ω t + 90°)
2
LEARNING EXAMPLE
cos(ω t )
Leads by 45 degrees
Lags by 315
cos(ω t + 45°)
cos(ω t + 45 − 360)
− cos(ω t + 45°)
cos(ω t + 45 ± 180)
Leads by 225 or lags by 135
LEARNING EXAMPLE
v1 (t ) = 12 sin(1000t + 60°), v2 (t ) = −6 cos(1000t + 30°)
FIND FREQUENCY AND PHASE ANGLE BETWEEN VOLTAGES
Frequency in radians per second is the factor of the time variable ω = 1000 sec −1
f ( Hz ) =
ω
= 159.2 Hz
2π
To find phase angle we must express both sinusoids using the same
trigonometric function; either sine or cosine with positive amplitude
take care of minus sign with cos(α ) = − cos(α ± 180°)
− 6 cos(1000t + 30°) = 6 cos(1000t + 30° + 180°)
Change sine into cosine with cos(α ) = sin(α + 90°)
6 cos(1000 t + 210°) = 6 sin(1000 t + 210° + 90°)
We like to have the phase shifts less than 180 in absolute value
6 sin(1000 t + 300°) = 6 sin(1000 t − 60°)
v1 (t ) = 12 sin(1000t + 60°)
(1000t + 60°) − (1000t − 60°) = 120°
v2 (t ) = 6 sin(1000t − 60°)
(1000t − 60°) − (1000t + 60°) = −120°
v1 leads v2 by 120°
v 2 lags v1 by 120°
LEARNING EXTENSION
i1 (t ) = 2 sin(377 t + 45°)
i2 (t ) = 0.5 cos(377 t + 10°)
i3 (t ) = −0.25 sin(377 t + 60°)
i1 leads i2 by_____?
i1 leads i3 by_____?
cosα = sin(α + 90°)
0.5 cos(377 t + 10°) = 0.5 sin(377 t + 10° + 90°)
(377 t + 45°) − (377 + 100°) = −55°
sin α = − sin(α ± 180°)
i1 leads i2 by − 55°
− 0.25 sin(377 t + 60°) = 0.25 sin(377 t + 60° − 180°)
(377 t + 45°) − (377 t − 120°) = 165°
i1 leads i3 by 165°
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
Learning Example
KVL : L
di
( t ) + Ri (t ) = v (t )
dt
In steady state i (t ) = A cos(ω t + φ ), or
i (t ) = A1 cos ω t + A2 sin ω t
*/ R
If the independent sources are sinusoids di
*/ L
(t ) = − A1ω sin ω t + A2ω cos ω t
of the same frequency then for any
dt
variable in the linear circuit the steady
state response will be sinusoidal and of (− LωA1 + RA2 ) sin ω t + ( LωA2 + RA1 ) cos ω t =
the same frequency
= VM cos ω t
− LωA1 + RA2 = 0 algebraic problem
LωA2 + RA1 = VM
To determine the steady state solution
we only need to determine the parameters A1 = RVM , A2 = ωLVM
2
2
2
2
R
L
+
(
ω
)
R
+
(
ω
L
)
B,φ
v (t ) = A sin(ω t + θ ) ⇒ i SS ( t ) = B sin(ω t + φ )
Determining the steady state solution can
be accomplished with only algebraic tools!
FURTHER ANALYSIS OF THE SOLUTION
The solution is i (t ) = A1 cos ω t + A2 sin ω t
The applied voltage is v (t ) = VM cos ω t
For comparison purposes one can write i (t ) = A cos(ω t + φ )
A1 = A cos φ , A2 = − A sin φ
A1 =
RVM
ωLVM
,
A
=
2
R 2 + (ωL) 2
R 2 + (ωL) 2
A=
i (t ) =
A = A12 + A22 , tan φ = −
A2
A1
VM
−1 ωL
,
φ
=
tan
R
R 2 + (ωL) 2
VM
−1 ωL
cos(
ω
t
−
tan
)
2
2
R
R + (ωL)
For L ≠ 0 the current ALWAYS lags the voltage
If R = 0 (pure inductor) the current lags the voltage by 90°
SOLVING A SIMPLE ONE LOOP CIRCUIT CAN BE VERY LABORIOUS
IF ONE USES SINUSOIDAL EXCITATIONS
TO MAKE ANALYSIS SIMPLER ONE RELATES SINUSOIDAL SIGNALS
TO COMPLEX NUMBERS. THE ANALYSIS OF STEADY STATE WILL BE
CONVERTED TO SOLVING SYSTEMS OF ALGEBRAIC EQUATIONS ...
… WITH COMPLEX VARIABLES
ESSENTIAL IDENTITY : e jθ = cosθ + j sin θ (Euler identity)
v (t ) = VM cos ω t → y (t ) = A cos(ω t + φ )
v (t ) = VM sin ω t → y ( t ) = A sin(ω t + φ ) * / j (and add)
y (t )
VM e jω t → Ae j (ωt +φ ) = Ae jθ e jω t
If everybody knows the frequency of the sinusoid
then one can skip the term exp(jwt)
VM → Ae jθ
Learning Example
R − jωL = R + (ωL ) e
2
IMe
v ( t ) = V M e jω t
Assume i (t ) = I M e ( jω t +φ )
di
KVL : L ( t ) + Ri (t ) = v (t )
dt
di
(t ) = jωI M e ( jω t +φ )
dt
di
L (t ) + Ri (t ) = jωLI M e ( jω t +φ ) + RI M e ( jω t +φ )
dt
= ( jωL + R) I M e ( jω t +φ )
= ( jωL + R) I M e jφ e jωt
( j ω L + R ) I M e jφ e j ω t = V M e jω t
VM
R − jωL
I M e jφ =
*/
jωL + R
R − jωL
I M e jφ =
VM ( R − jωL)
R 2 + (ωL) 2
IM =
jφ
=
2
VM
R 2 + (ωL )
VM
R 2 + (ωL ) 2
− tan −1
e
2
ωL
R
− tan −1
ωL
R
, φ = − tan −1
ωL
R
v (t ) = VM cos ω t = Re{VM e jω t }
⇒ i (t ) = Re{ I M e ( jω t −φ ) } = I M cos(ω t − φ )
C↔ P
x + jy = re jθ
r = x 2 + y 2 , θ = tan −1
x = r cosθ , y = r sin θ
x
y
PHASORS
ESSENTIAL CONDITION
ALL INDEPENDENT SOURCES ARE SINUSOIDS OF THE SAME FREQUENCY
BECAUSE OF SOURCE SUPERPOSITION ONE CAN CONSIDER A SINGLE SOURCE
u(t ) = U M cos(ω t + θ )
THE STEADY STATE RESPONSE OF ANY CIRCUIT VARIABLE WILL BE OF THE FORM
y (t ) = YM cos(ω t + φ )
SHORTCUT 1
u(t ) = U M e j (ω t +θ ) ⇒ y (t ) = YM e
Re{U M e j (ω t +θ ) } ⇒ Re{YM e
j ( ω t +φ )
j ( ω t +φ )
}
U M e j (ω t +θ ) = U M e jθ e jωt u = U M e jθ ⇒ y = YM e jφ
SHORTCUT IN NOTATION
NEW IDEA:
INSTEAD OF WRITING u = U M e jθ WE WRITE u = U M ∠θ
... AND WE ACCEPT ANGLES IN DEGREES
U M ∠θ IS THE PHASOR REPRESENTATION FOR U M cos(ω t + θ )
u(t ) = U M cos(ω t + θ ) → U = U M ∠θ ⇒ Y = YM ∠φ → y ( t ) = Re{YM cos(ω t + φ )}
SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF
THE RESPONSE GIVEN THE INPUT PHASOR(S)
Learning Extensions
Learning Example
It is essential to be able to move from
sinusoids to phasor representation
A cos(ωt ± θ ) ↔ A∠ ± θ
A sin(ωt ± θ ) ↔ A∠ ± θ − 90°
V = VM ∠0
v = Ve jωt
I = I M ∠φ
jω t
di
i
=
Ie
L (t ) + Ri ( t ) = v
dt
L( jωIe jωt ) + RIe jωt = Ve jωt
In terms of phasors one has
jωLI + RI = V
V
I=
R + jωL
The phasor can be obtained using
only complex algebra
We will develop a phasor representation
for the circuit that will eliminate the need
of writing the differential equation
v (t ) = 12 cos(377 t − 425°) ↔ 12∠ − 425°
y (t ) = 18 sin( 2513t + 4.2°) ↔ 18∠ − 85.8°
Given f = 400 Hz
V1 = 10∠20° ↔ v1 (t ) = 10 cos(800π t + 20°)
V2 = 12∠ − 60° ↔ v 2 (t ) = 12 cos(800π t − 60°)
Phasors can be combined using the
rules of complex algebra
(V1∠θ1 )(V2∠θ 2 ) = V1V2∠(θ1 + θ 2 )
V1∠θ1 V1
= ∠(θ1 − θ 2 )
V2∠θ 2 V2
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS
RESISTORS v ( t ) = Ri ( t )
VM e ( jω t +θ ) = RI M e ( jω t +θ )
VM e jθ = RI M e jθ
V = RI Phasor representation for a resistor
Phasors are complex numbers. The resistor
model has a geometric interpretation
The voltage and current
phasors are colineal
In terms of the sinusoidal signals this
geometric representation implies that
the two sinusoids are “in phase”
INDUCTORS
d
( I M e ( jω t +φ ) )
dt
= jωLI M e ( jω t +φ )
VM e ( jω t +θ ) = L
Relationship between sinusoids
VM e jθ = jωLI M e jφ
V = jωLI
Learning Example
The relationship between L = 20mH , v ( t ) = 12 cos(377 t + 20°). Find i ( t )
phasors is algebraic
For the geometric view
use the result
j = 1∠90° = e j 90°
V = ωLI∠90°
ω = 377
12∠20°
( A)
V = 12∠20° I =
ωL∠90°
V
12
I=
I
=
∠ − 70°( A)
jωL
377 × 20 × 10−3
i (t ) =
The voltage leads the current by 90 deg
The current lags the voltage by 90 deg
12
cos(377 t − 70°)
−3
377 × 20 × 10
CAPACITORS
I M e ( jω t +φ ) = C
d
(VM e ( jω t +θ ) )
dt
Relationship between sinusoids
I M e jφ = jωCe jθ
I = ωCV∠90°
I = jωCV
Learning Example
C = 100 µ F , v (t ) = 100 cos(314t + 15°). Find i ( t )
The relationship between
phasors is algebraic
In a capacitor the
current leads the
voltage by 90 deg
The voltage lags
the current by 90 deg
ω = 314
V = 100∠15°
I = ωC × 1∠90° × 100∠15°
I = jωCV
I = 314 × 100 × 10−6 × 100∠105°( A)
i (t ) = 3.14 cos(314t + 105°)( A)
LEARNING EXTENSIONS
L = 0.05 H , I = 4∠ − 30°( A), f = 60 Hz
Find the voltage across the inductor
ω = 2π f = 120π
V = jωLI
V = 120π × 0.05 × 1∠90° × 4∠ − 30°
V = 24π∠60°
v (t ) = 24π cos(120π + 60°)
Now an example with capacitors
C = 150 µ F , I = 3.6∠ − 145°, f = 60 Hz
Find the voltage across the inductor
ω = 2π f = 120π
I = jωCV ⇒ V =
V=
I
jω C
3.6∠ − 145°
120π × 150 × 10−6 × 1∠90°
200
V=
∠ − 235°
π
v (t ) =
200
π
cos(120π t − 235°)
IMPEDANCE AND ADMITTANCE
For each of the passive components the relationship between the voltage phasor
and the current phasor is algebraic. We now generalize for an arbitrary 2-terminal
element
Z (ω ) = R(ω ) + jX (ω )
R(ω ) = Resistive component
X (ω ) = Reactive component
| Z |= R 2 + X 2
θ z = tan −1
X
R
(INPUT) IMPEDANCE
V V ∠θ V
Z = = M v = M ∠(θ v − θ i ) =| Z | ∠θ z
I I M ∠θ i I M
(DRIVING POINT IMPEDANCE)
The units of impedance are OHMS
Impedance is NOT a phasor but a complex
number that can be written in polar or
Cartesian form. In general its value depends
on the frequency
Element Phasor Eq. Impedance
R
L
C
V = RI
V = jωLI
1
V=
I
jωC
Z=R
Z = jωL
1
Z=
jωC
KVL AND KCL HOLD FOR PHASOR REPRESENTATIONS
+ v2 (t ) −
+
+
v1 ( t )
v3 ( t )
−
−
i0 (t )
i1 (t )
i2 (t )
i3 (t )
KVL: v1(t ) + v2 (t ) + v3 (t ) = 0
KCL : − i0 (t ) + i1 (t ) + i2 (t ) + i3 (t ) = 0
vi (t ) = VMie j (ω t +θi ) , i = 1,2,3
ik (t ) = I Mk e j (ω t +φk ) , k = 0,1,2,3
In a similar way, one shows ...
KVL : (VM 1e jθ1 + VM 2e jθ 2 + VM 3e jθ 3 )e jωt = 0
VM 1∠θ1 + VM 2∠θ 2 + VM 3∠θ 3 = 0
V1 + V2 + V3 = 0 Phasors!
+ V2 −
+
+
V1
V3
−
−
− I 0 + I1 + I 2 + I 3 = 0
I0
I1
I2
I3
The components will be represented by their impedances and the relationships
will be entirely algebraic!!
SPECIAL APPLICATION:
IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED
FOR RESISTORS
I
I
+ V1 −
Z1
+ V2 −
+
I
Zs = Z1 + Z2
Z2
I
Z2 V
Z1
V
−
−
1
1
=∑k
Zp
Zk
Z s = ∑ k Zk
LEARNING EXAMPLE
+
Zp =
Z1Z 2
Z1 + Z 2
f = 60 Hz , v (t ) = 50 cos(ω t + 30°)
Compute equivalent impedance and current
ω = 120π , V = 50∠30°, Z R = 25Ω
ZR = R
Z L = j ωL
ZC =
1
jωC
1
j120π × 50 × 10−6
Z L = j 7.54Ω, Z C = − j 53.05Ω
Z L = j120π × 20 × 10−3 Ω , Z C =
Z s = Z R + Z L + Z C = 25 − j 45.51Ω
I=
V
50∠30°
50∠30°
=
( A) =
( A)
51.93∠ − 61.22°
Z s 25 − j 45.51
I = 0.96∠91.22°( A) ⇒ i (t ) = 0.96 cos(120π t + 91.22°)( A)
LEARNING EXTENSION
FIND i (t )
ω = 377
Z R = 20Ω
Z L = j 377 × 40 × 10 −3 = j15.08Ω
V = 120∠(60 − 90)°
ZC =
Z eq = ZC || ( Z R + Z L )
I=
j
= − j 53.05
377 × 50 × 10−6
Z eq = 30 .5616 + j 4 .9714 = 30.963∠9.239°
V
120∠ − 30°
=
= 3.876∠ − 39.924°( A)
Z eq 30.963∠9.239°
Parallel Combination of Admittances
(COMPLEX) ADMITTANCE
Y p = ∑ Yk
1
= G + jB (Siemens)
Z
G = conductanc e
B = Suceptanc e
Y =
k
YR = 0.1S
0.1S
Element Phasor Eq. Impedance
L
C
V = jωLI
1
V=
I
jωC
Series Combination of Admittances
1
1
=∑
Ys k Yk
R
R2 + X 2
−X
B= 2
R + X2
G=
V = RI
1
= j1( S )
− j1
Y p = 0.1 + j1( S )
1
1
R − jX
R − jX
×
= 2
=
Z R + jX
R − jX R + X 2
R
YC =
Z=R
Z = jωL
Z=
1
j ωC
Admittance
1
Y = =G
R
1
Y=
j ωL
Y = j ωC
− j 0.1S
1
1
1
=
+
Ys 0.1 − j 0.1
= 10 + j10
(0.1)(− j 0.1) 0.1 + j 0.1
×
0.1 − j 0.1 0.1 + j 0.1
1
10 − j10
Ys =
=
10 + j10
200
Ys = 0.05 − j 0.05 S
Ys =
LEARNING EXAMPLE VS = 60∠45°(V )
LEARNING EXTENSION
FIND Y p , I
Y p = YR + YL
= 0.5 − j 0.25
Y p = 0.5 − j 0.5 + j1 + 0.25 = 0.75 + j 0.5( S )
Y p = 0.9014∠33.69°( S )
2 × j4
2 + j4
Yp =
= 0.5 − j 0.25( S )
2 + j4
j8
I = Y pV = 0.9014∠33.69° × 10∠20°
I = Y pV = (0.5 − j 0.25) × 60∠45°( A)
I = 9.014∠53.79°( A)
Zp =
I = 0.559∠ − 26.565° × 60∠45°( A)
I = 33.54∠18.435°( A)
LEARNING EXAMPLE
SERIES-PARALLEL REDUCTIONS
1
2 − j4
=
2 + j 4 (2) 2 + (4) 2
1
4 + j2
=
Y34 =
4 − j2
20
Y2 =
Z3 = 4 + j 2
Y4 = − j 0.25 + j 0.5 = j 0.25
Z 4 = 1 / Y4 = − j 4
1× (− j 2)
1− j2
1
Z1 =
1 + j 0.5
Z1 =
1 − j 0.5
Z1 =
1 + (0.5) 2
Z1 = 0.8 − j 0.4(Ω)
Z4 =
j 4 × (− j 2) 8
=
j4 − j2
j2
Y2 = 0.1 − j 0.2( S )
Y34 = 0.2 + j 0.1
Z2 = 2 + j6 − j2 = 2 + j 4
Z 34 = 4 − j 2
Z 234 =
Y234 = 0.3 − j 0.1( S )
Z 234 =
1
Y234
=
1
0.3 + j 0.1
=
0.3 − j 0.1
0.1
Z 2 Z 34
= 3 + j1
Z 2 + Z 34
Z eq = Z1 + Z 234 = 3.8 + j 0.6Ω = 3.847∠8.973°
LEARNING EXTENSION FIND THE IMPEDANCE Z T
Z1 = 4 + j 6 − j 4
Z1 = 4 + j 2
Y12 = Y1 + Y2
( R → P ) Z1 = 4.472∠26.565°
Y1 = 0.224∠ − 26.565°
( P → R)Y1 = 0.200 − j 0.100
Y12 = Y1 + Y2 = 0.45 − j 0.35
( R → P )Y12 = 0.570∠ − 37.875°
Z12 = 1.754∠37.875°
( P → R) Z12 = 1.384 + j1.077
Z 2 = 2 + j 2 ( R → P ) Z 2 = 2.828∠45°
Y2 = 0.354∠ − 45°
( P → R)Y2 = 0.250 − j 0.250
1
4 − j2
ZT = 2 + (1.384 + j1077) = 3.383 + j1.077
Y1 =
= 2
4 + j 2 (4) + (2) 2
1
2 − j2
Y2 =
= 2
2 + j 2 (2) + (2) 2
1
1
0.45 + j 0.35
Z12 =
=
=
Y12 0.45 − j 0.35
0.325
1
Z12 =
Y12
PHASOR DIAGRAMS
Display all relevant phasors on a common reference frame
Very useful to visualize phase relationships among variables.
Especially if some variable, like the frequency, can change
LEARNING EXAMPLE
SKETCH THE PHASOR DIAGRAM FOR THE CIRCUIT
Any one variable can be chosen as reference.
For this case select the voltage V
KCL : I S =
V
V
+
+ jωCV
R j ωL
ω ↑ (capacitive)
| I L |>| I C |
| I L |<| I C |
I C = jωCV
IL =
V
jωl
ω ↑ (inductive)
INDUCTIVE CASE
CAPACITIVE CASE
LEARNING EXAMPLE
DO THE PHASOR DIAGRAM FOR THE CIRCUIT
ω = 377( s −1 )
2. PUT KNOWN NUMERICAL VALUES
| VL − VC |=| VR |
VR = RI
VL = jωLI
It is convenient to select
1
the current as reference
VC =
I
j ωC
VS = VR + VL + VC
1. DRAW ALL THE PHASORS
DIAGRAM WITH REFERENCE VS = 12 2∠90°
VL = 18∠135°(V )
Read values from
diagram!
∴ I = 3∠45°( A)
VR = 12∠45°(V )
(Pythagoras)
| VL |>| VC |
VC = 6∠ − 45°
FIND THE FREQUENCY AT WHICH v (t ) AND i (t )
LEARNING BY DOING
C
+
L
v (t )
−
ARE IN PHASE
i.e., the phasors for i (t ), v (t ) are co - lineal
1
V=
I + jωLI + RI
jωC
PHASOR DIAGRAM
R
jωLI
1
I
jωC
V=
RI
V and I are co - lineal iff jωL +
ω2 =
Notice that I was
chosen as reference
1
I + jωLI + RI
jωC
I
1
1
= 0 ⇒ω2 =
LC
jωC
1
9
4
=
10
⇒
ω
=
3
.
162
×
10
(rad / s )
−3
−6
10 × 10
f =
ω
= 5.033 × 103 Hz
2π
LEARNING EXTENSION
Draw a phasor diagram illustrating all voltages and currents
− j4
4∠ − 90°
I=
4∠45°
2 − j4
4.472∠ − 63.435°
I1 = 3.578∠18.435°( A)
I1 =
I2 =
Current
divider
1
2∠0°
I=
4∠45°
2 − j4
4.472∠ − 63.435°
I 2 = 1.789∠108.435° Simpler than I 2 = I − I1
V = 2 I1 = 7.156∠18.435°(V )
DRAW PHASORS. ALL ARE
KNOWN. NO NEED TO SELECT
A REFERENCE
BASIC ANALYSIS USING KIRCHHOFF’S LAWS
PROBLEM SOLVING STRATEGY
For relatively simple circuits use
Ohm' s law for AC analysis; i.e., V = IZ
The rules for combining Z and Y
KCL AND KVL
Current and voltage divider
For more complex circuits use
Node analysis
Loop analysis
Superposition
Source transformation
Thevenin' s and Norton' s theorems
MATLAB
PSPICE
LEARNING EXAMPLE
COMPUTE ALL THE VOLTAGES AND CURRENTS
Compute I1
Use current divider for I2 , I3
Ohm' s law for V1 , V2
V1 = 6∠90° I 2
Z eq = 4 + ( j 6 || 8 − j 4)
V1 = 16.26∠78.42°(V )
24 + j 48 32 + j8 + 24 + j 48
Z eq = 4 +
=
8 + j2
8 + j2
V2 = 7.28∠15°(V )
56 + j 56
79.196∠45°
=
= 9.604∠30.964°(Ω)
8 + j2
8.246∠14.036°
V
24∠60°
I1 = S =
= 2.498∠29.036°( A)
Z eq 9.604∠30.964°
Z eq =
j6
6∠90°
I1 =
2.498∠29.036( A)
8 + j2
8.246∠14.036
8 − j4
8.944∠ − 26.565°
I2 =
I1 =
2.498∠29.036°( A)
8 + j2
8.246∠14.036°
I3 =
I1 = 2.5∠29.06°
I 2 = 2.71∠ − 11.58°
V2 = 4∠ − 90° I 3
I 3 = 1.82∠105°
LEARNING EXTENSION
IF VO = 8∠45°, COMPUTE VS
THE PLAN...
COMPUTE I3
COMPUTE V1
COMPUTE I2 , I1
COMPUTE VS
VO
( A) = 4∠45°( A)
2
V1 = (2 − j 2) I 3 = 8∠ − 45° × 4∠45°
V1 = 11.314∠0°(V )
I3 =
I2 =
V1 11.314∠0°
=
= 5.657∠ − 90°( A)
j2
2∠90°
I1 = I 2 + I 3 = 5.657∠ − 90° + 4∠45°
I1 = − j 5.657 + (2.828 + j 2.828)( A)
I1 = 2.828 − j 2.829( A)
VS = 2 I1 + V1 = 2(2.828 − j 2.829) + 11.314∠0°
VS = 16.97 − j 5.658(V )
VS = 17.888∠ − 18.439°
ANALYSIS TECHNIQUES
PURPOSE: TO REVIEW ALL CIRCUIT ANALYSIS TOOLS DEVELOPED FOR
RESISTIVE CIRCUITS; I.E., NODE AND LOOP ANALYSIS, SOURCE SUPERPOSITION,
SOURCE TRANSFORMATION, THEVENIN’S AND NORTON’S THEOREMS.
COMPUTE I0
V2 − 6∠0°
V
− 2∠0° + V2 + 2 = 0
1 + j1
1 − j1
1 
6
 1
+1+
=
2
+
V2 
1 − j1
1 + j1
1 + j1
V2
(1 − j1) + (1 + j1)(1 − j1) + (1 + j1) 2(1 + j1) + 6
=
(1 + j1)(1 − j1)
1 + j1
V2
1. NODE ANALYSIS
V1
V
V
− 2∠0° + 2 + 2 = 0
1 + j1
1 1 − j1
V1 − V2 = −6∠0°
I0 =
V2
( A)
1
4
= 8 + j2
1− j
V2 =
( 4 + j )(1 − j )
2
3
5
I 0 =  − j ( A)
2
2
I 0 = 2.92∠ − 30.96°
NEXT: LOOP ANALYSIS
2. LOOP ANALYSIS
ONE COULD ALSO USE THE SUPERMESH
TECHNIQUE
SOURCE IS NOT SHARED AND Io IS
DEFINED BY ONE LOOP CURRENT
LOOP 1 : I1 = −2∠0°
I2
I 0 = − I3
LOOP 2 : (1 + j )( I1 + I 2 ) − 6∠0° + (1 − j )( I 2 + I 3 ) = 0
LOOP 3 : (1 − j )( I 2 + I 3 ) + I 3 = 0
CONSTRAINT : I1 − I 2 = −2∠0°
SUPERMESH : (1 + j ) I1 + 6∠0° + ( I 2 − I 3 ) = 0
MUST FIND I3
2 I 2 + (1 − j ) I 3 = 6 − (1 + j )(−2)
(1 − j ) I 2 + (2 − j ) I 3 = 0
((1 − j )
)
/* (1 − j )
/* (−2)
− 2(2 − j ) I 3 = (1 − j )(8 + 2 j )
5 3
10 − 6 j
I
=
−
+ j ( A)
I3 =
0
2 2
−4
2
MESH 3 : ( I 3 − I 2 ) + (1 − j ) I 3 = 0
I 0 = I 2 − I3
NEXT: SOURCE SUPERPOSITION
Circuit with voltage source
set to zero (SHORT CIRCUITED)
SOURCE SUPERPOSITION
I
I L2
1
L
=
1
L
V
+
Circuit with current
source set to zero(OPEN)
Due to the linearity of the models we must have
I L = I L1 + I L2
VL = VL1 + VL2
Principle of Source Superposition
The approach will be useful if solving the two circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient
VL2
3. SOURCE SUPERPOSITION
I 0' = 1∠0°( A)
Z ' = (1 + j ) || (1 − j ) =
(1 + j )(1 − j )
=1
(1 + j ) − (1 − j )
COULD USE SOURCE TRANSFORMATION
TO COMPUTE I"0
V1"
Z"
Z"
"
= "
6∠0°(V ) I 0 = "
6∠0°( A)
Z +1+ j
Z +1+ j
1− j
1− j
I 0" =
6
2
−
j
j
j
(
1
−
)
+
3
+
I 0" =
6 ( A)
1− j
6 6
"
+1+ j
I
=
− j ( A)
0
2− j
4 4
5 3 
I 0 = I 0' + I 0" =  − j ( A)
2 2 
Z"
Z " = 1 || (1 − j )
NEXT: SOURCE TRANSFORMATION
Source transformation is a good tool to reduce complexity in a circuit ...
WHEN IT CAN BE APPLIED!!
“ideal sources” are not good models for real behavior of sources
A real battery does not produce infinite current when short-circuited
ZV
+
-
RV
VS
a
ZI
a
RI
b
IS
THE MODELS ARE EQUIVALENTS WHEN
RV = RI = R
b
VS = RI S
Improved model
Improved model
for voltage source for current source
Source Transformationcan be used to determine the Thevenin or Norton Equivalent...
BUT THERE MAY BE MORE EFFICIENT TECHNIQUES
4. SOURCE TRANSFORMATION
IS =
8+2j
1+ j
Z = (1 + j ) || (1 − j ) = 1Ω
V '= 8 + 2 j
Now a voltage to current transformation
NEXT: THEVENIN
I0 =
I S 4 + j ( 4 + j )(1 − j ) 5 − 3 j
=
=
=
2 1 + j (1 + j )(1 − j )
2
THEVENIN’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
ZTH
+
−
RTH
vTH
+
i
a
vO
b
_
+
i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
PART B
PART A
Phasor
Thevenin Equivalent Circuit
for PART A
vTH
Thevenin Equivalent Source
RTH
Thevenin Equivalent Resistance
Impedance
5. THEVENIN ANALYSIS
Voltage Divider
VOC =
1− j
10 − 6 j
(8 + 2 j ) =
2
(1 + j ) + (1 − j )
ZTH = (1 + j ) || (1 − j ) = 1Ω
8+2j
I0 =
5−3j
( A)
2
NEXT: NORTON
NORTON’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
+
i
a
vO
b
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
Phasors
ZN
iN
RN
+
i
a
LINEAR CIRCUIT
vO
_
b
PART B
PART A
Norton Equivalent Circuit
for PART A
iN
Thevenin Equivalent Source
RNZ N Thevenin Equivalent Resistance
Impedance
6. NORTON ANALYSIS
ZTH = (1 + j ) || (1 − j ) = 1Ω
I0 =
Possible techniques: loops, source
transformation, superposition
BY SUPERPOSTION
I SC = 2∠0° +
6∠0° 8 + 2 j
=
( A)
1+ j
1+ j
I SC 4 + j ( 4 + j )(1 − j ) 5 − 3 j
=
=
=
2
1 − j (1 + j )(1 − j )
2
LEARNING EXAMPLE FIND V0 USING NODES, LOOPS, THEVENIN, NORTON
WHY SKIP SUPERPOSITION AND TRANSFORMATION?
Supernode constraint : V1 − V3 = 12∠0°
KCL @ Supernode
− 4∠0° +
V3 − V0 V3 − V2 V1 − V2 V3
+
+
+ =0
1
1
−j
j
KCL@V2
V2 − V1
V − V3
− 2I x + 2
=0
−j
1
NODES
KCL@ V0
V0 V0 − V3
+
+ 4∠0° = 0 ⇒ V3 = 2V0 + 4
1
1
V1 = V3 + 12
Controlling variable
V1 = 2V0 + 16
V3 − V0
Ix =
1
V3 − V0 = V0 + 4
j (V2 − 2V0 − 16) − 2(V0 + 4) + (V2 − 2V0 − 4) = 0
Notice choice of ground
− j (V2 − 2V0 − 16) − (V2 − 2V0 − 4) + (V0 + 4) − j ( 2V0 + 4) = 4
8+ 4 j
Adding : V0 = −
1+ 2 j
LOOP ANALYSIS
MESH CURRENTS DETERMINED BY SOURCES
I 2 = −4∠0°
I3 = 2 I x
⇒ I 3 = 2( I 4 + 4)
MESH 1 :
− jI1 + 12∠0° + 1( I1 − I 3 ) = 0
MESH 4 :
1( I 4 − I 2 ) + 1× I 4 + j ( I 4 − I 3 ) = 0
CONTROLLING VARIABLE : I x = I 4 − I 2
VARIABLE OF INTEREST : V0 = 1× I 4 (V )
MESH CURRENTS ARE ACCEPTABLE
I 4 + 4 + I 4 + j ( I 4 − 2( I 4 + 4)) = 0
( 2 − j ) I 4 = −( 4 − 8 j ) ⇒ I 4 = −
V0 = −
8+4j
1+ 2 j
4−8 j j
×
2− j
j
Alternative procedure to compute Thevenin
impedance:
1. Set to zero all INDEPENDENT sources
2. Apply an external probe
THEVENIN
ZTH = −
I "x
FOR OPEN CIRCUIT VOLTAGE
Vtest
I "x
KVL
Vtest = − I "x + jI "x ⇒ ZTH = 1 − j (Ω)
I x' = 4∠0°
ZTH = 1 − j
2I x
8∠0°
V0 =
VOC = −4 + 8 j (V )
1
(−4 + 8 j )(V )
2− j
Supernode constraint
⇒ V1 = V3 + 12
V1 − V 3 = 12 ∠ 0 °
NORTON
KCL@ Supernode
V3 V3 V3 − V2 V1 − V2
+ +
+
− 4∠0° = 0 /× j
1
j
1
−j
V2 − V3 V2 − V1
+
= 0 /× ( − j )
1
−j
V
Controlling Variable : I x''' = 3
1
KCL@ V2 : − 2 I X''' +
I SC
2 jV3 − j (V2 − V3 ) + (V2 − V3 − 12) = 0
(1 − j )V2 − (1 − 3 j )V3 = 12
(1 + j )V3 + jV3 − jV2 − (V3 + 12) + V2 = 4 j
I x''' =
V3
( A)
1
I SC = I x''' − 4
USE NODES
(1 − j )V2 + 2 jV3 = 12 + 4 j
4j
−4+8j
(1 − j )V3 = 4 j ⇒ V3 =
⇒ I SC =
1− j
1− j
I SC =
(−4 + 8 j ) j
8+4j
=−
(1 − j ) j
1+ j
Now we can draw the Norton
Equivalent circuit ...
NORTON’S EQUIVALENT CIRCUIT
ZTH
I SC
V0 = (1) I 0 (V ) =
1− j  8 + 4 j 
−
(V ) Current Divider
2 − j  1+ j 
EQUIVALENCE OF SOLUTIONS
12 − 4 j
(8 + 4 j )(1 − j )
=−
3+ j
(1 + 2 j )(1 − j )
Using Norton’s method
V0 = −
Using Thevenin’s
−4+8j j
×
V0 =
j
2− j
Using Node and Loop methods V0 = −
8+4j
1+ 2 j
LEARNING EXTENSION
COMPUTE V0
USE THEVENIN
USE NODAL ANALYSIS
V1
4j
VOC
V1 − 12∠30° V1 V1 V1 − V0
+ + +
= 0 /× 2 j
2
1 j2
−j
3 = 4 j = 4 j (2 − 6 j )
2 + j2 2 + 6 j
40
3
1 || j 2
j2
=
12∠30°=
12∠30°
2 + (1 || j 2)
2(1 + 2 j ) + 2 j
24∠120° 12∠120°
=
=
2+6j
1+ 3 j
ZTH = 2 || 1 || j 2 =
VOC
V0 − V1 V0
+ = 0 ⇒ V1 = (1 − j )V0
−j
1
ZTH
+
j (V1 − 12∠30°) + 2 jV1 + V1 − 2(V1 − V0 ) = 0
2V0 + (1 − 2 + 2 j + j )(1 − j )V0 = j12∠30°
(2 + (−1 + 3 j )(1 − j ))V0 = 1∠90° × 12∠30°
12∠120° 12∠120°
V0 =
=
= 2.12∠75°(V )
4+4j
5.66∠45°
− j1
VOC
1Ω
+
-
V0
−
V0 =
ZTH
1
VOC
+1− j
LEARNING EXTENSION COMPUTE V0 USING MESH ANALYSIS
USING NODES
V1
V1 − 24∠0°
V
− 2∠90° + 1 = 0
2
2−2j
2
V0 =
V1
2−2j
USING SOURCE SUPERPOSITION
CONSTRAINT
− I1 + I 2 = 2∠90° ⇒ I1 = I 2 − 2 j
SUPERMESH
− 24∠0° + 2 I1 − 2 jI 2 + 2 I 2 = 0
2( I 2 − 2 j ) + (2 − 2 j ) I 2 = 24⇒ (4 − 2 j ) I 2 = 24 + 4 j
V0 = 2 I 2 =
24 + 4 j
24.33∠9.46°
=
= 10.86∠36.03°
2− j
2.24∠ − 26.57°
V0V =
2
24∠0°
2+2−2j
V0I = 2 ×
2
2∠90°
4−2j
V0 =V0V +V0I
LEARNING EXTENSION
COMPUTE V0
V0"
V0 = V0' + V0"
1. USING SUPERPOSITION
2 || ( 2 − 2 j )
V2 =
V1
V1 =
2 || (2 − 2 j )
(−12∠0°)
j 2 + (2 || 2 − 2 j )
V0'
V0' =
2
V1
2−2j
( 2 j ) || (2 − 2 j )
24∠0°
2 + (2 j || (2 − 2 j )
V0" =
2
V2
2−2j
2. USE SOURCE TRANSFORMATION
−2j
Z
I eq
I1
+
2 V0
−
I eq = 12∠0° − 6∠ − 90° = 12 + 6 j
I1 =
V0 = 2I1
− j 2Ω
12∠0° 2Ω
I1 2Ω
j 2Ω 6∠ − 90°
Z = 2 || j 2
Z
I eq
Z +2−2j
+
V0
−
USE NORTON’S THEOREM
− j2
I1
I SC
ZTH = 2 || j 2
ZTH
I1 =
+
2 V0
−
ZTH
I SC
ZTH + 2 − 2 j
V0 = 2 I1
12∠0°
− 6∠ − 90°
I SC
LEARNING EXAMPLE
Find the current i(t) in steady state
The sources have different frequencies!
For phasor analysis MUST use source superposition
SOURCE 1: FREQUENCY 10 r/s
SOURCE 2: FREQUENCY 20r/s
Principle of superposition
Frequency domain