the bott-duffin synthesis answers a basic question in electrical

The Bott-Duffin synthesis

The Bott-Duffin synthesis answers a basic questionin electrical engineering:

J. Applied Physics, Vol. 210, No. 8, 1949

The Bott-Duffin synthesis

The Bott-Duffin synthesis answers a basic questionin electrical engineering:

J. Applied Physics, Vol. 210, No. 8, 1949

How to make an electrical circuit that doeswhat you want?

Bott on electrical circuits and differential

geometryText

around. But Duf-fin was definitelya master at it.And of course, asI said, he workedin many differentareas. For in-stance, there issomething calledthe Duffin basisin physics forspinors. He hadalso worked ex-tensively in com-plex variable the-ory. He was anartist in a way. He

wrote beautifully written, short papers. He wasnot a specialist at all, and that impressed mefrom the start.

Notices: You wrote in a Notices article that youtried to emulate his way of being a “mathemati-cal samurai”.

Bott: Yes, that’s the point. It’s the problem yougo after rather than the field. You have to trustyour instincts and hope that sometimes you willhit upon a subject to which you can maybe makea contribution.

Notices: Did you come into contact with JohnNash at Carnegie Tech?

Bott: Yes, indeed. He was in my class. In fact,in this class there was Nash and also Hans Wein-berger, a very good applied mathematician nowat Minnesota, and maybe two or three others.Duffin was teaching us a very amusing course onHilbert spaces. One of Duffin’s principles wasnever to prepare a lecture! So we were allowed tosee him get confused, and part of the fun was tosee whether we could fix things up. We were read-ing von Neumann’s book on quantum mechanics,which developed Hilbert spaces at the same time.And it soon became clear that Nash was ahead ofall of us in understanding the subtleties of infi-nite-dimensional phenomena.

Notices: Was he an undergraduate?Bott: He was an undergraduate, yes, and the

rest of us were graduate students. I was friendswith Nash; he didn’t have any close friends, re-ally, but we often talked about this and that.When he later got sick and had a really bad bout,he would sometimes send me a postcard withsome very strange associations, usually with re-ligious overtones. My closest contact with Johnwas at Carnegie Tech. When I came to the Insti-tute in Princeton, he came to Princeton as a grad-uate student, and then I only saw him casually.Later when he came to MIT and started his workin geometry, I unfortunately wasn’t at Harvard yet.I would have been glad to have been part of thedevelopment of geometry by Ambrose and Singerat MIT at that time. However, this whole devel-

376 NOTICES OF THE AMS VOLUME 48, NUMBER 4

opment turned Nash off. Eventually he went to Am-brose and asked for a “real problem”. And then ofcourse Nash proved his remarkable embeddingtheorems. But I was at Michigan at that time. Un-fortunately, Nash’s great gifts were marred by histerrible disease.

Notices: After Carnegie Tech, you went to the In-stitute in 1949. You had been doing things associ-ated with engineering up to that point. How did yourperspective on mathematics change when you wentto the Institute?

Bott: Well, I felt like a kid in a candy store. Firstof all, the people around me were so outstanding!It was a sort of Valhalla, with all these semigodsaround. Amazingly enough, we mathematicianshave a type of negative feedback built into us: Ifwe don’t understand something, it makes us wantto understand it all the more. So I went to lectures,most of them completely incomprehensible, andmy gut reaction was: I want to understand this. Os-tensibly I was at the Institute to write a book onnetwork theory, but after I found out I didn’t haveto do that, I went to an incredible number of lec-tures and just absorbed the atmosphere. I didn’twrite a single paper in my first year there. So I wasvery delighted when Marston Morse called me upat the end of that year and said, “Do you want tostay for another year?” And I said, “Of course,yes!” He said, “Is your salary enough?” It was $300a month. I said, “Certainly!” because I was so de-lighted to be able to stay another year. My wife tooka dimmer view! But we managed.

Notices: So this was a big change for you, to gofrom an environment where you had been work-ing on the engineering side to a place where therewas 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 mathe-matical thought; only the concepts were different.But the actual finding of something new seems tome the same. And you see, the algebraic aspectsof network theory were an ideal introduction to dif-ferential geometry and the de Rham theory and towhat Hermann Weyl was studying at the time, thatis, harmonic theory. In effect, networks are a dis-crete version of harmonic theory. So when I cameto the Institute, the main seminar I attended wasHermann Weyl’s, and Kodaira and de Rham werelecturing on harmonic forms. Weyl wanted to have,finally in 1949, a proof of Hodge’s theorem thathe could live with. Hodge’s theorem was proved inthe 1930s, but in a somewhat sloppy way. The de-tails were cleaned up in this seminar by two quitedifferent people from different points of view. Sothis didn’t seem strange to me; it was within mydomain of thinking. It then led to topology, andthere my course with Steenrod was the dominantexperience for my future development.

Raoul Bott lecturing at Universität Bonn,1969.

Phot

ogra

ph

cou

rtes

y of

F. H

irze

bru

ch.

fea-bott.qxp 2/27/01 3:52 PM 76

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(!),

defined as follows:

If you impose a current (using a current gen-erator)

I(t) = ej!t

through the port, the response voltage is

V (t)Z(!)ej!t.

Such a circuit has an impedance Z(!),

defined as follows:

If you impose a current (using a current gen-erator)

I(t) = est


V (t) = Z(s)est.

The function Z(s) is a PRF

(positive real function), i.e.

it is a rational function with real coe!cientsmapping the right half plane

Res ! 0

to itself.

Such a circuit has an impedance Z(s),

defined as follows:

If you impose a current (using a currentgenerator)

I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.

To each edge e is associated a current ie anda voltage drop ve (the sign of both depends

on the orientation of e).

These are subject to Kirckho"’s current law:

The sum of the currents at a node is zero

and to Kirckho"’s voltage law:

The sum of the voltage drops around anyloop is zero


defined as follows:


I(t) = est


V (t) = Z(s)est.

Equivalently, if you impose the voltageV (t) = est the resulting current is

I(t) = Y (s)est.

The function Y (s) = 1Z(s) is called the

admittance of the 1-port.

The physically meaningful values of s arethe purely imaginary values s = j!, where

! is a frequency.




Res ! 0

to itself.








defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.

According to Bott, the first step tobecoming an electrical engineeris to write

!"1 = j.




Res # 0

to itself.



These are subject to Kirchho"’s currentlaw:


and to Kirchho"’s voltage law:


Let E be the set of edges, and let I $ RE

be the space of currents allowed by the


defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.

The function Y (s) = 1z(s) is called the



! is a frequency.




Res ! 0

to itself.








defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.




Res ! 0

to itself.








defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.

According to Bott, the first step tobecoming an electrical engineeris to write

!"1 = j.




Res # 0

to itself.







Let E be the set of edges, and let I $ RE

be the space of currents allowed by the

Ending the synthesis

b) The function

W1(s) =sW (s)! s2

0

s!W (s)

is a PRF with rankW1(s) " rankW (s).

c) If W (j!) = j!, then W1 has a pole at j!.

a) The existence of s0 follows from theintermediate value theorem. The uniquenessfollows from Schwarz’s lemma: a map fromthe right halfplane to itself that has more

than one fixed point is the identity.

b) The map

"(s) =s! s0

s + s0

maps the right half-plane to the unit disc,and "(s0) = 0. So

f = " #W # "!1

maps the unit disk to itself with f(0) = 0.

By Schwarz’s lemma, so does g(z) = f(z)/z.Thus

W1 = "!1 # g # "

is a PRF. If you work it out, you will find

W1(s) =sW (s)! s2

0

s!W (s).

c) Clearly if W (j!) = j!, then

has a pole at j!.

Suppose that Z(j!) = Lj! for some L > 0.

Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

A bit of computation shows that

Z(s) =1

1LR(s) + 1

Ls

+1

R(s)Ls2

0+ s

Ls20

.

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).


Z(s) =1

1LR(s) + 1

Ls

+1

R(s)Ls2

0+ s

Ls20

.

One basic result of circuit theory is



defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.


The Bott-Duffin synthesis is the converse:


defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.


The Bott-Duffin synthesis is the converse:

Every PRF is the impedance of a passive 1-port.


defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.

An electrical engineer


1) Figures out the function g(jω) that he wantshis circuit to have



2) Finds a PRF f(s) such that f(jω) approximates g(jω)



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 approximatingCauchy integrals by Riemann sums to

Padé approximants.

In this phase, the engineer is going to have to compromise.He will only be able to approximate the real part (or the





There is a vast literature about this, from approximatingCauchy integrals by Riemann sums to

Padé approximants.

3) Applies the Bott-Duffin synthesis to make the circuit.

In this phase, the engineer is going to have to compromise.He will only be able to approximate the real part (or the


I will assume my audience has forgottenany circuit theory they ever knew

and start from scratch

Even from scratch, it is possible to givea complete proof in half an hour:

the original paper is only one page.

An electrical circuit is a graph, with each edge oriented,and each edge carrying a circuit element:

• A resistor

• a capacitor

• an inductor

Some basics

Text

Figure 1. Resistor, capacitor and inductor.

Figure 2. All the three in a cycle, and the ladder 10.3.1

A B

A

B

Figure 3. Ports in series and in parallel

Figure 4. Scheme 10.4.2

C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C

Figure 6. Scheme 10.4.4, two variants

1

Text



A B

A

B



C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C


1

Symbol

v = R i

Text



A B

A

B



C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C


1

Equation

C v′ = i

L i′ = v

Units

Ohms

Farads

Henrys


defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.




defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.








defined as follows:


I(t) = est


V (t) = Z(s)est.




Res ! 0

to itself.








defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.




Res ! 0

to itself.







Let E be the set of edges, and letI subsetRE be the space of currents allowedby the Kirchho" current law, and V " RE

be the space of voltage drops allowed by theKichho" voltage law.


defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.




Res ! 0

to itself.







Let E be the set of edges, and letI subsetRE be the space of currents allowedby the Kirchho" current law, and V " RE

be the space of voltage drops allowed by theKichho" voltage law.


defined as follows:


I(t) = est


V (t) = Z(s)est.


I(t) = Y (s)est.




! is a frequency.




Res ! 0

to itself.







Let E be the set of edges, and let I " RE

be the space of currents allowed by theKirchho" current law, and V " RE be the

space of voltage drops allowed by theKichho" voltage law.

Tellegen’s theorem. The spaces I and Vare orthogonal complements in RE.


Proof. They are respectively the kernel ofA and the image of A!, where A is the

incidence matrix:

ai,j =

!"#

"$

1 if node i is the end of edge j

!1 if node i is the origin of edge j

0 otherwise.



incidence matrix:

Using Kirchhoff’s laws and Ohm’s law for each resistor,we can eliminate many of the ve and ie



incidence matrix:

ai,j =

!"#

"$



0 otherwise.

Under quite general circumstances, theremaining variables from which all others

can be deduced algebraically are the

currents in the inductors (and currentgenerators), and the

voltages through the capacitors (andvoltage generators)

Tellegen’s theorem gives

!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.

Using undetermined coe!cients, we can findthe solution of the di"erential equationdescribing the circuit with iG = est as

vG = !Z(s)est

i! = I!est so that v! = L!I!se

st

v" = V"est so that i" = C"V"se

st

i# = I#est so that v#= R#I# est

Cancel the common factor |est|2 to get

Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.

Everything on the right is " 0 whenRes " 0.



incidence matrix:

ai,j =

!"#

"$



0 otherwise.






!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.

Under quite general circumstances,

–no loops of capacitors

–the inductors do not disconnect the circuit

the remaining variables from which allothers can be deduced algebraically are the



Form a vector x(t) whose entries are thesevariables.

The time evolution of the circuit isdescribed by an inhomogeneous linear

di!erential equation of the form

x" = Ax + f(t)

where A is a square matrix, and f(t) is theimposed current, expressed in the variables

of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.

Using undetermined coe"cients, we can findthe solution of the di!erential equationdescribing the circuit with iG = est as



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


A circuit of this sort is called normal



incidence matrix:

ai,j =

!"#

"$



0 otherwise.








x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st




incidence matrix:

ai,j =

!"#

"$



0 otherwise.








x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st




incidence matrix:

ai,j =

!"#

"$



0 otherwise.








x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st


Suppose that a 1-port is normal, and



incidence matrix:

ai,j =

!"#

"$



0 otherwise.







form a vector x(t) whose entries are thesevariables.



x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


The eigenvalues of the matrix A are nonnegative



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.

Di!erentiate the positive definite Hermitianform

E(i,v) =1

2

%

!

L!|i!|2 +1

2

%

"

C"|v"|2.



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


E(i,v) =1

2

%

!

L!|i!|2 +1

2

%

"

C"|v"|2.inductors capacitors



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


E(i,v) =1

2

%

!

L!|i!|2 +1

2

%

"

C"|v"|2.

The letters !, " and # are traditional forinductors, capacitors and resistors

respectively.

kinetic energy potential energy

We findd

dtE

!i(t),v(t)

"

=1

2

d

dt

#$

!

L!|i!(t)|2 +$

"

C"|v"(t)|2%

=$

!

L!Re(i!i!!) +

$

"

C"Re(v"v!")

=$

!

Re(i!v!) +$

"

C"Re(v"i")

= !$

#

Re(i#v#) = !$

#

R#|i#|2 " 0.


!iGvG =$

!

i!v! +$

"

i"v" +$

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s$

!

L!|I!|2+s$

"

C"|V"|2+$

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

#$

!

L!|I!|2 +$

"

C"|V"|2%

+$

#

R#|I#|2.

Everything on the right is # 0 whenRes # 0.

We findd

dtE

!i(t),v(t)

"

=1

2

d

dt

#$

!

L!|i!(t)|2 +$

"

C"|v"(t)|2%

=$

!

L!Re(i!i!!) +

$

"

C"Re(v"v!")

=$

!

Re(i!v!) +$

"

C"Re(v"i")

= !$

#

Re(i#v#) = !$

#

R#|i#|2 " 0.


!iGvG =$

!

i!v! +$

"

i"v" +$

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s$

!

L!|I!|2+s$

"

C"|V"|2+$

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

#$

!

L!|I!|2 +$

"

C"|V"|2%

+$

#

R#|I#|2.


This certainly proves that the eigenvalues of A are nonpositive, and further that the purely imaginary

eigenvalues are simple.

it also implies that if a 1-port is driven by a currenti(t) = est with Re s>0,

then the system of differential equations



incidence matrix:

ai,j =

!"#

"$



0 otherwise.








x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st


can be solved by undetermined coefficients, setting



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


E(i,v) =1

2

%

!

L!|i!|2 +1

2

%

"

C"|v"|2.


respectively.

xi(t) = aiest



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


E(i,v) =1

2

%

!

L!|i!|2 +1

2

%

"

C"|v"|2.


respectively.

xi(t) = aiest

and solving for the ai after cancelling acommon factor est.

Z(s) is a PRF

generator inductors capacitors resistors



incidence matrix:

ai,j =

!"#

"$



0 otherwise.

Telegen’s theorem gives

!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st


Cancel the common factor |est|2



incidence matrix:

ai,j =

!"#

"$



0 otherwise.










x" = Ax + f(t)


of x.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


Proving the first half of the fundamental theoremof circuit theory



incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st

i# = I#est so that v# = R#I#e

st



incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st


Insert into



incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st


Cancel the common factor |est|2



incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.




incidence matrix:

ai,j =

!"#

"$



0 otherwise.


!iGvG =%

!

i!v! +%

"

i"v" +%

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s%

!

L!|I!|2+s%

"

C"|V"|2+%

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

&%

!

L!|I!|2 +%

"

C"|V"|2'

+%

#

R#|I#|2.


Every PRF is the impedance of some 1-

port

Proving the second half of the fundamental theorem of circuit theory

Bott and Duffin give a straightforward constructionof an appropriate circuit.

Series and ParallelTe

t



A B

A

B



C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C


1

If two 1-ports are connected in seriestheir impedances add

Te

t



A B

A

B



C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C


1

If two 1-ports are connected in paralleltheir admittances add

We findd

dtE

!i(t),v(t)

"

=1

2

d

dt

#$

!

L!|i!(t)|2 +$

"

C"|v"(t)|2%

=$

!

L!Re(i!i!!) +

$

"

C"Re(v"v!")

=$

!

Re(i!v!) +$

"

C"Re(v"i")

= !$

#

Re(i#v#) = !$

#

R#|i#|2 " 0.


!iGvG =$

!

i!v! +$

"

i"v" +$

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s$

!

L!|I!|2+s$

"

C"|V"|2+$

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

#$

!

L!|I!|2 +$

"

C"|V"|2%

+$

#

R#|I#|2.


Z(s) = ZA(s) + ZB(s)

Y (s) = YA(s) + YB(s)1

Z(s)=

1

ZA(s)+

1

ZB(s)

We findd

dtE

!i(t),v(t)

"

=1

2

d

dt

#$

!

L!|i!(t)|2 +$

"

C"|v"(t)|2%

=$

!

L!Re(i!i!!) +

$

"

C"Re(v"v!")

=$

!

Re(i!v!) +$

"

C"Re(v"i")

= !$

#

Re(i#v#) = !$

#

R#|i#|2 " 0.


!iGvG =$

!

i!v! +$

"

i"v" +$

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s$

!

L!|I!|2+s$

"

C"|V"|2+$

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

#$

!

L!|I!|2 +$

"

C"|V"|2%

+$

#

R#|I#|2.



Y (s) = YA(s) + YB(s)1

Z(s)=

1

ZA(s)+

1

ZB(s)

We findd

dtE

!i(t),v(t)

"

=1

2

d

dt

#$

!

L!|i!(t)|2 +$

"

C"|v"(t)|2%

=$

!

L!Re(i!i!!) +

$

"

C"Re(v"v!")

=$

!

Re(i!v!) +$

"

C"Re(v"i")

= !$

#

Re(i#v#) = !$

#

R#|i#|2 " 0.


!iGvG =$

!

i!v! +$

"

i"v" +$

#

i#v#.


vG = !Z(s)est


st


st



Z(s) = s$

!

L!|I!|2+s$

"

C"|V"|2+$

#

R#|I#|2.

Take real parts:

ReZ(s) = Res

#$

!

L!|I!|2 +$

"

C"|V"|2%

+$

#

R#|I#|2.



Y (s) = YA(s) + YB(s)1

Z(s)=

1

ZA(s)+

1

ZB(s)

Conditions on PRF’s(1) a) infRes>0 ReZ(s) = R > 0

b) infRes>0 ReY (s) = 1/R > 0

(2) a) The function Z(s) has a pole onthe imaginary axis or at !

b) The function Z(s) has a pole onthe imaginary axis or at !



(4) a) There exists ! > 0 and L > 0 withZ(j!) = Lj!

b) There exists ! > 0 and C > 0 withY (j!) = Cj!

(1) a) infRes>0 ReZ(s) = R > 0




































Reduction in case 1If a PRF satisfies either of conditions 1a or 1b







we can write Z(s) = R + Z1(s) or

Y (s) = R + Y1(s)








Y (s) = 1R + Y1(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.

Reduction in case 2This case uses partial fractions:

Since Z and Y are PRF’s, any poles must be simple, with positive residues.








Y (s) = 1R + Y1(s)

If W is a PRF, we can write

W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s

and W1 is a PRF which is bounded on theright half-plane. Proposition.

–The PRF k0s is the impedance of a

capacitor of capacitance 1k0

.

–The PRF k!s is the impedance of aninductor of inductance k!.

–The PRF 2kss2+!2 is the impedance of a

capacitor or capacitance C = 12k in parallel

with an inductor of inductance L = 2k!2 .

Richard’s theorem. Let W (s) be a PRFwith no zeros or poles on the imaginary axis

or at infinity.

a) There then exists a unique s0 " R+ withW (s0) = s0.

b) The function

W1(s) =sW (s)# s2

0

s#W (s)

is a PRF with rankW1(s) $ rankW (s).








Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s




.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)









Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s




.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)









Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s




.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)









Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s




.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)









Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s




.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)









Y (s) = 1R + Y1(s)


W (s) = PW (s) + W1(s)

with

PW (s) =k0

s+

m!

i=0

2kis

s2 + !2i

+ k!s

with k0 " 0, all ki " 0 and k! " 0




.






or at infinity.

a) There then exists a unique s0 # R+ withW (s0) = s0.








Y (s) = 1R + Y1(s)

Proposition.



.












Y (s) = 1R + Y1(s)

Proposition.



.












Y (s) = 1R + Y1(s)

Proposition.



.





Thus if we can find circuits that realize the summands of PW , and if we can realize W1 , we

can realize W (as an impedance or as an admittance).

Case 3 This case requires Richard’s theorem








Y (s) = 1R + Y1(s)

Proposition.



.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)


c) If W (j!), then W1 has a pole at j!.








Y (s) = 1R + Y1(s)

Proposition.



.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)


c) If W (j!), then W1 has a pole at j!.








Y (s) = 1R + Y1(s)

Proposition.



.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)










Y (s) = 1R + Y1(s)

Proposition.



.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)



Richard’s theorem follows from Schwarz’s lemma








Y (s) = 1R + Y1(s)

Proposition.



.






or at infinity.


b) The function

W1(s) =sW (s)# s2

0

s#W (s)





b) The map

!(s) =s! s0

s + s0

maps the right half-plane to the unit disc,and !(s0) = 0. So

f = ! "W " !!1



W1 = !!1 " g " !.

If you work it out, you will find

W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !.


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).

c) Clearly if W (j") = j", then

has a pole at j".

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j".

b) The map

!(s) =s! s0

s + s0


f = ! "W " !!1



W1 = !!1 " g " !


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j".This completes the proof of Richard’s theorem.

Ending the synthesis

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).


Z(s) =1

1LR(s) + 1

Ls

+1

R(s)Ls2

0+ s

Ls20

.

b) The function

W1(s) =sW (s)! s2

0

s!W (s)





b) The map

"(s) =s! s0

s + s0


f = " #W # "!1



W1 = "!1 # g # "


W1(s) =sW (s)! s2

0

s!W (s).


has a pole at j!.


Set

Z1(s) =Z(s)

L,

and

R(s) =sZ1(s)! s2

0

s! Z1(s).


Z(s) =1

1LR(s) + 1

Ls

+1

R(s)Ls2

0+ s

Ls20

.



A B

A

B



C1

C2

C3

C4


i0

i1

L i4

i3

R

i2

i5

C


1

Suppose that the circuits C1, C2, C3, C4 haveadmittances

1

LR(s),

1

Ls,

R(s)

Ls20

,s

Ls20

.

Suppose that the circuits C1, C2, C3, C4 haveadmittances

1

LR(s),

1

Ls,

R(s)

Ls20

,s

Ls20

.

Then the circuit below has impedance Z(s).

the bott-duffin synthesis answers a basic question in electrical

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