5/14/20151Chapter 3Microwave EngineeringMicrowave Network Analysis1Multiport NetworksA general circuit can be representedby a multi-port network, where the ports are defined as access terminals at which we can define voltages and currents.Examples:1)One-port network2) Two-port network2Note: Equal and opposite currents are assumed on the two wires of a port. 1 I1 V1 I1 V1 I 1 I2 I2 I1 V 1 V2 V2 V3) N-port NetworkTo represent multi-port networks we use:- Z (impedance) parameters- Y (admittance)parameters- h (hybrid) parameters- ABCD parameters- S (scattering) parametersNot easily measurable at high frequencyMeasurable at high frequencyMultiport Networks (cont.)3+++++1I2I3ImINI1V2V3VmVNV-----5/14/20152Poynting Theorem (Phasor Domain)4( )( ) 2s f d m eP P j e = + + P W W( )*2 22 2121 12 21 124 4iV ScVcVE J dV S n dSE H dVj H E dVec ee c = | |'' '' + + |\ .| |' ' + |\ .} }}}

The last term is the VARS consumed by the region.The notation < > denotes time-average.Consider a general one-port networkComplex power delivered to network:Self Impedance( ) ( )*1 112122 iSm e n dP P E H n j W d WV Is e + == =} Average power dissipated in [W]Average magnetic energy (in [J]) stored inside VAverage electric energy (in [J]) stored inside V, E H nSV1V1I+-5d dP = Pm mW = We eW = W( )**1 11 122 2121221 11111121 124 (221)22diinin inminenim endR jXPRIV IW WXV I PII IP j W WVZIIIee== =+====+ Define Self Impedance (Zin), E H nSV1V1I+-65/14/20153We can show that for physically realizable networks the following apply:( ) ( )( ) ( )( ) ( )( )( )*1 1*1*1inin ininZ ZV VIXIRe eeee eee ee ===is an even function ofis an odd function ofSelf Impedance (cont.)Please see the Pozarbook for a proof.( ) ( ) ( )in in inZ R jX e e e = +Note: Frequencyis usually defined as a positive quantity. However, we consider the analytic continuation of the functions into the complex frequency plane. 7|| ||| |1 11 1 12 22 21 1 21 11 12 12 21 22 22 2V Z I Z IV Z Z IV Z Z IV Z IV Z I Z I (= += + (( = ((( =Consider a general 2-port linear network:In terms of Z-parameters,we have (from superposition)Impedance(Z) matrixTwo-Port Networks1 2+-+-1V1I2I2V8Elements of Z-Matrix: Z-Parameters(open-circuit parameters)0kiijj I k jVZI= ==111122 02222 0IIVZIVZI====Port 2 open circuited Port 1 open circuited221111 02211 0IIVZIVZI====1 2+-+-1V1I 2I2V91 11 1 12 22 21 1 22 2V Z I Z IV Z I Z I= += +5/14/20154Z-Parameters (cont.)100 kiijj I k jVZI= ==+jIiV -We inject a current into port j and measure the voltage(with an ideal voltmeter) at port i. All ports are open-circuited except j. N-port networkZ-parametersare convenient for series connected networks.1 1 12 2 2A BA BI I II I I = == =SeriesZ-Parameters (cont.)1AI1BI 2BI2AI2AV2BV1AV1BV1V 2V2I1I11( )| |( )1 1 12 2 21212A BA BA A B BA BA BA BV V VV V VZ I Z IZ Z IIZ ZIIZ ZI ((( = + ((( (((( = + (( = + ( (( = + ( ( ( = + ( 1 1 11 11 12 122 2 21 21 22 22A B A BA B A BV I Z Z Z ZV I Z Z Z Z ( + +((=(((+ + | | ||||1 11 12 12 21 202 2kiijj V k jI Y Y VI Y VIYY Y VVI= = (((= = ((( = Admittance (Y) ParametersConsider a 2-port network:AdmittancematrixShort-circuit parameters1 21V1I+-2I2V+-121 11 1 12 22 21 1 22 2I Y V Y VI Y V Y V= += +or5/14/20155Y-Parameters (cont.)130 kiijj V k jIYV= ==j ViI+-We apply a voltage across port j and measure the current (with an ideal current meter) at port i. All ports are short-circuited except j.N-port network1 1 12 2 21 11 11 12 122 21 21 22 22A BA BA B A BA B A BI I II I IV Y Y Y YV Y Y Y Y ((( = + ((( ( + +(=((+ + 1 1 12 2 2A BA BV V VV V V = == =Parallel1AI1AV1I1V1BV2V2I2BI2AI2BV2AV1BIAdmittance (Y) ParametersY-parametersare convenient for parallel connected networks.14|| ||1Y Z=Admittance (Y) ParametersRelation between [Z] and [Y] matrices:|| ||| | V Z I =| | |||| I Y V =|| || |||| ( )|||| ( )||V Z Y VZ Y V==|||| | | Z Y U = = IdentityMatrixHenceTherefore155/14/20156Reciprocal NetworksIf a network does not contain non-reciprocal devices or materials* (i.e. ferrites, or active devices), then the network is reciprocal.( )|| ||ij ji ij jiZ Z Y YZ Y = = and are symmetric* A reciprocal material is one that has reciprocal permittivity and permeability tensors. A reciprocal device is one that is made from reciprocal materialsNote: The inverse of a symmetric matrix is symmetric.16Example of a nonreciprocal material: a biased ferrite (This is very useful for making isolators and circulators.)Reciprocal MaterialsD EB c= = x xx xy xz xy yx yy yz yz zx zy zz zD ED ED Ec c cc c cc c c ((( (((=((( ((( x xx xy xz xy yx yy yz yz zx zy zz zBBB ((( (((=((( ((( Ferrite:0000 0 1jjo o ( (= ( ( is not symmetric!Reciprocal: ,ij ji ij jic c = =17We can show that the equivalent circuits for reciprocal 2-port networks are:T-equivalentPi-equivalent11 21Z Z 22 21Z Z 21Z21Y 11 21Y Y +22 21Y Y +Reciprocal Networks (cont.)185/14/20157ABCD-ParametersThere are defined only for 2-port networks.1 2'1 2V V A BI I C D (((= ((( ' '2 22 21 12 2 0 01 1' '2 2 0 0I IV VV IA CV VV IB DI I= == == == =1 2+ +1I1V2V'2I '2 2I I =19Cascaded Networks''2 1 11 1 21122A AAA ABABBA BBV V VABCDI I IVABCDIVABCD ABCDI ((( ( = =((( ( ( ( = ( ( (( =( ( 1 2'1 2ABV VABCDI I (( ( = (( 1 2 1 21AI'2AI1AV2AV+ + +1BV 2BV1I'2I1V2V+ +1BI'2BI+ 20A nice property of the ABCD matrix is that it is easy to use with cascaded networks: you simply multiply the ABCD matrices together. At high frequencies, Z, Y, h & ABCD parametersare difficult (if not impossible) to measure.o V and I are not uniquely definedo Even if defined, V and I are extremely difficult to measure (particularly I).o Required open and short-circuit conditions are often difficult to achieve. Scattering (S) parametersare often the bestrepresentationfor multi-port networks at high frequency. Scattering Parameters215/14/20158S-parametersare definedassuming transmission lines are connectedto each port.On each transmission line:( ) ( ) ( )( )( ) ( )0 00 0ii iiz zi i i ii i i iiii i iiiiV z Vz Ve V eV z V zI zZ ZV z + + + + = +==+1, 2 i =1 21z2z1a1b2b2a01 1, Z 02 2, Z Scattering Parameters (cont.)22Local coordinates( ) ( )( ) ( )00i i i i ii i i i ia z V z Zb z V z Z+ Incoming wave functionOutgoing wave functionFor a One-Port Network ( )( )( )( )01011111110000LZZVVSba+I===1l1a1b01ZLI( ) ( )( )1 111 10 00Lb aSa = I=23( ) ( )( ) ( )00i i i i ii i i i ia z V z Zb z V z Z+ Incoming wave functionOutgoing wave functionFor a one-port network, S11is defined to be the same as IL.Scatteringmatrix( ) ( ) ( )( ) ( ) ( )( )( )( )( )| | | || |1 11 1 12 22 21 1 221 11 12 12 21 22 220 0 00 000000b S a S ab S a Sb S S ab S ab S Saa ((( = = ((( = += +For a Two-Port Network 1 21z2z1a1b2b2a01 1, Z 02 2, Z 245/14/20159Scattering ParametersOutput is matchedInput is matchedinput reflection coef. w/ output matchedreverse transmission coef.w/ input matchedforward transmission coef.w/ outputmatchedoutput reflection coef.w/ input matched( ) ( ) ( )( ) ( ) ( )1 11 1 12 22 21 1 22 20 0 00 0 0b S a S ab S a S a= += +( )( )( )( )( )( )( )( )21211111011220221102222000000000aaaabSabSabSabSa========Output is matchedInput is matched25Scattering Parameters (cont.)26jaibFor a general multiport network:All ports except j are semi-infinite (or matched)( )( )000kiijja k jbSa= ==PortjPort iSemi-infiniteN-port networkScattering Parameters (cont.)2a2b3a3b1a1bIllustration of a three-port network275/14/201510For reciprocal networks, the S-matrix is symmetric.ij jiS S i j = =Scattering Parameters (cont.)Note: If all lines entering the network have the same characteristicimpedance, then( )( )000kiijjV k jVSV++= ==28Why are the wave functions (a and b) defined as they are? Note:( ) ( ) ( )( ) *2001 10 Re 0 02 2ii i iiVP V IZ++ + + ( = = ( ) ( )( ) ( )020 010 02i i ii ia V ZP a++= =Scattering Parameters (cont.)1 21z2z1a1b2b2a01 1, Z 02 2, Z (assuming lossless lines)29Similarly,( )( )( )202 010 0212ii iiVP bZ= =Also,Scattering Parameters (cont.)( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )2 222 2201 1021 120202iiiiiiiili i i i ili i i i ili i ili i iPVl a l a el V eV ll beP l b eVoo+ + +++ = = = = = =305/14/201511Example1a1b2a2b0Z 0ZZ2V1V + + 1z 2zFind the S parameters for a series impedance Z. Note that two different coordinate systems are being used here!31Example (cont.)( )( )( )( )( )( )02 21 1 0111 10000000 00 0inina ab V Z ZSa V Z ZZ Z ZZ Z Z+= == = =++ =+ +1102ZSZ Z=+By symmetry:32S11Calculation:1a1b 2b0Z 0ZZ2V1V + + inZ1z 2zSemi-infinite22 11S S =( )( )( )( )22221102100000aabSaVV=+===( ) ( )( ) ( )( ) ( )( ) ( ) ( )2 2 202 101 1 0 1102 2 1 0 1100 0 00 00 10 0 1a V VZV VZ ZV a Z SZV V a Z SZ Z= =| |= |+\ .= +| | = = + |+\ .Example (cont.)33S21Calculation:1a1b2b0Z 0ZZ2V1V + + inZ1z 2zSemi-infinite( ) ( )1 1 00 0 V a Z+=5/14/201512( ) ( )( )( )01 0 110211 00 0 0 0110 0 0 0 00 102 21 12 2Za Z SZ ZSa ZZ Z Z Z Z ZSZ Z Z Z Z Z Z Z Z Z| |+ |+\ .=| | | || | | | | | += + = + = || |||+ + + + +\ . \ .\ . \ . \ .021022ZSZ Z=+HenceExample (cont.)341a1b2b0Z 0ZZ2 V1V + + inZ1z 2zSemi-infinite12 21S S =Example0Z 0Z0 ,s sZ |Lz 2z1zFind the S parameters for a length L of transmission line. Note that three different coordinate systems are being used here!35Example (cont.)2220 02201110 0 1( )in ain a aZ ZSbZ ZSa== == = =+bysymmetry( )( )( )( )220 00 0 2 00 01 tantan 1ssj LL s sin s s j L as s Le Z jZ LZ Z ZZ jZ L e|||| =+ I += =+ I0 00 0sLsZ ZZ ZI =+0Z 0Z0 ,s sZ |Lz 2z1z( ) 10 V ( ) 20 VLI( ) Vz36Semi-infiniteS11Calculation:5/14/201513Example (cont.)( )( )( )( )0 00 00 011 220 00 00 0tantantantans sss ss sss sZ jZ LZ ZZ jZ LS SZ jZ LZ ZZ jZ L||||++= =+++20 0 0 11 22 00s in aZ Z Z Z S S== = = = Note: If0Z 0Z0 ,s sZ |Lz 2z1z( )10 V ( ) 20 VLI( ) VzHence37( )( )222 221101 00 000a abSaV ZV Z+= == =( ) ( ) ( )1 1 110 0 1 V V S+= +0 00 0sLsZ ZZ ZI =+Example (cont.)0Z 0Z0 ,s sZ |Lz 2z1z( )10 V ( ) 20 VLI( ) Vz( )( )1111001VVS+=+We now try to put the numerator of the S21equation in terms of V1 (0).Hence, for the denominator of the S21equation we have38S21Calculation:Semi-infinite( ) ( ) ( )( ) ( ) ( ) ( )( )( )( )211220 0 10010 1ss ss ss j z j zj L j LLj L j LLLVV z V e eV L V e eVVe e| || || |+ +++ + + = = + I =+ I= + IExample (cont.)0Z 0Z0 ,s sZ |Lz 2z1z( )10 V ( ) 20 VLI( ) Vz( ) ( ) ( ) ( ) ( )2 20 0 0 0 1LV V V V += = = + INext, use( )( )( ) ( )12 200 11s sL j L j LLVVe e| |+ = + I+ IHence, we have395/14/201514( )( )( ) ( )22 1121 2100 1 10 1ssj LLj LLaV S eSV e|| +=+ + I= =+ IExample (cont.)0Z 0Z0 ,s sZ |Lz 2z1z( )10 V ( ) 20 VLI( ) VzTherefore, we have( ) ( )1121 12 21 11ssj LLj LLS eS Se||+ + I= =+ Iby symmetry( )( ) 1111001VVS+=+( )( )( )( )12 200 11s sL j L j LLVVe e| |+ = + I+ Iso40Special cases:0 0 11 2221 12) 0, 0ss Lj La Z Z S SS S e| = = = I == =| |00ssj Lj LeSe|| (=( 20 11 22 0212)2 201 1sg gsgin aj Lb L LZ Z S Se S| t| t== = = = = == = | |0 11 0S(=( 0 ,s sZ |LExample (cont.)4102 011102 0101 0222 1102 01Z ZSZ ZZ ZS SZ Z=+= = +( )( )( )( )22222110210100000 0aabaVZVSZ+=== =ExampleFind the S parameters for a step-impedance discontinuity. 01Z 02Z425/14/201515( ) ( ) ( ) ( ) ( )2 222 2 1 1 110 000 0 0 0 1a aaV V V V S += === = = +01 0221 1201 022Z ZS SZ Z= =+Example (cont.)1V2V 01Z02ZBecause of continuity of the voltage across the junction, we have:( )( )( ) ( )( )2 2202 021 101 0101021110 1 00 0a aVZ ZV VZ ZVSS+ += =+= =+=02 011102 010202 011 12Z ZSZ ZZZ Z+ = ++=+( )0121 11021ZS SZ= +Henceso43Semi-infiniteS21Calculation:For reciprocal networks, the S-matrix is symmetric.For lossless networks, the S-matrix is unitary.Identity matrix| | | |TS S =| | | | | | | | | |* * T TS S S S U = =* *1 1( , )N NTik kj ki kj ijk ki j S S S S o= = = = Take element1 ;0;iji ji jo= = =Properties of the S MatrixEquivalently, | | | |* 1 TS S =Notation:| | | | | | * H TS S S = =44|| || 1S S = so|||| | ||||| | |AB UB A U==Note :IfthenN-port networkExample:||11 12 1321 22 2331 32 33S S SS S S SS S S ( (= ( ( * * *11 11 21 21 31 31* * *12 12 22 22 32 32* * *13 13 23 23 33 33* * *11 12 21 22 31 32* * *11 13 21 23 31 33* * *12 13 22 23 32 33111000S S S S S SS S S S S SS S S S S SS S S S S SS S S S S SS S S S S S + + =+ + =+ + =+ + =+ + =+ + =UnitaryProperties of the S Matrix (cont.) The column vectors form an orthogonal set.The rows also form orthogonal sets (see the note on the previous slide).455/14/201516Example5002 20 020 02j jjSjO ( ( (( ( = ( ( ( ( 50S ONot unitary Not lossless1) Find the input impedance looking into port 1 when ports 2 and 3 are terminated in 50 [O] loads.2) Find the input impedance looking into port 1 when port 2 is terminated in a 75 [O] load and port 3 is terminated in a50 [O] load.46(For example, column 2 doted with the conjugate of column three is not zero.11 11 110 50[ ]in inbS Za I = = = = O1 If ports 2 and 3 are terminated in 50 []:Example (cont.)(a2 = a3 = 0)20 I =2a2b3a3b30 I =50S O1a1b1 inI471 11 1 12 2 13 3b S a S a S a = + +22275 50 175 50 5ab I = = =+2) If port 2 is terminated in 75 [] and port 3 in 50 []:Example (cont.)215I =2a2b3a3b30 I =50S O1a1b1 inI485/14/201517( )3 1 21 11 12 131 1 1212 2 12 2 2111 15 10 2 2inb j jSa b aS S Sa a aS Sa| | | | | | | |= I = I = = |||\

. \ . \ .|\ . I = = + +2 2 2a b = I11144.55[1501]inininZ =| | + I =| |I\ .OExample (cont.)1 12 23 302 20 020 02a bb aj jjjb a ( ( (( ((( (( = (( ((( ( ( 493 22 1 21 22 231 1/a ab a S S Sa a| | | |= + + ||\ . \ .215I =2a2b3a3b30 I =50S O1a1b1 inITransfer (T) MatrixFor cascaded 2-port networks:||21 11 12 21 21 22 22a T T bb TbTTaa (((= ((( (= ( ||2221 2111 11 221221 211 SS STS S SSS S( ( ( = ( ( 1Aa2Aa2Ba1Ba1Bb2Bb2Ab 1AbT Matrix: | |2122 22212 121122 221 TT TST TTT T( ( ( = ( ( 50(Derivation omitted)1 21 2112 12 12 1111 2A AAA AAAAAA BAAA BBBBBBa bTb aaTbb aaababbTb aT ((= ((

(( ( = (( ( ( = ( ( ( = ( ( ( ( (( ButTransfer (T) Matrix (cont.)HenceThe T matrix of a cascaded set of networks is the product of the Tmatrices.ABT( 515/14/201518Conversion Between Parameters52Example53Derive Sijfrom the Z parameters.(The result is given inside row 1, column 2, of the previous table.) 11 21Z Z 22 21Z Z 21Z 0Z 0Z11S011 10inininZ ZSZ Z= I =+( ) ( ) ( )11 21 21 22 21 0||inZ Z Z Z Z Z Z = + +( S11Calculation:Example (cont.)54( ) ( ) ( )( )( )( ) ( ) ( )11 21 21 22 21 021 22 0 2111 2122 011 21 22 0 21 22 0 2122 0211 22 11 0 21 22 21 0 21 22 21 0 2122 0211 22 11 0 2122 0||inZ Z Z Z Z Z ZZ Z Z ZZ ZZ ZZ Z Z Z Z Z Z ZZ ZZ Z ZZ ZZ Z Z Z Z ZZ ZZ ZZ Z ZZ ZZ Z= + +( + = ++ + + + =++ + + =++ =+11 21 Z Z 22 21Z Z 21 Z 0Z 0Z11S5/14/201519Example (cont.)55( )211 0 22 2122 0inZ Z Z ZZZ Z+ =+11 21 Z Z 22 21Z Z 21 Z 0Z 0Z11S0110ininZ ZSZ Z=+( ) ( )( ) ( )211 0 22 21 0 0 2211 211 0 22 21 0 0 22Z Z Z Z Z Z ZSZ Z Z Z Z Z Z+ +=+ + +soExample (cont.)5611 21 Z Z 22 21Z Z 21 Z 0Z 0Z11S( ) ( )( ) ( )( ) ( )( ) ( )211 0 22 21 0 0 2211 211 0 22 21 0 0 222 211 0 11 22 21 0 0 222 211 0 11 22 21 0 0 2220 22 11 0 2120 22 11 0 21Z Z Z Z Z Z ZSZ Z Z Z Z Z ZZ Z Z Z Z Z Z ZZ Z Z Z Z Z Z ZZ Z Z Z ZZ Z Z Z Z+ +=+ + ++ =+ + ++ =+ + Example (cont.)5711 21 Z Z 22 21Z Z 21 Z 0Z 0Z11S( ) ( )( ) ( )20 22 11 0 2111 20 22 11 0 21Z Z Z Z ZSZ Z Z Z Z+ =+ + Note: to get S22, simply let Z11 Z22in the previous result. ( ) ( )( ) ( )20 11 22 0 2122 20 11 22 0 21Z Z Z Z ZSZ Z Z Z Z+ =+ + 5/14/201520Example (cont.)5811 21 Z Z 22 21Z Z 21 Z 0Z 0Z11ScV ( ) 2 0 V ( ) 10 V( )1 110 1 V S = +Assume ( )10 1 [ V] V +=( )( ) ( )( ) ( ) ( )21 22 21 0111 21 21 22 21 0||0||cZ Z Z ZV VZ Z Z Z Z Z| | +( =| | + +( \ .Use voltage divider equationtwice:( )( )0222 21 00cZV VZ Z Z| |=| | +\ .( ) ( )21 2 20 0 S V V= =S21Calculation:Example (cont.)59Hence( )( ) ( )( ) ( ) ( ) ( )21 22 21 0 021 1122 21 0 11 21 21 22 21 0||1||Z Z Z Z ZS SZ Z Z Z Z Z Z Z Z| | +( | | = +|| || + + +(\ . \ .11 21 Z Z 22 21 Z Z 21 Z 0Z 0Z11ScV ( ) 20 V ( ) 10 VAfter simplifying,we should get the result in the table. (You are welcome to check it!)