Topics in Electrical and Computer Engineering
Phasors
Douglas Wilhelm Harder
Department of Electrical and Computer Engineering
University of Waterloo
Copyright © 2008-10 by Douglas Wilhelm Harder. All rights reserved.
Phasors
Outline
In this topic, we will look at
–
–
–
–
–
–
–
–
The necessary background
Sums of sinusoidal functions
The trigonometry involved
Phasor representation of sinusoids
Phasor addition
Why phasors work
Phasor multiplication and inverses
Phasors for circuits
Phasors
Background
Consider any periodic sinusoid with
T = 1/f = 2p/w
f = 1/T = w/2p
w = 2pf = 2p/T
– Period
– Frequency
– Angular frequency
It is possible to write such a sinusoid as
v cos(wt + f)
where
– v is the amplitude
– f is the phase shift
f
v
Phasors
Background
As v cos(wt + f + 2p) = v cos(wt + f), restrict –p < f ≤ p
Engineers throw an interesting twist into this formulation
– The frequency term wt has units of radians
– The phase shift f has units of degrees: –180° < f ≤ 180°
For example,
V cos(377t + 45°)
V cos(377t – 90°)
Phasors
Background
A positive phase shift causes the function to lead of f
For example, –sin(t) = cos(t + 90°) leads cos(t) by 90°
Compare cos(377t) and cos(377t + 45°)
Blue peaks first
- Blue is leading
Phasors
Background
A negative phase shift causes the function to lag by f
For example, sin(t) = cos(t – 90°) lags cos(t) by 90°
Compare cos(377t) and cos(377t – 90°)
Red peaks first
- Blue lags behind
Phasors
Background
If the phase shift is 180°, the functions are out of phase
E.g., –cos(t) = cos(t – 180°) and cos(t) are out of phase
Compare cos(377t) and cos(377t – 180°)
Phasors
Background
Suppose we add a number of sinusoidal voltages with:
– Equal frequencies
– Different amplitudes (voltages) and phase shifts
E.g., 3 cos377t  75  10 cos377t  50  4 cos377t  15 
5 cos377t  42  12 cos377t  90
It may not be obvious, but the result will be another
sinusoid of the form
A cos377t  f 
Phasors
Background
These are the five sinusoids and their sum
14.81cos377t  3.36 
Phasors
Derivations (The Hard Way)
We will now show that the sum of two sinusoids with
– The same frequency, and
– Possibly different amplitudes and phase shifts
is a sinusoid with the same frequency
Our derivation will use trigonometric formula familiar to
all high-school students
– Later, we will see how exponentials with complex powers
simplify this observation!
Phasors
Example Derivation (The Hard Way)
Consider the sum of two sinusoids:

3 coswt   9 2 cos wt  45

Use the rule
cosa  b  cosa cosb  sin a sin b
Thus we expand the terms
9 2 coswt  p2   9 2 coswt  cos45   9 2 sin wt sin 45 
 9 coswt   9 sin wt 
Phasors
Example Derivation (The Hard Way)
Therefore
3 coswt   9 2 coswt  45   12coswt   9 sinwt 
We wish to write this as
v coswt  f   v coswt cosf  v sinwt sin f
We therefore deduce that
v cosf  12
v sin f  9
Phasors
Example Derivation (The Hard Way)
Given
v cosf  12
v sin f  9
Square both sides and add:
v cosf   v sin f 
2
2
 122  9 2
v 2 cos2 f  sin 2 f   225
2
2
Because cos f  sin f  1 it follows that
v  225  15
Phasors
Example Derivation (The Hard Way)
Again, given
v cosf  12
v sin f  9
Take the ratio:
v sin f
9
 tanf 
 0.75
v cosf
12
Therefore f
≈ 36.87°
Phasors
Example Derivation (The Hard Way)
It follows that
3 coswt   9 2 coswt  45   15coswt  36.87 
This is independent of the frequency:
w2
w 5
Phasors
Full Derivation (The Hard Way)
Consider the sum of two sinusoids:
v1 coswt  f1   v2 coswt  f2 
Using the rule
cosa  b  cosa cosb  sin a sin b
we can expand
v1 cos wt  1   v2 cos wt  2 
 v1 cos wt  cos 1   v1 sin wt  sin 1  
v2 cos wt  cos 2   v2 sin wt  sin 2 
 cos wt   v1 cos 1   v2 cos 2   
sin wt   v1 sin 1   v2 sin 2  
Phasors
Full Derivation (The Hard Way)
Now, given
v1 coswt  f1   v2 coswt  f2 
 coswt v1 cosf1   v2 cosf2   sin wt v1 sin f1   v2 sin f2 
Suppose we can write this in the form
v coswt  f 
First, v coswt  f   v coswt cosf  v sinwt sin f
and thus v cosf   v1 cosf1   v2 cosf2 
v sinf   v1 sinf1   v2 sin f2 
Phasors
Full Derivation (The Hard Way)
First, given
v cosf   v1 cosf1   v2 cosf2 
v sin f   v1 sin f1   v2 sin f2 
square both sides and add
v 2  v12  2v1v2 cosf1 cosf2  sin f1 sin f2   v2 2
 v  v12  2v1v2 cos f1 cos f 2  sin f1 sin f 2   v2 2
Phasors
Full Derivation (The Hard Way)
Similarly, given
v cosf   v1 cosf1   v2 cosf2 
v sin f   v1 sin f1   v2 sin f2 
take the ratio
sin f v1 sin f1  v2 sin f2
tanf 

cosf v1 cosf1  v2 cosf2
v1 sin f1  v2 sin f 2 

f  tan 
 v1 cosf1  v2 cosf 2 
1 
Phasors
Full Derivation (The Hard Way)
Adding n sinusoids must be done one pair at a time:
v1 cos wt  f1   v2 cos wt  f2   v3 cos wt  f3   


v1, 2 coswt  f1, 2 
 v3 coswt  f3   

v1, 2,3 coswt  f1, 2,3 

Phasors
Using Phasors
Rather than dealing with trigonometric identities, there is
a more useful representation: phasors
Assuming w is fixed, associate
v coswt  f   vf
where vf is the complex number
with magnitude v and argument f
The value vf is a phasor and
is read as vee phase fee
Phasors
Using Phasors
The addition of sinusoids is equivalent to phasor addition
Function of Time
n
v
k
coswt  fk 
k 1
trigonometry
v coswt  f 
Phasors
n
 v f
k
k
k 1
complex number
addition
vf
We transform the problem of trigonometric addition
into a simpler problem of complex addition
Phasors
Using Phasors
Recalling Euler’s identity
vf  ve jf  vcosf  j sin f 
Therefore
n
n
 v f   v
k
k 1
k
k 1
k
cosfk  j
n
v
k 1
k
sin fk
Phasors
Using Phasors
Given our motivating example 3 coswt   9 2 coswt  45 
– Using phasors:
30  9 245  3  9  9 j
 12  9 j

– The result is 15cos wt  36.87


 1536.87
Phasors
Using Phasors
For example, given our first sum:
3 cos377t  75  10 cos377t  50  4 cos377t  15 
5 cos377t  42  12 cos377t  90
Using phasors



375  1050  4  15  542  12  90
 3e
75 j
 10e
50  j
 4e
15 j
 5e
42  j
 3e
 14.78  0.87 j
 14.813.36
we get the sum 14.81cos377t  3.36 

90  j

Phasors
Using Phasors
These graphs show:
– The individual phasors
– The sum of the phasors
Phasors
Using Phasors
A plot of the sum
3 cos377t  75  10 cos377t  50  4 cos377t  15 
5 cos377t  42  12 cos377t  90
is identical to a plot of 14.81cos377t  3.36 
Phasors
Why Phasors Work
To understand recall that
ve j wt f   v coswt  f   jv sinwt  f 

j wt f 
Therefore v coswt  f    ve
Plot the complex exponential
14.81e

j 377t 3.36

The real part:
14.81 cos(377t + 3.36°)
The imaginary part:
14.81sin(377t + 3.36°)

Phasors
Why Phasors Work
Note that
v1e j wt f1   v2e j wt f2   v1e jwt e jf1  v2 e jwt e jf2


 v1e jf1  v2e jf2 e jwt
 v1f1  v2f2 e jwt
and because w  z  w z, it follows
v1 coswt  f1   v2 coswt  f2   v1e j wt f1   v2 e j  wt f2  
 v1e jwt e jf1  v2 e j wt e jf2 
 v1e jwt e jf1  v2 e jwt e jf2 
 v1f1  v2 f2 e jwt 
Phasors
Derivation (The Easy Way)
Previously, we used trigonometry to show that the sum
of two sinusoids with the same frequency must continue
to have the same frequency
Using complex exponentials, this becomes obvious
v1e
j wt f1 
 v e
2
j  wt f2 
 v f  v f e 
jwt
1
1
2
2
Phasors
Phasor Multiplication
Multiplication may be performed directly on phasors:
– The product of two phasors r1f1 and r2 f2 is the phasor
r r f  f 
1 2
1
2
– For example,
545  2  60   10 15



– This generalizes to products of n phasors:
r f    r f   r    r f
1
1
n
n
1
n
1
     fn 
Phasors
Phasor Multiplication
Calculating the inverse is also straight-forward:
– The inverse of the phasor rf is the phasor
– For example,
1
  f 
r
545 
 1
 0.2  45 
– Using multiplication, we note the product is 1:
 1 
  1
rf     f    r f   f   10
 r 
  r
Phasors
Linear Circuit Elements
When dealing with alternating currents (AC):
– The generalization of the resistance R is
the complex impedance Z = R + jX
– The generalization of the conductance G is
the complex admittance Y = G + jB
Ohm’s law easily generalizes:
V = IZ
Phasors
Linear Circuit Elements
With AC, the linear circuit elements behave like phasors
Inductor with inductance L
Z  jw L  w L90
Resistor with resistance R
Z  R0
1
1

  90
Capacitor with capacitance C Z 
jwC wC
– Note that inductance and capacitance
are frequency dependant
We can use this to determine the voltage
across an linear circuit with AC: V = IZ
Phasors
Linear Circuit Elements
Consider three circuit elements in series:
with
R= 2W
L = 50 mH
C = 750 mF
The total impedance is:
1
Z  20  w 0.05090 
  90
0.750w
• E.g., if w  1, it follows that
Z  2.376  32.69
Phasors
Linear Circuit Elements
If the current across this circuit is I = 10 cos(t) A,
we may compute the voltage:
V  IZ
 100    2.376  32.69
 23.76  32.69
or V = 23.76 cos(t – 32.69°) V

Phasors
Linear Circuit Elements
If the current across this circuit is I = 10 cos(377t) A,
we need to recalculate the impedance:
Z  20  377  0.05090  0.7501377   90
 18.9583.94
Now
V  IZ
 100   18.9583.94
 189.583.94
or V = 189.5 cos(377t + 83.94°) V

Phasors
Linear Circuit Elements
Consider three circuit elements in parallel with
R= 2W
L = 50 mH
C = 750 mF
The total admittance is:
1
1
1
1
Y 

 90
Z 20 w 0.05090 w 0.750
E.g., if w  1, it follows that
Y  0.50  20  90  0.7590
 18.95  83.94
 Z  0.0527683.94
Phasors
Linear Circuit Elements
If the current across this circuit is I = 10 cos(t) A
we may compute the voltage:
V  IZ
 100    0.0527683.94
 0.527683.94
or V = 0.5276 cos(t + 83.94°) V

Phasors
Linear Circuit Elements
If the current across this circuit is I = 10 cos(377t) A
we need to recalculate the impedance:
1
1
1
Y 


Z 20 377  0.05090
 282.789.90
or Z  0.003537  89.90
1
1
3770.750   90
Phasors
Linear Circuit Elements
Thus, with the current I = 10 cos(377t) A and
impedance Z  0.003537  89.90
we may now calculate
V  IZ
 100    0.003537  89.90

 0.03537  89.90
or V = 0.03537 cos(377t – 89.90°) V
This should be intuitive, as the capacitor
acts as essentially a short-circuit for high frequency AC
Phasors
Linear Circuit Elements
A consequence of what we have seen here is that any
linear circuit with alternating current
may be replaced with
– A single resistor, and
– A single complex impedance
• A capacitor or an inductor
R22  w 2 X 22
R22  w 2 X 22
R1 
, w X1 
wX2
R2
Phasors
Final Observations
These techniques do not work if the frequencies differ
The choice of 377 in the examples is deliberate:
– North American power is supplied at 60 Hz:
2p 60 ≈ 376.99111843
– Relative error: 0.0023%
– European power is supplied at 50 Hz:
2p 50 = 100p ≈ 314.159265358
Phasors
Final Observations
Mathematicians may note the peculiar mix of radians
and degrees in the formula
10cos377t  50
Justification:
– The phase shifts are 180° or less (p radians or less)
– Which shifts are easier to visualize?
120°
60°
45°
30°
2.094 rad or 2p/3 rad
1.047 rad or p/3 rad
0.785 rad or p/4 rad
0.524 rad or p/6 rad
– Is it obvious that 2.094 rad and 0.524 rad are orthogonal?
– Three-phase power requires the third roots of unity:
• These cannot be written easily with radians (even using p)
Phasors
Summary
In this topic, we will look at
–
–
–
–
Sum of sinusoidal functions
The trigonometry is exceptionally tedious
The phasor representation v coswt  f   vf
The parallel between phasor addition and trigonometric
summations
– Phasor multiplication and inverses
– Use of phasors with linear circuit elements
Phasors
Acknowledgments
• I would like to thank Prof. David Nairn for assistance in
answering questions and Hua Qiang Cheng for pointing
out errors in a previous version of these slides
Usage Notes
• These slides are made publicly available on the web for anyone to
use
• If you choose to use them, or a part thereof, for a course at another
institution, I ask only three things:
– that you inform me that you are using the slides,
– that you acknowledge my work, and
– that you alert me of any mistakes which I made or changes which you
make, and allow me the option of incorporating such changes (with an
acknowledgment) in my set of slides
Sincerely,
Douglas Wilhelm Harder, MMath
dwharder@alumni.uwaterloo.ca