Thetransmission(ABCD)matrix

WatcharapanSuwansan6suk

#7EIE/ENE450AppliedCommunica6onsandTransmissionLines

KingMongkut’sUniversityofTechnologyThonburi

Wewillcoverthesetopics•  Thetransmission(ABCD)matrix

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APerthislecture,youwillbeableto•  Statethedefini6onofthetransi6on(ABCD)matrix•  Computethetransi6onmatrixforasimplenetwork•  Compute the transi6on matrix for a series (cascade) ofnetworks

•  Use a table to iden6fy the transi6onmatrix for basic circuitelements

•  Usethetransi6onmatrixtocalculatebasicquan66essuchas–  thevoltageandcurrent–  theinputimpedance–  thepowerdeliveredtoasec6onofnetwork

•  Use tables to convert between the impedance, admiTance,scaTering,andtransi6onmatricesforany2-portnetwork

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4.4THETRANSMISSION(ABCD)MATRIX

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Anothercharacteriza6onofatwo-portnetworkisatransmission(ABCD)matrix

•  Mostnetworkshave2ports(input,output)•  TheABCDcancharacterizeonly2-portnetworks– while the [Z] and [S]matrices can characterize a networkwithanynumberofports

•  (Benefit) The ABCD matrix of a series connec6on of twonetworksiseasytofind(bymul6plyingmatrices)

a2-portnetwork

a2-portnetwork

Aseries(cascade)connec6onofnetworks

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ABCDmatrixrelatesthetotalvoltageandtotalcurrentattheinputtothoseattheoutput

•  Defini6on:TheABCDmatrixisa2-by-2matrixsuchthatorinthelinearequa6onform:

Vn=totalvoltageatportnIn=totalcurrentatportn(n=1,2)

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I3

A2 B2

C2 D2

�

Thechangeinsignconven6onmakesiteasytofindtheABCDmatrixforacascadednetwork

Wechangethesignconven6onofI2whendealingwiththeABCDmatrix

I2I1

ForABCDmatrix,I2flowsoutoftheport

I2

[Z]or[S]

I1

For[Z]and[S]matrices,I2flowsintotheport

A1 B1

C1 D1

�

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Example1:givenI2,thesearetheflowsofcurrent

Pictureifthenetworkischaracterizedby...

theABCDmatrix The[Z]and[S]matrices

I2 = �3 A

I2 = 1 A port2+

−V2

1 A

port2+

−V2

�3 Aor

3 A

port2+

−V2

1 A

port2+

−V2

�3 Aor

3 A

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Whatis?

9

I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

10\60� V10\30� V

10\90� V 10\120� V

V1

1 j200 1

�

Howareandrelated?

10

I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

I2 V2

I2 =1

50V2 I2 = 50V2

1 j200 1

�

I2 =1

j20V2 I2 = j20V2

Whatisthevalueof?

11

I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

V2

7\� 178� V

7\� 38� V

9\178� V

9\38� V

1 j200 1

�

Youwillgetthiserrorifyouuse60radiansinsteadof60°

Whatisthevalueof?

12

I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

1 j200 1

�

I2

180\38� mA 90\� 38� mV

18\38� mA0.18\38� V

Whatisthevalueof?

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I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

1 j200 1

�

180\38� mA

I1

fromthepreviousslide

9\38� mA 9\� 52� mA

180\� 52� mV

Whatistheinputimpedanceseenlookingintothenetwork?

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I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) (B)

(C) (D)

1 j200 1

�

Zin

50 ⌦ 50 + j20 ⌦

j20 ⌦40� j20 ⌦

AnaccurateanswerwhenkeepingenoughdecimalpointsforandI1V1

Whatistheaveragepowerdeliveredtothissec6onofthenetwork?

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I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

(A) −900mW (B) 130mW

(C) 280mW (D) 860mW

1 j200 1

�

AnaccurateanswerwhenkeepingenoughdecimalpointsforandI1V1

0 20 40 60 80 100−10

−5

0

5

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Whichgraphisthe6me-domainvoltage?

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I1

+

−V1

+

−V2

ZL = 50 ⌦

I2

+

−10\60� V

1 j200 1

�frequencyf=20Hz

v2(t)

Voltage(V)

Timet(ms)

(B) (C)

(A)

TheABCDmatrixofacascadeconnec6onofnetworksisamatrixmul6plica6on

•  Subs6tu6ongivesusV1

I1

�=

A1 B1

C1 D1

� A2 B2

C2 D2

� V3

I3

�

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keyresults

Acascadeconnec6onoftwonetworksisequivalenttoonenetwork

•  No6ce that the order of mul6plica6on is the same as theorderinwhichthenetworksappear

I3

A2 B2

C2 D2

�I2I1

A1 B1

C1 D1

�

I3I1

A1 B1

C1 D1

� A2 B2

C2 D2

�

V2

V1 V3

V3V1

+

−

+

−

+

−

+

−

+

−

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WhatistheABCDmatrixofthiscascadednetwork?

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I3I1

+

−

1 j50 1

� 0.5 00 2

�

cascadednetwork

V1 V3

+

−

0.5 j100 2

�0.5 j2.50 2

�

1 j50 1

� 1.5 j50 3

�

(A) (B)

(C) (D)

ParametersA,B,C,Daregivenby...

A =V1

V2

���I2=0

I1

A BC D

�

I2 = 0 (opencircuit)

+

−V1

+

−V2

B =V1

I2

���V2=0

I1

A BC D

�+

−V1

+

−

I2

V2 = 0

C =I1V2

���I2=0

I1

A BC D

�

I2 = 0 (opencircuit)

+

−V1

+

−V2

(shortcircuit)

I1

A BC D

�+

−V1

+

−

I2

V2 = 0

(shortcircuit)

D =I1I2

���V2=0

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Example2:findtheABCDmatrixofthecircuitbelow

•  Answer:Wewillshowthat

+

−

+

−

I2I1

Z

impedance

V2V1

A BC D

�=

1 Z0 1

�

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Example2(solu6onforA)

•  ForparametersAandB,wenotethat•  Hence,parameterAcanbefoundfromanequa6onwhichcorrespondstotheabovecircuit

•  UsingKCL,soand

V1 = AV2 +BI2

A =V1

V2

���I2=0

+

−

+

−

I1

Z

impedance

V2V1

I2 = 0

opencircuit

I1 = I2 = 0 V1 = V2 A = 1

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Example2(solu6onforB)

+

−

+

−

Z

impedance

V1

shortcircuit

•  ParameterBisgivenbywhichcorrespondstothecircuitabove

•  ByOhm’slaw,so

B =V1

I2

���V2=0

I2

V2 = 0

V1

I2= ZV1 = I2Z

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Example2(solu6onforC)

•  ForparametersAandB,wenotethat•  Hence,parameterCisgivenbywhichcorrespondstothecircuitabove

•  UsingKCL,so

I1 = CV2 +DI2

C =I1V2

���I2=0

+

−

+

−

I1

Z

impedance

V2V1

I2 = 0

opencircuit

I1 = I2 = 0 C = 0

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Example2(solu6onforD)

•  ParameterDisgivenbywhichcorrespondstothecircuitabove

•  UsingKCL,so

D =I1I2

���V2=0

+

−

+

−

Z

impedance

V1

shortcircuit

I2

V2 = 0

I1

I1 = I2 D = 1

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ABCDparametersofsomecommoncircuitsarebelow

impedance

port1isthisside

port2isthisside

admiTance

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keyresults

(con6nued)

port1isthisside

port2isthisside

losslesstransmissionline

transformer

Nota6on:

N/M

N : M

M/N

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N

M=

# of turns of the coil attaching to port 1

# of turns of the coil attaching to port 2

keyresults (con6nued)

port1isthisside

port2isthisside

admiTancesintheπconfigura6on

impedancesintheTconfigura6on

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keyresults

RELATIONTOIMPEDANCEMATRIX

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Aconversionfromthe[Z]matrixtotheABCDmatrixisstraighoorward

•  Supposethatthenetworkischaracterizedbytheimpedancematrix

•  Then,theABCDparametersforthenetworkare

⇥Z11 Z12Z21 Z22

⇤

[Z] =

Z11 Z12

Z21 Z22

�port1 port1

A =Z11

Z21

C =1

Z21D =

Z22

Z21

B =Z11Z22 � Z12Z21

Z21

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HereisareasonwhytheABCDparametersareasstated

•  Takethetotalvoltagesandtotalcurrentsasshown•  Then,bythedefini6onoftheimpedancematrix,

V2V1

+

−

+

−

I1

V1

+

−

I2

V2

+

−[Z] =

Z11 Z12

Z21 Z22

�

V1

V2

�=

Z11 Z12

Z21 Z22

� I1�I2

�=)

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(con6nued)deriva6on•  TheABCDparametersaregivenby

0

0

= Z11Z22

Z21� Z12

V2 = I1Z21 � I2Z22

0=) I1

I2=

Z22

Z21

V1 = I1Z11 � I2Z12

V1 = I1Z11 � I2Z12

V2 = I1Z21 � I2Z22

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(con6nued)deriva6on

0V2 = I1Z21 � I2Z22 =) I1

V2=

1

Z21

V2 = I1Z21 � I2Z22 =) I2Z22 = I1Z21 =) I1I2

=Z22

Z21

0

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CONVERSIONSBETWEEN2-PORTNETWORKPARAMETERS

(formulasinthissec6oncomefromp.192ofthetextbook)

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S Z Y ABCD

A

B

C

D

step1:giventheZparameters

step2:herearethecorrespondingA,B,C,andDparameters

Thisishowtousethetableinp.192ofthetextbook:

GiventhescaTeringmatrix,theconversionsare...

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Totheimpedancematrix[Z]

TotheadmiTancematrix[Y]

Tothetransi6onmatrix(ABCD)

Y0 = 1/Z0keyresults

Giventheimpedancematrix,theconversionsare...

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TothescaTeringmatrix[S]

TotheadmiTancematrix[Y]

Tothetransi6onmatrix(ABCD)

keyresults

GiventheadmiTancematrix,theconversionsare...

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TothescaTeringmatrix[S]

Totheimpedancematrix[Z]

Tothetransi6onmatrix(ABCD)

keyresults

GiventheABCDmatrix,theconversionsare...

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TothescaTeringmatrix[S]

Totheimpedancematrix[Z]

TotheadmiTancematrix[Y]

keyresults

WhatisthescaTeringparameterofthisnetwork?

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1:2

port1 port2

S11

astep-uptransformer

(A) −0.6 (B) 0.6

(C) −0.8 (D) 0.8

AssumetheT-linesconnec6ngtoports1and2havethecharacteris6cimpedance Z0 = 50 ⌦

Summary•  Thetransmission(ABCD)matrix– Defini6onoftheABCDmatrix– ABCDparametersforsomeusefultwo-portnetworks– ConversionsbetweenS,Z,Y,Z, andABCD parameters for two-portnetworks

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