ECE 5317-6351 Microwave Engineering Fall 2011 Prof. David R. Jackson Dept. of ECE Notes 14 Network Analysis Multiport Networks 1 Multiport Networks A general circuit can be represented by a multi-port network, where the “ports” are defined as access terminals at which we can define voltages and currents. Note: Equal and opposite currents are assumed on the two wires of a port. Examples: 1) One-port network I1 + V1 I1 + V1 R - - 2) Two-port network I2 I2 I1 + V1 + V2 - I1 - + V1 - + V2 - 2 Multiport Networks (cont.) IN 3) N-port Network + VN - I1 Im V1 +- To represent multi-port networks we use: + - Vm I2 - Z (impedance) parameters Y (admittance) parameters h (hybrid) parameters ABCD parameters Not easily measurable at high frequency - S (scattering) parameters Measurable at high frequency V2 + - I3 + V 3 3 Poynting Theorem (Phasor Domain) V 2 1 EJ i* dV S nˆ dS S 1 c E 2 V 2 1 2 j H 4 V 1 H 2 2 1 4 2 dV 2 c E dV The last term is the VARS consumed by the region. The notation < > denotes time-average. Ps P f Pd j 2 Wm We 4 Self Impedance Consider a general one-port network S I1 V1 + nˆ E,H - V Complex power delivered to network: Pin 1 2 1 2 E H nˆ S V1 I 1 d s Pd j 2 W m W e * Average power dissipated in [W] Pd Pd Average electric energy (in [J]) stored inside V W e We Average magnetic energy (in [J]) stored inside V W m Wm 5 Define Self Impedance (Zin) 1 Z in V1 I1 V1 I 1 I1 * 2 2 1 2 R in jX in V1 I 1 I1 * 2 X in 1 I1 2 2 Pd j 2 W m W e 1 2 R in Pin I1 2 2 Pd I1 2 S I1 4 (W m W e ) I1 2 V1 + - nˆ E,H V 6 Self Impedance (cont.) We can show that for physically realizable networks the following apply: Please see the Pozar book for a proof. V1 V1 * Z in Z in * I1 I1 * R in is a n e v e n fu n ctio n o f Z in R in jX in X in is a n o d d fu n ctio n o f Note: Frequency is usually defined as a positive quantity. However, we consider the analytic continuation of the functions into the complex frequency plane. 7 Two-Port Networks Consider a general 2-port linear network: I2 I1 V1 + - 1 + 2 - V2 In terms of Z-parameters, we have (from superposition) Impedance (Z) matrix V1 Z 11 I 1 Z 12 I 2 V 2 Z 21 I 1 Z 2 2 I 2 V1 Z 11 V 2 Z 21 Z 12 I 1 Z 22 I 2 V Z I 8 Elements of Z-Matrix: Z-Parameters (open-circuit parameters ) V1 Z 11 I 1 Z 12 I 2 V 2 Z 21 I 1 Z 22 I 2 Port 2 open circuited Z 11 Z 21 Port 1 open circuited Z ij V1 I1 Vi Ij Ik 0 Z 12 k j I2 0 V2 I1 Z 22 I2 0 V1 - I2 I1 0 V2 I2 I1 0 I2 I1 + V1 1 2 + - V2 9 Z-Parameters (cont.) N-port network Z ij Ij Vi Ij Ik 0 k j Vi + - 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. 10 Z-Parameters (cont.) Z-parameters are convenient for series connected networks. A B V1 V1 V1 A B V 2 V 2 V 2 + A A B B Z I Z I A B Z Z I I1 Z Z I2 Z A A B I1A I1 V1A I 2A + - 1 A 2 + - I2 V2A + V2 V1 I 2B I1B - V1B I1 Z I2 + - 1 B 2 + - V2B - B V1 Z 11 Z 11 A B V Z Z 2 21 21 A B A B Z 12 Z 12 I 1 A B Z 22 Z 22 I 2 S e rie s I1 I1 I1 A B I2 I2 I2 A B 11 Admittance (Y) Parameters Consider a 2-port network: I2 I1 V1 + 1 - + 2 - I 1 Y11V1 Y12V 2 Admittance matrix I 2 Y21V1 Y 22V 2 or I 1 Y11 I 2 Y 21 Yij V2 Y12 V1 I Y V Y 2 2 V 2 Ii Vj Short-circuit parameters Vk 0 k j 12 Y-Parameters (cont.) N-port network Yij Vj Ii Vj +Vk 0 k j Ii 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. 13 Admittance (Y) Parameters Y-parameters are convenient for parallel connected networks. I1 + V1 - I1A V1A + - 1 A 2 I1B B 1 Y11 Y11 A B Y Y 21 21 A B V Y12 Y12 V1 A B Y 22 Y 22 V 2 A I2 + A - V2 V2 + - I 2B + B 1 - I1 I I A B I2 I2 I2 A 1 I 2A 1 B + B - V2 2 B P a ra lle l V1 V1 V1 A B V2 V2 V2 A B 14 Admittance (Y) Parameters Relation between [Z] and [Y] matrices: V Z I I Y V Hence V Z Y V Z Y V Z Y U Identity M atrix Therefore Y Z 1 15 Reciprocal Networks If a network does not contain non-reciprocal devices or materials* (i.e. ferrites, or active devices), then the network is “reciprocal.” Z ij Z ji Z a n d Y Y ij Y ji Note: The inverse of a symmetric matrix is symmetric. a re sym m e tric * A reciprocal material is one that has reciprocal permittivity and permeability tensors. A reciprocal device is one that is made from reciprocal materials Example of a nonreciprocal material: a biased ferrite (This is very useful for making isolators and circulators.) 16 Reciprocal Materials D E B Η D x xx D y yx D z zx xy xz E x yy yz E y zy zz E z B x xx B y yx B z zx xy yy zy xz x yz y zz z Reciprocal: ij ji , ij ji Ferrite: 0 j 0 j 0 0 0 1 is not symmetric! 17 Reciprocal Networks (cont.) We can show that the equivalent circuits for reciprocal 2-port networks are: Z 22 Z 21 Z11 Z 21 T-equivalent Z 21 Y21 Pi-equivalent Y11 Y21 Y22 Y21 18 ABCD-Parameters There are defined only for 2-port networks. V 1 A I1 C I1 B V 2 ' D I2 V1 I 2' 1 2 V 2 I2 I2 ' A V1 V2 B ' I2 0 V1 V2 D ' I2 I1 C V2 0 ' I2 0 I1 ' I2 V2 0 19 Cascaded Networks I1 I1A V1 V1A I 1 A 2 V A' 2 A 2 I B 1 B 1 V I 1 B 2 A V 2 V 1 V 1 A A A B C D A ' I1 I1 I 2 B' 2 V B 2 I 2' V2 A V 1 A A B C D B I1 B V 2 A B A B C D A B C D ' B I 2 B V1 I1 ABCD AB A nice property of the ABCD matrix is that it is easy to use with cascaded networks: you simply multiply the ABCD matrices together. V 2 ' I 2 20 Scattering Parameters At high frequencies, Z, Y, h & ABCD parameters are difficult (if not impossible) to measure. o V and I are not uniquely defined o 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) parameters are often the best representation for multi-port networks at high frequency. 21 Scattering Parameters (cont.) S-parameters are defined assuming transmission lines are connected to each port. a1 b1 Z 01 , 1 1 Z 02 , 2 2 z1 a2 b2 z2 Local coordinates On each transmission line: Vi zi Vi 0 e I i zi Vi i zi zi Z 0i Vi 0 e Vi i zi zi Vi zi Vi zi i 1, 2 Z 0i Incom ing w ave function a i z i Vi z i Z 0i O utgoing w ave function bi z i Vi z i Z 0i 22 For a One-Port Network L L 0 V1 0 V1 Z 01 a1 Z 01 Z 01 b1 b1 0 a1 0 l1 b1 0 L a 1 0 S 11 a 1 0 For a one-port network, S11 is defined to be the same as L. S11 Incom ing w ave function a i z i Vi z i Z 0i O utgoing w ave function bi z i Vi z i Z 0i 23 For a Two-Port Network a1 Z 01 , 1 1 Z 02 , 2 2 b1 z1 b2 z2 b1 0 S 11 a 1 0 S 12 a 2 0 b2 0 S 21 a 1 0 S 22 a 2 0 b1 0 S 11 b2 0 S 21 a2 Scattering matrix S 12 a 1 0 b S a S 22 a 2 0 24 Scattering Parameters b1 0 S 11 a 1 0 S 12 a 2 0 b2 0 S 21 a 1 0 S 22 a 2 0 S11 S12 S 21 S 22 b1 0 a1 0 Output is matched a2 0 b1 0 a2 0 a1 0 b2 0 a1 0 a2 0 b2 0 a2 0 Input is matched Output is matched Input is matched a1 0 input reflection coef. w/ output matched reverse transmission coef. w/ input matched forward transmission coef. w/ output matched output reflection coef. w/ input matched 25 Scattering Parameters (cont.) For a general multiport network: S ij All ports except j are semi-infinite (or matched) bi 0 a j 0 ak 0 k j N-port network Semi-infinite aj Port i bi Port j 26 Scattering Parameters (cont.) Illustration of a three-port network a1 b1 2 1 3 a2 b2 a3 b3 27 Scattering Parameters (cont.) For reciprocal networks, the S-matrix is symmetric. S ij S ji i j Note: If all lines entering the network have the same characteristic impedance, then S ij 0 V j 0 V Vi k 0 k j 28 Scattering Parameters (cont.) Why are the wave functions (a and b) defined as they are? a1 Z 01 , 1 1 Z 02 , 2 2 b1 z1 Pi 0 a2 b2 z2 1 R e V i 0 I i 2 * 1 Vi 0 2 0 2 (assuming lossless lines) Z 0i Note: a i 0 Vi 0 1 Pi 2 0 Z 0i ai 0 2 29 Scattering Parameters (cont.) Similarly, Pi 0 1 Vi 2 0 2 Z 0i 1 2 bi 0 2 Also, Vi Vi Pi Pi li V i 0 e li V i 0 e li li 1 2 1 2 a i li 2 bi l i 2 i li i li 1 2 1 2 ai 0 e 2 bi 0 e 2 2 i li 2 i li 30 Example Find the S parameters for a series impedance Z. a1 Z0 b1 V1 Z V 2 z1 Z0 a2 b2 z2 Note that two different coordinate systems are being used here! 31 Example (cont.) Semi-infinite a1 V1 Z0 b1 Z V 2 Z in Z0 b2 z2 z1 S11 Calculation: S 11 b1 0 a1 0 0 V1 0 V1 a2 0 S 11 Z Z 2Z0 Z in Z 0 Z in Z 0 a2 0 Z Z Z0 Z0 Z0 Z0 By symmetry: S 22 S 11 32 Example (cont.) S21 Calculation: Semi-infinite a1 Z0 b1 V1 Z V 2 Z in b2 0 a1 0 a2 0 V2 a2 0 0 V1 0 a V1 b2 z2 z1 S 21 Z0 V2 0 a1 0 2 0 0 V2 0 Z0 V 2 0 V1 0 Z Z0 V1 0 a 1 Z 0 1 S 11 Z0 V2 0 V 2 0 a1 Z0 Z 0 1 S 11 Z Z 0 33 Example (cont.) Semi-infinite a1 Z0 V1 b1 Z S 21 2 Z in z1 a1 0 V Z0 b2 z2 Z0 Z 0 1 S 11 Z Z 0 a1 0 Z0 Z0 Z0 2Z 2Z0 Z0 Z 1 S 11 1 Z Z Z 2 Z Z Z Z 2 Z Z Z 0 0 0 0 0 Hence S 21 2 Z0 Z 2 Z0 S 12 S 21 34 Example Find the S parameters for a length L of transmission line. L Z0s , s Z0 z1 Z0 z z2 Note that three different coordinate systems are being used here! 35 Example (cont.) L S11 Calculation: L L V1 0 Z0 + Z0s , s - + V z - z z1 S 11 Z in a2 0 b1 a1 a2 0 Z in Z in a2 0 a2 0 Z0 Z0 Z 0 jZ 0 s tan s L Z 0s Z 0 s jZ 0 tan s L + - V2 0 z2 Z 0 Z 0s Z 0 Z 0s Z0 Semi-infinite S 22 ( by sym m etry ) Z 0s 1 1 e L e L j2s L j2s L 36 Example (cont.) L L V1 0 Z0 + Z0s , s - + V z - z z1 + - V2 0 Z0 z2 Hence Z 0s S 11 S 22 Z 0s N o te : If Z 0 jZ 0 s tan s L Z0 Z jZ tan L 0s 0 s Z 0 jZ 0 s tan s L Z0 Z 0 s jZ 0 tan s L Z 0s Z 0 Z in a2 0 Z0 S 11 S 22 0 37 Example (cont.) S21 Calculation: L L L Z0 V1 0 + Z0s , s - 0 V1 0 b2 a1 - a2 0 V2 V1 0 V1 V2 0 Z 0 Z 0s Z0 Semi-infinite z2 z z1 S 21 + + V z - Z 0 Z 0s Z0 Z0 a2 0 0 1 S 11 Hence, for the denominator of the S21 equation we have V1 0 V1 0 1 S 11 We now try to put the numerator of the S21 equation in terms of V1 (0). 38 Example (cont.) L L V1 0 Z0 + Z0s , s - - V2 0 Z0 z2 z z1 V2 + + V z - 0 V 2 0 V 0 V 0 1 L Next, use V z V 0 e j V1 0 V L V V 0 s 1 z 0 e j j s L 1 L e s j2s z L 1 e L j2s L Hence, we have V1 0 e e L j2s L V 2 0 V1 0 e j s L 1 L e j2s L 1 L 39 Example (cont.) L L V1 0 Z0 + - Z0s , s + V z - z z1 + - V2 0 Z0 z2 V2 0 V1 0 e Therefore, we have V1 0 V1 0 S 21 j s L V2 1 L e 0 j2s L 1 L V1 0 1 S 11 1 S 11 1 L e j L s a2 0 1 Le j2sL so 1 S 11 1 L e j L s S 21 1 Le j2sL S 12 by sym m etry 40 Example (cont.) L Special cases: a) Z0s , s Z 0 s Z 0 S 11 S 22 0, S 21 S 12 e b) L g 2 Z in e j s L j s L s L a2 0 Z0 1 L 0 2 g g 2 0 S j s L e e j s L 0 S 11 S 22 0 0 S 1 1 0 S 21 1 41 Example Find the S parameters for a step-impedance discontinuity. S 11 S 22 Z 02 Z 01 Z 02 Z 01 Z 01 Z 02 Z 02 Z 01 Z 01 Z 02 S 11 V2 S 21 b2 0 a1 0 a2 0 0 Z 02 V1 0 Z 01 a2 0 42 Example (cont.) Semi-infinite S21 Calculation: + V1 Because of continuity of the voltage across the junction, we have: V2 0 a 2 0 V2 0 V2 S 21 a2 0 0 V1 Z 02 V1 V1 0 0 Z 01 V1 a2 0 0 1 1 S 11 1 Z 02 V1 + - Z 02 V2 0 1 S 11 S11 0 Z 01 a2 0 - Z 01 a2 0 Z 02 Z 01 Z 02 Z 01 2 Z 02 Z 02 Z 01 so S 21 1 S 11 Z 01 Z 02 Hence S 21 S 12 2 Z 01 Z 02 Z 01 Z 02 43 Properties of the S Matrix For reciprocal networks, the S-matrix is symmetric. S S T N o te : A B U If For lossless networks, the S-matrix is unitary. S S T Equivalently, * S S U * T th e n B A U Identity matrix Notation: S S S † S T* S 1 so H S † S 1 N-port network N T a k e ( i , j ) e le m e n t k 1 N S T ik S * kj k 1 S ki S kj ij * 1 ; i j ij 0 ; i j 44 T* Properties of the S Matrix (cont.) Example: S 11 S S 21 S 31 U n ita ry S 12 S 22 S 32 S 13 S 23 S 33 S 11 S 11 S 21 S 21 S 31 S 31 1 * S 12 S * 12 * S 22 S * 22 * S 32 S * 32 1 S 13 S 13 S 23 S 23 S 33 S 33 1 * S 11 S * 12 * S 21 S * 22 * S 31 S * 32 0 S 11 S 13 S 21 S 23 S 31 S 33 0 * * The column vectors form an orthogonal set. * The rows also form orthogonal sets (see the note on the previous slide). S 12 S 13 S 22 S 23 S 32 S 33 0 * * * 45 Example 0 j S 50 2 j 2 j 2 0 0 j 2 0 0 1 S50 2 3 Not unitary Not lossless (For example, column 2 doted with the conjugate of column three is not zero. 1) Find the input impedance looking into port 1 when ports 2 and 3 are terminated in 50 [] loads. 2) Find the input impedance looking into port 1 when port 2 is terminated in a 75 [] load and port 3 is terminated in a 50 [] load. 46 Example (cont.) 1 If ports 2 and 3 are terminated in 50 [Ω]: (a2 = a3 = 0) b1 S 11 a1 S 12 a 2 S 13 a 3 in 1 in1 a1 b1 b1 a1 S 11 0 2 1 S50 3 Z in 1 50 [ ] a2 b2 a3 b3 2 0 3 0 47 Example (cont.) 2) If port 2 is terminated in 75 [Ω] and port 3 in 50 [Ω]: 2 in1 a1 b1 a2 b2 75 50 75 50 2 1 S50 3 1 5 a2 b2 a3 b3 2 1 5 3 0 48 Example (cont.) 0 b1 j b 2 2 b3 j 2 j 2 0 0 in 1 j 2 0 0 b1 a1 a1 a 2 a 3 in1 S 11 S 12 a2 a1 S 13 a1 b1 a3 2 1 S50 3 a2 b2 a3 b3 2 1 5 3 0 a a b 2 / a1 S 21 S 22 2 S 23 3 a1 a1 a1 b 1 j 1 j S 12 2 2 S 12 2 S 21 a 5 10 2 2 1 a 2 2 b2 Z in 1 1 in 1 50 1 in 1 44.55 [ ] 49 Transfer (T) Matrix For cascaded 2-port networks: 1 A 1 a b1A 2 A a b2A T12 b2 T 22 a 2 b2 T a2 1 2 b B a1B T Matrix: a 1 T11 b1 T 21 B 1 A 2 T 1 S 21 S 11 S 21 b2B a2B S 22 S 21 S S S 12 11 22 S 21 T 21 T 22 S 2 T12 T11 T 22 1 T 22 T12 T 22 (Derivation omitted) 50 Transfer (T) Matrix (cont.) a1 b2 A A T A b1 a2 A B ut A The T matrix of a cascaded set of networks is the product of the T matrices. b a A B a2 b1 A 2 B 1 a1 a1 A A T B b1 b1 A B a1 b2 A B A T T B b1 a2 A Hence B T AB 51 Conversion Between Parameters 52 Example Derive Sij from the Z parameters. (The result is given inside row 1, column 2, of the previous table.) S11 Calculation: 1 Z 22 Z 21 Z11 Z 21 S11 Z0 2 Z0 Z 21 Semi infinite S 1 1 in 1 Z in Z 0 Z in Z 0 Z in Z 1 1 Z 2 1 Z 2 1 || Z 2 2 Z 2 1 Z 0 53 Example (cont.) 1 Z 22 Z 21 Z11 Z 21 S11 Z0 Z 21 2 Z0 Semi infinite Z in Z 1 1 Z 2 1 Z 2 1 || Z 2 2 Z 2 1 Z 0 Z 11 Z 21 Z 21 Z 22 Z 0 Z 21 Z 11 Z 21 Z 22 Z 22 Z 0 Z 0 Z 21 Z 22 Z 0 Z 21 Z 22 Z 0 Z 11 Z 22 Z 11 Z 0 Z 21 Z 22 Z 21 Z 0 Z 21 Z 22 Z 21 Z 0 Z 21 2 Z 22 Z 0 Z 11 Z 22 Z 11 Z 0 Z 21 2 Z 22 Z 0 54 Example (cont.) 1 Z 22 Z 21 Z11 Z 21 S11 Z0 2 Z0 Z 21 Semi infinite Z in Z 11 Z 0 Z 22 S11 so S11 Z 221 Z 22 Z 0 Z in Z 0 Z in Z 0 Z 11 Z 0 Z 22 Z 221 Z 0 Z 0 2 Z 22 Z 21 Z 0 Z 0 Z 22 Z 22 Z 11 Z 0 55 Example (cont.) 1 Z 22 Z 21 Z11 Z 21 S11 Z0 2 Z0 Z 21 Semi infinite S11 Z 11 Z 0 Z 22 Z 221 Z 0 Z 0 2 Z 22 Z 21 Z 0 Z 0 Z 22 Z 11 Z 0 Z 11 Z 0 Z 11 Z 22 Z 21 Z 0 Z 0 Z 22 2 Z 22 2 Z 11 Z 0 Z 11 Z 22 Z 21 Z 0 Z 0 Z 22 2 Z0 Z0 2 Z 1 1 Z 0 Z 221 2 Z 22 Z 11 Z 0 Z 21 Z 22 56 Example (cont.) 1 Z 22 Z 21 Z11 Z 21 S11 Z0 2 Z0 Z 21 Semi infinite S11 Z0 Z0 Z 11 Z 22 Z 11 Z 22 Z 221 2 Z 0 Z 21 Z0 Note: to get S22, simply let Z11 Z22 in the previous result. S 22 Z0 Z0 Z 11 Z 22 Z 0 Z 11 Z 22 Z 221 2 Z 0 Z 21 57 Example (cont.) S21 Calculation: S11 V1 0 1 Z11 Z 21 Z0 Vc 2 Z 22 Z 21 V2 0 Z0 Z 21 Semi infinite Assume V1 0 1 [ V ] V1 0 1 S 1 1 S 21 V 2 0 V2 0 Use voltage divider equation twice: Z 2 1 || Z 2 2 Z 2 1 Z 0 V c V1 0 Z Z Z || Z Z Z 11 21 21 22 21 0 Z0 V2 0 Vc Z Z Z 22 21 0 58 Example (cont.) V1 0 1 S11 Z0 Z11 Z 21 Vc Z 22 Z 21 2 V2 0 Z0 Z 21 Semi infinite Hence S 21 Z 21 || Z 22 Z 21 Z 0 1 S 11 Z Z Z || Z Z Z 11 21 21 22 21 0 Z0 Z Z Z 22 21 0 After simplifying, we should get the result in the table. (You are welcome to check it!) 59