3/4/2009 4_3 The Scattering Matrix 1/3
4.3 – The Scattering Matrix
Reading Assignment: pp. 174-183
Admittance and Impedance matrices use the quantities I (z), V (z), and Z (z) (or Y (z)).
Q: Is there an equivalent matrix for transmission line activity expressed in terms of V +(z ), V −(z ), and Γ()z ?
A: Yes! Its called the scattering matrix.
HO: THE SCATTERING MATRIX
Q: Can we likewise determine something physical about our device or network by simply looking at its scattering matrix?
A: HO: MATCHED, RECIPROCAL, LOSSLESS
EXAMPLE: A LOSSLESS, RECIPROCAL DEVICE
Q: Isn’t all this linear algebra a bit academic? I mean, it can’t help us design components, can it?
A: It sure can! An analysis of the scattering matrix can tell us if a certain device is even possible to construct, and if so, what the form of the device must be.
HO: THE MATCHED, LOSSLESS, RECIPROCAL 3-PORT NETWORK
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 4_3 The Scattering Matrix 2/3
HE ATCHED OSSLESS ECIPROCAL PORT ETWORK HO: T M , L , R 4- N
Q: But how are scattering parameters useful? How do we use them to solve or analyze real microwave circuit problems?
A: Study the examples provided below!
EXAMPLE: THE SCATTERING MATRIX
EXAMPLE: SCATTERING PARAMETERS
Q: OK, but how can we determine the scattering matrix of a device?
A: We must carefully apply our transmission line theory!
EXAMPLE: DETERMINING THE SCATTERING MATRIX
Q: Determining the Scattering Matrix of a multi-port device would seem to be particularly laborious. Is there any way to simplify the process?
A: Many (if not most) of the useful devices made by us humans exhibit a high degree of symmetry. This can greatly simplify circuit analysis—if we know how to exploit it!
HO: CIRCUIT SYMMETRY
EXAMPLE: USING SYMMETRY TO DETERMINING A SCATTERING MATRIX
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 4_3 The Scattering Matrix 3/3
Q: Is there any other way to use circuit symmetry to our advantage?
A: Absolutely! One of the most powerful tools in circuit analysis is Odd-Even Mode analysis.
HO: SYMMETRIC CIRCUIT ANALYSIS
HO: ODD-EVEN MODE ANALYSIS
EXAMPLE: ODD-EVEN MODE CIRCUIT ANALYSIS
Q: Aren’t you finished with this section yet?
A: Just one more very important thing.
HO: GENERALIZED SCATTERING PARAMETERS
EXAMPLE: THE SCATTERING MATRIX OF A CONNECTOR
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 1/13
The Scattering Matrix
At “low” frequencies, we can completely characterize a linear device or network using an impedance matrix, which relates the currents and voltages at each device terminal to the currents and voltages at all other terminals.
But, at microwave frequencies, it is difficult to measure total currents and voltages!
* Instead, we can measure the magnitude and phase of each of the two transmission line waves V + ()zVz and − ().
* In other words, we can determine the relationship between the incident and reflected wave at each device terminal to the incident and reflected waves at all other terminals.
These relationships are completely represented by the scattering matrix. It completely describes the behavior of a linear, multi-port device at a given frequency ω , and a given line impedance Z0.
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 2/13
Consider now the 4-port microwave device shown below:
− + V22()z Z V22(z ) 0 port zz= 2 22P + port 1 − V11()z port 3 V ()z 33 4-port Z0 microwave Z0 device − + zz11= P = V11()z zz33P V33()z port zz= 4 44P
Z0 + V ()z V − (z ) 44 44
Note that we have now characterized transmission line activity in terms of incident and “reflected” waves. Note the negative going “reflected” waves can be viewed as the waves exiting the multi-port network or device.
Æ Viewing transmission line activity this way, we can fully characterize a multi-port device by its scattering parameters!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 3/13
+ ≠ Say there exists an incident wave on port 1 (i.e., V11()z 0), while the incident waves on all other ports are known to be zero +++ (i.e., V22()zVzVz=== 33() 44( ) 0).
+ port 1 V11()z Say we measure/determine the + voltage of the wave flowing into + Z0 = V11()zz 1p port 1, at the port 1 plane (i.e., + − determine V11(zz= 1P )).
zz= − 11P V (z ) port 2 22 Say we then measure/determine +
the voltage of the wave flowing − V ()zz= Z0 out of port 2, at the port 2 22 2p plane (i.e., determine − V − zz= ). 22() 2P = zz22P
+ − The complex ratio between V11()zz= 1PP and Vzz 22(= 2 ) is know
as the scattering parameter S21:
−−− + j βz2P Vz()= zVe V ++jzβ () z Se===22 2P 02 02 21PP 21 +++− jzβ 1P Vz11()= z 1P Ve 01 V 01
Likewise, the scattering parameters S31 and S41 are:
− − Vz33()= z 3PPVz 44()= z 4 SS31 ==++ and 41 Vz11()== z 1PP Vz 11 () z 1
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 4/13
We of course could also define, say, scattering parameter S34 + as the ratio between the complex values V44()zz= 4P (the wave − into port 4) and V33()zz= 3P (the wave out of port 3), given that the input to all other ports (1,2, and 3) are zero.
Thus, more generally, the ratio of the wave incident on port n to the wave emerging from port m is:
− Vzmm()= z mP + SVzknmn ==≠+ (given that kk() 0 for all ) Vznn()= z nP
Note that frequently the port positions are assigned a zero
value (e.g., zz12PP==0, 0 ). This of course simplifies the scattering parameter calculation:
−−− + j β 0 Vzmm(0)= Ve0m V0 m Smn ===+++ − j β 0 Vznn(0)= Ve0n V0 n Microwave lobe We will generally assume that the port
locations are defined as znP = 0 , and thus use the above notation. But remember where this expression came from!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 5/13
Q: But how do we ensure that only one incident wave is non-zero ?
A: Terminate all other ports with a matched load!
Γ2L = 0
− + V22()z Z0 V22(z ) = 0
+ − V ()z V33()z 11 4-port Z0 microwave Z0 Γ3L = 0 device − + V11()z V ()z = 0 33
+ − V ()z = 0 Z0 V (z ) 33 44
Γ4L = 0
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 6/13
Note that if the ports are terminated in a matched load (i.e., ZZL = 0 ), then ΓnL = 0 and therefore:
+ Vnn(z ) = 0
In other words, terminating a port ensures that there will be no signal incident on that port!
Q: Just between you and me, I think you’ve messed this up! In all
previous handouts you said that if Γ = 0 , the wave in the minus L direction would be zero:
− V (z ) = 0 if ΓL = 0
but just now you said that the wave in the positive direction would be zero: + V (z ) = 0 if ΓL = 0
Of course, there is no way that both statements can be correct!
A: Actually, both statements are correct! You must be careful to understand the physical definitions of the plus and minus directions—in other words, the propagation directions of waves + − Vnn(z ) and Vnn(z )!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 7/13
For example, we originally analyzed this case:
V + (z )
Γ − ( ) = Γ = Z0 L V z 0 if L 0
V − ()z
In this original case, the wave incident on the load is V + ()z (plus direction), while the reflected wave is V − (z ) (minus direction).
Contrast this with the case we are now considering:
− port n Vnn(z )
N-port Microwave Z0 ΓnL Network
V + (z ) nn
For this current case, the situation is reversed. The wave − incident on the load is now denoted as Vnn(z ) (coming out of port n), while the wave reflected off the load is now denoted as + Vnn(z ) (going into port n ).
+ As a result, Vnn(z ) = 0 when ΓnL = 0!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 8/13
Γ Perhaps we could more generally state that for some load L :
reflected incident V (zz==LL) Γ V( zz= L)
For each case, you must be able to correctly identify the mathematical statement describing the wave incident on, and reflected from, some passive load.
Like most equations in engineering, the
variable names can change, but the physics
described by the mathematics will not!
Now, back to our discussion of S-parameters. We found that if
znP = 0 for all ports n, the scattering parameters could be + − directly written in terms of wave amplitudes V0n and V0m .
− V0m + SVzknmn ==≠+ (when kk() 0 for all ) V0n
Which we can now equivalently state as:
− V0m Smn = + (when all ports, except port n , are terminated in matched loads ) V0n
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 9/13
One more important note—notice that for the ports terminated in matched loads (i.e., those ports with no incident wave), the voltage of the exiting wave is also the total voltage!
+−−+j ββzjznn Vmm()zVe=+00 m Ve m
− + jzβ m =+0 Ve0m
− + jzβ m =Ve0m (for all terminated ports)
Thus, the value of the exiting wave at each terminated port is likewise the value of the total voltage at those ports:
+− Vmmm()0 =+VV00 − =+0 V0m − =V0m (for all terminated ports)
And so, we can express some of the scattering parameters equivalently as:
Vm ()0 Smi.e.mnmn =≠+ (for terminated port , , for ) V0n
You might find this result helpful if attempting to determine scattering parameters where mn≠ (e.g., S21, S43, S13), as we can often use traditional circuit theory to easily determine the total port voltage Vm ()0 .
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 10/13
However, we cannot use the expression above to determine the scattering parameters when mn= (e.g., S11, S22, S33).
Think about this! The scattering parameters for these cases are: − V0n Snn = + V0n
Therefore, port n is a port where there actually is some + incident wave V0n (port n is not terminated in a matched load!). And thus, the total voltage is not simply the value of the exiting wave, as both an incident wave and exiting wave exists at port n.
Γ2L = 0 +− − V11()000=+VV () 1 () V33()00=V ( )
+ V − ()z Z V (z ) = 0 22 0 22
+ V − ()z V11()z ≠ 0 33 4-port Z 0 microwave Z0 Γ3L = 0 device
− + V11()z = V33()z 0
+ − Z0 V44()z = 0 V44(z )
Γ4L = 0
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 11/13
Typically, it is much more difficult to determine/measure the scattering parameters of the form Snn , as opposed to scattering parameters of the form Smn (where mn≠ ) where there is only an exiting wave from port m !
We can use the scattering matrix to determine the solution for a more general circuit—one where the ports are not terminated in matched loads!
Q: I’m not understanding the importance scattering parameters. How are they useful to us microwave engineers?
A: Since the device is linear, we can apply superposition. The output at any port due to all the incident waves is simply the coherent sum of the output at that port due to each wave!
For example, the output wave at port 3 can be determined by (assuming znP = 0):
−++++ V03= SV34 04 +++ SV 33 03 SV 32 02 SV 31 01
More generally, the output at port m of an N-port device is:
N −+ V00mmnnnP==∑SV() z 0 n =1
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 12/13
This expression can be written in matrix form as:
VV− = S +
Where V− is the vector:
T −−−−−⎡⎤… V = ⎣⎦V,V,V,,V01 02 03 0N
and V+ is the vector:
T +++++⎡⎤… V = ⎣⎦V,V,V,,V01 02 03 0N
Therefore S is the scattering matrix:
⎡SS11… 1n ⎤ ⎢ ⎥ S = ⎢ ⎥ ⎢ ⎥ ⎣SSmmn1 ⎦
The scattering matrix is a N by N matrix that completely characterizes a linear, N-port device. Effectively, the scattering matrix describes a multi-port device the way that ΓL describes a single-port device (e.g., a load)!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 The Scattering Matrix 723 13/13
But beware! The values of the scattering matrix for a
particular device or network, just like ΓL , are frequency dependent! Thus, it may be more instructive to explicitly write:
⎡⎤SS11(ω) … 1n (ω) ⎢⎥ S ()ω = ⎢⎥ ⎢⎥ ⎣⎦SSmmn1 ()ω ()ω
Also realize that—also just like ΓL—the scattering matrix is dependent on both the device/network and the Z0 value of the transmission lines connected to it.
Thus, a device connected to transmission lines with
Z0 =Ω50 will have a completely different scattering matrix than that same device connected to transmission lines with Z0 =Ω100 !!!
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 1/9
Matched, Lossless, Reciprocal Devices
As we discussed earlier, a device can be lossless or reciprocal. In addition, we can likewise classify it as being matched.
Let’s examine each of these three characteristics, and how they relate to the scattering matrix.
Matched
A matched device is another way of saying that the input impedance at each port is equal to Z0 when all other ports are terminated in matched loads. As a result, the reflection coefficient of each port is zero—no signal will be come out of a port if a signal is incident on that port (but only that port!).
In other words, we want:
− + Vmmmm==SV0 for all m a result that occurs when:
Smmm = 0 for all if matched
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 2/9
We find therefore that a matched device will exhibit a scattering matrix where all diagonal elements are zero.
Therefore: ⎡ 00.10.2j ⎤ ⎢ ⎥ S = ⎢ 0.1 0 0.3 ⎥ ⎢ ⎥ ⎣⎢ j 0.2 0.3 0 ⎦⎥ is an example of a scattering matrix for a matched, three port device.
Lossless
For a lossless device, all of the power that delivered to each device port must eventually find its way out!
In other words, power is not absorbed by the network—no power to be converted to heat!
Recall the power incident on some port m is related to the + amplitude of the incident wave (V0m ) as:
+ 2 + V0m Pm = 2Z0
While power of the wave exiting the port is:
− 2 − V0m Pm = 2Z0
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 3/9
Thus, the power delivered to (absorbed by) that port is the difference of the two: + 22− + − VV0m 0m ∆PPPmmm= − = − 22ZZ00
Thus, the total power incident on an N-port device is:
NN ++1 +2 PP==∑∑m V0m mm==112Z0
Note that: N +++2 H ∑ V0m = ()VV m=1 where operator H indicates the conjugate transpose (i.e., H Hermetian transpose) operation, so that (VV++) is the inner product (i.e., dot product, or scalar product) of complex vector V+ with itself.
Thus, we can write the total power incident on the device as:
N ++H ++1 2 (VV) PV==∑ 0m 22ZZ00m=1
Similarly, we can express the total power of the waves exiting our M-port network to be: N −−H −−1 2 (VV) PV==∑ 0m 22ZZ00m=1
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 4/9
Now, recalling that the incident and exiting wave amplitudes are related by the scattering matrix of the device:
VV− = S +
Thus we find: HH (VV−−) ( V++) SSH V P − == 22ZZ00
Now, the total power delivered to the network is:
M ∆PPPP= ∑ ∆ = + − − m=1 Or explicitly: ∆PP= + − P− HH ()VV++() V +SSH V + = − 22ZZ00 1 H = ()VV++()ISS− H 2Z0 where I is the identity matrix.
Q: Is there actually some point to this long, rambling, complex presentation?
A: Absolutely! If our M-port device is lossless then the total power exiting the device must always be equal to the total power incident on it.
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 5/9
If network is lossless, then PP+ = − .
Or stated another way, the total power delivered to the device (i.e., the power absorbed by the device) must always be zero if the device is lossless!
If network is lossless, then ∆P = 0
Thus, we can conclude from our math that for a lossless device:
1 H ∆P = ()VVV+++()ISS− H = 0 for all 2Z0
This is true only if:
ISS− HH= 0 ⇒ SS = I
Thus, we can conclude that the scattering matrix of a lossless device has the characteristic:
If a network is lossless , then SS H = I
Q: Huh? What exactly is this supposed to tell us?
A: A matrix that satisfies SS H = I is a special kind of matrix known as a unitary matrix.
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 6/9
If a network is lossless , then its scattering matrix S is unitary .
Q: How do I recognize a unitary matrix if I see one?
A: The columns of a unitary matrix form an orthonormal set!
⎡⎤S S S S ⎢⎥11 12 13 14 ⎢⎥S S S S S = 21 22 23 22 ⎢⎥S S S S ⎢⎥31 33 33 33 ⎣⎦⎢⎥S41 S42 S43 S44
matrix
columns
In other words, each column of the scattering matrix will have a magnitude equal to one:
N 2 ∑ Smn = 1 for all n m =1 while the inner product (i.e., dot product) of dissimilar columns must be zero.
N ∗∗ ∗∗" ∑SSniSSnj=+12 i12 j S i S j ++= S Ni SNj 0 for all ij ≠ n =1
In other words, dissimilar columns are orthogonal.
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 7/9
Consider, for example, a lossless three-port device. Say a signal is incident on port 1, and that all other ports are terminated. The power incident on port 1 is therefore:
+ 2 + V01 P1 = 2Z0 while the power exiting the device at each port is:
−−22 − VSV0101mm 2 + PSPmm== =1 1 22ZZ00
The total power exiting the device is therefore:
−−−− PPPP=++123 222+++ =++SP11111 SP 21 SP 31 222+ =++( SSSP11 21 31 ) 1
Since this device is lossless, then the incident power (only on − + port 1) is equal to exiting power (i.e, PP= 1 ). This is true only if: 222 SSS131211++=1
Of course, this will likewise be true if the incident wave is placed on any of the other ports of this lossless device:
SSS222++=1 122322 222 SSS13++= 23 33 1
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 8/9
We can state in general then that:
3 2 ∑ Snmn = 1 for all m =1
In other words, the columns of the scattering matrix must have unit magnitude (a requirement of all unitary matrices). It is apparent that this must be true for energy to be conserved.
An example of a (unitary) scattering matrix for a lossless device is:
1 3 ⎡⎤002 j 2 ⎢⎥ 1 0 0 j 3 S = ⎢⎥22 3 1 ⎢⎥j 220 0 ⎢⎥ 3 1 ⎣⎦00j 2 2
Reciprocal
Recall reciprocity results when we build a passive (i.e., unpowered) device with simple materials.
For a reciprocal network, we find that the elements of the scattering matrix are related as:
SSmn= nm
Jim Stiles The Univ. of Kansas Dept. of EECS 02/23/07 Matched reciprocal lossless 723 9/9
= For example, a reciprocal device will have SS21 12 or SS32= 23 . We can write reciprocity in matrix form as:
SST = if reciprocal
where T indicates (non-conjugate) transpose.
An example of a scattering matrix describing a reciprocal, but lossy and non-matched device is:
⎡⎤0.10−0.40 −j 0.20 0.05 ⎢⎥ −0.40j 0.20 0j 0.10 S = ⎢⎥ ⎢⎥−−−jj0.200 0.10 0.30 0.12 ⎢⎥ ⎣⎦0.05j 0.10 − 0.12 0
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example A Lossless Reciprocal Network 1/4
Example: A Lossless, Reciprocal Network
A lossless, reciprocal 3-port device has S-parameters of
S11 = 12, S31 = 12, and S33 = 0 . It is likewise known that all scattering parameters are real.
Æ Find the remaining 6 scattering parameters.
Q: This problem is clearly impossible—you have not provided us with sufficient information!
A: Yes I have! Note I said the device was lossless and reciprocal!
Start with what we currently know:
1 ⎡ 2 SS12 13 ⎤ ⎢ ⎥ S = ⎢SSS21 22 23 ⎥ ⎢ 1 ⎥ ⎣ 2 S32 0 ⎦
Because the device is reciprocal, we then also know:
1 SS21= 12 SS13= 31 = 2 SS32= 23
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example A Lossless Reciprocal Network 2/4
And therefore:
11 ⎡ 2 S21 ⎤ ⎢ 2 ⎥ S = ⎢SSS21 22 32 ⎥ ⎢ 1 ⎥ ⎣ 2 S32 0 ⎦
Now, since the device is lossless, we know that:
222 1 =++SSS11 21 31
2 2 2 11 =++()2 S21 ()2
2 1 =++SSS22 Columns have 12 22 32 222 unit magnitude. =++SSS21 22 32
222 1 =++SSS13 23 33
2 2 2 11 =+()2 S32 +()2 and:
0 =+SS∗ SS∗∗ + SS 11 12 21 22 31 32 11∗ ∗∗ =+2 SSS21 21 22 +2 S 32 Columns are ∗∗ ∗orthogonal. 0 =+SS11 13 SS 21 23 + SS 31 33
11∗ 1 =++2 ()22SS21 32 ()0
∗∗ ∗ 0 =+SS12 13 SS 22 23 + SS 32 33
1 ∗ =++SSSS21()2 22 32 32 ()0
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example A Lossless Reciprocal Network 3/4
These six expressions simplify to:
1 S21 = 2
222 1 =++SSS21 22 32
1 S32 = 2
11 0 =+2 SSS21 21 22 +2 S 32
0 = 1 + SS ()22 21 32
1 0 =+SSS21( 2 ) 22 32 where we have used the fact that since the elements are all ∗ real, then SS21= 21 (etc.).
Q Q: I count the expressions and find 6 equations yet only a paltry 3 unknowns. Your typical buffoonery appears to have led to an over-constrained condition for which there is no solution!
A: Actually, we have six real equations and six real unknowns, since scattering element has a magnitude and phase. In this case we know the values are real, and thus D D j 0 the phase is either 0 or 180 (i.e., e = 1 or e jπ =−1); however, we do not know which one!
From the first three equations, we can find the magnitudes:
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example A Lossless Reciprocal Network 4/4
S = 1 S = 1 S = 1 21 2 22 2 32 2
and from the last three equations we find the phase:
1 1 1 S21 = 2 S22 = 2 S32 = − 2
Thus, the scattering matrix for this lossless, reciprocal device is: 111 ⎡ 22 ⎤ ⎢ 2 ⎥ S = 111− ⎢ 22 2 ⎥ ⎢ 11− ⎥ ⎣ 220 ⎦
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 MLR 3 port network 1/2
A Matched, Lossless Reciprocal 3-Port Network
Consider a 3-port device. Such a device would have a scattering matrix :
⎡SSS11 12 13 ⎤ S = ⎢ ⎥ ⎢SSS21 22 23 ⎥ ⎢ ⎥ ⎣SSS31 32 33 ⎦
Assuming the device is passive and made of simple (isotropic) materials, the device will be reciprocal, so that:
SS21=== 12 SS 31 13 SS 23 32
Likewise, if it is matched, we know that:
SS11= 22== S 33 0
As a result, a lossless, reciprocal device would have a scattering matrix of the form:
⎡ 0 SS21 31 ⎤ S = ⎢ ⎥ ⎢SS210 32 ⎥ ⎢ ⎥ ⎣SS31 32 0 ⎦
Just 3 non-zero scattering parameters define the entire matrix!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 MLR 3 port network 2/2
Likewise, if we wish for this network to be lossless, the scattering matrix must be unitary, and therefore:
2 2 ∗ SS21+= 311 SS 31 32 = 0 2 2 ∗ SS21+= 321 SS 21 32 = 0 22 ∗ SS31+= 321 SS 21 31 = 0
Since each complex value S is represented by two real numbers (i.e., real and imaginary parts), the equations above result in 9
real equations. The problem is, the 3 complex values S21, S31 and
S32 are represented by only 6 real unknowns.
We have over constrained our problem ! There are no solutions to these equations !
As unlikely as it might seem, this means that a matched, lossless, reciprocal 3- port device of any kind is a physical impossibility !
You can make a lossless reciprocal 3- port device, or a matched reciprocal 3- port device, or even a matched, lossless (but non-reciprocal) 3-port network.
But try as you might, you cannot make a lossless, matched, and reciprocal three port component!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 MLR 4 port network 1/3
The Matched, Lossless, Reciprocal 4-Port Network
Gu ess what! I have determined that—unlike a 3-port device—a matched, lossless, reciprocal 4-port device is physically possible! In fact, I’ve found two general solutions!
The first solution is referred to as the symmetric solution:
⎡ 00α j β ⎤ ⎢ ⎥ α 0 0 j β S = ⎢ ⎥ ⎢j β 0 0 α ⎥ ⎢ ⎥ ⎣ 00j β α ⎦
Note for this symmetric solution, every row and every column of the scattering matrix has the same four values (i.e., α, jβ, and two zeros)!
The second solution is referred to as the anti-symmetric solution: ⎡00α β ⎤ ⎢α 00−β ⎥ S = ⎢ ⎥ ⎢β 00α ⎥ ⎢ ⎥ ⎣00−βα ⎦
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 MLR 4 port network 2/3
Note that for this anti-symmetric solution, two rows and two columns have the same four values (i.e., α, β, and two zeros), while the other two row and columns have (slightly) different values (α, -β, and two zeros)
It is quite evident that each of these solutions are matched and reciprocal. However, to ensure that the solutions are indeed lossless, we must place an additional constraint on the values of α, β. Recall that a necessary condition for a lossless device is:
N 2 ∑ Snmn = 1 for all m =1
Applying this to the symmetric case, we find:
αβ22+=1
Likewise, for the anti-symmetric case, we also get
αβ22+=1
It is evident that if the scattering matrix is unitary (i.e., lossless), the values α and β cannot be independent, but must related as:
αβ22+=1
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 MLR 4 port network 3/3
α ≥ β Generally speaking, we will find that . Given the constraint on these two values, we can thus conclude that:
0 ≤≤β 1 and 1 ≤ α ≤ 1 2 2
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 1/6
Example: The Scattering Matrix
Say we have a 3-port network that is completely characterized at some frequency ω by the scattering matrix:
⎡0.0 0.2 0.5⎤ ⎢ ⎥ S = ⎢0.5 0.0 0.2⎥ ⎣⎢0.5 0.5 0.0⎦⎥
A matched load is attached to port 2, while a short circuit has been placed at port 3:
ZZ= 0
zP 2 = 0
Z + − 0 V (z) V2 (z) 2 port 2 z = 0 + − P 1 zP 3 = 0 V1 (z) V3 (z) 3-port Z 0 port 1 microwave port 3 Z0 Z = 0 device
− + V1 (z) V3 (z)
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 2/6
Γ = Because of the matched load at port 2 (i.e., L 0 ), we know that: + + V22(0)z = V02 −−= = 0 Vz22(0)= V 02
and therefore: + V02 = 0
You’ve made a terrible mistake!
Fortunately, I was here to
correct it for you—since Γ=0 , L − + the constant V02 (not V02 ) is equal to zero.
− NO!! Remember, the signal V2 ()z is incident on the matched + + load, and V2 ()z is the reflected wave from the load (i.e., V2 ()z + is incident on port 2). Therefore, V02 = 0 is correct!
Likewise, because of the short circuit at port 3 (Γ=−L 1):
+ + V33(0)zV= 03 −−= =−1 Vz33(0)= V 03 and therefore:
+ − V03= −V 03
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 3/6
Problem: a) Find the reflection coefficient at port 1, i.e.:
− V01 Γ1 + V01
b) Find the transmission coefficient from port 1 to port 2, i.e.,
− V02 T21 + V01
I am amused by the trivial problems that you apparently find so difficult. I know that:
− V01 Γ=111+ =S =0.0 V01 and
− V02 T21===+ S 21 0.5 V01
NO!!! The above statement is not correct!
−+ Remember, V01VS 01= 11 only if ports 2 and 3 are terminated in matched loads! In this problem port 3 is terminated with a short circuit.
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 4/6
Therefore: − V01 Γ=111+ ≠S V01 and similarly:
− V02 T21=≠+ S 21 V01
To determine the values T21 and Γ1 , we must start with the three equations provided by the scattering matrix:
−++ V01=+02.V 02 05 .V 03
−+ + V02=+05.V 01 02 .V 03
−++ V03=+05.V 01 05 .V 02
and the two equations provided by the attached loads:
+ V02 = 0
+ − V03= −V 03
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 5/6
+ We can divide all of these equations by V01 , resulting in:
−++ VVV01 02 03 Γ=1 +++=+02.. 05 VVV01 01 01
− + VV02 03 T21 ==+ 05.. + 02 + VV01 01
−+ VV03 02 ++=+05.. 05 VV01 01
+ V02 + = 0 V01
+− VV03 03 ++=− VV01 01
Look what we have—5 equations and 5 unknowns! Inserting equations 4 and 5 into equations 1 through 3, we get:
− + VV01 03 Γ=1 ++=−05. VV01 01
− + VV02 03 T21 ==−+ 05.. 02 + VV01 01
− V03 + = 05. V01
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example The Scattering Matrix 6/6
Solving, we find:
Γ1 =−05..( 05) =− 025 .
T21 =−05.... 02() 05 = 04
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Scattering Parameters 1/4
Example: Scattering Parameters
Consider a two-port device with a scattering matrix (at some specific frequency ω0 ):
⎡ 01.j. 07⎤ S ()ωω==0 ⎢ ⎥ ⎣j 07..− 02⎦
and Z0 =Ω50 .
Say that the transmission line connected to port 2 of this device is terminated in a matched load, and that the wave incident on port 1 is:
+ − j βz1 V11(zje) =− 2
where zz12PP==0 .
Determine:
1. the port voltages V11(zz= 1P ) and V22(zz= 2P ).
2. the port currents Iz11( = z 1P ) and Iz22( = z 2P ).
3. the net power flowing into port 1
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Scattering Parameters 2/4
1. Since the incident wave on port 1 is:
+ − j βz1 V11(zje) =− 2
we can conclude (since z1P = 0 ):
+ − j βz1P V11()zz==− 1P je2 =−je2 − j β ()0 =−j 2 and since port 2 is matched (and only because its matched!), we find:
−+ V11()zz== 1PP SVzz 1111( = 1) =−01.j() 2 =−j.02 The voltage at port 1 is thus:
+ − V11()zz== 1PPP Vzz 1( 1 =+ 1) Vzz 1( 1 = 1) =−j.20 − j. 02
=−j.22
− j π = 22.e 2 Likewise, since port 2 is matched:
+ V22(zz= 2P ) = 0
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Scattering Parameters 3/4
And also:
−+ V22()zz== 2PP SVzz 2111( = 1) =−j.07() j 2 = 14. Therefore:
+− V22()zz== 2PPP Vzz 2( 2 =+ 2) Vzz 2( 2 = 2) =+014.
= 14. = 14.e− j 0
2. The port currents can be easily determined from the results of the previous section.
+ − Iz11()== z 1PPP I 1( z 1 =− z 1) I 1( z 1 = z 1) Vz+−()== z Vz() z =−11 1PP 11 1 ZZ00 20.. 02 =−jj + 50 50 18. =−j 50 =−j.0036
− j π = 0036.e2 and:
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Scattering Parameters 4/4
Iz== z I+− z =− z I z = z 22() 2PPP 2( 2 2) 2( 2 2) Vz+−()== z Vz() z =−22 2PP 22 2 ZZ00 014. =− 50 50 =−0. 028 = 0.e 028 + jπ
3. The net power flowing into port 1 is:
+− ∆=PP11 − P 1 22 VV+− =−01 01 22ZZ 00 22 ()202− (. ) = 250() = 00396. Watts
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Determining the Scattering Matrix 1/5
Example: Determining the Scattering Matrix
Let’s determine the scattering matrix of this two-port device:
Z0 2Z0 Z0
z z1 2 = z1P = 0 z2P 0
The first step is to terminate port 2 with a matched load, and then determine the values:
− − V11()zz= P 1 and V22(zz= P 2)
+ in terms of V11()zz= P 1.
+ +
Z0 V ()z 2Z0 = Z0 11 V22()z 0 − −
z1
z = 0 z = 0 1P 2P
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Determining the Scattering Matrix 2/5
Recall that since port 2 is matched, we know that:
+ V22(zz= 2P ) = 0 And thus:
+ − V22()zVzVz==000 2( 2 =+) 2( 2 =) − =+00Vz22() = − ==Vz22()0
In other words, we simply need to determine V22(z = 0) in order − to find V22()z = 0 !
− However, determining V11(z = 0) is a bit trickier. Recall that:
+ − V11()zVzVz= 1( 1) + 1( 1)
− Therefore we find V11()zVz= 00≠= 1( 1 ) !
Now, we can simplify this circuit:
+ 2 Z0 V11()z Z 3 0 −
z1
z1P = 0
And we know from the telegraphers equations:
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Determining the Scattering Matrix 3/5
V ()zVzVz= + ( ) + − ( ) 11 1 1 1 1 +−− j β+βzjz11 =+Ve01 Ve 01
Since the load 23Z0 is located at z1 = 0 , we know that the boundary condition leads to:
+ −βjz11 +β jz V11()zVe=+ 01( ΓL e ) where: (2 )ZZ− Γ = 3 00 L 2 ()3 ZZ00+ ()2 − 1 = 3 2 ()3 + 1 − 1 = 3 5 3 = −02. Therefore:
++−βj z1 −+ +βj z1 V11()zVe= 01 and V11(zV) =− 01( 02 .e) and thus: +++−βj (0) V1()zVeV 1= 0 == 01 01
−++βj (0) + V1()zV.e.V 1==00202 01( −) =− 01
We can now determine S11 !
− + Vz11( = 0) −02.V01 S.11 ===−++02 Vz11()= 0 V 01
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Determining the Scattering Matrix 4/5
− = Now its time to find V22(z 0)!
Again, since port 2 is terminated, the incident wave on port 2 must be zero, and thus the value of the exiting wave at port 2 is equal to the total voltage at port 2:
− V22(zVz= 00) == 22( )
This total voltage is relatively easy to determine. Examining the circuit, it is evident that V11(zVz= 00) == 22( ).
+ + Z 0 V11()z = 0 V22()z = 0 Z0 2Z0 − −
z1
z1P = 0 z2P = 0 Therefore:
V22()zVz==00 11( =) + −βjj()00 +β () =−Ve01 ()02 .e
+ =−V.01 ()102 + =V.01 ()08
And thus the scattering parameter S21 is:
− + Vz22( = 0) 08.V01 S.21 ===++08 Vz11()= 0 V 01
Jim Stiles The Univ. of Kansas Dept. of EECS 2/23/2007 Example Determining the Scattering Matrix 5/5
Now we just need to find S12 and S22 .
Q: Yikes! This has been an awful lot of work, and you mean that we are only half-way done!?
A: Actually, we are nearly finished! Note that this circuit is symmetric—there is really no difference between port 1 and port 2. If we “flip” the circuit, it remains unchanged!
Z 2Z Z 0 0 0
z2 z1
z = 0 z1P = 0 2P Thus, we can conclude due to this symmetry that:
SS11= 22 =−02 . and:
SS21= 12 = 08 .
Note this last equation is likewise a result of reciprocity.
Thus, the scattering matrix for this two port network is:
⎡−02.. 08⎤ S = ⎢ ⎥ ⎣ 08..− 02⎦
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 1/14
Circuit Symmetry
One of the most powerful concepts in for evaluating circuits is that of symmetry. Normal humans have a conceptual understanding of symmetry, based on an esthetic perception of structures and figures.
On the other hand, mathematicians (as they are wont to do) have defined symmetry in a very precise and unambiguous way. Using a branch of mathematics called Group Theory, first developed by the young genius Évariste Galois (1811-1832), symmetry is defined by a set of operations (a group) Évariste Galois that leaves an object unchanged.
Initially, the symmetric “objects” under consideration by Galois were polynomial functions, but group theory can likewise be applied to evaluate the symmetry of structures.
For example, consider an ordinary equilateral triangle; we find that it is a highly symmetric object!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 2/14
Q: Obviously this is true. We don’t need a mathematician to tell us that!
A: Yes, but how symmetric is it? How does the symmetry of an equilateral triangle compare to that of an isosceles triangle, a rectangle, or a square?
To determine its level of symmetry, let’s first label each corner as corner 1, corner 2, and corner 3.
2
1 3
First, we note that the triangle exhibits a plane of reflection symmetry:
2
1 3
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 3/14
Thus, if we “reflect” the triangle across this plane we get:
2
3 1
Note that although corners 1 and 3 have changed places, the triangle itself remains unchanged—that is, it has the same shape, same size, and same orientation after reflecting across the symmetric plane!
Mathematicians say that these two triangles are congruent.
Note that we can write this reflection operation as a permutation (an exchange of position) of the corners, defined as: 13→ 22→ 31→
Q: But wait! Isn’t there is more than just one plane of reflection symmetry?
A: Definitely! There are two more:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 4/14
2 1
12→ 21→ 1 3 33→ 2 3
2 3 11→ 23→ 1 3 32→ 1 2
In addition, an equilateral triangle exhibits rotation symmetry!
Rotating the triangle 120D clockwise also results in a congruent triangle:
2 1 12→ 23→ 31→ 1 3 3 2
Likewise, rotating the triangle 120D counter-clockwise results in a congruent triangle:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 5/14
2 13→ 3 21→ 32→
1 3 2 1
Additionally, there is one more operation that will result in a congruent triangle—do nothing!
2 2 11→ 22→ 33→ 1 3 1 3
This seemingly trivial operation is known as the identity operation, and is an element of every symmetry group.
These 6 operations form the dihedral symmetry group D3 which has order six (i.e., it consists of six operations). An object that remains congruent when operated on by any and
all of these six operations is said to have D3 symmetry.
An equilateral triangle has D3 symmetry!
By applying a similar analysis to a isosceles triangle, rectangle, and square, we find that:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 6/14
An isosceles trapezoid has D1 symmetry, a D1 dihedral group of order 2.
A rectangle has D2 symmetry, a dihedral group D 2 of order 4.
A square has D symmetry, a dihedral group of D 4 4 order 8.
Thus, a square is the most symmetric object of the four we have discussed; the isosceles trapezoid is the least.
Q: Well that’s all just fascinating—but just what the heck does this have to do with microwave circuits!?!
A: Plenty! Useful circuits often display high levels of symmetry.
For example consider these D1 symmetric multi-port circuits:
Port 1 Port 2 12→ 50Ω 21→ 200Ω 200Ω
34→ 100Ω 43→ Port 3 Port 4
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 7/14
→ Port 1 13 50Ω Port 2 24→ 100Ω 200Ω
31→ 50 Ω 42→
Port 3 Port 4
Or this circuit with D2 symmetry:
Port 1 50Ω Port 2 Ω 200Ω 200
50 Ω
Port 3 Port 4
which is congruent under these permutations:
13→ 12→ 14→ 24→ 21→ 23→
31→ 34→ 32→ 42→ 43→ 41→
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 8/14
Or this circuit with D4 symmetry:
Port 1 50Ω Port 2
50 Ω 50 Ω
50 Ω
Port 3 Port 4
which is congruent under these permutations:
13→ 12→ 14→ 14→ 11→ 24→ 21→ 23→ 22→ 23→
31→ 34→ 32→ 33→ 32→ 42→ 43→ 41→ 41→ 44→
The importance of this can be seen when considering the scattering matrix, impedance matrix, or admittance matrix of these networks.
For example, consider again this symmetric circuit:
Port 1 Port 2 50Ω 200Ω 200Ω
100Ω
Port 3 Port 4
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 9/14
This four-port network has a single plane of reflection symmetry (i.e., D1 symmetry), and thus is congruent under the permutation: 12→ 21→
34→ 43→
So, since (for example) 12→ , we find that for this circuit:
SS11=== 22 Z 11 Z 22 YY 11 22 must be true!
Or, since 12→ and 34→ we find:
SS13===24 ZZ 13 24 YY 13 24
SS31===42 ZZ 31 42 YY 31 42
Continuing for all elements of the permutation, we find that for this symmetric circuit, the scattering matrix must have this form:
⎡SSSS11 21 13 14 ⎤ ⎢SSSS⎥ S = ⎢ 21 1114 13 ⎥ ⎢SSSS3141 33 43 ⎥ ⎢ ⎥ ⎣SSSS4131 43 33 ⎦
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 10/14
and the impedance and admittance matrices would likewise have this same form.
Note there are just 8 independent elements in this matrix. If we also consider reciprocity (a constraint independent of symmetry) we find that SS31= 13 and SS41= 14 , and the matrix reduces further to one with just 6 independent elements:
⎡S11 S21 S31 S41 ⎤ ⎢S S S S ⎥ S = ⎢ 21 11 41 31 ⎥ ⎢S31 S41 S33 S43 ⎥ ⎢ ⎥ ⎣S41 S31 S43 S33 ⎦
Or, for circuits with this D1 symmetry:
Port 1 50 Ω Port 2 → 13 100Ω 200Ω 24→
31→ 50 Ω 42→ Port 3 Port 4
⎡S11 S21 S31 S41 ⎤ ⎢S S S S ⎥ S = ⎢ 21 22 41 31 ⎥ ⎢S31 S41 S11 S21 ⎥ ⎢ ⎥ ⎣S41 S31 S21 S22 ⎦
Q: Interesting. But why do we care?
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 11/14
A: This will greatly simplify the analysis of this symmetric circuit, as we need to determine only six matrix elements!
For a circuit with D2 symmetry:
Port 1 50Ω Port 2 200Ω 200Ω
50 Ω
Port 3 Port 4
we find that the impedance (or scattering, or admittance) matrix has the form:
⎡Z11 Z21 Z31 Z 41 ⎤ ⎢ ⎥ Z Z Z Z Z = ⎢ 21 11 41 31 ⎥ ⎢Z31 Z 41 Z11 Z21 ⎥ ⎢ ⎥ ⎣Z 41 Z31 Z21 Z11 ⎦
Note here that there are just four independent values!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 12/14
For a circuit with D4 symmetry:
Port 1 50Ω Port 2
50 Ω 50 Ω
50 Ω
Port 3 Port 4
we find that the admittance (or scattering, or impedance) matrix has the form:
⎡Y11 YY21 21 Y41 ⎤ ⎢ ⎥ YYY Y Y = ⎢ 2111 41 21 ⎥ ⎢YY21Y41 Y11 21 ⎥ ⎢ ⎥ ⎣Y41 YY21 21 Y11 ⎦
Note here that there are just three independent values!
One more interesting thing (yet another one!); recall that we earlier found that a matched, lossless, reciprocal 4-port device must have a scattering matrix with one of two forms:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 13/14
⎡⎤00α j β ⎢⎥ α 0 0 j β S = ⎢⎥ The “symmetric” solution ⎢⎥j β 0 0 α ⎢⎥ ⎣⎦00j β α
⎡⎤00α β ⎢⎥α 00−β S = ⎢⎥ The “anti-symmetric” solution ⎢⎥β 00α ⎢⎥ ⎣⎦00−βα
Compare these to the matrix forms above. The “symmetric solution” has the same form as the scattering matrix of a
circuit with D2 symmetry!
⎡ 0 α j β 0 ⎤ ⎢ ⎥ α 0 0 j β S = ⎢ ⎥ ⎢j β 0 0 α ⎥ ⎢ ⎥ ⎣ 0 j β α 0 ⎦
Q: Does this mean that a matched, lossless, reciprocal four- port device with the “symmetric” scattering matrix must
exhibit D2 symmetry?
A: That’s exactly what it means!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/4/2009 Circuit Symmetry 14/14
Not only can we determine from the form of the scattering matrix whether a particular design is possible (e.g., a matched, lossless, reciprocal 3-port device is impossible), we can also determine the general structure of a possible solutions (e.g. the circuit must have D2 symmetry).
Likewise, the “anti-symmetric” matched, lossless, reciprocal four-port network must exhibit D1 symmetry!
⎡0 α β 0 ⎤ ⎢α 0 0 −β ⎥ S = ⎢ ⎥ ⎢β 0 0 α ⎥ ⎢ ⎥ ⎣0 −β α 0 ⎦
We’ll see just what these symmetric, matched, lossless, reciprocal four-port circuits actually are later in the course!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Using Symmetry to Determine S 1/3
Example: Using Symmetry to Determine a Scattering Matrix
Say we wish to determine the scattering matrix of the simple two-port device shown below:
port port β Z , β Z , β Z0 , 0 0 1 2
z =−A z = 0
We note that that attaching transmission lines of
characteristic impedance Z 0 to each port of our “circuit” forms a continuous transmission line of characteristic impedance Z 0 .
Thus, the voltage all along this transmission line thus has the form: +−−+j βzjzβ V ()zVe=+00 Ve
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Using Symmetry to Determine S 2/3
=−A We begin by defining the location of port 1 as z1P , and the port location of port 2 as z2P = 0:
We can thus conclude:
++− j βz V10()zVe= ( z≤−A)
−−+ jzβ V10()zVe= ( z≤−A )
+−+ jzβ V20()zVe=≥() z0
−+− jzβ V20()zVe=≥() z0
+ V +e − j βz V −(z ) V1 ()z 0 2
− + + V ()z − + j βz 1 V e V2 (z ) V ()z 0
−
z =−A z = 0 Say the transmission line on port 2 is terminated in a matched − load. We know that the –z wave must be zero (V0 = 0), and so the voltage along the transmission line becomes simply the +z wave voltage: + − j βz V (zVe) = 0 and so:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Using Symmetry to Determine S 3/3
V ++()zVe==≤−− j βz Vz −( ) 0 ( z A) 10 1
+−+− jzβ V220()zVzVez==≥00 () ()
Now, because port 2 is terminated in a matched load, we can determine the scattering parameters S11 and S21 :
− − Vzz11()= P Vz()=−A 0 S11 ====+++ −−jβ()A 0 Vzz11()==−P + Vz()A + Ve0 V2 =0 V2 =0
−−+ − jβ (0) Vzz22()==P Vz 2( 0) Ve0 1 − jβA Se21 =====+++ −−jβ()A + jβA Vzz11()==−P + Vz 1()A + Ve0 e V2 =0 V2 =0
From the symmetry of the structure, we can conclude:
SS22= 11 = 0
And from both reciprocity and symmetry:
− j βA SSe12== 21
⎡ 0 e − j βA ⎤ Thus: S = ⎢ − j βA ⎥ ⎣⎢e 0 ⎦⎥
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 1/10
Symmetric Circuit Analysis
Consider the following D1 symmetric two-port device: 200Ω
I I1 100Ω 100Ω 2
+ + 50Ω V2
-
Q: Yikes! The plane of reflection symmetry slices through two resistors. What can we do about that?
A: Resistors are easily split into two equal pieces: the 200Ω resistor into two 100Ω resistors in series, and the 50Ω resistor as two 100 Ω resistors in parallel. 100Ω 100Ω
I I1 100Ω 100Ω 2
+ + 100Ω 100Ω V2 V1 - -
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 2/10
Recall that the symmetry of this 2-port device leads to simplified network matrices:
⎡⎤S11 S21 ⎡ZY11ZY21⎤⎡⎤ 11 21 S ==⎢⎥ Z ⎢ ⎥⎢⎥ Y = ⎣⎦S21 S11 ⎣Z21 Z11⎦⎣⎦Y21 Y 11
Q: Yes, but can circuit symmetry likewise simplify the procedure of determining these elements? In other words, can symmetry be used to simplify circuit analysis?
A: You bet!
First, consider the case where we attach sources to circuit in a way that preserves the circuit symmetry:
100Ω 100Ω
I I1 100Ω 100Ω 2
+ + + + 100Ω V Vs V s - V1 100Ω 2 - - -
Or,
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 3/10
100Ω 100Ω
I I1 100Ω 100Ω 2
+ + 100Ω V Is Is V1 100Ω 2 - -
Or, 100Ω 100Ω
I I1 100Ω 100Ω 2
Z0 Z0 + + + + 100Ω V Vs Vs - V1 100Ω 2 - - -
But remember! In order for symmetry to be preserved, the
source values on both sides (i.e, Is,Vs,Z0) must be identical!
Now, consider the voltages and currents within this circuit under this symmetric configuration:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 4/10
I1a I2a
+ V1a - - V2a + I1 I2 I1d I I2b I1b 2d
+ V1b - - V2b + + + + + + + V V V V2 - Vs s - V1 1c 2c - - - - I 1c I2c
Since this circuit possesses bilateral (reflection) symmetry (1221→→, ), symmetric currents and voltages must be equal:
II12= V12=V II12aa= V12aa=V II12bb= V12bb=V II12cc= V12cc=V II12dd=
Q: Wait! This can’t possibly be correct! Look at currents I1a
and I2a, as well as currents I1d and I2d. From KCL, this must be true:
II12aa=− II12dd =−
Yet you say that this must be true:
II12aa== II12dd
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 5/10
There is an obvious contradiction here! There is no way that both sets of equations can simultaneously be correct, is there?
A: Actually there is! There is one solution that will satisfy both sets of equations:
II12aa==00 II12dd ==
The currents are zero!
If you think about it, this makes perfect sense! The result says that no current will flow from one side of the symmetric circuit into the other.
If current did flow across the symmetry plane, then the circuit symmetry would be destroyed—one side would effectively become the “source side”, and the other the “load side” (i.e., the source side delivers current to the load side).
Thus, no current will flow across the reflection symmetry plane of a symmetric circuit—the symmetry plane thus acts as a open circuit!
The plane of symmetry thus becomes a virtual open!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 6/10
+ V1a - - V2a + I I2 1 I2b I1b
+ V1b - - V2b + + + + + + + V Vs Vs - V1 V1c V2c 2 - - - - - I1c I 2c
Virtual Open I=0 Q: So what?
A: So what! This means that our circuit can be split apart into two separate but identical circuits. Solve one half- circuit, and you have solved the other!
V12=VV= s
I1a V ==V 0 12aa V12bbs==VV2 + V1a - I1 I1b V12ccs==VV2
+ V1b - + + + IIV12==s 200 Vs - V V1c 1 II==0 - 12aa - I1c
IIV12bbs==200
IIV12ccs==200
II12dd==0
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 7/10
Now, consider another type of symmetry, where the sources are equal but opposite (i.e., 180 degrees out of phase).
100Ω 100Ω
I I1 100Ω 100Ω 2
+ + + + V 100Ω 100Ω V2 - -Vs s - V1 - -
Or, Ω 100 100Ω
I I1 100Ω 100Ω 2
+ + 100Ω V -Is Is V1 100Ω 2 - -
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 8/10
Ω Or, 100 100Ω
I I1 Ω 100Ω 2 100
Z0 Z0 + + + + V 100Ω 100Ω V2 -Vs s - V1 - - -
This situation still preserves the symmetry of the circuit— somewhat. The voltages and currents in the circuit will now posses odd symmetry—they will be equal but opposite (180 degrees out of phase) at symmetric points across the symmetry plane.
I1a I2a
+ V - - V + 1a 2a I I1 I 2 I I1d I2d 2b 1b
+ V1b - - V2b + + + + + + + V -Vs Vs - V1 V1c V2c 2 - - - - - I 1c I2c
II12= − V12=−V II12aa=− V =−V 12aa II=− V =−V 12bb 12bb II=− V =−V 12cc 12cc =− II12dd
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 9/10
Perhaps it would be easier to redefine the circuit variables as:
I1a I2a
+ V - + V - 1a 2a I I1 I2b 2 I1b I1d I2d
+ V1b - + V2b - + + - - + + V -Vs Vs - V1 V1c V2c 2 - - - + + I1c I 2c
II12= V12=V II12aa= V12aa=V II12bb= V12bb=V II12cc= V12cc=V II12dd=
Q: But wait! Again I see a problem. By KVL it is evident that:
V12cc= −V
Yet you say that V12cc=V must be true!
A: Again, the solution to both equations is zero!
V12cc=V = 0
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Symmetric Circuit Analysis 10/10
For the case of odd symmetry, the symmetric plane must be a plane of constant potential (i.e., constant voltage)—just like a short circuit!
Thus, for odd symmetry, the symmetric plane forms a virtual short.
I1a I2a
+ V - + V - 1a 2a I I1 I2b 2 I1d I I1b 2d
+ V1b - + V2b - + + - - + + V V V V2 - -Vs s - V1 1c 2c - + + - I 1c I2c
Virtual short V=0 This greatly simplifies things, as we can again break the circuit into two independent and (effectively) identical circuits!
I1a
+ V1a - IV= 50 I1 V =V 1 s I1b I1d 1 s IV1as= 100 V1as=V IV1b = s 100 + V1b - V1b =Vs I = 0 + + 1c + V1c = 0 V V IV= s 100 s - V1 1c 1d - - I1c
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 1/10
Odd/Even Mode Analysis
Q: Although symmetric circuits appear to be plentiful in microwave engineering, it seems unlikely that we would often encounter symmetric sources . Do virtual shorts and opens typically ever occur?
A: One word—superposition!
If the elements of our circuit are independent and linear, we can apply superposition to analyze symmetric circuits when non-symmetric sources are attached.
For example, say we wish to determine the admittance matrix of this circuit. We would place a voltage source at port 1, and a short circuit at port 2—a set of asymmetric sources if there ever was one!
100Ω 100Ω
I I1 100Ω 100Ω 2
+ + + + 100Ω V Vs2=0 Vs1=V s - V1 100Ω 2 - - -
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 2/10
Here’s the really neat part. We find that the source on port 1 can be model as two equal voltage sources in series, whereas the source at port 2 can be modeled as two equal but opposite sources in series.
+ Vs - 2 + - Vs + Vs - 2
Vs + 2 - + - 0 + − Vs 2 -
Therefore an equivalent circuit is: 100Ω 100Ω
I I1 100Ω 100Ω 2
+ Vs + Vs - 2 - 2 100Ω 100Ω
+ Vs + Vs − - 2 2 -
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 3/10
Now, the above circuit (due to the sources) is obviously asymmetric—no virtual ground, nor virtual short is present. But, let’s say we turn off (i.e., set to V =0) the bottom source on each side of the circuit: 100Ω 100Ω
I1 Ω I2 100Ω 100
Vs + + V s 2 - - 2 Ω 100 100Ω
Our symmetry has been restored! The symmetry plane is a virtual open.
This circuit is referred to as its even mode, and analysis of it is known as the even mode analysis. The solutions are known as the even mode currents and voltages!
Evaluating the resulting even mode half circuit we find: 100Ω
e I1 100Ω
eeVVss1 II12=== + 2 200 400 Vs/2 - 100Ω
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 4/10
Now, let’s turn the bottom sources back on—but turn off the top two! 100Ω 100Ω
I I1 100Ω 100Ω 2
100Ω 100Ω + V + Vs − s - 2 2 -
We now have a circuit with odd symmetry—the symmetry plane is a virtual short!
This circuit is referred to as its odd mode, and analysis of it is known as the odd mode analysis. The solutions are known as the odd mode currents and voltages!
Evaluating the resulting odd mode half circuit we find:
o 100Ω I1
100Ω ooVVss1 II12= ==− + 250 100 V s 2 - 100Ω
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 5/10
Q: But what good is this “even mode” and “odd mode” analysis? After all, the source on port 1 is Vs1 =Vs, and the source on port 2 is Vs2 =0. What are the currents I1 and I2 for these sources?
A: Recall that these sources are the sum of the even and odd mode sources:
VV VV V ==+VVss ==−0 ss ss1222 s 22 and thus—since all the devices in the circuit are linear—we
know from superposition that the currents I1 and I2 are simply the sum of the odd and even mode currents !
eo eo II11=+ I 1 II 22 =+ I 2
100Ω 100Ω
oe oe II=+ I II22=+ I 2 11 1 100Ω 100Ω
Vs + + V s 2 - - 2 100Ω 100Ω
+ Vs + Vs − - 2 2 -
Thus, adding the odd and even mode analysis results together:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 6/10
II=+eo I II =+ eo I 11 1 22 2 VV VV =+ss =− ss 400 100 400 100 VV3 = ss=− 80 400
And then the admittance parameters for this two port network is:
I1 Vs 11 Y11 === VVss1 80 80 Vs 2 =0
I2 3Vs 13− Y21 ==−= VVss1 400 400 Vs 2 =0
And from the symmetry of the device we know:
1 Y ==Y 22 11 80
−3 Y ==Y 12 21 400
Thus, the full admittance matrix is:
1 −3 ⎡ 80 400⎤ Y = ⎢−3 1 ⎥ ⎣ 400 80 ⎦
Q: What happens if both sources are non-zero? Can we use symmetry then?
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 7/10
A: Absolutely! Consider the problem below, where neither source is equal to zero: 100Ω 100Ω
I1 I2 100Ω 100Ω
+ + + + V 100Ω 100Ω V2 Vs2 s1 - V1 - - -
In this case we can define an even mode and an odd mode source as:
VV+− VV V eo==ss12V ss 12 ss22
+ e - Vs + =+eo Vs1 Vsss1 VV - + V o - s
e + V s - eo + Vsss2 =−VV - Vs 2 − o + Vs -
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 8/10
We then can analyze the even mode circuit:
100Ω 100Ω
I1 Ω I2 100Ω 100
e + + e V V s - - s Ω 100 100Ω
And then the odd mode circuit:
Ω 100 100Ω
I I1 100Ω 100Ω 2
100Ω 100Ω + o o + V −Vs - s -
And then combine these results in a linear superposition!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 9/10
Q: What about current sources? Can I likewise consider them to be a sum of an odd mode source and an even mode source?
A: Yes, but be very careful! The current of two source will add if they are placed in parallel—not in series! Therefore:
II+− II IIeo==ss12 ss 12 ss22
o e eo I Is Is III=+ s1 sss1
−I o I e =−eo Is2 s s IIIsss2
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Odd Even Mode Analysis 10/10
One final word (I promise!) about circuit symmetry and even/odd mode analysis: precisely the same concept exits in electronic circuit design!
Specifically, the differential (odd) and common (even) mode analysis of bilaterally symmetric electronic circuits, such as differential amplifiers!
Hi! You might remember differential and common mode analysis from such classes as “EECS 412- Electronics II”, or handouts such as “Differential Mode Small-Signal Analysis of BJT Differential Pairs”
BJT Differential Pair
Differential Mode Common Mode
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Odd Even Mode Circuit Analysis 1/5
Example: Odd-Even Mode Circuit Analysis
Carefully (very carefully) consider the symmetric circuit below.
- Z = 50 Ω + 4.0 V 0
50 Ω λ 50 Ω + 50 Ω 50 Ω v1
λ - 2
50 Ω 50 Ω Z0 = 50 Ω
The two transmission lines each have a characteristic impedance of .
Use odd-even mode analysis to determine the value of
voltage v1 .
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Odd Even Mode Circuit Analysis 2/5
Solution
To simplify the circuit schematic, we first remove the bottom (i.e., ground) conductor of each transmission line:
Z = 50 Ω 50 Ω 0
Ω λ + 50 4.0 V - + 50 Ω 50 Ω v 1 λ - 2
50 Ω 50 Ω Z0 = 50 Ω
Note that the circuit has one plane of bilateral symmetry:
50 Ω λ 50 Ω
+ 50 Ω 50 Ω v1 -
50 Ω λ 50 Ω 2
Thus, we can analyze the circuit using even/odd mode analysis (Yeah!).
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Odd Even Mode Circuit Analysis 3/5
The even mode circuit is:
50 Ω 50 Ω λ
+ 2.0 V + 2.0 V - + - Ω Ω e 50 50 v 1 -
Ω λ 50 Ω 50 2
I=0
Whereas the odd mode circuit is:
50 Ω λ 50 Ω
2.0 V + + - + -2.0 V - 50 Ω 50 Ω o v1 -
50 Ω λ 50 Ω 2 V =0 We split the modes into half-circuits from which we can e o determine voltages v1 and v1 :
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Odd Even Mode Circuit Analysis 4/5
λ 2 50 Ω A = λ Recall that a 2 transmission line terminated in an open 2.0 V + + - circuit has an input impedance 50 Ω e v1 of Zin = ∞ —an open circuit! - λ Likewise, a transmission line A = 4 λ 4 50 Ω terminated in an open circuit has an
input impedance of Zin = 0—a short circuit! 50 Ω
Therefore, this half-circuit simplifies to: + e 2.0 V And therefore the voltage v1 is easily + - 50 Ω e determined via voltage division: v1 -
e ⎛⎞50 v1 ==21.0⎜⎟V ⎝⎠50+ 50 50 Ω
Now, examine the right half-circuit of the odd mode: λ 2 50 Ω λ Recall that a A = 2 transmission line terminated in a short -2.0 V + + - circuit has an input impedance Ω 50 o Z = 0 v1 of in — a short circuit! - λ Likewise, a transmission line A = 4 λ 4 50 Ω terminated in an short circuit has an
input impedance of Zin = ∞ — an open circuit!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Example Odd Even Mode Circuit Analysis 5/5 50 Ω Æ This half-circuit simplifies to + -2.0 V It is apparent from the circuit + - o 50 Ω o that the voltage v1 = 0 ! v1 -
50 Ω
Thus, the superposition of the odd and even modes leads to the result: eo v1 =+=v1 v1 1.0 += 0 1.0V
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 1/8
Generalized Scattering Parameters Consider now this microwave network:
− + V22()z Z02 V22(z )
port zz= 2 22P + port 1 − V11()z port 3 V33()z
4-port
Z01 microwave Z 03 device
− + zz11= P = V11()z zz33P V33()z
port zz= 4 44P
Z04 + − V44()z V44(z )
Q: Boring! We studied this before; this will lead to the definition of scattering parameters, right?
A: Not exactly. For this network, the characteristic impedance of each transmission line is different (i.e.,
ZZZZ01≠≠≠ 02 03 04 )!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 2/8
Q: Yikes! You said scattering parameters are dependent on transmission line characteristic impedance Z 0 . If these values are different for each port, which Z 0 do we use?
A: For this general case, we must use generalized scattering parameters! First, we define a slightly new form of complex wave amplitudes:
+ − VV00nn abnn== ZZ00nn
So for example:
+ − VV01 03 ab13== ZZ01 03
The key things to note are:
A variable a (e.g., aa,,") denotes the complex a 12 amplitude of an incident (i.e., plus) wave.
A variable b (e.g., bb,,") denotes the complex b 12 amplitude of an exiting (i.e., minus) wave.
We now get to rewrite all our transmission line knowledge in terms of these generalized complex amplitudes!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 3/8
First, our two propagating wave amplitudes (i.e., plus and minus) are compactly written as:
+− V00n ==aZnn0 V n bZ n0n
And so:
+ − j βzn Vnn()ze= aZnn0
− + j βzn Vnn()ze= bZn 0n
bn + jz2β n Γ=()zen an
Likewise, the total voltage, current, and impedance are:
−+jzββnn jz Vnn()zZaebe=+0 nn( n )
−+jzββnn jz aenn− be Iznn()= Z0n
−+ββ aejznn+ be jz Zz()= nn n −+jzββnn jz aenn− be
Assuming that our port planes are defined with znP = 0, we can determine the total voltage, current, and impedance at port n as:
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 4/8
abnn− Vnnn Vz()==00 Z0 nnn() a + b I n Iz nn() == Z0n
abnn+ ZZznn ()==0 abnn−
Likewise, the power associated with each wave is:
+−222 2 +−VV00nnan bn PPnn== == 22ZZ0n 220n
As such, the power delivered to port n (i.e., the power absorbed by port n) is: 2 2 ab− PP=−=+− P n n n nn 2
This result is also verified:
1 ∗ PVInnn= Re{} 2
1 ∗∗ =+−Re{}()ababnn() nn 2
1 ∗∗∗∗ =+−−Re{}aann ba nn ab nn bb nn 2
1 2 ∗∗∗ 2 =+Re ababnnnnnn−−()ab 2 {}
1 2 ∗ 2 =−Reajbabnn+ Im{}nn 2 {} 2 2 a − b = nn 2
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 5/8
Q: So what’s the big deal? This is yet another way to express transmission line activity. Do we really need to know this, or is this simply a strategy for making the next exam even harder? ab+ Z = 11 1 ab11−
A: You may have noticed that this notation
(abnn, ) provides descriptions that are a bit “cleaner” and more symmetric between current and voltage.
However, the main reason for this notation is for evaluating the scattering parameters of a device with dissimilar
transmission line impedance (e.g., ZZZZ01≠ 02≠≠ 03 04 ).
For these cases we must use generalized scattering parameters:
− V0m Z0n + SVzknmn ==≠+ (when kk() 0 for all ) V0n Z0m
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 6/8
Note that if the transmission lines at each port are identical
(ZZ00mn= ), the scattering parameter definition “reverts − + back” to the original (i.e., SVVmn= 00 m n if ZZ00mn= ). E.G.:
− V02 + SV21= + when 02 = 0 V01
+ port 1 port 2 − V11()z V ()z 22 2-port Z01 =50 Ω microwave Z02 = 50 Ω device − + V11(z ) V22()z
But, if the transmission lines at each port are dissimilar
(ZZ00mn≠ ), our original scattering parameter definition is not −+ correct (i.e., SVVmn≠ 00 m n if ZZ00mn≠ )! E.G.:
− V02 + SV21≠ + when 02 = 0 V01
+ V ()z port 1 port 2 − 11 V22()z 2-port
Z01 =50 Ω microwave Z02 = 75 Ω device
− + V11(z ) V ()z 22
− V02 50 + SV21==+ when 02 0 V01 75
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 7/8
Note that the generalized scattering parameters can be more compactly written in terms of our new wave amplitude notation:
− V0m Z0n bm Sknmn ==+ (when ak = 0 for all ≠ ) V0n Z0m an
Remember, this is the generalized form of scattering parameter—it always provides the correct answer, regardless
of the values of Z0m or Z0n !
Q: But why can’t we define the scattering parameter as −+ SVVmn= 00 m n , regardless of Z0m or Z0n ?? Who says we must
define it with those awful Z0n values in there?
A: Good question! Recall that a lossless device is will always have a unitary scattering matrix. As a result, the scattering parameters of a lossless device will always satisfy, for example: M 2 1 = ∑ Smn m =1
This is true only if the scattering parameters are generalized!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 Generalized Scattering Parameters 8/8
The scattering parameters of a lossless device will form a unitary matrix only if defined as Sbamn= m n . If we use −+ SVVmn= 00 m n , the matrix will be unitary only if the connecting transmission lines have the same characteristic impedance.
Q: Do we really care if the matrix of a lossless device is unitary or not?
A: Absolutely we do! The:
lossless device ⇔ unitary scattering matrix relationship is a very powerful one. It allows us to identify lossless devices, and it allows us to determine if specific lossless devices are even possible!
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 The Scattering Matrix of a Connector 1/4
Example: The Scattering Matrix of a Connector
First, let’s consider the scattering matrix of a perfect connector—an electrically very small two-port device that allows us to connect the ends of different transmission lines together. Port Port 1 2 Iz11() Iz22()
+ + Z Z V11()z 0 0 V22()z − −
z1 z2
z1 = 0 z2 = 0
If the connector is ideal, then it will exhibit no series inductance nor shunt capacitance, and thus from KVL and KCL:
V11()()zVz==00 22 = IzIz 11( ==−= 0) 22() 0
Terminating port 2 in a matched load, and then analyzing the resulting circuit, we find that (not surprisingly!):
− − + V01 = 0 and V02=V 01
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 The Scattering Matrix of a Connector 2/4
V + = 0 From this we conclude that (since 02 ):
− − + V01 0 VV02 01 S11 ===++00. S21 ===++10. VV01 01 VV01 01
This two-port device has D2 symmetry (a plane of bilateral symmetry), meaning:
SS22== 11 00. and SS21= 12 = 10.
The scattering matrix for such this ideal connector is therefore: ⎡01⎤ S = ⎢ ⎥ ⎣10⎦
As a result, the perfect connector allows two transmission lines of identical characteristic impedance to be connected together into one “seamless” transmission line.
Z Z 0 0
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 The Scattering Matrix of a Connector 3/4
Now, however, consider the case where the transmission lines connected together have dissimilar characteristic impedances
(i.e., ZZ01≠ ):
Z Z01 02
Port Port 1 2
Q: Won’t the scattering matrix of this ideal connector remain the same? After all, the device itself has not changed!
A: The impedance, admittance, and transmission matrix will remained unchanged—these matrix quantities do not depend on the characteristics of the transmission lines connected to the device.
But remember, the scattering matrix depends on both the device and the characteristic impedance of the transmission lines attached to it.
After all, the incident and exiting waves are traveling on these transmission lines!
The ideal connector in this case establishes a “seamless” interface between two dissimilar transmission lines.
Jim Stiles The Univ. of Kansas Dept. of EECS 3/6/2009 The Scattering Matrix of a Connector 4/4
Z01 Z02
Remember, this is the same structure that we evaluated in an earlier handout!
+ In that analysis we found that—when V02 = 0 :
− − V01ZZ 02− 01 V022Z 02 + = and + = V01ZZ 02+ 01 V01ZZ 02+ 01
And so the (generalized) scattering parameters S11 and S21 are:
− − VZZ01Z01 02− 01 V02 ZZZ012 01 02 S11 ==+ and S21 ==+ VZZ01Z01 02+ 01 VZZ01Z02 02+ 01
As a result we can conclude that the scattering matrix of the ideal connector (when connecting dissimilar transmission lines) is: ⎡⎤ZZ− 2 ZZ ⎢⎥02 01 01 02 ZZ++ ZZ S = ⎢⎥02 01 01 02 ⎢⎥ 2 ZZ − ⎢⎥01 02 ZZ01 02 ⎣⎦⎢⎥ZZ01++ 02 ZZ 01 02
Jim Stiles The Univ. of Kansas Dept. of EECS