Electromagnetics & Antennas interview questions

1. A.electric field vector rotating at constant magnitude, produced by two orthogonal linear components of equal amplitude in phase quadrature, tolerant of receiver orientation
2. B.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode
3. C.cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave
4. D.incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised

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1. A.Brewster angle
2. B.Poynting vector
3. C.Rectangular waveguide cutoff
4. D.Friis transmission equation

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1. A.Poynting vector — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
2. B.Poynting vector — received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
3. C.Poynting vector — matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
4. D.Poynting vector — cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave

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1. A.resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd
2. B.ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short
3. C.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
4. D.ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω

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1. A.Brewster angle
2. B.Friis transmission equation
3. C.Intrinsic impedance of free space
4. D.Voltage standing wave ratio

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1. A.Intrinsic impedance of free space — cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave
2. B.Intrinsic impedance of free space — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
3. C.Intrinsic impedance of free space — TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode
4. D.Intrinsic impedance of free space — electric field vector rotating at constant magnitude, produced by two orthogonal linear components of equal amplitude in phase quadrature, tolerant of receiver orientation

Answer + AI explanation with Pro

1. A.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
2. B.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode
3. C.resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd
4. D.ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short

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1. A.Voltage standing wave ratio
2. B.Brewster angle
3. C.Quarter-wave transformer
4. D.Half-wave dipole

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1. A.Half-wave dipole — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
2. B.Half-wave dipole — resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd
3. C.Half-wave dipole — received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
4. D.Half-wave dipole — ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short

Answer + AI explanation with Pro

1. A.ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
2. B.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
3. C.electric field vector rotating at constant magnitude, produced by two orthogonal linear components of equal amplitude in phase quadrature, tolerant of receiver orientation
4. D.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode

Answer + AI explanation with Pro

1. A.Friis transmission equation
2. B.Intrinsic impedance of free space
3. C.Circular polarisation
4. D.Half-wave dipole

Answer + AI explanation with Pro

1. A.Circular polarisation — received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
2. B.Circular polarisation — TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode
3. C.Circular polarisation — electric field vector rotating at constant magnitude, produced by two orthogonal linear components of equal amplitude in phase quadrature, tolerant of receiver orientation
4. D.Circular polarisation — matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency

Answer + AI explanation with Pro

1. A.ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
2. B.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
3. C.ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short
4. D.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode

Answer + AI explanation with Pro

1. A.Half-wave dipole
2. B.Intrinsic impedance of free space
3. C.Voltage standing wave ratio
4. D.Friis transmission equation

Answer + AI explanation with Pro

1. A.Voltage standing wave ratio — matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
2. B.Voltage standing wave ratio — incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised
3. C.Voltage standing wave ratio — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
4. D.Voltage standing wave ratio — ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short

Answer + AI explanation with Pro

1. A.resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd
2. B.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
3. C.received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
4. D.incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised

Answer + AI explanation with Pro

1. A.Brewster angle
2. B.Quarter-wave transformer
3. C.Voltage standing wave ratio
4. D.Rectangular waveguide cutoff

Answer + AI explanation with Pro

1. A.Quarter-wave transformer — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
2. B.Quarter-wave transformer — electric field vector rotating at constant magnitude, produced by two orthogonal linear components of equal amplitude in phase quadrature, tolerant of receiver orientation
3. C.Quarter-wave transformer — matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
4. D.Quarter-wave transformer — incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised

Answer + AI explanation with Pro

1. A.cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave
2. B.ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
3. C.received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
4. D.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode

Answer + AI explanation with Pro

1. A.Circular polarisation
2. B.Intrinsic impedance of free space
3. C.Half-wave dipole
4. D.Rectangular waveguide cutoff

Answer + AI explanation with Pro

1. A.Rectangular waveguide cutoff — incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised
2. B.Rectangular waveguide cutoff — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
3. C.Rectangular waveguide cutoff — ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short
4. D.Rectangular waveguide cutoff — TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode

Answer + AI explanation with Pro

1. A.TE10 dominant mode propagates only above fc = c/(2a), below which fields are evanescent; hollow guides support no TEM mode
2. B.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
3. C.received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
4. D.cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave

Answer + AI explanation with Pro

1. A.Intrinsic impedance of free space
2. B.Circular polarisation
3. C.Brewster angle
4. D.Friis transmission equation

Answer + AI explanation with Pro

1. A.Friis transmission equation — incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised
2. B.Friis transmission equation — resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd
3. C.Friis transmission equation — ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short
4. D.Friis transmission equation — received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency

Answer + AI explanation with Pro

1. A.matching section of characteristic impedance √(Z0·ZL) and length λ/4, inherently narrowband because the electrical length is correct at one frequency
2. B.received power Pt·Gt·Gr·(λ/4πd)², so free-space path loss grows 6 dB for every doubling of distance or frequency
3. C.incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised
4. D.resonant antenna about 0.48λ long with radiation resistance near 73 Ω and directivity of 2.15 dBi, the reference for gain quoted in dBd

Answer + AI explanation with Pro

1. A.Brewster angle
2. B.Circular polarisation
3. C.Half-wave dipole
4. D.Quarter-wave transformer

Answer + AI explanation with Pro

1. A.Brewster angle — ratio (1 + |Γ|)/(1 − |Γ|) of maximum to minimum line voltage, equal to 1 for a perfect match and infinite for an open or short
2. B.Brewster angle — cross product E × H giving instantaneous power flow per unit area in W/m², directed along the propagation of the wave
3. C.Brewster angle — ratio of E to H for a plane wave in vacuum, √(μ0/ε0), approximately 377 Ω or 120π Ω
4. D.Brewster angle — incidence angle tan⁻¹√(ε2/ε1) at which a parallel-polarised wave is fully transmitted, so the reflected wave is purely perpendicular polarised

Answer + AI explanation with Pro