Circulator
In a circulator, we have four ports which are connected only to the next clockwise
terminal. For ex. Port 1 is connected to port 2 only and not to port 3 and port 4.
Similarly port 2 is connected only to the port 3 and so on…..
The S matrix for 4 ports circulator is
0
0
0 S14
S = S12 0
0 0
0 S32 0 0
0
0
S43 0
0
1
S = 1
0
0 0
…………(4)
0
1
0 0
0
0
1 0
If we assume 3 port circulator, then its S-matrix is
0
0 S13
S = S21 0
0
…………(4)
0
S32 0
S
=
0
0
0
0
…………(4)
1
0
0
0
1
1
…………(4)
0
Insertion loss- The important specification of a circulator is its insertion loss,
which is the loss in the signal level as it travel in the direction that is supposed to go
(or Insertion loss arises due to insertion of the component over a line).
For example if allt he ports of a circulator are matched, then a signal applied to one
port, for instance to port 1 will emerge from port 2 with a loss called the Insertion
loss.
Insertion loss = - 20 log IS21I , dB
Isolation- A small part of the input signal also emerges from port 3 and the ratio of
this signal to the input signal is called isolation, provided port 2 is terminated with
matched load.
Isolation = -20 log IS12I , dB
Nu. A three port circulator has an insertion loss of 1 dB, isolation 30 dB and VSWR
1.5. Find the S-matrix.
Sol. Since 3 port S matrix is
S11 S12 S13
S = S21 S22 S23
…………(1)
S31 S32 S33
Insertion loss = 1 dB = -20logIS21I = 20 log(1/I S21I)
S21 = 10-(1/20)
But insertion loss between port 1 & 2, 2 & 3 and 3 & 1
Thus IS21I = IS32I = IS13I = 0.89
Isolation between the ports is 30 dB = -20 log IS12I
IS12I = 10-(30/20) = 10-1.5 = 0.032
But insertion loss between port 2 & 1, 3 & 2 and 1 & 3
Thus IS12I = IS23I = IS31I = 0.032
Since VSWR (S) = 1.5
 Reflection coefficient  = (S-1)/(S+1) = 0.2 = IS11I = IS22I = IS33I
Thus S matrix becomes
0.20 0.032 0.89
S = 0.89 0.20
0.032
…………(2)
0.032 0.89
0.20
Nu. Determine the S-matrix of a 3 port circulator , given insertion loss of 0.5 db,
isolation of 20 dB and VSWR of 2.
Ferrite Circulator
Phase shifter 1 and 2 gives different phase shift for forward and backward motion
of the wave, because these phase shifters are non-reciprocal. Both the 3 dB
couplers are reciprocal, because they have phase shift of 900 in forward and
backward direction.
When wave is entering in port 1, phase shifter 1 gives 1800 phase shift and the
wave reaches to port 2. The couple 1 gives phase shift of 900, but phase shifter 2
will not give any phase shift to this wave. Coupler 2 gives 900 phase shift. Thus wave
passing in this way gets the phase shift of (900 + 900 = 1800). Hence these two waves
passing through different directions and reaches at port 2 will get added because of
same phase shift. So port 1 is coupled to port 2.
Port 1 cannot couple port 4 because
(i) The wave entering in port 1 gets the phase shift of 900, while passing through
coupler 1. The phase shifter 2 gives 00 phase shift to this wave. Thus wave reaching
at port 4 through this path gets the total phase shift of 900 + 00 = 900.
(ii) The phase shifter 1 gives the phase shift of 1800 and after this the coupler 2
gives 900 phase shift to this wave. Thus total phase shift in this way is 900 + 1800 =
2700.
These two waves are out of phase 900 and 2700 so they get cancelled at port 4. So
port 1 does not couple port 4.
Similarly input wave in port 2 will couple port 3 and input wave in port 3 will
couple port 4 only.
Nu. Prove that it is impossible to construct a perfectly matched, lossless, reciprocal
three port junction.
Sol. A perfectly matched three port junction has a scattering matrix
0 S12 S13
S = S12 0
S23
S13 S23 0
For a lossless junction S-matrix is unitary matrix
0 S12 S13
0 S12 S13
S12 0
S23
S12 0
S23 = I
S13 S23
0
S13 S23 0
From R1C1
S12S12 + S13S13 = 1
…………(1)
From R2C2
S12S12 + S23S23 = 1
…………(2)
From R3C3 S13S13 + S23S23 = 1
…………(3)
From R1C2
S13S23 = 0
From R1C3
S12S23 = 0 …………(4)
From R2C3
S12S13 = 0
As we know the reciprocal coefficients are S12, S13, and S23
If reciprocal coefficient S12 is not equal to zero, the equation(4)  S13 = 0
and the equation(4)  S23 = 0
But the reciprocal phenomenon S12 ≠ 0 is not satisfied by the equation (3)
Similarly, if reciprocal coefficient S13 is not equal to zero, the equation(4)  S23 = 0
and the equation(4)  S12 = 0
But the reciprocal phenomenon S13 ≠ 0 is not satisfied by the equation (2)
Thus,
If reciprocal S12 (or S13 or S32) is not equal to zero, the fourth gives S13 = 0 = S23.
But this does not satisfy the third equation. Therefore, a reciprocal lossless three
port junction cannot be perfectly matched.
Isolator
Isolator is a two port, unidirectional and nonreciprocal device that permits
unaffected transmission port 1 to port 2, but provides high attenuation from
transmission from port 2 to port 1. All the signal power from the generator can be
delivered to the load and reflections from the load do not get transmitted back to
generator. Thus generator appears a matched load and hence power output
variations and frequency pulling can be avoided.
Ferrite Isolator:- To enter the signal from port 1 we must need electric field E
perpendicular to resistive card. Otherwise signal is absorbed by resistive card.
A signal propagating from port 1 to port 2 is rotated by twist of 450 and the
dominant mode TE10 does not get attenuated. The signal is rotated 450 at the twist
and enters the circular waveguide transition as the TE11 mode. Now the length of
the ferrite rod is selected so as to obtain Faraday rotation 450 at the output and
regain its original polarization. The signal will come out from the port 2 with proper
polarization.
When signal propagates from port 2 to port 1, it will have plane of polarization
rotated by 900 and will make electric field parallel to resistive card. Hence gets
absorbed. Thus nonreciprocal isolation action takes place.
For a proper operation,
1. Insertion loss in forward direction is less than 1 dB and approximately
equal to 0.5 dB.
Insertion Loss = -20 logIS21I
2. Isolation exists in between 20 to 30 dB and
Isolation = -20 logIS12I
3. For an ideal lossless and matched Isolator
IS21I = 1 and IS11I = IS12I = IS22I = 0
Therefore S-matrix of an Isolator is
Nu. An Isolator has insertion loss of 0.5 dB, an isolation of 25 dB and VSWR of 2,
find the S-matrix.
Sol. Insertion loss = 0.5 dB = -20 logIS21I  IS21I = 10-(0.5/20) = 10- 0.025
Isolation = 25 dB = -20 logIS12I  IS12I = 10-(25/20) = 10- 1.2
Since VSWR = 1.5, so reflection coefficient = (S-1)/(S+1) = 0.2
 S11 = S22 = 0.2
Therefore S-matrix of an Isolator is
Directional Coupler
It is a four port device commonly used for coupling a known fraction of the
microwave power to a port (coupling port) in the auxiliary line while flowing from
the input port to the output port in the main line. The remaining is an isolated port
and matched terminated.
Construction:- directional coupler has four ports, primary waveguide 1-2 and
secondary waveguide 4-3.
Port 1 and port 4 are isolated
Port 2 and port 3 are isolated
If the signal is incident in port 1, it is distributed in between port 2 and port 3, but
not on port 4 due to perfect isolation.
Characteristics of directional Coupler:1. Coupling and Coupling Factor:- C = 10 log(P1/P3) , dB
2. Transmission loss or Insertion loss :- T = 10 log(P1/P3) , dB
3. Directivity:- D = 10 log (P3/P4), dB
4. Return Loss:- R = 10 log (Pi/Pr), dB
5. Isolation:- I = 10 log(P1/P4), dB
6. Directivity D = Isolation - C
Nu. The input power in a two hole directional coupler is 1 mW. The Coupler has a
coupling factor of 15 dB and a directivity of 30 dB. Calculate the power in all
the ports.
Sol. C = 10 log(P1/P3) = 15  P1/P3 = antilog(3/2) = 31.62
 P3 = P1/31.62 = 1 mW/31.62 = 32 W = 0.032 mW
D = 10 log(P3/P4) = 30  P3/P4 = antilog (3) = 103
 P4 = P3/103 = 32 W/103 = 0.000032 mW
Thus power in port 2
P2 = Input power – (power in coupled port + power in isolated port)
= P1 – (P3 + P4) = 1 – (0.032 + 0.000032) = 0.968 mW
Nu. The input power in a Two Hole Directional Coupler is 10 mW The Coupler has a
coupling factor of 10 dB, a directivity of 40 dB and transmission loss of 1 dB.
Calculate the power in all the ports.
Sol. C = 10 log(P1/P3) = 10  P1/P3 = antilog(1) =
 P3 = P1/ antilog(1) =
D = 10 log(P3/P4) = 40  P3/P4 = antilog (4) =
 P4 = P3/ antilog (4) =
T = 10 log(P1/P2) = 1  P1/P2 = antilog (0.1) =
 P2 = P1/ antilog (0.1) =
Nu. The incident power for 30 dB coupler is 560 mW. Calculate the power in the
main arm and auxiliary arm.
Sol.
Nu. In a directional coupler the output terminal of main arm is terminated with
matched load. The coupling factor is 18 dB and ouput power from coupler is
13.5 dBm. If 2 dB attenuator is placed in an auxiliary arm then calculate power
(Watt) in coupler.
Sol.
Nu. Input end of 23 dB coupler is shorted. When 45 mW power is applied to output
end of coupler then calculate the power in dBm in auxiliary arm.
Sol.
Nu.