Circuit Symmetry 1/16 Present Jim Stiles the Univ

Home , Reflection symmetry

is ahighly For example,consider anordinary understanding of symmetry, basedon an that ofsymmetry. One ofthemostpowerfulconcepts 3/4/2009 Circuit Symmetry 1/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles of structuresandfigures. Évariste Galois

Circuit Symmetry symmetric

evaluate thesymmetry of polynomial functions Initially, thesymmetric “objects”un operations (agroup)that leaves anobject genius of mathematics called defined symmetryin averyprecise an On theotherhand, Normal object! Évariste Galois humanshave a equilateral triangle inforevaluatingcircuitsis mathematicians , but grouptheory can likewise be applied to (1811-1832), Group Theory esthetic conceptual structures perception

(astheyarewonttodo)have symmetry ; wefindthatit der considerationbyGaloiswere , firstdevelopedbythe young . d unambiguous way.Using abranch unchanged isdefined byasetof .

A: Q: 3/4/2009 Circuit Symmetry 2/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles First, wenotethat thetr corner To determineitslevelofsymmetry, let’sfirstlabeleachcorneras compare Yes,but

Obviously 3 . to that ofanisosceles triangle,arectangle,orsquare?

how thisistrue. Wedon’tneed symmetric isit?How doesthesym iangle exhibitsa planeof 1 1

a mathematician totellusthat! 2 2

3 3

reflection symmetry metry ofanequilateraltriangle : 1 , corner 2 , and

Thus, ifwe“ 3/4/2009 Circuit Symmetry 3/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles unchanged Note that although corners 1and 3havech Note that wecanwritethis Mathematicians saythatthesetwo triangles are reflecting acrossthesymmetric plane! position) ofthe corners, definedas:

—that is,ithas the same reflect ” the triangleacross reflectionoperation as a 3 shape

this planeweget: , same 2 13 31 22 anged places,thetriangle itselfremains

→ → →

1 size congruent.

, andsame permutation orientation (anexchangeof after clockwise alsoresults ina A: Q: 3/4/2009 Circuit Symmetry 4/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

In addition,anequilateral triangleexhibits Definitely!Thereare

But wait!Isn’tthere is 1 1 1 2 2 2 two more

congruent more 3 3 3

thanjust : triangle:

12 12 11 33 21 31 23 32 23 one rotation symmetry!

→ → →

→ → → → → → planeofreflectionsymmetry?

2 1 3 1

3 1

Rotating 3 2 2

thetriangle 120 D

These 6operations formthe every nothing Likewise, rotatingthetriangle 3/4/2009 Circuit Symmetry 5/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles This seemingly Additionally, thereis and allof thesesixoperations issaid tohave

consists ofsixoperations). An objectthatremains

symmetrygroup. ! 1 1 1 trivial 2 operation isknown asthe 2 2 one more 3 An equilateral trianglehas

3 3

dihedral symmetry group 120 operationthatwillresultin a congruent triangle—do D

counter-clockwise

13 11 32 21 33 22 → → → → → → D 3 identity operation symmetry.

congruent D 3 resultsinacongruenttriangle: symmetry! 2

2 D 1

which has 3 3 whenoperated on byany 2

, andisanelement of 1 1 order six six order

(i.e., it

microwave circuits!?! A: Q: By applying asimilaranalysistoisoscelestr 3/4/2009 Circuit Symmetry 6/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles trapezoid isthe

Thus, asquare is the Plenty!

Well that’salljust fascinat D D D 1 2 4

Useful circuits A rectanglehas A squarehas An isoscelestrapezoidhas

least

most most . oftendisplayhigh levelsof symmetric objectofthefourwe D 4

symmetry,adihedralgroupof

D ing—but justwhat theheck doesthishave to dowith 2 symmetry, adihedralgroupof D 1 symmetry,adihedralgroupof iangle, rectangle, and symmetry havediscussed;theisosceles order 8 order 4 . square, wefindthat: . . order 2 order .

For example,considerthese 3/4/2009 Circuit Symmetry 7/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

12 34 21 43 13 31 24 42 → → → → → → → →

Port Port

D 1 200 100 symmetricmulti-portcircuits: Ω Ω

50 100 50 50 Ω Ω Ω Ω

200 200 Ω Ω

Port

Port Port

Port

which is Or thiscircuitwith 3/4/2009 Circuit Symmetry 8/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

congruent

Port Port

1 3

underthesepermutations: D 2 symmetry: 200 13 31 24 42 → Ω → → →

50 50 Ω Ω

12 34 21 43 200 → → → → Ω

Port Port 14 32 23 41

→

→ →

→ 4

The matrix, oradmittance matrixofthesenetworks. which is Or thiscircuitwith 3/4/2009 Circuit Symmetry 9/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

importance

congruent

Port Port

ofthiscanbeseen whenconsider 3

13 31 24 42 → → → → underthesepermutations: D 4

symmetry: 50 Ω 12 34 21 43

→ → → → 50 50

Ω Ω

14 32 23 50 41 → → → → Ω

ing thescattering matrix,impedance

Port Port 14 33 22 41 →

→ →

→

11 32 23 44 → → → →

So, since (for example) example) (for since So, must For 3/4/2009 Circuit Symmetry 10/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

This four-port network hasasingle planeof thus iscongruent under thepermutation:

example betrue!

, consideragainthis

Port Port

1 3

12 →

, we findthat forthis circuit: SZZYY Z Z SS 200 12 12 122 11 22 11 22 11 symmetric circuit = Ω

50 100 12 34 21 43 Ω → → → Ω → reflection symmetry

200

: Ω

Port

(i.e., 2

D 1

symmetry), and the scatteringmatrix Continuing for Note thereare just reciprocity Or, since 3/4/2009 Circuit Symmetry 11/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

the matrixreduces furthertoonewithjust and the

impedance 12 (aconstraintindependent ofsymmetry)wefind that → all and elements ofthepermutation,we find

and 34 8 independentelements inthismatrix. Ifwe alsoconsider → must admittance wefind: have have this S S matriceswould likewise have thissameform. = = form:

⎢ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ ⎣ ⎢ ⎢ ⎢

SSSS S SSSS SSSS S S SZZYY Z Z SS SZZYY Z Z SS SSSS S 11 13 33 11 31 21 31 2 143 41 4 31 13 13 13 13 31 31 31 12 13 21 11 11 1 1

= ===

S S S S 133 31 21 31 41 143 41 6 11 42 24 24 24 24 42 42 42

independent elements:

S S S S

33 43 14 31 41

S S S S 3 43 14 31 41 thatfor thissymmetric circuit, 3 == ⎥ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ ⎦ ⎥ ⎥ ⎥

SS 113 31 =

and

SS 114 41 = , and

A: Q: Or, forcircuitswith 3/4/2009 Circuit Symmetry 12/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

determine

This willgreatly Interesting. But 13 31 24 42 only → → → → sixmatrixelements!

simplify

why dowecare this

Port Port

3 D

1 the analysisofthissymmetr

symmetry: S 100

= ?

⎣ ⎢ ⎢ ⎢ ⎢ ⎡

Ω S S S S 31 21 4 11

S S S S 50 50 2 31 21 41

2 Ω

S S S S

21 31 41 11

200

S S S S 2 2 31 41 Ω 2 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎤

ic circuit,aswe need to

Port

we findthatthe impedance(orscatteri For acircuitwith 3/4/2009 Circuit Symmetry 13/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

Note here thatthereare just

Port Port

1 3

D 2

symmetry: 200 Ω four

Z 50 independent values! 50 =

⎣ ⎢ ⎢ ⎢ ⎢ ⎡

Z Z Z Z Ω

ng, oradmittance)matrixhasthe form: 3 21 41 11

Z Z Z Z 200 31 21 41 11

Z Z Z Z

21 31 41 11

Z Z Z Z 31 21 41 1

Port ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ Port

we findthatthe admittance (or scatteri For acircuitwith 3/4/2009 Circuit Symmetry 14/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles Note here thatthereare just

Port Port

1 3

symmetry: 50 three Ω

Y independent values! = 50 50

⎣ ⎢ ⎢ ⎢ ⎢ ⎡

Y YY YY Y Ω Ω 121 21 21 21 ng, orimpedance)matrixhasthe form: 4 11 1

Y YY YY Y 121 21 12 21 41 11

Y Y 41 11 Ω 1

Y Y 41 11 ⎦ ⎥ ⎥ ⎥ ⎥ ⎤

Port

forms

Compare One moreinteresting thing(yet 3/4/2009 Circuit Symmetry 15/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

as the scatteringmatrixofacircuit with matched, lossless,reciprocal

: thesetothematrix formsabove. S S = =

⎣⎦ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎡⎤ ⎣⎦ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎡⎤

00 00 j

00 α 00 β

−

00 00 α

βα

0 0 α

β β

4-port

α 0 0 β

− α

another β

j α

The “anti-symmetric” solution = β

device musthave ascatteringmatrixwithone of

⎣ ⎢ ⎢ ⎢ ⎢ ⎡

j The “symmetric” solution

0 α 0 β one!);recallthatwe earlierfoundthata D

The“symmetric solution”hasthe

2 j

α 0 0 β

symmetry!

α 0 0

0 α 0 β ⎦ ⎥ ⎥ ⎥ ⎥ ⎤

same form two

“symmetric” scatteringmatrix actually We’ll seejustwhat thesesym Not onlycanwedetermine from the also determinethe design ispossible (e.g.,amatched,lossless,reci exhibit Likewise, the“anti-symmetric” matched, lossless, reciprocal four-portnetwork symmetry). A: Q: 3/4/2009 Circuit Symmetry 16/16 present Jim Stiles The Univ. of Kansas Dept. of EECS of Dept. ofKansas Univ. The Jim Stiles

That’s Does thismeanthatamatched,lossless,reciprocal four-port device withthe D are 1

exactly symmetry! laterinthecourse! what itmeans! general structure

metric, matched, lossless, reciprocalfour-portcircuits must

S form exhibit = ofapossiblesolutions(

⎢ ⎣ ⎢ ⎢ ⎢ ⎡

α 0 β 0 ofthescattering matrix

−

0 0 α D

procal 3-portdeviceisimpossible), wecan symmetry?

α 0 0 β

−

0 α 0 β ⎦ ⎥ ⎥ ⎥ ⎥ ⎤

e.g. thecircuitmusthave whether aparticular must