Microwave Network Analysis - Chapter 4

174 Chapter 4: Microwave Network Analysis 4.2 IMPEDANCE AND ADMITTANCE MATRICES In the previous section we have seen how equivalent voltages and currents can be defined for TEM and non-TEM waves. Once such voltages and currents have been defined at vari-...

174 Chapter 4: Microwave Network Analysis 4.2 IMPEDANCE AND ADMITTANCE MATRICES In the previous section we have seen how equivalent voltages and currents can be defined for TEM and non-TEM waves. Once such voltages and currents have been defined at vari- ous points in a microwave network, we can use the impedance and/or admittance matrices of circuit theory to relate these terminal or port quantities to each other, and thus to essen- tially arrive at a matrix description of the network. This type of representation lends itself to the development of equivalent circuits of arbitrary networks, which will be quite useful when we discuss the design of passive components such as couplers and filters. (The term port was introduced by H. A. Wheeler in the 1950s to replace the less descriptive and more cumbersome phrase “two-terminal pair” [2, 3].) We begin by considering an arbitrary N -port microwave network, as depicted in Figure 4.5. The ports in Figure 4.5 may be any type of transmission line or transmission line equivalent of a single propagating waveguide mode. If one of the physical ports of the network is a waveguide supporting more than one propagating mode, additional electri- cal ports can be added to account for these modes. At a specific point on the nth port, a terminal plane, tn , is defined along with equivalent voltages and currents for the incident (Vn+ , In+ ) and reflected (Vn− , In− ) waves. The terminal planes are important in providing a phase reference for the voltage and current phasors. Now, at the nth terminal plane, the total voltage and current are given by Vn = Vn+ + Vn− , (4.24a ) In = In+ − In− , (4.24b ) as seen from (4.8) when z = 0. The impedance matrix [Z ] of the microwave network then relates these voltages and currents:       V1 Z 11 Z 12 · · · Z 1N I1  V2   ..     I   Z 21.   2 ..  =  ...  . , .  . .. ..  VN Z N1 · · · · · · Z N N IN V 2,– – I 2– + + V 3, I3 + + V 2, I2 V 3,– – I 3– t2 t3 + + V 1, I1 S + + V 4, I4 t1 – – V 1, – I 1– t4 V 4, – I 4– tN + + VN, IN VN,– – IN– FIGURE 4.5 An arbitrary N -port microwave network.  4.2 Impedance and Admittance Matrices 175 or in matrix form as [V ] = [Z ][I ]. (4.25) Similarly, we can define an admittance matrix [Y] as       I1 Y11 Y12 · · · Y1N V1  I2  ..       Y21.   V2  ..  =   .. , .   ......   .  IN YN 1 · · · · · · YN N VN or in matrix form as [I ] = [Y ][V ]. (4.26) Of course, the [Z ] and [Y ] matrices are the inverses of each other: [Y ] = [Z ]−1. (4.27) Note that both the [Z ] and [Y ] matrices relate the total port voltages and currents. From (4.25), we see that Z i j can be found as Vi Zi j =. (4.28) Ij Ik =0 for k= j In words, (4.28) states that Z i j can be found by driving port j with the current I j , open- circuiting all other ports (so Ik = 0 for k = j), and measuring the open-circuit voltage at port i. Thus, Z ii is the input impedance seen looking into port i when all other ports are open-circuited, and Z i j is the transfer impedance between ports i and j when all other ports are open-circuited. Similarly, from (4.26), Yi j can be found as Ii Yi j = , (4.29) Vj Vk =0 for k= j which states that Yi j can be determined by driving port j with the voltage V j , short- circuiting all other ports (so Vk = 0 for k = j), and measuring the short-circuit current at port i. In general, each Z i j or Yi j element may be complex. For an arbitrary N -port network, the impedance and admittance matrices are N × N in size, so there are 2N 2 independent quantities or degrees of freedom. In practice, however, many networks are either recipro- cal or lossless, or both. If the network is reciprocal (not containing any active devices or nonreciprocal media, such as ferrites or plasmas), we will show that the impedance and admittance matrices are symmetric, so that Z i j = Z ji , and Yi j = Y ji. If the network is lossless, we can show that all the Z i j or Yi j elements are purely imaginary. Either of these special cases serves to reduce the number of independent quantities or degrees of freedom that an N -port network may have. We now derive the above characteristics for reciprocal and lossless networks. Reciprocal Networks Consider the arbitrary network of Figure 4.5 to be reciprocal (no active devices, ferrites, or plasmas), with short circuits placed at all terminal planes except those of ports 1 and 2. Let Ē a , H̄a and Ē b , H̄b be the fields anywhere in the network due to two independent sources, 176 Chapter 4: Microwave Network Analysis a and b, located somewhere in the network. Then the reciprocity theorem of (1.156) states that   Ē a × H̄b · d s̄ = Ē b × H̄a · d s̄, (4.30) S S where S is the closed surface along the boundaries of the network and through the terminal planes of the ports. If the boundary walls of the network and transmission lines are metal, then Ē tan = 0 on these walls (assuming perfect conductors). If the network or the transmis- sion lines are open structures, like microstrip line or slotline, the boundaries of the network can be taken arbitrarily far from the lines so that Ē tan is negligible. Then the only nonzero contribution to the integrals of (4.30) come from the cross-sectional areas of ports 1 and 2. From Section 4.1, the fields due to sources a and b can be evaluated at the terminal planes t1 and t2 as Ē 1a = V1a ē1 , H̄1a = I1a h̄ 1 , (4.31a) Ē 1b = V1b ē1 , H̄1b = I1b h̄ 1 , (4.31b) Ē 2a = V2a ē2 , H̄2a = I2a h̄ 2 , (4.31c) Ē 2b = V2b ē2 , H̄2b = I2b h̄ 2 , (4.31d) where ē1 , h̄ 1 and ē2 , h̄ 2 are the transverse modal fields of ports 1 and 2, respectively, and the V s and I s are the equivalent total voltages and currents. (For instance, Ē 1b is the transverse electric field at terminal plane t1 of port 1 due to source b.) Substituting the fields of (4.31) into (4.30) gives

(V1a I1b − V1b I1a ) ē1 × h̄ 1 · d s̄ + (V2a I2b − V2b I2a ) ē2 × h̄ 2 · d s̄ = 0, (4.32) S1 S2 where S1 and S2 are the cross-sectional areas at the terminal planes of ports 1 and 2. As in Section 4.1, the equivalent voltages and currents have been defined so that the power through a given port can be expressed as VI∗ /2; then, comparing (4.31) to (4.6) implies that C1 = C2 = 1 for each port, so that

ē1 × h̄ 1 · d s̄ = ē2 × h̄ 2 · d s̄ = 1. (4.33) S1 S2 This reduces (4.32) to V1a I1b − V1b I1a + V2a I2b − V2b I2a = 0. (4.34) Now use the 2 × 2 admittance matrix of the (effectively) two-port network to eliminate the I s: I1 = Y11 V1 + Y12 V2 , I2 = Y21 V1 + Y22 V2. Substitution into (4.34) gives (V1a V2b − V1b V2a )(Y12 − Y21 ) = 0. (4.35) Because the sources a and b are independent, the voltages V1a , V1b , V2a , and V2b can take on arbitrary values. So in order for (4.35) to be satisfied for any choice of sources, we must have Y12 = Y21 , and since the choice of which ports are labeled as 1 and 2 is arbitrary, we have the general result that Yi j = Y ji. (4.36) Then if [Y ] is a symmetric matrix, its inverse, [Z ], is also symmetric.  4.2 Impedance and Admittance Matrices 177 Lossless Networks Now consider a reciprocal lossless N -port junction; we will show that the elements of the impedance and admittance matrices must be pure imaginary. If the network is lossless, then the net real power delivered to the network must be zero. Thus, Re{Pavg } = 0, where 1 1 1 Pavg = [V ]t [I ]∗ = ([Z ][I ])t [I ]∗ = [I ]t [Z ][I ]∗ 2 2 2 1 = (I1 Z 11 I1∗ + I1 Z 12 I2∗ + I2 Z 21 I1∗ + · · ·) 2 1 N N = Im Z mn In∗. (4.37) 2 n=1 m=1 We have used the result from matrix algebra that ([ A][B])t = [B]t [ A]t. Because the In are independent, we must have the real part of each self term (In Z nn In∗ ) equal to zero, since we could set all port currents equal to zero except for the nth current. So, Re{In Z nn In∗ } = |In |2 Re{Z nn } = 0, or Re{Z nn } = 0. (4.38) Now let all port currents be zero except for Im and In. Then (4.37) reduces to   Re (In Im∗ + Im In∗ )Z mn = 0, since Z mn = Z nm. However, (In Im∗ + Im In∗ ) is a purely real quantity that is, in general, nonzero. Thus we must have that Re {Z mn } = 0. (4.39) Then (4.38) and (4.39) imply that Re {Z mn } = 0 for any m, n. The reader can verify that this also leads to an imaginary [Y ] matrix. EXAMPLE 4.3 EVALUATION OF IMPEDANCE PARAMETERS Find the Z parameters of the two-port T-network shown in Figure 4.6. Solution From (4.28), Z 11 can be found as the input impedance of port 1 when port 2 is open-circuited: V1 Z 11 = = Z A + ZC. I1 I2 =0 ZA ZB + + Port V1 ZC V2 Port 1 2 – – FIGURE 4.6 A two-port T-network. 178 Chapter 4: Microwave Network Analysis The transfer impedance Z 12 can be found measuring the open-circuit voltage at port 1 when a current I2 is applied at port 2. By voltage division, V1 V2 ZC Z 12 = = = ZC. I2 I1 =0 I2 Z B + Z C The reader can verify that Z 21 = Z 12 , indicating that the circuit is reciprocal. Finally, Z 22 is found as V2 Z 22 = = Z B + ZC. I2 I1 =0 4.3 THE SCATTERING MATRIX We have already discussed the difficulty in defining voltages and currents for non-TEM lines. In addition, a practical problem exists when trying to measure voltages and currents at microwave frequencies because direct measurements usually involve the magnitude (inferred from power) and phase of a wave traveling in a given direction or of a standing wave. Thus, equivalent voltages and currents, and the related impedance and admittance matrices, become somewhat of an abstraction when dealing with high-frequency networks. A representation more in accord with direct measurements, and with the ideas of incident, reflected, and transmitted waves, is given by the scattering matrix. Like the impedance or admittance matrix for an N -port network, the scattering matrix provides a complete description of the network as seen at its N ports. While the impedance and admittance matrices relate the total voltages and currents at the ports, the scattering matrix relates the voltage waves incident on the ports to those reflected from the ports. For some components and circuits, the scattering parameters can be calculated using net- work analysis techniques. Otherwise, the scattering parameters can be measured directly with a vector network analyzer; a photograph of a modern network analyzer is shown in Figure 4.7. Once the scattering parameters of the network are known, conversion to other matrix parameters can be performed, if needed. Consider the N -port network shown in Figure 4.5, where Vn+ is the amplitude of the voltage wave incident on port n and Vn− is the amplitude of the voltage wave reflected from port n. The scattering matrix, or [S] matrix, is defined in relation to these incident and reflected voltage waves as       V1− S11 S12 · · · S1N V1+       V−  ..  V+   2   S21.   2       , ..  =     .   SN 1 · · · S N N  ...       .. V− N. V + N or [V − ] = [S][V + ]. (4.40) A specific element of the scattering matrix can be determined as Vi− Si j =. (4.41) V j+ Vk+ =0 for k= j In words, (4.41) says that Si j is found by driving port j with an incident wave of voltage V j+ and measuring the reflected wave amplitude Vi− coming out of port i. The incident  4.3 The Scattering Matrix 179 FIGURE 4.7 Photograph of the Agilent N5247A Programmable Network Analyzer. This instru- ment is used to measure the scattering parameters of RF and microwave networks from 10 MHz to 67 GHz. The instrument is programmable, performs error correc- tion, and has a wide variety of display formats and data conversions. Courtesy of Agilent Technologies. waves on all ports except the jth port are set to zero, which means that all ports should be terminated in matched loads to avoid reflections. Thus, Sii is the reflection coefficient seen looking into port i when all other ports are terminated in matched loads, and Si j is the transmission coefficient from port j to port i when all other ports are terminated in matched loads. EXAMPLE 4.4 EVALUATION OF SCATTERING PARAMETERS Find the scattering parameters of the 3 dB attenuator circuit shown in Figure 4.8. Solution From (4.41), S11 can be found as the reflection coefficient seen at port 1 when port 2 is terminated in a matched load (Z 0 = 50 ): (1) V1− Z in − Z 0 S11 = =  (1) |V + =0 = , V1+ 2 (1) Z in + Z 0 V2+ =0 Z 0 on port 2 8.56 Ω 8.56 Ω Port Port 141.8 Ω 1 2 FIGURE 4.8 A matched 3 dB attenuator with a 50  characteristic impedance (Example 4.4). 180 Chapter 4: Microwave Network Analysis (1) but Z in = 8.56 + [141.8(8.56 + 50)]/(141.8 + 8.56 + 50) = 50 , so S11 = 0. Because of the symmetry of the circuit, S22 = 0. We can find S21 by applying an incident wave at port 1, V1+ , and measuring the outcoming wave at port 2, V2−. This is equivalent to the transmission coeffi- cient from port 1 to port 2: V2− S21 =. V1+ V2+ =0 From the fact that S11 = S22 = 0, we know that V1− = 0 when port 2 is terminated in Z 0 = 50 , and that V2+ = 0. In this case we have that V1+ = V1 and V2− = V2. By applying a voltage V1 at port 1 and using voltage division twice we find V2− = V2 as the voltage across the 50  load resistor at port 2:    41.44 50 V2− = V2 = V1 = 0.707V1 , 41.44 + 8.56 50 + 8.56 where 41.44 = 141.8(58.56)/(141.8 + 58.56) is the resistance of the parallel com- bination of the 50  load and the 8.56  resistor with the 141.8  resistor. Thus, S12 = S21 = 0.707. If the input power is |V1+ |2 /2Z 0 , then the output power is |V2− |2 /2Z 0 = |S21 V1+ |2 /2Z 0 = |S21 |2 /2Z 0 |V1+ |2 = |V1+ |2 /4Z 0 , which is one-half (−3 dB) of the input power. We now show how the scattering matrix can be determined from the [Z ] (or [Y ]) matrix and vice versa. First, we must assume that the characteristic impedances, Z 0n , of all the ports are identical. (This restriction will be removed when we discuss generalized scattering parameters.) Then, for convenience, we can set Z 0n = 1. From (4.24) the total voltage and current at the nth port can be written as Vn = Vn+ + Vn− , (4.42a) In = In+ − In− = Vn+ − Vn−. (4.42b) Using the definition of [Z ] from (4.25) with (4.42) gives [Z ][I ] = [Z ][V + ] − [Z ][V − ] = [V ] = [V + ] + [V − ], which can be rewritten as ([Z ] + [U ]) [V − ] = ([Z ] − [U ]) [V + ], (4.43) where [U ] is the unit, or identity, matrix defined as   1 0 ··· 0 ..  0 1. [U ] = ... . ...  0 ··· 1  4.3 The Scattering Matrix 181 Comparing (4.43) to (4.40) suggests that [S] = ([Z ] + [U ])−1 ([Z ] − [U ]) , (4.44) giving the scattering matrix in terms of the impedance matrix. Note that for a one-port network (4.44) reduces to z 11 − 1 S11 = , z 11 + 1 in agreement with the result for the reflection coefficient seen looking into a load with a normalized input impedance of z 11. To find [Z ] in terms of [S], rewrite (4.44) as [Z ][S] + [U ][S] = [Z ] − [U ], and solve for [Z ] to give [Z ] = ([U ] + [S]) ([U ] − [S])−1. (4.45) Reciprocal Networks and Lossless Networks As we discussed in Section 4.2, the impedance and admittance matrices are symmetric for reciprocal networks, and are purely imaginary for lossless networks. The scattering matrices for these particular types of networks also have special properties. We will show that the scattering matrix for a reciprocal network is symmetric, and that the scattering matrix for a lossless network is unitary. By adding (4.42a) and (4.42b) we obtain 1 Vn+ = (Vn + In ), 2 or 1 [V + ] = ([Z ] + [U ])[I ]. (4.46a) 2 By subtracting (4.42a) and (4.42b) we obtain 1 Vn− = (Vn − In ), 2 or 1 [V − ] = ([Z ] − [U ])[I ]. (4.46b) 2 Eliminating [I ] from (4.46a) and (4.46b) gives [V − ] = ([Z ] − [U ])([Z ] + [U ])−1 [V + ], so that [S] = ([Z ] − [U ])([Z ] + [U ])−1. (4.47) Taking the transpose of (4.47) gives [S]t = {([Z ] + [U ])−1 }t ([Z ] − [U ])t. Now [U ] is diagonal, so [U ]t = [U ], and if the network is reciprocal, [Z ] is symmetric. so that [Z ]t = [Z ]. The above equation then reduces to [S]t = ([Z ] + [U ])−1 ([Z ] − [U ]), 182 Chapter 4: Microwave Network Analysis which is equivalent to (4.44). We have thus shown that [S] = [S]t , (4.48) so the scattering matrix is symmetric for reciprocal networks. If the network is lossless, no real power can be delivered to the network. Thus, if the characteristic impedances of all the ports are identical and assumed to be unity, the average power delivered to the network is 1 1 Pavg = Re{[V ]t [I ]∗ } = Re{([V + ]t + [V − ]t )([V + ]∗ − [V − ]∗ )} 2 2 1 = Re{[V + ]t [V + ]∗ − [V + ]t [V − ]∗ + [V − ]t [V + ]∗ − [V − ]t [V − ]∗ } 2 1 1 = [V + ]t [V + ]∗ − [V − ]t [V − ]∗ = 0, (4.49) 2 2 since the terms −[V + ]t [V − ]∗ + [V − ]t [V + ]∗ are of the form A − A∗ , and so are purely imaginary. Of the remaining terms in (4.49), (1/2)[V + ]t [V + ]∗ represents the total inci- dent power, while (1/2)[V − ]t [V − ]∗ represents the total reflected power. So, for a lossless junction, we have the intuitive result that the incident and reflected powers are equal: [V + ]t [V + ]∗ = [V − ]t [V − ]∗. (4.50) Using [V − ] = [S][V + ] in (4.50) gives [V + ]t [V + ]∗ = [V + ]t [S]t [S]∗ [V + ]∗ , so that, for nonzero [V + ], [S]t [S]∗ = [U ], (4.51) or [S]∗ = {[S]t }−1. A matrix that satisfies the condition of (4.51) is called a unitary matrix. The matrix equation of (4.51) can be written in summation form as  N Ski Sk∗j = δi j , for all i, j, (4.52) k=1 where δi j = 1 if i = j, and δi j = 0 if i = j, is the Kronecker delta symbol. Thus, if i = j, (4.52) reduces to  N ∗ Ski Ski = 1, (4.53a) k=1 while if i = j, (4.52) reduces to  N Ski Sk∗j = 0, for i = j. (4.53b) k=1 In words, (4.53a) states that the dot product of any column of [S] with the conjugate of that same column gives unity, while (4.53b) states that the dot product of any column with the  4.3 The Scattering Matrix 183 conjugate of a different column gives zero (the columns are orthonormal). From (4.51) we also have that [S] [S]∗t = [U ] , so the same statements can be made about the rows of the scattering matrix. EXAMPLE 4.5 APPLICATION OF SCATTERING PARAMETERS A two-port network is known to have the following scattering matrix:   0.15 0◦ 0.85 −45◦ [S] = 0.85 45◦ 0.2 0◦ Determine if the network is reciprocal and lossless. If port 2 is terminated with a matched load, what is the return loss seen at port 1? If port 2 is terminated with a short circuit, what is the return loss seen at port 1? Solution Because [S] is not symmetric, the network is not reciprocal. To be lossless, the scattering parameters must satisfy (4.53). Taking the first column [i = 1 in (4.53a)] gives |S11 |2 + |S21 |2 = (0.15)2 + (0.85)2 = 0.745 = 1, so the network is not lossless. When port 2 is terminated with a matched load, the reflection coefficient seen at port 1 is  = S11 = 0.15. So the return loss is RL = −20 log || = −20 log(0.15) = 16.5 dB. When port 2 is terminated with a short circuit, the reflection coefficient seen at port 1 can be found as follows. From the definition of the scattering matrix and the fact that V2+ = −V2− (for a short circuit at port 2), we can write V1− = S11 V1+ + S12 V2+ = S11 V1+ − S12 V2− , V2− = S21 V1+ + S22 V2+ = S21 V1+ − S22 V2−. The second equation gives S21 V2− = V +. 1 + S22 1 Dividing the first equation by V1+ and using the above result gives the reflection coefficient seen at port 1 as V1− V2− S12 S21 = = S11 − S12 = S11 − V1+ V1+ 1 + S22 (0.85 −45◦ )(0.85 45◦ ) = 0.15 − = −0.452. 1 + 0.2 So the return loss is RL = −20 log || = −20 log(0.452) = 6.9 dB. An important point to understand about scattering parameters is that the reflection coefficient looking into port n is not equal to Snn unless all other ports are matched (this