Lecture – 24 Date: 31.10.

2017

• Multi-port networks (Contd.), Scattering Matrix

• Matched, Lossless, and Reciprocal Networks
Scattering Matrix
• At “low” frequencies, a linear device or network can be fully characterized
using an impedance or admittance matrix, which relates the currents and
voltages at each device terminal to the currents and voltages at all other
terminals.
• But, at high frequencies, it is not feasible to measure total currents and
voltages!
• Instead, we can measure the magnitude and phase of each of the two
transmission line waves V+(z) and V−(z) → enables determination of
relationship between the incident and reflected waves at each
device terminal to the incident and reflected waves at all other
terminals
• These relationships are completely represented by the scattering matrix
that completely describes the behavior of a linear, multi-port device at a
given frequency ω, and a given line impedance Z0
Scattering Matrix (contd.)
Note that we have now V2 ( z2 ) V2 ( z2 )
characterized transmission line
activity in terms of incident and
“reflected” waves. The negative Z0
going “reflected” waves can be Port-2 z2  z2P
viewed as the waves exiting the
multi-port network or device. Port-1 Port-3

V ( z1 ) 4-port V3 ( z3 )
1
Linear
Z0 Z0

Microwave V3 ( z3 )
V ( z1 )
1 Network
z1  z1P z3  z3P
Viewing transmission line
activity this way, we can fully Port-4 z4  z4P
characterize a multi-port Z0
device by its scattering V4 ( z4 ) V4 ( z4 )
parameters!
Scattering Matrix (contd.)
• Say there exists an incident wave on port 1 (i.e., V1+ (z1) ≠ 0), while the
incident waves on all other ports are known to be zero (i.e., V2+(z2)
=V3+(z3) =V4+(z4) =0).
V1 ( z1 )
 Say we measure/determine the voltage of
Z0 V1 ( z1  z1P ) the wave flowing into port 1, at the port 1
 plane (i.e., determine V1+(z1 = z1P)).
z1  z1P
V2 ( z2 )
Say we then measure/determine the voltage 
of the wave flowing out of port 2, at the V2 ( z2  z2 P )
Z0
port 2 plane (i.e., determine V2−(z2 =z2P)). 
z2  z2P

The complex ratio between V1+(z1 = z1P) and V2−(z2 = z2P) is known
as the scattering parameter S21
Scattering Matrix (contd.)
V2 ( z2  z2 P ) V2e  j z
2P
V2  j  z  z1 P 
Therefore: S21      j z   e 2P

V1 ( z1  z1P ) V1 e 1P
V1

V3 ( z3  z3 P ) V4 ( z4  z4 P )
Similarly: S31   S 41  
V1 ( z1  z1P ) V1 ( z1  z1P )

• We of course could also define, say, scattering parameter S34 as the ratio
between the complex values V3−(z3 = z3P) (the wave out of port 3) and
V4+(z4 = z4P) (the wave into port 4), 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:
Vm ( zm  zmP )
S mn   Vk ( zk )  0 for all k ≠ n
Vn ( zn  znP )
Scattering Matrix (contd.)
• Note that, frequently the port positions
Vm ( zm  0) Vm e j 0 Vm
are assigned a zero value (e.g., z1P=0, S mn      j 0  
z2P=0). This of course simplifies the Vn ( zn  0) Vn e Vn
scattering parameter calculation:
• 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!

Q: How do we ensure that only one

incident wave is non-zero ?

A: Terminate all other ports with a matched

load!
Scattering Matrix (contd.) • Note that if the ports are
 0 2  0 terminated in a matched
load (i.e., ZL =Z0), then
V2 ( z2 ) V2 ( z2 )  0 (Γ0)n = 0 and therefore:
Z0
Vn ( zn )  0

V1 ( z1 ) V3 ( z3 )
4-port
Linear  0 3  0
Z0 Z0
Microwave
Network
V1 ( z1 )
V3 ( z3 )  0
In other words, terminating a
Z0 port ensures that there will
V4 ( z4 )  0 V4 ( z4 ) be no signal incident on that
port!
 0 4  0
Scattering Matrix (contd.)

Just between you and me, I think

you’ve messed this up! In all previous
slides you said that if Γ0 = 0 , the wave
in the minus direction would be zero:
V−(z) = 0 if Γ0 = 0
but just now you said that the wave in the positive
direction would be zero:
V+(z) = 0 if Γ0 = 0

Obviously, there is no way that both statements can be correct!

Scattering Matrix (contd.)
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 Vn+ (zn) and Vn− (zn)!

For example, we originally analyzed this case:

V1 ( z1 )

Z0 0 V−(z ) = 0 if Γ0 = 0

V1 ( z1 )

In this original case, the wave incident on the load is V+(z) (plus direction),
while the reflected wave is V−(z) (minus direction).
Scattering Matrix (contd.)
Contrast this with the case we are now considering:

Vn ( zn )
n-port
Linear
Microwave Z0  0 n
Network
Vn ( zn )

• For this current case, the situation is reversed. The wave incident on the
load is now denoted as Vn−(zn) (coming out of port n), while the wave
reflected off the load is now denoted as Vn+(zn) (going into port n ).
Scattering Matrix (contd.)
• back to our discussion of S-parameters. We Vm 
S
found that if znP = 0 for all ports n, the mn V 
 Vk ( zk )  0
n
scattering parameters could be directly written
in terms of wave amplitudes Vn+ and Vm− for all k ≠ n

• Which we can now equivalently state as:

Vm
S mn   (for all ports, except port n, are terminated in matched loads)
Vn

• 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!
For all
  j z   j z   j z   j z
Vm ( zm )  Vm em
 Vm e
m
 0  Vm e
m
 Vm e
m
terminated
ports!
Scattering Matrix (contd.)
• 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!
• 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 V3  S34V4  S33V3  S32V2  S31V1
(assuming znP = 0 ):
• More generally, the output at • This expression of Scattering
port m of an N-port device is: parameter can be written in
N matrix form as:
V   S mnVn
m

znP = 0 V - = SV +
n 1
Scattering Matrix (contd.)
 S11 S12 S1n 
S 
V - = SV + S   21 
Scattering Matrix  
 
 Sm1 Sm 2 Smn 
• The scattering matrix is N by N matrix that completely characterizes a
linear, N-port device. Effectively, the scattering matrix describes a multi-
port device the way that Γ0 describes a single-port device (e.g., a load)!
• The values of the scattering  S11 ( ) S12 ( ) S1n ( ) 
matrix for a particular device or  S ( ) 
S(  )   21 
network, like Γ0, are frequency  
dependent! Thus, it may be more  
 m1
S ( ) S m2 ( ) S mn ( ) 
instructive to explicitly write:
• Also realize that—also just like Γ0—the scattering matrix is dependent on
both the device/network and the Z0 value of the cable connected to it.
• Thus, a device connected to cables with Z0 =50Ω will have a completely
different scattering matrix than that same device connected to
transmission lines with Z0 =100Ω
Matched, Lossless, Reciprocal Devices
• A device can be lossless or reciprocal. In addition, we can also classify it as
being matched.
• Let’s examine each of these three characteristics, and how they relate to
the scattering matrix.

Matched Device
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 come out
from a port if a signal is incident on that port (but only that port!).
• In other words: Vm  SmmVm  0 For all m When all the ports ‘m’
are matched
• It is apparent that a matched 0 0.1 𝑗0.2
device will exhibit a S= 0.1 0 0.3
scattering matrix where all 𝑗0.2 0.3 0
diagonal elements are zero.
Matched, Lossless, Reciprocal Devices (contd.)
Lossless Device
• For a lossless device, all of the power that is 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!
• The power incident on some port m is related to the
2
Vm
amplitude of the incident wave (Vm+) as: Pm 
2Z 0
2
Vm
• The power of the wave exiting the port is: Pm 
2Z 0

 2  2
• power absorbed by that port is the difference  
V V
Pm  P  P  
m m
of the incident power and reflected power: m m
2Z 0 2Z 0
Matched, Lossless, Reciprocal Devices (contd.)
• For an N-port device, the total incident power is:
(V+)H is the conjugate
Vm   V +  V +
N N
1
P    Pm   m  2 2 H
V transpose of the row
m1 2Z0 m1
vector V+
Similarly, the total
P  P

N


V  + H
V+
reflected power N V   H
V
m 1
m
2Z 0 P    Pm 
m 1 2Z 0

• Recall that the incident and reflected wave amplitudes -

are related by the scattering matrix of the device as: V = SV +

• Therefore: P 
 V   H
V 


 V 
 H
S H SV 
2Z 0 2Z 0
• Therefore the total power delivered to the N-port device is:

 V  H
V 
 V   H
S SV H 
V 
 H

P  P  P 

2Z 0

2Z 0
 P 
2Z 0
 I  S S V
H 
Matched, Lossless, Reciprocal Devices (contd.)
V 
H

• For a lossless device: ∆P=0 

2Z 0
 I  S S V
H 
0 For all V+

• Therefore: I  SH S  0  SH S  I

a special kind of matrix known as a unitary matrix

If a network is lossless, then its scattering matrix S is unitary
• How to recognize a unitary matrix?
The columns of a unitary matrix form an orthonormal set!
 S11 S12 S13 S14  each column of the scattering matrix
Example: S S S S  will have a magnitude equal to one
S   21 22 23 24
N
 S31 S34   Smn 1
2
S32 S33 For all n
  m 1
 S41 S42 S43 S44 
inner product (i.e., dot product) of
dissimilar columns must be zero
dissimilar columns N

are orthogonal
 Smi Smj*  S1i S1*j  S2i S2* j  ....  S Ni S Nj*  0 For all i≠j
m 1
Matched, Lossless, Reciprocal Devices (contd.)
• For example, for a lossless three-port device: say a signal V  2

P 
1
is incident on port 1, and that all other ports are 1
2Z 0
terminated. The power incident on port 1 is therefore:
• and the power exiting the device at each  Vm
2 2
S m1V1 
  
2
port is: Pm S m1 P1
2Z 0 2Z 0
• The total power exiting the device is therefore:
P  P1  P2  P3  S11 P1  S21 P1  S31 P1
2 2 2


 P   S11  S 21  S31
2 2 2
P 1


• Since this device is lossless, the incident power

(only on port 1) is equal to exiting power (i.e, S11  S21  S31  1
2 2 2

P− =P1+). This is true only if:

• Of course, this will be true if the incident wave S12  S22  S32  1
2 2 2

is placed on any of the other ports of this

S13  S23  S33  1
2 2 2

lossless device:
Matched, Lossless, Reciprocal Devices (contd.)
N 2

• We can state in general then that:  Smn 1 For all n

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.
 0 1/ 2 j 3/2 0 
• An example of a (unitary)  
 1/ 2 0 0 j 3 / 2
scattering matrix for a 4-port S 
lossless device is: j 3/2 0 0 1/ 2 
 0 0 
 j 3/2 1/ 2
Reciprocal Device
• 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: S S
mn nm
Matched, Lossless, Reciprocal Devices (contd.)
• For example, a reciprocal device will have S21 = S12 or S32 =S23. We can
write reciprocity in matrix form as:
ST = S where T indicates 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 

 0.40 j 0.20 0 j 0.10 
S 
  j 0.20 0 0.10  j 0.30 0.12
 
 0. 05 j 0.10  0.1 2 0 
Example – 3
• A lossless, reciprocal 3-port device has S-parameters of 𝑆11 = 1 2, 𝑆31 =
1 , and 𝑆33 = 0. It is likewise known that all scattering parameters are
√2
real.
→ 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!
Example – 3 (contd.) 1 𝑆12 𝑆13
2
• Start with what we currently know: S= 𝑆21 𝑆22 𝑆23
1 𝑆32 0
√2
• As the device is reciprocal, we then also know:
𝑺𝟏𝟐 = 𝑺𝟐𝟏 𝑺𝟏𝟑 = 𝑺𝟑𝟏 = 𝟏 √𝟐
𝑺𝟑𝟐 = 𝑺𝟐𝟑

1 𝑆21 1
2 √2
• And therefore: S= 𝑆21 𝑆22 𝑆32
1 𝑆32 0
√2
• Now, since the device is lossless, we know that:
S11  S21  S31  1
2 2 2
(1 / 2) 2  S21  (1 / 2) 2  1
2

Columns have
S12  S22  S32  1
2 2 2
S21  S22  S32  1
2 2 2

unit magnitude
S13  S23  S33  1
2 2 2
(1 / 2) 2  S32  (1 / 2) 2  1
2
Example – 3 (contd.)
1 * 1 *
0  S S  S S  S S  S12  S 21S 22 
*
11 12
*
21 22
*
31 32
*
S32
2 2
1 1 1 Dissimilar columns
0  S11S13*  S 21S 23*
 S31S33
*
  S 21S32
*
 (0)
2 2 2 are orthogonal
 1 
0  S12 S13*  S 22 S 23
*
 S32 S33
*
 S 21  
 22 32  S32 (0)
S S *

 2
We can simplify these expressions and can further simplify them by using
the fact that the elements are all real, and therefore 𝑆21 = 𝑆21 ∗ (etc.).

Q: I count the simplified 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!
Example – 3 (contd.)
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 the phase is either 0° or 180°(i.e., 𝑒 𝑗0 = 1 or 𝑒 𝑗𝜋 =
− 1); however, we do not know which one!

1 1 1
• the scattering matrix for the given 2 2 √2
lossless, reciprocal device is: 1 1 − 1 √2
S= 2 2
1 −1 0
√2 √2