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KAVOSHCOM RF Communication Circuits

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Impedance and Admittance matrices Impedance matrix Admittance matrix For n ports network we can relate the voltages and currents by impedance and admittance matrices where

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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. Lossless networks can be shown that Z ij or Y ij are imaginary Refer to text book Pozar pg193-195

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Example Find the Z parameters of the two-port T –network as shown below Solution V1V1 V2V2 I1I1 I2I2 Port 2 open-circuited Port 1 open-circuited Similarly we can show that This is an example of reciprocal network!!

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S-parameters Microwave device Port 1 Port 2 V i1 V r1 V t2 V i2 V r2 V t1 Transmission and reflection coefficients Input signal reflected signal transmitted signal

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S-parameters Voltage of traveling wave away from port 1 is Voltage of Reflected wave From port 1 Voltage of Transmitted wave From port 2 Voltage of transmitted wave away from port 2 is Let V b1 = b 1, V i1 =a 1, V i2 =a 2, Then we can rewrite

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S-parameters Hence In matrix form S-matrix S 11 and S 22 are a measure of reflected signal at port 1 and port 2 respectively S 21 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. Logarithmic form S 11 =20 log( ) S 22 =20 log( ) S 12 =20 log( 12 ) S 21 =20 log( 21 )

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S-parameters V r2 =0 means port 2 is matched V r1 =0 means port 1 is matched

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Multi-port network network Port 1 Port 2 Port 3 Port 4 Port 5

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Example Below is a matched 3 dB attenuator. Find the S-parameter of the circuit. Solution Z 1 =Z 2 = 8.56 and Z 3 = 141.8 By assuming the output port is terminated by Z o = 50 , then Because of symmetry, then S 22 =0

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Continue From the fact that S 11 =S 22 =0, we know that V r1 =0 when port 2 is matched, and that V i2 =0. Therefore V i1 = V 1 and V t2 =V2 V1V1 V2V2 Therefore S 12 = S 21 = 0.707 VoVo

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Lossless network For lossless n-network, total input power = total output power. Thus Where a and b are the amplitude of the signal Putting in matrix form a t a * = b t b * =a t S t S * a * Thus a t (I – S t S * )a * =0This implies that S t S * =I Note that b t =a t S t and b * =S * a * In summation form Called unitary matrix

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Conversion of Z to S and S to Z where

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

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Example A certain two-port network is measured and the following scattering matrix is obtained: 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 |S 11 | 2 + |S 12 | 2 = (0.1) 2 + (0.8) 2 = 0.65 Since the summation is not equal to 1, thus it is not a lossless network. For i=j

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continue Reflected power at port 1 when port 2 is shorted can be calculated as follow and the fact that a 2 = -b 2 for port 2 being short circuited, thus b 1 =S 11 a 1 + S 12 a 2 = S 11 a 1 - S 12 b 2 b 2 =S 21 a 1 + S 22 a 2 = S 21 a 1 - S 22 b 2 (1) (2) From (2) we have a2a2 -a 2 =b 2 Short at port 2 Dividing (1) by a 1 and substitute the result in (3),we have (3) Return loss

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ABCD parameters Network V1V1 V2V2 I1I1 I2I2 Voltages and currents in a general circuit This can be written as Or A –ve sign is included in the definition of D In matrix form Given V 1 and I 1, V 2 and I 2 can be determined if ABDC matrix is known.

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Cascaded network a b I 1a V 1a I 2a V 2a V 1b I 1b I 2b V 2b However V 2a =V 1b and –I 2a =I 1b then Or just convert to one matrixWhere The main use of ABCD matrices are for chaining circuit elements together

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Determination of ABCD parameters Because A is independent of B, to determine A put I 2 equal to zero and determine the voltage gain V1/V2=A of the circuit. In this case port 2 must be open circuit. for port 2 open circuitfor port 2 short circuit for port 2 open circuit for port 2 short circuit

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ABCD matrix for series impedance Z I1I1 I2I2 V1V1 V2V2 for port 2 open circuitfor port 2 short circuit for port 2 open circuit for port 2 short circuit V 1 = V 2 hence A=1 V 1 = - I 2 Z hence B= Z I 1 = - I 2 = 0 hence C= 0I 1 = - I 2 hence D= 1 The full ABCD matrix can be written

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ABCD for T impedance network Z1Z1 Z2Z2 Z3Z3 V1V1 I1I1 I2I2 V2V2 for port 2 open circuit then therefore

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Continue for port 2 short circuit Solving for voltage in Z 2 But Hence I2I2 Z1Z1 Z3Z3 Z2Z2 VZ2VZ2

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Continue for port 2 open circuit I1I1 I2I2 Z1Z1 Z3Z3 V2V2 Therefore Analysis

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Continue for port 2 short circuit I2I2 Z1Z1 Z3Z3 Z2Z2 VZ2VZ2 I1I1 I 1 is divided into Z 2 and Z 3, thus Hence Full matrix

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ABCD for transmission line Input V1V1 I1I1 V2V2 I2I2 ZoZo Transmission line z =0 z = - For transmission line f and b represent forward and backward propagation voltage and current Amplitudes. The time varying term can be dropped in further analysis.

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

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continue From Eq. (1) and (3) we have Note that for port 2 short circuit For V 2 = 0, Eq. (3) implies –V f = V b = V o. From Eq. (1) and (4) we have

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continue for port 2 open circuit For I 2 =0, Eq. (4) implies V f = V b = V o. From Eq.(2) and (3) we have for port 2 short circuit For V 2 =0, Eq. (3) implies V f = -V b = V o. From Eq.(2) and (4) we have

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continue The complete matrix is therefore When the transmission line is lossless this reduces to Note that Where = attenuation k=wave propagation constant Lossless line = 0

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Table of ABCD network Transmission line Series impedance Shunt impedance Z Z

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Table of ABCD network T-network -network Ideal transformer n:1 Z1Z1 Z2Z2 Z3Z3 Z3Z3 Z1Z1 Z2Z2

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Short transmission line Lossless transmission line If << then cos(k ) ~ 1 and sin (k ) ~ k then

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Embedded short transmission line Z1Z1 Z1Z1 Transmission line Solving, we have

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

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Comparison with series and shunt Series If Z o >> Z 1 then the series impedance This is an inductance which is given by Where c is a velocity of light Shunt If Z o << Z 1 then the series impedance This is a capacitance which is given by

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Equivalent circuits ZoZo ZoZo Z oc ZoZo ZoZo Z oL Z o >> Z 1 Z o << Z 1

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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

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Conversion S to ABCD For conversion of ABCD to S-parameter For conversion of S to ABCD-parameter Z o is a characteristic impedance of the transmission line connected to the ABCD network, usually 50 ohm.

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MathCAD functions for conversion For conversion of ABCD to S-parameter For conversion of S to ABCD-parameter

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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

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Example 1 The matrix contains the odd and even parts Since the network is symmetry, Instead of 4 ports, we can only analyze 2 port Edge coupled line

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continue We just analyze for 2 transmission lines with characteristic Z e and Z o 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 Using conversion

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continue Taking Then (equivalent to quarter-wavelength transmission line)

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continue S 11 S 21 S 12 S 22 2-port network matrix Convert to S 11 S 21 S 12 S 22 S 11 S 21 S 12 S 22 S 11 S 21 S 12 S 22 S 11 S 21 S 12 S 22 S 33 S 44 S 34 S 43 S 13 S 14 S 23 S 24 S 32 S 31 S 42 S 41 4-port network matrix Odd + even

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continue Assuming ev = od = Then For perfect isolation (I.e S 41 =S 14 =S 32 =S 23 =0 ),we choose Z ev and Z od such that Z ev Z od =Z o 2. S 11 S 21 S 12 S 22 S 13 S 23 S 14 S 24 S 31 S 41 S 32 S 42 S 33 S 43 S 34 S 44 ev+ od ev- od Follow symmetrical properties

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continue Similarly we have Equal to zero if Z ev Z od =Z o 2. S 11 S 21 S 12 S 22 S 13 S 23 S 14 S 24 S 31 S 41 S 32 S 42 S 33 S 43 S 34 S 44 ev+ od ev- od

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continue We have if Z ev Z od =Z o 2. S 11 S 21 S 12 S 22 S 13 S 23 S 14 S 24 S 31 S 41 S 32 S 42 S 33 S 43 S 34 S 44 ev+ od ev- od

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continue if Z ev Z od =Z o 2. S 11 S 21 S 12 S 22 S 13 S 23 S 14 S 24 S 31 S 41 S 32 S 42 S 33 S 43 S 34 S 44 ev+ od ev- od

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

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continue For 3 dB coupler or Rewrite we have In practice Z ev > Z od so However the limitation for coupled edge (Gap size ) also ev and od are not pure TEM thus not equal

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A /4 branch line coupler Symmetrical line Odd Even

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Analysis Stub odd (short circuit) Stub even (open circuit) The ABCD matrices for the two networks may then found : stub Transmission line

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continue Convert to S For perfect isolation we require Thus or From previous definition

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continue Substituting into S-parameter gives us and Therefore for full four port And

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continue For power conservation and quadrature conditions to be met Equal split S or And If Z o = 50 then Z 2 = 35.4

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