AC Steady State Analysis Notes | EduRev

: AC Steady State Analysis Notes | EduRev

 Page 1


Electronic Instrumentation  ENGR-4300 
Susan Bonner - 1 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
AC Steady State Analysis 
 
What if we are given a circuit and we wish to find the output of that circuit for a range of 
inputs?  We could wire the circuit, attach the function generator to the input, and hook up 
the scope to the output.  Then, we could alter the input systematically and observe how 
the output responds to these changes.  If we could use this information to find a 
mathematical function that relates the input to the output, then we could use the function 
to predict how the circuit will behave for any given input, as shown in figure 1 below.  
What if we could use the components in the circuit to derive this function?  Then we 
could predict the output of any circuit for any given input. 
 
Figure 1 
 
 
A. What is a Transfer Function? 
 
A transfer function relates the output and the input of a circuit.  In order for a transfer 
function to be useful, it must be simple to use and easy to find using the circuit diagram.  
Therefore, let us define a transfer function, H, as the ratio of the output to the input of a 
circuit.   
in
out
V
V
H =
    [equation 1] 
 
In a circuit containing only resistors, transfer functions can be easily found using voltage 
dividers.  In figure 2, a voltage divider is used to find the transfer function at Vout. 
 
 
Figure 2 
3 2 1
3 2
*
R R R
R R
V V
in out
+ +
+
=
K K K
K K
V V
in out
3 2 1
3 2
*
+ +
+
=
6
5
= =
in
out
V
V
H
( ) Vin f Vout =
Page 2


Electronic Instrumentation  ENGR-4300 
Susan Bonner - 1 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
AC Steady State Analysis 
 
What if we are given a circuit and we wish to find the output of that circuit for a range of 
inputs?  We could wire the circuit, attach the function generator to the input, and hook up 
the scope to the output.  Then, we could alter the input systematically and observe how 
the output responds to these changes.  If we could use this information to find a 
mathematical function that relates the input to the output, then we could use the function 
to predict how the circuit will behave for any given input, as shown in figure 1 below.  
What if we could use the components in the circuit to derive this function?  Then we 
could predict the output of any circuit for any given input. 
 
Figure 1 
 
 
A. What is a Transfer Function? 
 
A transfer function relates the output and the input of a circuit.  In order for a transfer 
function to be useful, it must be simple to use and easy to find using the circuit diagram.  
Therefore, let us define a transfer function, H, as the ratio of the output to the input of a 
circuit.   
in
out
V
V
H =
    [equation 1] 
 
In a circuit containing only resistors, transfer functions can be easily found using voltage 
dividers.  In figure 2, a voltage divider is used to find the transfer function at Vout. 
 
 
Figure 2 
3 2 1
3 2
*
R R R
R R
V V
in out
+ +
+
=
K K K
K K
V V
in out
3 2 1
3 2
*
+ +
+
=
6
5
= =
in
out
V
V
H
( ) Vin f Vout =
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 2 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
Since H is 5/6 for this circuit, we can find the output at Vout by multiplying any input, 
Vin, by 5/6.  What if the circuit contains components other than resistors? 
 
 
Figure 3 
 
The equations that govern the behavior of capacitors and inductors are not linear, like 
those for resistors.  Voltage dividers are based on these simple linear relationships.  How 
can we find a simple transfer function for a circuit with voltage relationships that have 
derivatives and integrals?  We could use differential equations, but that would not be the 
simple solution we are looking for.  We need to find a way to get rid of these integrals 
and derivatives so that we can get back to a simple linear relationship.  This is the 
fundamental motivation for steady state analysis. 
 
 
B. A Model for Steady State Analysis 
 
Steady state analysis is a way to analyze AC circuits mathematically.  We can do this 
using linear transfer functions (and without the use of differential equations) if we do two 
important things.  The first is to take advantage of the fact that AC signals are sinusoids 
that do not change in frequency.  The second is to map the input from the time domain 
into a domain based on the amplitude and frequency of the signals involved. 
 
B.1 Steady State Sinusoids 
 
The term steady state refers to the concept that a circuit, when attached to an AC input 
signal, will, after a finite period of time, reach a state in which the signal across each 
device in the circuit behaves like a sinusoid of the same frequency as the input.  The 
signal across each device will have a constant amplitude and phase relative to the input 
that does not change once steady state has been reached.  Once a circuit has reached 
steady state, the output at any point in the circuit will look like a sinusoid with the same 
frequency as the input, a constant amplitude, and a phase that does not change relative to 
the input phase.  It is this feature that we will exploit to find transfer functions. 
Sinusoids have a special mathematical property in that the derivative of a sinusoid is just 
another sinusoid with the same frequency, but shifted in phase.  The same is true for the 
V
out
Page 3


Electronic Instrumentation  ENGR-4300 
Susan Bonner - 1 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
AC Steady State Analysis 
 
What if we are given a circuit and we wish to find the output of that circuit for a range of 
inputs?  We could wire the circuit, attach the function generator to the input, and hook up 
the scope to the output.  Then, we could alter the input systematically and observe how 
the output responds to these changes.  If we could use this information to find a 
mathematical function that relates the input to the output, then we could use the function 
to predict how the circuit will behave for any given input, as shown in figure 1 below.  
What if we could use the components in the circuit to derive this function?  Then we 
could predict the output of any circuit for any given input. 
 
Figure 1 
 
 
A. What is a Transfer Function? 
 
A transfer function relates the output and the input of a circuit.  In order for a transfer 
function to be useful, it must be simple to use and easy to find using the circuit diagram.  
Therefore, let us define a transfer function, H, as the ratio of the output to the input of a 
circuit.   
in
out
V
V
H =
    [equation 1] 
 
In a circuit containing only resistors, transfer functions can be easily found using voltage 
dividers.  In figure 2, a voltage divider is used to find the transfer function at Vout. 
 
 
Figure 2 
3 2 1
3 2
*
R R R
R R
V V
in out
+ +
+
=
K K K
K K
V V
in out
3 2 1
3 2
*
+ +
+
=
6
5
= =
in
out
V
V
H
( ) Vin f Vout =
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 2 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
Since H is 5/6 for this circuit, we can find the output at Vout by multiplying any input, 
Vin, by 5/6.  What if the circuit contains components other than resistors? 
 
 
Figure 3 
 
The equations that govern the behavior of capacitors and inductors are not linear, like 
those for resistors.  Voltage dividers are based on these simple linear relationships.  How 
can we find a simple transfer function for a circuit with voltage relationships that have 
derivatives and integrals?  We could use differential equations, but that would not be the 
simple solution we are looking for.  We need to find a way to get rid of these integrals 
and derivatives so that we can get back to a simple linear relationship.  This is the 
fundamental motivation for steady state analysis. 
 
 
B. A Model for Steady State Analysis 
 
Steady state analysis is a way to analyze AC circuits mathematically.  We can do this 
using linear transfer functions (and without the use of differential equations) if we do two 
important things.  The first is to take advantage of the fact that AC signals are sinusoids 
that do not change in frequency.  The second is to map the input from the time domain 
into a domain based on the amplitude and frequency of the signals involved. 
 
B.1 Steady State Sinusoids 
 
The term steady state refers to the concept that a circuit, when attached to an AC input 
signal, will, after a finite period of time, reach a state in which the signal across each 
device in the circuit behaves like a sinusoid of the same frequency as the input.  The 
signal across each device will have a constant amplitude and phase relative to the input 
that does not change once steady state has been reached.  Once a circuit has reached 
steady state, the output at any point in the circuit will look like a sinusoid with the same 
frequency as the input, a constant amplitude, and a phase that does not change relative to 
the input phase.  It is this feature that we will exploit to find transfer functions. 
Sinusoids have a special mathematical property in that the derivative of a sinusoid is just 
another sinusoid with the same frequency, but shifted in phase.  The same is true for the 
V
out
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 3 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
integral os a sinusoid.  So when we need to take the derivative (or integral) of a 
sinusoidal signal, we can think of it simply as the application of some change in 
amplitude and some shift in phase. 
 
Mathematically, the derivative of a sine wave of frequency ? and amplitude A, 
) sin( f ? + = t A Vin , is given by a cosine wave of frequency ? and amplitude A?, 
) cos( f ? ? + = t A
dt
dVin
.  The output of the derivative operation is a new sine wave with a 
different amplitude and a phase shifted by and additional 90 degrees, 
)
2
sin(
p
f ? ? + + = t A
dt
dVin
.  Mathematically, this means that 
2
' ' ) ' sin( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  The derivative of a cosine (or any 
other sinusoid) results in the same changes.  For example, if ) cos( f ? + = t A Vin , then 
)
2
cos( )
2
cos( ) sin( ) sin(
p
f ? ?
p
p f ? ? p f ? ? f ? ? + + = - + + = + + = + - = t A t A t A t A
dt
dVin
 and, therefore, 
2
' ' ) ' cos( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  Similarly, the 
integral of a sinusoid has a new amplitude of A/? and an additional phase shift of -90 
degrees.  So, if ) sin( f ? + = t A Vin , then 
2
' ' ) ' sin( '
p
f f
?
f ? - = = + =
?
and
A
A where t A dt Vin . 
 
If we can somehow model our input and output waves so that we are concerning 
ourselves just with the changes in amplitude and phase brought about by the integrals and 
derivatives involved, then we can eliminate the need to use differential equations.  Since 
the phase is an angle, it seems logical to begin with polar coordinates. 
 
 
B.2 Polar Coordinates 
Polar coordinates are a simple re-mapping of the traditional rectangular (x-y) coordinates 
on a plane into another set of variables, r and ?, where r is the distance from the origin 
and ? is the angle measured from the positive x axis in counter-clockwise direction.  This 
is shown in figure 4. 
 
To convert a point between the polar and x-y coordinate systems, trigonometric identities 
must be used.  The distance to the origin can be found using the Pythagorean theorem: 
2 2
y x r + = and the angle from the x axis is simply found by 
x
y
= ? tan or 
?
?
?
?
?
?
=
-
x
y
1
tan ? .  To go the other way, we can use x = r cos ?  and y = r sin ?. 
Page 4


Electronic Instrumentation  ENGR-4300 
Susan Bonner - 1 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
AC Steady State Analysis 
 
What if we are given a circuit and we wish to find the output of that circuit for a range of 
inputs?  We could wire the circuit, attach the function generator to the input, and hook up 
the scope to the output.  Then, we could alter the input systematically and observe how 
the output responds to these changes.  If we could use this information to find a 
mathematical function that relates the input to the output, then we could use the function 
to predict how the circuit will behave for any given input, as shown in figure 1 below.  
What if we could use the components in the circuit to derive this function?  Then we 
could predict the output of any circuit for any given input. 
 
Figure 1 
 
 
A. What is a Transfer Function? 
 
A transfer function relates the output and the input of a circuit.  In order for a transfer 
function to be useful, it must be simple to use and easy to find using the circuit diagram.  
Therefore, let us define a transfer function, H, as the ratio of the output to the input of a 
circuit.   
in
out
V
V
H =
    [equation 1] 
 
In a circuit containing only resistors, transfer functions can be easily found using voltage 
dividers.  In figure 2, a voltage divider is used to find the transfer function at Vout. 
 
 
Figure 2 
3 2 1
3 2
*
R R R
R R
V V
in out
+ +
+
=
K K K
K K
V V
in out
3 2 1
3 2
*
+ +
+
=
6
5
= =
in
out
V
V
H
( ) Vin f Vout =
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 2 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
Since H is 5/6 for this circuit, we can find the output at Vout by multiplying any input, 
Vin, by 5/6.  What if the circuit contains components other than resistors? 
 
 
Figure 3 
 
The equations that govern the behavior of capacitors and inductors are not linear, like 
those for resistors.  Voltage dividers are based on these simple linear relationships.  How 
can we find a simple transfer function for a circuit with voltage relationships that have 
derivatives and integrals?  We could use differential equations, but that would not be the 
simple solution we are looking for.  We need to find a way to get rid of these integrals 
and derivatives so that we can get back to a simple linear relationship.  This is the 
fundamental motivation for steady state analysis. 
 
 
B. A Model for Steady State Analysis 
 
Steady state analysis is a way to analyze AC circuits mathematically.  We can do this 
using linear transfer functions (and without the use of differential equations) if we do two 
important things.  The first is to take advantage of the fact that AC signals are sinusoids 
that do not change in frequency.  The second is to map the input from the time domain 
into a domain based on the amplitude and frequency of the signals involved. 
 
B.1 Steady State Sinusoids 
 
The term steady state refers to the concept that a circuit, when attached to an AC input 
signal, will, after a finite period of time, reach a state in which the signal across each 
device in the circuit behaves like a sinusoid of the same frequency as the input.  The 
signal across each device will have a constant amplitude and phase relative to the input 
that does not change once steady state has been reached.  Once a circuit has reached 
steady state, the output at any point in the circuit will look like a sinusoid with the same 
frequency as the input, a constant amplitude, and a phase that does not change relative to 
the input phase.  It is this feature that we will exploit to find transfer functions. 
Sinusoids have a special mathematical property in that the derivative of a sinusoid is just 
another sinusoid with the same frequency, but shifted in phase.  The same is true for the 
V
out
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 3 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
integral os a sinusoid.  So when we need to take the derivative (or integral) of a 
sinusoidal signal, we can think of it simply as the application of some change in 
amplitude and some shift in phase. 
 
Mathematically, the derivative of a sine wave of frequency ? and amplitude A, 
) sin( f ? + = t A Vin , is given by a cosine wave of frequency ? and amplitude A?, 
) cos( f ? ? + = t A
dt
dVin
.  The output of the derivative operation is a new sine wave with a 
different amplitude and a phase shifted by and additional 90 degrees, 
)
2
sin(
p
f ? ? + + = t A
dt
dVin
.  Mathematically, this means that 
2
' ' ) ' sin( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  The derivative of a cosine (or any 
other sinusoid) results in the same changes.  For example, if ) cos( f ? + = t A Vin , then 
)
2
cos( )
2
cos( ) sin( ) sin(
p
f ? ?
p
p f ? ? p f ? ? f ? ? + + = - + + = + + = + - = t A t A t A t A
dt
dVin
 and, therefore, 
2
' ' ) ' cos( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  Similarly, the 
integral of a sinusoid has a new amplitude of A/? and an additional phase shift of -90 
degrees.  So, if ) sin( f ? + = t A Vin , then 
2
' ' ) ' sin( '
p
f f
?
f ? - = = + =
?
and
A
A where t A dt Vin . 
 
If we can somehow model our input and output waves so that we are concerning 
ourselves just with the changes in amplitude and phase brought about by the integrals and 
derivatives involved, then we can eliminate the need to use differential equations.  Since 
the phase is an angle, it seems logical to begin with polar coordinates. 
 
 
B.2 Polar Coordinates 
Polar coordinates are a simple re-mapping of the traditional rectangular (x-y) coordinates 
on a plane into another set of variables, r and ?, where r is the distance from the origin 
and ? is the angle measured from the positive x axis in counter-clockwise direction.  This 
is shown in figure 4. 
 
To convert a point between the polar and x-y coordinate systems, trigonometric identities 
must be used.  The distance to the origin can be found using the Pythagorean theorem: 
2 2
y x r + = and the angle from the x axis is simply found by 
x
y
= ? tan or 
?
?
?
?
?
?
=
-
x
y
1
tan ? .  To go the other way, we can use x = r cos ?  and y = r sin ?. 
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 4 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
 
Figure 4 
 
If we think of our signal as having a phase offset of f and an amplitude of A, we can 
model a sinusoid (of angular frequency, ?) as pictured in figure 5.  Note that the sinusoid 
begins at f and cycles around the origin one time for each cycle (?t).  The ray has a 
length of A, the amplitude of the sinusoid.  If we want to get back the original sinusoid, 
we have only look at either the x coordinate, x = A cos(?t+f), or the y coordinate, y = 
Asin(?t+f).  Note that ?t represents a complete cycle around the circle (2p) at a rate of ? 
radians per second.  This corresponds to the constant cycling of the sinusoid in time.  We 
could just as easily drop the ?t, since it is a simple multiple of 2p, and we would have an 
x coordinate, x = A cos(f), and a y coordinate, y = A sin(f). 
 
 
Figure 5 
 
Page 5


Electronic Instrumentation  ENGR-4300 
Susan Bonner - 1 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
AC Steady State Analysis 
 
What if we are given a circuit and we wish to find the output of that circuit for a range of 
inputs?  We could wire the circuit, attach the function generator to the input, and hook up 
the scope to the output.  Then, we could alter the input systematically and observe how 
the output responds to these changes.  If we could use this information to find a 
mathematical function that relates the input to the output, then we could use the function 
to predict how the circuit will behave for any given input, as shown in figure 1 below.  
What if we could use the components in the circuit to derive this function?  Then we 
could predict the output of any circuit for any given input. 
 
Figure 1 
 
 
A. What is a Transfer Function? 
 
A transfer function relates the output and the input of a circuit.  In order for a transfer 
function to be useful, it must be simple to use and easy to find using the circuit diagram.  
Therefore, let us define a transfer function, H, as the ratio of the output to the input of a 
circuit.   
in
out
V
V
H =
    [equation 1] 
 
In a circuit containing only resistors, transfer functions can be easily found using voltage 
dividers.  In figure 2, a voltage divider is used to find the transfer function at Vout. 
 
 
Figure 2 
3 2 1
3 2
*
R R R
R R
V V
in out
+ +
+
=
K K K
K K
V V
in out
3 2 1
3 2
*
+ +
+
=
6
5
= =
in
out
V
V
H
( ) Vin f Vout =
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 2 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
Since H is 5/6 for this circuit, we can find the output at Vout by multiplying any input, 
Vin, by 5/6.  What if the circuit contains components other than resistors? 
 
 
Figure 3 
 
The equations that govern the behavior of capacitors and inductors are not linear, like 
those for resistors.  Voltage dividers are based on these simple linear relationships.  How 
can we find a simple transfer function for a circuit with voltage relationships that have 
derivatives and integrals?  We could use differential equations, but that would not be the 
simple solution we are looking for.  We need to find a way to get rid of these integrals 
and derivatives so that we can get back to a simple linear relationship.  This is the 
fundamental motivation for steady state analysis. 
 
 
B. A Model for Steady State Analysis 
 
Steady state analysis is a way to analyze AC circuits mathematically.  We can do this 
using linear transfer functions (and without the use of differential equations) if we do two 
important things.  The first is to take advantage of the fact that AC signals are sinusoids 
that do not change in frequency.  The second is to map the input from the time domain 
into a domain based on the amplitude and frequency of the signals involved. 
 
B.1 Steady State Sinusoids 
 
The term steady state refers to the concept that a circuit, when attached to an AC input 
signal, will, after a finite period of time, reach a state in which the signal across each 
device in the circuit behaves like a sinusoid of the same frequency as the input.  The 
signal across each device will have a constant amplitude and phase relative to the input 
that does not change once steady state has been reached.  Once a circuit has reached 
steady state, the output at any point in the circuit will look like a sinusoid with the same 
frequency as the input, a constant amplitude, and a phase that does not change relative to 
the input phase.  It is this feature that we will exploit to find transfer functions. 
Sinusoids have a special mathematical property in that the derivative of a sinusoid is just 
another sinusoid with the same frequency, but shifted in phase.  The same is true for the 
V
out
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 3 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
integral os a sinusoid.  So when we need to take the derivative (or integral) of a 
sinusoidal signal, we can think of it simply as the application of some change in 
amplitude and some shift in phase. 
 
Mathematically, the derivative of a sine wave of frequency ? and amplitude A, 
) sin( f ? + = t A Vin , is given by a cosine wave of frequency ? and amplitude A?, 
) cos( f ? ? + = t A
dt
dVin
.  The output of the derivative operation is a new sine wave with a 
different amplitude and a phase shifted by and additional 90 degrees, 
)
2
sin(
p
f ? ? + + = t A
dt
dVin
.  Mathematically, this means that 
2
' ' ) ' sin( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  The derivative of a cosine (or any 
other sinusoid) results in the same changes.  For example, if ) cos( f ? + = t A Vin , then 
)
2
cos( )
2
cos( ) sin( ) sin(
p
f ? ?
p
p f ? ? p f ? ? f ? ? + + = - + + = + + = + - = t A t A t A t A
dt
dVin
 and, therefore, 
2
' ' ) ' cos( '
p
f f ? f ? + = = + = and A A where t A
dt
dVin
.  Similarly, the 
integral of a sinusoid has a new amplitude of A/? and an additional phase shift of -90 
degrees.  So, if ) sin( f ? + = t A Vin , then 
2
' ' ) ' sin( '
p
f f
?
f ? - = = + =
?
and
A
A where t A dt Vin . 
 
If we can somehow model our input and output waves so that we are concerning 
ourselves just with the changes in amplitude and phase brought about by the integrals and 
derivatives involved, then we can eliminate the need to use differential equations.  Since 
the phase is an angle, it seems logical to begin with polar coordinates. 
 
 
B.2 Polar Coordinates 
Polar coordinates are a simple re-mapping of the traditional rectangular (x-y) coordinates 
on a plane into another set of variables, r and ?, where r is the distance from the origin 
and ? is the angle measured from the positive x axis in counter-clockwise direction.  This 
is shown in figure 4. 
 
To convert a point between the polar and x-y coordinate systems, trigonometric identities 
must be used.  The distance to the origin can be found using the Pythagorean theorem: 
2 2
y x r + = and the angle from the x axis is simply found by 
x
y
= ? tan or 
?
?
?
?
?
?
=
-
x
y
1
tan ? .  To go the other way, we can use x = r cos ?  and y = r sin ?. 
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 4 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
 
Figure 4 
 
If we think of our signal as having a phase offset of f and an amplitude of A, we can 
model a sinusoid (of angular frequency, ?) as pictured in figure 5.  Note that the sinusoid 
begins at f and cycles around the origin one time for each cycle (?t).  The ray has a 
length of A, the amplitude of the sinusoid.  If we want to get back the original sinusoid, 
we have only look at either the x coordinate, x = A cos(?t+f), or the y coordinate, y = 
Asin(?t+f).  Note that ?t represents a complete cycle around the circle (2p) at a rate of ? 
radians per second.  This corresponds to the constant cycling of the sinusoid in time.  We 
could just as easily drop the ?t, since it is a simple multiple of 2p, and we would have an 
x coordinate, x = A cos(f), and a y coordinate, y = A sin(f). 
 
 
Figure 5 
 
Electronic Instrumentation  ENGR-4300 
Susan Bonner - 5 - Revised: 1/16/2006  
Rensselaer Polytechnic Institute  Troy, New York, USA 
 
 
Polar coordinates give us a good way to express the important features of our sinusoid, 
but we still need a way to operate on the sinusoids to simulate the behavior of the circuit.  
For this, we will need to rethink our model in terms of complex numbers. 
 
B.3 Complex Numbers 
Complex numbers are numbers expressed as ordered pairs, consisting of a “real” part and 
an “imaginary” part. In other words, a complex number is a point on a plane called the 
complex plane.  A complex number z can be written as  z = x + j y.  In a complex 
number:  x is called the “real” part, y is called the “imaginary” part, and 1 - = j . (Note 
that in common mathematical notation, i is used for 1 - , however, in electrical 
engineering, since i is commonly used for current, we denote 1 - with j.)   The term 
“imaginary” comes from the fact that 1 - = j is not a “real” number.  Complex 
numbers are generally represented on a plane where y is plotted against x.  The x 
coordinate consists of the basic number line containing all real numbers.  The y 
coordinate is the same set of numbers multiplied by 1 - .  Each complex number, z, is a 
unique point on the complex plane, shown in figure 6, where the x coordinate is the real 
part and the y coordinate is the imaginary part. 
 
 
Figure 6 
 
Before we continue there are two useful identities to remember about j: 
 
1 1 1 1
2 2
- = - = - · - = · = j j j j
 
 
and        
j
j
j
j
j j
j
j
- = - =
-
=
·
·
=
1
1
1 1
. 
 
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