The transmission (ABCD) matrix...
Transcript of The transmission (ABCD) matrix...
Thetransmission(ABCD)matrix
WatcharapanSuwansan6suk
#7EIE/ENE450AppliedCommunica6onsandTransmissionLines
KingMongkut’sUniversityofTechnologyThonburi
Wewillcoverthesetopics• Thetransmission(ABCD)matrix
2
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
3
4.4THETRANSMISSION(ABCD)MATRIX
4
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
5
ABCDmatrixrelatesthetotalvoltageandtotalcurrentattheinputtothoseattheoutput
• Defini6on:TheABCDmatrixisa2-by-2matrixsuchthatorinthelinearequa6onform:
Vn=totalvoltageatportnIn=totalcurrentatportn(n=1,2)
6
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
�
7
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
8
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?
13
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?
14
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?
15
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
10
Whichgraphisthe6me-domainvoltage?
16
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
�
17
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
+
−
+
−
+
−
+
−
+
−
18
WhatistheABCDmatrixofthiscascadednetwork?
19
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
20
Example2:findtheABCDmatrixofthecircuitbelow
• Answer:Wewillshowthat
+
−
+
−
I2I1
Z
impedance
V2V1
A BC D
�=
1 Z0 1
�
21
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
22
Example2(solu6onforB)
+
−
+
−
Z
impedance
V1
shortcircuit
• ParameterBisgivenbywhichcorrespondstothecircuitabove
• ByOhm’slaw,so
B =V1
I2
���V2=0
I2
V2 = 0
V1
I2= ZV1 = I2Z
23
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
24
Example2(solu6onforD)
• ParameterDisgivenbywhichcorrespondstothecircuitabove
• UsingKCL,so
D =I1I2
���V2=0
+
−
+
−
Z
impedance
V1
shortcircuit
I2
V2 = 0
I1
I1 = I2 D = 1
25
ABCDparametersofsomecommoncircuitsarebelow
impedance
port1isthisside
port2isthisside
admiTance
26
keyresults
(con6nued)
port1isthisside
port2isthisside
losslesstransmissionline
transformer
Nota6on:
N/M
N : M
M/N
27
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
28
keyresults
RELATIONTOIMPEDANCEMATRIX
29
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
30
HereisareasonwhytheABCDparametersareasstated
• Takethetotalvoltagesandtotalcurrentsasshown• Then,bythedefini6onoftheimpedancematrix,
V2V1
+
−
+
−
I1
V1
+
−
I2
V2
+
−[Z] =
Z11 Z12
Z21 Z22
�
V1
V2
�=
Z11 Z12
Z21 Z22
� I1�I2
�=)
31
(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
32
(con6nued)deriva6on
0V2 = I1Z21 � I2Z22 =) I1
V2=
1
Z21
V2 = I1Z21 � I2Z22 =) I2Z22 = I1Z21 =) I1I2
=Z22
Z21
0
33
CONVERSIONSBETWEEN2-PORTNETWORKPARAMETERS
(formulasinthissec6oncomefromp.192ofthetextbook)
34
S Z Y ABCD
A
B
C
D
step1:giventheZparameters
step2:herearethecorrespondingA,B,C,andDparameters
Thisishowtousethetableinp.192ofthetextbook:
GiventhescaTeringmatrix,theconversionsare...
35
Totheimpedancematrix[Z]
TotheadmiTancematrix[Y]
Tothetransi6onmatrix(ABCD)
Y0 = 1/Z0keyresults
Giventheimpedancematrix,theconversionsare...
36
TothescaTeringmatrix[S]
TotheadmiTancematrix[Y]
Tothetransi6onmatrix(ABCD)
keyresults
GiventheadmiTancematrix,theconversionsare...
37
TothescaTeringmatrix[S]
Totheimpedancematrix[Z]
Tothetransi6onmatrix(ABCD)
keyresults
GiventheABCDmatrix,theconversionsare...
38
TothescaTeringmatrix[S]
Totheimpedancematrix[Z]
TotheadmiTancematrix[Y]
keyresults
WhatisthescaTeringparameterofthisnetwork?
39
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
40