12. Questions On Waveguides 14442c

Waveguides Types - Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Types of Waveguides”. 1. The phenomenon employed in the waveguide operation is a) Reflection b) Refraction c) Total internal reflection d) Adsorption View Answer Answer: c Explanation: The waveguides use total internal reflection phenomenon to transmit the waves ing through it. Thus the acceptance angle and critical angle are important for effective transmission. 2. The dominant mode in waveguide is the mode which has a) Highest frequency b) Highest wavelength c) Lowest phase constant d) Highest attenuation View Answer Answer: b Explanation: The dominant mode is the mode which has the minimum frequency or maximum wavelength available for propagation of the waves. 3. The modes are calculated from which parameter? a) Frequency b) Wavelength c) Phase constant d) V number View Answer Answer: d Explanation: The modes are calculated from the V number of the waveguides. It is given by M= V2/2. 4. The circular waveguides use which function in the frequency calculation? a) Laplace function

b) Schottky function c) Bessel function d) Transfer function View Answer Answer: c Explanation: The circular or cylindrical waveguides use the Bessel function for the frequency calculation of a particular mode. 5. The scattering parameters are used to indicate the a) Permittivity and permeability b) Electric and magnetic field intensities c) Reflection and transmission coefficients d) Frequency and wavelength View Answer Answer: c Explanation: The scattering matrix consists of the transmission coefficients in the main diagonal and the reflection coefficients in the opposite diagonal. 6. Which of the following two parameter models cannot be used to represent a transmission line? a) H parameter model b) T parameter model c) ABCD parameter model d) S parameter model View Answer Answer: a Explanation: The T, ABCD and S parameter models are used in the transmission line modelling. The h parameter is not used for the same. 7. For the matched line, the parameters S12 and S21 are a) 1 b) 0 c) -1 d) ∞ View Answer Answer: b Explanation: The parameters S12 and S21 are the reflection coefficients. For a matched line, the reflection coefficients are zero. Thus the parameters S12 and S21 are also zero. 8. The waveguides are materials with characteristics of a) Low bulk resistivity b) High bulk resistivity c) High conductivity

d) Low conductivity View Answer Answer: a Explanation: Generally, the waveguides are made of materials with low bulk resistivity like brass, copper, silver etc. But if the interior walls are properly plated, it is possible with poor conductivity materials too. It is even possible to make plastic waveguides. 9. The parameters S11 and S22 indicate the transmission coefficients. State true/false. a) True b) False View Answer Answer: a Explanation: In a scattering matrix, the parameters S11 and S22 indicate the transmission coefficients and the parameters S21 and S12 indicate the reflection coefficients. 10. The waveguides increase the transmission of the electromagnetic waves. State true/false. a) True b) False View Answer Answer: a Explanation: The waveguides aid in effective transmission of the electromagnetic power from the source antenna to the destination antenna.

Waveguides Properties - Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Waveguides”. 1. The waveguide is employed in the transmission lines, when operated at the range of a) Hz b) KHz c) MHz d) GHz View Answer Answer: d Explanation: Waveguides are employed for effective transmission, when the lines carry electromagnetic waves in the GHz range.

2. The cut off frequency for a waveguide to operate is a) 3 MHz b) 3 GHz c) 6 MHz d) 6 GHz View Answer Answer: d Explanation: The cut off frequency of the waveguide is 6 GHz. This is the frequency at which the waveguide will start to operate. 3. In rectangular waveguides, the dimensions a and b represent the a) Broad wall dimensions b) Broad wall and side wall dimension respectively c) Side wall and broad wall dimension respectively d) Side wall dimensions View Answer Answer: b Explanation: In rectangular waveguide, the a parameter is the broad wall dimension of the waveguide and the b parameter is the side wall dimension of the waveguide. Always, a > b in a waveguide. 4. The Bessel function is denoted by a) Jn(ha) b) Jm(ha) c) Jn(hb) d) Jm(hb) View Answer Answer: a Explanation: The Bessel function is used in the circular waveguides. Normally Jn(ha) = 0. Here n is the order of the Bessel function. 5. In a waveguide, always which condition holds good? a) phase velocity = c b) phase velocity greater than c c) phase velocity lesser than c d) group velocity = c View Answer Answer: b Explanation: In air medium, the phase velocity is assumed to be the speed of light. For waveguides, the phase velocity is always greater than the speed of the light.

6. The group wavelength is greater than the wavelength at any point. State true/false. a) True b) False View Answer Answer: a Explanation: In a waveguide, the phase velocity is greater than the velocity of light. Thus the group velocity will be less. This implies the group wavelength will be greater than the wavelength at any point. 7. Find the group wavelength of a wave, given that the group phase constant is 6.28 units. a) 2 b) 3.14 c) 6.28 d) 1 View Answer Answer: d Explanation: The group wavelength is given by λg = 2π/βg, where βg is the group wavelength of the wave. On substituting for βg = 6.28, we get group wavelength as unity. 8. The phase velocity of a wave with frequency of 15 radian/sec and group phase constant of 2 units is a) 30 b) 15 c) 7.5 d) 2/15 View Answer Answer: c Explanation: The phase velocity of a wave is given by Vp = ω/βg. on substituting for ω = 15 and βg = 2, we get phase velocity as 15/2 = 7.5 units. 9. The modes in a material having a V number of 20 is a) 20 b) 400 c) 200 d) 100 View Answer Answer: c Explanation: The relation between the modes and the V number is given by m = v2/2. Given that v = 20, we get m = 202/2 = 200 modes. 10. The number of modes is given as 50, find the V number. a) 5

b) 10 c) 50 d) 2500 View Answer Answer: b Explanation: The relation between the modes and the V number is given by m = v2/2. Given that m = 50, we get v2 = 2 x 50 = 100. The V number is 10.

Cut-off Frequency & Wavelength Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Cut off Frequency and Wavelength”. 1. The real part of the propagation constant is the a) Attenuation constant b) Phase constant c) Permittivity d) Permeability View Answer Answer: a Explanation: The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant. 2. The phase constant of a wave is given by a) ω√(LC) b) ω√(L/C) c) ω√(C/L) d) ω√(1/LC) View Answer Answer: a Explanation: The phase constant of a wave in a transmission line is given by β = ω√(LC), where L and C are the specifications of the line. 3. The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is a) 4.5 GHz

b) 5 GHz c) 5.5 GHz d) 6 GHz View Answer Answer: d Explanation: The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz. 4. The cut off frequency of the TE01 mode will be a) mc/2a b) mc/2b c) nc/2a d) nc/2b View Answer Answer: d Explanation: The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light. 5. The condition which will satisfy the dimensions of the waveguide is a) a = b b) a > b c) a < b d) ab = 0 View Answer Answer: b Explanation: The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true. 6. The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is a) 0.1 b) 1 c) 10 d) 0.01 View Answer Answer: a Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units. 7. The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is a) 1 b) 2

c) 3 d) 4 View Answer Answer: b Explanation: The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm. 8. The sin θ in the waveguide refers to the ratio of the a) Frequency to wavelength b) Wavelength to frequency c) Cut off frequency to frequency d) Frequency to cut off frequency View Answer Answer: c Explanation: The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide. 9. Is the transmission of a frequency 5 GHz possible in waveguides? a) Yes b) No View Answer Answer: a Explanation: The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide. 10. The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is a) 2 b) 4 c) 6 d) 8 View Answer Answer: b Explanation: The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units. 11. The waveguides are used in a transmission line for a) Increasing transmission coefficient b) Increasing reflection coefficient c) Decreasing transmission coefficient d) Decreasing reflection coefficient View Answer

Answer: a Explanation: The waveguides are used to increase the transmission efficiency of the waves travelling through it. 12. The attenuation coefficient of the wave having a resistance of 15 ohm in a 50 ohm line is a) 15 b) 1.5 c) 0.15 d) 0.015 View Answer Answer: c Explanation: The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Zo. On substituting for R = 15 and Zo = 50, we get α = 15/(2 x 50) = 0.15 units.

Electromagnetic Theory Problems Sanfoundry by staff10 This set of Electromagnetic Theory Problems focuses on “Transverse Electric Waves(TE)”. 1. In transverse electric waves, which of the following is true? a) E is parallel to H b) E is parallel to wave direction c) E is transverse to wave direction d) H is transverse to wave direction View Answer Answer: c Explanation: In TE waves, the electric field strength will be transverse to the wave direction. Thus the TE waves are also called H waves. 2. The dominant mode in rectangular waveguide is a) TE01 b) TE10 c) TM01 d) TM10 View Answer Answer: b Explanation: TE10 is the dominant mode in the rectangular waveguide. This is because it gives the minimum cut off frequency required for transmission.

3. The number of modes in a waveguide having a V number of 10 is a) 10 b) 25 c) 100 d) 50 View Answer Answer: d Explanation: The number of modes is given by m = V2/2, where V is the v number. On substituting for V = 10, we get m = 100/2 = 50. 4. Does the waveguide with dimensions 3 cm x 5.5 cm exist? a) Yes b) No View Answer Answer: b Explanation: For a waveguide, the dimension a should be greater than b. Here a = 3 and b = 5.5, thus such waveguide does not exist. 5. The mode which has the highest wavelength is called a) Dominant mode b) Evanescent mode c) Generate mode d) Degenerate mode View Answer Answer: a Explanation: Dominant modes are the modes having least cut off frequency. This implies they have highest cut off wavelength. 6. The intrinsic impedance of a TE wave having a cut off frequency of 6 GHz at a frequency of 7.5 GHz in air is a) 628.33 b) 338.62 c) 498.76 d) 342.24 View Answer Answer: a Explanation: The intrinsic impedance of a TE wave is given by ηTE = η/cos θ, where cos θ is given by √(1- (fc/f)2). On substituting for fc = 6 GHz, f = 7.5 GHz and η = 377, we get the intrinsic impedance as 628.33 units. 7. The cut off frequency of a rectangular waveguide of dimensions 3 cm x 1.5 cm is a) 12 GHz

b) 6 GHz c) 4 GHz d) 5 GHz View Answer Answer: d Explanation: The cut off frequency in dominant mode will be fc = mc/2a. On substituting for c = 3 x 108 and a = 0.03, we get the cut off frequency as 5 GHz. 8. The propagation constant for a lossless transmission line will be a) Real b) Complex c) Real and equal to phase constant d) Complex and equal to phase constant View Answer Answer: d Explanation: The propagation constant is given by γ = α + jβ, where α and β are the attenuation and phase constants respectively. For a lossless line, the attenuation constant is zero. Thus γ = jβ. It is clear that γ is complex and equal to β. 9. The attenuation of a 50 ohm transmission line having a resistance of 100 ohm is a) 0.01 b) 0.1 c) 1 d) 10 View Answer Answer: c Explanation: The attenuation of a wave is given by α = R/2Z0. On substituting for R = 100 and Z0 = 50, we get α = 100/(2 x 50) = 1 unit. 10. The cut off frequency of a TE wave with waveguide dimension of a= 3.5 cm in a medium of permittivity 2.2 is a) 2.88 GHz b) 3.32 GHz c) 4.5 GHz d) 2.12 GHz View Answer Answer: a Explanation: The cut off frequency of a TE wave in any other medium is mc/2a√εr. On substituting for a = 0.035 and εr = 2.2, we get the cut off frequency as 2.88 GHz. 11. The phase constant and frequency are related by a) Phase constant α ω

b) Phase constant α 1/ω c) Phase constant α 1/2ω d) Phase constant α ω/2 View Answer Answer: a Explanation: The phase constant is given by β = ω√LC. Thus the relation is β is directly proportional to ω. 12. Which of the following parameter is non zero for a lossless line? a) Attenuation b) Resistance c) Conductance d) Phase constant View Answer Answer: d Explanation: The attenuation constant, resistance and conductance are zero for a lossless line. Only the phase constant is non zero.

Transverse Magnetic Waves Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Transverse Magnetic Waves”. 1. In transverse magnetic waves, which of the following is true? a) E is parallel to H b) H is parallel to wave direction c) H is transverse to wave direction d) E is transverse to H View Answer Answer: c Explanation: In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves. 2. The dominant mode in the TM waves is a) TM01 b) TM10

c) TM20 d) TM11 View Answer Answer: d Explanation: The modes TM10, TM01 and TM20 does not exist in any waveguide. The TM11 mode is the dominant mode in the waveguide. 3. The modes in a waveguide having a V number of 20 is a) 400 b) 200 c) 100 d) 40 View Answer Answer: b Explanation: The number of modes in a waveguide is given by m = V2/2. On substituting for V = 20, we get m = 400/2 = 200 modes. 4. The v number of a waveguide having 120 modes is a) 15.5 b) 18 c) 24 d) 12 View Answer Answer: a Explanation: The number of modes in a waveguide is given by m = V2/2. On substituting for m = 120, we get V = √(2 x 120) = 15.5. 5. The intrinsic impedance of a TM wave will be a) Greater than 377 ohm b) Equal to 377 ohm c) Lesser than 377 ohm d) Does not exist View Answer Answer: c Explanation: The intrinsic impedance of the transverse magnetic wave is given by ηTM = η √(1(fc/f)2). Here the term √(1-(fc/f)2) is always lesser than unity. Thus the intrinsic impedance of the TM wave is lesser than 377 ohms. 6. The modes TMmo and TMon are called a) Generate modes b) Degenerate modes c) Dominant modes

d) Evanescent modes View Answer Answer: d Explanation: The modes TMmo and TMon does not exist. These modes are said to be evanescent mode. 7. Two modes with same cut off frequency are said to be a) Generate modes b) Dominant modes c) Degenerate modes d) Regenerate modes View Answer Answer: c Explanation: Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution. 8. Does the mode TM30 exists? a) Yes b) No View Answer Answer: b Explanation: Modes in the format of TMmo and TMon does not exist. The given mode is in the form of TMmo, which is does not exist. It is an evanescent mode. 9. The boundary between the Fresnel and Fraunhofer zones having a length of 12 units and a wavelength of 3 units is a) 96 b) 48 c) 192 d) 36 View Answer Answer: a Explanation: The Fresnel- Fraunhofer boundary is related by the wavelength as R = 2L2/λ. On substituting for L = 12 and λ = 3, we get R = 2 x 122/3 = 96 units. 10. The reflection coefficient, when a resonant cavity is placed between the waveguide is a) 0 b) 1 c) -1 d) Infinite View Answer

Answer: b Explanation: When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves. 11. The distance between two terminated plates is given by the a) Guided wavelength b) 2(guided wavelength) c) Guided wavelength/2 d) (Guided wavelength)/4 View Answer Answer: c Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. 12. Find the guided wavelength if the distance between the two conducting plates in the waveguide is 2 cm. a) 4cm b) 2cm c) 1cm d) 0.5cm View Answer Answer: a Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. On substituting for Vmin = 2cm, we get λg = 2Vmin = 2 x 0.02 = 4cm. Electromagnetic Theory Questions and Answers – Transverse Electric Magnetic Waves(TEM) This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Transverse Electric Magnetic Waves”. 1. In a transverse electric magnetic wave, which of the following will be true? a) E is transverse to H b) E is transverse to wave direction c) H is transverse to wave direction d) E and H are transverse to wave direction View Answer Answer: d Explanation: In the transverse electric magnetic wave (TEM wave), both the electric and magnetic field strengths are transverse to the wave propagation. 2. The cut off frequency of the TEM wave is a) 0 b) 1 GHz c) 6 GHz

d) infinity View Answer Answer: a Explanation: The TEM waves have both E and H perpendicular to the guide axis. Thus its cut off frequency is zero. 3. Which component is non zero in a TEM wave? a) Ex b) Hz c) Ez d) Attenuation constant View Answer Answer: a Explanation: In a TEM wave, the wave propagates along the guided axis. Thus the components Ez and Hz are zero. The attenuation is also zero. The non-zero component will be Ex. 4. TEM wave can propagate in rectangular waveguides. State true/false. a) True b) False View Answer Answer: b Explanation: The rectangular waveguide does not allow the TEM wave. TEM mode can exist only in two conductor system and not in hollow waveguide in which the centre conductor does not exist. 5. The cut off wavelength in the TEM wave will be a) 0 b) Negative c) Infinity d) 1/6 GHz View Answer Answer: c Explanation: The cut off frequency in a TEM wave is zero. Thus the cut off wavelength will be infinity. 6. The guided wavelength of a TEM wave in a waveguide having a wavelength of 5 units is a) 0 b) Infinity c) 5 d) 1/5 View Answer Answer: c Explanation: The guided wavelength is same as the wavelength of the waveguide with a TEM wave. Thus the guided wavelength is 5 units.

7. The guided phase constant of a TEM wave in a waveguide with a phase constant of 2.8 units is a) 2.8 b) 1.4 c) 0 d) Infinity View Answer Answer: a Explanation: The guided phase constant is same as the phase constant of the waveguide. For the given data, the guided phase constant is 2.8 units. 8. Which type of transmission line accepts the TEM wave? a) Copper cables b) Coaxial cable c) Rectangular waveguides d) Circular waveguides View Answer Answer: b Explanation: Hollow transmission lines TE and TM waves only. The TEM wave is possible only in the coaxial cable transmission line, which is not hollow. 9. For a TEM wave to propagate in a medium, the medium has to be a) Air b) Insulator c) Dispersive d) Non dispersive View Answer Answer: d Explanation: The medium in which the TEM waves propagate has to be non- dispersive. This implies the phase velocity and the characteristic impedance has to be constant over a wide band. 10. Stripline and parallel plate waveguides the TEM wave. State true/false. a) True b) False View Answer Answer: a Explanation: The stripline and parallel plate waveguides are not hollow and the dielectric is lossless. The medium is non dispersive. Thus the statement is true.

Phase & Group Velocity - Electromagnetic Theory Questions and Answers - Sanfoundry by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Phase and Group Velocity”. 1. In a waveguide, which of the following condition is true always? a) phase velocity = c b) group velocity = c c) phase velocity > c d) phase velocity < c View Answer Answer: c Explanation: The phase velocity is always greater than the speed of light in waveguides. This implies the group velocity is small. 2. The term cos θ is given by 2.5. Find the phase velocity. a) 3 b) 5 c) 7.5 d) 2.5 View Answer Answer: c Explanation: The phase velocity is given by Vp = c cos θ. On substituting for cos θ = 2.5 and the speed of light, we get the phase velocity as 7.5 x 108 m/s. 3. The cut off wavelength and the guided wavelength are given by 0.5 and 2 units respectively. Find the wavelength of the wave. a) 0.48 b) 0.32 c) 0.45 d) 0.54 View Answer Answer: a Explanation: The cut off wavelength and the guided wavelength are related as (1/λ)2 = (1/λc)2 + (1/λg)2. On substituting for λc = 0.5 and λg = 2, we get λ = 0.48 units. 4. The cut off wavelength of the rectangular waveguide in dominant mode with dimensions 6 cm x 4 cm is a) 12cm b) 6cm c) 4cm d) 2cm View Answer

Answer: a Explanation: The cut off wavelength in the dominant mode is given by λc = 2a/m, where a is the broad wall dimension. On substituting for m = 1 and a = 6cm, we get the cut off wavelength as 12cm. 5. The product of the phase and the group velocities is given by the a) Speed of light b) Speed of light/2 c) 2 x Speed of light d) (speed of light)/4 View Answer Answer: d Explanation: The product of the phase and the group velocities is given by the square of the speed of the light. Thus Vp x Vg = c2 is the relation. 6. The phase velocity of a wave having a group velocity of 6 x 106 is (in order of 108 m/s) a) 2.4 b) 3 c) 15 d) 150 View Answer Answer: d Explanation: We know that the phase and the group velocities are given by Vp x Vg = c2. On substituting for Vg = 6 x 106 and the speed of light, we get Vp = 150 x 108 m/s. 7. The group velocity of a wave with a phase velocity of 60 x 109 is (in 106 order) a) 1.5 b) 2 c) 2.5 d) 3 View Answer Answer: a Explanation: We know that the phase and the group velocities are given by Vp x Vg = c2. On substituting for Vp = 60 x 109 and the speed of light, we get Vg = 1.5 x 106 m/s. 8. The phase velocity of a wave having a phase constant of 4 units and a frequency of 2.5 x 109 radian/sec is (in 108 order) a) 3.25 b) 3.75 c) 6.25 d) 6.75 View Answer

Answer: c Explanation: The phase velocity and the phase constant are related by Vp = ω/βg. On substituting for ω = 2.5 x 109 and β = 4, we get the phase velocity as 6.25 units. 9. The guided wavelength and the phase constant are related by a) 2π/βg = λg b) 1/βg = λg c) 1/2πβg = λg d) βg = λg View Answer Answer: a Explanation: The guided wavelength and the phase constant are related by 2π/βg = λg, where βg is the guided phase constant and λg is the guided wavelength. 10. The phase velocity refers to a group of waves and the group velocity refers to a single wave. State true/false. a) True b) False View Answer Answer: b Explanation: The phase velocity refers to a single wave and the group velocity refers to a group of waves. 11. The phase and group velocities does not depend on which of the following? a) Frequency b) Wavelength c) Phase constant d) Attenuation constant View Answer Answer: d Explanation: The phase and the group velocities are directly related by the frequency, wavelength and the phase constant. It is independent of the attenuation constant. 12. The distance between two successive points in a waveguide is the a) Guided wavelength b) 2 x guided wavelength c) Guided wavelength/2 d) (guided wavelength)/4 View Answer Answer: c Explanation: The distance between two successive points in a waveguide is equal to half of the guided wavelength.

Basic Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Basic Electromagnetic Theory Questions and Answers focuses on “Waveguide Current and Excitation”. 1. The source voltage of a 75ohm transmission line is given by 150V. Find the load current. a) 0.5 b) 2 c) 4 d) 1 View Answer Answer: b Explanation: The load current is given by IL = VS/Z0. On substituting for VS = 150 and Z0 = 75, we get IL = 150/75= 2A. 2. The guided terminations are used to a) Increase reflection b) Increase transmission c) Eliminate reflection loss d) Eliminate attenuation View Answer Answer: c Explanation: The guided termination refers to the waveguide shorted by conducting plates. This is done in order to eliminate the reflection losses. 3. Which type of wave does the resonant cavity produce? a) Standing waves b) Guided waves c) Transmitted waves d) Attenuated waves View Answer Answer: a Explanation: Resonant cavity is the waveguide shorted by a conducting plate. This is to reduce the reflection losses. Such arrangement leads to standing waves. 4. Which of the following parameter cannot be calculated from the standing waves? a) Peak voltage and peak current b) SWR

c) Reflection and transmission coefficients d) Attenuation constant View Answer View AnswerAnswer: d Explanation: The peak voltage and current can be directly measured from the standing waves. The standing wave ratio, reflection coefficient and the transmission coefficient can also be calculated from it. Only the attenuation constant cannot be calculated directly. 5. For efficient transmission, the characteristic impedance of the transmission line has to be a) 50 ohm b) 75 ohm c) Either 50 or 75 ohm d) Neither 50 nor 75 ohm View Answer Answer: c Explanation: Generally, for ideal transmission lines, the characteristic impedance should be either 50 ohm or 75 ohm. 6. The cavity resonators are used in the klystron amplifiers for a) Amplifying RF signals b) Amplifying microwave signals c) Attenuating RF signals d) Attenuating microwave signals View Answer Answer: b Explanation: The cavity resonators are employed in the klystron amplifiers for amplifying the microwave signals. 7. The cavity resonators used in the reflex klystron oscillators are for a) Generating RF signals b) Generating microwave signals c) Amplifying RF signals d) Amplifying microwave signals View Answer Answer: b Explanation: Oscillators are devices that generate signal waveforms. The reflex klystron oscillator is used to generate microwave signals. 8. One of the applications of the cavity resonators is duplexer in RADAR systems. State true/false. a) True

b) False View Answer Answer: a Explanation: Cavity resonator is used in duplexers of RADAR systems, as resonant cavity in transmit receive (TR) tubes and antitransmit receive (ART) tubes. 9. Cavity wave meters are used to measure which parameter of the wave? a) Wavelength b) Reflection factor c) Phase d) Frequency View Answer Answer: d Explanation: Cavity resonators are used in cavity wave meters for the measurement of frequency of the microwave signals. 10. The waveguides are terminated by the procedure of a) Trimming b) Polishing c) Tapering d) Pruning View Answer Answer: c Explanation: A waveguide is terminated by concept of tapered or exponential line and uses a dielectric having considerable conductivity to provide power absorbing properties. This will eliminate the reflection losses.

Transients - Electromagnetic Theory Questions and Answers - Sanfoundry by staff10 This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Transients”. 1. The resonant circuit in a waveguide refers to the a) Tank circuit b) RC circuit c) Bridge circuit d) Attenuator circuit View Answer

Answer: a Explanation: The resonant circuit refers to the tank circuit. It is parallel combination of an inductor and a capacitor. 2. When a waveguide is terminated, the mode of the guided termination will be a) zero b) non-zero c) infinite d) does not exist View Answer Answer: b Explanation: A waveguide mode with a guided termination is represented by TEmnp and TMmnp. Here m and n are the orders of the waveguide and p is the order of the resonant cavity. It is always a non-zero value. 3. The cut off frequency of a waveguide with resonant cavity is given by a) V√((m/a)2 + (n/b)2 + (p/d)2)/2 b) V√((m/a)2 + (n/b)2 + (p/d)2) c) 2V√((m/a)2 + (n/b)2 + (p/d)2) d) V√((m/a) + (n/b) + (p/d))/2 View Answer Answer: a Explanation: The cut off frequency of the waveguide of dimensions a x b with the resonant cavity of dimension d is given by fc = √((m/a)2 + (n/b)2 + (p/d)2)/2. Here m and n are the orders of the waveguide, p is the order of the cavity and v is the velocity. 4. The power of a wave in a transmission line, when the current and the resistance are 5A and 120 ohm respectively is a) 3000 b) 4000 c) 2000 d) 1500 View Answer Answer: a Explanation: The power of a line is given by P = I2R. On substituting for I = 5 and R = 120, we get P = 52 x 120 = 3000 units. 5. The power of a wave having a magnetic field intensity of 2.5 units is a) 1.17 kilo watt b) 1.17 mega watt c) 1.17 watt d) 11.7 watt View Answer

Answer: a Explanation: The power of a wave is given by P = 0.5 η H2, where η is the intrinsic impedance and H is the magnetic field intensity. On substituting for η = 377 and H = 2.5, we get P = 0.5 x 377 x 2.52 = 1.17 kilo watts. 6. The power of a wave having an electric field strength of 12.8 units is a) 0.217 b) 0.721 c) 0.127 d) 0.172 View Answer Answer: a Explanation: The power of a wave is given by P = 0.5 E2/η, where E is the electric field intensity and η is the intrinsic impedance. On substituting for E = 12.8 and η = 377, we get P = 0.5 x 12.82/377 = 0.217 units. 7. The form or mode of propagation is determined by which factors? a) Type of excitation device b) Location of excitation device c) Type and location of the excitation device d) Waveguide characteristics View Answer Answer: c Explanation: The form and the mode of propagation of the wave in a waveguide is determined by the type and location of the excitation device. 8. The phase of the wave after the installation of the guided terminations will be a) 0 b) 45 c) 90 d) 180 View Answer Answer: a Explanation: Using a guided termination, the guides serves as a reflector. If the distance between the exciter and the wall is properly adjusted, the transmitted and the reflected wave will be in phase. 9. The exciter of the waveguide in a transmission line is the a) Transmitter b) Receiver c) Transponder d) Antenna View Answer

Answer: d Explanation: The waveguide is usually excited by the antenna rod. The reflection depends on the phase of excitation and the antenna current. 10. A waveguide imitates which type of filter characteristics? a) Low filter b) High filter c) Band filter d) Band reject filter View Answer Answer: b Explanation: A waveguide es only high frequency waves (in GHz range) and attenuates low frequencies. This is the characteristic of a high filter.