Copyright © 2012 by Wayne Stegall
Updated October 27, 2012. See Document History at end for
details.
Understanding the SDomain
Introduction
Because I often create and solve sdomain equations in my articles to
analyze circuits, I thought it well to present here a brief explanation
of the sdomain.
Mathematical Difficulties
Circuit analysis would be a simple case of adding series and parallel
resistors and doing related calculations as in DC analysis if not for
the addition of the energy storage components: the inductor and
capacitor. In
figure 1
below the inductor charges current at a
rate proportional to the voltage drop across it. Likewise in
figure 2 below the capacitor
charges voltage at a rate proportional to
its input current.
Figure
1: Current increases through inductor as it charges.


Figure
2: Voltage increases across capacitor as it charges




Unfortunately these component behaviors are mathematically modeled by
the calculus operation of differentiation, or integration if the
equations are reversed with respect to voltage and current.
Differentiation represented in
figures
1 and
2 above is
simply the
instantaneous slope of the input function.
Direct circuit analysis with calculus level math then requires putting
these operations together to solve them in a form called the
differential equation:
(3)

D(t) = a_{0}f(t) + a_{1} 
d f(t)
dt

+ a_{2} 
d^{2} f(t)
dt^{2}

+ a_{3} 
d^{3} f(t)
dt^{3} 
+ … + a_{n} 
d^{n} f(t)
dt^{n} 
where

d^{n} f(t)
dt^{n} 
represents the function f(t)
differentiated n times.

Arbitrary differential equations are only sensible if they represent
functions that have the same form only altered in scale when
differentiated or integrated. Because exponential and sinusoid
functions always follow this rule, it is usually assumed that the
solution to such a differential equation is a function of this
sort.
If it is further assumed that sinusoids are derived from an exponential
with an imaginary power (i.e. Euler's Law: e
^{jx} =
cos(x) +
jsin(x)),
additional generalization of the solution function will result.
(4)

f(t) = e^{st}

^{a} where s = σ + jω 
Each level of differentiation multiplies the solution equation by s.
(5)

d e^{st}
dt

= se^{st} 
If this generalized solution equation is substituted for f(t) a
considerable simplification can be affected.
(6)

D(t) = a_{0}e^{st}
+ a_{1} 
d e^{st}
dt^{1}

+^{2}a_{2} 
d^{2} e^{st}
dt^{2}

+^{3}a_{3} 
d^{3} e^{st}
dt^{3} 
+ … +^{n}a_{n} 
d^{n} e^{st}
dt^{n} 
Differentiating each term multiplies it by a power of s of the same
degree as the derivative.
(7)

D(t) = a_{0}e^{st}
+ a_{1}se^{st}
+ a_{2}s^{2}e^{st} + a_{3}s^{3}e^{st}
+ … + a_{n}s^{n}e^{st}

Now dividing the entire equation by the generalized solution
leaves a
simple polynomial in the sdomain that is much easier manipulated and
solved.
(8)

D(s) = a_{0} + a_{1}s
+
a_{2}s^{2} + a_{3}s^{3} + … + a_{n}s^{n}

This transformation of a differential equation to an algebraic one in a
complex variable is known generally as the Laplace
Transformation. Where F(s) is the Laplace transform of f(t) the
following integral defines this transformation.
(9)

F(s) =

∫ 
∞
0 
e^{st}f(t) 
Simplified AC Analysis
Performing the same reductions steps on the defining differential
equations 1 and
2
for the inductor and capacitor gives sdomain terms for their impedance
that can be used to solve circuit equations with simple algebra rather
than the calculus on which they are based. Because the Laplace
transform is linear, the transforms of the individual components can be
added together.
These simple sdomain representations of otherwise complicated calculus
can be used in serial, parallel, and other DC equations and then
solved with regular algebra.
(12)

Z_{SERIES} = Z_{1}
+ Z_{2}


(13) 
Z_{PARALLEL} = 
Z_{1}Z_{2}
Z_{1} + Z_{2} 
(14)

Y_{PARALLEL} = Y_{1}
+ Y_{2} 

(15)

Y_{SERIES} = 
Y_{1}Y_{2}
Y_{1} + Y_{2} 
The simple reduced sdomain equation of
equation 8 above contains only
zeros when factored and is likely only to occur in a series RL
impedance or a parallel RC conductance. In other cases reduction
of an sdomain equation will yield an equation with zeros in the
numerator and poles in the denominator. Zeros are zero at their
characteristic svalue and poles represent infinity at theirs.
(16)

H(s) = 
a_{0} + a_{1}s
+
a_{2}s^{2} + a_{3}s^{3} + … + a_{m}s^{m}
b_{0} + b_{1}s
+
b_{2}s^{2} + b_{3}s^{3} + … + b_{n}s^{n}

(17)

H(s) = 
(s  z_{1})(s  z_{2})(s

z_{3}) … (s  z_{m})
(s  p_{1})(s  p_{2})(s

p_{3})
… (s  p_{n}) 
What does s actually
represent?
The Math
S is a twodimensional number, a vector, in a form commonly used in
mathematics called a
complex number.
Because
complex
numbers
have
unique
properties
beyond
a
mere
vector some prefer
to call them phasors instead. In
the
form
s
=
σ
+
jω,
σ is a vector component oriented at
zero
degrees and
jω a vector
vector component oriented at ninety degrees.
j
is just a 90º angle marker allowing the vector to be easily represented
as
an addition or a subtraction in which case
j would represent 270º.
Complex numbers have the following properties:
The 90º vector
j is defined
as:
therefore
and
They have a magnitude:
(21) 
(a + jb)^{2}= 
a^{2} + b^{2}

and a phase angle:
(22) 
Phase(a + jb) = arctan 
b
a

Note: This is a four quadrant arctangent. A calculator
arctangent will only be valid for complex numbers with a positive real
part. If the real part is negative, the arctangent will have to
be adjusted by adding 180º.
(23) 
Phase(a + jb) = 180º + arctan 
b
a

They also have complex conjugate:
They can be represented in rectangular form
or polar form:
(26) 
a + jb ∠ Phase(a + jb) 
Addition and subtraction are done individually in the real and
imaginary parts:
(27)

(a + jb) + (c + jd) = (a + c) + j(b + d) 
(28) 
(a + jb)  (c + jd) = (a  c) + j(b  d) 
In multiplication and division, the real and imaginary parts interact
to produce the following
results:
(29)

(a + jb) × (c + jd) = (acbd) + j(ad+bc) = (a + jb × c + jd) ∠ (Phase(a + jb) + Phase(c + jd)) 
(30) 
(a + jb)
(c + jd) 
=

(a + jb) × (c  jd)
(c + jd) × (c  jd) 
=

(ac+bd)^{2}+ j(bcad)
c^{2} + d^{2}

=

a + jb
c + jd 
∠ (Phase(a + jb)  Phase(c + jd)) 
The Time Domain Significance
Apart from the complex math, an s variable represents a signal in the
form the product of an exponential and a sinusoid as the following
derivations show.
(31)

f(t) = e^{st}

(32) 
f(t) = e^{(σ+jω)t}

(33) 
f(t) = e^{σt}(cos(ωt) + jsin(ωt)) 
Thus σ represents damping and ω a frequency component in natural
frequency units of radians/s.
To this point the derivation leaves a paradox: the timebased
signal has an imaginary component, something it should not have.
In order to resolve the dilemma, try again with the signal divided into
normal and complexconjugate parts.
(34)

f(t) = ½e^{st} + ½e^{s*t}

(35) 
f(t) = ½e^{(σ+jω)t}+
½e^{(σjω)t} 
(36) 
f(t) = ½e^{σt}(cos(ωt) + jsin(ωt))
+
½e^{σt}(cos(ωt)
 jsin(ωt)) 
(37) 
f(t) = e^{σt}cos(ωt)

For this reason, all real signals and systems represent frequencies
with positive and negative parts and have poles and zeros in conjugate
pairs.
Figure 3 below shows what kind of signals are represented in different
locations in the sdomain. Those on the real axis represent only
exponential decay and lack a resonant frequency. Those on the
imaginary axis represent only a sinusoidal signal without an
exponential component. Those left of the imaginary axis but off
of the real one are damped sinusoids. Those right of the
imaginary axis are all unstable due to exponential increase.
Figure
3:
Sdomain
showing
time
signal
representations.


The Frequency Significance of the Sdomain
If
jω for a range of frequencies [ω
_{1} = 2πf
_{1}
… ω
_{n} = 2πf
_{n}] is substituted for s in a full
sequation like
equation 16
above, each frequency point will calculate
to a concrete svalue representing magnitude and phase the total of
which will plot the magnitude or phase frequency response.
Equation 16 reduces to the
following following equation on the way to
yielding concrete numbers. Note that the coefficients now
alternate in sign as a result of the way powers of
j progress.
(38)

H(ω) = 
(a_{0 } a_{2}ω^{2}
+ a_{4}ω^{4}  …) + j(a_{1}ω  a_{3}ω^{3}
+ a_{5}ω^{5}  …)
(b_{0}  b_{2}ω^{2}
+ b_{4}ω^{4}  …) + j(b_{1}ω  b_{3}ω^{3}
+ b_{5}ω^{5}  …) 
Even though s itself is twodimensional, the plot of an sequation is
three dimensional (or even four dimensional if the complex nature of
the results are considered) as if σ and ω defined the base of the plot
and
the magnitude or phase plots the altitude of a mathematical
landscape. Reducing
the sequation with
jω slices this threedimensional plot along
the imaginary axis yielding back a two dimensional frequency plot
spanning both positive and negative frequencies. Poles and zeros
on this landscape do not affect resonance or frequency response as you
would expect at this point. You would expect frequency affects to
directly correspond their imaginary component
jω. Instead frequency effects
are correlated to the vector magnitude of their place in the
sdomain.
Consider the multiplication of two conjugate poles in a
second order lowpass equation.
(39)

H(s) =

k
(s +^{2}a+jb)(s + a jb)

=

k
s^{2} + 2as + a^{2}+b^{2}

Compare this with the standard equation.
Solving for the resonance frequency returns the magnitude of the
conjugate poles a result arising from the nature of complex math.
(41) 
ω^{2} =

a^{2} + b^{2} 

(42)

ω^{2}=

a^{2} + b^{2} 
=^{2}a + jb

For completeness at this point, lets solve for Q.
(43)

Q
ω 
=

1
2a





(44)

Q

=

ω^{2}
2a 
=

a^{2} + b^{2}
2a 
=

1^{2 }
2cos(θ_{pole}
180º) 
Or for simpler calculations Q reduces to:
(45)

Q =

1
2cos(θ_{pole})

How convenient! Q is a function of the angle of the pole.
In this light, all the poles in the sdomain pole plot of
figure 4 below represent the same
frequency as a result of having all the same magnitude. The
complex conjugate pairs each correspond to a resonant frequency f
_{0}
and
the real pole on the σ axis to a 3dB cutoff frequency f
_{c}.
Figure
4:
Sdomain
plot
showing
normalized
fifthorder
lowpass
Butterworth
response


Figure
5: Bode plot shows magnitude response of transfer function
represented in polezero plot of figure 4 above scaled to 1kHz.

Figure
6: Broken down by pole pairs, the bode plot shows how the
nearness of a pole to the jω axis raises Q.




Color


Q


Corresponding
normalized
pole(s)

yellow


NA


1 + j0

green


0.618


0.809 ± j0.5878

red


1.618


0.309 ± j0.9511


Examples
Examples of the use and manipulation of sdomain equations may be found
in the following articles.
Phono Equalization Calculations
Phono Termination
Calculations and
Calculator
Active Inductor Load
Warp Filter
Last Note
I may return to clarify any points that seem unclear at any point.
Document History
October 26, 2012 Created.
October 27, 2012 Added Q to legend of figure 6 and another link
under examples.