Download The Bott-Duffin synthesis answers a basic question in electrical

The Bott-Duffin
synthesis
J. Applied Physics,Vol. 210, No. 8, 1949
The Bott-Duffin synthesis answers a basic question
in electrical engineering:
The Bott-Duffin
synthesis
J. Applied Physics,Vol. 210, No. 8, 1949
The Bott-Duffin synthesis answers a basic question
in electrical engineering:
How to make an electrical circuit that does
what you want?
lly written, short papers. He was
st at all, and that impressed me
to understand it all the more. So I went to lectures,
most of them completely incomprehensible, and
my gut reaction was: I want to understand this. Ostensibly I was at the Institute to write a book on
network theory, but after I found out I didn’t have
to do that, I went to an incredible number of lectures and just absorbed the atmosphere. I didn’t
write a single paper in my first year there. So I was
very delighted when Marston Morse called me up
at the end of that year and said, “Do you want to
stay for another year?” And I said, “Of course,
yes!” He said, “Is your salary enough?” It was $300
a month. I said, “Certainly!” because I was so delighted to be able to stay another year. My wife took
a dimmer view! But we managed.
Notices: So this was a big change for you, to go
from an environment where you had been working on the engineering side to a place where there
was so much mathematics.
Bott: I didn’t think of it that way.
Notices: It wasn’t such a contrast for you?
Bott: No, because the actual work is just the
same. When I worked with Duffin, it was mathematical thought; only the concepts were different.
But the actual finding of something new seems to
me the same. And you see, the algebraic aspects
of network theory were an ideal introduction to differential geometry and the de Rham theory and to
what Hermann Weyl was studying at the time, that
is, harmonic theory. In effect, networks are a discrete version of harmonic theory. So when I came
to the Institute, the main seminar I attended was
Hermann Weyl’s, and Kodaira and de Rham were
lecturing on harmonic forms. Weyl wanted to have,
finally in 1949, a proof of Hodge’s theorem that
Bott on electrical
circuits and differential
geometry
u wrote in a Notices article that you
e his way of being a “mathemati-
at’s the point. It’s the problem you
than the field. You have to trust
and hope that sometimes you will
ect to which you can maybe make
you come into contact with John
gie Tech?
deed. He was in my class. In fact,
ere was Nash and also Hans Weingood applied mathematician now
and maybe two or three others.
ching us a very amusing course on
. One of Duffin’s principles was
re a lecture! So we were allowed to
nfused, and part of the fun was to
could fix things up. We were readnn’s book on quantum mechanics,
ed Hilbert spaces at the same time.
came clear that Nash was ahead of
derstanding the subtleties of infinal phenomena.
s he an undergraduate?
s an undergraduate, yes, and the
e graduate students. I was friends
didn’t have any close friends, reften talked about this and that.
got sick and had a really bad bout,
etimes send me a postcard with
nge associations, usually with re-
Text
A passive 1-port is a box containing
resistors, capacitors and inductors
(passive circuit elements),
with two wires sticking out (a port).
Such
a
circuit
has
an
impedance
Z
Such
a
circuit
has
an
impedance
Z(ω),
Such a circuit has an impedance Z(s),
defined
as
follows:
defined as follows:
defined as follows:
If
you
impose
a
current
(using
a
cu
If you
impose
a
current
(using
a
current
genIf you impose a current (using a current
erator)
erator)
generator)
jωt
st
I(t) = e st I(t) = e
I(t) = e
through
the port, voltage
the response
volt
through the port,
the response
is
through the port, the response
voltage
is
jωt
st
V (t) = Z(s)e stV. (t)Z(ω)e .
V (t) = Z(s)e .
V (t) = Z(s)e .
st
V (t) = Z(s)e .
Equivalently, if you impose the voltage
Equivalently,
st if you impose the voltage
V (t) = est the resulting current is
V (t) = e the resulting current
is
st
I(t) = Y (s)e
.
st
I(t) = Y (s)e .
The
The
1
1
function
Y
(s)
=
function Y (s) = Z(s)Z(s)
is
is called
called
the the
admittanceofofthethe
1-port.
admittance
1-port.
The physically
physically meaningful
values
of sof
ares are
The
meaningful
values
the purely
purely imaginary
s =sjω,
where
the
imaginaryvalues
values
= jω,
where
ωωisisa afrequency.
frequency.
According to Bott, the first step to
The
function
Z(s)
is
a
PRF
becoming an electrical engineeris to write
√
(positive real
i.e.
−1 =function),
j.
V (t) = Z(s)e .
st
V (t) = Z(s)e .
Equivalently,
if
you
impose
the
voltage
Equivalently,
if
you
impose
the
voltage
st
Equivalently,
if
you
impose
the
voltage
stthe
VV(t)
=
e
resulting
current
is
(t) = est the resulting
current
is
st
V (t) = e I(t)
the =
resulting
current
is
st
Y
(s)e
.
I(t) = Y (s)e .
st
I(t) = Y (s)e .
1
1 is called the
The
function
Y
(s)
=
1z(s)
is called
The
function
Y
(s)
=
The function Y (s) = Z(s)Z(s)
is called
the the
admittance
of
the
1-port.
admittanceofofthethe
1-port.
admittance
1-port.
The
physically
meaningful
values
of
s
are
The
physically
meaningful
values
of
s
are
The physically meaningful values of s are
the
purely
imaginary
values
s
=
jω,
where
the
purely
imaginary
values
s
=
jω,
where
the purely imaginary values s = jω, where
ωωωisisaaafrequency.
frequency.
frequency.
According
to Bott,Z(s)
the first
step
to
The
function
is
a
PRF
The
function
Z(s)
is
a
PRF
becoming an electrical engineeris to write
√
(positive
real
function),
i.e.
(positive real
i.e.
−1 =function),
j.
sW (s) − s0
s−W
(s)
has
a
pole
at
jω.
= Suppose that
. Z(jω)
Suppose
that
Z(jω)
=
Ljω
for
some
=
Ljω
for
some
L
>
0.
has
a
pole
at
jω.
s
−
W
(s)
c) Clearly
if
W
(jω)
=
jω,
then
Suppose
that Z(jω)
= Ljω
for then
some L > 0.
c) Clearly
if W (jω)
= jω,
SetL > 0.
Set
Suppose
that
Z(jω)
=
Ljω
for
some
pole
at jω. Set
f W (jω)has
= ajω,
then
Z(s)
has a pole at
jω.
Z(s)
Z
(s)
=
,
1
Set
Z
(s)
=
,
Z(s)
thatatZ(jω)
= Ljω1 for some
L
>
0.
L
asepole
jω.
L
Z1 (s)
Suppose that Z(jω)
==
LjωZ(s)
for, some L > 0.
L
and
Z
(s)
=
,
and
1 0.
Set L >
ω) = Ljω for some
L 2
2
Set
and
sZ
(s)
−
s
1
0
sZ
(s)
−
s
Z(s)
1
0
R(s)
=
.
2
and
R(s)
=
.
Z(s)
Set Z1 (s) =
, sZ1 (s) − s0
s
−
Z
(s)
1
s
−
Z
(s)
Z
(s)
=
,
L
1 − s.2
1 = sZ (s)
R(s)
1 Z (s) 0
Z(s)
L
s
−
1
.
(s) =
, and R(s) = s − Z
(s)
and
1
L A bit of computation
shows
that
2
sZ1of(s)computation
− s0
2
A
bit
shows
that
sZ
(s)
−
s
andR(s) =
1 .1
0 1
.
s=− Z=
(s)
Z(s)
+
.
1
2 R(s)
1
1
1
1
R(s)
sZ1 (s) −Z(s)
s0 =
s
s
−
Z
(s)
1
+
2 + Ls2 .
LR(s)
Ls1 + Ls
=
.
1
R(s)
s0
0
+
+
s − Z1 (s)
2
LR(s)
Ls
Ls0
Ls20
Ending the synthesis
One basic result of circuit theory is
st
V (t) = Z(s)e .
One basic result of circuit theory is
The function Z(s) is a PRF
(positive real function), i.e.
it is a rational function with real coefficients
mapping the right half plane
Res ≥ 0
to itself.
st
V (t) = Z(s)e .
One basic result of circuit theory is
The function Z(s) is a PRF
(positive real function), i.e.
it is a rational function with real coefficients
mapping the right half plane
Res ≥ 0
to itself.
The Bott-Duffin synthesis is the converse:
st
V (t) = Z(s)e .
One basic result of circuit theory is
The function Z(s) is a PRF
(positive real function), i.e.
it is a rational function with real coefficients
mapping the right half plane
Res ≥ 0
to itself.
The Bott-Duffin synthesis is the converse:
Every PRF is the impedance of a passive 1-port.
An electrical engineer
An electrical engineer
1) Figures out the function g(jω) that he wants
his circuit to have
An electrical engineer
1) Figures out the function g(jω) that he wants
his circuit to have
2) Finds a PRF f(s) such that f(jω) approximates g(jω)
An electrical engineer
1) Figures out the function g(jω) that he wants
his circuit to have
2) Finds a PRF f(s) such that f(jω) approximates g(jω)
In this phase, the engineer is going to have to compromise.
He will only be able to approximate the real part (or the
imaginary part, or the modulus) of g
An electrical engineer
1) Figures out the function g(jω) that he wants
his circuit to have
2) Finds a PRF f(s) such that f(jω) approximates g(jω)
In this phase, the engineer is going to have to compromise.
He will only be able to approximate the real part (or the
imaginary part, or the modulus) of g
There is a vast literature about this, from approximating
Cauchy integrals by Riemann sums to
Padé approximants.
An electrical engineer
1) Figures out the function g(jω) that he wants
his circuit to have
2) Finds a PRF f(s) such that f(jω) approximates g(jω)
In this phase, the engineer is going to have to compromise.
He will only be able to approximate the real part (or the
imaginary part, or the modulus) of g
There is a vast literature about this, from approximating
Cauchy integrals by Riemann sums to
Padé approximants.
3) Applies the Bott-Duffin synthesis to make the circuit.
I will assume my audience has forgotten
any circuit theory they ever knew
and start from scratch
Even from scratch, it is possible to give
a complete proof in half an hour:
the original paper is only one page.
Some basics
An electrical circuit is a graph, with each edge oriented,
and each edge carrying a circuit element:
Symbol
Equation
Units
v = Ri
Ohms
• A resistor Text
Farads
C v′ = i
Text
• a capacitor
Figure 1. Resistor, capacitor and induc
an inductor Text
Henrys
L i′ = v
•
Figure 1. Resistor, capacitor and inductor.
stor, capacitor and inductor.
to
itself.
Res
≥
0
to
itself.
totoitself.
itself.
to
itself.
To
each
edge
e
is
associated
aa current
iiee and
To
each
edge
e
is
associated
current
and
To
each
edge
e
is
associated
a
current
i
and
eie and
To
each
edge
e
is
associated
a
current
aa voltage
drop
v
(the
sign
of
both
depends
e
voltage
drop
v
(the
sign
of
both
depends
e
each
edge
e
is
associated
a
current
ie and
aTo
voltage
drop
v
(the
sign
of
both
depends
eve (the sign of both depends
a voltage on
drop
the
orientation
of
e).
on
the
orientation
of
e).
a voltage on
drop
v
(the
sign
of
both
the
orientation
of
e).
on thee orientation of e). depends
onsubject
the orientation
of e). current
These are
to Kirchhoff’s
Theseare
aresubject
subjecttotoKirckhoff’s
Kirchhoff’scurrent
currentlaw:
These
law:
These are subject tolaw:
Kirckhoff’s current law:
The sum of the currents at a node is zero
The
sum
of
the
currents
at
a
node
is
zero
The sum
sum of
of the
the currents
currents at
at aa node
node isis zero
zero
The
and to Kirckhoff’s voltage law:
and to
to Kirckhoff’s
Kirchhoff’s voltage
voltage law:
law:
and
and to Kirchhoff’s voltage law:
The sum of the voltage drops around any
The sum
sum of
of the
the loop
voltage
drops
around
any
The
voltage
drops
around
any
is zero
The sum of the voltage
drops around any
loop is
is zero
zero
loop
loop is zero
Let E be the set of edges, and let
loop is zero
Let E be the set of edges, and let I ⊂ R
be the space of currents allowed by the
E
Kirchhoff current law, and V ⊂ R be the
space of voltage drops allowed by the
Kichhoff voltage law.
E
Tellegen’s theorem. The spaces I and V
E
are orthogonal complements in R .
Tellegen’s theorem. The spaces I and V
E
are orthogonal complements in R .
Tellegen’s theorem. The spaces I and V
E
Proof.
They
are
respectively
the
kernel
of
are orthogonal complements
in
R
.
!
A and the image of A , where A is the
Proof. They are
respectively
the kernel of
incidence
matrix:
!
A and the image of A , where A is the
incidence
matrix:


if node i is the end of edge j
 1
ai,j = −1
if node i is the origin of edge j

 0
otherwise.
!
0
otherwise.
ai,jA=and−1
the image
of Ai is, where
A is of
theedge j
if node
the origin
incidence
matrix:

0
otherwise.


incidence
matrix:
Using Kirchhoff’s
laws
and
Ohm’s
law for each resistor,
0
otherwise.


1
if
node
i
is
the
end
of
edge
j

we
can
eliminate
many
of
the
v
and
i

Under
quite
general
circumstances,
e
e j
1
if
node
i
is
the
end
of
edge
 quite general circumstances,
Under
a
=
−1
if
node
i
is
the
origin
of
edge
j
i,j
der quite
general
circumstances,
ai,j =
−1
if
node
i
is
the
origin
of
edge
j


Under
quite
general
circumstances,
–no
loops
of
capacitors

0 loops
otherwise.
of
capacitors
–no
0
otherwise.
–no loops of capacitors
–no
loops
of
capacitors
–the
inductors
do
not
disconnect
the
circuit
–the Under
inductors
dogeneral
not disconnect
the circuit
quite
circumstances,
the
Under
quite general
circumstances,
the
uctorsremaining
do
not disconnect
the
circuit
variables
from
which
all
others
–the
inductors
do
not
disconnect
the
circuit
the
remaining
variables
from
which
all
remaining
variables
from
which
all
others
the can
remaining
variables
from
which
all
be
deduced
algebraically
are
the
others
can
be
deduced
algebraically
are
the
emaining
variables
from
which
all
can
be
deduced
algebraically
are
the
othersthe
canremaining
be deduced
algebraically
are
the
variables
from
which
all
an be deduced
are the
currentsalgebraically
in the inductors
(and current
currents
in
the
inductors
(and
current
others
can
be
deduced
algebraically
are
the
currents
in
the
inductors
(and
current
currents in generators),
the inductorsand
(and
thecurrent
generators),
and
the
generators),
and
the
nts in the inductors
(and
current
generators),
and
the
currents
in
the
inductors
(and
current
voltages
through
the
capacitors
(and
generators),
and
the
voltages
through
the
capacitors
(and
generators),
and
the
voltages
through
the
capacitors
(and
voltagethe
generators)
voltages
through
capacitors
(and
voltage
generators)
voltage
generators)
ages through the
capacitors
(and
voltage
generators)
A
circuit
of
this
sort
is
called
normal
voltages
through
the
capacitors
(and
Tellegen’s
theorem
gives
voltage generators)
Tellegen’s
theorem
gives
%
%
%
generators), and the
voltages
through
the
capacitors
(and
voltages
through
the
capacitors
(and
voltagesthrough
throughthe
thecapacitors
capacitors(and
(and
voltages
generators)
Supposevoltage
that
a
1-port
is normal, and
voltage
generators)
voltagegenerators)
generators)
voltage
form
a
vector
x(t)
whose
entries
are
these
Form
a
vector
x(t)
whose
entries
are
these
Form
a
vector
x(t)
whose
entries
are
these
Form a vector x(t) whose entries are these
variables.
variables.
variables.
variables.
The
time
evolution
ofof
the
circuit
isis
The
time
evolution
ofof
the
circuit
isis
The
time
evolution
the
circuit
The
time
evolution
the
circuit
described
by
an
inhomogeneous
linear
described
by
an
inhomogeneous
linear
described
by
an
inhomogeneous
linear
described
by
an
inhomogeneous
linear
differential
equation
ofof
the
form
differential
equation
ofof
the
form
differential
equation
the
form
differential
equation
the
form
" "
"
x
" x ==Ax
Ax++f (t)
f (t)
Ax++f f(t)
(t)
xx ==Ax
where
AAis
isisaaasquare
square
matrix,
and
(t)
isisthe
the
whereA
squarematrix,
matrix,and
andff (t)
f (t)is
the
where
where
A
is
a
square
matrix,
and
f
(t)
is
the
imposed
current,
expressed
in
the
variables
imposedcurrent,
current,expressed
expressedininthe
thevariables
variables
imposed
imposed current, expressed
in
the
variables
of
x.
ofofx.x.
of x.
Tellegen’s
theorem
gives
Tellegen’stheorem
theoremgives
gives
Tellegen’s
where A is a square matrix, and f (t) is the
"
xexpressed
=
Ax + in
f (t)
imposed
current,
thethe
variables
imposed
current,
expressed
in
variables
The eigenvalues of the
matrix
A are nonnegative
of of
x.
x.
where A is a square matrix, and f (t) is the
imposed current,
expressed
in Hermitian
theHermitian
variables
Differentiate
thethe
positive
definite
Differentiate
positive
definite
of x.
form
form
inductors
capacitors
Differentiate%
the
positive
definite
Hermitian
%
%
%
1 1
2 2 1 1
2 2
E(i,
v)v)
==
LλL
|iλform
| λ+
CγC
|vγγ|v
| γ. | .
E(i,
|i
| +
2 2λ
2 2γ
γ
λ
%
%
1
1
2
2
E(i, v) =
L
|i
|
+
C
|v
|
.
λ
λ
γ
γ
kinetic
energy
potential
energy
2
2
λ
γ
The letters λ, γ and ρ are traditional for
inductors, capacitors and resistors
respectively.
We
find
We find
""
dd !!
EE i(t),
v(t)
v(t)
dtdt i(t),
##
%%
$
11dd $
$
$
2
2
==
LLλ |i|iλ (t)|
C
|v
(t)|
2+
2
γ
γ
Cγ |vγ (t)|
λ λ (t)| +
22dtdt λ
γ
γ
λ
$
$
$
$
!
!
== LLλ Re(i
i
C
Re(v
v
!) +
!)
λ
γ
γ
λ
γ
Cγ Re(v γ vγ )
λ Re(iλ iλ ) +
γ
λ
γ
λ
$
$
$
$
== Re(i
v
)
+
C
Re(v
i
)
λ
λ
γ
γ
γ
Re(iλ vλ ) +
Cγ Re(v γ iγ )
γ
λ
γ $
λ
$
$
$
2
==−− Re(i
v
)
=
−
R
|i
|
2≤ 0.
ρ
ρ
ρ
ρ
Re(iρ vρ ) = −
Rρ |iρ | ≤ 0.
ρ
ρ
ρ
ρ
imposed
current,
expressed
in
the
variables
imposed current, expressed in the variables
x. capacitors (and
ofofx.
voltages through
the
This certainly proves
that
the eigenvalues of A are
voltage
generators)
Differentiate
the
positive
definite
Hermitian
Differentiate
the
positive
definite
Hermitian
nonpositive, and further that the purely imaginary
form
form
Form a vector
x(t)
whose
entries are these
eigenvalues
are simple.
%
%variables.
%
1 1%
1
1
2 2
2 2
E(i,
v)v)==
LλL|iλλ|i|λ |++
CγC|vγ |v
E(i,
γ |γ.| .
2 2 λ if a 1-port is2driven
it also implies that
by
a
current
2
The time evolution
of theγ circuit
is
γ
λ st
i(t) = e with Re s>0,
described by an inhomogeneous linear
The letters
λ,
γ
and
ρ
are
traditional
for
differential
of traditional
theequations
form for
The
letters
λ, γequation
and
ρ are
then
the system
of differential
inductors, capacitors and resistors
inductors,xcapacitors
and
resistors
"
= Ax + f (t)
respectively.
respectively.
st coefficients, setting
canwhere
be solved
by
undetermined
A is a square
xi (t) =matrix,
ai e st and f (t) is the
x
(t)
=
a
e
i
i
imposed current, expressed in the variables
and solving for theofai x.
after cancelling a
st
common factor e .
Tellegen’s theorem gives
Proof. They are respectively the kernel of
differential
equation
! of the form
A and the image of A , where A is the
Proving the first half of the fundamental theorem
incidence
matrix:
"
x circuit
= Ax +
f (t)

of
theory

if node i is the end of edge j
 1
where A is a square matrix, and f (t) is the
ai,j = −1
if
node
i
is
the
origin
of
edge
j
imposed
current,
expressed
in
the
variables

 0
otherwise.
of x.
Z(s) is a PRF
Tellegen’stheorem
theorem gives
Telegen’s
gives
%
%
%
%
%
%
−iG
iλiλvvλλ +
−ivGGv=
+
iiγγvvγγ++ iρ viρ.vρ .
G =
λλ
generator inductors
γ
capacitors
ρ ρ
resistors
Using
undetermined
coefficients,
we
can
find
Using undetermined coefficients, we can find
thethe
solution
of
the
differential
equation
solution of the differential equation
st st
λ
γ
ρ
Using undetermined coefficients, we can find
the solution
of the differential
Using
undetermined
coefficients,equation
we can find
st
describing
theofcircuit
with iG =equation
e as
the solution
the differential
st
describing the circuit with iG = e as
st
vG = −Z(s)e
st
st
st
v
=
−Z(s)e
iλG= Iλ e
so that vλ = Lλ Iλ se
st
st
st
st
that iγv=
I
se
λ=
λ
λ
vγiλ==VIγλee
sosothat
CL
V
se
γ γ
st
st
st
st
that vρiγ==RCρ IγρVγ ese
ivρ γ==IρVeγ e
sosothat
st
iρ = Iρ e
st
so that vρ = Rρ Iρ e
stst
v
=
−Z(s)e
st
G
vG =iλ−Z(s)e
= Iλ e
soiλthat
stvλ = Lλ Iλ se
= Iλ e
so
st
st that vλ = L
stλ e
stse
i
=
I
so
that
v
=
L
I
λ
λ
λ
λ
st
st
iλ =vγIλ=
e Vγ eTelegen’s
so that
v
=
L
I
se
theorem
gives
Insert
into
st
λ
λ
λ
so
that
i
=
C
V
se
γ
γ
γ
v
=
V
e
so
that
i
=
C
γ
γ
γ
st
st
%
%
%
st
st
v
=
V
e
so
that
i
=
C
V
se
γC V γseγ
vγ =γ−i
Vγ e vγ =
i
=
st so that
st .
st v
γ
γi
γ +
i
v
+
i
v
= Iρ e γvργ= Rso
iρ =G IGρste
λ so
λiρthat
ρe ρ vρ = R
ρ
stIρthat
stρ e
st e
i
=
I
so
that
v
=
R
I
ρ
ρ
ρ
ρ
i =I e
so that v = R I e
ρ
ρ γ
λ
ρ
ρ
ρ ρ
st 2
2
st
Cancel
the
common
factor
|e
|
2
st
Cancelthethe
common
factor
|e
|
to
get
2
Cancel
common
factor
|
to
get
st|e
%
% find
Cancel
common
factor
|e
|
to
get
Usingthe
undetermined
coefficients,
we
can
%
%
2 %
2
%
%
%
Z(s)
=s %
%
2
2| +
2%
2Lλ |Iλ2| +s 2 Cγ |Vγ
the
solution
of
the
differential
equation
Z(s)
=
s
L
|I
|
+s
|V
|
+
|I
|
.
2λ
2C
2 ρ| R
Z(s)
=
s
L
|I
|
+s
C
|V
|
+
R
|I
.
λ
γ
γ
ρ
ρ
λ
λ
γ
γ
ρ
Z(s) = s
L |I | +s
C |V | +
R |I | . γ
λ
λ
γ
γ
λ
ρ
ρ
st
eρ
describing
theγcircuitγ withρ i =
as
λ
Take real parts:
Take
real
parts:
&
Take
real
parts:
Take
real parts:
&
'
%
%
& %
'
&
2'
%
%
= Res% 2 L%
% %
λ |Iλ | + %
2%
2C
st ReZ(s)
2 λ| + 2
2 γ|
2 ρ| .
ReZ(s)
=
Res
L
|I
C
|V
+
R
|I
v
=
−Z(s)e
λ
γ
ρ
2
G
ReZ(s)
= Res
+γ |Vγ R
γ |Vγ | λC
ρ | γ.
ReZ(s)
= Res Lλ λ |Iλ | L+λ |Iλ | Cγ+
|ρ ρ |I+
st
st
γ
ρ
λ
the
iλ = Iλ e
that vλγon
=L
λ soEverything
λ Iright
λ se is ≥ρ 0
Everything
on
the
right
is
≥
0
when
Everything onstthe right is ≥ 0 when Res ≥ 0. st
on
the
right
is
≥
0
when
vEverything
so
that
i
=
C
V
se
≥
0.
γ = Vγ e ResRes
γ
γ
γ
≥ 0.
λ λ
γ
ρ
G
Proving the second half of the fundamental theorem
of circuit theory
Every PRF is the
impedance of some 1port
Bott and Duffin give a straightforward construction
of an appropriate circuit.
st 22
common
Take
realfactor
parts: |e st| to
Cancel
Cancel the
the #
common factor |e | to get
get%
$
$
$ $2 $
$ $2 $
2+
2+s
Z(s)
=
s
L
|I
|
C
|V
|
R
γ
γ
ρ |I
λ
λ
2
2
Z(s)
=
s
L
|I
|
+s
C
|V
|
+
R
|
γ
γ
ρ
λ
λ
ReZ(s)
=
Res
|I
|
+
C
|V
|
+
ure 2. All the three
in a cycle,
and
the ladderL10.3.1
λ
λ
γ
γ
ρ
γ
λ
acitor and inductor.
Series and Parallel
λ
Te
A
B
t
γ
λ
γ
ρ
Take
real
parts:
real
parts:
A Take
If
two
1-ports
are
connected
in
series
Everything#
on
the
right
is
≥
0
when
%
#$
%
$
$
$
their
impedances
Res ≥ 0.22 add
2
B
2 +
ReZ(s)
=
Res
L
|I
|
+
C
|V
|
λ
λ
γ
γ
ReZ(s) = Res
Lλ |Iλ | +
Cγ |Vγ |
Z(s) =λλZA (s) + ZB (s)
γ
γ
Figure 3. Ports in series and in parallel
ycle, and the ladder 10.3.1
Everything
right
≥
Y (s)on
= the
Ythe
+ Yis
A (s)
B
Everything
on
right
is(s)
≥00 when
when
Res
≥
0.
1
1
1
Res
≥
0.
If two 1-ports
in parallel
= are connected
+
Z(s)
Z
(s)
Z
(s)
A
B
Z(s)
=
+
Z
(s)
A
B
their=admittances
Z(s)
Z (s) + Z add
(s)
Te
A
A
B
YY(s)
Bt
(s)==YYAA(s)
(s)++YYBB(s)
(s)
11
11
11
Figure 4. Scheme 10.4.2
ries and in parallel
C1
==
++
Z(s)
ZZAA(s)
ZZBB(s)
Z(s)
(s)
(s)
C3
Conditions
on
PRF’s
(1)
a)
inf
ReZ(s)
=
R
>
0
(1) a) inf
ReZ(s) = R > 0
Res>0
Res>0
(1)
a)
inf
ReZ(s)
=
R
>
0
(1)
a)
inf
ReZ(s)
=
R
>
0
Res>0
Res>0
(1) a)
a) inf
infRes>0
ReZ(s) =
=R
R>
> 00
Res>0 ReZ(s)
(1)
b) inf
infRes>0
ReY (s)
(s) =
= 1/R
1/R >
> 00
Res>0 ReY
b)
b)inf
infRes>0
ReY(s)
(s)=
=1/R
1/R>
>000
b)
Res>0ReY
b)
inf
ReY
(s)
=
1/R
>
Res>0 ReY (s) = 1/R > 0
b) inf Res>0
(2) a)
a) The
The function
function Z(s)
Z(s) has
has aa pole
pole on
on
(2)
(2)
a)
The
function
Z(s)
has
a
pole
on
(2)
a)
The
function
Z(s)
has
a
pole
on
(2) a)
a)
The
functionaxis
Z(s)
has∞aa pole
pole on
on
the
imaginary
or
at
(2)
The
function
Z(s)
has
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the imaginary
imaginary axis
axis or
or at
at ∞
∞
the
b)
The
function
Z(s)
has
a
pole
on
b)
The
function
Z(s)
has
a
pole
on
b)
The
function
Z(s)
has
a
pole
on
b)
The
function
Z(s)
has
a
pole
on
b)
The
function
Z(s)
has∞aa pole
pole on
on
the
imaginary
axis
or
at
b)
The
function
Z(s)
has
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the imaginary axis or at ∞
(3) a)
a) There
exists ω Z(s)
> 0 and
La >pole
0 with
(3)
The
function
has
on
(3)
a)
There
exists
ω
>
0
and
L
>
0
with
(3)
a)
The
function
Z(s)
has
a
pole
on
(3)
a)
The
function
Z(s)
has
a
pole
on
Z(jω)
=
Ljω
(3) a)
The
function
Z(s)
has
a
pole
on
the
imaginary
axis or
or at
at ∞
∞
Z(jω)
=
Ljω
the
imaginary
axis
the imaginary
imaginary axis
axis or
or at
at ∞
∞
the
b)
There
exists
ω
>
0
and
C
>
0
with
b)
The
function
Z(s)
has
a
pole
on
b)
There
exists
ω
>
0
and
C
>
0
with
b)
The
function
Z(s)
has
a
pole
on
b)
The
function
Z(s)
has
a
pole
on
Y
(jω)
=
Cjω
b)
The
function
Z(s)
has
a
pole
on
the
imaginary
axis
or
at
∞
Y
(jω)
=
Cjω
the
imaginary
axis
or
at
∞
the
imaginary
axis
or
at
∞
the imaginary axis or at ∞
(3)
a)
There
exists
ω
>
0
and
L
>
0
with
(3) a) There exists ω > 0 and L > 0 with
Z(jω)
=
Ljω
Z(jω) = Ljω
Reduction
in
case
1
b)
There
exists
ω
>
0
and
C
>
0
with
b) There exists ω > 0 and C > 0 with
Y
(jω)
=
Cjω
Y (jω)
= Cjω
If a PRF
satisfies
either of conditions 1a or 1b
we
can
write
Z(s)
=
R
+
Z
(s)
or
1
we can write Z(s) = R + Z (s) or
1
Y (s)
1
Y
(s)
=
+
Y
(s)
1
= R + YR1 (s)
and if we can find a circuit with impedance Z1 or
admittance Y1
then putting it either in series (or in parallel)
with a resistance R will realize Z (or Y).
The PRF Z1 (or Y1) belongs to case 2 or 3.
Z(jω) = LjωY (jω) = Cjω b) There exists ω > 0 and C >
1 ω >
There
b)
exists
There
exists
0
and
ω
C
>
>
0
and
0
with
C
>
0
with
exists ω > 0 b)
and
C
>
0
with
+
Y
(s)
Y
(s)
=
1(jω) = Cjω
Y
R
we
can
write
Z(s)
=
R
+
Z
(s)
or
b)
There
exists
ω
>
0
and
C
>
0
with
1
Y (jω) = YCjω
(jω) = Cjω
Cjω
Y (jω) = Cjω
1 can
If W is a YPRF,
we
can
write
we
write
Z(s)
=
R
+
Z
(s)
1
(s)
=
+
Y
(s)
1 Z=
R
we
can
write
we
can
Z(s)
write
=
R
Z(s)
+
Ror
+ Z1 (s) or
te Z(s) = R + This
Z1 (s)case
or uses partial fractions:
1 (s)
1
we can write Z(s)
=
R
+
Z
(s)
or
1
Y
(s)
=
+
Y
(s)
1
W
(s)
=
P
(s)
+
W
(s)
1
1
If
W
is
a
PRF,
we
can
write
W any poles
1
R simple, with
Since
PRF’s,
must
be
1 Z and Y are
Y
(s)
=
Y
(s)
=
+
Y
(s)
+
Y
(s)
s) = R + Y1 (s) 1
1
1
R
R
positive
residues.
Y (s) = R +
Y
(s)
1
If W+ is
PRF, we can write
W (s) with
= PW (s)
W1a(s)
If write
W is aIfPRF,
W iswe
a PRF,
can write
we can write
a PRF, we can
If W is a PRF, we can!
write
m
with W (s) = PW (s) + W1 (s)
k0 = PW
2k
s
W (s)
(s)
=
+
P
W
(s)
+
W
(s)
i
= PW (s) + W1 (s)
W(s)
W
1
1
m
P
(s)
=
+
+
k
s
!
W PW (s) + W
∞
W (s) =
(s)
with
k
2k
s
2
1
2
0
i
s
s
+
ω
i
with
with
P
(s)
=
+
+
k
s
with
W
∞
i=0
m
2 + ω2
!
s
s
k
2k
s
with
m
m
i
0
i
m
i=0
!
!
!
k
k
2k
s
2ki s 2
P
(s)
=
+
+ k∞
k0
2ki s !
0
W0 i
m
2
P
(s)
=
P
(s)
+
=
+
+
k
s
+
k
s
s
s
+
ω
+
+
k
s
W
W
∞
∞
k
2k
s
i
∞
0
i
2
2
and
W
is
a
PRF
which
is
bounded
on
the
2
2
with
k
and
k
≥
0
20 ≥ 0, sall ki ≥ 0
i=0
1
2
∞s + ωi
s
s
+
ω
sPW (s)
s
+
ω
=
+
+
k
s
i
i
i=0 ∞
i=0
2
2
i=0
right
half-plane.
Proposition.
s
s
+
ω
i W1 is a PRF which is bounded
and
i=0
and
aand
is
bounded
on
the
andW
aPRF
PRF
Won
is
which
a
PRF
is
bounded
which
is
bounded
on
the
on
the
1bounded
RF which
isW
the
1is is
1which
k0
right
half-plane.
Proposition
–The
PRF
is
the
impedance
of
a
W
is
a
PRF
which
is
bounded
on
the
1
right
half-plane.
Proposition.
s half-plane.
right half-plane.
right
Proposition.
Proposition.
plane. Proposition.
1
k
0
right half-plane.capacitor
Proposition.
of
capacitance
. impedance
–The
PRF
isk the
Reduction in case 2
Cjω
YY(jω)
Y(jω)
(jω)
===
Cjω
Cjω
Thus
ifcan
we
can Z(s)
find
circuits
that
realize
we
can
write
Z(s)
we
we
can
write
write
Z(s)
===
RRR
+++
ZZ
(s)
(s)
oror
or the
11(s)
1Z
summands of PW , and
if
we
can
realize
W
,
we
1
1
1 1 + Y (s)
Y
(s)
=
Y
(s)
Y
(s)
=
=
+
+
Y
Y
(s)
(s)
1
1
1
R
R impedance or as an
can realize W (asR an
admittance).
Proposition.
Proposition.
Proposition.
kk00
k
0
–The
PRF
the
impedance
–The
–The
PRF
PRFs ssisisis
the
the
impedance
impedance
ofof
of
a aa
11
1
capacitor
capacitance
capacitor
capacitor
ofof
of
capacitance
capacitancek0k.k00..
–The
PRF
kk∞
the
impedance
–The
–The
PRF
PRF
k∞
s∞sis
s isis
the
the
impedance
impedance
ofof
of
anan
an
inductor
inductance
kk∞
inductor
inductor
ofof
of
inductance
inductance
k∞
.∞..
2ks
2ks
2ks is the impedance of a
–The
PRF
–The
–The
PRF
PRFs2s+ω
is
the
the
impedance
impedance
of
of
a
a
22+ω
2is
2
2
s +ω
11
1
capacitor
capacitance
parallel
capacitor
capacitor
oror
or
capacitance
capacitance
CCC===2k2k2kinin
in
parallel
parallel
2k
2k
2k
with
inductor
inductance
with
with
anan
an
inductor
inductor
ofof
of
inductance
inductance
LLL
===ω2ωω.22..
–The
is
2s+ω
2 +ω
2 the impedance of a
22
–ThePRF
PRFs2ss+ω
2 +ω
2 is the impedance of a
111
1
capacitor
capacitororor
orcapacitance
capacitance
capacitanceCC
inparallel
parallel
parallel
capacitor
inin
capacitor
or
capacitance
CC==
==2k2k
in
parallel
2k
2k
2k
2k
2k
2k
with
withan
an
aninductor
inductor
inductor
of
of
inductance
inductance
L
L
=
=
.
.
with
of
inductance
L
=
.
2
2
2
with
an
inductor
of
inductance
L
=
.
This case requires Richard’s
2 theorem
ωω
ωω
Richard’s
Richard’stheorem.
theorem.
theorem. Let
Let
LetW
W
W
(s)
(s)be
be
beaaaaPRF
PRF
PRF
Richard’s
(s)
Richard’s
theorem.
Let
W
(s)
be
PRF
with
withno
no
nozeros
zeros
zerosoror
orpoles
poles
poleson
on
onthe
the
theimaginary
imaginary
imaginaryaxis
axis
axis
with
with
no
zeros
or
poles
on
the
imaginary
axis
oratat
atinfinity.
infinity.
infinity.
oror
or
at
infinity.
a)There
There
Therethen
then
thenexists
exists
existsaaaaunique
unique
uniquesss0s00∈
with
with
a)a)
RR
with
0∈
+
a)
There
then
exists
unique
∈∈
R+R+
with
+
W
W
)=
W
(s(s
sss0s.00.0. .
00)
0)=
0)
W
(s(s
==
Case 3
b)
b)The
The
Thefunction
function
function
b)
b)
The
function
222
2
sW
sW
(s)
(s)
−
sW
(s)
−−
sss0s000
sW
(s)
−
W
W
(s)
W
==
11(s)
1(s)
1 (s)
W
== ss−−W
W
(s)
(s)
ss−−WW
(s)
(s)
PRF
PRFwith
with
withrankW
rankW
rankW
(s)
rankW
rankW
(s).
(s).
isis
≤≤
rankW
(s).
11(s)
1(s)
1 (s)
isisaaaaPRF
PRF
with
rankW
≤≤
rankW
(s).
W
(jω)
(jω),
jω,
then
then
W
has
aa apole
a pole
at
at
jω.
c)c)
Ifc)
WIf
(jω)
==
jω,
then
pole
jω.
11 W
1 has
1 has
c)If
IfW
W
(jω),
then
WW
has
pole
atatjω.
jω.
is a PRF with rankW1 (s) ≤ rankW (s).
Richard’s
theorem
Schwarz’s
lemma
c) If W (jω)
= jω,follows
then Wfrom
has
a
pole
at
jω.
1
a) The existence of s0 follows from the
intermediate value theorem. The uniqueness
follows from Schwarz’s lemma: a map from
the right halfplane to itself that has more
than one fixed point is the identity.
The map
φ(s) =
b) Theb)
map
b)
The
map
b)
b)
The
b)
The
The
map
map
map
s
−
s
b)
The
map
0
b) The map φ(s)s=− s0s − s0
s
−
s
φ(s)
=
maps
the
right
half-p
0
φ(s)
=
s
−
s
−
s
s
s
−
s
0
b) The map
0
0
0
s
−
s
s
+
s
0
0
φ(s)
=
− s0 map
s
+
s
b)s The
s
+
s
0
φ(s)
φ(s)
φ(s)
=
=
0 s=
φ(s)
=
and
φ(s
)
+
s
φ(s)
=
0
0
s
+
s
+
s
s
s
+
s
000 0unit disc,
s
−
s
0
maps
the
right
half-plane
to
the
s
+
s
0
sthe
+ shalf-plane
right
half-plane
tounit
thedisc,
unit disc,
s
−
s
0
maps=maps
the
right
to
the
0
φ(s)
maps
the
right
half-plane
to
the
unit
disc,
φ(s)
=
f
=
φ
◦
W
and
φ(s
)
=
0.
So
maps
maps
maps
the
the
right
the
right
right
half-plane
half-plane
half-plane
to
to
the
to
the
unit
the
unit
unit
disc,
disc,
disc,
s
+
s
0
maps
the
right
half-plane
to
the
unit
disc,
0
and
φ(s
)
=
0.
So
and
)
=
0.
So
right half-plane
to
the
unit
disc,
0
0
sφ(s
+
s
0 φ(s0 ) = 0. So
and
and
and
φ(s
and
φ(s
)
=
)
=
0.
)
0.
=
So
0.
So
So
−1
00φ(s
0
0
0
and
φ(s
)
=
0.
So
t the
half-plane
to
the
unit
disc,
−1 maps the unit disk to
=
φ
◦
W
◦
φ
andright
φ(s0half-plane
) =f 0.=fSo
−1
=
φthe
◦φW
◦ φ−1
unit
disc,
φff◦ to
W
◦
−1
−1
−1 −1
=
φ
◦
W
◦
φ
−1
nd φ(s0 )and
= 0.φ(s
So) −1
f
f
=
=
φ
f
◦
φ
=
W
◦
φ
W
◦
φ
W
◦
φ
◦
φ
f
=
φ
◦
W
◦
φ
=
0.
So
maps
the
unit
disk
to
itself
with
f
(0)
=
0.
f
=
φ
◦
W
◦
φ
0
By
Schwarz’s
lemma,
s
maps
the
unit
disk
to
itself
with
f
(0)
=
0.
maps maps
the unit
disk
to
itself
with
f
(0)
=
0.
−1
the
unit
disk
to
itself
with
f
(0)
=
0.
= φ maps
◦maps
W
◦
φ
−1
maps
maps
the
the
unit
the
unit
unit
disk
disk
disk
to
to
itself
to
itself
itself
with
with
with
f
(0)
f
(0)
f
=
(0)
=
0.
0.
=
0.
Th
the
unit
disk
to
itself
with
f
(0)
=
0.
f
=
φ
◦
W
◦
φ
unitBydisk
to
itself
with
f
(0)
=
0.
Schwarz’s
lemma,
so
does
g(z)
=
f
(z)/z.
By Schwarz’s
lemma,
so g(z)
does=g(z)
= f (z)/z.
By
Schwarz’s
lemma,
so
does
f
(z)/z.
−
disk
to
itself
with
f
(0)
=
0.
By
Schwarz’s
lemma,
so
does
g(z)
=
f
(z)/z.
W
=
φ
Thus
By
By
Schwarz’s
By
Schwarz’s
Schwarz’s
lemma,
lemma,
lemma,
so
so
does
does
so
does
g(z)
g(z)
g(z)
=
=
f
(z)/z.
f
=
(z)/z.
f
(z)/z.
1
the
unit
disk
to
itself
with
f
(0)
=
0.
By
Schwarz’s
lemma,
so
does
g(z)
=
f
(z)/z.
Thus=Thus
’s lemma, so does g(z)
f
(z)/z.
Thus
−1
Thus
Thus
Thus
Thus
mma,
so
does
g(z)
=
f
(z)/z.
−1
is
a
PRF.
If
you
work
W
=
φ
◦
g
◦
φ
−1
Thus
1
W
φ
◦
g
◦
φ
hwarz’s lemma,
g(z)
=
f
(z)/z.
W1so=does
φ
◦
g
◦
φ
1 =
−1
−1
−1
−1◦−1
=
φ
g
◦
φ.
−1
Thus −1 ThusWW
1
W
=
W
=
φ
φ
=
◦
φ
g
◦
◦
g
◦
φ.
◦
g
φ.
◦
φ.
1
1
1
1
W
=
φ
◦
g
φ.
sW
1
is
a
PRF.
If
you
work
it
out,
you
will
find
W
=
φ
◦
g
◦
φ
1 is a PRF.
Ifwork
you it
work
ityou
out,will
youfind
will find
is a PRF.
IfIf you
out,
W
(s)
=
−1
1
you
work
it
out,
you
will
find
W1 = φ W◦ gIf=If
◦Ifyou
φ.
2
−1
you
If
work
you
work
work
it
it
out,
out,
it
out,
you
you
will
you
will
find
will
find
find
s
you
work
it
out,
you
will
find
2
sW
(s)
−
s
◦ gyou
◦
φ.
2 −0 s
If you work
will
find
1 itφout,
sW
(s)
sW
(s)
−
s
02
W
(s)
=
.
0
2
2
2
2
1
sW
(s)
−
s
2
W
(s)
=
.
W
=
.
ork
it
out,
you
will
0
1find
sW
sW
(s)
sW
(s)
−
(s)
−
s
s
−
s
2
1 (s)
0
sW
(s)
−
s
s
−
W
(s)
0
0
0
W
(s)
=
.
0
sWit(s)
−
s
If you work
out,
you
will
find
s
−
W
(s)
1
s
−
W
(s)
W
W
(s)
W
(s)
=
(s)
=
=
.
.
.
1110(s)
1
1
W
=
.
2
(s)
W1 (s) =
.
ss−
s−−
W
sW
W
−
(s)
(s)
W (s)
2
is
a
PRF.
If
you
work
it
out,
you
will
find
sW
(s)
−
s
hwarz’s lemma, so does g(z) = f (z)/z.
0
W1 (s) =
.
2
Thus
s − W (s) sW (s) − s0
W1 (s) =
.
−1
s − W (s)
W1 = φ ◦ g ◦ φ
c) Clearly if W (jω) = jω, then
PRF. If you work it out,
will iffind
c) you
Clearly
W (jω) = jω, then
has a pole
at jω.
2
sW (s) − s0
W1 (s) =
. has a pole at jω.
s − W (s)
c) Clearly if W (jω) = jω, then
has completes
a pole at jω.
This
the proof of Richard’s theorem.
sW (s) − s0
s−W
(s)
has
a
pole
at
jω.
= Suppose that
. Z(jω)
Suppose
that
Z(jω)
=
Ljω
for
some
=
Ljω
for
some
L
>
0.
has
a
pole
at
jω.
s
−
W
(s)
c) Clearly
if
W
(jω)
=
jω,
then
Suppose
that Z(jω)
= Ljω
for then
some L > 0.
c) Clearly
if W (jω)
= jω,
SetL > 0.
Set
Suppose
that
Z(jω)
=
Ljω
for
some
pole
at jω. Set
f W (jω)has
= ajω,
then
Z(s)
has a pole at
jω.
Z(s)
Z
(s)
=
,
1
Set
Z
(s)
=
,
Z(s)
thatatZ(jω)
= Ljω1 for some
L
>
0.
L
asepole
jω.
L
Z1 (s)
Suppose that Z(jω)
==
LjωZ(s)
for, some L > 0.
L
and
Z
(s)
=
,
and
1 0.
Set L >
ω) = Ljω for some
L 2
2
Set
and
sZ
(s)
−
s
1
0
sZ
(s)
−
s
Z(s)
1
0
R(s)
=
.
2
and
R(s)
=
.
Z(s)
Set Z1 (s) =
, sZ1 (s) − s0
s
−
Z
(s)
1
s
−
Z
(s)
Z
(s)
=
,
L
1 − s.2
1 = sZ (s)
R(s)
1 Z (s) 0
Z(s)
L
s
−
1
.
(s) =
, and R(s) = s − Z
(s)
and
1
L A bit of computation
shows
that
2
sZ1of(s)computation
− s0
2
A
bit
shows
that
sZ
(s)
−
s
andR(s) =
1 .1
0 1
.
s=− Z=
(s)
Z(s)
+
.
1
2 R(s)
1
1
1
1
R(s)
sZ1 (s) −Z(s)
s0 =
s
s
−
Z
(s)
1
+
2 + Ls2 .
LR(s)
Ls1 + Ls
=
.
1
R(s)
s0
0
+
+
s − Z1 (s)
2
LR(s)
Ls
Ls0
Ls20
Ending the synthesis
A
B
B
Suppose
that
the
circuits
C
,
C
,
C
,
C
have
1
2
3
4
Suppose that the circuits C1 , C2 , C3 , C4 have
Figure 3.admittances
Ports in series and in parallel
admittances
11
1 1 R(s)
ss
R(s)
, , , , 2 , , 2. .
LR(s)
Ls
Ls
2 Ls0 2
0
LR(s) Ls Ls0 Ls0
Then the circuit below has impedance Z(s).
Figure 4. Scheme 10.4.2
C1
C3
C2
C4
Figure 5. Scheme 10.5.2
i0