Download REVIEW OF COMPLEX NUMBERS

REVIEW OF COMPLEX NUMBERS
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Complex numbers are widely used to facilitate
computations involving ac voltages and currents
j = (-1); j2 = -1
A complex number C has a real and imaginary part
C = a + jb
a is the real part, b is the imaginary part
Can also use C = (a, b)
C = a + jb (rectangular form)
C = M θ (polar form)
C = Mcosθ + jMsinθ
Im
M  a  jb
b = Msinθ
θ =tan-1(b/a)
Re
a =Mcosθ
Complex Algebra and
Phasors
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ARITHMETIC OPERATIONS
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(a + jb) + (c + jd) = (a + c) + j(b + d)
(a + jb) - (c + jd) = (a - c) + j(b - d)
Polar form: M11 M 22   M1M 21  2 
(a + jb)  (c + jd) = (ac – bd) + j(bc + ad)
Polar form: M 11  M 1    
M 2  2
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M2
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2
Complex conjugate C’ = a – jb = M 
CC’ = a2 + b2
Special case reciprocal 1/j:
1 10

 1  90   j
j 190
Complex Algebra and
Phasors
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PHASORS (1)
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A phasor is a mathematical representation of an ac
quantity in polar form
A phasor can be treated as the polar form of a complex
number, so it can be converted to an equivalent
rectangular form
To represent an ac voltage or current in polar form, the
magnitude M is the peak value of the voltage or current
The angle θ is the phase angle of the voltage or current
Examples: vt   170 sin 377t  40 V  17040 V
i t   0.05 sin ωt  A  0.050 A


vt   10 3 sin 106 t  120 A  10 3   120 V
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The frequency of the phasor waveform does not
appear in its phasor representation, because we
assume that all voltages and currents in a problem
have the same frequency
Phasors and phasor analysis are used only in circuit
problems where ac waveforms are sinusoidal
Complex Algebra and
Phasors
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PHASORS (2)
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We can also use phasors to convert waveforms
expressed as sines or cosines to equivalent waveforms
expressed as cosines and sines respectively
Im
cos
cos (ωt - 30°)
=sin(ωt + 60°)
-30°
60°
Re
sin
-sin
45°
-135°
-sin(ωt + 45°)
=sin(ωt - 135°)
-cos
Complex Algebra and
Phasors
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WORKED EXAMPLE
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Take the two voltage waveforms:
v1  10 sin ω  100 V
v2  20 sin ωt  60  2060 V
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Converting to rectangular form and adding
v1  10cos 0  j sin 0  10  j 0
v2  20cos 60  j sin 60  10  j17.32
v1  v2  20  j17.32
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The polar form of v1 + v2 is
17.32 

20 2  17.32 2  tan 1
  26.4640.9
20 

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Finally converting the polar form to sinusoidal form
v1  v2  26.46 sin ωt  40.9 V
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Example: Find i1 – i2 if i1 = 1.5sin(377t + 30°) A and i2 =
0.4sin(377t - 45°) A
Complex Algebra and
Phasors
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PHASOR FORM OF RESISTANCE
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The voltage v(t) = Vpsin(ωt + θ) V is in phase with the
current i(t) = Ipsin(ωt + θ) A when across a resistor
By Ohm’s Law
R
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Converting to phasors we have
R
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vt  V p sin ωt  θ 

it  I p sin ωt  θ 
V p θ
I p θ

Vp
Ip
0 ohms
We can regard resistance as a phasor whose
magnitude is the resistance in ohms and whose angle
is 0°
In the complex plane, resistance is a phasor that lies
along the real axis
The rectangular form of resistance is R + j0
Example: The voltage across a 2.2kΩ resistor is v(t) =
3.96sin(2000t + 50°) V. Use phasors to find the current
through the resistor. Draw a phasor diagram showing
the voltage and current
Complex Algebra and
Phasors
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PHASOR FORM OF CAPACITIVE
REACTANCE
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The current through a capacitor leads the voltage
across it by 90°
When v(t) = Vpsin(ωt + θ) V, i(t) = Ipsin(ωt + θ + 90°) A
Applying Ohm’s Law for capacitive reactance:
v(t) = XCi(t) or
V p sin ωt  θ 
vt 
XC 

i t  I p sin ωt  θ  90
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Converting to phasor form
XC 
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V p θ
I p θ  90

Vp
Ip
 90  
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 90 
ωC
Capacitive reactance is regarded as a phasor whose
magnitude is |XC| = 1/ωC ohms, whose angle is -90°
Capacitive reactance is plotted down the negative
imaginary axis
The rectangular form is XC = 0 – j|XC|
Example: The current through a 0.25F capacitor is
i(t) = 40sin(2104t + 20°) mA. Use phasors to find the
voltage across the capacitor. Draw a phasor diagram
showing the voltage and current
Complex Algebra and
Phasors
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PHASOR FORM OF INDUCTIVE
REACTANCE
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The voltage across an inductor leads the current
through it by 90°
When i(t) = Ipsin(ωt + θ) A, v(t) = Vpsin(ωt + θ + 90°) V
Applying Ohm’s Law: v(t) = Xli(t) or
XL 
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Converting so phasor form
XL 
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vt  V p sin ωt    90

i t 
I p sin ωt   
V p θ  90
I p θ

Vp
Ip
90   ωL90 
Inductive reactance is regarded as a phasor whose
magnitude is |XL| = ωL ohms with angle 90°
Inductive reactance is plotted up the imaginary axis
The rectangular form is XL = 0 + j|XL|
Example: The voltage across an 8mH inductor is
v(t) = 18sin(2π106t + 40°) V. Use phasors to find the
current through the inductor. Draw a phasor diagram
showing the voltage and current.
Complex Algebra and
Phasors
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