RF Communication Circuits: S-Parameters and Impedance Matrices

RF Communication Circuits Lecture 3: S- Parameters Dr. Fotowat-Ahmadi Sharif University of Technology Fall-1391 Prepared by: Siavash Kananian & Alireza Zabetian Prepared by: Siavash Kananian & Alireza Zabetian Dr. Fotowat-Ahmadi Sharif University of Technology Fall-1391 KAVOSHCOM

Impedance and Admittance matrices For n ports network we can relate the voltages and currents by impedance and admittance matrices Impedance matrix Admittance matrix . . Y Y Y V . . I Z Z Z I 11 21 1 1 n V 1 11 21 1 1 n 1 . . Y Y Y V . . I Z Z Z I 12 22 2 2 n . V 2 12 22 2 2 n . 2 = . . . . . = . . . . . . . . . . . . . . . . . . . . . I . . Y Y Y V V n . . Z Z Z I 1 2 n n nn n n 1 2 n n nn n 1 = Z Y where

Reciprocal and Lossless Networks Reciprocal networks usually contain nonreciprocal media such as ferrites or plasma, or active devices. We can show that the impedance and admittance matrices are symmetrical, so that. = = Z Z or Y Y ij ji ij ji Lossless networks can be shown that Zij or Yij are imaginary Refer to text book Pozar pg193-195

Example Find the Z parameters of the two-port T network as shown below ZA ZB I1 I2 V1 V2 ZC Solution Similarly we can show that Port 2 open-circuited V V 1 2 = = + = = Z Z Z Z Z 11 21 A C C I I 1 1 =0 =0 I I 2 2 This is an example of reciprocal network!! Port 1 open-circuited Z Z V V V ( ) 1 2 C C = = = + = 2 = = + Z Z Z Z Z Z Z 12 B C C 22 B C + + I I Z Z Z Z I 2 2 B C B C = 0 2 I =0 I 1 1

S-parameters Port 1 Port 2 Microwave device Vi2 Vi1 Vr1 Input signal reflected signal transmitted signal Vr2 Vt1 Vt2 Transmission and reflection coefficients V V t r = = V V i i

S-parameters Voltage of traveling wave away from port 1 is V V 2 1 t r = + V V V 1 1 i 2 b i V V 1 i 2 i Voltage of Reflected wave From port 1 Voltage of Transmitted wave From port 2 Voltage of transmitted wave away from port 2 is V V 1 2 t r = + V V V 2 1 i 2 b i V V 1 i 2 i V V V 1 t = 2 1 t r = , = , 21 Let Vb1= b1 , Vi1=a1 , Vi2=a2 , 1 12 V V V 1 i 1 i 2 i V 2 r = and 2 Then we can rewrite V 2 i

S-parameters = + S11and S22 are a measure of reflected signal at port 1 and port 2 respectively S21 is a measure of gain or loss of a signal from port 1 to port 2. S12 ia a measure of gain or loss of a signal from port 2 to port 1. b a a 1 1 1 12 2 Hence = + b a a 2 21 1 2 2 b a 1 1 12 1 = In matrix form b a 2 21 2 2 Logarithmic form S11=20 log( ) S22=20 log( ) S12=20 log( 12) S21=20 log( 21) b S S a 1 11 12 1 = S-matrix b S S a 2 21 22 2

S-parameters V 2 t = S 12 V V = V 1 S r 2 i 11 2= 0 V r 1 i = 0 V 2 r Vr2=0 means port 2 is matched V V 1 2 t r = = S S 21 22 V V 1 i 2 i 1= 1= 0 0 V V r r Vr1=0 means port 1 is matched

Multi-port network Port 5 Port 1 network Port 4 S S S S S a b 11 12 13 14 15 1 1 S S S S S a b 21 22 23 24 25 2 2 = S S S S S a b 31 32 33 34 35 3 3 b S S S S S a 4 41 42 43 44 45 4 b S S S S S a 5 51 52 53 54 55 5

Example Below is a matched 3 dB attenuator. Find the S-parameter of the circuit. 8.56 8.56 Z1=Z2= 8.56 and Z3= 141.8 141.8 Solution Z Z V 1 in o r = = = S 11 + V Z Z 1 i in o = 0 V 2 r By assuming the output port is terminated by Zo = 50 , then + + = 56 . 8 ( 8 . 141 56 . 8 ) o = + + //( Z Z Z Z Z 1 3 2 in + + = 50 ) 141 /( 8 . . 8 56 50 ) 50 50 50 = = 0 S 11 Because of symmetry , then S22=0 + 50 50

Continue 8.56 8.56 V V1 2 t = S V2 141.8 Vo 21 V 2 i 2= 0 V r From the fact that S11=S22=0 , we know that Vr1=0 when port 2 is matched, and that Vi2=0. Therefore Vi1= V1 and Vt2=V2 // Z Z Z + Z + Z + 2 3 o o = = = V V V V 2 2 1 t o // Z Z Z Z Z Z 2 44 . + 3 1 3 3 o o 41 50 + = = . 0 707 V V 1 1 41 44 . . 8 56 50 . 8 56 0 . 0 707 S = . 0 Therefore S12 = S21 = 0.707 707 0

Lossless network For lossless n-network , total input power = total output power. Thus n n = i = i * * = Where a and b are the amplitude of the signal a a b b i i i i 1 1 Putting in matrix form at a* = bt b* Note that bt=atSt and b*=S*a* =at St S* a* Called unitary matrix Thus at (I St S* )a* =0 This implies that St S* =I In summation form = n 1 for i j = k * = S S ki kj 0 for i j 1

Conversion of Z to S and S to Z S ( Z + ( Z ) ) 1 = U U Z ( U ( U + ) ) 1 = S S 1 0 . 0 where 0 . . . U = . . 1 . 0 . . 1

Reciprocal and symmetrical network t U = U Since the [U] is diagonal , thus For reciprocal network t Z = Since [Z] is symmetry Z Thus it can be shown that t S = S

Example A certain two-port network is measured and the following scattering matrix is obtained: 90 8 . 0 o o 1 . 0 0 8 . 0 90 = S o o 2 . 0 0 From the data , determine whether the network is reciprocal or lossless. If a short circuit is placed on port 2, what will be the resulting return loss at port 1? Solution Since [S] is symmetry, the network is reciprocal. To be lossless, the S parameters must satisfy For i=j = n 1 for i j = k |S11|2 + |S12|2 = (0.1)2 + (0.8)2 = 0.65 * = S S ki kj 0 for i j 1 Since the summation is not equal to 1, thus it is not a lossless network.

continue Reflected power at port 1 when port 2 is shorted can be calculated as follow and the fact that a2= -b2 for port 2 being short circuited, thus (1) b1=S11a1 + S12a2 = S11a1 - S12b2 Short at port 2 a2 -a2=b2 (2) b2=S21a1 + S22a2 = S21a1 - S22b2 From (2) we have (3) S + 21 S = b a 2 1 1 22 Dividing (1) by a1 and substitute the result in (3) ,we have ( )( 2 . 0 + ) 8 . 0 j 8 . 0 j b b S S 1 2 12 + 21 = = = = = 1 . 0 . 0 633 S S S 11 12 11 1 1 a a S 1 1 22 ( . 0 ) = = Return loss 20 log 20 log 633 . 3 97 dB

ABCD parameters I1 I2 V2 V1 Network Voltages and currents in a general circuit In matrix form I V V V I I 2 2 1 2 1 2 V V A B 1 2 I This can be written as = I C D 1 2 + V V I I V I 1 2 2 1 2 2 Given V1 and I1, V2 and I2 can be determined if ABDC matrix is known. Or = I CV DI = V AV BI 1 2 2 1 2 2 A ve sign is included in the definition of D

Cascaded network I2a I1a I1b I2b V1a b V2a V1b V2b a V A B V V A B V 1 2 1 2 b b b b a a a a = = I C D I I C D I 1 2 1 2 b b b b a a a a However V2a=V1b and I2a=I1b then The main use of ABCD matrices are for chaining circuit elements together V A B A B V 1 2 a a a b b b = I C D C D I 1 2 a a a b b b Or just convert to one matrix Where V V A B A B A B A B 1 2 a a b b a b = = C D C D I I C D C D 1 2 a a b b a b

Determination of ABCD parameters BI AV V = I = CV DI 1 2 2 1 2 2 Because A is independent of B, to determine A put I2 equal to zero and determine the voltage gain V1/V2=A of the circuit. In this case port 2 must be open circuit. V A for port 2 open circuit V 1 I = B 1 = for port 2 short circuit V 2 2= 0 V 2 2= 0 I I I for port 2 short circuit 1 I = D 1 = C for port 2 open circuit V 2 2= 0 V 2 2= 0 I

ABCD matrix for series impedance I1 I2 Z V2 V1 V V 1 1 I = = A B for port 2 open circuit for port 2 short circuit V 2 2 2= 2= 0 0 I V V1= - I2 Z hence B= Z V1= V2 hence A=1 I I 1 I = D for port 2 short circuit 1 = C for port 2 open circuit V 2 2= 0 V 2 2= 0 I I1 = - I2 = 0 hence C= 0 I1 = - I2 hence D= 1 1 Z The full ABCD matrix can be written 0 1

ABCD for T impedance network I1 I2 Z1 Z2 V1 Z3 V2 V 1 = A for port 2 open circuit V 2 therefore 2= 0 I Z + + Z Z V Z 3 then = V V 1 3 1 1 = = = + 1 A 2 1 Z Z V Z Z 1 3 2 3 3

Continue V 1 I Z1 = B for port 2 short circuit I2 2 2= 0 V VZ2 Z2 Solving for voltage in Z2 Z3 Z Z 2 + 3 Z Z 2 3 Z = VZ V 1 Z 2 2 + 3 Z + Z 1 Z 2 3 Hence V Z Z But 1 I 1 Z 2 = = + + B Z Z 2 1 = VZ I Z 2 3 2 2 2

Continue Z1 I1 I2 I 1 = C for port 2 open circuit V 2 2= 0 I V2 Z3 Analysis I = I 2 1 Therefore 1 I = = V I Z I Z 1 = = C 2 2 3 1 3 V Z 2 3

Continue I 1 I = D Z1 for port 2 short circuit I2 I1 2 2= 0 V VZ2 Z2 Z3 I1 is divided into Z2 and Z3, thus Z + 3 Z = I I 2 1 Z Full matrix 2 3 Z Z Z Hence 1 1 Z 2 + + + 1 Z Z 1 2 Z 2 3 I Z 1 I 2 = = + 1 D 1 Z Z 2 + 1 2 3 Z Z 3 3

ABCD for transmission line I1 I2 V1 V2 Input Transmission line Zo z =0 z = - j t z j t z + V e e V e e f b For transmission line j t z j t z = + ( ) V z V e e V e e V V f f b b = = Z o I I f b ( V ) 1 j t z j t z = ( ) I z e e V e e f b Z o f and b represent forward and backward propagation voltage and current Amplitudes. The time varying term can be dropped in further analysis.

continue At the input z = - ( V ) 1 (2) = = + ( ) I I e V e (1) = = + ( ) V V V e V e 1 f b 1 f b Z o At the output z = 0 ( ) 1 (4) = = ) 0 ( I I V V = = + (3) ) 0 ( V V V V 2 f b Z 2 f b o Now find A,B,C and D using the above 4 equations V 1 = A for port 2 open circuit V 2 2= 0 I For I2 =0 Eq.( 4 ) gives Vf= Vb=Vo giving

continue Note that From Eq. (1) and (3) we have x x + ( ) e e + = ( ) V e e cosh( ) x o = = cosh( ) A 2 2 V o x x ( ) e e = sinh( ) x 2 V 1 I = B for port 2 short circuit 2 2= 0 V For V2 = 0 , Eq. (3) implies Vf= Vb = Vo . From Eq. (1) and (4) we have ( ) Z V e e o o = = sinh( ) B Z o 2 V o

continue I 1 = C for port 2 open circuit V 2 2= 0 I For I2=0 , Eq. (4) implies Vf = Vb = Vo . From Eq.(2) and (3) we have ( ) sinh( ) V e e o = = C 2 V Z Z o o o I 1 I = D for port 2 short circuit 2 2= 0 V For V2=0 , Eq. (3) implies Vf = -Vb = Vo . From Eq.(2) and (4) we have + ( ) Z V e e o o = = cosh( ) D 2 Z V o o

continue Note that The complete matrix is therefore cosh( ) sinh( ) Z = + jk o sinh( ) cosh( ) Where = attenuation k=wave propagation constant Z o When the transmission line is lossless this reduces to Lossless line = 0 cos( ) sin( ) k jZ k o = cosh( ) cos( ) jk k sin( ) k cos( ) j k = sinh( ) sin( ) jk j k Z o

Table of ABCD network cosh( ) sinh( ) Z o Transmission line sinh( ) cosh( ) Z o Z 1 Z Series impedance 0 1 1 0 Shunt impedance Z 1 Z 1

Table of ABCD network Z Z Z Z1 1 1 Z 2 + + + Z2 1 Z Z 1 2 Z 2 3 1 Z T-network 2 + Z3 1 Z Z 3 3 Z3 Z 3 + 1 Z 3 Z Z1 2 Z2 -network 1 Z 1 Z Z 3 Z 3 + + + 1 Z Z Z 1 2 1 2 1 0 n 0 1 Ideal transformer n n:1

Short transmission line cos( ) sin( ) k jZ k o sin( ) k = ABCD Lossless transmission line cos( ) j k tline Z o If << then cos(k ) ~ 1 and sin (k ) ~ k then 1 jZ k o 1 = ABCD 1 j k tlineshort Z o

Embedded short transmission line Transmission line Z1 Z1 1 jZ k 1 0 1 0 o 1 Z 1 Z 1 = ABCD 1 1 1 j k embed Z 1 1 o Solving, we have jZ k o + 1 jZ k o Z 1 = ABCD 2 Z jZ k jZ k k embed o o + + + 1 j 2 Z Z Z 1 1 o 1

Comparison with -network Z 3 + 1 Z 3 Z 2 = ABCD net 1 Z 1 Z Z 3 Z 3 + + + 1 Z Z Z 1 2 1 2 1 jZ k o + 1 jZ k o Z 1 = ABCD 2 Z jZ k jZ k k embed o o + + + 1 j 2 Z Z Z 1 1 o 1 It is interesting to note that if we substitute in ABCD matrix in -network, Z2=Z1 and Z3=jZok we see that the difference is in C element where we have extra term i.e Z k Z 2 1 k j o So the transmission line exhibit a -network k o Both are almost same if Z Z o

Comparison with series and shunt Series = Z jZ k If Zo >> Z1 then the series impedance o Z o = L This is an inductance which is given by c Where c is a velocity of light Shunt k = Z j If Zo << Z1 then the series impedance Z o = C This is a capacitance which is given by Z c o

Equivalent circuits ZoL Zo Zo Z o = L Zo >> Z1 c Zoc Zo Zo Zo << Z1 = C Z c o

Transmission line parameters It is interesting that the characteristic impedance and propagation constant of a transmission line can be determined from ABCD matrix as follows B Zo= C 1 1 ( ) A 1 2 = = cosh ln 1 A A

Conversion S to ABCD For conversion of ABCD to S-parameter ( B ) 2 S S + 2 1 2 Z A B Z C Z D Z AD BC 11 12 o o o o = 2 2 S S + + + + + Z Z A Z C Z D Z A B Z C Z D 21 22 o o o o o o o For conversion of S to ABCD-parameter ( )( ) ( ( )( ) ) + + + + 1 1 1 1 S S S S Z S S S S 11 S 22 S 12 S 21 S 11 22 12 S 21 o A B 1 S 1 = ( ( )( ) ) ( )( ) + + 1 1 1 1 S S S C D 2 11 22 12 21 11 22 12 21 21 Z o Zo is a characteristic impedance of the transmission line connected to the ABCD network, usually 50 ohm.

MathCAD functions for conversion For conversion of ABCD to S-parameter ( + ) . + . Z . . Z . . . 2 . . . Z 1 , 1 A 2 , 1 A Z Z 1 , 2 A Z 2 , 2 A Z 1 , 1 A 2 , 2 A 2 , 1 A 1 , 2 A 1 = ( ) S A + + + + . 2 . . Z 1 , 1 A 2 , 1 A Z 1 , 2 A Z 2 , 2 A . . . . Z 1 , 1 A 2 , 1 A Z Z 1 . 2 A Z 2 , 2 A For conversion of S to ABCD-parameter ( 1 )( 1 . ) ( ( 1 )( 1 . ) ) + + + + . . . 1 , 1 S 2 , 2 S 2 , 1 S 1 , 2 S Z 1 , 1 S 2 , 2 S 2 , 1 S 1 , 2 S 1 = ( ( 1 )( 1 . ) ) ( 1 )( 1 . ) ( ) . A S 1 + + . . . 1 , 1 S 2 , 2 S 2 , 1 S 1 , 2 S 1 , 1 S 2 , 2 S 2 , 1 S 1 , 2 S . 2 1 , 2 S Z o

Odd and Even Mode Analysis Usually use for analyzing a symmetrical four port network Equal ,in phase excitation even mode Equal ,out of phase excitation odd mode (1) Excitation (2) Draw intersection line for symmetry and apply short circuit for odd mode Open circuit for even mode (3) Also can apply EM analysis of structure Tangential E field zero odd mode Tangential H field zero even mode (4) Single excitation at one port= even mode + odd mode

Example 1 1 2 Edge coupled line Line of symmetry 4 3 The matrix contains the odd and even parts + + S S S S S S S S 11 11 12 12 13 13 14 14 ev od ev od ev od ev od + + S S S S S S S S 1 S 21 21 22 22 23 23 24 24 ev od ev od ev od ev od = + + S S S S S S S S 2 31 31 32 32 33 33 34 34 ev od ev od ev od ev od + + S S S S S S S S 41 41 42 42 43 43 44 44 ev od ev od ev od ev od Since the network is symmetry, Instead of 4 ports , we can only analyze 2 port

continue We just analyze for 2 transmission lines with characteristic Ze and Zo respectively. Similarly the propagation coefficients e and o respectively. Treat the odd and even mode lines as uniform lossless lines. Taking ABCD matrix for a line , length l, characteristic impedance Z and propagation constant ,thus cos( ) sin( ) jZ = Z sin( ) ABCDtline cos( ) j Using conversion ( B ) 2 + 2 1 2 Z A B Z C Z D Z AD BC S o o o o = 2 2 + + + + + Z Z A Z C Z D Z A B Z C Z D o o o o o o o

continue 2 2 Z Z Z o sin 2 j Z o 1 S = 2 2 2 2 Z + Z Z Z Z o o + 2 sin Z j 2 cos sin Z j o 2 Z = Taking (equivalent to quarter-wavelength transmission line) 2 Then 2 2 o 2 Z 1 2 Z Z j ZZ S o = 2 2 o 2 2 o + j ZZ Z Z Z o

continue S13 S14 S23 S24 S11 S12 S11 S12 Odd + even Convert to S21 S22 S21 S22 S34 S11 S12 S11 S12 S11 S12 S21 S22 S33 S21 S22 S21 S22 S31 S44 S41 S42 S32 S43 4-port network matrix 2-port network matrix

continue Follow symmetrical properties S11 S12 S13 S14 ev- od ev+ od S21 S22 S23 S24 S31 S32 S33 S34 ev- od S41 S42 S43 S44 ev+ od Assuming ev = od = Then 2 jZ Z Z o ev od = = = = S S S S 41 14 32 23 2 ev ( 2 o 2 od Z 2 o 2 + + ) Z Z Z Z Z Z 2 o Z Z ( ) jZ Z ev Z o od od ev = 2 ev 2 o 2 od 2 o 2 + + ( ) ( ) Z Z For perfect isolation (I.e S41=S14=S32=S23=0 ),we choose Zev and Zod such that Zev Zod=Zo2.

continue S11 S12 S13 S14 ev- od ev+ od S21 S22 S23 S24 S31 S32 S33 S34 ev- od S41 S42 S43 S44 ev+ od Similarly we have = 22 11 S S 2 ev 2 o 2 od 2 o 1 Z Z Z Z Z = = = + S S 33 44 2 ev Z 2 o 2 od 2 o 2 + 2 od + Z 2 ev Z Z Z Z Z 4 o 1 Z = 2 ev 2 o 2 od 2 o 2 + + ( )( ) Z Z Equal to zero if Zev Zod=Zo2.

continue S11 S12 S13 S14 ev- od ev+ od S21 S22 S23 S24 S31 S32 S33 S34 ev- od S41 S42 S43 S44 ev+ od We have 2 ev 2 o 2 od 2 o 1 Z Z Z Z = = = = S S S S 31 13 24 42 2 ev ( 2 ev 2 o 2 od ) 2 od 2 o 2 + + Z Z Z Z 2 ev Z 2 od Z 2 o 2 o Z Z Z = 2 o + + ( )( ) Z Z Z Z ev od = + if Zev Zod=Zo2. Z Z ev od

continue S11 S12 S13 S14 ev- od ev+ od S21 S22 S23 S24 S31 S32 S33 S34 ev- od S41 S42 S43 S44 ev+ od jZ Z Z o ev od = = = = + S S S S 21 12 34 43 2 ev 2 o 2 od 2 o 2 + + Z Z Z Z 1 = jZ if Zev Zod=Zo2. o + Z Z ev od

continue This S-parameter must satisfy network characteristic: (1) Power conservation 2 2 2 2 + + + = 1 S S S S 11 21 31 41 Reflected power transmitted power to port 2 transmitted power to port 3 transmitted power to port 4 Since S11 and S41=0 , then 2 21 + S S 2 = 1 31 S 11 = Arg (2) And quadrature condition 2 S 21

continue For 3 dB coupler Z Z 1 2 ev od = Z Z 1 ev od or = 2 + Z Z ev od 2 + Z Z ev od Rewrite we have 1 od Z + ( 2 ) Z ev = = 3 2 2 1 ( 2 ) Z ev = + = 3 2 2 . 5 83 In practice Zev > Zod so Z od However the limitation for coupled edge Z (Gap size ) also ev ev and od are not pure TEM thus not equal 2 Z od