Basics in Systems and Circuits Theory
BSC Modul 2: Complex Algebra and AC Circuits
•
•
•
Michael E.Auer
Complex Algebra
Sinusoids and Phasors
Steady-State Analysis
01.11.2011
BSC02
Basics in Systems and Circuits Theory
BSC Modul 2: Complex Algebra and AC Circuits
• Complex Algebra
•
•
Michael E.Auer
Sinusoids and Phasors
Steady-State Analysis
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Complex Numbers
A complex number is a number of the form
z = x + j⋅ y
Where a and b are real numbers and j is the imaginary unit defined by
j = -1
a – real part of the complex number
b – imaginary part of the complex number
z = Re{z}+ j ⋅ Im{z}
In applications, complex numbers often are dimension, in which
case the real part and the imaginary part must have the same
dimension!
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Polar Form of a Complex Number
A complex number, when written as
z = x + j⋅ y
is said to be expressed in rectangular form, also known as
cartesian coordinates.
But it also may be expressed in polar form:
z = r ∠φ
where
Michael E.Auer
r=
x2 + y2
φ = tan −1
y
x
Magnitude
Cotangent
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Conjugate of a Complex Number
The conjugate z* of a complex number z is obtained by
changing the sign of the imaginary part of the number.
Conjugate of z = z * = Re{z}− Im{z}= r ∠ (−φ )
Conjugation is distributive over addition, multiplication and division:
( z1 + z 2 )* = z1* + z 2*
( z1 ⋅ z 2 )* = z1* ⋅ z 2*
*
 z1 
z1*
  = *
z2
 z2 
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Complex Arithmetic
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 ∠ φ1 + φ2
4.
Division
z1 r1
= ∠φ1 − φ 2
z 2 r2
5.
Reciprocal
1 1
= ∠ −φ
z
r
6.
Square root
7.
Complex conjugate
Michael E.Auer
z = r ∠φ 2
z ∗ = x − jy = r ∠ − φ
01.11.2011
BSC02
Basics in Systems and Circuits Theory
The Complex Plane
Im ( z )
Complex numbers represented by points
3+ j4
−3 + j 3
j2
−4
Re ( z )
3− j2
Operations in a complex plane
−3 − j 3
Im ( z )
Im ( z )
z1 + z2
jz
z
z
z1
θ
θ
z2
Re ( z )
θ
−z
z*
(b)
(a)
Michael E.Auer
Im ( z )
01.11.2011
π
z
π
2
π
2
Re ( z )
j2 z = −z
2
Re ( z )
π
2
j3 z = − j z
(c)
BSC02
Basics in Systems and Circuits Theory
BSC Modul 2: Complex Algebra and AC Circuits
•
Complex Algebra
• Sinusoids and Phasors
•
Michael E.Auer
Steady-State Analysis
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Introduction
How can we determine
i(t) and v(t)?
vs = 10V (DC !)
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Sinusoids (1)
•
•
A sinusoid is a signal that has the form of the sine or cosine
function.
A general expression for the sinusoid is:
v(t ) = V sin(ωt
m
+ φ)
where
Vm = the amplitude of the sinusoid
ω = the angular frequency in radians/s
Ф = the phase
Michael E.Auer
01.11.2011
ω = 2πf
f =
1
T
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Basics in Systems and Circuits Theory
Sinusoids (2)
T=
f =
•
•
Michael E.Auer
1
T
2π
ω
ω = 2πf
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.
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Sinusoids (4)
Phasor
Time functions
Michael E.Auer
01.11.2011
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Basics in Systems and Circuits Theory
Sinusoids (5)
Example:
A ⋅ cos ωt + B ⋅ sin ωt = C ⋅ cos(ωt + Θ)
where
C = A2 + B 2 and
Michael E.Auer
Θ = arctan
3 ⋅ cos ωt − 4 ⋅ sin ωt = 5 ⋅ cos(ωt + 53.1°)
B
A
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Basics in Systems and Circuits Theory
Sinusoids Example (1)
Given a sinusoid
5 sin( 4πt − 60 o )
calculate its amplitude, phase, angular frequency, period,
and frequency.
Solution:
Amplitude = 5, phase = –60o, angular frequency = 4π rad/s,
period = 0.5 s, frequency = 2 Hz.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Sinusoids 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:
Since sin (ωt+90o) = cos ωt
i2 = 5 sin(377t − 40o + 90o ) = 5 sin(377t + 50o )
i1 = −4 sin(377t + 25o ) = 4 sin(377t + 180o + 25o ) = 4 sin(377t + 205o )
therefore, i1 leads i2 155o.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Differention and Integration of Sinusoids
lags 90°
leads 90°
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
v – i –Relations in the Time Domain (1)
vR = R ⋅ i
di
vL = L ⋅
dt
1
vC = ⋅ ∫ i ⋅ dt + vC (t0 )
C
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Circuit Analysis in the Time Domain (1)
R
Example
vR = R ⋅ i
vS
C
dv
i =C⋅
dt
The time domain analysis of electrical circuits results in
differential equation systems!
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Circuit Analysis in the Time Domain (2)
•
We can derive the differential equations for the following circuit in
order to solve for vo(t) in phase domain Vo.
d 2 vo 5 dv0
400
o
20
v
sin(
4
t
15
)
+
+
=
−
−
0
2
dt
3 dt
3
•
However, the derivation may sometimes be very tedious.
Is there any quicker and more systematic methods to do it?
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Phasor (1)
•
A phasor is a complex number that
represents the amplitude and phase of
a sinusoid.
•
It can be represented in one of the
following three forms:
a. rectangular
b. polar
A = x + jy = r (cos φ + j sin φ )
where
r = x2 + y2
A = re jφ
A = r ∠φ
φ = arctan
and
y
y
= tan −1
x
x
x = r ⋅ cos φ
y = r ⋅ sin φ
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Phasor (3)
•
Transform a sinusoid to and from the time domain to the phasor domain:
v(t ) = Vm cos(ωt + φ )
(time domain)
V = Vm ∠φ
(phasor domain)
•
Amplitude and phase difference are two principal concerns in the
study of voltage and current sinusoids.
•
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.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Phasor (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 x = cos (x+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor => V = 4∠140° V
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Phasor (5)
Example 5:
Transform the sinusoids corresponding to phasors:
a. V = − 10∠30° V
b. I = j(5 − j12) A
Solution:
a) v(t) = 10cos(ωt + 210o) V
5
b) Since I = 12 + j5 = 12 + 5 ∠ tan ( ) = 13∠ 22.62°
12
2
2
−1
i(t) = 13cos (ωt + 22.62°) A
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Phasor (6)
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.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Differentiation and Integration of a Phasor (1)
A = r ⋅e
j (ωt +φ )
dA
= A ⋅ jω
dt
dA
= r ⋅ e j (ωt +φ ) ⋅ jω = A ⋅ jω = Aω ⋅ e j⋅90°
dt
A = r ⋅ e j (ωt +φ )
∫ A ⋅ dt = A ⋅
∫ A ⋅ dt = r ⋅ e
Michael E.Auer
j (ωt +φ )
1
j
1
A A − j⋅90°
⋅
=
= ⋅e
jω jω ω
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Differentiation and Integration of a Phasor (2)
V = V∠φ
v(t )
Michael E.Auer
dv
dt
j ωV
∫ vdt
V
jω
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Circuit Analysis
•
We can derive the differential equations for the following circuit in
order to solve for vo(t) in phase domain Vo.
d 2 vo 5 dv0
400
+
+
v
=
−
20
sin( 4t − 15o )
0
2
dt
3 dt
3
Is there any quicker and more systematic methods to do it?
The answer is YES!
Instead of first deriving the differential equation and then transforming it into
phasor to solve for Vo, we can transform all the RLC components into
phasor first, then apply the KCL laws and other theorems to set up a phasor
equation involving Vo directly.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Resistor
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Inductor
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Capacitor
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Summary of Voltage-Current Relationship
Element
Michael E.Auer
Time domain
Frequency domain
R
v = Ri
V = R⋅I
L
di
v=L
dt
V = j ωL ⋅ I
C
dv
i=C
dt
I
01.11.2011
V=
j ωC
BSC02
Basics in Systems and Circuits Theory
Example
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
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Impedance and Admittance (1)
• The impedance Z of a circuit is the ratio of the phasor voltage V to
the phasor current I, measured in ohms Ω.
V
Z = = R + jX
I
where R = Re(Z) is the resistance and X = Im(Z) is the reactance.
Positive X is for L and negative X is for C.
• The admittance Y is the reciprocal of impedance, measured in
siemens (S).
I
Y = = = G + jB
Z V
1
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Impedance and Admittance (2)
• The impedance and admittance can be represented in rectangular or
polar coordinates
Michael E.Auer
V
Z = = R + jX = Z ∠φ
I
where:
I
Y = = G + jB = Y ∠γ
V
where:
Z = Z = R2 + X 2
φ = arctan
01.11.2011
X
R
Y = Y = G2 + B2
γ = arctan
B
= −φ
G
BSC02
Basics in Systems and Circuits Theory
Impedance and Admittance (3)
Impedances and admittances of passive elements
Element
Michael E.Auer
Impedance
Admittance
R
Z=R
1
Y=
R
L
Z = j ωL
Y=
C
1
Z=
j ωC
01.11.2011
1
j ωL
Y = j ωC
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Basics in Systems and Circuits Theory
Impedance and Admittance (4)
ω = 0; Z = 0
Z = j ωL
ω → ∞; Z → ∞
ω = 0; Z → ∞
Z=
Michael E.Auer
1
j ωC
ω → ∞; Z = 0
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Kirchhoff’s Law in the Frequency Domain
• Both KVL and KCL are hold in the phasor domain or
more commonly called frequency domain.
• Moreover, the variables to be handled are phasors,
which are complex numbers.
• All the mathematical operations involved are now in
complex domain.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Impedance Combinations
• The following principles used for DC circuit analysis all
apply to AC circuit.
• For example:
• voltage division
• current division
• circuit reduction
• impedance equivalence
• Y-Δ transformation
• …
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Series Impedances
Z eq = Z1 + Z 2 + ⋅ ⋅ ⋅ Z N
V1 Z1
=
V2 Z 2
Michael E.Auer
01.11.2011
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Basics in Systems and Circuits Theory
Parallel Impedances
1
Yeq = Y1 + Y2 + ⋅ ⋅ ⋅ YN =
Z eq
I1 Y1 Z 2
= =
I 2 Y2 Z1
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Example
R
R
C
L
1
j ωC
Z eq = R + ( jωL
Michael E.Auer
01.11.2011
1
j ωC
j ωL
)
BSC02
Basics in Systems and Circuits Theory
BSC Modul 2: Complex Algebra and AC Circuits
•
•
Complex Algebra
Sinusoids and Phasors
• Steady-State Analysis
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Circuits in the Phasor Domain (1)
After we know how to convert RLC components
from time to phasor domain, we can transform
a time domain circuit into a phasor/frequency
domain circuit.
Hence, we can apply the KCL laws and other
theorems to directly set up phasor equations
involving our target variable(s) for solving.
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Circuits in the Phasor Domain (2)
Time Domain
Phasor Domain
Differential
Equations
Algebraic
Equations
complex !
Transformation
Solution of the
Differential
Equations
Solution of the
Algebraic
Equations
Solution
Solution
Retransformation
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Basic Approach
1.
2.
3.
Transform the circuit to the phasor or frequency
domain.
Solve the problem using circuit techniques (nodal
analysis, mesh analysis, superposition, etc.).
Transform the resulting phasor to the time domain.
Time to Freq
Michael E.Auer
Solve
equations in
Freq domain
01.11.2011
Freq to Time
BSC02
Basics in Systems and Circuits Theory
Basic Example (1)
Time domain
1
j ωC
Frequency domain
Michael E.Auer
Vs
V
s
01.11.2011
V0
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Basics in Systems and Circuits Theory
Basic Example (2)
Time domain
Frequency domain
Michael E.Auer
Vs
01.11.2011
j ωL
1
j ωC
V0
BSC02
Basics in Systems and Circuits Theory
Thevenin and Norton Equivalent Circuits (1)
Thevenin transform
Norton transform
Michael E.Auer
01.11.2011
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Basics in Systems and Circuits Theory
Thevenin and Norton Equivalent Circuits (2)
R = 1Ω
Example
v(t ) = 5 cos t V
Michael E.Auer
C = 1F
j ωL
R
V
L = 1H
1
j ωC
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Thevenin and Norton Equivalent Circuits (3)
Find the Thevenin equivalent at a - b:
30 < 20° V
Solution:
Michael E.Auer
ZTh = 12.4 – j3.2 ; VTh = 18.97 < -51.57° V
01.11.2011
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Basics in Systems and Circuits Theory
Source Transformation (1)
Michael E.Auer
01.11.2011
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Basics in Systems and Circuits Theory
Source Transformation (2)
Use source transformation
to obtain Vx:
5,52 < -28° V
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01.11.2011
BSC02
Basics in Systems and Circuits Theory
Mesh Analysis
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Nodal Analysis
Michael E.Auer
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Superposition (1)
Michael E.Auer
•
for linear circuits only
•
applies to ac circuits the same way it applies to dc circuits
•
circuits with more than one source
•
sources operating at different frequencies possible
01.11.2011
BSC02
Basics in Systems and Circuits Theory
Superposition (2)
Example
dc voltage source
Michael E.Auer
ac voltage source
01.11.2011
ac current source
BSC02