JRE SCHOOL OF Engineering PUT EXAMINATION SET-A MAY 2015 Subject Name Microwave Engineering Subject Code EEC 603 Roll No. of Student Max Marks 100 Max Duration 3 hrs Date 02/05/2015 Time 10:00 a.m. to 1:00 p.m. For Branches: EC Branch only (6th sem) Q. 1 Attempt any FOUR from the following. All question carry equal marks. (5 X 4 = 20) a) A TE11 wave is propagating in a air-filled circular waveguide of diameter 12cm at 2.5GHz, find the cutoff frequency, guide wavelength, wave impedance in the guide. b) Show that TM01 and TM10 modes do not exist in a rectangular waveguide. c) What is a microstrip line? Compare microstrip lines with striplines. Write advantages and disadvantages of both. Microstrip Transmission Line: It is also called open strip line because of the openness of its structure. It has very simple geometry. It is an unsymmetrical strip line that is nothing but a parallel plate transmission line having dielectric substrate, the on face of which is metallised ground and the other (top) face, has thin conducting strip of certain width ‘w’ and thickness ‘t’. The top ground plate is not present and so cover plate is used for shielding purpose. Modes are only quasi TEM, thus the theory of TEM coupled lines applies only. Losses: (i) Dielectric loss in substrate (ii) ohmic skin losses in conductor strip and ground plane. Advantages: (i) Simple construction (ii) easier integration with semiconductor device (iii) fabrication cosh is lower (iv) package and unpacked semiconductor chips can be attached to these lines. Disadvantages: (i) Radiate from irregularities, (ii) Both Q and power handling ability are lower (iii) Losses are significant above 100 GHz. Application; Used to interconnect high speed logic ckts in digital computers. d) Define quality factor of any resonator. Design a rectangular cavity to have resonant frequency of 10.2GHz having dimensions a = d and b = a/2. e) Derive the TEmn mode field equations in a rectangular waveguide. TE MODE IS RECRANGULAR WAVEGUIDES The co-ordinate system for TE modes in rectangular wavelength is shown in figure. The TE (transverse electric modes) are denoted by TEmn. Here, m= Number of half wave of electric or magnetic intensities in X- direction n= Number of half waves of electric or magnetic intensities in Y- direction. As its name indicates there exists a transverse electric field. This transverse electric field is denoted by Enormal. Since Enormal exists, the transverse magnetic field Hnormal is zero. In this case the wave is assumed to be propagating in the positive Z direction. Since Enormalexists, ( is the component of electric field in Z direction. That means, this is the component parallel to waveguide. This component is also denoted by . Thus the boundary conditions for TE waves are. and Consequently from the given Helmholtz’s equation … (1) A solution in the form of … (2) These are determined in accordance with the given boundary conditions, where, … (3) And are replaced. … (4) The wave equations are obtained by applying the Maxwell’s curl equations and then by applying the boundary condition. We have Maxwell’s curl equations … (5) … (6) In a rectangular co-ordinate it becomes, … (7) … (8) … (9) … (10) … (11) … (12) But as we know that the electric component in Z direction, that is is zero. Now putting , And consider We get , …(13) … (14) … (15) … (16) … (17) … (18) Now we will solve all these equation in terms of will give the TE mode field equation in rectangular waveguide. To find out equation for Put the value of from equation (17) into equation (13), we get, But … … (19) Similarly we can write … (20) … (21) … (22) Now we have the operation for as … (23) Now differentiating equation (23) w.r.t. ‘x’ and ‘y’ we get, And Substituting the value of and in the solution of equation (13) through equation (18) gives a new set of field equation. The boundary conditions are applied to newly found equation in such a manner that the tangent E or the normal H field vanishes at the surface of conductor. Since then =0 at y=0 and b Hence Since then Hence Therefore the magnetic field in the positive Z- direction is given by: … (24) Hence is the amplitude constant. Substituting the values in equation (19) through equation (23) gives the field equations in rectangular waveguide … (25) … (26) … (27) … (28) … (29) Here Q. 2 Attempt any FOUR from the following. All question carry equal marks. (5 X 4 = 20) a) Explain the working of a multi-hole directional coupler and define the following terms in reference to a multi-hole directional coupler - i) Directivity ii) Coupling factor. DIRECTIONAL COUPLERS Directional couplers are used to sample the part of energy passing through the main waveguide. The directional coupler is a four port device. They are also used to check whether the signal passing though two arms in phase as well as amplitude or not. There are no reflections at the junction of these four ports. Directional coupler having such property is shown in figure. When the power is incident from port (1) then it is passed to port (2) and port (3), but it is not appears on port (4). Thus port (4) is uncouple when incident power is fed from port (1). Similarly when the power is incident from port (4) then this power is coupled to port (2) and port (3) but not coupled to port (1). The directional coupler can also be made from two waveguides. In such cases one waveguide is straight but other is having a curve shape. Here an input power comes out from other end of main arm when it passes through the main arm as shown in fig. The holes are located at quarter wavelength. So through this hole some of the power passes from main arm to auxiliary arm. Other end of auxiliary arm is terminated. So if there are any reflections occur from output end of auxiliary arm then it is avoided. Directional Coupler Parameters The performance of directional couplers depends on the following parameters; (i) Coupling factors. It is defined as the ratio of incident power to the power to the power output from auxiliary arm. Coupling factor is always given in decibles. It is denoted by C Where, Incident power Power coming out from auxiliary arm. If coupling power is small it indicated that the value of coupling factor (C) is high. It is considered here that the directional coupler does not couple any power in the backward direction. The coupling factor also gives the amount of attenuation taking place in the output power. The total input power into the coupler is sum of output power from coupler and the output power from auxiliary arm. (ii) Directivity. If in case of directional coupler the power is incident from output end, there should not be any leakage of power into the auxiliary arm. For this an input end of coupler is terminated with a matched load. The power incident from output end is called reverse power. It is defined as the ratio of power coupled in the forward direction to reverse direction in the auxiliary arm. the directivity is denoted by D. it is given in decibels. All other arms which are not used should be terminated with the match load. Where, = power coupled in auxiliary arm due to power in forward direction = power coupled in auxiliary arm due to power in reverse direction is mor it indicated that the leakage of power in the auxiliary arm is less. So the leakage auxiliary arm takes place because of the reverse power flow. b) What is Circulator? How can a four port Circulator be realized using two magic tee & a Gyrator? Generally circulator is a four port (four terminals) device. Circulator with other number of ports can also be design. The schematic of four port circulator is as shown in fig. 5.10. In such cases each port is connected to the next port in a clockwise direction, i.e., port 1 is connected only to the Port-2 and not to the port.3 and port 4. Circulators are generally used to provide the isolation between input and output terminals in case of two terminal amplifier devices. Now we will discuss the different types of circulators in the following section. c) Draw the constructional details of isolator and explain the working of isolator. Faraday Rotational Isolator Principle of operation: Basically this type of isolator consists of two resistive attenuators (resistive pack). The ferrite core is placed in between two resistive pads. This situation is as shown in fig. Now consider that input E waves enters into the first resistive pad. The direction of this wave is perpendicular to resistive pad 1. So this wave will not be attenuated by resistive pad 1. A permanent magnet is placed near the ferrite rod. The ferrite rod has tapered ends. So it will not produce any reflections of input wave. Now the ferrite rod gives 45° attenuation to the incoming forward wave. This is as shown in fig. The second resistive pad is placed after the ferrite rod. This resistive pad 2 is rotated at 45° and so the forward wave becomes perpendicular to the resistive pad 2. So this wave will not he attenuated. Thus the forward wave passing through this structure is not affected by any resistive pad. Now if this wave is reflected from the output end then it will first enter into resistive pad 2 and will passes through the ferrite rod and gives another 45° rotation. Rotation given by ferrite rod depends on the dc magnetic field and not on direction of wave propagation through it.