3.1
Electric Circuits 1301ENG
Module 3: Sinusoidal AC Analysis
References
[1] R. L. Boylestad, Introductory Circuit Analysis, Prentice
Hall, 8th ed., 1997, chapters 13 and 14.
[2] A. R. Hambley, Electrical Engineering: Principles and
Applications, Prentice Hall, 1997, chapter 5.
3.2
3.1 Introduction
We have so far considered dc (direct current) signals whose
amplitudes do not change with time
Now consider ac (alternating current) signals whose
amplitudes change with time
• ac means sinusoidally varying voltage or current
• ac voltage means a sinusoidally varying voltage
(whether or not there is any current)
Sinusoidal ac signals are important because
• electrical power is normally generated and distributed in
ac form eg. power is distributed to household
powerpoints as a 50-Hz 240-V ac voltage
• many practical signals are sinusoids or can be expressed
as a sum of sinusoids
• sinusoidal “carriers” are used for modulation in radio
and other forms of telecommunications
• circuits are commonly specified and tested using
sinusoidal signals with a range of different frequencies
3.2 Sinusoidal waveform basics
Definitions
Figure 1 shows a sinusoidal voltage waveform as a
function of time t
− vertical scaling is in volts (v)
− horizontal scaling is always in units of time (eg. second
(s))
3.3
v (t )
T1
Vm
0
t Vp-p
Vm
T2
T3
Figure 1
1s
(a)
1s
T = 0.4 s
(b)
1s
T = 0.5 s
(c)
Figure 2
Peak amplitude Vm : maximum value measured from
average or mean. Mean value is zero in Fig. 1.
Peak-to-peak value Vp-p : sum of positive and negative peak
amplitudes. Vp-p = 2Vm in Fig. 1.
Period T: time interval between successive repetitions of
similar points. Period T1 = T2 = T3 in Fig. 1. Thus,
sinusoidal waveforms are periodic, repeating same pattern
of values in each period.
Cycle: portion of waveform in one period.
Frequency f:
3.4
• Number of cycles in one second
1
cycles
2
per second in Fig. 2(b) where T = 0.4 s , 2 cycles per
second in Fig. 2(c) where T = 0.5 s .
⇒ Large frequency means large number of cycles per
second
⇒ Fig. 2(b) shows a higher frequency waveform than
those in Figs. 2(a) and (c)
• Frequency is 1 cycle per second in Fig. 2(a), 2
• Frequency has unit of hertz
1 hertz (Hz) = 1 cycle per second
• Different frequencies in everyday life applications
Sound (human ear): 15 Hz−20 kHz
FM radio: 88 MHz−108 MHz
AM radio: 540 kHz−1600 kHz
TV: 54 MHz−890 MHz
• Frequency f is inversely proportional to period T
f = Hz
1
T
T = seconds (s)
1
T=
f
eg. A 50-Hz ac voltage has a period of 20 ms.
f =
Radians/Degrees
3.5
Unit of measurement for horizontal axis (abscissa) of Figs.
1 and 2 are sometimes expressed in radians (rad) or degrees
There are 2π rad around a 360o circle (see Fig. 3)
2 π rad = 360o
π ≅ 314
.
1 rad = 57.3o
In sinusoidal analysis, angle is measured as positive from
horizontal axis of a circle (ie. from 0o ).
90o
π 2 rad
(1.57 rad)
2
o
180
π rad
(3.14 rad)
3
1 rad
0.28 rad
4
5
6
270o 3π 2 rad
(4.71 rad)
Figure 3
Conversion between radians and degrees:
0o
2π rad
(6.28 rad)
3.6
 π 
Radians = 
 × (degrees)
 180o 
 180o 
 × ( radians)
Degrees = 

 π 
Example 1
90°:
π
rad :
3
Radians =
π
(90o ) =
π
rad
2
180o
180o  π 
o
Degrees =
  = 60
π  3
Mode in your calculator may be in RAD or DEG.
Angular velocity
Angular velocity ω is velocity of radius vector rotating
about center of circle (see Fig. 4).
angular distance (radians) θ
Angular velocity ω =
=
time taken (seconds)
t
ω
θ
Figure 4
3.7
Time required to complete one revolution (or cycle) is
period t = T seconds and angular distance travelled in this
time interval is θ = 2 π radians. Thus,
ω=
2π
T
( rad s )
Thus, smaller T gives larger ω (ie. rotating vector takes
less time to complete one cycle of 2π radians and hence it
must travel at a faster speed).
Also, angular frequency ω is proportional to frequency f
ω = 2 πf
( rad s )
eg. angular velocity of a 50-Hz sinusoidal ac voltage is
ω = 2 π (50) = 314 rad s .
3.3 Sinusoidal waveforms
By convention, cosine (rather than sine) waveform is used
in sinusoidal analysis.
A sinusoidal voltage is generally given by:
v (t ) = Vm cos(ωt + θ)
Vm = peak amplitude of voltage (v)
ω = 2 π T = 2 πf = angular frequency (rad/s)
θ = phase (or displacement) angle (rad or degrees)
3.8
Lowercase letter represents time-varying or ac signal (eg.
v (t ) ). Capital letter represents amplitude (eg. Vm )
Case 1: v1(t ) = Vm cos(ωt ) for θ = 0
For one cycle (ie. 0 ≤ ωt ≤ 360o ), Fig. 5(a) shows that v1( t )
has:
• maxima of Vm at ωt = 0 and ωt = 360o
• minima of − Vm at ωt = 180o
• zero value at ωt = 90o and ωt = 270o
• thus, these maxima, minima and zero value occur at
ωt =
v1(t ) =
0 90o 180o 270o 360o
↑ ↑
↑
↑
↑
Vm 0 − Vm 0 Vm
(
)
Case 2: v2 (t ) = Vm cos ωt + 90o for θ = 90o
Similar to Case 1, maxima, minima and zero value of v2 (t )
for one cycle occur at:
ωt + 90o =
0
ωt = − 90o
↑
v2 (t ) =
Vm
90o 180o 270o
0
90o 180o
↑
↑
↑
0 − Vm
0
360o
270o
↑
Vm
The dotted curve of Fig. 5(b) shows a plot of v2 (t ) .
Compared to v1( t ) (solid curve of Fig. 5(b)), waveform
v2 (t ) (dotted curve) is shifted to left of time axis by 90o .
3.9
Thus, v2 (t ) leads v1(t ) by 90o (or v1(t ) lags v2 (t ) by 90o )
because peaks of v2 (t ) occur before peaks of v1(t ) .
Note that v2 (t ) = − sin(ωt ) and hence
(
sin(ωt ) = −Vm cos ωt + 90o
(
)
)
Case 3: v3 (t ) = Vm cos ωt − 90o for θ = −90o
Compared to v1(t ) (solid curve of Fig. 5(c)), waveform
v3 (t ) (dotted curve) is shifted to right of time axis by 90o .
v3 (t ) lags v1(t ) by 90o (or v1(t ) leads v3 (t ) by 90o )
because peaks of v3 (t ) occur after peaks of v1(t ) .
Note that v3 (t ) = sin(ωt ) and hence
(
sin(ωt ) = Vm cos ωt − 90o
)
We can thus deduce that:
− Cosine wave leads sine wave by 90o
− Sine wave lags cosine wave by 90o
− Cosine and sine waves are 90o out of phase
Remarks
• θ > 0 means phase lead
• θ < 0 means phase lag
• For v (t ) = Vm cos(ωt + θ) and given v (t ) = V we have:
3.10
V 
ωt = cos−1  − θ
 Vm 
(a) θ = 0
v (t )
v1 ( t ) = Vm cos( ω t )
Vm
180 o
− 90o
90o
0
270o
360o
ωt
450o
− Vm
(b) θ = 90o
v (t )
(
v 2 ( t ) = V m cos ω t + 90 o
90o
)
90o
v1 ( t ) = Vm cos(ω t )
Vm
− 180 o
− 90o
0
90 o
180o
− Vm
90o
Figure 5
(c) θ = −90o
270o
360o
450 o
ωt
3.11
(
v3 ( t ) = Vm cos ωt − 90o
v(t )
o
90
0
90o
v1( t ) = Vm cos(ωt )
Vm
− 90o
)
90o
180o
270o
360o
450o
540o
ωt
−Vm
90o
Figure 5 continued
Relationship between Cosine and Sine
Figure 6 shows geometric relationship between various
forms of sine and cosine functions
• Starting at + sinα position, + cosα is an additional 90o
in counterclockwise direction. Thus,
(
cos α = sin α + 90o
)
• Starting at + sinα position, we must travel 180o in
counterclockwise ( + 180o ) or clockwise ( − 180o ) to get to
− sinα . Thus,
(
− sin α = sin α ± 180o
)
3.12
• Using Fig. 6, various sine-cosine relationships can be
obtained:
( )
sin α = cos(α − 90o )
− sin α = sin(α ± 180o )
− cos α = sin(α + 270o ) = sin(α − 90o )
cos α = sin α + 90o
etc.
Also, sine is an odd function and cosine is an even
function:
sin( −α ) = − sin α
cos( −α ) = cos α
+ cosα
− sinα
+ sinα
− cosα
Figure 6
Example 2
Given
α
and
using
cos( −α ) = cos α we have:
(
sin α = cos α − 90o
)
and
3.13
α = 0:
sin 0o = cos( −90o ) = cos(90o ) = 0
α = 30o :
sin 30o = cos( −60o ) = cos(60o ) = 0.5
α = 45o :
sin 45o = cos( −45o ) = cos( 45o ) = 1
α = 60o :
sin 60o = cos( −30o ) = cos(30o ) = 3 2 = 0.866
α = 90o :
sin 90o = cos(0) = 1
2 = 0.707
3.4 Average value
Average value of any current or voltage is value indicated
on a dc meter
• average value is equivalent dc value over one cycle
When both dc and ac voltage sources are applied to an
electrical circuit
• need to know dc and ac components of voltage or
current in various parts of circuit
dc (or average value) =
Sum of areas over one period
one period T
dc value of rectangular wave
Determine dc components of waveforms of Fig. 7.
In Fig. 7(a):
By inspection, area above axis equals area below over one
period giving dc = 0 V. Also, using the above dc equation:
3.14
(10 V)(1 ms) − (10 V)(1 ms)
2 ms
=0V
dc =
In Fig. 7(b):
(14 V)(1 ms) − (6 V)(1 ms)
dc =
2 ms
=4V
In fact, waveform of Fig. 7(b) is square wave of Fig. 7(a)
with a dc shift of 4 V:
v2 = v1 + 4 V
(a)
v1(V)
(Square wave)
10 V
0
1
2
3
− 10 V
1 period
Figure 7(a)
4
t (ms)
3.15
v2 (V)
dc value = 4 V
14 V
4V
0
1
2
3
4
t (ms)
−6 V
1 period
Figure 7(b)
dc value of sinusoidal waveforms
Area under one positive (or negative) half of a sine wave
(see Fig. 8) is:
π
Area = ∫ Am sin α dα
0
= Am[ − cos α ] 0π
[
]
= − A [ − 1 − ( +1) ]
= − Am cos π − cos 0o
m
o
= 2 Am
average value =
2 Am
= 0.637 Am
π
dc or average value of a pure sinusoidal waveform over
one period is zero
3.16
Am
π
0
Figure 8
Example 3
Determine dc value of sinusoidal waveform of Fig. 9.
Peak-to-peak value = 16 mV + 2 mV = 18 mV
Peak value = 18 mV/2 = 9 mV
Thus, dc value is 9 mV down from 2 mV (or 9 mV up from
-16 mV) ⇒ dc = -7 mV.
v
+ 2 mV
t
0
dc = −7 mV
− 16 mV
Figure 9
3.17
Instrumentation to measure dc or average value
dc (or average value) of any waveform can be measured
using a digital multimeter (DMM) or an oscilloscope.
• DMM can measure voltage or current levels of any
waveform (dc or ac signal) by setting to dc mode
• Oscilloscope can only measure voltage levels
Oscilloscope to measure dc levels of dc circuits
1. Choose GND (ground) from DC-GND-AC mode
− GND mode blocks any signal into oscilloscope
− A horizontal line is shown
− Set horizontal line in middle of screen (see Fig. 10(a))
2. Apply oscilloscope probe to voltage to be measured and
switched to DC mode
− If dc voltage is present, horizontal line will shift up
(positive voltage) or down (negative voltage). (Fig. 10(b))
− Positive voltage ≡ higher potential at red or positive lead
− Negative voltage ≡ lower potential at red or positive lead
3. Measure voltage using:
Vdc = ( vertical shift in div. ) × ( vertical sensitivity in V div.)
For waveform of Fig. 10(b),
Vdc = (2.4 div. )(50 mV div.) = 120 mV
3.18
Shift=2.4div.
Vertical sensitivity=50 mVdiv.
(a)
(b)
Figure 10
Oscilloscope to measure dc or average level of any
waveform
1. Using GND, reset horizontal line to middle of screen
2. Switch to AC
− only passes ac (alternating or changing) components of
signal (see Fig. 11(a))
− blocks dc components of signal
3. Then switch to DC
− passes both dc and ac components of signal
− ac waveform of Fig. 11(a) is shifted up to give waveform
consisting of dc and ac components of signal (see Fig.
11(b)
− Average value of waveform of Fig. 11(b) is:
Vaverage = Vdc = (0.8 div. )(5 V div.) = 4 V
3.19
Reference
Shift=0.8div.
level
(a)
(b)
Figure 11
3.5 Effective values (or root mean square (rms))
To determine amplitude of a sinusoidal ac current required
to deliver same power as a dc current using Fig. 12.
iac = Im sinωt
Thermometer
Switch 2
v
Switch 1
R
Water bath
Idc
dc source
ac generator
V
Figure 12
If switch 1 only is closed, dc power is delivered to resistor
R by dc current I dc
If switch 2 only is closed, ac power is delivered to resistor
by ac peak current I m
3.20
For same temperature, average power delivered to resistor
by ac source is same as that delivered by dc source.
The ac power delivered to resistor at any instant of time is:
2
Pac = (iac )2 R = ( I m sin ωt )2 R = ( I m
sin2 ωt ) R
1
sin2 ωt = (1 − cos 2ωt )
2
2
Im
R
Pac =
−
2
(trigonometric identity)
2
Im
R
cos 2ωt
2
Average power delivered by ac source is just first term
since average value of a sine wave is zero.
Equating average power delivered by ac source to that
delivered by dc source:
Pav(ac) = Pdc
2
Im
R
2
R and I m = 2 I dc
= I dc
2
I
I dc = m = 0.707 I m
2
The equivalent dc value of a sinusoidal current or voltage
is 1 2 or 0.707 of its maximum value regardless of signal
frequency
3.21
Equivalent dc value = effective value = root mean square
(rms) value
In summary:
I
I rms = I eff = I eq dc = m = 0.707 I m
2
I m = 2 I rms = 2 I eff
Example 4
An ac current with peak value of 2 (10) = 14.14 A is
required to deliver same power to resistor in Fig. 12 as a dc
current of 10 A.
rms or effective value of any time-varying waveform (not
necessarily a sinusoidal wave) is:
T
2
∫ i (t )
I rms = I eff =
0
T
area(i 2 (t ))
I rms = I eff =
T
Note that the above rms or effective value equations are
also applicable to time-varying voltages.
Example 5
Find the rms or effective value of waveform of Fig. 13(a).
3.22
v 2 ( t ) curve is shown in Fig. 13(b).
Vrms = Veff =
(100)(2) + (16)(2) + (4)(2)
= 4.899 V
10
v (V)
(a)
1 Period
4
0
−2
2
4
6
8
10
t (s)
− 10
v 2 (V)
(b)
100
16
4
0
t (s)
2
4
6
8
10
Figure 13
3.6 ac meters and instruments
Digital Multimeters (DMMs)
dc meters can be used to measure sinusoidal ac voltages
and currents by using a bridge rectifier which consists of
four diodes or electronic switches (see Fig. 14).
DMMs operate using this technique.
3.23
+
D3
+
−
vi
D1
v movement
D4
D2
−
Figure 14
v movement
vi
Vm
Vm
0
π
2π
α
0
− Vm
Vdc = 0.637Vm
π
2π
α
(b )
(a)
Figure 15
vi > 0 for 0 ≤ α ≤ π (see Fig. 15(a))
• Diodes D1 and D2 are conducting (ie. they are equivalent
to short circuits) and D3 and D4 are not conducting (ie.
open circuits)
• vmovement = vi for 0 ≤ α ≤ π (see Fig. 15(b)).
vi < 0 for π ≤ α ≤ 2 π (see Fig. 15(a))
• D1 and D2 are open circuits and D3 and D4 short circuits
3.24
• vmovement = − vi for π ≤ α ≤ 2 π ; negative portion of
input is “flipped over” (see Fig. 15(b)).
Zero average value of Fig. 15(a) has been replaced by a
waveform of Fig. 15(b) having an average or dc value of:
Vdc =
2Vm + 2Vm
= 0.637Vm
2π
Movement of pointer is directly related to peak value of
signal by factor 0.637.
Ratio between rms and dc levels:
Vrms 0.707Vm
=
≅ 111
.
Vdc 0.637Vm
Vrms = 111
. Vdc
Meter indication = 1.11 (dc or average value)
dc or average reading multimeter is then scaled 1.11 to read
rms value (voltage or current).
Example 6
Determine the reading of a DMM placed across a resistor
with Vm=20Vac.
Solution:
Vrms = 0.707Vm = 0.707(20 V) = 14.14 V
3.25
Oscilloscopes
In general, a signal consists of both dc and ac components
Switch or knob with DC/GND/AC mode (see Fig. 16)
DC (Direct Coupling) mode
• DC means Direct Coupling but not measuring dc (or
average) value of quantity
• DC mode directly couples complete signal consisting of
dc and ac components to display
AC (Alternating Current) mode
• only passes ac component of signal
• dc component of signal is blocked by capacitor
GND (Ground)
• Input signal is blocked by direct ground connection
− A single horizontal line will be shown on display
AC
Oscilloscope
Capacitor
Display
AC
GND
GND
DC
DC
Figure 16
Input signal
3.26
3.7 Complex Numbers
3.7.1 Introduction
How do we find the algebraic sum of voltages and currents
that are varying sinusoidally?
One solution: find algebraic sum on a point-to-point basis
− A long and tedious process
Phasor analysis
• A quick, direct and accurate technique for finding
algebraic sum of sinusoidal ac waveforms
• Allows same techniques used for dc networks to be used
for sinusoidal ac networks
• Makes use of complex numbers
− need to consider complex numbers first which provide a
mathematical tool for phasor analysis (Section 3.9)
Complex number
• represents a point in a two-dimensional plane (Fig. 17)
− real numbers are on real (Re) horizontal axis
− complex numbers on imaginary (Im) vertical axis
Im( j)
+
−
+
−
Figure 17
Re
3.27
Two forms to represent complex numbers
• Rectangular form: represents a point in the plane
• Polar form: represents a radius vector from origin to point
Definition
• The operator j is a point on imaginary axis and has a 90o
counterclockwise rotation
j 2 = −1 or j = − 1
3.7.2 Rectangular form
A complex number in rectangular form (see Fig. 18(a)) is:
C = A + jB
where A and B are real numbers. See example in Fig. 18(b)
Im
Im
C = A + jB
B
− 10
Re
0
A
C = −10 − j 20
− 20
( b)
(a )
Figure 18
Re
3.28
3.7.3 Polar form
A complex number in polar form (see Fig. 19(a)) is:
C = C ∠θ
∠θ = e jθ = cos θ + j sin θ
where C is magnitude of C and θ is a positive angle
measured counterclockwise from positive real axis.
A negative sign in front of polar form gives a complex
number directly opposite to complex number with a
positive sign (see Fig. 19(b)):
− C = − C ∠θ = C ∠θ ± π
Im
Im
C
C
θ
C
π
θ
Re
Re
−π
−C
(b)
(a )
Figure 19
3.29
Example 7
Im
Im
5
C = 5∠30o
θ = 30o
Re
Re
θ = −120o
7
C = 7∠−120o
(a )
( b)
Figure 20
3.7.4 Conversion between forms
The two forms are related by the following equations.
Rectangular to polar form (See Fig. 21).
C=
A2 + B 2
θ = tan −1
B
A
Polar to rectangular form (See Fig. 21).
A = C cosθ
B = C sinθ
3.30
Im
C = C ∠θ = A + jB
C
B
θ
Re
A
Figure 21
Example 8
Convert the following from rectangular to polar form:
C = 3 + j4
(Fig. 22(a))
Solution:
C = 32 + 42 = 25 = 5
 4
θ = tan −1  = 5313
. o
 3
Therefore
C = 5∠5313
. o
Example 9
Convert the following from polar to rectangular form:
C = 10∠45o
(Fig. 22(b))
Solution:
A = 10 cos 45o = (10)(0.707) = 7.07
Therefore
B = 10 sin 45o = (10)(0.707) = 7.07
C = 7.07 + j 7.07
3.31
Im
Im
C = 10∠45o
C = 3 + j4
10
C
45o
4
θ
Re
Re
3
(a)
(b)
Figure 22
If complex number is in 2nd, 3rd, or 4th quadrant
• determine angle of vector in that quadrant
• then determine angle of vector relative to horizontal real
axis
Example 10
Convert the following from rectangular to polar form:
C = −6 + j 3
(Fig. 23(a))
Solution:
C = 62 + 32 = 6.71
 3
β = tan −1  = 26.57o
 6
Therefore
θ = 180o − 26.57o = 15353
. o
C = 6.71∠153.43o
3.32
C = −6 + j3
C
3
β
Im
Im
θ = 230o
A
θ
Re
β
Re
6
B
10
(a)
C = 10∠230o
(b)
Figure 23
Example 11
Convert the following from polar to rectangular form:
C = 10∠230o
(Fig. 23(b))
Solution:
A = C cos β = 10 cos(230o − 180o ) = 10 cos 50o = 6.428
B = C sin β = 10 sin 50o = 7.66
Therefore
C = −6.428 − j 7.66
3.8 Mathematical operations with complex numbers
Similar to real numbers, we can also perform addition,
subtraction, multiplication and division on complex
numbers.
Consider the following definitions for the symbol j
associated with imaginary part of complex numbers:
j 2 = −1 or j = − 1
3.33
j 3 = j 2 j = −1 j = − j
Thus
j 4 = j 2 j 2 = ( −1)( −1) = +1
j5 = j
1  1  j 
j
j
Also,
=    =
=
j  j   j  j2 − 1
1
=−j
Thus
j
3.8.1 Complex conjugate
Conjugate or complex conjugate of a complex number can
be found by:
• changing the sign of imaginary part in rectangular form
• putting a negative of the angle in polar form
Example 12
The conjugate of C = 2 + j 3 is 2 − j 3 . See Fig. 24(a).
The conjugate of C = 2∠30o is 2∠ − 30o . See Fig. 24(b)
C = 2 + j3
Im
Im
2
3
2
Re
−3
C
30o
− 30 o
Re
2
Conjugate of C
Conjugate of C
2 − j3
(b)
(a)
Figure 24
3.34
3.8.2 Addition
To add two or more complex numbers, just add real and
imaginary parts separately. For example, if
C1 = ± A1 ± jB1 and C2 = ± A2 ± jB2
then C1 + C2 = ( ± A1 ± A2 ) + j ( ± B1 ± B2 ) (see Fig. 25(a))
Note in Fig. 25(a) that the resultant vector C1 + C2 is
obtained by adding the head of one vector (eg. C1) with the
tail of another vector (eg. C2 ).
Example 13
(a) Add C1 = 2 + j 4 and C2 = 3 + j1 (see Fig. 25(a))
Solution: C1 + C2 = (2 + 3) + j (4 + 1) = 5 + j5
(b) Add C1 = 3 + j 6 and C2 = −6 + j 3
Solution: C1 + C2 = (3 − 6) + j (6 + 3) = −3 + j 9
3.8.3 Subtraction
In subtraction, real and imaginary parts are again
considered. For example, if
C1 = ± A1 ± jB1 and C2 = ± A2 ± jB2
C1 − C2 = [ ± A1 − ( ± A2 )] + j[ ± B1 − ( ± B2 )]
then
(see Fig. 25(b))
3.35
Im
Im
C1 + C2
6
4
C1
4 C2
2
C1 − C2
2
C2
0
C1
6
2
4
Re
6
−2 0
Re
2
4
6
− C2
(b)
(a)
Figure 25
Example 14
(a) Subtract C2 = 1 + j 4 from C1 = 4 + j 6 (see Fig. 25(b))
Solution: C1 − C2 = (4 − 1) + j (6 − 4) = 3 + j 2
(b) Subtract C2 = −2 + j5 from C1 = 3 + j 3
Solution: C1 − C2 = [ 3 − ( −2)] + j (3 − 5) = 5 − j 2
3.8.4 Addition or subtraction in polar form
Addition or subtraction cannot be performed in polar form
unless the complex numbers have the same angle θ or
differ only by multiples of 180o .
Example 15
(a) 2∠45o + 3∠45o = 5∠45o (see Fig. 26(a))
(b) 2∠0o − 4∠180o = 6∠0o
(see Fig. 26(b))
3.36
Im
2
Im
45
5
3
− 4 ∠ 1 8 0o
o
Re
4∠180
o
Re
2
6
(a)
(b)
Figure 26
3.8.5 Multiplication
Rectangular form
Multiply real and imaginary parts of one in turn by real and
imaginary parts of the other. For example, if
C1 = A1 + jB1 and C2 = A2 + jB2
C1 ⋅ C2 = ( A1 + jB1)( A2 + jB2 )
then
=
=
A1( A2 + jB2 ) + jB1( A2 + jB2 )
A1 A2 + jA1B2 + jB1 A2 − B1B2
C1 ⋅ C2 = ( A1 A2 − B1B2 ) + j ( A1B2 + B1 A2 )
Example 16
Find C1 ⋅ C2 if C1 = −2 − j 3 and C2 = 4 − j 7
3.37
Solution:
C1 ⋅ C2 = [( −2)(4) − ( −3)( −7)] + j[( −2)( −7) + ( −3)(4)]
= −29 + j 2
Polar form
In polar form, multiply magnitudes and add angles. For
example, if C1 = C1 ∠θ1 and C2 = C2 ∠θ2 then
C1 ⋅ C2 = C1 C2 ∠θ1 + θ2
Example 17
Find C1 ⋅ C2 if C1 = −5∠30o and C2 = 10∠ − 20o
Solution: C1 ⋅ C2 = −50∠10o
3.8.6 Division
Rectangular form
Need to multiply numerator and denominator by the
conjugate of denominator and then collect real and
imaginary parts. If C1 = A1 + jB1 and C2 = A2 + jB2 then
C1 ( A1 + jB1)( A2 − jB2 )
=
C2 ( A2 + jB2 )( A2 − jB2 )
( A A + B1B2 ) + j ( − A1B2 + B1 A2 )
= 1 2
A22 + B22
C1 ( A1 A2 + B1B2 )
( − A1B2 + B1 A2 )
=
+j
C2
A22 + B22
A22 + B22
3.38
Example 18
Find C1 C2 if C1 = −1 + j 4 and C2 = 4 − j5
C1 ( −1)(4) + ( 4)( −5)
− ( −1)( −5) + (4)(4)
=
+j
2
2
2
2
C2
4
5
4
5
+
+
Solution:
11
− 24
=
+ j ≅ −0.585 + j 0.268
41
41
Polar form
• Divide magnitude of numerator by magnitude of
denominator
• Subtract angle of denominator from angle of numerator
If C1 = C1 ∠θ1 and C2 = C2 ∠θ2 then
C1 C1
=
∠θ1 − θ2
C2 C2
Also
1
1
= ∠−θ
C∠θ C
Example 19
Find C1 C2 if C1 = 15∠ − 10o and C2 = −2∠7o .
C
15
∠ − 10o − 7o = −7.5∠ − 17o
Solution: 1 =
C2 − 2
3.39
3.9 Phasor Analysis
Having considered complex numbers, we are now ready to
consider phasors and to show how phasors can be used to
solve sinusoidal ac circuit problems.
A sinusoidal waveform (eg. voltage)
v (t ) = Vm sin α = Vm sin ωt
can be generated through the vertical projection of a radius
vector rotating at an uniform angular velocity ω (see Fig.
27).
Similarly, a sinusoidal waveform
v (t ) = Vm sin(ωt + θ); θ = 45o
can also be generated (see Fig. 28(b)) through the vertical
projection of a rotating vector (see Fig. 28(a)).
This rotating vector, having a constant magnitude Vm with
one end fixed at the origin, is called a phasor in sinusoidal
ac circuits.
3.40
Figure 27
3.41
v(t ) = Vm sin(ωt + θ); θ = 45o
ω
Vm
o
θ = 45
Vm
Vm sin45o
θ = 45o
0
45o
135o
225o
315o
ωt
t=0
( b)
( a)
Figure 28
During its rotational development of the sine wave, the
phasor will, at the instant t = 0 , have the position shown in
Fig. 28(a) which is called a phasor diagram.
Note that the vertical projection of the rotating vector (ie.
Vm sin45o ) in Fig. 28(a) corresponds to the amplitude of
v (t ) at time t = 0 (ie. v (0) = Vm sin 45o ).
Thus, the phasor diagram is a “snapshot” of the rotating
vectors at time t = 0 .
Hence, the phasor representation of a sinusoidal ac
waveform v (t ) is given by Vm∠ ± θ . That is,
v (t ) = Vm sin(ωt ± θ) ⇒ Vm∠ ± θ
3.42
Example 20
Figure 29 shows an example of the addition of two current
waveforms using phasors.
Note that the vertical heights of the waveforms in Fig.
29(b) at t = 0 are determined by the vertical projections of
the radius vectors in Fig. 29(a).
Imagine trying to add these waveforms “by hand” without
using phasors!
Figure 29