• 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 network analysis techniques.
• Otherwise, the scattering parameters can be measured directly with a vector network analyzer.Once the scattering parameters of the network are known, conversion to othermatrix parameters can be performed, if needed.
• Matrices are N$\times$ N in size, so there are 2N 2 independent quantities or degrees of freedom. In practice, however, many networks are either reciprocal or lossless, or both.
• If the network is reciprocal (not containing any active devices or nonreciprocal media, such as ferrites or plasmas), it can proved that the impedance and admittance matrices are symmetric, so that$Z_{i j} $=$Z _{ji}$, and$Y_{i }j = Y _{ji}$.
• If the network is lossless, it can shown that all the$Z_{i j} or Y_{i 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.
• Consider the arbitrary network of Figure to be reciprocal (no active devices, ferrites, or plasmas), with short circuits placed at all terminal planes except those of ports 1 and 2.

SYMMETRIC PROPERTY:

• Let$\hat E_ a, \hat H_a and \hat E_ b, \hat H_b$ be the fields anywhere in the network due to two independent sources, a and b, located somewhere in the network. Then the reciprocity theorem of (1.156) states that

$\int_{a}^{b}\hat E_ a \times\hat H_{b} .\hat{ds}= \int_{a}^{b}\hat E_ b \times\hat H_{a} .\hat{ds}$

• If the boundary walls of the network and transmission lines are metal, then E (tan) = 0 on these walls (assuming perfect conductors).
• If the network or the transmission lines are open structures, like microstrip line or slotline, the boundaries of the network can be taken arbitrarily far from the lines so that E tan is negligible.
• Then the only nonzero contribution to the integrals above come from the cross-sectional areas of ports 1 and 2.The fields due to sources a and b can be evaluated at the terminal planes t1 and t2 as

$\hat E _{1a} = V_{1a}\hat e 1; $

$ \hat E _{1b} = V_{1b}\hat e 1; $

$ \hat E _{2a} = V_{2a}\hat e 2; $

$ \hat E _{1b} = V_{1b}\hat e 2;$

$\hat H _{1a} = V_{1a}\hat h 1; $

$ \hat H _{1b} = V_{1b}\hat h 1; $

$ \hat H _{2a} = V_{2a}\hat h 2; $

$ \hat H_{1b} = V_{1b}\hat h 2;$

• where$ \hat e 1,\hat h 1 ,\hat e 2, \hat h 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. Substituting the above values in the above equations we have

$(V_{1a} I_{1b} - V_{1b} I_{1a})
\int_{a}^{b}\hat e 1 \times \hat h 1 .\hat {ds} + (V_{2a} I_{2b} - V_{2b} I_{2a})
\int_{a}^{b}\hat e 2 \times \hat h 2 ·\hat{ ds} = 0$

• Since

                                                  $\int_{a}^{b}\hat e 1 \times \hat h 1 ·\hat {ds} =
  \int_{a}^{b}\hat e 2 \times\hat h 2 ·\hat{ ds} = 1$

• we have

                                                $(V_{1a} I_{1b}- V_{1b} I_{1a})
  + (V_{2a} I_{2b} - V_{2b} I_{2a})
  = 0$

• We know that

$I_1 = Y_{11}V_{1} + Y_{12}V_{2}, $

$I_{2} = Y_{21}V_1 + Y_{22}V_{2}.$

• Substituting in the equation we have

                                                               $(V_{1a}V_{2b} -V_{1b}V_{2a})(Y_{12} +Y_{21}) = 0$

   

• Because the sources a and b are independent, the voltages V1a, V1b, V2a, and V2b can take on arbitrary values.Thus

$Y_{ij}=Y_{ji}$

• Then if [Y ] is a symmetric matrix, its inverse, [Z], is also symmetric.

UNIIARY PROPERTY:

• 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

$P_ {avg} = \dfrac1 2 [V]^t[I]^* = \dfrac1 2 ([Z][I])^t[I]^* = \dfrac1 2 [I]^t[Z][I]^*$

$ \dfrac1 2 (I_1Z_{11}I_1^* + I_1Z_{12}I_2^*+ I_2Z_{21}I_1^* + ....)$

• Because the In are independent, we must have the real part of each self term (In Znn In∗) equal to zero, since we could set all port currents equal to zero except for the nth current. So,

$Re({I_n Z_{nn} I_n^*}) = |I_n|^2 Re({Z_{nn}}) = 0$

• Now let all port currents be zero except for Im and In.

$Re (I_{n} I_{m} ^* + I_{m} I_{n}^*)Z_{mn} = 0$

• since$Z _{mn} = Z_{nm}.$
• However,$(I_{n} I_{m}^ * + I_m I_n^*) $is a purely real quantity that is, in general, nonzero. Thus we must have that

$Re({Z_{mn}})=0$