Q.1 What is the dominant mode of propagation in a rectangular waveguide with dimensions a > b?
TE10
TE01
TM11
TEM
Explanation - In a rectangular waveguide the TE10 mode has the lowest cutoff frequency when a > b, making it the dominant (first) mode.
Correct answer is: TE10
Q.2 The cutoff frequency of the TE10 mode in a rectangular waveguide of width a is given by:
c / (2a)
c / a
c / (2b)
c / (a + b)
Explanation - For TE_mn modes, f_c = (c/2) * sqrt[(m/a)^2 + (n/b)^2]; for TE10, m=1, n=0, so f_c = c/(2a).
Correct answer is: c / (2a)
Q.3 Which of the following waveguide modes can support a longitudinal electric field component?
TM modes
TE modes
TEM modes
All of the above
Explanation - TM (Transverse Magnetic) modes have no longitudinal magnetic field, but possess a longitudinal electric field component.
Correct answer is: TM modes
Q.4 In a circular waveguide, the dominant mode is:
TE11
TM01
TE01
TM11
Explanation - The TE11 mode has the lowest cutoff frequency in a circular waveguide, thus it is the dominant mode.
Correct answer is: TE11
Q.5 The phase velocity in a waveguide is always:
Greater than the speed of light in vacuum
Less than the speed of light in vacuum
Equal to the speed of light in vacuum
Dependent on the material filling the waveguide
Explanation - Phase velocity v_p = ω/β = c / sqrt(1 - (f_c/f)^2) > c for frequencies above cutoff.
Correct answer is: Greater than the speed of light in vacuum
Q.6 Group velocity in a lossless waveguide is:
Less than the speed of light in vacuum
Greater than the speed of light in vacuum
Equal to the phase velocity
Zero at cutoff frequency
Explanation - Group velocity v_g = c sqrt(1 - (f_c/f)^2) and is always less than c for f > f_c.
Correct answer is: Less than the speed of light in vacuum
Q.7 Which of the following statements is true about TEM mode in a waveguide?
TEM mode cannot exist in a single-conductor hollow waveguide
TEM mode is the dominant mode in rectangular waveguides
TEM mode has a cutoff frequency of zero
Both A and C
Explanation - TEM mode requires two conductors to support electric field lines; it also has zero cutoff frequency, but cannot exist in a hollow waveguide.
Correct answer is: Both A and C
Q.8 The quality factor (Q) of a resonant cavity is defined as:
2π × (Stored Energy) / (Energy Lost per Cycle)
Frequency / Bandwidth
Both A and B are equivalent
None of the above
Explanation - Q = ω₀ * (Stored Energy) / (Power Loss) = f₀ / Δf; both definitions are equivalent.
Correct answer is: Both A and B are equivalent
Q.9 In a rectangular cavity resonator, the resonant frequency for the TE₁₀₁ mode is determined by:
Both the length and height of the cavity
Only the width of the cavity
Only the length of the cavity
Only the height of the cavity
Explanation - TE_mnp mode frequency depends on all three dimensions: f = (c/2) sqrt[(m/a)² + (n/b)² + (p/d)²]. For TE₁₀₁, m=1, n=0, p=1.
Correct answer is: Both the length and height of the cavity
Q.10 Which boundary condition is applied to the tangential electric field at the conducting walls of a waveguide?
E_t = 0
H_t = 0
∂E_t/∂n = 0
∂H_t/∂n = 0
Explanation - At a perfect electric conductor, the tangential component of the electric field must vanish.
Correct answer is: E_t = 0
Q.11 For a wave propagating in a waveguide, the relationship between phase velocity (v_p) and group velocity (v_g) is:
v_p * v_g = c²
v_p = v_g
v_p + v_g = c
v_p - v_g = c
Explanation - In lossless waveguides, the product of phase and group velocities equals the square of the speed of light in the medium.
Correct answer is: v_p * v_g = c²
Q.12 Which of the following is NOT a type of waveguide cavity resonator?
Rectangular cavity
Cylindrical cavity
Spherical cavity
Planar microstrip cavity
Explanation - Planar microstrip structures are not classified as 3‑D waveguide cavities; the other three are standard cavity geometries.
Correct answer is: Planar microstrip cavity
Q.13 The dispersion relation for the TE₁₀ mode in a rectangular waveguide is:
β = sqrt[(ω/c)² - (π/a)²]
β = sqrt[(ω/c)² + (π/a)²]
β = (ω/c) - (π/a)
β = (ω/c) + (π/a)
Explanation - For TE₁₀, the propagation constant β = sqrt(k² - (π/a)²) where k = ω/c.
Correct answer is: β = sqrt[(ω/c)² - (π/a)²]
Q.14 In a waveguide, the term "cutoff wavelength" (λ_c) is:
The wavelength at which the mode ceases to propagate
The wavelength at which the waveguide becomes lossless
The wavelength of the dominant mode
The wavelength of the TE mode only
Explanation - When λ > λ_c (or f < f_c), the mode becomes evanescent and does not propagate.
Correct answer is: The wavelength at which the mode ceases to propagate
Q.15 Which of the following statements about a pillbox cavity is correct?
Its fundamental mode is TM₀₁₀
It cannot support TM modes
Its resonant frequency is independent of its radius
It operates only in TE modes
Explanation - The pillbox (cylindrical) cavity's lowest mode is TM₀₁₀, having an axial electric field and no azimuthal variation.
Correct answer is: Its fundamental mode is TM₀₁₀
Q.16 The skin depth in a good conductor at microwave frequencies:
Decreases with increasing frequency
Increases with increasing frequency
Is independent of frequency
Becomes infinite at high frequencies
Explanation - Skin depth δ = sqrt(2/ (ωμσ)); higher frequency ω reduces δ.
Correct answer is: Decreases with increasing frequency
Q.17 A waveguide with a gradual taper is used to:
Match impedances between sections of different size
Increase the operating frequency
Create a standing wave pattern
Eliminate higher order modes
Explanation - A tapered transition provides a smooth impedance transformation, reducing reflections.
Correct answer is: Match impedances between sections of different size
Q.18 In a rectangular waveguide, the electric field distribution of the TE10 mode varies as:
Sin(πx/a) along the broad dimension
Cos(πx/a) along the narrow dimension
Uniform across the cross‑section
Sin(πy/b) along the narrow dimension
Explanation - TE10 has one half-wave variation across the wide wall (x‑direction) and no variation in the y‑direction.
Correct answer is: Sin(πx/a) along the broad dimension
Q.19 The resonant frequency of a rectangular cavity resonator is inversely proportional to:
Its linear dimensions
Its volume
The square root of its volume
The square of its dimensions
Explanation - f ∝ 1/(2) * sqrt[(m/a)² + (n/b)² + (p/d)²]; each dimension appears in the denominator, so increasing any linear dimension reduces f.
Correct answer is: Its linear dimensions
Q.20 Which parameter primarily determines the bandwidth of a waveguide filter?
Coupling aperture size
Waveguide material conductivity
Operating temperature
Waveguide length
Explanation - In coupled‑cavity filters, the aperture (or iris) size controls the coupling coefficient, which directly affects bandwidth.
Correct answer is: Coupling aperture size
Q.21 For a TM₁₁ mode in a circular waveguide, the cutoff frequency depends on:
The radius of the waveguide
The length of the waveguide
The material permittivity only
Both radius and permittivity
Explanation - f_c = (χ₁₁ / (2πa)) * (c/√ε_r), where χ₁₁ is the first root of J₁, a is radius, and ε_r is relative permittivity.
Correct answer is: Both radius and permittivity
Q.22 In a cavity resonator, the term "mode degeneracy" refers to:
Two or more modes having the same resonant frequency
A mode that does not radiate
A mode with zero field inside the cavity
A mode that exists only at cutoff
Explanation - Degeneracy occurs when distinct field patterns share the same eigenfrequency due to symmetry.
Correct answer is: Two or more modes having the same resonant frequency
Q.23 A waveguide filled with a dielectric of relative permittivity ε_r will have its cutoff frequency:
Reduced by a factor of √ε_r
Increased by a factor of √ε_r
Unchanged
Reduced by ε_r
Explanation - Cutoff frequency f_c ∝ 1/√(μ_r ε_r); increasing ε_r lowers f_c by √ε_r.
Correct answer is: Reduced by a factor of √ε_r
Q.24 The dominant mode in a coaxial line is:
TEM
TE11
TM01
TE01
Explanation - Coaxial lines support a TEM mode with no cutoff frequency, making it the dominant mode.
Correct answer is: TEM
Q.25 In a waveguide, the attenuation constant due to conductor losses varies with frequency as:
α_c ∝ √f
α_c ∝ 1/√f
α_c ∝ f
α_c is independent of f
Explanation - Conductor loss α_c ≈ (R_s / 2Z₀) where surface resistance R_s ∝ √f, leading to α_c ∝ √f.
Correct answer is: α_c ∝ √f
Q.26 The field configuration of the TM₁₁ mode in a rectangular waveguide has:
One half‑wave variation in both dimensions
A full wave variation in the broad dimension only
No variation in either dimension
Two half‑wave variations in the narrow dimension only
Explanation - TM₁₁ has m=1, n=1, giving sin(πx/a) and sin(πy/b) variations.
Correct answer is: One half‑wave variation in both dimensions
Q.27 Which of the following cavity shapes is most commonly used in particle accelerators?
Pillbox (cylindrical) cavity
Rectangular cavity
Spherical cavity
Elliptical cavity
Explanation - Elliptical cavities provide higher shunt impedance and better field flatness, making them standard in modern accelerators.
Correct answer is: Elliptical cavity
Q.28 The shunt impedance (R_sh) of a cavity resonator is defined as:
V² / P_loss
P_loss / V²
V / I
Q × ω₀ × L
Explanation - Shunt impedance quantifies how effectively the cavity converts RF power into accelerating voltage.
Correct answer is: V² / P_loss
Q.29 In a waveguide, the term "evanescent mode" refers to:
A mode with frequency below cutoff that decays exponentially
A mode with zero phase velocity
A mode that propagates with no attenuation
A mode that only exists in dielectric‑filled guides
Explanation - Below cutoff, the propagation constant becomes imaginary, leading to exponential decay along the guide.
Correct answer is: A mode with frequency below cutoff that decays exponentially
Q.30 A waveguide bend introduces which type of loss?
Radiation loss due to mode conversion
Conductor loss only
Dielectric loss only
No loss if the bend radius is large enough
Explanation - Bends cause part of the power to couple into higher order modes and radiate, creating loss.
Correct answer is: Radiation loss due to mode conversion
Q.31 For a resonant cavity, the bandwidth (Δf) is related to its Q factor by:
Δf = f₀ / Q
Δf = Q / f₀
Δf = f₀ × Q
Δf = √(f₀ / Q)
Explanation - By definition, Q = f₀ / Δf, therefore Δf = f₀ / Q.
Correct answer is: Δf = f₀ / Q
Q.32 Which waveguide mode has the highest cutoff frequency for a given waveguide geometry?
The mode with the highest indices (m, n)
TE10
TM01
TEM
Explanation - Cutoff frequency increases with increasing m and n values, so the highest-index mode has the highest f_c.
Correct answer is: The mode with the highest indices (m, n)
Q.33 In a dielectric‑filled waveguide, the phase velocity can be:
Less than c
Exactly equal to c
Greater than c
Both A and C depending on frequency
Explanation - Below cutoff, v_p > c; above cutoff in a dielectric, v_p = c/√ε_r, which can be less than c if ε_r > 1.
Correct answer is: Both A and C depending on frequency
Q.34 The method of moments (MoM) is commonly used to compute:
Fields in complex waveguide structures
Thermal conductivity of waveguide walls
Mechanical stress in waveguide supports
Propagation speed in free space
Explanation - MoM is a numerical technique for solving integral equations for electromagnetic fields, widely used for waveguide analysis.
Correct answer is: Fields in complex waveguide structures
Q.35 A waveguide filter designed with three coupled resonators will typically exhibit:
A 3‑pole response
A 2‑pole response
A 1‑pole response
No filtering effect
Explanation - Each resonator contributes one pole; three coupled resonators give a third‑order (3‑pole) filter.
Correct answer is: A 3‑pole response
Q.36 In a rectangular waveguide, the dominant mode’s electric field has its maximum at:
The center of the broad wall
The walls of the waveguide
The narrow wall
Uniform across the cross‑section
Explanation - For TE10, the sin(πx/a) variation peaks at x = a/2 (center of the broad wall).
Correct answer is: The center of the broad wall
Q.37 The term "standing wave ratio" (SWR) in a waveguide is a measure of:
Impedance mismatch between the guide and load
Propagation speed of the wave
Dielectric loss tangent
Physical dimensions of the guide
Explanation - SWR = (1 + |Γ|) / (1 - |Γ|), where Γ is the reflection coefficient caused by mismatch.
Correct answer is: Impedance mismatch between the guide and load
Q.38 Which of the following is true about the TM₀₁₀ mode in a cylindrical cavity?
It has an electric field only along the axis
It has a magnetic field only along the axis
Both electric and magnetic fields are transverse
It cannot exist in a perfect conductor
Explanation - TM₀₁₀ features an axial electric field with no azimuthal variation; magnetic field is circumferential.
Correct answer is: It has an electric field only along the axis
Q.39 A waveguide with dimensions a = 2 cm and b = 1 cm operates at 10 GHz. Which mode is most likely to propagate?
TE10
TE01
TM11
TE20
Explanation - Cutoff for TE10: f_c = c/(2a) ≈ 7.5 GHz. Since 10 GHz > 7.5 GHz, TE10 propagates. TE01 cutoff is higher (≈15 GHz).
Correct answer is: TE10
Q.40 In a waveguide, the term "dispersion" refers to:
Frequency‑dependence of phase velocity
Loss of power due to skin effect
Radiation from bends
Mode conversion at discontinuities
Explanation - Dispersion describes how different frequencies travel at different phase velocities, causing pulse spreading.
Correct answer is: Frequency‑dependence of phase velocity
Q.41 The first zero of the Bessel function J₁(x) is approximately:
3.832
2.405
1.841
5.331
Explanation - The first root of J₁(x) used for TM₁₁ cutoff in circular waveguides is 3.832.
Correct answer is: 3.832
Q.42 A cavity resonator is said to be "over‑coupled" when:
External coupling loss exceeds internal loss
Internal loss exceeds external coupling loss
Both losses are equal
There is no external coupling
Explanation - Over‑coupling occurs when the power extracted through the coupler is larger than the intrinsic dissipative loss.
Correct answer is: External coupling loss exceeds internal loss
Q.43 In a waveguide filter, increasing the iris size generally:
Increases bandwidth
Decreases bandwidth
Has no effect on bandwidth
Turns the filter into an amplifier
Explanation - Larger irises increase coupling between resonators, widening the passband.
Correct answer is: Increases bandwidth
Q.44 The field inside a perfectly conducting rectangular waveguide is:
Zero at the walls
Maximum at the walls
Uniform everywhere
Only magnetic fields exist
Explanation - Tangential electric fields are forced to zero at perfect conductor boundaries.
Correct answer is: Zero at the walls
Q.45 Which material property predominantly determines dielectric loss in a waveguide?
Loss tangent (tan δ)
Conductivity
Permeability
Thermal expansion coefficient
Explanation - Dielectric loss is proportional to tan δ, representing the imaginary part of permittivity.
Correct answer is: Loss tangent (tan δ)
Q.46 A waveguide operating well above its cutoff frequency will exhibit:
Low dispersion
High dispersion
No propagation
Mode conversion
Explanation - As f ≫ f_c, the term (f_c/f)² becomes small, making β ≈ k and reducing dispersion.
Correct answer is: Low dispersion
Q.47 In a rectangular cavity, the mode TE₁₀₁ has field variations along which axes?
x‑ and z‑axes
y‑ and z‑axes
Only x‑axis
Only z‑axis
Explanation - Indices (1,0,1) indicate half‑wave variation in x and z directions, none in y.
Correct answer is: x‑ and z‑axes
Q.48 The characteristic impedance of a rectangular waveguide in the TE₁₀ mode is:
Z_TE = (η₀ / √(1 - (f_c/f)²))
Z_TE = η₀ √(1 - (f_c/f)²)
Z_TE = η₀ (f/f_c)
Z_TE = η₀ (f_c/f)
Explanation - For TE modes, Z_TE = η₀ / √(1 - (f_c/f)²), where η₀ is the intrinsic impedance of free space.
Correct answer is: Z_TE = (η₀ / √(1 - (f_c/f)²))
Q.49 Which phenomenon enables coupling of energy into a cavity through a small aperture?
Magnetic coupling (loop)
Electric coupling (probe)
Both A and B
Neither; energy cannot couple through a small aperture
Explanation - A loop couples magnetically, while a probe (electric) couples electrically; both are used depending on mode orientation.
Correct answer is: Both A and B
Q.50 If the quality factor Q of a cavity is 10,000 at 2 GHz, what is its -3 dB bandwidth?
200 kHz
20 kHz
2 kHz
200 Hz
Explanation - Δf = f₀ / Q = 2 × 10⁹ Hz / 10⁴ = 2 × 10⁵ Hz = 200 kHz.
Correct answer is: 200 kHz
Q.51 The term "higher order mode" (HOM) in accelerator cavities refers to:
Any mode other than the fundamental accelerating mode
Only TE modes
Only TM modes with n > 0
Modes below cutoff
Explanation - HOMs are undesired resonances that can affect beam stability; they include both TE and TM modes above the fundamental.
Correct answer is: Any mode other than the fundamental accelerating mode
Q.52 A waveguide with a rectangular cross‑section can be transformed into a circular one using:
A mode converter
A dielectric lens
A ferrite rod
A step‑change in length
Explanation - Mode converters (e.g., using irises or probes) can transition fields between rectangular and circular guides.
Correct answer is: A mode converter
Q.53 The power handling capability of a waveguide is primarily limited by:
Breakdown electric field of the interior medium
Length of the waveguide
Operating temperature
Number of modes supported
Explanation - When the electric field exceeds the breakdown strength (e.g., of air), arcing occurs, limiting power.
Correct answer is: Breakdown electric field of the interior medium
Q.54 In the context of waveguides, the term "iris" usually denotes:
A thin conducting plate with an opening used for coupling
A reflective coating on the interior wall
A dielectric slab placed inside the guide
A temperature sensor
Explanation - Iris plates introduce controlled apertures to couple resonators or adjust impedance.
Correct answer is: A thin conducting plate with an opening used for coupling
Q.55 For a given waveguide, increasing the operating frequency well above cutoff tends to:
Reduce the waveguide’s attenuation per unit length
Increase the waveguide’s attenuation per unit length
Leave attenuation unchanged
Cause the wave to become evanescent
Explanation - Conductor loss per unit length decreases as frequency moves farther from cutoff because fields become more confined.
Correct answer is: Reduce the waveguide’s attenuation per unit length
Q.56 A rectangular waveguide with dimensions a = 2.28 cm and b = 1.02 cm is used at 9 GHz. What is the approximate wavelength inside the guide for the TE10 mode?
≈ 30 mm
≈ 20 mm
≈ 10 mm
≈ 5 mm
Explanation - Free‑space λ = c/f ≈ 33 mm. Inside the guide λ_g = λ / √(1 - (λ/λ_c)²); λ_c = 2a = 45.6 mm, so λ_g ≈ 30 mm.
Correct answer is: ≈ 30 mm
Q.57 Which of the following is a common method to suppress unwanted higher‑order modes in a cavity?
Installing HOM couplers/dampers
Increasing the cavity temperature
Enlarging the cavity volume
Reducing the operating frequency
Explanation - HOM couplers extract energy from undesired modes, reducing their Q and preventing beam interaction.
Correct answer is: Installing HOM couplers/dampers
Q.58 The stored energy in a cavity resonator is proportional to:
V² / (2 × R_sh)
R_sh / V²
V × I
Q / f₀
Explanation - Stored energy W = V² / (2R_sh) for a resonant cavity, where V is the peak voltage.
Correct answer is: V² / (2 × R_sh)
Q.59 In a waveguide, the term "cut‑off wave number" (k_c) is related to cutoff frequency (f_c) by:
k_c = 2πf_c / c
k_c = f_c / c
k_c = √(μ₀ε₀) f_c
k_c = (π/a) + (π/b)
Explanation - Wave number k = 2π/λ; at cutoff λ_c = c/f_c, so k_c = 2πf_c / c.
Correct answer is: k_c = 2πf_c / c
Q.60 The term "waveguide dispersion diagram" typically plots:
Frequency versus propagation constant (β)
Voltage versus current
Power versus distance
Loss versus temperature
Explanation - Dispersion curves show f(β) for each mode, illustrating cutoff and propagation characteristics.
Correct answer is: Frequency versus propagation constant (β)
Q.61 A waveguide filled with a ferrite material can be used to achieve:
Non‑reciprocal behavior (isolator)
Zero loss transmission
Superconducting propagation
Frequency doubling
Explanation - Ferrite under magnetic bias introduces gyrotropic properties, enabling isolation or circulator action.
Correct answer is: Non‑reciprocal behavior (isolator)
Q.62 The dominant TE mode in a rectangular waveguide has its magnetic field oriented:
Predominantly along the broad wall (y‑direction)
Along the propagation direction (z‑direction)
Radially outward
Uniform in all directions
Explanation - For TE10, H has a strong y‑component and varies sinusoidally across the x‑dimension.
Correct answer is: Predominantly along the broad wall (y‑direction)
Q.63 In a cavity, the term "detuning" refers to:
Changing the resonant frequency away from the drive frequency
Increasing the Q factor
Reducing the physical size of the cavity
Adding more coupling ports
Explanation - Detuning shifts the cavity resonance, often used to control beam‑cavity interaction.
Correct answer is: Changing the resonant frequency away from the drive frequency
Q.64 The characteristic impedance of a rectangular waveguide in the dominant TE10 mode is approximately:
377 Ω / √(1 - (f_c/f)²)
377 Ω × √(1 - (f_c/f)²)
188.5 Ω
600 Ω
Explanation - Z_TE = η₀ / √(1 - (f_c/f)²), with η₀ ≈ 377 Ω.
Correct answer is: 377 Ω / √(1 - (f_c/f)²)
Q.65 The presence of a dielectric slab of thickness t placed centrally in a rectangular waveguide primarily affects:
The cutoff frequency (lowers it)
The waveguide’s physical dimensions
The magnetic field only
The temperature stability
Explanation - Increasing ε_r reduces the effective wavelength, thereby lowering the cutoff frequency.
Correct answer is: The cutoff frequency (lowers it)
Q.66 Which numerical technique is best suited for modeling complex waveguide junctions with arbitrary geometry?
Finite Element Method (FEM)
Method of Images
Geometric Optics
Ray Tracing
Explanation - FEM can handle arbitrary 3‑D boundaries and material inhomogeneities, making it ideal for waveguide junctions.
Correct answer is: Finite Element Method (FEM)
Q.67 In accelerator physics, the term "shunt impedance per unit length" (R′) is important because:
It determines the accelerating gradient achievable for a given RF power
It defines the mechanical strength of the cavity
It indicates the thermal conductivity of the cavity walls
It measures the magnetic field uniformity
Explanation - Higher R′ means more voltage per unit length for the same input power, increasing acceleration efficiency.
Correct answer is: It determines the accelerating gradient achievable for a given RF power
Q.68 A rectangular waveguide with a 45° bend and radius of curvature R will have additional loss compared to a straight guide. This loss scales approximately as:
(a/R)², where a is the broad wall dimension
(R/a)²
(a·R)
Independent of R
Explanation - Bend loss is proportional to the square of the curvature, i.e., (a/R)².
Correct answer is: (a/R)², where a is the broad wall dimension
Q.69 If a rectangular waveguide operates at its cutoff frequency, the fields inside the guide are:
Purely reactive (standing wave)
Propagating with zero attenuation
Uniform and static
Completely absent
Explanation - At cutoff, the propagation constant β = 0, leading to standing (non‑propagating) fields.
Correct answer is: Purely reactive (standing wave)
Q.70 The term "mode converter" in waveguide technology refers to a device that:
Transforms one guided mode into another (e.g., TE10 to TE01)
Converts RF power into DC power
Amplifies the propagating wave
Filters out all higher order modes
Explanation - Mode converters use discontinuities or irises to couple power between different field configurations.
Correct answer is: Transforms one guided mode into another (e.g., TE10 to TE01)
Q.71 Which of the following best describes the field pattern of the TE01 mode in a circular waveguide?
Azimuthally symmetric magnetic field with a radial electric field
Uniform electric field along the axis
Azimuthal electric field with no radial component
Zero fields everywhere
Explanation - TE01 has no axial electric field; the magnetic field circles the axis while the electric field points radially.
Correct answer is: Azimuthally symmetric magnetic field with a radial electric field
Q.72 A waveguide cavity is said to be "critical coupling" when:
External coupling loss equals internal loss, giving maximum power transfer
External coupling loss is zero
Internal loss is infinite
Both couplings are zero
Explanation - Critical coupling occurs when the loaded Q is half the unloaded Q, optimizing power extraction.
Correct answer is: External coupling loss equals internal loss, giving maximum power transfer
Q.73 The group delay (τ_g) through a waveguide of length L is given by:
τ_g = L / v_g
τ_g = L × v_p
τ_g = v_p / L
τ_g = L / c
Explanation - Group delay is the time taken for the envelope of a pulse to travel the distance L, equal to L divided by group velocity.
Correct answer is: τ_g = L / v_g
Q.74 In a rectangular waveguide, the TE20 mode has a cutoff frequency that is:
Twice that of TE10
Equal to TE10
Half of TE10
Independent of waveguide dimensions
Explanation - f_c(TE20) = 2c/(2a) = 2f_c(TE10) because m = 2 instead of 1.
Correct answer is: Twice that of TE10
Q.75 When a waveguide is operated at a frequency slightly above its cutoff, the phase constant β is:
Small but non‑zero
Zero
Negative
Imaginary
Explanation - Just above cutoff, β = sqrt(k² - k_c²) is real but approaches zero as f → f_c⁺.
Correct answer is: Small but non‑zero
Q.76 A waveguide cavity used for electron acceleration typically operates in which mode?
TM010
TE101
TE111
TM011
Explanation - The TM010 mode provides an axial electric field ideal for accelerating particles.
Correct answer is: TM010
Q.77 Which of the following statements about waveguide losses is correct?
Conductor loss dominates at low frequencies, dielectric loss at high frequencies
Dielectric loss dominates at low frequencies, conductor loss at high frequencies
Both losses are independent of frequency
Losses are negligible in all practical waveguides
Explanation - Skin effect (conductor loss) decreases with higher frequencies, while dielectric loss (∝ f) rises.
Correct answer is: Conductor loss dominates at low frequencies, dielectric loss at high frequencies
Q.78 In a waveguide filter, the term "Chebyshev response" refers to:
A filter with ripple in the passband and a steep roll‑off
A filter with a flat passband and gentle roll‑off
A filter that only passes a single frequency
A filter that converts TE to TM modes
Explanation - Chebyshev filters trade passband ripple for a sharper transition compared to Butterworth.
Correct answer is: A filter with ripple in the passband and a steep roll‑off
Q.79 The term "waveguide iris coupling coefficient" (k) quantifies:
Strength of coupling between adjacent resonators
Thermal conductivity of the iris material
Electric field magnitude at the iris
Mechanical rigidity of the waveguide
Explanation - Coupling coefficient k = (π/2) (Δf / f₀) for narrowband filters, determined by iris aperture.
Correct answer is: Strength of coupling between adjacent resonators
Q.80 A rectangular waveguide terminated with a short circuit will exhibit which type of standing wave pattern for the dominant TE10 mode?
Voltage maximum at the short, current minimum
Current maximum at the short, voltage minimum
Both voltage and current maxima at the short
No standing wave; all energy is absorbed
Explanation - A short forces the tangential electric field to zero, creating a current antinode (maximum) at the termination.
Correct answer is: Current maximum at the short, voltage minimum
Q.81 In the context of waveguide cavities, the term "multipacting" refers to:
Resonant electron multiplication leading to breakdown
Mode conversion between TE and TM
Thermal expansion of the cavity walls
Magnetic field saturation
Explanation - Multipacting is a phenomenon where electrons oscillate and multiply, potentially causing RF breakdown.
Correct answer is: Resonant electron multiplication leading to breakdown
Q.82 The effective relative permittivity of a waveguide partially filled with a dielectric slab is:
Weighted average based on the field distribution
Simply the dielectric's ε_r
Always equal to 1 (free space)
Independent of the slab thickness
Explanation - The fields see a composite medium; the effective ε_eff depends on how much of the mode resides in the dielectric.
Correct answer is: Weighted average based on the field distribution
Q.83 The transverse electric (TE) mode designation indicates:
No electric field component in the direction of propagation
No magnetic field component in the direction of propagation
Both fields are purely transverse
The waveguide is filled with a dielectric
Explanation - In TE modes, E_z = 0, while H_z may be non‑zero.
Correct answer is: No electric field component in the direction of propagation
Q.84 For a rectangular waveguide, the dominant mode TE10 has a cutoff wavelength of:
2a
a
b
2b
Explanation - λ_c = 2a for TE10 (since f_c = c/(2a)).
Correct answer is: 2a
Q.85 When a waveguide is filled with a magnetized ferrite, which property becomes direction‑dependent?
Permeability (μ)
Permittivity (ε)
Conductivity (σ)
Thermal conductivity
Explanation - Ferrites under bias exhibit gyromagnetic behavior, making μ a tensor that varies with direction.
Correct answer is: Permeability (μ)
Q.86 A cavity with a very high Q factor will:
Store energy for many cycles with low loss
Radiate all its energy immediately
Have a very wide bandwidth
Be insensitive to frequency changes
Explanation - High Q implies low fractional energy loss per cycle and narrow bandwidth.
Correct answer is: Store energy for many cycles with low loss
Q.87 In a waveguide, the term "evanescent field" is associated with:
Fields that decay exponentially with distance
Fields that propagate without attenuation
Fields that have constant amplitude
Fields that increase in amplitude
Explanation - Evanescent fields appear when the operating frequency is below cutoff, causing exponential attenuation.
Correct answer is: Fields that decay exponentially with distance
Q.88 A rectangular waveguide is used as a high‑power RF transmission line at 12 GHz. If the waveguide dimensions are a = 2.54 cm, b = 1.27 cm, what is the approximate cutoff frequency for the TE10 mode?
5.9 GHz
7.5 GHz
11.8 GHz
14.9 GHz
Explanation - f_c = c/(2a) ≈ 3×10⁸ m/s / (2×0.0254 m) ≈ 5.9 GHz.
Correct answer is: 5.9 GHz
Q.89 The field distribution of the TM₁₁ mode in a circular waveguide has its first zero at a radius equal to:
Approximately 0.61 times the guide radius
Exactly the guide radius
Half the guide radius
Zero (field is uniform)
Explanation - The first zero of J₁(x) occurs at x ≈ 3.832; for TM₁₁, the field zero radius is r = (3.832 / χ₁₁)·a ≈ 0.61a.
Correct answer is: Approximately 0.61 times the guide radius
Q.90 Which of the following best describes the effect of adding a tuning slug (metal rod) into a cavity resonator?
It changes the resonant frequency by altering the stored electric energy
It increases the Q factor indefinitely
It converts TE modes to TM modes
It eliminates higher order modes completely
Explanation - Inserting a metallic slug changes the effective volume and field distribution, thus shifting the resonant frequency.
Correct answer is: It changes the resonant frequency by altering the stored electric energy
Q.91 In a waveguide filter composed of coupled cavities, increasing the number of cavities generally:
Sharpens the filter skirt (steeper roll‑off)
Widenes the passband
Reduces the filter's Q factor
Eliminates all reflections
Explanation - More poles (cavities) give higher filter order, providing a steeper transition between passband and stopband.
Correct answer is: Sharpens the filter skirt (steeper roll‑off)
Q.92 The characteristic impedance of a coaxial line is given by Z₀ = (60/√ε_r) ln(b/a). Which parameter does NOT affect Z₀?
Length of the line
Inner conductor radius a
Outer conductor radius b
Relative permittivity ε_r
Explanation - Z₀ depends only on geometry (a, b) and dielectric constant, not on physical length.
Correct answer is: Length of the line
Q.93 A rectangular waveguide with a width a = 2 cm is operated at 8 GHz. What is the wavelength inside the guide for the TE10 mode?
≈ 30 mm
≈ 15 mm
≈ 50 mm
≈ 10 mm
Explanation - Free‑space λ = c/f ≈ 37.5 mm. λ_c = 2a = 40 mm. Guided wavelength λ_g = λ / √(1 - (λ/λ_c)²) ≈ 30 mm.
Correct answer is: ≈ 30 mm
Q.94 In accelerator cavities, the term "beam loading" refers to:
Energy extracted from the RF fields by the particle beam
Heating of the cavity walls due to RF loss
Mechanical deformation caused by the beam
Reflection of RF power at the input coupler
Explanation - Beam loading is the reduction in cavity voltage due to power transferred from the fields to the accelerated particles.
Correct answer is: Energy extracted from the RF fields by the particle beam
Q.95 The phase shift per unit length (β) of a TE mode in a rectangular waveguide can be expressed as:
β = √(k² - (π/a)² - (π/b)²)
β = k + (π/a) + (π/b)
β = k / √(1 - (f_c/f)²)
β = (π/a)·(π/b) / k
Explanation - General expression for TE_mn: β = √[k² - (mπ/a)² - (nπ/b)²]; for TE10, n=0.
Correct answer is: β = √(k² - (π/a)² - (π/b)²)
Q.96 Which of the following best describes the effect of surface roughness on waveguide losses?
Increases conductor loss due to higher effective surface resistance
Decreases dielectric loss
Has no effect on losses
Improves power handling capability
Explanation - Rough surfaces increase the effective path for currents, raising the surface resistance and thus losses.
Correct answer is: Increases conductor loss due to higher effective surface resistance
Q.97 A rectangular waveguide is bent by 90° with a radius equal to three times its broad wall dimension (R = 3a). The additional insertion loss due to the bend is approximately:
0.1 dB
0.5 dB
1 dB
Negligible
Explanation - For gentle bends (R ≥ 3a), the loss is small, typically ~0.1 dB for a 90° turn.
Correct answer is: 0.1 dB
Q.98 The term "mode hunting" in waveguide measurements refers to:
The process of locating the resonant frequency of a particular mode
Searching for mechanical defects in the waveguide
Finding the optimal coupling aperture size
Identifying the dielectric constant of the filling material
Explanation - Mode hunting involves sweeping frequency and observing resonance peaks to identify specific modes.
Correct answer is: The process of locating the resonant frequency of a particular mode
Q.99 If a waveguide cavity is over‑coupled, the reflection coefficient at resonance is:
Negative (phase inversion)
Zero
Positive (same phase)
Undefined
Explanation - Over‑coupling leads to a reflection coefficient of magnitude less than one with a 180° phase shift.
Correct answer is: Negative (phase inversion)
Q.100 In a rectangular waveguide, the dominant mode TE10 has a magnetic field maximum at:
The centre of the narrow wall
The centre of the broad wall
The waveguide walls
Uniform across the cross‑section
Explanation - For TE10, H_y peaks at the centre of the narrow dimension (y‑direction).
Correct answer is: The centre of the narrow wall
Q.101 Which of the following statements about the dispersion relation of a waveguide is true?
It is linear for all frequencies above cutoff
It is nonlinear; phase velocity decreases with frequency
It is nonlinear; phase velocity decreases with frequency while group velocity increases
Both phase and group velocities are constant
Explanation - As frequency increases, v_p approaches c from above, while v_g rises from zero toward c, showing opposite trends.
Correct answer is: It is nonlinear; phase velocity decreases with frequency while group velocity increases
Q.102 A waveguide cavity resonator operating at 1.3 GHz with Q = 2×10⁴ stores 1 J of energy. What is the average power dissipated in the cavity?
≈ 65 W
≈ 130 W
≈ 260 W
≈ 520 W
Explanation - P_loss = ωW/Q = (2π·1.3×10⁹)·1 J / 2×10⁴ ≈ 65 W.
Correct answer is: ≈ 65 W
Q.103 Which type of waveguide is commonly used for millimeter‑wave applications (30‑300 GHz) due to low loss?
Rectangular WR‑28 (7.0 mm × 3.5 mm)
Coaxial cable
Twisted pair
Optical fiber
Explanation - WR‑28 waveguide is standard for 26.5‑40 GHz and provides low loss at mm‑wave frequencies.
Correct answer is: Rectangular WR‑28 (7.0 mm × 3.5 mm)
Q.104 When a waveguide is filled with a material of relative permeability μ_r > 1, the cutoff frequency:
Decreases by 1/√μ_r
Increases by √μ_r
Remains unchanged
Becomes zero
Explanation - Cutoff frequency f_c ∝ 1/√(μ_r ε_r); increasing μ_r reduces f_c.
Correct answer is: Decreases by 1/√μ_r
Q.105 A waveguide cavity is said to be "critically coupled" when:
External Q equals internal Q
External Q is much larger than internal Q
External Q is much smaller than internal Q
There is no coupling aperture
Explanation - Critical coupling occurs when the loaded Q is half the unloaded Q, maximizing power transfer.
Correct answer is: External Q equals internal Q
Q.106 In the design of a high‑power waveguide transmission line, the primary limitation on maximum power is:
Electric field breakdown in the guide interior
Mechanical strength of the waveguide walls
Thermal conductivity of the metal
Magnetic saturation of the material
Explanation - When the electric field exceeds the dielectric strength of the medium (usually air), arcing occurs, limiting power.
Correct answer is: Electric field breakdown in the guide interior
Q.107 Which of the following is NOT a typical method to tune the resonant frequency of a cavity?
Adjusting the cavity dimensions mechanically
Changing the temperature of the cavity walls
Varying the magnetic bias in a ferrite‑filled cavity
Altering the external RF source frequency
Explanation - Changing the source frequency does not tune the cavity itself; tuning involves altering cavity properties.
Correct answer is: Altering the external RF source frequency
Q.108 The term "overmoded waveguide" refers to:
A waveguide operating well above the cutoff of many higher order modes
A waveguide with only the dominant mode propagating
A waveguide with a dielectric filling
A waveguide that is physically oversized
Explanation - Overmoded guides support many modes; they are used for low‑loss power transmission at high frequencies.
Correct answer is: A waveguide operating well above the cutoff of many higher order modes
Q.109 In a rectangular waveguide, the TE01 mode has its electric field maximum at:
The centre of the narrow wall
The centre of the broad wall
Both walls equally
Zero everywhere
Explanation - TE01 varies sinusoidally across the narrow dimension (b) and is uniform across the broad dimension (a).
Correct answer is: The centre of the narrow wall
Q.110 When a waveguide cavity is "detuned" away from the beam frequency, the effect on beam stability is:
Reduced beam‑cavity interaction, improving stability
Increased beam‑cavity interaction, causing instability
No effect on the beam
Complete loss of acceleration
Explanation - Detuning moves the resonance away from the beam's frequency, decreasing the coupling and mitigating instabilities.
Correct answer is: Reduced beam‑cavity interaction, improving stability
Q.111 The resonant frequency of a rectangular cavity with dimensions a = b = d is given by:
f = (c/2) √(m² + n² + p²) / a
f = (c/2a) (m + n + p)
f = (c/π) (m/a + n/b + p/d)
f = (c/2) (m/a + n/b + p/d)
Explanation - For a cube (a=b=d), f_mnp = (c/2a) √(m² + n² + p²).
Correct answer is: f = (c/2) √(m² + n² + p²) / a
Q.112 In waveguide theory, the term "scattering parameters" (S‑parameters) are used to describe:
Reflection and transmission characteristics of discontinuities
Thermal conductivity of the waveguide material
Mechanical stress distribution
Electric field magnitude only
Explanation - S‑parameters (S₁₁, S₂₁, etc.) quantify how waves are reflected and transmitted at junctions.
Correct answer is: Reflection and transmission characteristics of discontinuities
Q.113 A waveguide resonator with a Q of 10⁴ and a resonant frequency of 5 GHz is coupled to an external transmission line with a coupling coefficient β = 0.5. What is the loaded Q (Q_L)?
6 667
5 000
20 000
15 000
Explanation - Q_L = Q_unloaded / (1 + β) = 10⁴ / (1 + 0.5) ≈ 6 667.
Correct answer is: 6 667
Q.114 Which of the following phenomena is most likely to limit the maximum electric field in a high‑power waveguide?
Multipactor discharge
Skin effect
Mode conversion
Thermal expansion
Explanation - Multipactor can cause electron avalanches and breakdown at high fields, limiting power.
Correct answer is: Multipactor discharge
Q.115 The cutoff wavenumber (k_c) for the TE10 mode in a rectangular waveguide is:
π / a
π / b
2π / a
2π / b
Explanation - k_c = √[(mπ/a)² + (nπ/b)²]; for TE10, m=1, n=0 ⇒ k_c = π/a.
Correct answer is: π / a
Q.116 In a waveguide, the term "standing wave ratio" (SWR) of 1:1 indicates:
Perfect matching (no reflections)
Maximum reflection
Half the power transmitted
Zero transmitted power
Explanation - SWR = 1 corresponds to reflection coefficient Γ = 0, meaning all power is absorbed/transmitted.
Correct answer is: Perfect matching (no reflections)
Q.117 A waveguide cavity is designed to operate in the TM011 mode. Which field component is zero at the cavity axis?
Axial electric field (E_z)
Azimuthal magnetic field (H_φ)
Radial electric field (E_r)
All components are non‑zero
Explanation - In TM011, the axial electric field has a node at the center, while the magnetic field is non‑zero.
Correct answer is: Axial electric field (E_z)
Q.118 The effect of a small metallic post placed off‑center in a rectangular waveguide is to:
Introduce mode coupling and perturb the field distribution
Eliminate all higher order modes
Increase the cutoff frequency of the dominant mode
Make the waveguide operate in TEM mode
Explanation - A post acts as a perturbation, scattering energy into other modes and altering the field pattern.
Correct answer is: Introduce mode coupling and perturb the field distribution
Q.119 In a waveguide filter, a "half‑wave resonator" has a physical length approximately equal to:
Half the guided wavelength (λ_g/2)
Quarter of the free‑space wavelength (λ/4)
Full free‑space wavelength (λ)
Two times the guided wavelength (2λ_g)
Explanation - A half‑wave resonator supports a standing wave with a node at each end, requiring length = λ_g/2.
Correct answer is: Half the guided wavelength (λ_g/2)
Q.120 The parameter "β" in the context of waveguide dispersion is:
Propagation constant along the guide axis
Attenuation constant due to losses
Phase shift per unit length for a TEM line
Magnetic flux density
Explanation - β is the real part of the complex propagation constant, representing phase change per unit length.
Correct answer is: Propagation constant along the guide axis
Q.121 A waveguide with dimensions a = 2.54 cm, b = 1.27 cm is used at 15 GHz. Which higher order mode is most likely to be above cutoff?
TE20
TE01
TM11
TE30
Explanation - TE20 cutoff f_c = c/a ≈ 5.9 GHz; at 15 GHz it is well above cutoff. TE01 cutoff ≈ 11.8 GHz (also above), but TE20 is the lowest higher order mode.
Correct answer is: TE20
Q.122 The term "surface wave" in waveguide contexts most accurately describes:
A mode that propagates bound to the waveguide wall with exponential decay away from it
A TEM mode in a coaxial line
Radiation from an open waveguide end
A wave that only exists in free space
Explanation - Surface waves are bound to conducting surfaces, decaying perpendicular to the surface.
Correct answer is: A mode that propagates bound to the waveguide wall with exponential decay away from it
Q.123 For a rectangular waveguide, the wave impedance Z_TE for TE modes is given by:
Z_TE = η₀ / √(1 - (f_c/f)²)
Z_TE = η₀ √(1 - (f_c/f)²)
Z_TE = η₀ (f/f_c)
Z_TE = η₀ (f_c/f)
Explanation - Wave impedance for TE modes is Z_TE = η₀ / √(1 - (f_c/f)²).
Correct answer is: Z_TE = η₀ / √(1 - (f_c/f)²)
Q.124 A waveguide cavity is designed for use at 2.45 GHz (ISM band). If the cavity is over‑coupled, the reflected power at resonance is:
Non‑zero, with a 180° phase shift
Zero (perfectly matched)
Maximum (total reflection)
Indeterminate without knowing Q
Explanation - Over‑coupling leads to a reflection coefficient of magnitude less than one and a phase inversion.
Correct answer is: Non‑zero, with a 180° phase shift
Q.125 In a waveguide, the term "modal loss factor" quantifies:
Loss per unit length associated with a specific mode
Coupling strength between modes
Mechanical strain in the waveguide walls
Temperature rise per watt of loss
Explanation - Modal loss factor expresses attenuation (Np/m) for a particular mode due to conductor and dielectric losses.
Correct answer is: Loss per unit length associated with a specific mode
Q.126 If the radius of a circular waveguide is increased, the cutoff frequency for the TE11 mode:
Decreases
Increases
Remains the same
Becomes zero
Explanation - f_c ∝ 1/a; larger radius reduces cutoff frequency.
Correct answer is: Decreases
Q.127 A waveguide with a perfect electric conductor (PEC) wall has which of the following boundary conditions for the magnetic field at the wall?
The tangential component of H is zero
The normal component of H is zero
Both tangential and normal components are zero
No restriction on H at a PEC wall
Explanation - At a PEC, the normal component of magnetic flux density B (and thus H) must be zero; the tangential H can be non‑zero.
Correct answer is: The normal component of H is zero
Q.128 Which of the following is an advantage of using overmoded waveguides for high‑power transmission?
Lower surface electric field for a given power level
Higher cutoff frequency
Simpler mode control
Reduced mechanical size
Explanation - Larger cross‑section spreads the power, reducing peak electric fields and allowing higher power handling.
Correct answer is: Lower surface electric field for a given power level
Q.129 The term "cavity detuning" in RF accelerators is often achieved by:
Changing the temperature of the cavity walls
Adjusting the RF source phase
Increasing the input power
Adding a second cavity in series
Explanation - Thermal expansion changes cavity dimensions, shifting the resonant frequency (detuning).
Correct answer is: Changing the temperature of the cavity walls
Q.130 For a rectangular waveguide operating in the TE10 mode, the ratio of the electric field amplitude at the center of the broad wall to that at the wall itself is:
1 (maximum at centre)
0 (zero at centre)
√2
2
Explanation - The TE10 electric field varies as sin(πx/a) and peaks at x = a/2 (centre), giving maximum amplitude.
Correct answer is: 1 (maximum at centre)
Q.131 A waveguide cavity resonator is said to be "underdamped" when:
The damping (loss) is small compared to the stored energy, yielding a high Q
The damping is large, yielding a low Q
The cavity is over‑coupled
The cavity has no external coupling
Explanation - Underdamped systems have low losses relative to stored energy, resulting in high Q and narrow bandwidth.
Correct answer is: The damping (loss) is small compared to the stored energy, yielding a high Q
Q.132 If a rectangular waveguide operates at a frequency exactly twice its TE10 cutoff frequency, the phase velocity is:
Approximately 1.15 c
Exactly c
Less than c
Infinite
Explanation - v_p = c / √(1 - (f_c/f)²). With f = 2f_c ⇒ v_p = c / √(1 - 0.25) = c / √0.75 ≈ 1.155c.
Correct answer is: Approximately 1.15 c
Q.133 In a waveguide, the term "mode conversion" typically occurs at:
Discontinuities such as steps, bends, or irises
Uniform sections of waveguide
At the waveguide termination only
Only in dielectric‑filled guides
Explanation - Geometric changes cause part of the power to couple into other modes.
Correct answer is: Discontinuities such as steps, bends, or irises
Q.134 A waveguide cavity resonator used for particle acceleration must have a high shunt impedance because:
It provides more accelerating voltage per unit RF power
It reduces the cavity size
It eliminates beam loading
It increases the operating frequency
Explanation - Higher shunt impedance means the cavity converts RF power into accelerating voltage more efficiently.
Correct answer is: It provides more accelerating voltage per unit RF power
Q.135 The electric field in the TE10 mode of a rectangular waveguide is primarily oriented:
Along the narrow dimension (y‑direction)
Along the broad dimension (x‑direction)
Along the propagation direction (z‑direction)
Radially outward
Explanation - TE10 has E_y dominant, varying sinusoidally across the wide wall and directed along the narrow dimension.
Correct answer is: Along the narrow dimension (y‑direction)
Q.136 In a waveguide filter, the term "bandstop" refers to:
A filter that rejects a specific frequency band
A filter that passes all frequencies equally
A filter that only passes the dominant mode
A filter that converts TE to TM modes
Explanation - Bandstop (notch) filters attenuate a narrow frequency range while allowing others to pass.
Correct answer is: A filter that rejects a specific frequency band
Q.137 The presence of an external magnetic field applied to a ferrite‑filled waveguide primarily influences:
The permeability tensor, leading to non‑reciprocal behavior
The dielectric constant
The physical dimensions of the guide
The conductivity of the walls
Explanation - A bias field makes μ anisotropic, enabling devices like isolators and circulators.
Correct answer is: The permeability tensor, leading to non‑reciprocal behavior
Q.138 A waveguide with a rectangular cross‑section of a = 3 cm, b = 1.5 cm, operating at 5 GHz, will support which of the following modes?
Only TE10
TE10 and TE01
TE10, TE01, and TE20
No propagating modes
Explanation - f_c(TE10)=c/(2a)≈5 GHz; f_c(TE01)=c/(2b)≈10 GHz. At 5 GHz only TE10 is at cutoff (just at cutoff) while TE01 is above cutoff, so only TE10 propagates.
Correct answer is: Only TE10
Q.139 The term "waveguide junction" typically refers to:
A region where two waveguides of different dimensions meet
The end of a waveguide
A cavity resonator
A dielectric slab inside a waveguide
Explanation - Junctions are discontinuities such as steps, bends, or T‑junctions that connect waveguides.
Correct answer is: A region where two waveguides of different dimensions meet
Q.140 A cavity resonator is said to be "critically coupled" when the reflection coefficient at resonance is:
Zero
One
−1
0.5
Explanation - Critical coupling yields perfect power transfer; the reflected wave cancels, giving Γ = 0.
Correct answer is: Zero
Q.141 In a rectangular waveguide, the TE10 mode cutoff wavelength is 2a. If a = 1.5 cm, what is the cutoff frequency?
10 GHz
5 GHz
15 GHz
20 GHz
Explanation - λ_c = 2a = 3 cm → f_c = c/λ_c ≈ 3×10⁸ m/s / 0.03 m = 10 GHz.
Correct answer is: 10 GHz
Q.142 The term "over‑moded waveguide" is advantageous for:
High‑power transmission with reduced electric field stress
Low‑frequency operation only
Simplifying mode control
Eliminating the need for cooling
Explanation - Larger cross‑sections lower surface fields, allowing higher power before breakdown.
Correct answer is: High‑power transmission with reduced electric field stress