THE RLC, BY AN ME (student)
INTRODUCTION
This guide was written by a frustrated mechanical engineering student, who,
intrigued by the material, (but not enough to change fields) thought that others in the
same situation as he, desiring of a simple guide to understanding the components of the
externally driven resistor-inductor-capacitor network and their effects on the network’s
behavior would enjoy and perhaps profit from a concise elaboration of the factors.
PREREQUISITES
This guide assumes some familiarity with differential equations, but I will attempt
to explain the wackiness involved with the phasor transformations in a way that will
make sense both before and after taking the diff. eq. class.
THE CRUX
The first thing to export from diff. eq. into our circuits class is the transformation
into s-space we learned about from the Laplace transform. To the best of my knowledge,
the Laplace transform (and her relative, the Fourier series expansion) take a function in
the time-domain and move it into an imaginary domain where all we consider is the
behavior of the system for one oscillation.
We perform an operation to move the expressions given on our circuit schematic
from the time domain into the s-domain, in smooth parallel with what we know from
differential equations. We must now calculate the contribution of each element in the
network to the circuit’s oscillatory behavior. There are two important (reciprocal) values
for this: reactance (Y) and impedance (Z). Y=1/Z, Z=1/Y. We refer to the time domain as
“t-space” and the frequency domain as “s-space”.
t-space
Driving Current/Voltage A⋅cos(ωt+θ)
Resistor
R(Ω)
Capacitor
C(F)
Inductor
L(H)
s-space
A∠θ
Z=R
Comments
The resistor does not have any time dependent
behavior, even in an oscillating circuit and so
translates simply into s-space.
Z=1/jWC If the circuit fluctuates rapidly around a capacitor, it
has a low impedance and acts as (almost) a short
circuit. This should explain why capacitors’ impedance
decreases with frequency.
Z=jWL
We know that the magnetic field from current through
an inductor resists changing current through the
inductor. At high frequencies, the circuit will try to
change the direction of current through the inductor
many times per second, which will create large voltage
drops across the inductor. This should explain why
inductors’ impedance grows with frequency.
Table I: Circuit Element Transformations Into S-Space. j's are imaginary numbers, which are discussed below.
The transformation into s-space allows us to talk about resistors, inductors and
capacitors behavior in the circuit as it oscillates. Essentially we are describing their
contribution to the oscillatory behavior. Because we want to describe the whole circuit’s
behavior and the behavior of voltage and current through individual branches, we need
some common value to compare between elements. This is reactance and impedance.
A fundamental relationship between voltage, current and resistance is useful in
moving forward. We know (from doing oodles of homework problems) that v(t)=i(t)⋅R.
We have transformed our circuit from the time domain into the frequency domain, and
obtained a value Z. A clue in understanding this circuit is that R translates directly into sspace as Z=R. We transform the v/i/r relationship into s-space as follows: v(t)=i(t)⋅R 
V=IZ where V is the phasor representation of the driving voltage, I is the s-space
transformation of the current and Z is the impedance.
TWO IMPORTANT AND USEFUL TOOLS WITH NO DERIVATION
We will find it useful from time to time to describe the behavior of the voltage
across a parallel RLC circuit and the current through a series RLC circuit.
Parallel
Series
Impedance/Reactance
Ytotal=ΣYi
Ztotal=ΣZi
Circuit Behavior
I=V/Z
V=ZI
Branch Behavior
Ii=V/Zi
Vi=ZiI
Table II: Circuit and Branch Behavior in S-Space
The branch behavior column shows that the current through an element in parallel
will be the voltage s-space representation divided by the element’s individual inductance.
An element in series will see a voltage drop across its terminals equal to the product of
the element’s inductance and the current through the whole series. From our experience
with loops and KCL/KVL, these two relationships should make sense: the voltage drop
across the sum of the three elements in series should equal the voltage (in s-space, at
least) across the series of elements, and the sum of the currents into our out of a node
should equal zero. The impedance/reactance conversions for each case are useful for
summing, so as to avoid fraction multiplication and related algebraic hoop-jumping.
Naturally, we will want to return from the frequency domain to the time domain
at some point, otherwise we’d only ever be able to design circuits that oscillated once
(jokes, jokes…). At this point it is useful to discuss the imaginary j’s mentioned above.
When performing operations on branches of the circuit in s-space, you will be
operating on parameters with a real component and an imaginary component (imaginary
numbers act exactly like real ones, its just that instead of an x-y coordinate plane, we
operate in a x-[imaginary] space). They will resemble: a+bj (no comments from the
peanut gallery, please). When you want to take two elements in a branch and combine
them into one element, you can add their real components and imaginary components. ∴
[a1+b1j]+[a2+b2j] = (a1+a2)+j(b1+b2), := a3+b3j.
Indubitably, you will want to perform operations in s-space related to V=IZ. You
will have a phasor from your voltage or current source of A⋅cos(ωt+θ) and an impedance
value with real and imaginary parts. Given an equation that looks like:
A∠θ
a3 + b3 j
…how on earth can we perform any mathematical operations at all on that beast?
The trick lies in remembering the equation:
2
2
tan −1
b
a
€ a + bj = a + b e
which should kick out answers in the format of [integer]∠[angle]. Remembering
rules for division and addition of exponentials, divide your integers in the numerator by
your integers in the denominator and subtract/add degrees as dictated by sign.
€
Remembering also that we wish to ultimately transform back into the time
domain, we recall that:
A∠θ = Aeωt +θ = Acos(ωt + θ )
…upon completion of these maths, we have returned to the time domain.
It is important to note at this point that we have come up with equations that
€ the branch current or voltage (A) in conjunction with an
describe the amplitude of
oscillating function of the same period as our driving frequency (whether voltage or
current, across a parallel set of elements or through a set in series) plus an offset term.
The conclusion to draw is that the whole circuit will resonate at the same frequency,
although individual elements may lag or lead the driving signal as dictated by their
impedances and values in the rest of the circuit.