Sinusoidal Functions,
Complex Numbers, and Phasors
Discussion D9.2
Sections 4-2, 4-3
Appendix A
3/26/2006
1
Sinusoids
A sinusoid is a signal that has the form of
the sine or cosine function.
XM
x(t)
0

2

3
2
XM
x(t)
2
t
3
2

0
2
2
x(t )  X M cos t
x(t )  X M sin t
period T  2 
t

sin

2
1
cos

2
0
2
Sinusoids
In general:
x(t )  X M sin t   
x(t )  X M cos t   
Note trig identities:
sin t     sin t cos   cos t sin 
cos t     cos t cos  sin t sin 
3
A sinusoidal current or voltage is usually referred to as an
alternating current (or AC) or voltage and circuits excited
by sinusoids are called ac circuits. Sinusoids are important
for several reasons:
1.
2.
3.
4.
5.
Nature itself is characteristically sinusoidal (the
pendulum, waves, etc.).
A sinusoidal current or voltage is easy to generate.
Through Fourier Analysis, any practical periodic signal
(one that repeats itself with a period T) can be
represented by a sum of sinusoids.
A sinusoid is easy to handle mathematically; its
derivative and integral are also sinusoids.
When a sinusoidal source is applied to a linear circuit,
the steady-state response is also sinusoidal, and we call
the response the sinusoidal steady-state response.
4
Complex Numbers
What is the solution of
X2 = -1
X   1   j
Complex Plane
j
imaginary
1
-1
j 2  1
j  1
Note:
real
1
j
j
  2 j
j jj j
-j
5
Complex Numbers
A  x  jy
imag
A  A cos j  jA sin j
y  A sin j
A  x2  y 2
j  tan
1
A
y
x
j
e jj  cos j  j sin j
e
j j  
j measured positive
counter-clockwise
A  Ae jj
Note:
real
x  A cos j
Euler's equation:

Complex
Plane
 e e  cos j     j sin j   
jj
j
6
Sinusoidal Functions and Phasors
t
t=0
0
0
e
t
jt
 cos t  j sin t
http://www.kineticbooks.com/physics/trialp
se/33_Alternating%20Current%20Circuits/1
3/sp.html
7
Relationship between sin and cos
Im

Im
cos(t   )
Re
Re
t=0
cos
t=0

sin(t   )
sin
0
Note that
0



sin t     cos  t      
2


or
Phasor projection
on the real axis
t


sin t     cos  t    
2

t
8
Relationship between sin and cos
Im

Im
cos(t   )
Re
Re
t=0
cos
t=0

sin
0
0
sin(t   )


sin  t     cos t
2

sin t      sin t


cos  t    sin t
2

Phasor projection
on the real axis
cos t      cos t
t
t
sin t     sin t cos   cos t sin 
cos t     cos t cos  sin t sin 
9
Comparing Sinusoids
-sin t
cos t      cos t
Im

-cos t
Re
cos t
2

Note: positive angles are counter-clockwise
Im

45
135
45



cos  t    sin t
2

sin t
cos t  45
sin t      sin t


sin  t     cos t
sin t
Re
cos t
cos t leads sin t by 90
cos t lags - sin t by 90



sin t  45  cos t  135
 leads cost by 45

and leads sint by 135
10
R
i(t)
VM cos t
+
KVL
+ VR
AC
-
VM cos t  Ri (t )  L
- +
VL
-
L
di (t )
dt
This is a differential equation we
must solve for i(t).
How?
Guess a solution and try it!
Note that
Re VM e jt   Re VM cos t  jVM sin t   VM cos t
It turns out to be easier to use VM e jt as the forcing function
rather than VM cos t and then take the real part of the solution.
de jt
 je jt
This is because
dt
which will allow us to convert the differential equation to an
algebraic equation. Let's see how.
11
Solve the differential equation for i(t).
i(t)
VM cos t
+
R
+ VR
AC
- +
VL
-
di (t )
VM cos t  Ri (t )  L
dt
L
Instead, solve
VM e jt  Ri (t )  L
di (t )
dt
(1)
and take the real part of the solution
Guess that
i(t )  I M e j t  
and substitute in (1)
VM e jt  RI M e j t    j LI M e j t    e jt  RI M e j  j LI M e j 
Divide by e j t
VM  RI M e j  j LI M e j
12
Solve the differential equation for i(t).
i(t)
VM cos t
+
R
VM e jt  Ri (t )  L
- +
+ VR
AC
L
VL
-
-
Rearrange (3)
Recall that
di (t )
dt
i(t )  I M e j t  
VM  RI M e j  j LI M e j
VM
IM e 
R  j L
j
R  j L  R 2   2 L2 e
(1)
(2)
(3)
(4)
 L 
j tan 1 

 R 
Therefore (4) can be written as
I M e j 
VM
R 2   2 L2
e

  L 
j   tan 1 

 R 

(5)
13
Solve the differential equation for i(t).
i(t)
VM cos t
+
R
+ VR
VM e jt  Ri (t )  L
- +
AC
L
VL
-
i(t )  I M e j t  
I M e j 
Therefore, from (5)
VM
R 2   2 L2
IM 
di (t )
dt

  L 
j   tan 1 

 R 

e
VM
2 2
(2)
(5)
 L 
   tan 

R


1
R  L
2
(1)
(6)
Substituting (6) in (2) and taking the real part
i(t ) 
L 

cos  t  tan 1

2
2 2
R


R  L
VM
14
Phasors
A phasor is a complex number that represents the amplitude
and phase of a sinusoid.
X M e jj  X M j  X
XM
j
j t  
Recall that when we substituted i(t )  I M e
j t
in the differential equations, the e
cancelled out.
We are therefore left with just the phasors
15
Solve the differential equation for i(t) using phasors.
i(t)
VM cos t
+
R
+ VR
- +
AC
L
VL
-
di (t )
dt
(1)
i(t )  I M e jt    Ie jt
(2)
V  VM 0
(3)
VM e jt  Ri (t )  L
-
Substitute (2) and (3) in (1)
Ve jt  RIe jt  j LIe jt
(4)
Divide by e j t and solve for I
I
V
 I M  
R  j L
i(t ) 
L 

   tan 1
R 
R 2   2 L2 
VM
L 

cos  t  tan 1

2
2 2
R


R  L
VM
(5)
16
By using phasors we have transformed the problem from
solving a set of differential equations in the time domain to
solving a set of algebraic equations in the frequency domain.
The phasor solutions are then transformed back to the time
domain.
A cos t    
 A  

A sin t    
 A   90

17
Impedance
Impedance Z 
i(t)
VM cos t
+
AC
-
V phasor voltage

I phasor current
R
+ VR
I
- +
VL
-
L
Units = ohms
V
R  j L
V
Z   R  j L  Z  z
I
Z  R  jX
R  resistance
Z  R2   2 L2
X  reactance
1  L
 z  tan
R
Note that impedance is a complex number containing a real, or
resistive component, and an imaginary, or reactive, component.
18
Admittance
1 I
phasor current
Admittance Y = = 
Z V phasor voltage
i(t)
VM cos t
+
Units = siemens
R
+ VR
AC
-
conductance G 
V
I
R  j L
- +
VL
-
R
R 2   2 L2
L
I
1
R  j L
Y= 
 G  jB  2
V R  j L
R   2 L2
susceptance
B
 L
R 2   2 L2
19
Capacitor Circuit
R
i(t)
VM cos t
+
+ VR
VM e jt  Ri (t ) 
-+
C
VC
AC
-
-
Ve
 RIe
(1)
i(t )  I M e jt    Ie jt
(2)
V  VM 0
(3)
Substitute (2) and (3) in (1)
jt
1
i (t )dt

C
jt
1

Ie jt
jC
(4)
Divide by e j t and solve for I
V
I
R
1
jC
i (t ) 
 I M  
VM
1
R  2 2
C
2
VM
R2 
1
 2C 2
1 

  tan 1
 RC 

1 

1
cos  t  tan


RC


(5)
20
Impedance
V phasor voltage
Z 
I phasor current
Impedance
i(t)
VM cos t
+
R
+ VR
-+
VC
AC
-
-
1
Z R  2 2
C
2
 z   tan 1
1
 RC
V
I
R
C
Z
1
jC
V
1
1
 R
 R j
 Z  z
I
jC
C
Z  R  jX
R  resistance
X  reactance
21
Im
V
V
I
I
Re
I in phase
with V
V
Im
I
V
Re
I lags V
V
I
Im
I
V
I
Re
I leads V
22
Expressing Kirchoff’s Laws in the Frequency Domain
KVL: Let v1, v2,…vn be the voltages around a closed loop.
KVL tells us that: v1  v2  ...vn  0
Assuming the circuit is operating in sinusoidal steady-state at
frequency  we have:
VM 1 cos t  1   VM 2 cos t  2   ...  VMn cos t  n   0
or
Re VM 1e j1 e jt   Re VM 2e j2 e jt   ...  Re VMn e jn e jt   0
j
Phasor Vk  VMk e k
Since e jt  0
Re  V1  V2  ...  Vn  e jt   0
V1  V2  ...  Vn  0
Which demonstrates that KVL holds for phasor voltages.
23
KCL: Following the same approach as for KVL, we can show that
I1  I 2  ...  I n  0
Where Ik is the phasor associated with the kth current entering a
closed surface in the circuit.
Thus, both KVL and KCL hold when working with
phasors in circuits operating in sinusoidal steady-state.
This implies that all of the circuit analysis methods (mesh
and nodal analysis, source transformations, voltage &
current division, Thevenin equivalent, combining
elements, etc,) work in the same way we found for
resistive circuits. The only difference is that we must
work with phasor currents & voltages and the
impedances &/or admittances of the elements.
24
I
Find Zin
DC
Z1
V
Z2
I1
Z3
I2
Z4
Zin
 Z1  Z2  Z3

 Z3

Z3   I1   V 
 



Z3  Z 4   I 2   0 
 Z1  Z2  Z3  I1  Z3I2  V
Z3I1   Z3  Z4  I 2  0
25
 Z1  Z2  Z3  I1  Z3I2  V
Z3I1   Z3  Z4  I 2  0
I
DC
Z1
V
Z2
I1
Z3
I2
Z4
Zin
Z3
I2 
I1
Z3  Z 4
I  I1
Z32
I1  V
 Z1  Z2  Z3  I1 
Z3  Z 4
2
Z

Z

Z
Z

Z

Z
 1 2 3  3 4  3
Z3  Z 4
I1  V
Z3Z 4
V
Zin   Z1  Z 2 
I
Z3  Z 4
26
I
Z3Z 4
Zin  Z1  Z 2 
Z3  Z 4
DC
Z1
V
Z2
I1
Z3
I2
Z4
Zin
We see that if we replace Z by R the impedances add like resistances.
Impedances in series add like resistors in series
Impedances in parallel add like resistors in parallel
27
Voltage Division
Z1
+ V1 DC
V
I
Z2
+
V2
-
V
I
Z1  Z 2
But
V1  Z1I
V2  Z2I
Therefore
Z1
V1 
V
Z1  Z 2
Z2
V2 
V
Z1  Z 2
28