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Circuit Analysis - ECE 2300
LECTURE NOTES
DAVE SHATTUCK
SET
Chapter 9 - Sinusoidal Steady-State Analysis
Now we take on two major paradigm shifts,
which are closely related. What we do in this
chapter is:
- Define what we mean by “steady state”
when we have sinusoidal sources.
- Define “phasors.”
- Introduce a new solution technique, called
phasor analysis.
- Introduce and define “RMS.”
First, I wish to introduce the first
engineering paradigm, Fourier’s Theorem.
Before we go on, the pronunciation.
Everyone repeat, after me, 4-E-A. Furriers are
people who make coats.
Fourier’s Theorem - Any physically
realizable signal can be represented by, and is
equivalent to, a summation of sinusoids of
different frequency, phase and amplitude.
This theorem has profound implications, and
has effectively shaped the way that we look at
almost all of electrical engineering. It is
remarkably powerful. It is profound. I am still
not sure that I believe it. But, apparently, it is
true.
Now, through special mathematical
techniques, called phasor analysis, the solution
of complex differential equations, that describe
circuits and their responses, can be found
easily (relatively speaking), as long as the
sources are sinusoidal. By Fourier’s Theorem,
ultimately, all sources are sinusoidal.
Thus, we have a simple technique with
universal application. This is almost too good
to be true.
There are two hitches. It may require an
infinite number of sinusoidal components to
v(t) = V m
cos (t + )
where V m
is the amplitude, is the frequency,
and is the phase of the sinusoid. Note that if
T is the period of the sinusoid (or any periodic
source), then
f = 1 / T
and
= 2πf. f.
The frequency f typically has units of [cycles/s]
or [Hertz] or [Hz] = [s
]. The variable is
called the angular frequency, and typically has
units of [radians/s], or [rad/s]. It is very
important to keep these two quantities straight.
Thus t has typical units of [radians]. The
phase shift has typical units of [degrees] or
[deg]. This can be awkward. We need to be
very careful, especially when we evaluate
sinusoids. Fortunately, we do not do this very
often in this course.
We now define RMS. This means Root-
Mean-Squared. It is a value that is associated
with periodic functions. It has some
particularly valuable information with regard to
average power calculations using periodic
functions. However, for the time being, we will
simply say that we can calculate the RMS
value of any periodic function by taking the
square root of the mean value of the squared
function. Some important notes:
- In order to get this value, we take the
inverse order. We square the function, then
take the mean value of that squared function,
and then take the square root of that mean
value. So, we do SMR.
- We take the RMS value of periodic
functions, that is, functions that vary with time.
However, since the RMS function includes a
mean value, the RMS value is not a function
of time.
i(t) = {-V
m
(R
L
1/2 } cos () e
(R/L) t
{V
m
(R
L
1/2 } cos (t + )
where
= arctan (L / R)
Let us examine this solution.
- The first term is not sinusoidal (with time).
Rather, it is a decaying exponential. It dies out
with time. After several ’s, it is gone. This is
called the transient response.
- The second term is sinusoidal (with time).
It does not die out. This called the steady-
state response.
We will only be able to use phasor analysis to
get the steady-state solution, but this may be
all we want. Some fun facts to know and tell
about the steady-state solution.
- The steady-state solution is a sinusoidal
function.
- The frequency of the steady-state solution
is the same as that of the source.
- The amplitude of the steady-state solution
can be different from that of the source.
- The phase of the steady-state solution can
be different from that of the source.
Thus, when we want to solve for the steady-
state solution with sinusoidal sources, we want
to get two things: the amplitude and the
phase. Everything else is already known.
Section 9.3 - The Phasor
Remember Euler’s relation:
e
j = cos + j sin
So, we can say that
v(t) = V m
cos (t + ) = V m
Re {e
j(
where Re means the real part of the
expression that follows.
We will always use an m subscript when we
use magnitude based phasors.
We can also define the Inverse Phasor
Transform. Here, if we put in the phasor, we
get the sinusoid out. Remember that we knew
the frequency from the beginning. When we
are in the phasor domain, it is understood that
the frequency is known. Also, we assume that
when we take the inverse transform, that we
know whether the magnitude of the phasor is
the zero to peak, or rms value, and convert
accordingly.
We will take phasors with magnitude values
sometimes, and with rms values sometimes.
We will use notation to make it clear which one
we are using.
Let us solve a problem the hard way, to make
a point. I am going to work through all the
steps. Often, in this class, we skip this
process. However, here we will try to meet two
goals:
- Several steps in the solution will make clear
what is going on in the simpler approach.
- We will see why we never want to do this
again.
Take the RL circuit from before. (Draw it
again, showing the series voltage source,
resistor and inductor.)
We want the steady-state solution for i(t),
which we will call i ss
(t). From our previous
analysis, we know that
i ss
(t) = I m
cos (t + ) = I m
Re {e
j(t e
j(
We want I m
and .
Now, v S
= Ri + L di/dt, and since i ss
(t) has got
to work in this equation, we can plug it in to
get:
V
m
Re {e
j(t e
j( = R I m
Re {e
j(t e
j( +
L d/dt [I
m
Re {e
j(t e
j(
Note that this equation is not a function of t.
Next, we factor out I m
and e
j( out of the
terms, and write
V
m
e
j( = I m
e
j( (R + jL)
We solve for I m
e
j( to get
V
m
e
j( / (R + jL)
= I
m
e
j( Equation
Now, how many equations do we have here?
Ans: There only appears to be one equation,
but it is a complex equation. Thus, the real
parts must be equal (1 equation) and the
imaginary parts must be equal (1 equation).
This is 2 equations. It is also true that the
magnitudes must be equal and the phases
must be equal, which give us two more
equations, but they are not independent. We
can pick either pair and solve for our two
unknowns.
So, we can solve Equation #1 for I m
and .
and we get the solution for
i ss
(t) = I m
cos (t + )
Our goal now is to find a quicker way to get to
Equation #1. That’s what phasor analysis will
give us. Notice that we can always divide
through by e
jt
. This will always give us an
equation which is independent of time. We
can rewrite Equation #1 in terms of phasors, as
_ _
V
R,m
= R I
max
If we take the ratio of these two, we get
_ _
V
R,m
/ I
R,m
= R which is called Complex
Ohm’s Law.
Inductor: Suppose that
i L
(t) = I max
cos (t + ),
then, by the defining equation of an inductor,
we have
v L
(t) = -L I max
sin (t + ).
By basic trig relationships, we can write
v L
(t) = -L I max
cos (t + - 90 ° ).
Now,
_
I
L,m
= I
max
and
_
V
L,m
= - L I
max
Putting this back in its exponential form, we
see that
V
L,m
= - L I
max
e
j e
j(-90 ° )
which is equal to
V
L,m
= - L I
max
e
j (-j) = jL I max
e
j
If we take the ratio of these two, we get
_ _
V
L,m
/ I
L,m
= jL.
If this has a name, I don’t know what it is.
Please be careful. It is not useful, nor typical
for us to take the ratio for an inductor in the
time domain. If we take
v L
(t) / i L
(t) = -L I max
sin(t + ) / I max
cos(t + )
= -L tan(t + ).
This function is sometimes zero, sometimes ,
and sometimes -. It is not equal to jL.
By a similar approach, we can find that
_ _
- Solve the resulting circuit using the
impedances in the same way that you would
use resistances.
- Convert the phasor voltage/current that yo
obtain back to a time domain voltage/current.
Do not skip step 4.
Take special care not to mix domains at any
time. There should never be a t (for time) in
the phasor domain. There should never be a j
(square root of -1) in the time domain. A single
diagram should never have components or
voltages or currents labelled from both
domains.
Let’s do some examples.
