Abstract
Component used in any circuit, as per functionality, can be classified into two basic categories, passive and active components. Passive component can’t provide energy to the circuit in the form of voltage or current i.e. provides no power gain to the circuit. As well as it can’t amplify or process any signal. In low frequency (considerable up to 3 GHz, parasitic effect and radiation loss increases fatally beyond that) circuit where the component dimension is very less than operating wavelength (λ), i.e. Lumped parameter circuit, Resistor, Capacitor, Inductor or Conventional PN Junction diode can be the typical example of passive circuit components. Passive device or component characterized with monotonic I-V curve which must be within 1st or 3rd (or both) quadrant with always positive differential resistance. In high frequency where the dimension of λ is in order of the component dimension, Distributed parameter concept is applicable. Microwave device is the most suitable device to operate in this range. The frequency range can be from 3 to 100 GHz or even more. But at very high frequency dielectric and Ohmic losses became fatal and manufacturing devices for higher frequency became very demanding. Different waveguide sections like Attenuator, Terminator, Filter, Coupler and Ferrite devices are passive microwave components. Whereas, different types of Oscillators, Amplifiers, Mixers, Multipliers and detectors used in microwave frequency are the typical examples of active devices. Passive device in Microwave frequency range can be characterized based on several aspects. Transfer characteristics of a passive device must be linear, continuous wave signal must not get distorted and the S- parameter of the device have to be independent of power. For a two-port passive device or component, reflection at both port are identical, i.e. |S11|2 = |S22|2. And most of the passive components used in microwave circuit are reciprocal except Ferrite isolators and circulators [1]. In this chapter, we will discuss about different types of passive circuit components used in microwave frequency range.
Access provided by Autonomous University of Puebla. Download chapter PDF
Similar content being viewed by others
Keywords
1 Waveguide TEE Junctions
Waveguide junction consists of three or more individual ports and is called a Tee junction. By connecting a section of waveguide with the main waveguide in series or parallel, a Tee junction can be formed. Based on structure, Tee junction can be of three types, such as.
1.1 H-Plane TEE
By connecting another rectangular waveguide section along the thinner side of the main waveguide, H-plane Tee is formed. The section which is connected to the main waveguide is called side arm and collinear arm is formed by both ends of the main waveguide. Here the plane containing the magnetic field of the main waveguide is parallel to the axis of the side arm; hence it is called H-plane Tee. Collinear arm contains port1 and port2 whereas side arm contains port3 (Figs. 1 and 2).
H-plane TEE
Ports of H-plane TEE
H-plane Tee is completely symmetric and port3 is perfectly matched. If any input signal is given at port3, get divided between port1 and port2 with equal magnitude and in phase components i.e. S13 = S23, provides both collinear arms having an absolutely equal length.
1.1.1 S-Matrix of H-Plane TEE
Scattering matrix of a three port device must be a 3 × 3 matrix.
So, scattering matrix can be the form of,
It is a symmetrical device i.e.
Input power from port3 gets divided into two equal, in-phase parts at port1 and port2. So,
Port3 is perfectly matched \(S_{33} = 0\),
From Eq. (1),
As per unitary property, [S][S*] = [I]
From Eq. (5),
But from Eq. (3) \(S_{13} = S_{23}\)
From Eqs. (6), (7) and (9) we have,
Again, from Eq. (5), R1*C3 => \(S_{11} S_{13}^*\) + \(S_{12} S_{23}^*\) = 0
Or, \(S_{13}^* \left( {S_{11} + S_{12} } \right) = 0\quad\) but \(\quad S_{13}^*\) ≠ 0;
Hence, \(S_{11}\) = \({-}S_{12}\).
And from Eq. (10) we have, \(S_{11}\) = \(\frac{1}{2}\) and \(S_{12}\) = \({-}\frac{1}{2}\).
Similarly, from Eq. (11) \(S_{22} = \frac{1}{2}\).
Finally, the scattering matrix for H-plane Tee can be written as,
1.1.2 Applications
-
(a)
As Power Divider: Incident signal at port3 get divided into two equal, in-phase part at port1 and port2. Let's consider the incident signal at port3 having amplitude A and power P, then as, \(S_{13}\) = \(S_{23}\) = \(\frac{1}{\sqrt 2 }\), signal appears at port1 and port2 is \(\frac{A}{\sqrt 2 }\) and power is \(\frac{P}{2}\). So it can act as a power divider. It is to be noted here that the input signal level at port3 reduced to port1 and port2 by a factor of \(\frac{1}{\sqrt 2 }\) and in dB scale, it is equivalent to 20log \(\left( {\frac{1}{\sqrt 2 }} \right)\) = −3 dB. For that reason, H-plane TEE is called a 3 dB signal splitter.
-
(b)
As Power Combiner: If the lengths of collinear arms are same, two in-phase input signals given through port1 and port2 get added and appear at port3. So, H-plane Tee can be used as a signal combiner.
Except that H-plane Tee can be used as a tuner or in duplexer assemblies of radar installation operating.
Illustrative Example 1:
A 48 mW signal is fed into one of the collinear arm of a H-plane Tee. Determine the power that appears at all the ports when ports are terminated by a perfectly matched load.
Solution: We know that scattering element, \(S_{ij}\) = \(\frac{v_i }{{v_{j } }}\) analogous to voltage ratio and power, P ∝ \(v^2\) so P ∝ \(S^2\) so \(P_o = [{\text{S}}][S^* ][P_i ]\).
Let's consider input and output power of port1, port2 and port3 are \(P_{i1} ,P_{i2} ,P_{i3} \;and\;P_{o1} ,P_{o2} ,P_{03}\) respectively.
We have, \(\left[ {\begin{array}{*{20}c} {P_{o1} } \\ {P_{o2} } \\ {P_{o3} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {S_{11}^2 } & {S_{12}^2 } & {S_{13}^2 } \\ {S_{21}^2 } & {S_{22}^2 } & {S_{23}^2 } \\ {S_{31}^2 } & {S_{32}^2 } & {S_{33}^2 } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {P_{i1} } \\ {P_{i2} } \\ {P_{i3} } \\ \end{array} } \right]\).
Given that \(P_{i1} = 48\,{\text{mW}}\), \(P_{i2} =\) 0, \(P_{i3} =\) 0; and we know that s-matrix of a H-plane Tee is given as,
So, the power output at port1, port2 and port3 are 12 mW, 12 mW and 24 mW respectively.
1.2 E-Plane TEE
By connecting another rectangular waveguide along the width of the main waveguide E-plane TEE can be formed. As the electric field of the main waveguide is parallel to the axis of the side arm, it is called E-plane TEE (Fig. 3).
E-plane TEE a schematic, b ports
Input power at port3 gets divided into two parts and comes out from port1 and port2 with an equal magnitude but one signal is 180° out of phase with others i.e. \(S_{23}\) = \({-}S_{13}\).
1.2.1 S-Matrix for E-Plane TEE
Scattering matrix for E-plane Tee can be a 3 × 3 matrix.
As side arm i.e. port3 is perfectly matched \(S_{33}\) = 0.
All the three ports are perfectly symmetric, hence \(S_{ij}\) = \(S_{ji}\).
Using the values from Eqs. (14) and (15) the scattering matrix can be modified to
As per unitary property, [S][S*] = [I]
From matrix multiplication,
From Eqs. (18), (19) and (21) we have,
Again, from Eq. (17), R1 × C3 => \(S_{11} S_{13}^*\) \({-}S_{12} S_{13}^*\) = 0
Hence, \(S_{11}\) = \(S_{12}\), putting this value in Eqs. (22) and (23) we have,
Substituting all these values in Eq. (16) we have
1.2.2 Applications
-
(a)
As Power Divider: If input signal of amplitude A and power P is given through port3 then output at ports1 and 2 having equal amplitude \(\frac{A}{\sqrt 2 }\) and power \(\frac{P}{2}\) but both outputs are 180° out of phase from one another i.e. \(S_{23} = - S_{23}\).
-
(b)
As Power Combiner: Conversely, when two input signals of opposite phase are given through ports1 and 2 it gets added and appears at port3. So, E-plane Tee can be used as a signal combiner.
Illustrative Example 2:
An E-plane Tee made of waveguide section of 50 Ω characteristics impedance. A signal power of 40 mW is applied to E-arm which is perfectly matched. Determine the amount of power that is delivered to the load of 75 and 100 Ω connected to port1 and port2 respectively.
Solution: Clearly both of the collinear arms are not matched properly, so having a reflection from there. Ideally the 40 mW should be equally distributed to both of the arms having 20 mW each with opposite phase. But reflection due to load mismatch some of the power gets reflected.
Here, ideally received power at port1 and port2 should be 20 mW each, but it should be less practical due to the reflection present there. Actual received power \(P_1\) and \(P_2\) can be calculated as:
Hence power received at port1 and port2 is 19.2 mW and 17.82 mW respectively.
1.3 Magic TEE or Hybrid TEE
The structural combination of E-plane TEE and H-plane TEE is called Hybrid TEE or Magic TEE. It is formed by connecting two side arms along the broad and narrow side wall of the main waveguide. Two arms of the main waveguide i.e. port1 and port2 called collinear arms. Other two arms are E-arm and H-arm can be considered as port3 and port4 respectively or vice-versa (Figs. 4 and 5).
Magic TEE
Illustration of ports of magic TEE
1.3.1 Characteristics of Magic TEE
-
(i)
If two waves with equal amplitude and phase incident on port1 and 2 respectively then there will be zero output at port3 and the sum of the two waves will appear at port4.
-
(ii)
If wave incident at port3, it gets equally divided between port1 and port2 but in opposite phase and no power appears at port4 i.e. \(S_{43}\) = 0.
-
(iii)
If wave incident at port4 then it gets equally distributed between port1 and 2 and both of the parts are in-phase and no power appears at port3 i.e. \(S_{34}\) = 0.
-
(iv)
If signal incident into one of the collinear arms then no output appears at other collinear arm as E-arm produces a phase delay and H-arm produce a phase advance to the wave. Hence \(S_{21}\) = \(S_{12}\) = 0.
1.3.2 Scattering Matrix of Magic TEE
Scattering matrix for any four port device can be given as
Magic Tee is a symmetric device, hence \(S_{ij}\) = \(S_{ji}\),
Putting the above values in scattering matrix,
As per unitary property, [S][S*] = [I]
From matrix multiplication,
Putting the values of \(S_{13}\) and \(S_{14}\) in Eqs. (32) and (33) we have \(S_{11} = 0\;and\;S_{22}\) = 0.
Replacing these values into Eq. (30) we have
It is to be noted here that port3 can also be considered at H-arm and port4 as E-arm. In that case, scattering matrix will be different as, \(S_{24}\) = \({-}S_{14}\) and \(S_{23}\) = \(S_{13}\) under that consideration.
1.3.3 Applications
-
(a)
Impedance Measurement: Microwave source can be connected at port4, null detector at port3 and two collinear arms can form a bridge which can be used to measure the impedance.
-
(b)
As a Duplexer: Duplexer is a circuit where a single antenna is used as transmitter and receiver. As both of the collinear arms are mutually isolated, it can be used as transmitter and receiver. Antenna is connected to E-arm and H-arm is perfectly matched.
-
(c)
As a Signal Mixer: E-arm is to be connected to an antenna and H-arm is to be connected to a local oscillator. One of the collinear ports is perfectly matched and the other collinear port consisting the mixer circuit which gets half of the total signal and local oscillator power to produce the intermediate frequency (IF).
Illustrative Example3:
If 1 W power is applied to the perfectly matched port3 of a magic Tee, what will be the power delivered to port1, port2 and port4 if it is terminated by a reflection of 0.5, 0.6 and 0.8 respectively?
Solution: Here port3 is perfectly matched but all other ports are having reflections, \(\rho_1\) = 0.5, \(\rho_2\) = 0.6, and \(\rho_4\) = 0.8
We know that if [a] and [b] are the normalized parameter for input and output voltage then, [b] = [S] [a]
Given that, \(P_3\) = 1 W, power delivered at port3, \(P_3\) = \(\frac{1}{2}\left| {a_3 } \right|^2\) \({-}\frac{1}{2}\left| {\rho_{3} a_{3} } \right|^{2}\)
where, \(\rho_3\) = 0, so \(P_3\) = \(\frac{1}{2}\left| {a_3 } \right|^2\) = 1 W or, \(a_3 = \sqrt 2\) V
Hence, \(a_3 = \sqrt 2\); \(b_1\) = \(\rho_1\) \(a_1\) = 0.5 \(a_1\); \(b_2\) = \(\rho_2\) \(a_2\) = 0.6 \(a_2 ; b_4\) = \(\rho_4\) \(a_4\) = 0.8 \(a_4\)
Using these values, we have,
Or, \(0.5a_1\) = \(\frac{1}{\sqrt 2 }\) \(\left( {a_3 + a_4 } \right)\); \(0.6a_2\) = \(\frac{1}{\sqrt 2 }\) \(\left( { - a_3 + a_4 } \right)\); \(\sqrt 2\) = \(\frac{1}{\sqrt 2 }\) \(\left( {a_{1} - a_{2} } \right)\)
And \(0.8a_4\) = \(\frac{1}{\sqrt 2 }\)\(\left( {a_1 + a_2 } \right)\).
Solving the above four equations we have \(a_1 = 0.928\); \(a_2 = - 1.07\); \(a_3 = 0.781;\)
\(a_{4 } = - 0.125\)
2 Directional Coupler
It is a 4-port passive waveguide device that can couple a small fraction of total microwave power for measurement. It can measure incident and reflected power and VSWR values etc. It is a 4-port device where the main waveguide contains port1 (i.e. input port) and port2 (i.e. output port). And secondary auxiliary waveguide contains port3 (i.e. isolated port) and port4 (i.e. coupled port). Power at the coupled port is called forward power and the power at the isolated port is called back power. For bi-directional device, input can be given from port2 and for that case, port3 is coupled port and port4 is isolated port as shown in Fig. 6.
Directional coupler
2.1 Properties of Directional Coupler
-
(i)
All ports are perfectly matched i.e. \(S_{11} = S_{22} = S_{33} = S_{44} = 0\).
-
(ii)
When power travels from port1 to port2, some portion of it gets coupled to port4 but no power appears at port3 i.e. \(S_{31} = 0\).
-
(iii)
For bi-directional coupler, when power travels from port2 to port1, some portion of it gets coupled at port3 and no power appears at port4 \(S_{42} = 0.\)
-
(iv)
Generally, same degree of coupling is used between port1 to port4 and port2 to port3.
-
(v)
Output from directional couplers are always in phase quadrature.
A directional coupler can be characterized by four parameters,
-
i.
Coupling Factor (C)
-
ii.
Directivity (D)
-
iii.
Isolation (I)
-
iv.
Insertion Loss (IL)
Coupling Factor (C): It is the fraction of input power that is coupled at the coupling port. If the input power is \(P_1\) and coupled power is \(P_4\) then coupling factor is given by,
Directivity (D): It is the ratio of forward power to the back power of a directional coupler.
Isolation (I): The ratio of incident power to the back power is called isolation.
Insertion Loss (IL): It is the ratio of input power to output power.
2.2 S-Matrix of Directional Coupler
Scattering matrix of any 4-port device can be written as,
As it is a symmetrical device, \(S_{ij} = S_{ji}\),
All ports are perfectly matched \(S_{ii}\) = 0.
Using Eqs. (42), (43) and (44), the above s-matrix reduced to
As per unitary property, [S][S*] = [I]
From matrix multiplication,
Multiplied both sides by \(S_{12}\), \(\left| {S_{12} } \right|^{2} \left[ {S_{23} + S_{23}^{*} } \right] = 0,\)
Or, \(S_{23}\) = \({-}S_{23}^*\) only possible when \(S_{23}\) is purely imaginary. Whereas \(S_{12} = S_{34}\) used to be real.
Let consider \(S_{14}\) = \(S_{23}\) = jq and \(S_{12}\) = \(S_{34}\) = p where p is purely real.
The scattering matrix of directional coupler used to be, using these values,
2.3 Applications
-
(a)
As Reflectometer: By connecting the power sensor at the coupled port, VSWR of the antenna system can be measured. Many RF devices need to have minimum VSWR or protection from excessive VSWR from the circuit.
-
(b)
As Signal Sampler: Coupled port provides a fraction of the power of the main waveguide. This fraction is called coupling factor. By using this spectrum analysis, waveform monitoring etc. of any RF system can be done.
-
(c)
As Reference Signal Generator: Signal of the coupled port can be used as a reference signal to control feedback circuitry.
Illustrative Example 4:
Two identical couplers are used in a waveguide to sample the incident power of 5 mW and reflected power as 0.16 mW. What will be the value of VSWR?
Solution: Here reflection coefficient, ρ = \(\sqrt {\frac{P_r }{{P_i }}}\) = \(\sqrt {\frac{0.16}{5}}\) = 0.179.
Now, Voltage Standing Wave Ratio (VSWR) = \(\frac{{1 + {\uprho }}}{{1 - {\uprho }}}\) = \(\frac{1 + 0.179}{{1 - 0.179}}\) ≈ 1.44.
So, VSWR = 1.44.
Illustrative Example 5:
Two identical 40 dB directional couplers are used to sample incident and reflected power in a waveguide. Voltage standing wave ratio is 1.5 and the output of the coupler sampling incident power is 6 mW. Calculate the value of the reflected power.
Solution: Here VSWR = \(\frac{{1 + {\uprho }}}{{1 - {\uprho }}}\) = 1.5
Or, 1 + ρ = 1.5– 1.5ρ.
Or, 2.5ρ = 0.5
Or, ρ = \(\frac{0.5}{{2.5}}\) = 0.2
Now, we know that ρ = \(\sqrt {\frac{P_r }{{P_i }}}\) = ρ = \(\sqrt {\frac{P_r }{6}}\).
Or, \({\uprho }^2\) = \(\frac{P_r }{6}\quad\) or, \(\frac{P_r }{6} = 0.04\quad\) or, \(P_r\) = 0.24 mW.
So, the reflected power is 0.24 mW.
Illustrative Example 6:
Scattering matrix of a directional is given by
Calculate Directivity, Coupling factor, Isolation and Insertion loss.
Solution: The port orientation of directional coupler is given by,
Directivity is given by, D = 10 log \(\frac{P_4 }{{P_3 }}\) dB = 10 log \(\left( {\frac{P_4 }{{P_1 }}.\frac{P_1 }{{P_3 }}} \right)\) dB = 20 log \(\frac{{\left| {S_{41} } \right|}}{{\left| {S_{31} } \right|}}\) dB.
D = 20 log \(\left( {\frac{0.03}{{0.10}}} \right)\) = –10.46 dB.
Coupling, C = 10 log \(\frac{P_1 }{{P_4 }}\) dB =–20 log \(\left( {\left| {S_{41} } \right|} \right)\) dB. = –20 log(0.03) = 30.46 dB.
Isolation, I = 10 log \(\frac{P_1 }{{P_3 }}\) dB = –20 log \(\left( {\left| {S_{31} } \right|} \right)\) = –20 log(0.10) = 20 dB.
Here Isolation, I = 30 dB = 30.46 + (–10.46) = C + D = Coupling + Directivity.
Insertion Loss, IL = 10 log \(\frac{P_1 }{{P_2 }}\) = –20 log \(\left( {\left| {S_{21} } \right|} \right)\) dB. = –20 log(0.85) = 1.41 dB.
3 Isolator
It is a passive, non-reciprocal two-port waveguide section that passes signals coming from one direction but blocks signals coming from other direction or any reflected signal (Fig. 7). Usually, an isolator is placed after the microwave source to pass the signal transmitted from the source but blocks any reflected wave to go to the source. That’s how it isolates the source from other parts of the circuit.
Isolator
Usually, an isolator consists of a waveguide section containing two slots of resistive card at both ends of it. A 45o anti-clockwise twist is used in between and after the twist a slot of ferrite rod so chosen that it can provide a 45o clockwise turn to the wave passing through it as illustrated in Fig. 8.
Illustration of the working of an Isolator
When wave travels from input to output port, resistive card at the input end blocks horizontally polarized wave and passes vertically polarized wave through it. Then this vertically polarized wave gets rotated by 45o anti-clockwise by the twist in the waveguide. Ferrite rod provides 45° further rotation to the wave so that it became vertically polarized again and gets easily passed through the resistive card at the output end.
But when wave travels from output side, resistive card at the output end passes vertically polarized wave but blocks horizontally polarized wave. Ferrite rod which provides rotation in the same direction to the wave coming from both directions provides 45° clockwise rotation to it. The twist in the waveguide provides further 45° clockwise (opposite due to wave coming from reverse side) rotation and ultimately it became horizontally polarized due to two consecutive turns of 45° each. This horizontally polarized wave gets blocked by a resistive card at the input side. Hence no power from output can appear at input side [2].
3.1 S-Matrix of Isolator
It is a two-port device. S-matrix of a two-port device can be written as,
As both of the ports are perfectly matched, \(S_{11} = 0\) and \(S_{22}\) = 0.
If signal is applied from port2 no power appears at port1, hence \(S_{12} = 0.\)
When signal is applied from port1, total output appears at port2 i.e. \(S_{21} = 1.\)
3.2 Applications
-
(a)
As Protecting Device: In the process of testing and measurement (T&M) it is very possible that any undesired reflection from device under test (DUT) can damage the original circuit. So, the Isolator is employed in between to suppress that undesired reflection.
Illustrative Example 7:
Determine the scattering matrix of an isolator having insertion loss of 0.5 dB and that provides an isolation of 40 dB. Consider all the ports are perfectly matched.
Solution: As we know that isolator is a two-port device, the generic form of scattering matrix will be, \([S] = \left[ {\begin{array}{*{20}c} {S_{11} } & {S_{12} } \\ {S_{21} } & {S_{22} } \\ \end{array} } \right]\).
Here, it is given that all the ports are perfectly matched, hence \(S_{11} = S_{22} = 0;\)
Insertion loss is given as 0.5 dB, hence \({-}20{\text{ log}}\left| {S_{21} } \right| = 0.5\)
Or, \(S_{21} = 0.994\).
Again isolation is 40 dB so, \(- 20{\text{log}}\left| {S_{12} } \right| = 40\).
Or, \(S_{12} = 0.01\).
So, the s-matrix will be, \([S] = \left[ {\begin{array}{*{20}c} 0 & {0.01} \\ {0.994} & 0 \\ \end{array} } \right]\).
4 Circulator
Circulator is a 3-port or 4-port device which can circulate RF signal in a particular direction (mainly clockwise but can be anti-clockwise as well). The structure is a combination of rectangular and circular waveguide and ferrite rod inserted in between, so that if input is given from port1 output can be obtained from port2 only, similarly if input is given from port2, output is available only at port3 and so on (Figs. 9 and 10).
Circulator
Four-port circulator
Let's Consider RF signal of the dominant TE10 mode inserted from port1 of the rectangular waveguide section and get converted to TM11 dominant mode when it approaches circular waveguide section. But the orientation of port3 is completely out of phase with port1 so no output appears at port3. The ferrite rod between port3 and port4 provides 45° clockwise rotations to the signal and so that port4 became completely out of phase hence no output appears at port4. The RF signal then enters to rectangular waveguide section at port2 as the dominant TE10 mode. So the complete output appears at port2 [3].
Similarly, any input applied through port3 does appear at port1 as it is completely out of phase. But after 45° clockwise rotations through the ferrite rod, it appears at port4 and as port2 is again completely out of phase, no output appears at port2. Likewise, any input applied through port2 can appear at port3 only (Fig. 11).
Internal structure of a circulator
4.1 S-Matrix
For any 4-port device S-matrix can be the form of,
As all the four ports are perfectly matched, \(S_{ii} = 0\).
Again from the property of isolation, \(S_{14}\) = \(S_{12}\) = \(S_{32}\) = \(S_{43}\) = 1 and all other elements are zero. So S-matrix reduce to the form of
For exactly similar reason S-matrix of a 3-port circulator will be
Here, \(S_{13 } = S_{21} = S_{32} = 1\) and all other elements are zero.
4.2 Applications
Due to its unique isolation property, circulator can be employed for several important applications like,
-
As Duplexer
-
As Reflection amplifier
-
In Radar systems
-
In Amplifier systems
-
In Antenna transmitting or receiving Systems
Illustrative Example 8:
Determine the scattering matrix of a 3-port circulator having an insertion loss of 0.5 dB, isolation of 40 dB and VSWR 0f 3.
Solution: The generic form of s-matrix of a 3-port device can be written as,
For circulator, insertion loss is given by \(S_{21} = S_{32} = S_{13}\)
And isolation is given by, \(S_{12} = S_{31} = S_{23}\).
And reflection coefficient which is related to VSWR is represented by \(S_{11} = S_{22} = S_{33}\). In this case, all ports are not matched perfectly.
Here, \(- 20{\text{log}}\left| {S_{21} } \right|\) = 0.5, or \(S_{21}\) = 0.994; So, \(S_{21}\) = \(S_{32}\) = \(S_{13}\) = 0.994.
Again, \(- 20{\text{log}}\left| {S_{12} } \right|\) = 40, or, \(S_{12} = 0.01\).
So, \(S_{12} = S_{31} = S_{23}\) = 0.01.
Now reflection coefficient, \(\rho = S_{11} = S_{22} = S_{33} = \frac{VSWR - 1}{{VSWR + 1}} = \frac{3 - 1}{{3 + 1}} = 0.5\)
So, \(S_{11} = S_{22} = S_{33} = 0.5\)
Putting all these values we have,
5 Gyrator
Gyrator is a passive, 2-port, non-reciprocal, ferrite device which can provide a phase shift of 180° for RF signal transmission in the forward direction and 0° phase shift in the reverse direction [4]. It is a linear and lossless device which is very similar like a transformer but the fundamental difference is, a transformer does not provide any phase shift i.e. if voltage is at the primary end, induced secondary signal will be in the form of voltage only. But in the case of gyrator due to 180° phase reversal it will be in current form. Reverse transmission is very similar for both (Fig. 12).
Gyrator
Gyrator consists of two rectangular waveguide sections in both ends of port1 and port2, circular waveguide in between which contains ferrite rod to provide 90° F rotation in counter clockwise direction. There is a twist of 90° between rectangular waveguide and circular waveguide at port1.
Consider the figure shown below, for wave that propagates from left to right; it passes the twist and gets rotated by 90° in a counter clockwise direction. Again the ferrite rod provides another 90° rotation; the total rotation will be 180° at port2. The wave that propagates from right to left, experiences Faraday rotation of 90° in a similar manner. But while passing through the twist, it gets another 90° of rotation in a direction that cancels the Faraday rotation. For that reason, in transmission from port2 to port1, there is no phase shift (Fig. 13).
Internal structure Gyrator
5.1 S-Matrix
S-matrix of a 2-port device having generic form of,
As it is a symmetrical device, \(S_{ii}\) = 0;
So, S-matrix for Gyrator is
5.2 Applications
Gyrator can be employed for several applications like,
-
As an Inductor
-
As BPF
-
In Telephony device that is connected to POTS
-
As parametric equalizer.
6 Rat Race Coupler
Rat race or hybrid ring is a 4-port passive device that is used to combine two in-phase signals or nullify the signals having path differences. Total circumference of the ring is 1.5λ, where port1, port2 and port3 are 0.5λ apart and the distance between port1 to port4 is 0.75λ (Fig. 14).
Rat race coupler
When input is applied through port1, it appears as two equal parts one at port2 in clockwise direction and other at port4 in counter clockwise direction. As λ/4 path difference corresponds to 90° phase shift, port2 and port4 became in-phase and output appears there. And no output appears at port3 as it is out of phase.
Similarly, when input signal is applied to port3 it gets equally divided between port2 and port4 and no output appears at port1. Again if two different signals are applied to port1 half of the sum appears at port2 and another half appears at port4 and the difference appears at port3.
6.1 S-Matrix
S-matrix of a 4-port device can be written as,
As it is a symmetrical device, \(S_{ij} = S_{ji}\),
All ports are perfectly matched \(S_{ii}\) = 0.
Using Eqs. (62) and (63) above s-matrix reduced to
Considering input from port1, \(S_{21} = - S_{41} \; {\text{and}}\;S_{31} = 0\);
For input from port2, \(S_{12}\) = \(S_{32}\) and \(S_{42}\) = 0;
For input from port3, \(S_{23}\) = \(S_{43}\) and \(S_{13}\) = 0;
For input from port4, \(S_{34}\) = \(- S_{14}\) and \(S_{24}\) = 0;
For perfect matching from the feed line to the ring, impedance at the port should be \(\frac{1}{\sqrt 2 }\) times of the impedance of the ring. For example, it can be 75 Ω for ring and 50 Ω for ports and it is imaginary by nature. Combining all these, Eq. (64) can be reduced as
6.2 Application
-
Combiner of signal
-
Splitter of signal.
7 Matched Termination
It is a two-port passive waveguide section which used to absorb all the incident power without any reflection or radiation from it. Though it is a two-port waveguide section port2 is perfectly matched and terminated by characteristics impedance (Fig. 15).
Matched terminator
Practical circuit includes a tapered section of waveguide with one end terminated and a lossy dielectric inside it. Any input from port1 is absorbed at port2 due to the presence of lossy dielectric which formed matched termination (Fig. 16).
Internal structure of matched terminator
7.1 S-Matrix
As port2 is terminated by characteristics impedance, Zin = Zout and there is no reflection from port2 as it is perfectly matched. Thus the reflection coefficient is Г = \(S_{11}\) = 0.
7.2 Application
The Matched Termination is used to terminate the waveguide transmission line with no reflection at all.
8 Attenuator
It is a passive waveguide device that is used to diminish the strength of the signal without affecting the characteristics impedance (Z0) of the waveguide. If characteristics impedance is not maintain fixed, there caused impedance discontinuity and hence undesired reflection. Generally, a resistive material is placed in parallel to the electric field line, current induce in the resistive material which introduced I2R loss that introduces attenuation (Fig. 17).
Variable attenuator
Basically, attenuator is of three types,
-
Fixed type attenuator
-
Electronically or Mechanically variable type
-
Series of fixed step type
A fixed slab of dielectric is placed inside the waveguide to provide a fixed amount of attenuation, in fixed type attenuator.
Variable attenuator is used to provide fixed or variable attenuation. The depth of attenuation depends on the insertion depth of the plate containing absorbing material into the waveguide. The attenuation is maximum when the dielectric slab is inserted totally into the waveguide. Variable attenuator can be of different types, like.
-
Resistive card type (flap type) attenuator
-
Slide vane attenuator
-
Rotary vane attenuator
Among these, rotary vane attenuator is the most widely employed attenuator. It is having two tapered sections of rectangular to circular waveguide along with an intermediate circular waveguide section which is free to rotate. All the three waveguide section contains thin resistive cards [5]. When dominant TE10 mode enters from rectangular to circular waveguide, input resistive card allows only perpendicular components to pass. Resistive card inside the circular waveguide is available to rotate and adjust its orientation as per attenuation required. Inside circular waveguide TE11 mode have parallel and perpendicular components. Parallel component absorbed through resistive card and perpendicular component passes through it. Output resistive card further attenuates parallel components and perpendicular components appear at the output. The output power can be controlled by rotating circular waveguide to change the orientation of the resistive card inside it which can change the attenuation (Fig. 18).
Rotary vane attenuator
As the basic property of the attenuator is to maintain characteristics impedance (Z0) fixed, it does not introduce any reflection to the waveguide, hence no additional scattering property for it.
8.1 Applications
It provides attenuation to the waveguide where a signal with lower strength is required.
9 Phase-Shifter
Microwave phase-shifter is a passive device which can alter the phase of oscillation of electromagnetic wave at the output of the transmission line with respect to the phase at the input. Microwave phase-shifter can be used as power divider, beam forming network, phase discriminator and in phase-array antenna. The main difference between phase-shifter and attenuator is, the phase-shifter alters the phase of RF signal in a desired manner without doing any change in signal strength, whereas attenuator changes signal strength without altering the phase of the signal. There are different types of phase-shifter available in the industry, among them the two most widely employed phase-shifter are,
-
Dielectric Phase-Shifter
-
Precision Rotary Phase-Shifter
The working principle of all the phase-shifter is fundamentally similar. If we consider two arbitrary points having phases \(\upvarphi\)1 and \(\upvarphi\)2 and distance L among them. The phase difference, Δ\(\upvarphi\) = (\(\upvarphi\)2 – \(\upvarphi\)1) = βL = \(\left( {\frac{{2\rm{\pi }}}{\lambda }} \right){\text{L}}\). Where β is phase constant. This means, by changing the distance between two points, phase alteration is possible. Alternatively, if the velocity of RF signal can get changed, that is equivalent to the change in distance travel considering the time as constant. This implies that any retardation in velocity can have a net effect on the phase of the signal. Applying this principle different phase-shifter is developed.
9.1 Dielectric Phase-Shifter
Inside a rectangular waveguide a slab of dielectric of thickness ‘t’ and height ‘h’ is inserted so that ‘h’ is in parallel orientation to the electric field. When the dominant TE10 mode is propagated through the waveguide, due to the presence of dielectric constant ε, the effective path length increased and hence velocity of propagation is reduced. Due to that, ultimately phase of the signal gets delayed [6].
9.2 Precision Rotary Phase-Shifter
This is one of the most widely employed phase-shifter, where a particular waveguide structure, one half-wave plate and two quarter-wave plate are used. Two rectangular to circular tapered section contains a quarter-wave plate inside it, mounted at 45° angles with the broad wall of the rectangular waveguide. The rotating circular waveguide contains half-wave plate at 0° default angle and precision over there (Fig. 19).
Precision Rotary Phase-Shifter
Both of the quarter-wave plates provides a phase shift of 90° each and a 180° phase shift for the half-wave plate. So when the signal of the dominant TE10 mode is applied to the input end it gets converted to TE11 mode inside the circular waveguide section and reaches the half-wave plate as a parallel and perpendicular component. Perpendicular component passes it with a net phase delay of (90° + 180°) = 270° for quarter-wave and half-wave plate respectively when half-wave plate is in 0th or default position. Ultimately it passes the quarter-wave with an accumulated phase of (270° + 90°) = 360°. So there is no phase alteration of the signal. But when the half-wave plate rotates from 0th position to an angle of θ, the output signal experienced a total phase delay of 2θ. So, by rotating the circular waveguide section, i.e. changing the orientation of half-wave plate, the output phase of the signal can be regulated in the desired manner.
9.3 Applications
Microwave phase-shifter can be employed in different types of communication systems, radar systems, microwave measurement systems and in different industrial operations.
10 Waveguide Bends and Twists
Sometimes it became necessary to have bends in the waveguide structure to direct the signal in the desired direction. But any abrupt variation in the size or shape of the waveguide can cause reflection and hence a loss in efficiency. When such a change is required, certain conditions must be satisfied to prevent unwanted reflection. In general, bends can be of four types,
-
Gradual E-Bend
-
Gradual H-Bend
-
Sharp E-Bend
-
Sharp H-Bend
Gradual E-Bend: It is a gradual bend that distorts the E-field only. The bend gradually follows a radius of curvature where radius r must satisfy the condition of r > \(2\lambda_g\) to avoid unwanted reflection (Fig. 20).
Gradual E-Bend
Gradual H-Bend: It is a gradual bend that distorts H-field only. Radius of curvature of the bend is greater than twice the wavelength i.e. r > \(2\lambda_g\) to avoid any unwanted reflection through it (Fig. 21).
Gradual H-Bend
Sharp E-Bend: Two sharp bends of 45° on E-field waveguide placed quarter-wave \(\left( {\frac{\lambda_g }{4}} \right)\) apart so that reflection caused by one sharp bend can be canceled by another, leaving no net reflection at all (Fig. 22).
Sharp E-Bend
Sharp E-Bend: Two sharp bends of 45° on H-field waveguide placed quarter-wave \(\left( {\frac{\lambda_g }{4}} \right)\) apart so that reflection caused by one sharp bend can be canceled by another, leaving no net reflection at all (Fig. 23).
Sharp H-Bend
Twist: For some particular use, sometimes it is required to rotate the RF signal in such a way that it can get the desired phase for phase adjustment at load. That can be achieved by using a twist in the waveguide which is gradual in nature and greater than 2λg (Fig. 24).
Waveguide twist
Some special kinds of lossy bends can also be used that consist of wound ribbons of conducting material like brass with chromium plated inner surface. It is used for a short section when no other solution is possible.
11 Cavity Resonator
Cavity resonator is a closed metallic structure that encloses electromagnetic energy. The structure can be hollow or filled with dielectric material. Cavity resonator can be formed by introducing metallic walls at a distance d apart both ends of a rectangular or circular waveguide along z direction. The microwave signal bounces back and forth inside the cavity to form a standing wave. Thus it can act like a Band Pass Filter (BPF) allowing a particular band of frequency to pass through it and blocking all other frequencies. Microwave cavity resonator acts as a very low loss resonant device at its resonance frequency and offers a quality factor as high as 106. At resonance stored electrical energy is equal to stored magnetic energy and impedance is purely real in nature. Here we will discuss about the cavity resonator made of rectangular i.e. rectangular cavity resonator and of circular waveguide i.e. cylindrical cavity resonator [7].
11.1 Rectangular Cavity Resonator
Rectangular waveguide closed from both ends by a metallic wall along z direction at a distance of d (Fig. 15). The transverse component of electric field can be written as (Fig. 25),
Rectangular cavity resonator
Putting boundary condition Et = 0 for z = 0;
We have, \({\hat{\text{e}}}({\text{x}},{\text{y}})\left\{ {{\text{A}}^+ + {\text{A}}^- } \right\} = 0\);
Second boundary condition is Et = 0 for z = d;
Or, \({\hat{\text{e}}}({\text{x}},{\text{y}})\left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} + {\text{A}}^- {\text{e}}^{ + {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}}} \right\} = 0;\)
Or, \(\left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} + {\text{A}}^- {\text{e}}^{ + {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}}} \right\} = 0;\)
Or, \(\left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} - {\text{A}}^+ {\text{e}}^{ + {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}}} \right\} = 0;\quad \quad\)\(\quad \quad [{\text{using}}\;{\text{Eq.}}\;(67)]\).
Or, \({\text{A}}^+ \left[ { - 2{\text{j}}\;\sin \left( {\rm{\beta }_{{\text{mn}}} {\text{d}}} \right)} \right] = 0\quad \quad\)\(\quad \quad \left[ {{\text{as}}\;\frac{{e^{jx} - e^{ - jx} }}{2j}\sin ({\text{x}})} \right]\).
That implies, βmnd = pπ where p = 0,1,2,3, …….
From Eq. (67) we have,
Now resonance frequency for different modes can be presented as,
For dominant mode i.e. TE101
Illustrative Example 9:
Determine the lowest resonant frequency of a rectangular cavity resonator having dimensions 3 cm × 4 cm × 5 cm. Also find the change in length required to have a resonant frequency of 1.5 times than the earlier one.
Solution: Here given that a = 3 cm; b = 4 cm; d = 5 cm;
Dominant mode is TE101 here.
Resonant frequency, \(f_r = \frac{c}{2}\sqrt { \left( \frac{m}{a} \right)^2 + \left( \frac{n}{b} \right)^2 + \left( \frac{p}{d} \right)^2 }\)
Now for the second part, consider the new length of cavity is d,
New resonant frequency to be \(f_r^{\prime}\) = 1.5 × \(f_r = 8.745\;{\text{GHz}}\).
So, \(f_r^{\prime} = \frac{c}{2}\sqrt { \left( \frac{1}{3} \right)^2 + \left( \frac{0}{4} \right)^2 + \left( {\frac{1}{{d^{\prime} }}} \right)^2 }\) or, 8.745 × \(10^9 = \frac{{3 \times 10^{10} }}{2}\sqrt { \frac{1}{9} + \frac{1}{{d^{\prime}\,^2 }}}\).
Or, \(d^{\prime} = 2.08\;{\text{cm}}\).
Length of the resonator must be reduced by \(\left( {5.00 - 2.08} \right)\) = 2.92 cm.
Illustrative Example 10:
Calculate the resonant frequency for TE111 mode of a rectangular cavity resonator of dimension 8 cm × 6 cm × 4 cm. What will be the new resonant frequency if the cavity is filled with a dielectric of \(\varepsilon_r = 2.2\)?
Solution: Here, a = 8 cm; b = 6 cm; d = 4 cm
Now, filling the cavity will reduce the resonant frequency of the cavity resonator.
New resonant frequency, \(f_r^{\prime} = \frac{f_r }{{\sqrt {\varepsilon_r } }} = \frac{4.88}{{\sqrt {2.2} }}\;{\text{GHz}}\) = 3.29 GHz.
Illustrative Example 11:
A rectangular waveguide cavity resonator filled with dielectric \(\varepsilon_r = 2.5\) and having a cross section of 5 cm × 4 cm. Find the length of the cavity resonator required to have the resonant frequency of 3 GHz at dominant TE101 mode.
Solution: Given as a = 5 cm; b = 4 cm and \(f_r = 3\;{\text{GHz}}\).
We know that \(f_r^{\prime} = \frac{c}{{2\sqrt {\varepsilon_r } }}\sqrt { \left( \frac{m}{a} \right)^2 + \left( \frac{n}{b} \right)^2 + \left( \frac{p}{d} \right)^2 }\).
For TE101 mode, \(f_r = 3\;{\text{GHz}} = 3 \times 10^9 = \frac{{3 \times 10^{10} }}{{2\sqrt {2.5} }}\sqrt { \frac{1}{25} + \frac{1}{d^2 }}\).
Or, \(\sqrt { \frac{1}{25} + \frac{1}{d^2 }} = \frac{{\sqrt {2.5} }}{5}\).
Or, \(\frac{1}{25} + \frac{1}{d^2 } = \frac{2.5}{{25}}\quad {\text{or}},\frac{1}{d^2 } = \frac{1.5}{{25}}\).
Or, \(d^2 = \frac{25}{{1.5}}{ }\) or, d = 4.08 cm.
So the length of the resonator should be 4.08 cm.
11.2 Cylindrical Cavity Resonator
It is made of a circular waveguide where both ends are shorted by a metallic wall along the z-axis at z = 0 and z = d (Fig. 26). At resonant frequency total electrical energy stored equals to the total stored magnetic energy. The transverse modes supported by cavity resonators are the TE and TM modes.
Cylindrical cavity resonator
The transverse component of the electric field can be written as,
Putting boundary condition Et = 0 for z = 0;
We have, \({\hat{\text{e}}}({\text{x}},{\text{y}})\left\{ {{\text{A}}^+ + {\text{A}}^- } \right\} = 0\);
Second boundary condition is Et = 0 for z = d;
Or, \({\hat{\text{e}}}({\text{x}},{\text{y}})\left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} + {\text{A}}^- {\text{e}}^{ + {\text{j}}\beta }_{{\text{mn}}} {\text{d}} = 0} \right\}\);
Or, \(\quad \left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} + {\text{A}}^- {\text{e}}^{ + {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}}} \right\} = 0\);
Or, \(\quad \left\{ {{\text{A}}^+ {\text{e}}^{ - {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}} - {\text{A}}^+ {\text{e}}^{ + {\text{j}}\rm{\beta }}_{{\text{mn}}} {\text{d}}} \right\} = 0\quad \quad\) \(\quad \quad [{\text{using}}\;{\text{Eq.}}\, (72)]\).
Or, \(\quad {\text{A}}^+ \left[ { - 2{\text{j}}\;\sin \left( {\rm{\beta }_{{\text{mn}}} {\text{d}}} \right)} \right] = 0\quad \quad\) \(\quad \quad \left[ {{\text{as}}\;\frac{{e^{jx} - e^{ - jx} }}{2j} = \sin ({\text{x}})} \right]\).
That implies, \(\quad \rm{\beta }_{{\text{mn}}} d = {\text{p}}\rm{\pi }\).
Or, \(\quad \beta_{{\text{mn}}} = \frac{{{\text{p}}\pi }}{d}\).
So, for TE mode \(\rm{\beta }_{{\text{mn}}} = \frac{{{\text{p}}\pi }}{d}\quad \quad {\text{where}}\;{\text{p}} = 1,2,3, \ldots \ldots\).
And for TM mode \(\rm{\beta }_{{\text{mn}}} = \frac{{{\text{p}}\pi }}{d}\quad \quad {\text{where}}\;{\text{p}} = 0,1,2, \ldots .\)
Dominant mode to be transmitted here is TE101 and TM010.
The resonant frequency can be given as,
Here the values of \(X_{mn}\) can be determined from Bessel function values. The following two tables (Tables 1 and 2) can be used to determine the same for TM as well as TE mode.
11.3 Applications
-
As microwave sensors
-
Microwave oscillator
-
Amplifier
-
Wave meter
-
Band pass filter at microwave frequency.
Illustrative Example 12:
A circular cavity resonator having a diameter of 14 cm and length of 6 cm. Determine the resonant frequency if the resonator is operating at TM012 mode.
Solution: The resonant frequency of the circular waveguide can be given as,
For TM012 mode \({\text{f}}_{\text{r}} = \frac{{\text{c}}}{2\pi }\sqrt { \left( {\frac{{X_{01} }}{7}} \right)^2 + \left( {\frac{2\pi }{6}} \right)^2 } \quad\) D = 14 cm; \(\rho = \frac{14}{2} = 7\;{\text{cm}}\)
\(X_{01}\) = 2.405 for m = 0, n = 1; for TM mode (See table given above)
Illustrative Example 13:
A circular waveguide cavity resonator has a radius of 5 cm and operating at TM011 mode. Determine the length of the resonator if it has to resonate at 12 GHz.
Solution: Given that ρ = 5 cm; fr = 12 GHz.
Resonant frequency can be calculated as, \({\text{f}}_{\text{r}} = \frac{{\text{c}}}{2\pi }\sqrt { \left( {\frac{{X_{mn} }}{\rho }} \right)^2 + \left( {\frac{p\pi }{d}} \right)^2 }\).
For TM011 mode \(X_{01}\) = 2.405 (See table of Bessel function above for TM mode),
and p = 1.
So, f011 = \(\frac{{3 \times 10^{10} }}{2\pi }\sqrt { \left( {\frac{2.405}{5}} \right)^2 + \left( {\frac{\pi }{d}} \right)^2 } = 12\, \times 10^9\)
Or, \(\sqrt { \left( {\frac{2.405}{5}} \right)^2 + \left( {\frac{\pi }{d}} \right)^2 } = \frac{{{ }2\rm{\pi }\,\times\,12\,\times\,10^9 }}{{3 \times 10^{10} }} = 2.51\)
Or, \(\left( {\frac{\pi }{d}} \right)^2 = 6.069\)
Or, d = 1.275 cm
So the length should be 1.275 cm.
Illustrative Example 14:
A cylindrical waveguide providing the same resonant frequency for TE as well as TM mode. Comment on the lowest possible mode of operation.
Solution: We know that the resonant frequency of circular waveguide,
where m = 0,1,2,3..; n = 1,2,3,..; and p = 1,2,3,…;
where m = 0,1,2,3..; n = 1,2,3,..; and p = 0,1,2,3,…;
As per the problem statement \(f_{mnp} = f^{\prime}_{mnp}\).
So, \(\frac{{\text{c}}}{2\pi }\sqrt { \left( {\frac{{X_{mn} }}{\rho }} \right)^2 + \left( {\frac{p\pi }{d}} \right)^2 } = \frac{{\text{c}}}{2\pi }\sqrt { \left( {\frac{{X^{\prime}_{mn} }}{\rho }} \right)^2 + \left( {\frac{p\pi }{d}} \right)^2 }\).
As the resonator is same, all the parameters like radius (ρ) and length (d) are same.
So the above condition must be satisfied if and only if \(X_{mn} = X^{\prime}_{mn}\) and a suitable value of p is chosen. From the above table of Bessel function, it can be easily observed that \(X_{mn} = X^{\prime}_{mn}\) satisfies under the condition given below.
\(X_{10} = X^{\prime}_{01}\) = 3.832; \(X_{12} = X^{\prime}_{02}\) = 7.016; \(X_{13} = X^{\prime}_{03}\) = 10.173;
and \(X_{14} = X^{\prime}_{04}\) = 13.324 (Please refer table given above).
So the lowest ever value is \(X_{10} = X^{\prime}_{01}\) = 3.832 and if p = 1 for both of the cases.
Hence the mode of operation is TM101 and TE011.
Here p can never be zero for TE mode so TE010 mode is not possible.
Very similar pattern can be observed for the higher order modes where \(X_{mn} = X^{\prime}_{mn}\). As per example \(X_{12} = X^{\prime}_{02}\) = 7.016, \(X_{13} = X^{\prime}_{03}\) = 10.173 and \(X_{14} = X^{\prime}_{04}\) = 13.324 etc.
Problems
-
1.
When input power is divided in the ratio of 2:1 in a T- junction coupler and the characteristic impedance of the two output lines is 150Ω and 75Ω, calculate the impedance of the input line.
-
2.
If a signal of power 32 mW is fed into one of the collinear arm of a H-plane Tee. Determine the power that appears at all the ports when ports are terminated by matched load.
-
3.
Design a lossless T-junction signal divider with a 80 Ω source impedance to give a 3:1 power split. Design quarter-wave matching transformers to convert the impedances of the output lines to 80 Ω. Determine the magnitude of the scattering parameters for this circuit, using a 80 Ω characteristic impedance.
-
4.
An E-plane Tee made of waveguide section of 50Ω characteristics impedance. A signal power of 20 mW is applied to E-arm which is perfectly matched. Determine the power delivered to the load of 60Ω and 75Ω connected to port1 and port2 respectively.
-
5.
If 500 mW power is applied to the perfectly matched port3 of a magic Tee, what will be the power delivered to port1, port2 and port4 if it is terminated by a reflection of 0.4, 0.6 and 0.7 respectively?
-
6.
Two couplers with identical natures are used in a waveguide to sample the incident power of 10 mW and reflected power as 0.20 mW. What will be the value of VSWR?
-
7.
Two 20 dB identical directional couplers are used to sample incident and reflected power in a waveguide. Voltage standing wave ratio is 1.5 and the output of the coupler sampling incident power is 8 mW. Calculate the value of the reflected power.
-
8.
A directional coupler has the scattering matrix given below. Find the return loss, coupling factor, directivity, and insertion loss. Assume that all ports are terminated by matched load.
$${\text{S}} = \left[ {\begin{array}{*{20}c} {0.1\angle 30^\circ } & {0.9\angle 90^\circ } & {0.18\angle 180^\circ } & {0.005\angle 90^\circ } \\ {0.9\angle 90^\circ } & {0.1\angle 30^\circ } & {0.005\angle 90^\circ } & {0.18\angle 180^\circ } \\ {0.18\angle 180^\circ } & {0.005\angle 90^\circ } & {0.1\angle 30^\circ } & {0.9\angle 90^\circ } \\ {0.005\angle 90^\circ } & {0.18\angle 180^\circ } & {0.9\angle 90^\circ } & {0.1\angle 30^\circ } \\ \end{array} } \right]$$
-
9.
A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35 dB, and an insertion loss of 0.5 dB. If all ports are matched, find the output powers (in dBm) at the through, coupled, and isolated ports.
-
10.
Two 40 dB directional couplers with identical properties are used for sampling incidents and reflected power in a waveguide. The value of VSWR is 7 and the output of the coupler sampling incident power is 5 mw. Find the reflected power.
-
11.
Design a single-section coupler with coupled line having a coupling of 19 dB, a system impedance of 60 Ω, and a center frequency of 8 GHz. If the coupler is to be made in strip line (edge-coupled), with r = 2.2 and b = 0.32 cm, find the necessary strip widths and separation.
-
12.
A four port directional coupler has a 4:1 power splitting ratio and has a dissipation loss of 3 dB. The coupler directivity is 40 dB. What fraction of input power P1 will go to port P2 (output port) and P3 (coupled port)?
-
13.
Design a field displacement isolator in an X-band waveguide to operate at 11 GHz. The ferrite has 4π Ms = 2500 G and r = 13. Ignore ferrite losses.
-
14.
A thin ferrite rod with 4π Ms = 600 G is magnetically biased along its axis. Find the external bias field strength required to produce a gyro-magnetic resonance at 2.52 GHz.
-
15.
A lossless circulator having a return loss of 12 dB. Find the value of isolation? What will be the isolation if the return loss changes to 24 dB?
-
16.
A two-port is known to have the following scattering matrix:
$$[{\text{S}}] = \left[ {\begin{array}{*{20}c} {0.15\angle 0^\circ } & {0.85\angle - 45^\circ } \\ {0.85\angle 45^\circ } & {0.2\angle 0^\circ } \\ \end{array} } \right]$$
Determine if the network is reciprocal and lossless. If port2 is terminated by matched load, then what is the return loss seen at port1?
-
17.
A circular waveguide, filled with air, has a radius of 3 cm and is acting as a resonator for TE01 mode at 10 GHz by placing two perfectly conducting plates at its two ends. Determine the minimum distance between the two end plates.
-
18.
A rectangular cavity resonator has dimensions of a = 5 cm, b = 2 cm, and d = 15 cm. Compute:
-
a.
The resonant frequency of the dominant mode for an air-filled cavity
-
b.
The resonant frequency of the dominant mode for a dielectric-filled cavity of εr = 2.56
-
a.
-
19.
Determine the lowest resonant frequency of a rectangular cavity resonator having dimensions 4 cm × 5 cm × 6 cm. Also find the change in length required to have a resonant frequency of 1.2 times than earlier one.
-
20.
Calculate the resonant frequency for TE111 mode of a rectangular cavity resonator of dimension 6 cm × 5 cm × 4 cm. What will be the new resonant frequency if the cavity is filled with a dielectric of \(\varepsilon_r = 4\)?
-
21.
A rectangular waveguide cavity resonator filled with dielectric \(\varepsilon_r = 2.2\) and having a cross section of 4 cm × 3 cm. Find the length of the cavity resonator required to have the resonant frequency of 3 GHz at dominant TE101 mode.
-
22.
A circular cavity resonator having diameter of 12 cm and length of 5 cm. Determine the resonant frequency if the resonator is operating at TM012 mode.
-
23.
A circular waveguide cavity resonator has a radius of 6 cm and operating at TM011 mode. Determine the length of the resonator if it has to resonate at 10 GHz.
-
24.
A circular waveguide providing same resonant frequency for TE as well as TM mode. Comment on all possible modes of operation.
References
Pozar DM (2012) Microwave engineering. Wiley, Hoboken, NJ
Rizzi PA (1988) Microwave engineering: passive circuits. Prentice Hall
Cohn SB (1968) Microwave band pass filters containing high-q dielectric resonators. IEEE Trans Microw Theory Tech 16:218–227
Bailey AE (ed) (1985) Microwave measurement. Peter Peregrinus, London
Montgomery CG, Dicke RH, Purcell EM (1948) Principles of microwave circuits, MIT radiation laboratory series, vol 8. McGraw-Hill, New York
Collin RE (2001) Foundations for microwave engineering, 2nd edn. Wiley–IEEE Press, Hoboken, N. J
Heaviside O (1950) Electromagnetic theory, vol 1, 1893. Reprinted by Dover. New York
Author information
Authors and Affiliations
Corresponding author
Editor information
Editors and Affiliations
Rights and permissions
Copyright information
© 2022 The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd.
About this chapter
Cite this chapter
Mistri, H. (2022). Brief Introduction to High Frequency Passive Circuits. In: Acharyya, A., Biswas, A., Inokawa, H. (eds) New Horizons in Millimeter-Wave, Infrared and Terahertz Technologies. Lecture Notes in Electrical Engineering, vol 953. Springer, Singapore. https://doi.org/10.1007/978-981-19-6301-8_11
Download citation
DOI: https://doi.org/10.1007/978-981-19-6301-8_11
Published:
Publisher Name: Springer, Singapore
Print ISBN: 978-981-19-6300-1
Online ISBN: 978-981-19-6301-8
eBook Packages: Physics and AstronomyPhysics and Astronomy (R0)