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Lecture 6
Chapter 3 Microwave Network Analysis
 3.1 Impedance and Equivalent Voltages and Currents
 3.2 Impedance and Admittance Matrices
 3.3 The Scattering Matrix
 3.4 The Transmission (ABCD) Matrix
Microwave network analysis
Advantages:
Extend circuit and network concepts in the lowfrequency circuits to handle microwave analysis and
design problems of interest.
Basic procedure:
(1) Field analysis and Maxwell equations  (2) obtain
quantities (β, z0, etc.)  (3) treat a TL or WG as a
distributed component  (4) interconnect various
components and use network and/or TL theory to analyze
the whole system (R, T, Loss, etc.)
3.1 Impedance and Equivalent Voltages and Currents
Equivalent voltages and Currents
(a) Two-conductor TEM line
(b) Non-TEM line
Example: TE10
E
Quasi-static fields in the transverse surface
V=
ò
+
E × dl; I =
ò
c
H × dl; Z 0 =
V
I
Non-uniform V, I along x
V=
ò
b
0
Ey dy = -
jwm a
p
Asin
px
a
e- jb z ò dy
y
Define equivalent voltages and currents in non-TEM line
C1: Voltage and current are defined only for a particular waveguide
mode, and are defined so that the voltage is proportional to the
transverse electric field, and the current is proportional to the
transverse magnetic field.
C2: In order to be used in a manner similar to voltages and currents of
circuit theory the equivalent voltages and currents should be
defined so that their product gives the power flow of the mode.
C3: The ratio of the voltage to the current for a single traveling wave should
be equal to the characteristic impedance of the line. This impedance
may be chosen arbitrarily, but is usually selected as equal to the wave
impedance of the line, or else normalized to unity.
Transverse fields for an arbitrary WG mode:
V(z) = (V +e- jb z +V -e jb z )
+
V
V
e
(x,
y)
+
j
b
z
j
b
z
(V e +V e ) (C1 = + = - )
Et (x, y, z) = et (x, y)(A+e- jb z + A-e jb z ) = t
A
A
Apply
C1
H t (x, y, z) = ht (x, y)(A+e- jb z - A-e jb z ) =
ht (x, y) + - jb z - jb z
(I e - I e )
C2
et , ht are the transverse field variations of the mode.
C1
I+ I(C2 = + = - )
A
A
I(z) = (I +e- jb z - I -e jb z )
Apply C2
Complex Power flow of the incident wave which should be equal to ½V+I+*:
1 +2
V + I +*
+
*
*
+I+*
=
½V
P = A òò e ´ h × zds =
e
´
h
×
zds
òò
2
2C1C2* s
s
C1C2* =
òò e ´ h
s
*
× zds
Integrate over the cross
section of the WG
Apply C3
Characteristic Impedance Z0 of a WG
V + V - C1
Z0 = + = - =
I
I
C2
If choose Z0 = Zw (ZTE or ZTM), C1/C2 = Zw.
If choose Z0 = 1 (normalized),
C1/C2 =1.
Together with C1C2* = òò e ´ h* × zds , C1 and C2 can be solved for a WG mode. Then,
s
equivalent voltages and
currents are defined
+
+
V(z) = (V +e- jb z +V -e jb z ) V = A C1, V = A C1
I(z) = (I +e- jb z - I -e jb z )
I + = A+C2 , I - = A-C2
Propagation constant β and the field intensity A+, A- are calculated from the field analysis.
Equivalent voltages and currents for higher WG modes
Higher order modes can be treated in the same way, so that a
general field in a waveguide can be expressed in the following
form:
Vn+ - jbn z Vn- jbn z
Et (x, y, z) = å(
e
+
e )en (x, y)
C1n
n=1 C1n
N
I n+ - jbn z I n- jbn z
H t (x, y, z) = å (
e
e )hn (x, y)
C2n
n=1 C2n
N
where V± and I±, are the equivalent voltages and currents for the n-th
mode, and C1n, and C2n arc the proportionality constants for each mode.
Example
-Equivalent voltage and current for TE10 mode of a rectangular WG
Waveguide fields
Transmission line model
Ey = (A+e- jb z + A-e jb z )sin(p x / a)
Hx =
-1 + - jb z
(A e - A-e jb z )sin(p x / a)
ZTE
P+ =
-1
ab
*
E
H
dx
dy
=
A
ò y x
2 S
4ZTE
+ 2
V + = A+C1, V - = A-C1
+
P =
ab A
+ 2
4ZTE
1 + +* 1 + 2
= V I = A C1C2*
2
2
V + C1
If choose Z0 = ZTE, then
=
= ZTE
+
I
C2
Vz = V +e- jb z +V -e jb z
I z = I + e- j b z - I - e jb z
=
1 + - jb z
(V e -V -e jb z )
Z0
1
P = V + I +*
2
I + = A+C2 , I - = A-C2
C1 =
C2 =
ab
2
1
ZTE
ab
2
Impedance
(1) Intrinsic impedance of the medium: h = m / e
(dependent only on the material parameters of the medium, and is
equal to the wave impedance for plane waves.)
(2) Wave impedance: Zw = Et / Ht =1/Yw
(a characteristic of the particular type of wave. TEM, TM, and TE waves
each have different wave impedances, which may depend on the type of
line or guide, the material, and the operating frequency.)
(3) Characteristic impedance: Z0 =1/ Y0 = V0 / I 0
(The ratio of voltage to current for a traveling wave on a transmission
line; uniquely defined for TEM waves; defined in various ways for TE
and TM waves.
Reflection of a rectangular WG discontinuity for TE10
εr =2.54
a = 2.286cm; b = 1.016 cm
Characteristic impedance
Reflection coefficient
3.2 Impedance and Admittance Matrices
Port: any type of transmission line or transmission line equivalent of a
single propagating waveguide mode (one mode, add one electric port).
tN: terminal plane providing
the phase reference for V, I
The total voltage and
current on the n-th port:
Vn = Vn+ + VnI n = I n+ - I nAn arbitrary N-port microwave network
• The impedance (admittance) matrix [Z] ([Y]) of the microwave network
relating these voltages and currents
é V
ê 1
ê V2
ê
ê
êë VN
ù é Z
ú ê 11
ú ê Z 21
ú=ê
ú ê
úû êë Z N1
é I
ê 1
ê I2
ê
ê
êë I N
ù é Y
ú ê 11
ú ê Y21
ú=ê
ú ê
úû êë YN1
Z12
Y12
Z1N ùé I1
úê
úê I 2
úê
úê
Z NN úûêë I N
ù
ú
ú
ú
ú
úû
[V]=[Z][I]
Y1N ùé V1
úê
úê V2
úê
úê
YNN úûêë VN
ù
ú
ú
ú
ú
úû
[I]=[Y][V]
-1
[Z] =[Y ]
[Z] or [Y] is complex number and totally 2N2 elements but many networks are
either reciprocal (symmetric matrix) or lossless (purely imaginary matrix), or both
and the total number will be substantially reduced.
Find the impedance matrix elements
Vi
Zij =
Ij
I k =0 for k¹ j
 Transfer impedance, Zij (between ports i and j):
Calculated by driving port j with the current Ij, open-circuiting all other
ports (so Ik = 0 for k ≠ j), and measure the open-circuit voltage at port i.
 Input impedance, Zii
Seen looking into port i when all other ports are open-circuited.
Find the admittance matrix elements
Ii
Yij =
Vj V =0
k
for k¹ j
 Transfer admittance, Yij (between ports i and j):
Calculated by driving port j with the voltage Vj, short-circuiting all other
ports (so Vk = 0 for k ≠ j), and measure the short-circuit current at port i.
 Input admittance, Yii
Seen looking into port i when all other ports are short-circuited.
Example: Evaluation of impedance parameters
Find the Z parameters of the two-port T-network shown below.
Solutions:
Z11 can be found as the input impedance
of port 1 when port 2 is open-circuited:
Z11 =
V1
= Z A + ZC
I1 I2 =0
Similarly, we have
Z22 =
V2
I2
= Z B + ZC
I1=0
The transfer impedance Z12 can be found
measuring the open-circuit voltage at port 1 when
a current I2 is applied at port 2. By voltage
division
Z12 =
V1
V ZC
= 2
= ZC
I 2 I1=0 I 2 Z B + ZC
It can be verified that Z21 = Z12, indicating the
circuit is reciprocal.
3.3 The scattering Matrix
The scattering matrix S 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
é ê V1
ê Vê 2
ê
ê Vë N
ù é
ú ê S11
ú ê S21
ú=ê
ú ê
ú ê SN1
û ë
S12
é
S1N ùê V1+
ú
úê V2+
úê
úê
SNN úûê VN+
ë
ù
ú
ú
ú
ú
ú
û
-
+
[V ]=[S][V ]
[S] is symmetric for reciprocal
network and unitary for
lossless network.
Find the admittance matrix elements
Vector network analyzer
-
Vi
Sij = +
Vj
(directly measure S-parameters)
Vk+ =0 for k¹ j
Sij is found by driving port j with an
incident wave of voltage V+ , and
measuring the reflected wave
amplitude V-, coming out of port i. The
incident 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.
Conversion between Scattering and impedance parameters
For a N-ports network, assuming every port has the same impedance
and set Z0n = 1, we have
Vn = Vn+ + VnI n = I n+ - I n- = Vn+ - Vn-
[Z ][I ] = [Z ][V + ] -[Z ][V - ] = [V ] = [V + ]+[V - ]
Þ ([Z ] +[U ])[V - ] = ([Z ] -[U])[V + ]
Þ [S] = [V - ][V + ]-1 = ([Z ] +[U ])-1 ([Z ] -[U])
1
0
Identity matrix [U ] = [
]
0
1
3.4 The transmission (ABCD) matrix
 Define a 2 x 2 transmission, or ABCD matrix, for each two-
port network according to the total voltages and currents
 The ABCD matrix of the cascade connection of two or more
two-port networks can be found by multiplying the ABCD
matrices of the individual two-ports.
V1 = AV2 + BI 2
I1 = CV2 + DI 2
é V ù é
ê 1 ú=ê A
êë I1 úû ë C
é V
ù
B ê 2
ú
D ûêë I 2
ù
ú
úû
ABCD matrix for a cascaded connection of two-port network
é V ù é A
ê 1 ú=ê 1
êë I1 úû êë C1
B1 ùé V2
úê
D1 úûêë I 2
é V ù é A
ê 1 ú=ê 1
êë I1 úû êë C1
ù
ú
úû
é V
ê 2
êë I 2
B1 ùé A2
úê
D1 úûêë C2
ù é A
ú=ê 2
úû êë C2
B2 ùé V3 ù
ú
úê
D2 úûêë I 3 úû
B2 ùé V3
úê
D2 úûêë I 3
é
ù é
For a N sections of two- ê V1 ú = ê A1
êë I1 úû êë C1
port networks
B1 ù
ú
D1 úû
ù
ú
úû
(The order of ABCD
matrix cannot be
changed)
é A
ê N -1
êë CN -1
BN -1 ùé VN
úê
DN-1 úûêë I N
ù
ú
úû
The ABCD parameters of some Useful two-Port circuits
Conversion between two-port network parameters