Microwave-Engineering - For Mid Sem
Microwave-Engineering - For Mid Sem
Microwave-Engineering - For Mid Sem
6 network theory
and passive devices
INTRODUCTION 6.1
A microwave network is formed when several microwave devices and components such as sources,
attenuators, resonators, filters, amplifiers, etc., are coupled by transmission lines or waveguides for the
desired transmission of a microwave signal. The point of interconnection of two or more devices is called a
junction.
For a low-frequency network, a port is a pair of terminals whereas for a microwave network, a port is a
reference plane transverse to the length of the microwave transmission line or waveguide. At low frequencies,
the physical length of the network is much smaller than the wavelength of the signal transmitted. Therefore,
the measurable input and output variables are voltage and current which can be related in terms of the
impedance Z-parameters, or admittance Y-parameters, or hybrid h-parameters, or ABCD parameters. For a
two-port network as shown schematically in Fig. 6.1, these relationships are given by
È V1 ˘ È Z11 Z12 ˘ È I1 ˘
ÍV ˙ = Í Z ˙Í ˙ Impedance (Z) parameters (6.1)
Î 2˚ Î 21 Z 22 ˚ Î I 2 ˚
È I1 ˘ ÈY11 Y12 ˘ È V1 ˘
Í I ˙ = ÍY ˙Í ˙ Admittance (Y) parameters (6.2)
Î 2˚ Î 21 Y22 ˚ ÎV2 ˚
ÈV1 ˘ È h11 h12 ˘ È I1 ˘
Í I ˙ = Íh ˙Í ˙ Hybrid (h) parameters (6.3)
Î 2˚ Î 21 h22 ˚ ÎV2 ˚
ÈV1 ˘ È A + B ˘ ÈV2 ˘
ÍI ˙ = Í ˙Í ˙ Transmission (ABCD) parameters (6.4a)
Î 1˚ ÎC + D ˚ Î I 2 ˚
In Eq. (6.4a), +I2 for current at Port 2 is outward and –I2 taken for current at Port 2 is inward. Here, Zij,
Yij, and A, B, C and D are suitable constants that characterize the junction. A, B, C and D parameters are
convenient to represent each junction when a number of circuits are connected together in cascade. Here
the resultant matrix, which describes the complete cascade connection, can be obtained by multiplying the
matrices describing each junction as follows:
ÈA B˘ È A1 B1 ˘ È A2 B2 ˘ È An Bn ˘
ÍC D ˙ = ÍC D ˙ ÍC ˙ ... Í ˙ (6.4b)
Î ˚ Î 1 1˚ Î 2 D2 ˚ ÎCn Dn ˚
Microwave Network Theory and Passive Devices 219
where S is the closed surface area of the conducting walls enclosing the junction and N ports in the absence
of any source. Since the integral over the perfectly conducting walls vanishes, the only non-zero integrals are
those taken over the reference planes of the corresponding ports, so that
N
Â Ú (Ei ¥ Hj – Ej ¥ Hi) ◊ dS = 0 (6.6)
n = 1 tn
Since Vn except Vi and Vj are zero, Eti = n ¥ Ei and Etj = n ¥ Ej are zero on all reference planes at the
corresponding ports except ti and tj, respectively. Therefore, Eq. (6.6) reduces to
Ú (Ei ¥ H j )◊ dS = Ú (E j ¥ Hi ) ◊ dS (6.7)
ti tj
220 Microwave Engineering
where a’s represent normalized incident wave amplitude and b’s represent normalized reflected wave
amplitude at the corresponding ports. Here, the total voltage wave is the sum of incident and emergent
voltage waves V + and V – respectively:
Therefore, in this normalization process, the characteristic impedance is normalized to unity. For a two-
port network (Fig. 6.1), the relation between incident and reflected waves are expressed in terms of scattering
parameters Sij’s:
b1 = S11 a1 + S12 a2 (6.18)
b2 = S21 a1 + S22 a2 (6.19)
The normalization process leads to a symmetrical scattering matrix for reciprocal structures. The physical
significance of S-parameters can be described as follows:
S11 = (b1 /a1 )a2 = 0 = Reflection coefficient G1 at Port 1 when Port 2 in terminated with a
matched load (a2 = 0)
S22 = (b2 /a2 )a1 = 0 = Reflection coefficient G2 at Port 2 when Port 1 in terminated with a
matched load (a1 = 0)
S12 = (b1 /a2 )a1 = 0 = Attenuation of wave travelling from Port 2 to Port 1 when a1 = 0
S21 = (b2 /a1 )a2 = 0 = Attenuation of wave travelling from Port 1 to Port 2 when a2 = 0
In general, since the incident and reflected waves have both amplitude and phase, the S-parameters are
complex numbers.
For multiport (N) networks or components, the S-parameter equations are expressed by
È b1 ˘ È S11 S12 ... S1N ˘ È a1 ˘
Í ˙ Í ˙Í ˙
b
Í 2˙ S S ... S2 N ˙ Í a2 ˙
Í 21 22
Í ˙ Í ˙Í ˙
Í ˙ = Í ˙Í ˙ (6.20)
Í ˙ Í ˙Í ˙
Í ˙ Í ˙Í ˙
Í ˙ Í ˙Í ˙
ÍÎbN ˙˚ ÎÍ SN 1 SN 2 ... SNN ˚˙ ÎÍ aN ˚˙
bn = Vn- / Z on = I n- Z on (6.20b)
Net input power at port n
1 2 2
Pninp = ( an - bn ) (6.20c)
2
Power dissipated in a termination at port n is
1 2 2 1 2 2
Pndissp = ( bn - an ) = bn (1 - G n ) (6.20d) Fig. 6.2 A mul port (N) linear
2 2
network
bn
Gn = (6.20e)
an
is the magnitude of the load reflection coefficient at port n. Above relations hold for all the ports with proper
suffices. The total voltage and current at port n are
Vn = Vn+ + Vn- = Z on (an + bn ) (6.20f)
1
In = I n+ - I n- = (an - bn ) (6.20g)
Z on
222 Microwave Engineering
Solving the above equations for an and bn in terms of the terminal voltage and current yields
1Ê V ˆ
an = Á n + I n Z on ˜ (6.20h)
2 Ë Z on ¯
1 Ê Vn ˆ
bn = Á - I n Z on ˜ (6.20i)
2 Ë Z on ¯
In microwave devices or circuits, it is important to express several losses in terms of S-parameters when
the ports are match terminated. In a two-port network, if power fed at Port 1 is Pi, power reflected at the same
port is Pr and the output power at Port 2 is Po then following losses are defined in terms of S-parameters:
Pi | a |2
Insertion loss (dB) = 10 log = 10 log 1
Po | b2 |2
1 1
= 20 log = 20 log (6.21)
| S21 | | S12 |
Pi – Pr 1– | S11 |2
Transmission loss or attenuation (dB) = 10 log = 10 log (6.22)
Po | S12 |2
Pi 1
Reflection loss (dB) = 10 log = 10 log (6.23)
Pi – Pr 1 - | S11 |2
Pi
Return loss (dB) = 10 log
Pr
1 1
= 20 log = 20 log (6.24)
|G| | S11 |
Example 6.1 Find the S-matrix of a length l of a lossless transmission line terminated by matched
impedance.
Solution For a length l of a transmission line, there is no discontinuity at the two ends, so that S11 = S22
= 0. The output signals arise due to input:
b1 = a2 e–jbl
b2 = a1 e–jbl
È b1 ˘ È 0 e – j b l ˘ È a1 ˘
or, Íb ˙ = Í – jb l ˙ Í ˙ = [ S ] [ a]
Î 2˚ ÍÎe 0 ˙˚ Î a2 ˚ Fig. 6.3 Transmission lines of
Example 6.1
È 0 e– jbl ˘
\ [S] = Í ˙
ÍÎe – j b l 0 ˙˚
1. S-matrix is always a square matrix of order N ¥ N for an N-port network and its elements are
complex quantities (real and imaginary parts).
2. Phase-shift property of S-Matrix
At any given frequency and for a given positions of the reference planes, the elements of S-matrix, Sij, have
definite values.
(i) If the frequency is changed, these elements change values also.
(ii) At a given frequency, the complex S-parameters of a network are defined with respect to the
positions of the reference plan (ports). If these positions change, S-parameters of the network also
change. For an outward shift of reference plane (Fig. 6.4) by a distance lj, the phase shift occurs is
fj = bjlj and correspondingly new elements of the S-matrix become
S¢ij = Sij e- jf j ; i π j
- j 2f j
S¢ij = Sij e ;i = j
Power of exponent is doubles for i = j, since reflected wave travels 2lj distance.
Let the original S-matrix equation for the two-port network with reference planes 1, 2 be
[b] = [S][a] (6.25a)
Equation (6.25a) is modified to
[b] = [S¢][a] (6.25b)
where [S] is changed to [S¢] for outward shift in reference planes to 1¢, 2¢ but the incident and reflected wave
voltage can be still denoted by [a] and [b], respectively:
S¢11 = S11 e- j 2f1 , S22
¢ = S22 e- j 2f2 , S12
¢ = S12 e- j (f1 + f2 ) , S21
¢ = S21 e- j (f1 + f2 ) .
These can be proved as follows. For the above shift in reference planes, the new wave variables
becomes
- jf - jf
a1¢ = a1 e 1 , a2¢ = a2 e 2
b1¢ = b1 e- jf1 , b2¢ = a2 e- jf2 (6.25c)
where f = b1l1, f2 = b2l2 and –ve sign in the power of exponent arises due to outward shift of ports. For
inward shift, this sign would be +ve. Hence, we can write
È a1¢ ˘ Èe - jf1 0 ˘ È a1 ˘
Í ˙ = Í - jf2 ˙
Í ˙ (6.25e)
Îa2¢ ˚ Î 0 e ˚ Î a2 ˚
From (6.25a), (6.25b), (6.25d) and (6.25e), the new S-matrix becomes
(ii) The sum of the products of each term of any one row or any column of the S-matrix multiplied
by the complex conjugate of a different row or column, respectively, is zero (columns are ortho-
normal)
N
 Sni ◊ Snj* = 0; iπ j
n =1
� Proof
Let us consider a 4-port lossless Network (N = 4).
The S-matrix equation can be written as
[b] = [S][a]
or, b1 = S11a1 + S12a2 + S13a3 + S14a4
b2 = S21a2 + S22a2 + S23a3 + S24a4
N =4 N =4
 Â
2
or, an = Sn1 a1 + Sn2 a2 + + Snn an
n =1 n =1
N =4 2
or, 2
a1 + a2 + a3 + a4
2 2 2
= Â Sn1 a1 + Sn2 a2 + Sn3 a3 + Sn4 a4 (6.36)
n =1
If only one port (ith say) is excited and all other ports are match terminated, an = 0, n π i.
N 2 N 2
 Sn  Sn
2 2
In this case, ai = i
ai = ai i
n =1 n =1
N 2
or, Â Sn i
=1 (6.37)
n =1
226 Microwave Engineering
2
*
If Sni is complex, Sn
i
= Sni ◊ Sni (6.38)
2 2 2 2
= S1i ai + S1 j a j + S2i ai + S2 j a j + S3i ai + S3 j a j + S4i ai + S4 j a j (6.40a)
Since for any complex quantity |S|2 = S ◊ S*, the above expression may be multiplied out to give the
following results:
2
ai + a j
2
= [S + S + S + S \ a
11
2
21
2
31
2
41
2
i
2
2
+ [S + S \a
2 2 2 2
+ S + S
12 22 32 42 j
* * * *
+ {S11 S12 + S21 S22 + S31 S32 + S41 S42 } ai a*j
* * * *
+ {S12 S11 + S22 S21 + S32 S31 + S42 S41} a j ai*
Since ai and aj are independent signals, they may be chosen in any convenient manner, let ai = aj. Then
from (6.40c),
* * * *
{S11S12 + + S41S42 } + {S12 S11 + + S42 S41}= 0 (6.40d)
Microwave Network Theory and Passive Devices 227
[ S* ][ S ]t = [U ] or [ S* ] = [ S ]t-1 (6.41)
Therefore, (6.39a) and (6.40g) or (6.41) is the necessary condition for the network to be lossless and is
called unitary property of S-matrix. [S] is called a unitary matrix.
Fig. 6.5 Two-port junc ons: (a) waveguide step junc on, (b) coaxial to
waveguide transi on, and (c) a two-port network model
Therefore, for a mismatch load, input reflection coefficient G1 π S11. For a reciprocal network, S12 = S21
so that
2
S12 G2
G1 = S11 + (6.46)
1 – S22 G 2
Further, if the junction is lossless, from Eqs (6.39) and (6.40),
S11 S11* + S12 S12* = 1 (6.47)
S22 S22* + S12 S12* = 1 (6.48)
S11 S12* + S12 S22* = 0 (6.49)
Therefore, for a lossless, reciprocal two-port network, terminated by a mismatch load, Eqs 6.47) and
(6.48) give
|S11| = |S22| (6.50)
From Eqs (6.49) and (6.50),
|S12| = (1–| S11 |2 ) (6.51)
Microwave Network Theory and Passive Devices 229
n
= Â Z kj I j (6.53a)
j =1
+ 1
and
–
Ik = I k – I k = (Vk+ – Vk- ) (6.53b)
Z0
1 1Ê n ˆ
or, Vk- =
2
(Vk + Z 0 I k )= Á Â Z kj I j + Z 0 I k ˜ ;
2 ÁË j =1 ˜¯
(6.53c)
1Ê n ˆ
Similarly, Vk- = Á Â Z kj I j – Z 0 I k ˜ ; (6.53d)
2 ÁË j =1 ˜¯
1
= ([ Z ] + Z 0 [U ]) [ I ] (6.53e)
2
or, [I] = 2 ([ Z ] + Z 0 [U ])-1 [V + ] (6.53f)
1
Similarly, [V–] = ([ Z ] - Z 0 [U ]) [ I ]
2
230 Microwave Engineering
= ([ Z ] – Z 0 [U ]) ([ Z ] + Z 0 [U ])–1 [V + ] (6.53g)
È V– ˘ È V+ ˘
or, Í ˙ = ([ Z ] – Z 0 [U ]) ([ Z ] + Z 0 [U ])–1 Í ˙
ÍÎ Z 0 ˙˚ ÍÎ Z 0 ˙˚
1 1 È n ˘
Therefore, Vk+ = (VkY0 + I k ) = ÍVkY0 + Â (V jYkj )˙ ; (6.54c)
2Y0 2Y0 Í ˙˚
Î j =1
1 1 È n ˘
and Vk– = (VkY0 – I k ) = ÍVkY0 – Â (V jYkj )˙ ; (6.54d)
2Y0 2Y0 Í ˙˚
Î j =1
1
Hence, [V+] = ([Y ] + Y0 [U ]) [V ] ;
2Y0
-1
and [V–] = ([Y ] - Y0 [U ]) [V ] ; (6.54e)
2Y0
1
=
2Y0
(Y0[U ] – 2Y0[Y ])([Y ] + Y0[U ])–1 [V + ] (6.54f)
or, [b] = (Y0[U] – [Y]) ([Y] + Y0 [U])–1 [a] = [S] [a] (6.54g)
–1
Therefore, [S] = ([U ] – [Y ]/ Y0 ) ([U ] + [Y ]/ Y0 ) (6.54h)
–1
Similarly, [ Z ] = (U – S) (U + S); Z = Z/Z0, a normalized parameter (6.54i)
Thus, the following properties are found common for [S], [Z] and [Y]:
1. Number of elements are equal.
2. For reciprocal devices, both [Z] and [S] satisfy reciprocity properties.
Zij = Zji, Sij = Sji
3. If [Z] is symmetrical, [S] is also symmetrical.
4. The following are the advantages of [S] over [Z] or [Y].
(a) In microwave techniques, the source remains ideally constant in power, regardless of circuit
changes. Besides frequency measurements, the only other possible measurement parameters
are VSWR, power and phase. These are essentially measurements of b/a, |a|2 and |b|2. Such
a direct correspondence is not possible with [Z ] or [Y] representations.
Microwave Network Theory and Passive Devices 231
(b) The unitary property of [S] helps a quick check of the power balance of lossless structures. No such
immediate check is possible with [Z] or [Y].
(c) [S] is defined for a given set of reference planes only. If the reference planes are changed, the
S-coefficients vary only in phase. This is not the case in [Z] or [Y], because voltage and current are
functions of complex impedance and, therefore, both magnitude and phase change in [Z] and [Y].
When the characteristic impedances of two ports are different, the S-matrix of the junction can be derived
as given in Examples 6.2 and 6.3.
Example 6.2 A shunt impedance Z is connected across a transmission line with characteristic
impedance Z0. Find the S-matrix of the junction.
Solution For S-parameters, two ports are considered matched. Let the output line be match terminated.
So that a2 = 0.
Therefore,
b1 = S11a1, b2 = S12a1
Y0 – Yin Y0 - (Y0 + Y )
S11 = =
Y0 + Yin Y0 + (Y0 + Y )
–Y –1
= = = S22 for symmetry Fig. 6.6 Transmission lines for
2Y0 + Y 1 + 2 Z /Z 0
Example 6.2
Therefore, 2Z/Z0 = –1/S11 – 1
Z0
or Z= - (1/S11 + 1)
2
Z0 Ê 1ˆ
L= j 1+ ˜
2w ÁË
If Z = jwL,
Sn ¯
Now for the pure shunt element, the transmitted wave amplitude (for a2 = 0) can be expressed by b2 = a1
+ b1 = a1 + S11 a1 = a1 (1+ S11).
Therefore,
b 2Y0 2 Z /Z 0
S21 = 2 = 1 + S11 = S12 = =
a1 2Y0 + Y 1 + 2 Z /Z 0
1 È –1 2Y0 ˘ 1 È –1 2 Z /Z 0 ˘
\ [S] = Í ˙= Í ˙
2Y0 + Y Î2Y0 –1 ˚ 1 + 2 Z /Z 0 Î 2 Z /Z 0 –1 ˚
Alternative
ÈZ Z˘
V1 = Z I1 + Z I2 ; [Z] = Í ˙
ÎZ Z˚
V2 = ZI1 + ZI2;
–1
Ê 1 ˆ Ê 1 ˆ
[S] = Á [ Z ] - [U ]˜ Á Z [ Z ] + [U ]˜
Ë 0
Z ¯ Ë 0 ¯
–1 –1
= [ Z + U – 2] [ Z + U ] = [U ] – 2[ Z + U ]
232 Microwave Engineering
1 È –1 2 Z ˘
= Í ˙; S11 – S12 = –1
1 + 2 Z Î2 Z –1 ˚
If Y = jB, S matrix is Unitary.
Example 6.3 Two transmission lines of characteristic impedance Z1 and Z2 are joined at plane
pp¢. Express S-parameters in terms of impedances when each line is matched
terminated.
Solution The incident and scattered wave amplitude are related by [b] = [S] [a].
(i) Since the output line is matched (a2 = 0), the input
impedance Zin at the junction = Z2 = load for line Z1.
Therefore,
Z – Z1
S11 = 2 = reflection coefficient on the
Z 2 + Z1 input side.
Ê Z – Z1 ˆ Z1 2 Z 2 Z1
or, S21 = Á 1 + 2 =
Ë Z 2 + Z1 ˜¯ Z2 Z 2 + Z1
(iv) With the input line matched (a1 = 0), b2 = a2S22
Ê Z - Z2 ˆ Z 2 2 Z1Z 2
= Á1 + 1 =
Ë Z1 + Z 2 ˜¯ Z1 Z1 + Z 2
È Z –Z 2 Z1Z 2 ˘
Í 2 1
˙
Í Z 2 + Z1 Z1 + Z 2 ˙
Therefore, [S] = Í ˙
Í 2 Z1Z 2 Z1 – Z 2 ˙
Í Z +Z Z1 + Z 2 ˙˚
Î 1 2
Microwave Network Theory and Passive Devices 233
Example 6.4 A series reactance Z = jX is connected between two lines with different characteristic
impedances Z1 and Z2. Find the S-matrix of the junction.
Solution The normalized voltages and power inputs at Port 1 and 2 are
V1+ V2+
a1 = , a2 =
Z1 Z2
a12 a2
P1 = , P2 = 2
2 2
S-matrix is determined assuming that each of the lines are
terminated by its characteristic impedances. Fig. 6.8 Transmission lines for Example 6.4
V1– b Z – Z1 jX + Z 2 – Z1
Therefore, = 1 = S11 = in =
V1+ a1 Z in + Z1 jX + Z 2 + Z1
V2– b2 jX + Z1 – Z 2
and = = S22 =
V2+ a2 jX + Z1 + Z 2
Net input voltage for output line matched (a2 = 0)
a1 + b1 = a1 (1 + S11)
V1+ V1– V+
Input current I1 = – = (1 – S11 ) 1
Z1 Z1 Z1
For the continuity of current in a lossless series element jX, with output port matched (a2 = 0),
V1+
I2 = –I1 = –
Z1
(1 – S11 )= – I2–
V2–
Also I2– =
Z2
V1+ V-
\ (1 – S11 ) = 2
Z1 Z2
Thus, for symmetry
b2 V– V1+ V– Z1
S21 = S12 = = 2 = 2+
a1 Z2 Z1 V1 Z2
ÊZ ˆ Z Z2
= Á 2 ˜ (1 – S11 ) 1 = (1 – S11 )
Ë Z1 ¯ Z2 Z1
Z 2 2 Z1Z 2
or, S21 = S12 =
Z1 Z1 + Z 2 + jX
2Z0 jX
If Z1 = Z2 = Z0, S1 = S21 = (1 – S11)= =1-
2 Z 0 + jX jX + 2 Z 0
234 Microwave Engineering
Solution
Here, Z = jX
1
Let Y=
jX
\ I1 = YV1 – YV2
I2 = –YV1 + YV2
Fig. 6.9 Transmission lines for Example 6.5
ÈY –Y ˘
[Y] = Í
Î –Y Y ˙˚
1 È1 2Y ˘
= (U – Y )(U + Y )–1 = Í ˙ ; S11 + S12 = 1
1 + 2Y Î2Y 1˚
(Z + Z0 ) - Z0 Z
S11 = G1 a = =
2 =0 (Z + Z0 ) + Z0 Z + 2Z0
Z
S22 = G 2 a1 = 0
=
Z + 2Z0
2V1 2Z0 Z
S12 = = = 1-
Vg Z + 2Z0 Z + 2Z0
2V2 2Z0 Z
S21 = = = 1-
Vg Z + 2Z0 Z + 2Z0
È G 1 - G1 ˘
[S] = Í 1
G1 ˙˚
Hence,
Î1 - G 1 :
* * * *
If Z = jX, S11 2 + S21 2 = 1 , S12 2 + S22 2 = 1 , S11S12 + S21S22 = 0 and S12 S11 + S22 S21 = 0 . The last
equations satisfy unitary conditions. Therefore, S matrix of a series reactance (lossless) circuit is unitary.
Example 6.6 Determine the S-matrix of a 3 dB T-network attenuator shown in Fig. 6.10 terminated
in a 50-ohm matched load with Z1 = 17.12 ohms, Z2 = 141.78 ohms.
Solution
Z1 17.12
= = 8.56
2 2
V1– b1 Z in1 – Z 0
S11 = = =
V1+ V2+ = 0
a1 Z in1 + Z 0
Fig. 6.10 Network a enuator for Example 6.6
Microwave Network Theory and Passive Devices 235
Ê 41.44 ˆ Ê 50 ˆ
Therefore, V2– = V2 = V1 Á = 0.707 V1
Ë 41.44 + 8.56 ˜¯ ÁË 50 + 8.56 ˜¯
( j 0.6)2 0.36
= 0.2 – = 0.2 + = 0.5273
1 + 0.1 1.1
236 Microwave Engineering
Example 6.8 An ideal transformer of 1000 : 100 turns operates at 500 MHz and excited by voltage
generator of impedance by 50 ohms. Find the S-matrix.
Z 0 n2 - Z 0 n2 - 1 99
S11 = G = 2
= 2
= ;
a2 = 0
Z0 n + Z0 n +1 101
V2 n2 nV
Vg1 = 2
= 2
n + 1 Vg1
2V2 2n 20
\ S21 = = 2
=
Vg1 n +1 101
a2 = 0
( Z 0 /n 2 ) - Z 0 1 - n2 99
Similarly, S22 = G 2 2
= 2
=-
a1 = 0 ( Z 0 /n ) + Z 0 1+ n 101
V2 Z 0 /n 2 1 V /n
= = = 1
Vg2 ( Z 0 /n2 ) + Z 0 1 + n2 Vg2
2V1 2n 20
S12 = = =
Vg2 n2 + 1 101
a1 = 0
È 99 20 ˘
Í 100 101 ˙
\ [S] = Í ˙
Í 20 - 99 ˙
ÍÎ 101 101 ˙˚
Here we see that
1. S11 π S22 since the network is not symmetrical.
2. S12 = S21 (reciprocal)
Microwave Network Theory and Passive Devices 237
0.64
or G1 = 0.1 - = 0.1 - 0.53 = –0.43
1.2
Therefore, return loss at port 1 is
RL1 = –20 log(|G1|) = –20 log (0.43)
= 7.33 dB.
The above example shows that the reflection coefficient Gi at any port i is not equal to Sii:
Gi π Sii
Therefore, unless all ports are matched Sii π Gi. Similarly, it can be shown that the transmission coefficient
Tij from port j to Port i is not equal to Sij unless all other ports are matched. The scattering parameters of a
linear network are exclusive properties of the network itself and are defined under the conditions that all
ports are matched. If the terminations of the ports changed, S-parameters of the network do not change.
Only the external characteristics of the ports, such as reflection coefficient at a given port and transmission
coefficients between two ports are changed.
Example 6.10 (a) If a lossless two-port network is reciprocal, show that |S21|2 = 1 – |S11|2
(b) If the lossless two-port network is non-reciprocal, show that it is impossible to have unidirectional
transmission, where S12 = 0 and S21 π 0.
Solution
(a) For reciprocal network,
ÈS S ˘
[S] = Í 11 12 ˙ ; S12 = S21
Î S12 S22 ˚
For lossless network
1st row: |S11|2 + |S12|2 = 1
\ |S12|2 = 1 – |S11|2 or |S21|2 = 1 – |S11|2. Since S21 = S12
(b) For lossless non-reciprocal network (S12 π S21). If S12 = 0 π S21
ÈS 0 ˘
[S] = Í 11 ˙
S S
Î 21 22 ˚
First row: |S11|2 = 1; \ |S11| = 1
First column: |S11|2 + |S21|2 = 1
\ |S21|2 = 1 – |S11|2 = 1 – 1 = 0
\ for lossless nonreciprocal network, S12 = 0 = S21
Therefore, it is impossible to have unidirectional transmission.
V1 Z11Z 22 – Z12 Z 21
B= =
I2 V2 = 0
Z 21
I1 1
C= =
V2 I2 = 0
Z 21
I1 Z 22
D= = (6.55c)
I2 V2 = 0
Z 21
For a reciprocal network, Z12 = Z21 and AD – BC = 1. Equation (6.55c) show the relationship between
ABCD parameters and Z-parameters.
–Y11
D= (6.56b)
Y21
V+ V– V–
I= – ; b= ; (6.57b)
Z0 Z0 Z0
240 Microwave Engineering
Therefore, V= Z 0 ( a + b) (6.57c)
1
I= ( a – b) (6.57d)
Z0
Now for the two-port network shown
V1 = AV2 + BI2 Fig. 6.12 Two-port network between two lines
and I1 = CV2 + DI2 (6.57e) of unequal impedance Z01 and Z02
Z 01
a1 – b1 = C Z 01Z 02 (a2 + b2 ) + D (b2 – a2 ) (6.58b)
Z 02
Using following normalizations,
Z 02
A = A
Z 01
B
= B , C Z 01Z 02 = C and D Z 01 = D (6.58c)
Z 01Z 02 Z 02
we can express
a1 + b1 = A (a2 + b2) + B (b2 – a2) (6.58d)
a1 – b1 = C (a2 + b2) + D (b2 – a2) (6.58e)
The S-parameter equation is for the network given by
b1 = S11 a1 + S12 a2 (6.58f)
b2 = S21 a1 + S22 a2 (6.58g)
Solving for Eqs (6.58d – 6.58g),
A+ B–C – D
S11 = (6.58h)
A+ B+C + D
–A + B – C + D
S22 = (6.58i)
A+ B+C + D
2( AD – BC )
S12 = (6.58j)
A+ B+C + D
Microwave Network Theory and Passive Devices 241
2
S21 = (6.58k)
A+ B+C + D
(i) If the network is reciprocal,
S12 = S21, so that AD – BC = 1
2
and S12 = S21 = (6.59a)
A+ B+C + D
(ii) If the network is symmetrical,
S11 = S22 so that A = D
B–C
and S11 = S22 = (6.59b)
2A + B + C
(iii) If the symmetrical and reciprocat is placed in a common transmission line of characteristic
impedance Z0 = Z01 = Z02, then
A = A, B = B/Z 0 , C = CZ 0 , D = D
and
A + B/Z 0 – CZ 0 + D
S11 = = S22 (6.59c)
A + B/Z 0 + CZ 0 + D
2
S21 = = S12 (6.59d)
A + B/Z 0 + CZ 0 + D
Example 6.11 Find ABCD parameters of the shunt admittance Y in the transmission line shown in
Fig. 6.13.
Solution
ÈV1 ˘ È A B ˘ ÈV2 ˘
ÍI ˙ = Í ˙Í ˙
Î 1˚ ÎC D ˚ Î I 2 ˚
Here, V1 = V2; \ A = 1, B = 0, for the common node 1
I –I Fig. 6.13 Transmission lines for
and V1 = V2 = 1 2 , or I1 = V2Y + I2 Example 6.11
Y
\ C = Y, D = 1 from (6.55c)
È1 0˘
Hence, [ABCD] = Í
ÎY 1 ˙˚
Example 6.12 Find the ABCD parameters of the series reactance jX placed in the following
transmission line.
Solution For the current continuity, I1 = I2.
Therefore, C = 0, D = 1 from (6.55c)
Also,
V1 = I2 ( jX) + V2 leads to A = 1, B = jX.
È1 jX ˘
Fig. 6.14 Transmission line for
\ [ABCD] = Í
Î0 1 ˙˚ Example 6.12
242 Microwave Engineering
Example 6.13 In a lossless transmission line of characteristic impedance Z0, a series reactance jX
and two shunt susceptances jB are placed as shown in Fig. 6.15. Find the ABCD
parameters for the network.
Example 6.15 Find the ABCD matrix of a length of lossless microstrip transmission line terminated
in its characteristic impedance Z0.
Solution
It is seen that the condition AD–BC = 1 holds for all the above five circuits (Example 6.11–6.15) because
they are reciprocals.
Fig. 6.18 Coaxial cable: (a) flexible (b) semi-rigid (c) rigid
244 Microwave Engineering
The shielding effectiveness of the outer conductor is expressed in terms of transfer impedance ZT of the
cable, which is defined as
Longitudinal voltage Vi induced per unit
length on one side of the shield (outside) V
ZT = = i (6.60)
The leakage current I s flowing on the other side Is
(inside) of the shield
If both ends of a short cable are terminated into
matched loads, as shown in Fig. 6.19, the load
voltage is
V IZ
V0 = i = s T (6.61)
2 2
Therefore,
ZT = 2V0/Is (6.62)
For a thin-walled tubular shield, the transfer Fig. 6.19 Transfer impedance of a coaxial cable
impedance is given by
1 (1 + j )t /d
ZT = for t << b << d (6.63)
2p bs t sinh [(1 + j )t /d ]
where b is the inner radius of the shield, t is the shield wall thickness, s is the conductivity of the shield, d is
the skin depth in the shield. For a better shielding, ZT should be smaller.
Some standard coaxial cables are given in Table 6.1 with their radio guide (RG) and universal (U)
numbers and specifications.
75-ohm flexible cables used up to 1 GHz. The TNC (Threaded Navy Connector) is like BNC, except that,
the outer conductor has a thread to make firm contact in the mating surface to minimise radiation leakage at
higher frequencies. These connectors are used up to 12 GHz.
(a)
The SMA (Sub-Miniature A) connectors are used for thin flexible or semirigid cables. The higher
frequency is limited to 24 GHz because of generation of higher-order modes beyond this limit. All the above
connectors can be of male or female configurations except the APC-7(Amphenol Precision Connector-7
mm) which provides coupling without male or female configurations. The APC-7 is a very accurate 50-ohm,
low VSWR connector which can operate up to 18 GHz.
Another APC-3.5 connector is a high-precision 50-ohm, low VSWR connector which can be either the
male or female and can operate up to 34 GHz. It can mate with the oppositely sexed SM connector. Table 6.2
shows the type, dielectric in mating space and impedance of some of the above standard connectors.
246 Microwave Engineering
Fig. 6.21 Waveguide sec ons: (a) rectangular (b) circular (c) rectangular taper
The general theory of waveguides is given in Chapter 3. This section describes some practical aspects of
rectangular and circular waveguides for the construction of waveguide components.
Hence, the dimensions for dominant mode rectangular waveguides are to be selected according to
0 < b < l/2, l/2 < a < l (6.66)
To provide adequate power handling capability, usually the dimensions are chosen as b = 0.3l to 0.4l, a
= 0.7l to 0.8l. For a standard X-band guide a = 2.286 am, b = 1.016 cm.
Fig. 6.22 Circular waveguide modes (a) TE11 (b) TM01 (c) TE01
The electric currents on the waveguide walls for the TM01 mode are purely longitudinal, whereas for the
TE01 mode it flows along closed circular paths and does not have longitudinal components.
The attenuation in the circular waveguide for TE01 mode is very low and decreases with frequency as
f –3/2. Therefore, although TE01 mode is not dominant, it is used for long telecommunication lines and also
for resonant cavities displaying very high Q wavemeters. Figure 6.23 shows the distribution of the cut-off
wavelengths of a circular waveguide for a few modes.
248 Microwave Engineering
Selection of the guide radius is made for propagation of a single dominant TE11 mode such that
1.841 1.841
a£ f ≥ fcTE11 = (6.67)
2p fc (Hz) me 2p a me
For other higher modes, the dimensions are selected based on the corresponding root of Bessel functions
as described in Chapter 3 and by an appropriate excitation method.
Fig. 6.23 Distribu on of cut-off wavelength in a circular waveguide and a enua on characteris cs
Fig. 6.24 Choke flange: (a) side view (b) cross sec on
At the joint, the first waveguide is equipped with a choke flange and the second one with a plane flange.
The choke flange is frequency selective. At the edge of a 10–15% bandwidth, VSWR of the order of 1.02 to
1.05 can be achieved. Choke connections are used in microwave oscillator and amplifier tubes and coaxial
rotary joints.
Open-Ended Section
The following three phenomena are associated with a physical open end of a microstrip line:
(a) Fringing fields extended beyond the physical end of the strip
(b) Radiation from the open end due to discontinuity of current
(c) Launching of surface waves from the open end of the strip which disturbs the quasi-TEM mode of
propagations
The phenomenon (a) can be accounted for by modelling the fringing fields by means of an equivalent
shunt capacitance Cf. This capacitance is also equivalent to an extra length Dl << lg of the same line with
electrically open circuit end as shown in Fig. 6.26.
For a given structure the end effect length is given by
Ê e + 0.3 ˆ Ê w/h + 0.262 ˆ
Dl = 0.412h Á eff ˜ (6.68)
Ë e eff – 0.258 ¯ ÁË w/h + 0.813 ˜¯
Short Circuit
In a microstrip line, a short circuit can be achieved by a short-circuiting post or called via between the strip
and the ground plane as shown in Fig. 6.27, place through the substrate.
At frequencies below 3 GHz, this shorting post could be a short metallic wire bonded to the microstrip
and the ground plane.
Microwave Network Theory and Passive Devices 251
Microstrip line
Shorting post
Substrate
Ground plane
Fig. 6.26 Equivalent end effect length Fig. 6.27 Short circuit in a microstrip line
Right-angled Bend
In order to divert the microwave signal path, a right-angled microstrip bend is often used. The microstrip
bend is a discontinuity where it is seen that current flow around the corner is critical. The skin effect pushes
the current to the outside edges of the microstrip, and the current crowding towards the inside edge of the
bend. A schematic diagram of such a bend with electric current distributions is shown in Fig. 6.28 and
6.29(a). The charge accumulation at the outer corner and interruption of current result in null current at
the outer corner and the current maximum on the inner corner [Fig. 6.29(a)]. The result is an equivalent
T-network with series inductance Lb and a shunt capacitance Cb.
Different authors have given the expression of the bend capacitance and inductances based on their
works. The expressions given by K C Gupta, et al. are
Cb w
= [(9.5 er – 1.25) w/h + 5.2 er + 7.0]; pF/m (6.69a)
h h
Lb w
= 4 – 4.21; nH/m (6.69b)
h h
The expressions given by Konpang, et al. are modified and are given by
2
Ê wˆ Ê wˆ
Cb/h = (10.35e r + 2.5) Á ˜ + (2.6 e r + 5.64) Á ˜ pF/m (6.70a)
Ë h¯ Ë h¯
È Ê wˆ
1.39
˘
-0.18Á ˜
Í
Lb/h = 0.7 Î1.0 - 1.35e Ë h¯ ˙ nH/m (6.70b)
˚
252 Microwave Engineering
A typical variations of these equivalent parameters with w/h are shown in Fig. 6.29(b) where the results
of both the expressions given here are agreed well.
The bend can be metered at the outer corner with a simple 45° cut to reduce the bend effect without
affecting any major current distributions [Fig. 6.30(a)]. The capacitance has been reduced by 50% and
inductance may be doubled than those in the unmetered case. A very good VSWR is obtained for a cut where
the range of cut is b = 0.6w – 0.8w for er = 4.4, h = 1.6 mm. The results of VSWR vs frequency are shown
in Fig. 6.30(b).
Microwave Network Theory and Passive Devices 253
It is seen that there is improvement is VSWR in a metered bend. Over the frequency range 1–2 GHz,
VSWR ranges from 1.035–1.23 for uncut band and from 1.005–1.2 . . . for metered cut bend (b = 0.50 W,
er = 4.4).
Example 6.16 A waveguide termination having VSWR of 1.1 is used to dissipate 100 watts of
power. Find the reflected power.
Solution
S – 1 1.1 – 1 0.1
|G | = = = = 0.04762
S + 1 1.1 + 1 2.1
A choke plunger uses the impedance transformation properties of a quarter-wave transformer. In the
waveguide choke plunger, the front choke is a quarter-wave section. The second section is also a quarter
guide wavelength long. The widths of the plungers are uniform and slightly less than the interior guide width
of the broad wall. The height of the plungers are not uniform, the from one is b – 2b1 and the second one is
b – 2b2, where b1 is made as small as possible and b2 is made as large as possible. The back section makes a
sliding fit in the guide with almost zero gap. In this case, the contact resistance is in a current antinode and
the input impedance is zero at the plane of reflection. If Z01 and Z02 are the characteristic impedances of the
quarter wavelength coaxial sections b – 2b1 and b – 2b2, the impedance seen at the input plane AA¢ is
Zin = (Z01/Z02)2 Z¢i = (b1/b2)2 Z¢i (6.71)
where Z¢i is the input impedance at the back section plane BB¢. To make Zin Æ 0, Z01 << Z02 and accordingly
the gaps b1 and b2 are to be selected where b2 >> b1. If by good mechanical design, b2 is made equal to 5b1,
Z¢
Zin = i . Therefore, the shorting effect is improved 25 times as compared to a single non-choke type sliding
25
short. For a coaxial line, shorting plungers are used to provide impedance-transforming properties which do
not have physical contact with the line conductors.
The disadvantage of all non-contact plungers and choke plungers is the bandwidth limitation of 20 to
30% of the mid-frequency. In case of a circular waveguide in the TE01 mode, a short-circuit plunger does not
require any choke connections since there is no longitudinal current. A simple sliding metal disc is adequate.
The locus of reflection coefficient of this plunger is the outermost circle of the Smith chart (G = 1 p ).
transition should be quarter wavelength to avoid abrupt dimensional changes and generation of higher-order
modes. The usual length of complete section is kept ≥ 2lg.
Fig. 6.35 Waveguide (a) Twists, (b) Corners, (c) H-bend, and (d) E-bend
Such arrangements produce discontinuities and as a result, reflection of propagating waves takes place
at any point of the discontinuities. In order to cancel the reflected waves from both ends of the waveguide
corner, the mean length L between the junction of corners is kept an odd multiple of the quarter wavelength.
For waveguide bend, the discontinuity effects are reduced by keeping the minimum radius of curvature R =
1.5b for an E bend and R = 1.5a for an H bend.
A waveguide twist is used to change the polarization of the propagating wave by 90 degrees. The length
of the twist is again kept equal to an odd multiple of the quarter guide wavelength.
In order to have radiation in one propagating direction only, a short circuit is placed at a distance l
approximately equal to a quarter wavelength in the backward direction. By designing the values of l and d,
the input impedance Zin = Rin + jXin of the probe can be made pure resistive which is equal to the characteristic
impedance Z0 of the coaxial line feeding the signal to the guide:
2h0
Zin = R = sin2 b10 l tan2 (k0 d/2) (6.72)
abb10 k0
Since the electric field in the coaxial line for TEM mode and in the waveguide for TE10 mode in the vicinity
of the probe are orthogonal to each other, higher-order modes (evanescent) are also excited near the probe
which are highly attenuated within one wavelength distance. By making the probe diameter very small
(< 0.15 a) the reactance Xin of the probe resulting from higher order mode is made negligible small (or ideally
zero) with proper choice of l and d. E-field radiated from the probe toward the waveguide propagation (rhs)
becomes inphase with the E-field radiated back and reflected from the shorting end.
k0 h0
Zin = Z0 = R = (p/a)2 (pd2)2 sin2 b10l (6.74)
2 ab b10
Zin is made equal to Z0 by adjusting the diameter of the thin loop and l such that d < l0/10 and lg/2<l<lg/4.
When a small circular aperture of radius r0 << l0 exists in a transverse wall in a rectangular waveguide
operating in the TE10 mode, it offers an inductive susceptance B given by
–3 ab
B= 3
; Zw = guide wave impedance (6.76)
8 r0 b10 Z w
The transmission and reflection coefficients of the aperture are given by
j16 r03 k0 h0
T= (6.77)
3 ab Z w
l0
sin (pd/a) = , there will be zero power at Port 4 in the upper guide and input power at Port 1 will be
6a
coupled into Ports 2 and 3 only. The 4-port device is then called a directional coupler or Bethe hole coupler
Attenuators 6.4.14
Attenuators are passive devices used to control power levels in a microwave system by partially absorbing
the transmitted signal wave. Both fixed and variable attenuators are designed using resistive films (aquadag
coated dielectric sheet).
A coaxial fixed attenuator uses a film with losses on the centre conductor to absorb some of the power as
shown in Fig. 6.39(a). The fixed waveguide type [Fig. 6.39(b)] consists of a thin dielectric strip coated with
resistive film and placed at the centre of the waveguide parallel to the maximum E field. Induced current on
the resistive film due to the incident wave results in power dissipation, leading to attenuation of microwave
energy. The dielectric strip is tapered at both ends up to a length of more than half wavelength to reduce
reflections. The resistive vane is supported by two dielectric rods separated by an odd multiple of quarter
wavelength and perpendicular to the electric field [Fig. 6.39(b)].
Fig. 6.39 Microwave a enuator: (a) coaxial line fixed a enuator (b) and (c) waveguide a enuators
A variable-type attenuator can be constructed by moving the resistive vane by means of micrometer screw
from one side of the narrow wall to the centre where the E field is maximum [Fig. 6.39(b)] or by changing
the depth of insertion of a resistive vane at an E field maximum through a longitudinal slot at the middle of
the broad wall as shown in Fig. 6.39(c). A maximum of 90 dB attenuation is possible with VSWR of 1.05.
The resistance card can be shaped to give a linear variation of attenuation with the depth of insertion.
so that it has a negligible effect on the field perpendicular to it but absorbs any component parallel to it.
Therefore, a pure TE11 mode is excited in the middle section.
With reference to Fig. 6.40(b), if the resistive card in the centre section is kept at an angle q relative to
the E field direction of the TE11 mode, the component E cos q parallel to the card gets absorbed while the
component E sin q is transmitted without attenuation. This later component finally appears as electric field
component E sin2 q in the rectangular output guide. Therefore, the attenuation of the incident wave is
E 1 1
a= = =
E sin 2 q sin 2 q | S21 |
or, a (dB) = –40 log (sin q) = –20 log |S21| (6.79)
Therefore, the precision rotary attenuator produces attenuation which depends only on the angle of
rotation q of the resistive card with respect to the incident wave polarization. Attenuators are normally
matched reciprocal devices, so that
|S21| = |S12| (6.80)
VSWR–1
and |S11| or |S22| = << 0.1 (6.81)
VSWR + 1
where the VSWR is measured at the input or output port concerned. The S-matrix of an ideal precision rotary
attenuator is
È 0 sin 2 q ˘
[S] = Í ˙ (6.82)
ÍÎsin 2 q 0 ˙˚
2p l0
b2 = , l g0 = (6.84)
l g0 Êl ˆ
2
1- Á 0˜
Ë 2a ¯
Thus, the differential phase shift produced by the phase shifter is Df = (b1 ~ b0)l.
By adjusting the length l, different phase shifts can be produced.
Microwave Network Theory and Passive Devices 263
The principle of operation of the rotary phase shifter can be explained as follows. The TE11 mode incident
field Ei in the input quarter-wave section can be decomposed into two transverse components, one E1,
polarized parallel and other, E2 perpendicular to quarter-wave plate. After propagation through the quarter-
wave plate, these components becomes
- jb l - jb l 1
E1 = Ei cos 45° e 1 = E0 e 1 cos 45° = sin 45° = (6.86)
2
- j b2 l
E2 = Ei sin 45° e = E0 e - j b 2 l (6.87)
where, E0 = Ei / 2 . The length l is adjusted such that these two components will have equal magnitude but
a differential phase change of (b1 – b2) l = 90°. Therefore, after propagation through the quarter-wave plate,
these field components become
- jb l
E1 = E0 e 1 (6.88)
- jb1l jp /2 - j ( b1 l - p /2)
E2 = E0 e ◊e = E0 e (6.89)
Thus, the quarter-wave sections convert a linearly polarized TE11 wave to a circularly polarized wave and
vice versa. Equations (6.88) and (6.89) are input components to half-wave section C.
After emergence from the half-wave section, the field components parallel and perpendicular to the half-
wave plate can be represented as
- j 2 b2 l - j 3 b1l
E3 = (E1 cos q – E2 sin q ) e = E0 e–jq e (6.90)
- j 2 b2 l –jq - j 3 b1l –jp /2
E4 = (E2 cos q + E1 sin q) e = E0 e e e (6.91)
since, 2 (b1 – b2)l = p or –2 b2 l = p – 2 b1l (6.92)
After emergence from the half-wave section, the field components E3 and E4 may again be decomposed
into two TE11 modes, polarized parallel and perpendicular to the output quarter-wave plate B. At the output
end of this quarter-wave plate, the field components parallel and perpendicular to the quarter-wave plate can
be written as
- jb l - j 4 b1l
E5 = (E3 cos q + E4 sin q) e 1 = E0 e–j2q e ; (6.93)
- j b2 l - j 4 b1l
E6 = (E4 cos q – E3 sin q) e = E0 e–j2q e ; (6.94)
Therefore, the parallel component E5 and perpendicular component E6 at the output end of the quarter-
wave plate are equal in magnitude and in phase to produce a resultant field which is a linearly polarized TE11
wave
- j 4 b1l
Eout = 2 E0 e–i2q e
- j 4 b1l
= Ei e–j2q e = Ei e - j (2q + 4 b1l ) (6.95)
having the same direction of polarization as the incident field Ei with a phase change of 2q + 4b1l. Since
q can be varied and 4b1l is fixed at a given frequency and structure, a phase shift of 2q can be obtained by
rotating the half-wave plate precisely through an angle of q with respect to the quarter-wave plates. The
rotation angle can be calibrated to obtain phase shift.
Phase shift in a ferrite device is generated by magnetizing the ferrite inside the wave guide by RF current
such that the phase constant in the propagating line alters due to change in permeability with magnetization
(Ref: para 2.18.3 in Chapter 2).
Ferrite phase shifters may be latching or non-latching, depending upon whether continuous holding
current must be supplied to sustain the magnetic bias field. Figure 6.43 shows a schematic diagram of a
twin-toroid phase shifter. The twin toroid is a latching, nonreciprocal device. It uses either a closed magnetic
circuit or a magnetic circuit with very small air gaps. The relative permeability of the ferrite is controlled by
adjusting the magnetic flux level existing in the closed magnetic circuit.
The dielectric spacer is used to concentrate the rf energy in the center of the waveguide. The walls of
the ferrite toroids, which contact the dielectric spacer, are located in those regions of the waveguide which
support a circularly polarized magnetic field. If b + is the propagation constant when a positive bias field
saturates the ferrite and b – be the same for negative bias field, the maximum amount of phase shift per unit
length is (b + – b –). The variable phase shift less than this can be achieved by reducing the bias field level.
When the the direction of propagation changes, the values of b – and b + interchange.
The device geometry and the saturation magnetization of the ferrite influence the frequencies for practical
realization of these phase shifters. The conductive and dielectric losses are directly proportional to the length
or inversely proportional to the saturation magnetization while the magnetic loss varies approximately
directly with the saturation magnetization. The sum of these losses shows a minimum value in the frequency
range given by
w
0.2 £ m £ 0.6
w
where wm = 2pg (4pMs), g = 2.8 MHz/Oe, 4pMs is the saturation magnetization and w is the microwave
radian frequency. The latching phase shifters have been realized from about 2 GHz to 94 GHz.
The following tree shows the versions of ferrite MIC phase-shifters—analog and digital.
� Analog Control In analog phase shifter, the phase is controlled by varying either the magnitude or the
direction of the magnetization vector M using applied static magnetic field H0.
� Digital Control In digital phase shifter, the phase is controlled by using high remanence ferrite materials
and latching the substrate by a current pulse to change direction (±) of dc magnetic field digitally. The
remanence of magnetic materials is defined in terms of its residual magnetic induction when an externally
applied magnetizing current is reduced to zero as explained in the B–H curve in Fig. 6.44. A pulse current
of sufficient amplitude, through a single turn wire threading the ferrite, is used to magnetize the ferrite
substrate.
If n is the number of bits, then the least significant bit will have a f-shift of 360°/2n. A digital phase shifter
360∞ 360∞ 360∞
of n-bits will consist of n separately actuated sections giving 1
, 2 , º, n of f-shift. A series of
2 2 2
different Df = 180°, 90° and 45° can be cascaded in a single housing to form a 3-bit digital f-shifter.
m2 – K 2
mreff = (6.96a)
m
where m and K are the component of the permeability tensor (m = B/H, K = g 4pM/w, M = magnetic dipole
moment per unit volume)
When the applied magnetic field is small,
mreff ª 1– (wm/w)2 for w > g 4p Ms = wm (6.96b)
where 4pMs = ferrite magnetization, g = gyromagnetic ratio (2.8 MHz/oersted). When w Æ wm, the
magnetic loss cannot be neglected and mreff expression gets modified. The propagation constant of the TEM
mode on the ferrite substrate is given by
Fig. 6.47 Non-reciprocal phase shi er: (a) Top view of strip (b) Side view from the plane of the strip
By magnetizing the ferrite along the direction of the lines and reversing the direction of the magnetization,
a non-reciprocal differential interaction is obtained resulting in a non-reciprocal differential phase shift.
Digital/latching phase shifter can be obtained by using sufficiently thin ferrite and small wires through holes
in the substrate for dc currents I1 and I2. The magnitude and direction of magnetization can be altered by
current pulses. Typical parameters for such a phase shifter are the following:
Phase shift—1 bit (360° + 10°)
Microwave Network Theory and Passive Devices 269
Fig. 6.48 Waveguide tees: (a) E-Tee or Series-T (b) H-Tee or Shunt-T
E-Plane Tee
From considerations of symmetry and the phase relationship of the waves in each of the arms, it can be seen
that a wave incident at Port 3 will result in waves at Ports 1 and 2, which are equal in magnitude but opposite
in phase, i.e., S31 = S13 = – S23 = – S32, S12 = S21. If two in-phase input waves are fed into Ports 1 and 2 of the
collinear arm, the output waves at Port 3 will be opposite in phase and subtractive. Sometimes, this third port
is called the difference arm. By analogy with the voltage relationship in the series circuit, E-plane junction is
also called a series junction. All diagonal elements of the S-matrix of an E-plane T-junction cannot be zero
simultaneously since the tee junction cannot be matched to all the three arms simultaneously. Considering
the Port 3 as matched, the S-matrix of a E-plane T can be derived as follows.
270 Microwave Engineering
Denoting the incident and outgoing signal variables as the ith port by ai and bi, respectively, for an input
1 1
power at Port 3, the net input power to Port 3 is (|a3|2 – |b3|2) = |a3|2 (1 – |S33|2), and the output power
2 2
1
is (|b1|2 + |b2|2) = |a3|2 ◊ |S13|2, since |S31| = |S32| by symmetry. For lossless junction, input power must be
2
equal to the output power, i.e.,
1
|a3|2 (1 – |S33|2) = |a3|2 ◊ |S13|2 or, (1 – |S33|2) = 2 |S13|2
2
By suitable matching elements, we can make S33 = 0, so that |S13| = 1/ 2 . From the symmetry
characteristics described above,
S13 = S31 = 1/ 2 , S23 = S32 = – 1/ 2
After matching Port 3, if one attempts to match either Port 1 or 2 by a similar method, the matching
elements, such as irises or tuning screws will interact with each other and matching at Port 3 would be
disturbed since no port is isolated. Based on power consideration, it can also be shown that S11 = S22 = 1/2
and S12 = S21 = 1/2 for S33 = 0. Therefore, with matching at Port 3, the S-matrix of an E-plane T can be
expressed by real values with proper choice of reference plane:
È 1/2 1/2 1/ 2 ˘ È 1 1 2 ˘
Í ˙ 1Í ˙
[S] = Í 1/2 1/2 -1/ 2 ˙ = Í 1 1 - 2˙ (6.98)
Í ˙ 2Í ˙
ÍÎ1/ 2 -1/ 2 0 ˙˚ ÍÎ 2 - 2 0 ˙˚
There are three cases to describe below:
� Case 1 Input at port 3 and no inputs at port 1 and 2, a3 π 0, a1 = a2 = 0.
1
b1 = a3
2
1
b2 = – a3
2
b3 = 0
� Case 2 Input at port 1 and Port 2, and no input at port 3. a3 = 0, a1 = a2 = a.
a a
b1 = + = a
2 2
a a
b2 = + =a
2 2
b3 = 0
� Case 3 Input at port 1 and Port 2, and no input at port 3. a1 π 0, a3 = a2 = 0.
a1
b1 =
2
a
b2 = 1
2
a1
b3 = -
2
Similarly, we have all combinations of input and output.
Microwave Network Theory and Passive Devices 271
H-Plane Tee
In an H-plane tee, if two in-phase input waves are fed into ports 1 and 2 of the collinear arm, the output waves
at Port 3 will be in-phase and additive. Because of this, the third port is called the sum arm. Conversly, an
input wave at Port 3 will be equally divided into ports 1 and 2 in phase. Because the magnetic field loops
get divided into two arms 1 and 2 in a manner similar to currents between branches in the parallel circuit,
an H-plane junction is also called a shunt junction. For a symmetrical and lossless junction, in absence of
non-linear elements at the H-plane junction, the S-parameters are obtained in a similar manner as in the case
of E-plane junction:
Since it is a three-port junction, the scattering matrix can be derived as follows:
[S] Matrix of order 3 ¥ 3.
È S11 S12 S13 ˘
[S] = ÍS21 S22 S23 ˙ (6.99a)
Í ˙
ÍÎ S31 S32 S33 ˙˚
\ S12 = –S11
È 1 -1 2˘
1Í ˙
[S] = Í -1 1 2˙ ((6.99h)
2
Í 2 0 ˙˚
Î 2
Output at Port 3 is the addition of the two inputs at Port 1 and Port 2 and these are added in phase.
Example 6.17 Consider a lossless two-port network. (a) If the network is reciprocal, show that
|S21|2 = 1 – |S11|2. (b) If the network is nonreciprocal, show that it is impossible to
have unidirectional transmission, where S12 = 0 and S21 π 0.
È S11 S21 ˘
(a) [S] = Í ˙ S12 = S21 since reciprocal
ÎS21 S22 ˚
Microwave Network Theory and Passive Devices 273
ÈS S ˘
(b) [S] = Í 11 21 ˙ S12 π S21 since nonreciprocal
Î 0 S22 ˚
First row: |S11|2 + |S21|2 = 1
First Column: |S11|2 = 1
\ |S21| = 0
Example 6.18 A 20 mW signal is fed into one of the collinear (port 1) of a lossless H-plane
T-junction. Calculate the power delivered through each port when other ports are
terminated in matched load.
Solution Since Ports 2 and 3 are matched terminated, a2 = a3 = 0, |S11| = 1/2. The total effective power
input to Port 1 is
1
P1 = |a1|2 (1 – |S11|2)
2
= 20 (1 – 0.52) = 15 mW
The power transmitted to Port 3 is
1
P3 = |a1|2 |S31|2
2
= 20 ¥ (1/ 2 )2 = 10 mW
The power transmitted to Port 2 is
1
P2 = |a1|2 |S21|2
2
= 20 ¥ (1/2)2 = 5 mW.
Therefore, P1 = P3 + P2
Example 6.19 In an H-plane T-junction, compute power delivered to the loads of 40 ohms and 60
ohms connected to arms 1 and 2 when a 10 mW power is delivered to the matched
port 3.
Solution With Port 3 matched, the scattering matrix for H-plane T is
È 1 -1 2˘
Í ˙
[S] = 1/2 Í -1 1 2˙
Í ˙
ÍÎ 2 2 0 ˙˚
Therefore, input power at Port 3 is equally divided in arms 1 and 2. Since input at Port 3 = 10 mW =
0.01 W, power towards ports 1 and 2 = 0.005 W = (1/2) |b1|2 = (1/2) |b2|2. Considering first-order reflection,
reflected power from ports 1 and 2 are
(1/2) |G1b1|2 and (1/2) |G2b2|2
274 Microwave Engineering
Hybrid or Magic-T
A hybrid tee is formed with the combination of the E-plane and H-plane tees and is called a magic-T. It has
four ports as shown in Fig. 6.49(a) and 6.49(b).
The magic-T has the following characteristics when all the ports are terminated with a matched loads. Let
us consider the structure (a) with the ports as indicated.
1. If two waves of equal magnitude and equal phase are fed into ports 1 and 2, the output at Port 3 is
subtractive and becomes zero and total output will appear additively at the port 4. Hence, Port 3 is
called the difference or E-arm and 4, the sum or H-arm.
2. A wave incident at Port 3 (E-arm) divides equally between ports 1 and 2 but is opposite in phase
with no coupling to Port 4 (H-arm). Thus,
S13 = –S23, S43= 0 (6.100)
3. A wave incident at Port 4 (H-arm) divides equally between ports 1 and 2 in phase with no coupling
to port 3 (E-arm). Thus,
S14 = S41 = 1/ 2 = S24 = S42 and S34 = 0 (6.101a)
4.A wave fed into one collinear port, 1 or 2, will not appear in the other collinear Ports, 2 or 1,
respectively. Hence, two collinear ports 1 and 2 are isolated from each other, making
S12 = S21 = 0 (6.101b)
A magic-T can be matched by putting tuning screws suitably in the E and H-arms without destroying the
symmetry of the junctions. Therefore, for an ideal lossless magic-T matched at ports 3 and 4, S33 = S44 = 0.
The procedure of derivation of the S-matrix considers the symmetry property at the junction for which S14
Microwave Network Theory and Passive Devices 275
= S41 = S24 = S42, S31 = S13 = –S23 = –S32, S34 = S43 = 0, S12 = S21 = 0. Therefore, the S-matrix for a magic-T,
matched at ports 3 and 4 given by
È S11 S12 S13 S14 ˘
Í ˙
S S22 - S13 S14 ˙
[S] = Í 12 (6.102)
Í S13 - S13 0 0 ˙
Í ˙
Î S14 S14 0 0 ˚
From the unitary property applied to rows 1 and 2, we get
|S11|2 + |S12|2 + |S13|2 + |S14|2 = 1 (6.103)
2 2 2 2
|S12| + |S22| + |S13| + |S14| = 1 (6.104)
Subtracting these two equations:
|S11|2 – |S22|2 = 0 or, |S11| = |S22| (6.105)
Form the unitary property applied to rows 3 and 4,
2 |S13|2 = 1, or |S13| = 1/ 2 (6.106)
2
2 |S14| = 1, or |S14| = 1/ 2 (6.107)
Substituting these values in Eq. (6.103),
|S11|2 + |S12|2 + 1/2 + 1/2 = 1 or, |S11|2 + |S12|2 = 0 (6.108)
which is valid if S11 = S12 = 0 (6.109)
From Eqs (6.105) and (6.109), S22 = 0 (6.110)
È 0 0 S13 S13 ˘
Í ˙
0 0 - S13 S13 ˙
Therefore, [S] = Í (6.111)
Í S13 - S13 0 0 ˙
Í ˙
ÍÎ S13 S13 0 0 ˙˚
where |S13| = 1/ 2 = |S14|
By proper choice of reference planes in arms 3 and 4, it is possible to make both S13 and S14 real, resulting
in the final form of S-matrix of magic-T.
È0 0 1 1˘
Í ˙
0 0 -1 1˙
[S] = 1/ 2 Í (6.112a)
Í 1 -1 0 0˙
Í ˙
Î1 1 0 0˚
For the structure (b) where ports 3 and 4 are interchanged, the S-matrix becomes
È0 0 1 1˘
Í ˙
0 0 1 -1˙
[S] = 1/ 2 Í (6.112b)
Í1 1 0 0˙
Í ˙
Î 1 -1 0 0˚
Example 6.20 A magic-T is terminated at collinear ports 1 and 2 and difference port 4 by
impedances of reflection coefficients G1 = 0.5, G2 = 0.6 and G4 = 0.8, respectively.
If 1W power is fed at the sum port 3, calculate the power reflected at Port 3 and power transmitted to the
other three ports.
276 Microwave Engineering
Solution S-matrix for a matched magic-T with collinear ports 1 and 2 and sum and difference ports 3
and 4, respectively, is given by
È0 0 1 1˘
Í ˙
0 0 1 -1˙
[S] = 1/ 2 ÍÍ
1 1 0 0˙
Í ˙
Î 1 -1 0 0˚
If a1, a2, a3 and a4 be the normalized input voltages and b1, b2, b3, and b4 are the corresponding output
voltage at ports 1, 2, 3, and 4 respectively, then
a1 = G1b1, a2 = G2b2, a3 = input applied voltage at port 3, and
a4 = G4b4.
Now, Pi = |a3|2/2 = 1W, or, a3 = 2V
Therefore,
È b1 ˘ È0 0 1 1˘ È .5b1 ˘
Í ˙ Í ˙Í ˙
Í b2 ˙ = 1/ 2 Í0 0 1 -1˙ Í.6b2 ˙
Í b3 ˙ Í1 1 0 0˙ Í 2 ˙
Í ˙ Í ˙Í ˙
Îb4 ˚ Î 1 -1 0 0 ˚ ÍÎ.8b4 ˙˚
È 2 0 0 - 0.8 ˘
Í ˙
1 Í 2 2 0 0.8 ˙
Í ˙
2 Í 0 - 0.6 2 0 ˙
Í ˙
Î0 0.6 0 2 ˚ 2.0782
b1 = = = 0.928 V
È 2 0 0 - 0.8 ˘ 2.2381
Í ˙
1 Í 0 2 0 0.8 ˙
Í ˙
2 Í- 0.5 - 0.6 2 0 ˙
Í ˙
Î- 0.5 0.6 0 2 ˚
2.3981
Similarly, b2 = = 1.07 V
2.2381
1.7524
b3 = = 0.78 V
2.2381
–0.2827
b4 = = – 0.126 V
2.2381
Microwave Network Theory and Passive Devices 277
Therefore,
1
Power transmitted at Port 1 = |b1|2 = 0.4309 W
2
1
Power transmitted at Port 2 = |b2|2 = 0.5738 W
2
1
Power transmitted at Port 4 = |b4|2 = 0.00797 W
2
1
Power reflected at Port 3 = |b3|2 = 0.3065 W
2
Note: Power absorbed at Port i = 1/2 (|bi|2 – |ai|2); i = 1, 2, 4. Total power absorbed by the system
1
= (|a3|2 – |b3|2) .
2
Application of Magic-T
The magic-T has a number of applications in various microwave circuits, such as the E-H tuner for impedance
matching, balanced mixer in a microwave superheterodyne to balance out the local oscillator noise at the IF
amplifier input, power combiner and duplexer.
� E-H Tuner In an E-H tuner (Fig. 6.50), both the E and
H-arms are terminated by movable shorts which act as E-plane
and H-plane stubs. The position of the shorts can be adjusted so
that a wide range of load impedance may be matched to reduce
the VSWR of a waveguide system connected through the collinear
arms.
� Balanced Mixer In a balanced microwave mixer
configuration, an incoming signal is fed to the E-arm and a local
oscillator signal is fed to the H-arm as shown in Fig. 6.51. When Fig. 6.50 E-H tuners
these two signals enter the collinear arms, the crystal diodes
placed in these arms produce the IF signal or difference signal in
the following manner.
The local oscillator signal from the H-arm will arrive at the diodes in collinear arms in-phase, whereas
the incoming signal from the E-arm will arrive at the diodes out-of-phase. These signals are mixed in the
non-linear diodes and produce IF signals in the collinear arms which are out-of-phase by 180°. Since local
oscillator noise will be in-phase at the diodes, this gets cancelled at the balanced IF input to IF amplifier
whereas, the IF signals are added up in phase for amplification in IF amplifier. LO and RF signals are
uncoupled due to magic-T properties of E and H-arms. For equal power inputs at isolated ports 3 and 4, P3
1
= P4 = P = |a|2, where a3 = a4 = a. If Port 1 is matched, a1 = 0. Therefore,
2
b2 = (0 + 0 – a + a)/ 2 = 0.
and hence a2 = 0. Consequently b3 = b4 = 0. Thus
b1 = S13 a + S14 a = 2 a
1
Therefore, output power at Port 1 is P1 = |b1|2 = |a|2 = 2P = sum of two equal input powers.
2
In a duplexer circuit of a radar system, a common antenna is connected to Port 1 while the transmitter and
receiver are connected to isolated E and H-arms and a dummy load is connected at Port 2. Half of the power
transmitted is coupled to the antenna and 50 % of the received power gets into the receiver, remaining 50 %
power is absorbed in dummy load.
Circulators
A circulator is a multiport junction in which
the wave can travel from one port to the next
immediate port in one direction only as shown
in Fig. 6.52(a). Commonly used circulators are
three-port or four-port passive devices although
more number of ports is possible.
� Four-port Circulator A four-port circulator
can be constructed from two magic-T’s and a non-
reciprocal 180° phase shifter or a combination of
two 3 dB side hole directional couplers with two
non-reciprocal phase shifters as shown in Figs
6.52 (b) and (c).
In Fig. 6.52 (b), an input signal at Port 1 is split
into two in-phase and equal amplitude waves in Fig. 6.52(a) Schema c diagram of a four-port circulator
the collinear arms b and d of the magic-tee, T1 and
added up to emerge from Port 2 in the magic tee, T2. On the other hand a signal at Port 2 will split into two
equal amplitude and equiphase waves in the collinear arms of the magic-tee, T2 and appears at point b and d
out of phase due to presence of the non-reciprocal 180° phase shifter. These out-of-phase waves add up and
appear from Port 3 in the magic-tee, T1. In a similar manner, an input signal at Port 3 will emerge from 4, an
input at Port 4 will appear at Port 1. Thus, the circulator property is exhibited.
Microwave Network Theory and Passive Devices 279
In Fig. 6.52(c), each of the two 3 dB couplers introduces a 90° phase shift. An input signal at Port 1 is
split into two components by the coupler 1 and the coupled signals are again split into two components by
the coupler 2 with a 90° phase shift in each. Each of the two phase shifters produces additional phase shift
so that the signal components at Port 2 are in phase, and at Port 4, they are out of phase. Since Port 3 is the
decoupled port for the directional coupler, the input signal at Port 1 appears in Port 2. Similarly, signals from
Port 2 to Port 3, from Port 3 to Port 4 and from Port 4 to Port 1.
A perfectly matched, lossless, and non-reciprocal four-port circulator has S-matrix:
È0 0 0 1˘
Í ˙
1 0 0 0˙
[S] = Í (6.113)
Í0 1 0 0˙
Í ˙
Î0 0 1 0˚
� Three-port Circulator A three-port circulator is formed by a 120° H-plane waveguide or strip line
symmetrical Y-junction with a central ferrite post or disc. A steady magnetic field H0 is applied along the
280 Microwave Engineering
axis of the disc as shown in Fig. 6.53. Depending on the polarization of the incident wave and the direction
of H0, the microwave signal travels from one port to the immediate next one only.
For a perfectly matched, lossless, non-reciprocal three-port circulator, the S-matrix is
È 0 0 S13 ˘
Í ˙
[S] = Í S21 0 0 ˙ (6.114)
ÍÎ 0 S32 0 ˙˚
If the terminal planes are properly chosen to make the phase angles of S13, S21 and S32 zero,
S13 = S21 = S32 = 1
so that
È0 0 1 ˘
Í ˙
[S] = Í1 0 0 ˙ (6.115)
ÍÎ0 1 0 ˙˚
Fig. 6.53 Three-port circulator: (a) Waveguide type (b) Stripline type
The matching of the junction can be achieved by placing suitable tuning elements in each arm.
Since, in practice, losses are always present, the performance is limited by finite isolation and non-zero
insertion loss. Typical characteristics can be represented by
Insertion loss < 1 dB
Isolation ª 30–40 dB
VSWR < 1.5
Example 6.21 Prove that it is impossible to construct a perfectly matched, lossless, reciprocal
three-port junction.
Solution A perfectly matched three-port junction has a symmetric scattering matrix with zero diagonal
elements:
È 0 S12 S13 ˘
Í ˙
[S] = Í S12 0 S23 ˙
ÍÎ S13 S23 0 ˙˚
Microwave Network Theory and Passive Devices 281
Example 6.22 A three-port circulator has an insertion loss of 1 dB, isolation-30 dB and VSWR =
1.5. Find the S-matrix.
Solution The S-matrix of a three-port circulator is
È S11 S12 S13 ˘
Í ˙
[S] = Í S21 S22 S23 ˙
ÍÎ S31 S32 S33 ˙˚
Insertion loss = 1 dB = – 20 log |S21| Fig. 6.54 Three-port circulator for
–1/20
Example 6.22
or, |S21| = 10 = 0.89
For the same insertion loss between ports 1 and 2, 2 and 3, 3 and 1, |S21| = |S32| = |S13| = 0.89.
The isolation between the ports is 30 dB = – 20 log |S31|
or, |S31| = 10–30/20 = 10–1.5 = 0.032
= |S23| = |S12|
Since VSWR S = 1.5, reflection coefficient
S - 1 1.5 - 1
|G| = = = 0.2 = |S11|
S + 1 1.5 + 1
= |S22| = |S33|
By placing reference planes suitably to make the phase of S-parameters zero, the S-matrix becomes
È0.200 0.032 0.890 ˘
[S] = Í0.890 0.200 0.032 ˙
Í ˙
ÎÍ0.032 0.890 0.200 ˙˚
not done
� Microstrip Circulator A microstrip circulator is
constructed in the same form as strip line one without having
the top ground plane. The entire circulator can be made by
using the whole ferrite substrate. The basic design criteria
are
∑ Selection of radius R of ferrite disc
∑ Calculation of radius R of magnet cylinder for the
whole ferrite substrate.
The value of R is given by
1.84
R= (6.116a)
w 0 (e 0e r m0 meff )
Fig. 6.55 Microstrip circulator
282 Microwave Engineering
w
H0 = = Field required for gyromagnetic resonance in the infinite ferrite medium
|g |
For low magnetic loss, ferrite material selection is done such that
g 4pMs/w = 0.6 (6.116b)
The insertion loss between coupled ports is determined by
1. Copper loss of the strip and ground plane
2. Dielectric loss of the input/output strips
3. Magnetic loss of the ferrite disc
The isolation between the ports is primarily dependent on the mismatching between the ports and the
junction. Tuning screws, l/4 transformer or tapered strips can be used for matching and this can be the parts
of the mask used in photo-etching process. The magnetic disc can be located above or below the substrate.
Using YIG substrate of thickness h = 0.055¢¢ one can get Isolation > 20 dB, VSWR < 1.2, insertion loss
1 L < 0.8 dB, power handling capability of 60 W over the frequency range 8.5–9.9 GHz.
The power-handling capability of such a device can be increased by
1. Lowering the impedance or increasing intrinsic line width
2. Increasing substrate thickness h
3. Decreasing 4pMs of the material by substituting Al ions in the YIG material.
Isolators
An isolator is a two-port, non-reciprocal device which produces a minimum attenuation to wave propagation
in one direction and very high attenuation in the opposite direction. Thus, when inserted between a signal
source and load, almost all the signal power can be transmitted to the load and any reflected power from
the load is not fed back to the generator output port. This eliminates variations of source power output and
frequency pulling due to changing loads.
An isolator can be constructed in a rectangular waveguide (a ¥ b) operating in dominant mode as shown
in Fig. 6.56(a). The non-reciprocal characteristics are obtained by establishing a steady magnetic field H0 in
the y direction and placing a ferrite slab at any of the longitudinal planes x = x1 near and parallel to the narrow
waveguide wall, where the magnetic field exhibits circular polarization. This occurs at x1 = a/4 or, 3a/4.
For the propagation of waves in +z direction, direction of rotation of H in the planes at x1 = a/4 and 3a/4
are opposite to each other. The non-reciprocal characteristic is achieved by placing a ferrite slab at any one
of these two planes. The required steady state magnetic field H0 in the y-direction is established by placing
permanent magnetic poles between the two broad walls.
Microwave Network Theory and Passive Devices 283
It is known that the attenuation in ferrite for negative/clockwise circular polarization is very small whereas
for positive/counter clockwise circular polarization is very large at and near the resonance frequency f ª f0.
Therefore, the ferrite slab is placed in such a way that while transmission it encounters negative circular
polarization in the reverse direction. The steady magnetic field is set to be equal to the resonant value. The
isolation of the order of 20 – 30 dB in the backward direction and a transmission loss of 0.5 dB in the forward
direction can be achieved with a VSWR of the order of 1.1.
Since the reverse power is absorbed in the ferrite and dissipated as heat, the maximum power handling
capability of an isolator is limited. To increase the capacity of heat dissipation, two ferrite slabs of smaller
heights are used instead of one with a larger height.
The requirement of very high steady magnetic field (10,000 oersteds at l = 1 cm) is the main drawback
of waveguide resonance isolator at higher frequencies.
not done
� Faraday Rotation Isolator A Faraday rotation
isolator is a circular waveguide section axially loaded
with a ferrite rod of smaller diameter as shown in
Fig. 6.56(b).
The ferrite rod is subjected to a steady axial magnetic
field H0 of strength much smaller than the resonant
intensity so that dissipative loss in the ferrite is neglected.
The dominant TE11 mode in the circular section can
be decomposed into two oppositely rotating circularly
polarized waves of equal magnitude. These waves
encounter different permeabilities m¢+ and m¢– for the
clockwise and anticlockwise directions of field rotation
and exhibit changes in the phase velocities. This will
result in a change in the plane of polarization of the main
mode TE11 which will experience gradual rotation q
during propagation to the other end. The rotation angle q
is proportional to the length of the ferrite rod. In this case
for the reverse wave the direction of rotation remains the Fig. 6.56(b) Faraday rota on isolator
same confirming the non-reciprocal characteristics of the
ferrite.
The isolator input is a 45° twist where a tapered resistive card is mounted parallel to the broad wall of
the rectangular waveguide part. The dominant TE10 mode does not get attenuated while transmission and is
284 Microwave Engineering
rotated at 45° at the twist output and enters the circular waveguide through rectangular to circular waveguide
transition as the TE11 mode. The length of the ferrite rod is selected so as to obtain Faraday rotation q = 45°
at the output and regain its original polarization. The plane of polarization of the reflected wave from the
load is again rotated by the same angle q = 45° and at the emergence through the 45° twist becomes aligned
with the surface of the absorbing plate and gets absorbed. Thus, non-reciprocal isolation action lakes place.
Typical insertion loss and isolation are approximately 1dB and 20~30 dB, respectively, for these isolators.
Isolators are also available in the coaxial and microstrip forms.
For an ideal lossless, matched isolator
|S21| = 1, |S12| = |S11| = |S22| = 0 (6.117a)
È0 0 ˘
i.e., [S] = Í ˙ (6.117b)
Î1 0 ˚
Example 6.23 A matched isolator has insertion loss of 0.5 dB and an isolation of 25 dB. Find the
scattering coefficients.
Solution
Insertion loss = 0.5 dB = –20 log |S21|
or, S21 = 10–.5/20 = 10–0.025 = 0.9441
The isolation is 25 dB = –20 log |S12|
or, S12 = 10–25/20 = 10–1.2 = 0.0631
Since there is no reflection, S11 = S22 = 0. Therefore, the S-matrix for the isolator is
È 0 10 -1.2 ˘ È 0 0.0631˘
[S] = Í - 0.025 ˙=Í ˙
ÎÍ10 0 ˚˙ Î0.9441 0 ˚
not done � Microstrip Isolator A microstrip isolator can be formed by match terminating one port of a 3-port
circulator. Several termination methods are described with the help of Fig. 6.57(a–d).
(a) Method 1 A chromium layer is deposited on ferrite initially in all circuits for adherence. A tapered
Cr film of 0.6¢¢ long with a surface impedance of 270-ohm termination gives VSWR < 1.2 over the
frequency range 5.5–11.0 GHz.
(b) Method 2 Initial Cr adherence is done as Method 1. Then the Cr film acts as a lumped 50-ohm
resistor which is terminated in a quarter wavelength open circuit impedance. This arrangement
provides VSWR < 1.2 over the frequency range 4.5–6.0 GHz.
(c) Method 3 A 50-ohm lumped resistor shunt mounted across the strip line and terminated by a short
at the end.
(d) Method 4 An isolator in MIC form on ferrite substrate can be designed with a triangular patch
having a narrow transverse slot and corner cuts as shown in Fig. 6.57(d). The dc magnetic field
is applied perpendicular to the substrate plane. For X-band operation, the approximate size of the
isolator may be of the order of 10 mm × 8 mm × 1/.8” and slot size of approximately 6 mm ¥ 0.8
mm for best matching at 11 GHz. The insertion loss of nearly 2 dB and isolation of 18 dB could be
achieved. The metallization on the Ni–Zn ferrite is deposited by sputtering a seed layer of titanium
tungsten (TiW) prior to the final metallization.
Microwave Network Theory and Passive Devices 285
(a) (b)
(c) (d)
Fig. 6.59 Direc onal couplers: (a) Waveguide forward coupler (b) Strip and microstrip line
backward coupler, and (c ) Branch line coupler
Microwave Network Theory and Passive Devices 287
Fig. 6.60 Waveguide direc onal couplers: (a) Bethe-hole coupler (b) Mul hole coupler
288 Microwave Engineering
Bethe-hole Coupler
The Bethe-hole coupler is a single hole broad wall aperture coupled waveguide coupler, where the hole is
located at the centre of a common broad wall of two waveguides inclined at an angle q, or at an offset position
d of two parallel waveguides as shown in Fig. 6.60(a). If the aperture is small compared to the wavelength,
it can be considered as an electric dipole normal to the aperture with dipole moment proportional to the
normal component of the electric field in the main guide at the aperture, plus a magnetic dipole in the plane
of the aperture with dipole moment proportional to the tangential component of the exciting magnetic field
at the aperture. The coupling to the auxiliary guide is due to radiation from these dipoles. The electric dipole
radiates equally in both the directions longitudinally. But the magnetic dipole radiates asymmetrically in
longitudinal directions. By varying the angle q between the two waveguides, or by adjusting the distance d
of the hole from the broad wall edge, powers in Port 3 and Port 4 can be controlled. Ideally, power at Port 3
can be zero (isolated) whereas that at Port 4 is maximum to achieve directional coupling.
Bethe Centre Hole For the centre-coupling hole, since practically P3 π 0, the directivity is finite.
Coupling and directivity of this Bethe-hole coupler are given by
È 4 br 3 Ê l g2 ˆ ˘
C = –20 log Í o
Á cosq + ˜ ˙ dB (6.121)
Í 3 ab ÁË 2 l02 ˜¯ ˙
Î ˚
2 b 2 cosq + k0 2
D = 20 log dB (6.122)
2 b 2 cosq - k0 2
where r0 = Radius of hole
a ¥ b = Waveguide cross-sectional dimension
b = 2p/lg
For w/g excitation in dominant mode TE10
l0
lg =
È Ê l ˆ2˘
Í1 - Á 0 ˜ ˙
ÍÎ Ë 2a ¯ ˙˚
k0 = 2p/l0
Here, the guide wall thickness is assumed negligibly small. The optimum angle q (for P3 = 0) is obtained
Ê l2 ˆ
from the relation q = cos–1 Á g ˜ .
ÁË 2 l02 ˜¯
Bethe Off-set Hole For the off-set hole, the optimum forward coupling (P3 = 0) is obtained for the
off-set value d given by
l0 Ê l g2 ˆ
sin (pd/a) = and q = cos -1 Á 2 ˜ (6.123)
6a ÁË 2 l0 ˜¯
16p r03 Ê pd ˆ
where X= sin 2 Á
3abl g Ë a ˜¯
For non-zero finite wall thickness (t ª 0.51 mm), the coupling decreases and will be 1 to 2 dB less
compared to the zero thickness value.
Example 6.24 Design a centre-hole Bethe-hole directional coupler with air-filled rectangular
waveguide of dimensions 0.9 in ¥ 0.4 in at 9.8 GHz for 20 dB coupling and 40 dB
directivity, respectively.
Solution
a = 0.9 in = 2.286 cm, b = 0.4 in = 1.016 cm, f0 = 9.8 GHz
30 2p 2 ¥ 3.1415
l0 = c/f = = 3.06 cm, k0 = = = 2.0533 rad/cm
9.8 l0 3.06
1 1
lg/l0 = = = 1.3458
È1 - (l /2a ) 2˘ È1 - (3.06 /2 ¥ 2.286 )2 ˘
ÎÍ 0
˚˙ ÎÍ ˚˙
lg = 1.3458 ¥ 3.06 = 4.118 cm,
b = 2p/lg = 1.525 rad/cm
cosq = 1/2 (lg/l0)2 = 1/2 ¥ (1.3458)2 = 0.9055
q = cos–1 0.9055 = 25.09°
ÈÊ 4 1.525 ˆ 3˘
Therefore, C = 20 dB = – 20 log ÍÁË 3 ¥ 2.286 ¥ 1.016 ¥ 2 ¥ 0.9055˜¯ r0 ˙
ÍÎ ˙˚
= – 20 log (1.5855 r03)
or, log (1.5855 r03) = – 1
or, 1.8555 r03 = 10–1 = 0.1
or, r03 = 0.1/1.5855 = 0.0631
Therefore,
r0 = (0.0631)1/3 = 0.398 cm
d = a/2 = 1.143 cm
2 b 2 cosq + (2p / l0 )
2
D = 20 log
2 b 2 cosq - (2p / l0 )
2
2 È Ê ˆ
2 2˘
Ê 2p ˆ Ê 2p ˆ
2
Í1.811 2p Ê 2p ˆ ˙
1.811 ¥ Á ˜ +Á Á ˜ - Ál ˜ ˙
or, ˜ = 100 Í Ë l g1 ¯ Ë 01 ¯
Ë l g1 ¯ Ë l01 ¯
ÎÍ ˚˙
1.811 1 181.1 100
or, + = -
l g12 l012 l g12 l012
1 1
or, [1.811 – 181.1] = – (100 + 1)
lg12 l012
179.89 101
or, =
l g12 l012
l g12 179.89
or, = = 1.775
l012 101
1
or, = 1.775
È1 - [ l01 (2 ¥ 2.286)] 2 ˘
Î ˚
2 2
Ê l ˆ Ê 1 ˆ
1 – Á 01 ˜ = Á
Ë 1.775 ˜¯
or,
Ë 4.572 ¯
È Ê 1 ˆ2˘
or, l012 = Í1 - Á ˜ ˙ (4.575) = 9.1276
2
ÍÎ Ë 1.775 ¯ ˙˚
or, l01 = 3.02 cm
c 30
Therefore, f1 = = = 9.929 GHz
l01 3.02
f2 = 2f0 – f1 = 2 ¥ 9.8 – 9.929 = 9.671 GHz
Therefore, bandwidth for D > 40 dB is D f = f1 ~ f2 = 258 MHz
È 0 S12 0 S14 ˘
Í ˙
S 0 S23 0 ˙
[S] = Í 12 (6.126)
Í 0 S23 0 S34 ˙
Í ˙
ÎÍ S14 0 S34 0 ˚˙
Since for a loss network, the S-matrix is unitary,
S12 S12* + S14 S14* = 1 (6.127)
* *
S12 S12 + S23 S23 = 1 (6.128)
S23 S23* + S34 S34* = 1 (6.129)
* *
S14 S14 + S34 S34 = 1 (6.130)
S12 S23* + S14 S34* = 0 (6.131)
From the first two equations, we get
|S14| = |S23| (6.132)
and from the second and third equations, we get
|S12| = |S34| (6.133)
Further, by choosing reference plane of Port 1 with respect to that of Port 2 and the reference plane of Port
3 with respect to that of 4, we can make S-parameters real
S12 = S34 = a, say (6.134)
where a is a positive real number. Then from the fifth Eq. (6.131),
a (S23* + S14) = 0
Since a π 0, S23* + S14 = 0 (6.135)
Further, selecting the reference plane of Port 4 with respect to Port 1, we can make S14 real, so that
– S*23 = S14 = b, say real (6.136)
\ S23 = S23* = – b
È0 a 0 b˘
Í ˙
Therefore, [S] = Ía 0 -b 0˙ (6.137)
Í0 -b 0 a˙
Í ˙
Îb 0 a 0˚
where a2 + b2 = 1 for conservation of energy. Here, a is called the transmission factor and b is the coupling
factor.
From above, the coupling, directivity and transmission loss for a matched directional coupler can be
expressed in terms of S-parameters as
P1
C = 10 log = – 20 log |S41| (6.138)
P4
P4 S S
D = 10 log = 20 log 4 1 = 20 log 4 1 (6.139)
P3 S31 S42
T = 10 log P1/P2 = 20 log |S21| (6.140)
The amount of coupling (3 dB, 6 dB, 10 dB, 20 dB, 30 dB, etc.) and the directivity (30–40 dB) depend
upon the sizes and locations of the holes in the common wall.
Since the phase cancellation in the reverse direction between successive holes can occur only at the
designed frequency which satisfy d = lg0/4, the bandwidth and the frequency response of the coupler
are determined by designing the coupler following a suitable reverse voltage response function, such as,
maximally flat or binomial response, Chebyshev response, etc., as shown in Fig. 6.61.
Fig. 6.61 Coupling characteris cs of direc onal couplers: (a) Binomial (b) Chebyshev
For simplicity, we assume that the total power coupled is a small fraction of the incident power so that
the incident wave amplitude can be considered to be same at each coupling aperture and only the phase will
change at different holes. The total reverse and forward voltages at any frequency f can be represented for an
unit input voltage at Port 1, respectively by,
G¢ = G1 + G2e–j2q + G3e–j4q + + Gne–j2(n – 1)q
n
= Â Gk e–j2 (k –1)q (6.141)
k =1
n
and C¢ = e–j (n – 1)q  Gk (6.142)
k =1
2p d p
Here, q = , with q = at the design frequency f0. Coupling holes are assumed to be non-directional
lg 2
due to small sizes. Normally, the design is made keeping the structure symmetrical with respect to a centre
plane P as shown in Fig. 6.60(b) so that the size of the kth hole is the same as that of the (n + 1 – k)th one.
Under this condition, G1 = G n , G 2 = G n-1, etc. Therefore, by changing the reference plane to P, from (6.141)
and (6.142):
( n -1)/2
=2 Â G k cos{n - (2 k - 1)}q ; for n = odd number (6.143b)
k =1
jnq - jnq
since e + e = 2cos nq .
� The Forward or the Couple Wave
n -1
-j q
C= e 2 2 ÈG + G2 + + G n /2 ˘˚ ; n even
Î 1
n -1
-j q n /2
or, C= e 2 2
 G k ; n even (6.144a)
k =1
Ê n -1ˆ
- jÁ q È ˘
Ë 2 ¯˜
and, C= e 2 Í G1 + G 2 + + G n -1 + G n +1 ˙ ; n odd
ÍÎ 2 2 ˚
˙
È n -1 ˘
n -1
-j q Í 2 ˙
or, C= e 2 Í2 Â G k + G n +1 ˙ ; n odd (6.144b)
Í k =1 2 ˙
ÎÍ ˚˙
The coupling and directivity of the directional coupler can be defined from (6.143), and (6.144) as
Coupling C (dB) = –20log |C| (6.145)
G
Directivity D(dB) = –20log (6.146)
C
The multihole directional couplers are designed for a desired coupling and directivity from the type of
distribution by keeping the reverse voltage magnitude |G| under a specified value Gm over the frequency band
between the edges f1 and f2 (Fig. 6.61).
� Binomial Coupler Binomial couplers are designed by choosing the coupling coefficient Gk at a
hole k proportional to the corresponding binomial coefficient of (a + b)n–1 for total n holes. For a unit
input voltage, the total reverse coupled voltage G = 0 at f = f0 and increases slowly at either side of f0
with a maximally flat realisable response as shown in Fig. 6.61(a).
Here, Gk = n – 1Ck – 1G1 (6.147)
n–1 (n - 1)!
where Ck – 1 = binomial coefficient (6.148)
(k - 1)! (n - k )!
n -1 n -1 n -1
viz., for n holes G1 = C0 , G 2 = C1, G 3 = C2 …
Gk can be adjusted by varying the aperture size while maintaining its shape and position fixed. For the
reverse wave for f1 £ f £ f2,
È ˘
and |G| = 2G1 Í n - 1C0 cos(n - 1)q + n -1
C1 cos(n - 3)q + + n -1
C n -1 ˙
ÍÎ 2 ˚
˙
( n -1) / 2
= 2G 1 Â n -1
Ck -1 cos(n - 2 k -1)q + n -1C( n -1)/2 £ Gm; n odd (6.150)
k =1
Expanding cos mq in each term of the above equation, in terms of cos q, only one term containing cosn – 1
q with coefficient An – 1 does not cancel, so that
G = 2 G1 An–1 cosn–1 q (6.151)
where An – 1 is a constant determined from the expansion of cos (n – 1)q. Since all the derivatives up to the
(n – 2)th terms of G at mid-frequency f = f0 are zero, the binomial distribution results in a maximally flat
response without any ripple. Since the maximum value of G at the band edges is specified to be Gm,
n −1
Gm = 2Γ1 An −1 cos θ; q1 for f1, and q2 for f2
or
Gm
G1 max £ ; q1 £ q < q 2 (6.152)
2 An -1 cosn -1 q
since G = G m at f1 and f2 q 2 = p - q1 .
If the band-edge frequency ratio is defined by,
r = lg1/lg2 = q2/q1, and d = lg0 /4,
q1 = p /(1 + r) = 2pd/lg1
q2 = p r/(1 + r) = 2pd/lg2
The coupling coefficients for a binomial coupler having specified number of holes are given as follows.
No. of holes n Coefficients
2 a1 = 2 – 1C0 = 1 = a2,
3 a1 = 3 – 1C0 = 1 = a3, a2 = 3 – 1 C1 = 2
4–1
4 a1 = C0 = 1 = a4, a2 = 4 – 1 C1 = 3 = a3
5–1
5 a1 = C0 = 1 = a5, a2 = 5 – 1 C1 = 4 = a4,
a3 = 5 – 1 C2 = 6
The coefficient for other number of holes can be determined in a similar manner. G1 can be computed
from the desired coupling C for a given number of holes.
A major practical disadvantage of a binomial coupler is the wide variation between the coupling
coefficients of different holes of the coupler, especially for a large number of holes. This leads to fabrication
difficulties while maintaining equal-hole spacings of quarter wavelength.
Example 6.25 Design a maximally flat 20 dB directional coupler so that D ≥ 40 dB in the band
r = 2.
Let n = number of holes to be selected. If G1, G2, G3, …, are the coupled wave at corresponding holes
G12n – 1 cosn – 1 q1 £ Gm; q1 = p/(1 + 2) = p /3 (A)
C = 20 dB = –20 log C
Therefore, C = 10–1 = 0.1 = G1 + G2 + G3 + … + Gn (B)
For symmetrical coupler with binomial type,
G1 = Gn
G2 = (n – 1)G1 = Gn – 1
G3 = [(n – 1)(n – 2)/2] G1 = Gn – 2, etc.
Selecting n = 3, G1 = G3
G2 = (n – 1)G1 = 2G1
Therefore, (B) reduces to
C = 0.1 = G1 + 2G1 + G1 = 4G1
G1 = 0.1/4 = 0.025
Therefore, the left-hand side of (A) is
0.1 Êpˆ 0.1
¥ 22 cos2 Á ˜ = > Gm
4 Ë 3¯ 4
Therefore, n > 3
Selecting n = 5, G1 = G5
G2 = (n – 1)G1 = 4G1
G3 = [(n – 1)(n – 2)/2]G1 = 6G1,
Therefore, C = 0.1 = 2(G1 + G2) + G3 = 2(G1 + 4G1) + 6G1 = 16G1
or, G1 = 0.1/16
Therefore, the left-hand side of (A) is
0.1 Êpˆ
¥ 24 cos4 Á ˜ = 6.25 ¥ 10–3 > Gm
16 Ë 3¯
Therefore, n > 5
Selecting n = 6, G1 = G6
G2 = (n – 1)G1 = 5G1
G3 = [(n – 1)(n – 2)/2]G1 = 10G1,
Therefore, C = 0.1 = 2 (G1 + G2 + G3) = 2 (G1 + 5G1 + 10G1) = 32G1
or, G1 = 0.1/32
Therefore, the left-hand side of (A) is
0.1 Êpˆ
¥ 25 cos5 Á ˜ = 3.125 ¥ 10–3 > Gm
32 Ë 3¯
Therefore, n > 6
296 Microwave Engineering
Example 6.26 A rectangular waveguide binomial coupler coupled by 5 circular holes in a common
side/narrow wall to produce 30 dB coupling at 10 GHz. The guide width a = 2.5 cm
and the height b = 1.2 cm. The dominant input mode of unit amplitude radiates a field of amplitude
4 1
–j r03(p /a)2 in both directions in the other guide. Find the required hole radii and the frequency ratio
3 abb
for which the directivity is greater than 50 dB.
Solution Given,
4 3 Ê p ˆ2 1
|C| = r0 Á ˜ , for narrow-wall waveguide coupler with number of holes n = 5
3 Ë a ¯ abb
Binomial coupling C = 30 dB at f = f0 = 10 GHz. Waveguide dimension a = 2.5 cm and b = 1.2 cm
(a) To find C
Assuming symmetrical binomial-type coupler
C = G1 + G2 + G3 + G4 + G5 = 2 (G1 + G2) + G3
where G1 = G5
G2 = G4 = (n – 1) G1 = 4G1
G3 = [(n – 1)(n – 2)/2] G1 = 6G1
Therefore, C = 2 (G1 + G2) + G3 = 16 G1
Given, 30 dB = – 20 log C
C = 10–30/20 = 10–1.5 = 0.0316
Therefore, G1 = C/16 = 1.976 ¥ 10–3 = G5
G2 = G4 = 4G1 = 7.905 ¥ 10–3
G3 = 6G1 = 11.858 ¥ 10–3
(b) To find hole spacing
l0 = c/f = 30/10 = 3 cm
l0 3
lg0 = = = 3.75 cm
2
[1 - (l0 /2 a ) ] [1 - {3/(2 ¥ 2.5)}2 ]
Microwave Network Theory and Passive Devices 297
� Chebyshev Coupler The Chebyshev coupler response is an equal ripple characteristic where the
reverse coupling G is made proportional to a Chebyshev polynomial of order n – 1 (for number of holes = n)
Tp (x) = cos [p cos–1 (x) ]; – 1 £ x £ + 1, p = n – 1
= cosh [p cosh–1 (x)]; x < – 1, x > +1, p = n – 1 (6.153)
where x is the parameter for change of variable and is a function of frequency:
x = cos q /cos q1 = cos q /|cos q2| (6.154)
The polynomials have the following properties as shown in Chapter 5.
1. All polynomials pass through the point (1,1).
2. –1 £ Tp (x) £ +1, for –1 £ x £ 1.
298 Microwave Engineering
3. All roots occur within –1 £ x £ 1, and all maxima and minima have values of +1 and –1,
respectively.
Thus, we have within the band
|G| = 2 ÈÎG1 cos(n - 1)q + G 2 cos(n - 3)q + + G n /2 cosq ˘˚
n/2
= 2 Â Gk cos (n – 2 k -1 ) q;
k =1
( n - 1) / 2
= Â Gk cos [n – (2k – 1)]q + G(n + 1)/2;
k =1
Example 6.27 Design a five-hole 30 dB directional coupler with Chebyshev distribution for
wavelength ratio of 2 at the band edges.
= Gm T5 – 1(x)
= Gm (8x4 – 8x2 + 1)
Equating the coefficients
G1 = 8 Gm
G2 = 4 G1 – 8 Gm = 32 Gm – 8 Gm = 24 Gm
G3 + 2 G1 – 2 G2 = Gm or, G3 = Gm – 16 Gm + 48 Gm = 33 Gm
Now coupling
C = 2(G1 + G2) + G3 = 30 dB = 10–30/20 = 0.0316
or, 2(8 + 24) Gm + 33 Gm = 0.0316.
or, 97 Gm = 0.0316 or, Gm = 3.26 ¥ 10–4
Therefore, coupling values of the holes are
G1 = 8 Gm = 26.08 ¥ 10–4
G2 = 24 Gm = 78.24 ¥ 10–4
G3 = 33 Gm = 107.58 ¥ 10–4
Directivity D = – C + 20 log (1/Gm) = 39.7 dB
In case the coupling coefficients make the hole size large, holes may overlap to make the centre-to-centre
distance equal to the quarter wavelength. Otherwise, also for large holes, the coupling coefficients become
more frequency sensitive. Under such a situation, the number of holes are increased to meet the desired
specifications of coupling and directivity.
Under such a situation, the number of holes is increased to meet the desired specifications of coupling
and directivity.
� Coupled Co-axial Lines The electromagnetic field from port 1 of one coaxial transmission line can be
coupled to a second adjacent line when a narrow longitudinal slot is cut between the lines on the common
outer conductor joint as shown in Fig. 6.62. The electric field in the input or primary line induces an equal
and opposite charge on the centre conductor of the two lines.
This results in an electric field in the coupled or
secondary line directed oppositely to that in the primary
guide. But the magnetic field follows the same direction
in the both the lines. Hence, the directions of power
density flow P = E × H in the two guides are opposite.
Therefore, the coupled power flows in the backward
direction compared to forward power flow in the
primary guide. For this reason, this coupler is called the
backward coupler. The forward power in the coupled
line at Port 4 is ideally zero and this port is called the
isolated port. The coupling slot is made equal to the
quarter wavelength in length for achieving maximum
coupling as can be seen in the analysis section given
below. Fig. 6.62 Coupled co-axial lines
� Microstrip and Strip Line Coupler Of the various types of strip line and microstrip line directional
couplers, the edge-coupled parallel conductor configuration is most extensively used in practical circuits.
300 Microwave Engineering
Detailed analysis and design data on the characteristic impedances and effective dielectric constants of such
structures are derived by S.B. Cohn using the conformal mapping method and even and odd mode method
of analysis. A typical planer coupled transmission linear coupler is shown schematically in Fig. 6.63. The
lines are coupled over a length l with spacing S between the adjacent edges. The width w of the lines are
designed for desired impedance (typically 50 ohms). The lines are supported by dielectric substrate with a
bottom ground plane.
Since the spacing S is usually very small, it would be difficult to install miniature co-axial line signal
launching connectors at the end (port) of each line [Fig. 6.63(a)]. Each connector’s outer diameter could be
larger than the spacing S. Hence signals are fed /taken out of the ports through extra lengths of uncoupled
feed lines which are at right angled to the coupled length l as shown in Fig. 6.63(b).The right angled bends
are metered by a length d, nearly equal to the line width w to reduce impedance mismatch and charge
accumulation due to sharp discontinuities. For 50 ohm strip line, d = 1.131 w, and for a 50-ohm microstrip,
d = 1.194 w. As in a coaxial system, these planar couplers are also backward couplers.
(a) (b)
� Even and Odd Mode Analysis of the Transmission Line Coupler A pair of coupled lines forms a
4-port device with two orthogonal planes of reflection symmetry—PP¢ and QQ¢ as shown in Fig. 6.63. As a
result the scattering matrix of this four port symmetric device is
where
a = |S12| = |S21| = |S34| = |S43| (between through ports)
¸
b = |S14| = |S41| = |S23| = |S32| (between coupled ports 1-4, 4-1, 2-3, 3-2) ˝ (6.156c)
g = |S31| = |S13| = |S24| = |S42|
˛
To determine these S-parameters, we can apply a source to port 1 and then terminate all other ports with
matched loads as shown in Fig. 6.64(a). The main properties of the parallel line coupler can be analyzed
by decomposing the actual excitation into individual even and odd symmetry modes with reference to the
plane of symmetry PP’. Response for actual excitation can be obtained by superimposing the responses of
isolated even and odd modes for a symmetrical structure as shown in Fig. 6.64(b) and 6.64(c). In general,
the characteristic impedances of these modes are not equal due to conditions that the capacitive coupling
associated with these modes are different (i.e., Z e0 π Z o0). Here superscript ‘e’ will be used for even mode and
‘o’ for odd mode.
Fig. 6.64 Transmission-line couplers: (a) coupled transmission line (b) even-mode symmetric
(c) odd-mode symmetry (d) single equivalent circuit
302 Microwave Engineering
For odd mode of excitation, the instantaneous voltages at two strips are out of phase by 180° so that
a majority of the electric field lines start from one conductor at positive potential to the other at negative
potential. On the plane of symmetry PP¢, E is normal with zero tangential component, and H is tangential
with zero normal component. Therefore, PP¢ can be replaced by an electric wall (Et = 0).
For even mode of excitation, both the conductors are in phase of potential (positive, say) so that the E field
lines originate from both of them and terminate on the ground planes/plane. On the plane of symmetry PP¢,
the electric field lines are completely tangential and magnetic field lines are completely normal. Therefore,
PP’ can be replaced by a magnetic wall (Ht = 0).
These impedances are the major design parameters for any parallel coupled lines and are obtained in
terms of coupling C and the single-line terminating characteristic impedance Z0 in the following manner.
Let V + and V – represent input voltage towards a port and output voltage from that port, respectively,
having characteristic impedance Z0.
The S-matrix equation of the coupler is
[b] =[S][a] (6.156d)
where normalized voltages
ai = Vi+ / Z 0
e
Ê Vg ˆ Z
where V1+ e = Á ˜ e 1 (6.161b)
Ë 2 ¯ Z1 + Z 0
Ê Vg ˆ Z 0
V1+0 = Á ˜ 0 1 (6.161c)
Ë 2 ¯ Z 1 + Z0
ÊV ˆ 1
I1+ e = Á g ˜ (6.161d)
Ë 2 ¯ Z 1e + Z 0
Ê Vg ˆ 1
I1+0 = Á ˜ 0 (6.161e)
Ë ¯ Z 1 + Z0
2
Z 0 + jZ 0e tan q
Z1e = Z0e (6.161f)
Z 0e + jZ 0 tan q
Z 0 + jZ 00 tan q
Z10 = Z00 (6.161g)
Z 00 + jZ 0 tan q
where q = bl, the electrical length of the parallel coupled section of length l of a lossless (a = 0) line in which
it is assumed that q e = q 0 = q.
For a perfect match,
Ê Vg ˆ Z1e Ê Vg ˆ Z10
ÁË 2 ˜¯ Z e + Z + ÁË 2 ˜¯ Z 0 + Z
2 Z1e Z10 + Z 0 ( Z1e + Z10 )
Zin = Z0 = 1 0 1 0
= (6.162a)
Ê Vg ˆ 1 Ê Vg ˆ 1 Z1e + Z 10 + 2 Z 0
ÁË 2 ˜¯ Z e + Z + ÁË 2 ˜¯ Z 0 + Z
1 0 1 0
Under the assumption that q e ª q 0 = q and matching condition Z0 = Z 0e Z 00 , above equation leads to
È Ze Z 00 ˘
j Í 00 – ˙ sin q
Í Z0 Z 0e ˙
G e= –G 0 = Î ˚ (6.164b)
È Ze Z 00 ˘
2cosq + j Í 00 – ˙ sin q
Í Z0 Z 0e ˙
Î ˚
2
Te = T0 = (6.164c)
È Ze Z 00 ˘
2cosq + j Í 00 + ˙ sin q
Í Z0 Z 0e ˙
Î ˚
Then b1 = b3 = 0 (6.164d)
2
b2/a1 = T = S21 = (6.164e)
È Ze Z 00 ˘
2cosq + j Í 00 + ˙ sin q
Í Z0 Z 0e ˙
Î ˚
È Ze Z 00 ˘
j Í 00 – ˙ sin q
Í Z0 Z 0e ˙
C =
b4
= S41 = Î ˚ (6.165)
a1 È Ze Z 00 ˘
2cosq + j Í 00 + ˙ sin q
Í Z0 Z 0e ˙
Î ˚
These equations show that
(1) the coupled voltages at Ports 2 and 4 are out of phase by 90°;
(2) coupling is a function of frequency or q ; and
(3) coupled voltage at Port 4 is maximum, when q = odd multiple of p/2.
This corresponds to a minimum coupling length of l = lg0/4 at the midband of frequency f0.
If C0 is the midband coupling coefficient (for q = p/2, l = lg0/4)
b4 Z 0e - Z 00
C0 = = (6.166)
a1 Z 0e + Z 00
Therefore, the impedances required are
The coupling coefficient, transmission coefficient and directivity at any frequency can be written as
j C0 sin q
C = b4/a1 = 2
(6.169)
[(1 - C0 ) cosq + j sin q ]
Microwave Network Theory and Passive Devices 305
(1 - C02 )
T = b2/a1 = (6.170)
[(1 - C02 ) cosq + j sin q ]
D = b3/a4 = 0 (6.171)
Since T is 90° out of phase with both a1 and a4, parallel coupled line is called 90° hybrid or quadrature
coupler. D = 0 ideally, but in practice, q is not same for even and odd modes. Therefore, the relation
Z0 = Z 0e Z 00 is only approximate. Moreover, due to imperfection in matching, directivity becomes non-
zero. For loose coupling (more than 10 dB) the present approximation is sufficiently good. The frequency
response of this coupler is obtained by writing
2p l Ê 2 p l ˆ
q= = f (6.172)
l g ÁË u ˜¯
pl g 0
Since l = lg0/4, q = , when u is the velocity of propagation. Assuming u a constant, q is proportional
2u
to f. The ideal frequency response of this coupler is shown in
Fig. 6.65.
The closed form expression for the even and odd mode
characteristic impedances Z0e and Z00, respectively, of edge-
coupled homogeneous symmetric strip-line coupler with negligible
thickness strip conductors are given by
e
Z 0e e r = 30p K (k1 ) (6.173)
e
K (k )
Fig. 6.65 Typical frequency response
Z 00 er = 30p K (k10 ) of a coupler
(6.174)
K (k 0 )
Ê pwˆ Ê p w + Sˆ
ke = tanh Á tanh Á ◊
Ë 2b ˜¯
where (6.175)
Ë2 b ˜¯
Ê pwˆ Ê p w + Sˆ
k0 = tanh Á coth Á ◊ (6.176)
Ë 2b ˜¯ Ë2 b ˜¯
e
Here, K(k) and K(k1) are the complete elliptic integrals of the first kind. The Variation of Z 0 e r and
Z 00 e r as a function of the dimensional parameters w/b and S/b is shown in Fig. 6.66.
The values of Z0e and Z00 can be obtained from the coupling factor. The design data of w/b and S/b are
obtained from the graph in Fig. 6.66 for a given substrate (er ).
306 Microwave Engineering
Fig. 6.66 Z0e e r and Z00 e r with w/b and S/b as parameters f or strip-line coupler
Example 6.28 Design a 10 dB stripline coupler at 5 GHz midband frequency with single feedline
characteristic impedance Z0 = 50 ohms, substrate permittivity er = 9, substrate
thickness b = 1 mm.
1. Length of coupled section
c 30
= = 6 cm
f0 5
lg0 l0 l0 Fig. 6.67 Single feedline for
\ l= = = = 0.5 cm Example 6.28
4 4 er 4 9
Microwave Network Theory and Passive Devices 307
(1 + C0 )
Z0e = Z0 = 69.5 ohms
(1 - C0 )
(1 - C0 )
Z00 = Z0 = 36 ohms
(1 + C0 )
w
which satisfy Z 0 ª Z 0e Z 00 = 50.02 ohms; Z 0 e r = 150 ohms Æ = 0.25 for S/b = 0.85
b
3. Line spacing S and width w
S and w are obtained from impedance vs line dimensions graphs from Fig. 6.66. Hence,
S = 0.85 mm, w = 0.25 mm.
( Z 0ek )2 - 1
C0k = (6.179)
( Z 00k )2 + 1
The coupling coefficients can be determined from the binomial or Chebyshev distributions in a similar
manner as done in multisection quarterwave transformer or multihole directional couplers for a given
coupling.
Fig. 6.69 Parallel coupled Microstrip lines: (a) coupler, (b) line dimensions (c) even and odd mode fields
� Coupler Line Width and Spacing We see that in the design of coupled lines, the following parameters
are considered: w/h, S/h, er, eeff, Z0e, Z0e, Z0 and coupling. By using even- and odd-mode excitation as in the
figure, we find even and odd-mode impedances Z0e and Z00. Analysis yielding Z0e and Z00 as functions of the
shape ratio of w/h, S/h, and er has been carried out by several workers.
In design of coupled transmission lines we use the following approximate synthesis technique while
finding Z0e and Z00:
∑ Firstly, with given single line characteristic impedance Z0 ª Z 0e Z 00 relative dielectric constant er
of the substrate and coupling coefficient C of the coupled lines, the shape ratios w/h are determined
for equivalent single microstrip lines.
∑ Secondly, the shape ratio w/h and the spacing ratio S/h for the desired coupled microstrip are
determined by using the single line shape ratios found in first step.
From C(in dB) and Z0 , Z0e and Z00 are determined as follows:
-C ( dB ) 20
C = 10 (6.180a)
Microwave Network Theory and Passive Devices 309
Therefore, Z0e and Z00 for the coupled lines can be calculated from the above equations when absolute
value of coupling C is known.
If we consider Z0e and Z00 for single line as Z0es and Z00s;
Z 0e Z0
Z0es = and Z00S = 0 (6.180c)
2 2
es os
Ê wˆ Ê wˆ
To find Á from Z0es and Z00s, let us use single line equations: given in Chapter 3 for Z0
h ˜¯
and Á
Ë Ë h ˜¯
and eeff or the following:
Ï 8e A
Ô 2A for w / h < 2
w Ôe - 2
=Ì (6.180d)
h Ô2 È er - 1 Ï 0.61 ¸ ˘ for w / h > 2
Ô p Í B - 1 - ln (2 B - 1) + 2e Ìln ( B - 1) + 0.39 - e ˝ ˙
Ó ÍÎ r Ó r ˛˙ ˚
Z0 er + 1 er - 1 Ê 0.11ˆ
where A= + Á 0.23 + (6.180e)
60 2 er + 1 Ë e r ˜¯
377p
B= (6.180f)
2Z0 er
Here, Z0 = Z0es and w/h = (w/h)es for even mode and Z0 = Z00s, w/h = w/h)0s for odd mode in equation
(6.180d) – (6.180f)
The w/h and S/h for the desired coupled microstrip line are now determined using a family of approximate
equations given Edward Terry as follows:
2 Ê 2d - g + 1ˆ
(w/h)es = cos h -1 Á (6.180g)
p Ë g + 1 ˜¯
2 Ê 2d - g - 1ˆ 4 Ê w/h ˆ
(w/h)0s = cos h -1 Á + cos h -1 Á 1 + 2 ; er £ 6 (6.180h)
p Ë g - 1 ˜¯ p (1 + e r /2) Ë S / h ˜¯
2 Ê 2d - g - 1ˆ 1 Ê w/h ˆ
= cos h -1 Á ˜ + cos h -1 Á 1 + 2 ; er ≥ 6 (6.180i)
p Ë g -1 ¯ p Ë S / h ˜¯
Ê p Sˆ
where g = cos h Á (6.180j)
Ë 2 h ˜¯
Ê p Sˆ
and d = cos h Á p w / h + (6.180k)
Ë 2 h ˜¯
310 Microwave Engineering
� Coupled Region Length Up to this stage the cross-sectional parameters of the coupler are determined.
The effective permittivity needs to be determined to find the length of the coupling region.
In order to find the coupling region lengths, consideration is made that neither phase velocities nor the
effective permittivities are equal for even and odd modes due to unequal distributions of field and capacitances
for each mode for microstrip quasi-TEM configuration as shown in Fig. 6.70 specially for light coupling.
Fig. 6.70 Capacitances for microstrip-coupled lines: (a) Even mode (b) Odd mode
L
Z0air = ;
Cair
L remains unaltered for non-magnetic substrate m = m0 (6.180o)
1
or, Z0air = cL = ; c = velocity of light in air
cCair
Microwave Network Theory and Passive Devices 311
1
2
L c Cair
Combining Z0 = =
C C
1
or, Z0 = (6.180p)
c C Cair
1 1 c
v= ;C= ;v= (6.180q)
LC LCair e eff
c
or, e eff =
v
2
2 Ê ˆ
= ÊÁ c ˆ˜ = eeff = Á l0 ˜ ;
C
\ (6.180r)
Cair Ë v¯ Ë lg ¯
Z 0air
\ Z0 = ; (6.180s)
e eff
c
and v= (6.180t)
e eff
300
lg = l0 / e eff = mm ; (6.180u)
f (GHz) e eff
For wide line eeff ª er, all E-field almost confined with in dielectric substrate.
1 + er
For very thin line eeff ª , field is almost equally shared by air and dielectric substrate.
2
e +1
For medium width, eeff = 1 + q (er – 1); r £ eeff £ er
2
where q = filling factor ; 0.5 £ q £ 1.
For coupled lines
1
Z0 Z 0e Z 00 per strip = (6.180v)
c Cair e eff
and for even and odd mode coupled lines
1 1 Ce
Z0e = e e
= e e ; eeff = (6.180w)
cC air e eff c C C air e
Cair
1 1 C0
Z00 = = ; eeff = 0 (6.180x)
0
cCair 0
e eff c C 0 Cair
0 Cair
0
Ceair and Cair can be determined from the above equations for Z0e and Z00 and determining Ce and C0 from
Ce
e
eeff = (cCeZ0e)2 = e
Cair
C0
eeff
0
= (cC0 Z00)2 = 0
Cair
312 Microwave Engineering
e 0
e eff + e eff cC e Z 0e + cC 0 Z 00
\ e eff = = (6.180y)
2 2
It can be shown also that,
300 Ze 300 e
Cair
lge = ◊ e0 (mm) = (6.181a)
f (GHz) Z 0air f (GH z ) Ce
Cf er
C¢f = (6.181h)
1 + A (h /S ) tan h (8S / h ) e eff
where A = exp {–0.1 exp (2.333 – 2.53 w/h)} (6.181i)
Here, c is the free space velocity. Cga, is obtained by using an equivalent coplanar strip geometry,
K (k ¢) 1 È 1 + k¢ ˘
˙ ; 0 £ k £ 0.5;
2
Cga = e 0 = e 0 ln Í2 (6.181j)
K (k ) p ÎÍ 1 - k ¢ ˙˚
p
= e0 ; 0.5 £ k2 £ 1;
ln {2 (1 + k ) / (1 - k )}
where
S /h
k= (6.181k)
S / h + 2w / h
k¢ = 1 - k2 (6.181l)
Microwave Network Theory and Passive Devices 313
Fig. 6.71 Branch line coupler: (a) Coupler (b) Even symmetry (c) Odd symmetry
For matched terminated symmetrical network, the S-parameters are expressed in terms of ABCD
parameters in similar manner as described in 6.4.18 of coupled parallel coupler. The even and odd mode
[ABCD] for cascaded shunt, series and shunt sections can be expressed as
È 1 0˘ È 0 j YA ˘ È 1 0 ˘
[ABCD]e = Í ˙ Í ˙ Í ˙
Î jYB 1 ˚ Î jYA 0 ˚ Î jYB 1 ˚
(shunt l /8 oc) (series l /4) (shunt l /8 oc)
line line line
314 Microwave Engineering
È 1 0˘ È 0 j YA ˘ È 1 0˘
Similarly,[ABCD]0 = Í
Î – jYB 1 ˙˚ ÍÎ jYA 0 ˙˚ Í – jY
Î B 1 ˙˚
È YB /YA j/YA ˘
= Í ˙
( 2
Í j YA – YB /YA
Î ) YB /YA ˙
˚
(6.182b)
0
Ê 1 ˆ 1 YA
Á A + BY ˜ = = ; (6.182g)
Ë 0¯
YB YA + jY0 YA YB + jY0
1È YA YA ˘
\ S14 = Í – ˙
2 Î –YB + jY0 YB + jY0 ˚
Therefore, the coupling can be varied by varying characteristic admittances of series and shunt arms YA
and YB, respectively. If YA = 2 Y0, YB = Y0, then S14 = –1/ 2 . Therefore, under this condition the branch
line is a 3 dB forward coupler with –180° phase between ports 4 and 1.
The through coefficient
È e 0˘
e
S12 0
+ S12 1 ÍÊ 1 ˆ Ê 1 ˆ ˙
S12 = +
= 2 ÍÁ A + BY ˜ Á A + BY ˜ ˙
2 Ë 0¯ Ë 0¯
Î ˚
1 È YA YA ˘ 2 jYAY0 Y
= Í + ˙= =j 0 ; (6.183)
2 –
Î B Y + jY0 YB + jY0˚ –2YA
2 – YA
–j
If YA = 2 Y0, S12 = . So that the branch line coupler is a 3 dB forward coupler with quadrature phase
w.r.t. Port 2 and 1. 2
e 0
S11 – S11
S13 = = 0, port 3 is isolated. (6.184a)
2
For matched port and symmetry, the S-matrix of branch line coupler is
È 0 – j/ 2 0 -1/ 2 ˘ È 0 –j 0 –1˘
Í ˙ Í
Í – j/ 2 0 -1/ 2 0 ˙ 1 Í– j 0 –1 0 ˙˙
[S] = Í ˙= (6.184b)
Í 0 -1/ 2 0 – j/ 2 ˙ 2 Í 0 -1 1 - j ˙
Í ˙
Í ˙ Î -1 0 - j 0˚
Î -1/ 2 0 – j/ 2 0 ˚
The coupler is called 3 dB, 90° hybrid coupler in general.
of l/4 or 3l/4 TEM elements of the appropriate characteristic admittance. A few examples for hybrid or
directional coupler are given in the following paragraphs. The basic method of synthesis is to preselect the
admittance matrix of a hybrid or a directional coupler and manufacture such a matrix by branching quarter
wavelength elements to obtain prototypes.
� Synthesis of rat-Race Hybrid Coupler from Magic-tee The S-matrix for a matched magic-T with
collinear Ports 1 and 2, E and H Ports 4 and 3, respectively, is
È0 0 1 1˘
Í ˙
0 0 1 - 1˙
[S] = 1/ 2 Í (6.188)
Í1 1 0 0˙
Í ˙
Î1 -1 0 0˚
When all the reference planes are moved away from the junction by q = p/4 without altering the isolation
property between 1, 2, 3, and 4, now S-parameter is
S¢ = S12e–j(q1 + q2) = S12e–j(p/d + p/4)
= S12e–jp/2 = –j S12, etc. (6.189)
The new S-matrix becomes
È0 0 1 1˘
Í ˙
0 0 1 - 1˙
[S] = - j / 2 Í
Í1 1 0 0˙
Í ˙ (6.190)
Î1 - 1 0 0˚
Here, [S]* = – [S], [S]t = [S] (symmetry property). For a lossless network, the unitary property of [S] gives:
[S]* [S]t = [U] (6.191)
or, [S]2 = – [U] (6.192)
–1
Now, [Y] = Y0 [U – S] [U + S] ; (6.193)
By associative property of matrix operations
[Y] = Y0[U – S] [U + S] [U – S]–1 [U + S]–1
= [U – 2S + S2] [U – S2]–1 (6.194)
2
So that [Y] = –Y0 [S]; since [S] = [U]
È0 0 1 1˘
Í ˙
0 0 1 - 1˙
or, [Y] = + Y0 j / 2 Í (6.195)
Í1 1 0 0˙
Í ˙
Î 1 - 1 0 0˚
Thus, admittance matrix (6.195) may be synthesized by l/4 lines for the positive coefficients and 3l/4
lines for the negative coefficients of admittance:
Y0A/Y0 = 1/ 2 (6.196)
Here, the self-admittances parameters y11, y22, y33 and y44 = 0. Since y12 = y34 = 0, and y21 = y43 = 0, there
is no direct element between Ports 1 and 2 and between Ports 3 and 4. The following elements are required
to be inserted:
(i) Between Ports 1 and 3, a l/4 element of normalized characteristic admittance 1/ 2
(ii) Between 1 and 4, a l/4 element of normalized characteristic admittance 1/ 2
Microwave Network Theory and Passive Devices 317
Example 6.29 For the rat-race hybrid shown in Fig. 6.72, (a) calculate input admittance at Port 1
when all others are matched, (b) when input is fed at Port 1, find signal distributions
in all other ports, (c) if input is fed to Port 3 and all others are matched, calculate power distribution in all
ports.
Solution
(a) When all ports are matched, the admittance matrix equation is
È I1 ˘ È0 0 1 1 ˘ È V1 ˘
Í ˙ Í ˙Í ˙
0 0 1 - 1˙ ÍV2 ˙
Í I2 ˙ = j / 2 Í
Í I3 ˙ Í1 1 0 0 ˙ ÍV3 ˙
Í ˙ Í ˙Í ˙
ÎI4 ˚ Î1 - 1 0 0 ˚ ÎV4 ˚
Adding, I3 + I4 = j/ 2 (2V1) = j 2 V1
Substituting in the first equation above,
j j
I1 = - ◊ ◊ 2V1 = V1
2 2
Therefore, I1/V1 = 1 = normalized input admittance at Port 1. Thus, if all ports are matched, Port 1
also matched.
(b) When input is fed at Port 1, the S-matrix equation is
È b1 ˘ È0 0 1 1 ˘ È a1 ˘
Í ˙ Í ˙Í ˙
0 0 1 - 1˙ Í 0 ˙
Í b2 ˙ = - j / 2 Í
Í b3 ˙ Í 1 1 0 0˙ Í 0 ˙
Í ˙ Í ˙Í ˙
Î 1 - 1 0 0˚ Î 0 ˚
Îb4 ˚
Therefore, b1 = b2 = 0; P1 = P2 = 0
1
b3 = –ja1/ 2 ; P3 = |b3|2 = |a1|2/4
2
1
b4 = –ja1/ 2 ; P4 = |b4|2 = |a1|2/4
2
Thus, the input to Port 1 is split equally and in phase at Ports 3 and 4, with no power reflected back
to Port 1 and no output at Port 2. If there is mismatch at Port 4 it will not affect Port 3 and vice-versa
since S34 = 0.
(c) When input is fed to Port 3 and all others are matched, the S-matrix equation is
È b1 ˘ È0 0 1 1˘ È 0 ˘
Í ˙ Í ˙Í ˙
0 0 1 - 1˙ Í 0 ˙
Í b2 ˙ = - j / 2 Í
Í b3 ˙ Í1 1 0 0 ˙ Í a3 ˙
Í ˙ Í ˙Í ˙
Îb4 ˚ Î 1 - 1 0 0˚ Î 0 ˚
b1 = –ja3/ 2 , P1 = |a3|2/4
b2 = –ja3/ 2 , P2 = |a3|2/4
b3 = b4 = 0, P3 = P4 = 0
Thus, input at Port 3 is equally split with the same phase at Ports 1 and 2, with no power reflected
back to Port 3 and no output at Port 4.
A practical application of rat-race is shown in Fig. 6.73 as a hybrid transformer.
� Synthesis of 3 dB, 90° hybrid from magic-T By choosing the shift of reference planes 1 and 3 as
q1 = q3 = p/2; and those 2 and 4 as q2 = q4 = p, the S-matrix of the magic-T reduces to
È0 0 1 j ˘
Í0 0 j 1 ˙
[S] = - 1/ 2 Í ˙ (6.197)
Í1 j 0 0 ˙
Í ˙
Î j 1 0 0˚
such that isolation between 1 and 2, 3 and 4 are not violated. Here, new parameters are
b1 e - jq 1 b1 - j (q 1 + q 3)
S13 = = e ;
a3 e jq 3 a3
È 0 1 0 2˘
Í ˙
Í 1 0 2 0 ˙
= jÍ ˙ (6.202)
Í 0 2 0 1 ˙
Í ˙
Î 2 0 1 0 ˚
Thus, Y11 = Y13 = Y22 = Y24 = Y31 = Y33 = Y42 = Y44 = 0. The physical line elements to be inserted are as
follows:
1. Between ports (3, 4) and (1, 2): a l/4 element of characteristic admittance = Y0.
2. Between (2, 3) and (1, 4): a l/4 element of characteristic admittance = 2 Y0 .
The implementation of this hybrid is shown in Fig. 6.74. If power is fed at Port 1 with all the other ports
matched, power distribution in all ports can be found as follows:
È b1 ˘ È0 0 1 j ˘ È a1 ˘
Í ˙ Í ˙Í ˙
0 0 j 1˙ Í 0 ˙
Í b2 ˙ = - 1/ 2 Í (6.203)
Í b3 ˙ Í1 j 0 0 ˙ Í 0 ˙
Í ˙ Í ˙Í ˙
Îb4 ˚ Î j 1 0 0˚ Î 0 ˚
Therefore, b3 and b4 are 90∞ out-of-phase but the magnitudes of power are same. This coupler is called 3
dB, 90∞ hybrid coupler.
320 Microwave Engineering
Example 6.30 Find the coupling to Port 2 when signal is fed to Port 1 with 3 and 4 connected to
l/4 short-circuited line sections.
Solution Power distribution can be found from the matrix equation {b} = [S]{a}.
È b1 ˘ È0 0 1 j ˘ È a1 ˘
Í ˙ Í ˙Í ˙
0 0 j 1˙ Í 0 ˙
or, Í b2 ˙ = - 1/ 2 Í
Í b3 ˙ Í1 j 0 0 ˙ Í b3 ˙
Í ˙ Í ˙Í ˙
Îb4 ˚ Î j 1 0 0 ˚ Îb4 ˚
Since impedance of l/4 short circuit line = •, Port 3 & 4 are opened, i.e. G = 1
Therefore,
1 1
b1 = - (b3 + j b4 ) , b2 = - ( jb3 + b4 )
2 2
1 1
b3 = - a1 , b4 = - j a1
2 2
1 È 1 ˘ 1 È 1 a ˘
b1 = - Í- a1 + j ( - j / 2 ) a1 ˙ = - Í- a1 + 1 ˙ = 0
2 Î 2 ˚ 2 Î 2 2˚
1 1 È j j ˘
b2 = - [ jb3 + b4 ] = - Í- a1 - a1 ˙
2 2 Î 2 2 ˚
1 jp/2
= j 2 a1 = ja1 = a1e
2
Now, in terms of current and voltage representation V2 = –I2 for matched termination.
Now {I} = [Y] {V}
È I1 ˘ È 0 1 0 2 ˘ È V1 ˘
ÍI ˙ Í ˙Í ˙
Í 2˙ = j Í 1 0 2 0 ˙ Í - I2 ˙
or, Í ˙
Í0˙
Í 0 2 0 1 ˙ Í V3 ˙
Í ˙ Í ˙
Í ˙
Î ˚
0
Î 2 0 1 0 ˚ Î V4 ˚
Microwave Network Theory and Passive Devices 321
I2 = j [V1 + 2 V3]
0 = –V2 I2 + V4
0=+ 2 V1 + V3
Therefore, V3 = – 2 V1
and I2 = j (V1 – 2 V2V1) = j (V1 – 2V1) = – jV1
V4 = 2 I2 = – j 2 V1
Therefore,
I1 = j[–jV1 – j 2 ( - 2 )V1 + 2 (–j 2 )V1]
Normalized input admittance
Y1 = I1/V1 = j[–j + 2j – 2j] = 1
Therefore, the port must be matched also. Actual strip-line
configuration of the coupler is shown in Fig. 6.75.
If V1 and V2 are the input voltages at Ports 1 and 2, output
voltage from Ports 3 and 4 are –j (V1 + V2)/ 2 and –j (V1 – V2)/
2 2
2 or power (V1 + V2) /2Z0 and (V1 – V2) /Z0. Thus, sum and Fig. 6.75 Strip-branch line coupler
difference arms of magic-T are obtained.
waveguide can deliver power through any one of these waves for which E is parallel to the polarization of
the waves excited in the guide. If the polarization is somewhere in between, then both the ports 5 and 6 draw
power.
There are four planes of symmetry P1, P2, P3 and P4 and one axis of symmetry A65. Coupling between the
circular and rectangular waveguides is achieved through a circular hole at the junction in the plane transverse
to the circular guide. Plane P1 interchanges the fields at Ports 2 and 4, reverses the fields at 6, but leaves fields
Microwave Network Theory and Passive Devices 323
at 1, 3 and 5 unchanged. The reference planes are set to make the scattering coefficients real for a matched
turnstile junction:
Fig. 6.77 Turns le junc on: (a) Isometric view (b) Top view of 4-port hybrid junc on
È 0 1 0 1 2 0 ˘
Í ˙
Í 1 0 1 0 0 2 ˙
Í ˙
Í 0 1 0 1 - 2 0 ˙
[S] = 1/2 Í ˙ (6.205)
Í 1 0 1 0 0 - 2˙
Í 2 0 - 2 0 0 0 ˙
Í ˙
ÍÎ 0 2 0 - 2 0 0 ˙˚
Ports 5 and 6 can be matched by tuning screws in the corresponding arms independently, since S56 = 0.
Ports 1, 2, 3 and 4 are matched by inserting a tuning post along the axis A56 within the junction from the
bottom.
Application of Turnstile Junction
There are many applications of a turnstile junction as described below.
� Four port Hybrid Junction By placing a short circuit at Port 2 at distance q2 from the reference plane
and another at Port 4 at q4 = q2 + p/2, and with all other ports matched, the S-matrix equation for an input
at Port 1 is
È b1 ˘ È a1 ˘
Í ˙ Í j 2q 2 ˙
Í b2 ˙ Í - b2 e ˙
Í b3 ˙ Í 0 ˙
Í ˙ = [S] Í ˙
Íb4 ˙ Í b e j 2q 2 ˙ (6.206)
Íb ˙ Í 4 ˙
Í ˙ 5 Í 0 ˙
ÎÍ b6 ˚˙ Í ˙
Î 0 ˚
324 Microwave Engineering
If q2 = np, n = 0, 1, 2, ...; ejq2 = 1, the [S] becomes that of hybrid junction or magic-T with some phase
change due to negative sign in the elements (ej2np = 1):
È 0 0 1 -1˘
Í ˙
0 0 -1 -1˙
[S] = 1/ 2 Í (6.208)
Í 1 -1 0 0 ˙
Í ˙
Î -1 -1 0 0 ˚
Here, ports 3 and 4 are contained in the same physical circular waveguide through cross polarization
replacing the ports 5 and 6, respectively, of turnstile junction as shown in Fig. 6.62(b). An input signal at Port
1 produces equal and opposite signals at new ports 3 and 4. An input at port 2 produces equal and in-phase
signals at 3 and 4. Ports 1 and 2, 3 and 4 are isolated to each other. This satisfies the properties of a hybrid
four-port coupler.
� Matched Four-way Power Divider When an equal signal is fed into ports 5 and 6 in the turnstile
junction with all other ports match terminated, the output signal can be expressed by
È b1 ˘ È0˘ È a˘
Í ˙ Í ˙ Í ˙
Í b2 ˙ Í0˙ Í a˙
Í b3 ˙ Í0˙ Í- a˙
Í ˙ = [ S ] Í ˙ = 1/ 2 Í ˙ (6.209)
Íb4 ˙ Í0˙ Í- a˙
Íb ˙ Í a ˙ Í 0˙
Í 5˙ Í ˙ Í ˙
ÍÎ b6 ˙˚ ÍÎ a ˙˚ ÍÎ 0 ˙˚
This acts as a four-way (ports 1, 2, 3 and 4) equisignal distribution network. Here, the phase of signals b1
and b2 is opposite to that of b3 and b4.
� Matched Three-way Power Divider When a short circuit is placed in circular waveguide ports at a
distance q from the reference plane and input power is fed to Port 1, the output power distribution can be
expressed by
È a1 ˘ È - 2 b5 e j 2q ˘
È b1 ˘ Í ˙
Í ˙ Í ˙
Í 0 ˙ Í a1 - 2 b6 e j 2q ˙
Í b2 ˙ Í ˙
Í b3 ˙ Í 0 ˙
Í ˙ Í 2 b5 e j 2q ˙
Í ˙ = [ S ] 0 = 1/ 2 Í ˙ (6.210)
b
Í 4˙ Í ˙ 2 b6 e j 2q ˙
Í j 2q ˙
Í a1 +
Íb ˙ - b e Í ˙
Í ˙5 Í 5 ˙ 2 a1 ˙
Í ˙ Í
ÎÍ 6 ˚˙
j 2q
Î - b6 e ˚
b
ÍÎ 0 ˙˚
Substituting the elements of the S-matrix and solving the above equation yields
b1 = – a1ej2q/2, b2 = a1/2, b3 = a1ej2q/2, b4 = a1/2,
Microwave Network Theory and Passive Devices 325
b5 = a1/ 2 and b6 = 0
Since b5 and b6 are not used due to short-circuit terminations, and if q = np, b2, b3 and b4 are equiphase
and equi-amplitudes always. Since b1 is the reflected component at the input port, this configuration acts as
a three-way power divider. Since the reflection coefficient at Port 1 is ej2q/2,
a + b1
Input VSWR at Port 1 = 1 = 3.0
a1 - b1
Because of high VSWR, an auxiliary matching structure is required at the input port. After matching, the
power will be equally divided into the three rectangular waveguide Ports 2, 3 and 4.
review questions
6.1 A shunt susceptance j0.5 mho is connected across a lossless transmission line with characteristic
impedance 50 ohms, terminated by matched impedances. Find the S-matrix of the junction.
6.2 A series reactance j 40 ohm is connected between two lossless transmission lines of characteristic
impedances 50 ohms and 75 ohms. Find the S-matrix of the junction. Assume matched terminated at
both ends of these lines.
6.3 A reciprocal two-port microwave device has a VSWR of 1.5 and an insertion loss of 2 dB. Find the
magnitudes of S-parameters for the device.
6.4 A waveguide load has a VSWR of 1.1 and is used to absorb an average power of 5 W. Find the reflected
power and the return loss.
6.5 Find the scattering matrix for an ideal short-circuited section of waveguide with the reference ports
chosen. Show that the coefficients are real numbers. How does the S-matrix change if the short-circuit
position of Port 2 is variable?
6.6 A 10 dB attenuator having an input VSWR of 1.2 is terminated by matched load. Find the reflected
power, the absorbed power and the transmitted power for 1W input.
6.7 Plot the dB attenuation versus angle for an ideal rotary attenuator.
6.8 An ideal rotary phase shifter is terminated with a matched load. Find the scattering matrix for the device.
Express the transmitted voltage at the output in terms of the input for angles 0, 30, 45 and 90 degrees.
Assume that the input voltage is real.
6.9 In a precision rotary phase shifter, the output quarter-wave plate, the transition and the rectangular
waveguides are rotated by an angle of 30°. Show that phase of the transmitted wave will be changed by 30°.
6.10 The input power to the sum arm of an ideal matched magic-T is 1W. Find the output powers from the
other arms when matched terminated.
6.11 Find the magnitude of scattering coefficient for a directional coupler having a coupling coefficient of 3
dB and directivity of 25 dB and VSWR of 1.2 for the main guide Port 1.
6.12 The input power to a lossless matched directional coupler is 100 mW. If the coupling coefficient is 20 dB
and the directivity is 30 dB, find the output powers at other ports.
6.13 An ideal three-port circulator is fed at Port 1 with average power 100 W. If the power reflected by
the antenna at the next port 2 is 100 mW, find the power outputs at all ports assuming they are match
terminated.
6.14 A three-port circulator has an insertion loss of 1 dB, an isolation of 20 dB, and a VSWR of 1.2. Find the
output power at ports 2 and 3 for an input power of 100 mW at Port 1.
6.15 An E-plane tee is matched terminated at all the ports with an input power of 5 mW fed at Port 2 (E-arm).
Determine the power flow through the junction. What changes in the power distribution will occur if the
power is fed at the collinear arm 1.
6.16 A directional coupler of 10 dB coupling and 40 dB directivity produces a transmission loss of 1 dB. For
an input power of 10 mW at the input port of the main arm, determine the power at the other ports when
matched terminated.
326 Microwave Engineering
6.17 A microstrip edge-coupled directional coupler is designed for a coupling factor of 10 dB and a
characteristic impedance of 50 ohms. Determine the even and odd mode characteristic impedances.
6.18 A Bethe-hole directional coupler with a centre circular aperture using the rectangular waveguide of size
0.9 in ¥ 0.4 in is designed to operate at 10 GHz with 30 dB coupling. Find the aperture radius and the
frequency band over which the directivity remains greater than 20 dB.
6.19 Design a three-hole Chebyshev directional coupler using centre apertures in the common broad wall
between two rectangular waveguides of dimensions 0.9¢¢ ¥ 0.4¢¢ to be operated at 9 GHz. Find the aperture
radii, spacing and the bandwidth of the coupler for a coupling of 30 dB and directivity of 30 dB.
6.20 Design an air dielectric stripline edge-coupled 10 dB directional coupler for ground plane spacing of 1 cm
at 5 GHz. Find the strip width and spacing for obtaining input and output line characteristic impedances
equal to 50 ohms for air dielectric.
6.21 A branch-line 3 dB microstrip directional coupler at 5 GHz is designed with input and output line
impedances of 50 ohms. Find the characteristic impedance of the through lines and the branch lines
and their lengths in terms of the wavelength. The dielectric substrate is 20 mm thick and has a dielectric
constant of 10. Find the width of the lines and the lengths of the through lines and branch lines.
6.22 A microstrip hybrid ring is constructed on 1 mm thick substrate with a dielectric constant of 2.5 at 3
GHz. Determine the widths of the transmission line and the ring as well as the radius of the ring for input
and output line impedances of 50 ohms.
6.23 Show using S-matrix theory that a lossless non-reciprocal two-port microwave device cannot be
constructed.
references
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4. Bhat, Bharathi and Shiban K. Koul, Stripline-like Transmission Lines for Microwave Integrated Circuits,
Wiley Eastern limited, New Delhi, 1989.
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11. Gupta, K. C., et al, Microstrip Lines and Slotlines, 2nd ed., Artech House, 1996.
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14. Lance, A. L., Introduction to Microwave Theory and Measurements, New York, McGraw-Hill, 1964.
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Book Company, New York, 1948.
19. R. J. P. Douville and D. S. James, “Experimental Study of Symmetric Microstrip Bends and Their
Compensation,” IEEE Trans. Microwave Theory Tech., Vol. 26, No. 3, March 1978, pp. 175–181.
Microwave Network Theory and Passive Devices 327