Properties of S Matrices Shifting Reference Planes.
we saw that for reciprocal networks the Z and Y matrices are:
1. Purely imaginary for lossless networks, and
2. Symmetric about the main diagonal for reciprocal networks.
In these two special instances, there are also special properties of the S matrix which we will discuss in this lecture.
Reciprocal Networks and S Matrices
In the case of reciprocal networks, it can be shown that
[s] = [S]t
where [s]t indicates the transpose of [S]. In other words, (1) is a statement that [S] is symmetric about the main diagonal, which is what we also observed for the Z and Y matrices.
Lossless Networks and S Matrices
The condition for a lossless network is a bit more obtuse for S matrices. As derived in your text, if a network is lossless then
[s]* ={[s]t} -1
which is a statement that [S] is a unitary matrix. This result can be put into a different, and possibly more useful, form by pre-multiplying (2) by [s]t
[s]t .[s] * = [s]t {[s]t}-1 = [I]
[I ] is the unit matrix defined as
Expanding (3) we obtain
Three special cases –
. Take row 1 times column 1:
S11S11* + S21 S21* + …. +SN1SN1* = 1
Generalizing this result gives
åNK=1 Ski Ski* = 1
In words, this result states that the dot product of any column
of [S] with the conjugate of that same column equals 1 (for a
lossless network).
. Take row 1 times column 2:
S11 S12* + S21 S22* + …..+ SN1 SN2* = 0
Generalizing this result gives
In words, this result states that the dot product of any column
of [S] with the conjugate of another column equals 0 (for a
lossless network).
. Applying (1) to (7): If the network is also reciprocal, then [S]
is symmetric and we can make a similar statement concerning
the rows of [S].
That is, the dot product of any row of [S] with the conjugate
of another row equals 0 (for a lossless network).
Example N16.1 In a homework assignment, the S matrix of a
two port network was given to be
Is the network reciprocal? Yes, because [S]t =[S].
Is the network lossless? This question often cannot be answered
simply by quick inspection of the S matrix. Rather, we will systematically apply the conditions stated above to the columns of the S matrix:
. C1:C1* : (0.2 + j0.4)(0.2 - j0.4) + (0.8 - j0.4)(0.8 + j0.4) =1
. C2:C2* : Same = 1
. C1:C2* : (0.2 + j0.4)(0.8 + j0.4) + (0.8 - j0.4)(0.2 - j0.4) = 0
. C2:C1* : Same = 0
Therefore, the network is lossless. As an aside, in Example N15.1 of the text, which we saw in the last lecture,
This network is obviously reciprocal, and it can be shown that
it’s also lossy.
Example N16.2 (Text Example 4.4). Determine the S
parameters for this T-network assuming a 50-om system
impedance, as shown.
First, take a general look at the circuit:
. It’s linear, so it must also be reciprocal. Consequently, [S]
must be symmetric (about the main diagonal).
. The circuit appears unchanged when “viewed” from either
port 1 or port 2. Consequently, S11 = S 22. Based on these observations, we only need to determine S11 and S21 since S22 = S11 and S12 = S21 . Proceeding, recall that S11 is the reflection coefficient at port 1 with port 2 matched:
The input impedance with port 2 matched is Zin = 8.56 +141.8 8.56 + 50 om = 50.00 om which, not oincidentally, equals Z0 ! With this Zin :
S11 = Zin - Z0 / Zin + Z0 = 0
which also implies S22 = 0.
Next, for S21 we apply V1 + with port 2 matched and measure V2 - :
At port 1, which we will also assume is the terminal plane,
V1 =V1+ + V1- . However, with 50-om termination at port 2, V1- = 0
(since S11 = 0). Therefore, V1 =V 1+ . Similarly,V2 = V2- . These last findings imply we can simply use voltage division to determine V2 - (from V2 ):
And
V2 = 50/50+8.56 . VA = 0.8538.0.8288V1 = 0.7077V1
Therefore
V2- = 0.7077V1+ Þ S21 = ½ =s12
The complete S matrix for the given T-network referenced to 50-
om system impedance is therefore
Lastly, notice that when port 2 is matched
S21 = ½ =T21|V2 = 0
So that
|T21|v2 = 0|2 = 1/2
which says that half of the power incident from port 1 is
transmitted to port 2 when port 2 is matched. We can now see
why this T-network is called a 3-dB attenuator.
Shifting Reference Planes
Recall that when we defined S parameters for a network,
terminal planes were defined for all ports. These are arbitrarilychosen phase locations on TLs connected to the ports. It turns out that S parameters change very simply and
predictably as the reference planes are varied along lossless TLs.
This fact can prove handy, especially in the lab.
Referring to Fig 4.9:
To be specific, [S] is the scattering matrix of the network with
reference planes (i.e., ports) at tn, while [S’] is the scattering matrix of the network with the reference planes shifted to tn,. Applying the definition of the scattering matrix in these two situations yields
[V-] = [s].[ V+]
And
[V’-] = [s ’].[ V’+]
We’ve shifted the reference planes along lossless TLs. Hence,
these voltage amplitudes only change phase as
Vn’+ = Vn sup>+ e +jqn
And
Vn’- = Vn- e-jqn
where qn= bn ln . Remember, these are the phase shifts when the phase planes are all moved away from the ports.
It is easy to prove these phase shift relationships in (10) and
(11). First, we know that
Vn+(zn) =Vn+ +e -bnzn . Hence, V n+ (-ln) =
Vn+ e+bnzn . Therefore, Vn’+ ?Vn+ ( -ln) =Vn+ e+qnzn , which is (10).
Likewise, Vn - (zn =Vn - e bnzn so that Vn- (-ln) =Vn- e-bnzn .Therefore, Vn-‘ º
Vn-(-ln) =Vn-e-qnzn , which is (11).
Now, armed only with this information in (10) and (11), we can
express [S?] in terms of [S]. Writing (10) and (11) in matrix form and substituting these into
The inverse of a diagonal matrix is simply a diagonal matrix
with inverted diagonal elements. So, we can pre-multiply (12)
by the inverse of the first matrix (which is known, and is also
not singular) giving:
Comparing (13) with (9) we can immediately deduce that:
Multiplying out this matrix equation gives:
Smn’ = Smn e-j(qm + qn)
and when m = n,
Smn’ = Snn e -j2qn
The factor of two in this last exponent arises since the wave
travels twice the electrical distance q n : once towards the port and once back to the new terminal plane tn .Equations (15) and (16) provide the simple transformations for S parameters when the phase planes are shifted away from the ports.
Many times you’ll find that your measured S parameters differ
from simulation by a phase angle, even though the magnitude is
in good agreement. This likely occurred because your terminal
planes were defined differently in your simulations than was set
during measurement.
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