Nepal Engineering Council · Electronics, Communication & Information Engineering · Chapter 6
Electromagnetic and Communication System
Tap an option to check it. Wrong picks show the right answer and the hint.
216 questions in 6 syllabus topics.
6.1 Electric field and Magnetic field
34 questions · AEiE0601
1. Physically, the divergence of a vector field at a point represents:
Show hintHide hint
Think of a source or sink of field lines.
Show answerHide answer
Answer: D. The net outward flux per unit volume as the volume shrinks to that point
Divergence is defined as the limit of the net outward flux through a closed surface divided by the enclosed volume; it measures the source strength at the point.
2. The divergence theorem relates:
Show hintHide hint
Surface on one side, volume on the other.
Show answerHide answer
Answer: B. The closed-surface integral of a vector to the volume integral of its divergence
Divergence theorem: ∮ A·dS = ∫ (∇·A) dv over the volume enclosed by the surface. The closed-line/curl relation is Stokes' theorem.
3. A point charge of 2 nC is in free space. The electric field intensity at a distance of 3 m from it is approximately:
Show hintHide hint
Field falls as the square of distance.
Show answerHide answer
Answer: C. 2 V/m
E = Q/(4πε0 r²) = 9×10⁹ × 2×10⁻⁹ / 3² = 18/9 = 2 V/m.
4. In free space the electric field intensity at a point is 100 V/m. The electric flux density there is about:
Show hintHide hint
D = ε0E in free space.
Show answerHide answer
Answer: B. 0.885 nC/m²
D = ε0E = 8.854×10⁻¹² × 100 = 8.85×10⁻¹⁰ C/m² ≈ 0.885 nC/m².
5. A closed surface encloses charges of +5 nC and −2 nC, while a charge of +4 nC lies outside it. The total electric flux (Ψ = ∮D·dS) leaving the surface is:
Show hintHide hint
Charges outside the surface do not count.
Show answerHide answer
Answer: A. 3 nC
By Gauss's law, only enclosed charge counts: Ψ = 5 − 2 = 3 nC. The outside charge contributes zero net flux.
6. Given D = x²y ax + yz ay C/m², the volume charge density at the point (1, 2, 3) m is:
Show hintHide hint
Take ∂Dx/∂x + ∂Dy/∂y.
Show answerHide answer
Answer: A. 7 C/m³
ρv = ∇·D = ∂(x²y)/∂x + ∂(yz)/∂y = 2xy + z = 2(1)(2) + 3 = 7 C/m³.
7. The potential in a region is V = x²y volts (x, y in metres). The electric field intensity at (1, 2) m is:
Show hintHide hint
E is the negative gradient of V.
Show answerHide answer
Answer: D. −4 ax − ay V/m
E = −∇V = −(2xy ax + x² ay). At (1, 2): −(4 ax + 1 ay) = −4 ax − ay V/m.
8. For a static electric field, ∮E·dl = 0 around any closed path. This means:
Show hintHide hint
Work done moving a charge round a loop.
Show answerHide answer
Answer: A. The electrostatic field is conservative, i.e. ∇×E = 0
A zero closed-path line integral means the work done is path-independent, so E is conservative (curl-free) and can be written as −∇V.
9. The energy density stored in an electrostatic field in a linear dielectric of permittivity ε is:
Show hintHide hint
Compare with ½CV² for a capacitor.
Show answerHide answer
Answer: C. ½ εE² J/m³
wE = ½ D·E = ½ εE² joules per cubic metre.
10. The electric field in air is uniform at 1 MV/m. The energy stored per unit volume is about:
Show hintHide hint
Square the field before multiplying.
Show answerHide answer
Answer: A. 4.43 J/m³
w = ½ ε0E² = 0.5 × 8.854×10⁻¹² × (10⁶)² = 4.43 J/m³.
11. In a polarized dielectric with polarization P, the bound volume charge density is:
Show hintHide hint
Mind the sign; P·an is the surface version.
Show answerHide answer
Answer: A. ρb = −∇·P
Bound volume charge density is ρb = −∇·P; the bound surface charge density is ρsb = P·an.
12. A dielectric with εr = 5 is in a uniform electric field of 10 kV/m. The polarization P is about:
Show hintHide hint
Use the susceptibility χe = εr − 1.
Show answerHide answer
Answer: B. 0.354 µC/m²
P = ε0(εr − 1)E = 8.854×10⁻¹² × 4 × 10⁴ = 3.54×10⁻⁷ C/m² ≈ 0.354 µC/m².
13. The relative permittivity εr of a linear dielectric is related to its electric susceptibility χe by:
Show hintHide hint
Free space has χe = 0 and εr = 1.
Show answerHide answer
Answer: B. εr = 1 + χe
D = ε0E + P = ε0(1 + χe)E, so ε = ε0(1 + χe) and εr = 1 + χe.
14. For charges +Q and −Q separated by a distance d, the electric dipole moment is:
Show hintHide hint
Convention: the vector points toward the positive charge.
Show answerHide answer
Answer: B. Qd, directed from −Q to +Q
The dipole moment p = Qd, where d is the vector from the negative charge to the positive charge.
15. At a large distance r from an electric dipole, the potential and the field intensity vary respectively as:
Show hintHide hint
One power of r faster than a point charge.
Show answerHide answer
Answer: A. 1/r² and 1/r³
Dipole potential V = p cosθ/(4πε0 r²); taking the gradient adds one more power of r, so E ∝ 1/r³.
16. An electric dipole of moment 1 nC·m is in free space. The potential at a point 1 m away, at 60° from the dipole axis, is about:
Show hintHide hint
Use cosθ, measured from the dipole axis.
Show answerHide answer
Answer: C. 4.5 V
V = p cosθ/(4πε0 r²) = 9×10⁹ × 10⁻⁹ × cos60° / 1² = 9 × 0.5 = 4.5 V.
17. Under electrostatic conditions, at the surface of a perfect conductor:
Show hintHide hint
Field lines leave a conductor perpendicularly.
Show answerHide answer
Answer: C. Tangential E is zero and normal D equals the surface charge density
Inside a conductor E = 0; at its surface Et = 0 and Dn = ρs, so the field leaves the surface normally.
18. Two dielectrics with εr1 = 2 and εr2 = 6 meet at a plane boundary with no free surface charge. If the normal component of E in medium 1 is 30 V/m, the normal component of E in medium 2 is:
Show hintHide hint
Which component is continuous: normal D or normal E?
Show answerHide answer
Answer: B. 10 V/m
Normal D is continuous: εr1E1n = εr2E2n, so E2n = 2 × 30 / 6 = 10 V/m.
19. At the interface of two perfect dielectrics carrying no free surface charge, which quantity is continuous?
Show hintHide hint
Comes from ∮E·dl = 0 around a thin loop.
Show answerHide answer
Answer: B. Tangential component of E
Boundary conditions: E1t = E2t and D1n = D2n (with no free surface charge). Tangential D and normal E change by the permittivity ratio.
20. Free charges differ from bound charges in that free charges:
Show hintHide hint
Think conductor versus dielectric.
Show answerHide answer
Answer: D. Can move over macroscopic distances, like conduction electrons in a metal
Bound charges are displaced only slightly within atoms or molecules (polarization); free charges such as conduction electrons can move through the material.
21. According to the Biot-Savart law, the magnetic field intensity due to a current element I dl at distance R is:
Show hintHide hint
H is perpendicular to both dl and R.
Show answerHide answer
Answer: B. dH = I dl × aR / (4πR²)
The Biot-Savart law gives dH = I dl × aR/(4πR²): a cross product (H is perpendicular to both dl and R) with inverse-square dependence.
22. The SI unit of magnetic flux density B is:
Show hintHide hint
Flux per unit area.
Show answerHide answer
Answer: B. Tesla (Wb/m²)
B is flux per unit area: 1 T = 1 Wb/m². A/m is the unit of H, Wb is the unit of flux and H/m is the unit of permeability.
23. A long straight wire in air carries 10 A. The magnetic flux density at 5 cm from it is:
Show hintHide hint
Use B = µ0I/(2πr).
Show answerHide answer
Answer: A. 40 µT
B = µ0I/(2πr) = 4π×10⁻⁷ × 10 / (2π × 0.05) = 4×10⁻⁵ T = 40 µT.
24. An electron (charge 1.6×10⁻¹⁹ C) moves at 10⁶ m/s perpendicular to a uniform magnetic field of 0.1 T. The magnitude of the magnetic force on it is:
Show hintHide hint
F = qvB when v is perpendicular to B.
Show answerHide answer
Answer: D. 1.6×10⁻¹⁴ N
F = qvB sin90° = 1.6×10⁻¹⁹ × 10⁶ × 0.1 = 1.6×10⁻¹⁴ N.
25. A static magnetic field acting on a moving charged particle:
Show hintHide hint
Is the force ever along the velocity?
Show answerHide answer
Answer: A. Changes the direction of its velocity but not its kinetic energy
F = q(v × B) is always perpendicular to v, so it does no work: it only bends the path and the speed stays constant.
26. A straight conductor 0.2 m long carrying 4 A is placed at 30° to a uniform field of 0.5 T. The force on it is:
Show hintHide hint
Use the sine of the angle between the conductor and B.
Show answerHide answer
Answer: B. 0.2 N
F = BIL sinθ = 0.5 × 4 × 0.2 × sin30° = 0.4 × 0.5 = 0.2 N.
27. A single square loop of side 10 cm carries a current of 2 A. Its magnetic dipole moment is:
Show hintHide hint
Moment = current × area.
Show answerHide answer
Answer: D. 0.02 A·m²
m = IA = 2 × (0.1)² = 0.02 A·m², directed normal to the loop by the right-hand rule.
28. A 50-turn coil of area 0.01 m² carries 0.5 A in a uniform field of 0.2 T. The plane of the coil is parallel to B. The torque on the coil is:
Show hintHide hint
Find the angle between the coil's normal and B.
Show answerHide answer
Answer: A. 0.05 N·m
With the plane parallel to B, the moment (normal to the plane) is perpendicular to B, so T = NIAB = 50 × 0.5 × 0.01 × 0.2 = 0.05 N·m (maximum).
29. The torque on a magnetic dipole of moment m in a uniform field B is:
Show hintHide hint
Torque is a vector that tends to align m with B.
Show answerHide answer
Answer: A. T = m × B
Torque is the cross product T = m × B; −m·B is the dipole's potential energy, not the torque.
30. At distances much larger than its size, a small current loop behaves as:
Show hintHide hint
Compare it with a bar magnet.
Show answerHide answer
Answer: A. A magnetic dipole whose field falls off as 1/r³
A small loop is the magnetic dipole; its far field has the same form as an electric dipole's, varying as 1/r³.
31. Magnetization M of a material is defined as:
Show hintHide hint
It is the magnetic analogue of polarization P.
Show answerHide answer
Answer: C. Magnetic dipole moment per unit volume
M is the net magnetic dipole moment per unit volume (A/m), and B = µ0(H + M).
32. A linear magnetic material with µr = 101 has H = 50 A/m inside it. The magnetization M is:
Show hintHide hint
χm = µr − 1.
Show answerHide answer
Answer: C. 5000 A/m
M = χmH = (µr − 1)H = 100 × 50 = 5000 A/m.
33. At the boundary between two magnetic media, which quantity is always continuous?
Show hintHide hint
It follows from ∇·B = 0.
Show answerHide answer
Answer: C. The normal component of B
From ∮B·dS = 0, B1n = B2n always. Tangential H is continuous only when there is no surface current (H1t − H2t = K).
34. Region 1 (µr1 = 4) and region 2 (µr2 = 1) share a boundary carrying no surface current. In region 1, H1 = 5 at + 3 an A/m (at tangential, an normal). In region 2, H2 is:
Show hintHide hint
Apply Ht continuity and Bn continuity separately.
Show answerHide answer
Answer: D. 5 at + 12 an A/m
Tangential H is continuous (5). Normal B is continuous: µr1H1n = µr2H2n, so H2n = 4 × 3/1 = 12. Hence H2 = 5 at + 12 an.
6.2 Wave propagation and antenna
40 questions · AEiE0602
35. Maxwell added the displacement current density ∂D/∂t to Ampere's law mainly to:
Show hintHide hint
Take the divergence of both sides of ∇×H = J.
Show answerHide answer
Answer: B. Make it consistent with the continuity equation for time-varying fields
Taking the divergence of ∇×H = J gives ∇·J = 0, which contradicts ∇·J = −∂ρv/∂t for time-varying charge; adding ∂D/∂t removes the contradiction.
36. A capacitor is connected to an AC source. The current through the dielectric between its plates is:
Show hintHide hint
Current continuity must hold around the circuit.
Show answerHide answer
Answer: C. Displacement current, equal in magnitude to the conduction current in the leads
No charge crosses the gap; the changing D between the plates constitutes a displacement current ε dE/dt × A, equal to the lead current, which keeps Ampere's law consistent.
37. In free space E = 10 sin(10⁹ t) ax V/m. The amplitude of the displacement current density is about:
Show hintHide hint
Differentiate E with respect to time and multiply by ε0.
Show answerHide answer
Answer: A. 0.0885 A/m²
Jd = ε0 ∂E/∂t = ε0 × 10 × 10⁹ cos(10⁹t), so amplitude = 8.854×10⁻¹² × 10¹⁰ = 0.0885 A/m².
38. Which Maxwell equation in point form expresses Faraday's law of induction?
Show hintHide hint
Look for the curl of E.
Show answerHide answer
Answer: D. ∇×E = −∂B/∂t
Faraday's law states that a time-varying magnetic field produces a circulating electric field: ∇×E = −∂B/∂t.
39. The integral form of the Maxwell equation ∇·B = 0 is:
Show hintHide hint
Divergence goes with a closed surface.
Show answerHide answer
Answer: C. ∮B·dS = 0
By the divergence theorem, ∫(∇·B)dv = ∮B·dS = 0: the net magnetic flux leaving any closed surface is zero.
40. In a source-free, lossless, homogeneous medium, the electric field satisfies the wave equation:
Show hintHide hint
It should contain a second time derivative.
Show answerHide answer
Answer: B. ∇²E = µε ∂²E/∂t²
Taking the curl of Faraday's law and using Ampere-Maxwell with J = 0 gives ∇²E = µε ∂²E/∂t², with wave speed 1/√(µε).
41. A uniform plane wave is a transverse electromagnetic (TEM) wave, which means:
Show hintHide hint
Both fields are transverse.
Show answerHide answer
Answer: A. E and H are perpendicular to each other and to the direction of propagation
In a uniform plane wave neither E nor H has a component along the direction of propagation; E, H and the direction of propagation are mutually perpendicular.
42. A plane wave in free space has a magnetic field amplitude of 1 A/m. The electric field amplitude is about:
Show hintHide hint
Use the intrinsic impedance of free space.
Show answerHide answer
Answer: C. 377 V/m
E = η0H with η0 = 120π ≈ 377 Ω, so E ≈ 377 V/m.
43. The intrinsic impedance of a lossless non-magnetic dielectric with εr = 4 is about:
Show hintHide hint
η scales as 1/√εr.
Show answerHide answer
Answer: C. 188 Ω
η = η0√(µr/εr) = 377/√4 ≈ 188.5 Ω.
44. The phase velocity of a plane wave in a lossless non-magnetic dielectric with εr = 9 is:
Show hintHide hint
Divide c by the refractive index.
Show answerHide answer
Answer: B. 1×10⁸ m/s
vp = c/√(µrεr) = 3×10⁸/3 = 1×10⁸ m/s.
45. For a uniform plane wave in a lossless dielectric:
Show hintHide hint
With σ = 0 the intrinsic impedance is real.
Show answerHide answer
Answer: A. There is no attenuation and E and H are in time phase
With σ = 0, α = 0 and η is purely real, so E and H are in phase and the amplitude stays constant.
46. For a plane wave in a lossy dielectric (σ ≠ 0):
Show hintHide hint
Compute η with σ included.
Show answerHide answer
Answer: D. The intrinsic impedance is complex, so E and H are out of time phase
η = √(jωµ/(σ + jωε)) becomes complex, so H lags E by the angle of η, and α > 0 causes attenuation.
47. A medium behaves as a good conductor at a given frequency when:
Show hintHide hint
Compare conduction and displacement currents.
Show answerHide answer
Answer: D. σ ≫ ωε
When the loss tangent σ/(ωε) ≫ 1, conduction current dominates displacement current and the medium is a good conductor.
48. The skin depth of copper (σ = 5.8×10⁷ S/m, µr = 1) at 1 MHz is about:
Show hintHide hint
δ = 1/√(πfµσ).
Show answerHide answer
Answer: D. 66 µm
δ = 1/√(πfµσ) = 1/√(π × 10⁶ × 4π×10⁻⁷ × 5.8×10⁷) ≈ 6.6×10⁻⁵ m = 66 µm.
49. If the frequency of a wave incident on a good conductor is increased four times, the skin depth:
Show hintHide hint
Skin depth varies inversely with the square root of frequency.
Show answerHide answer
Answer: A. Becomes half
δ ∝ 1/√f, so a 4× increase in f reduces δ by √4 = 2.
50. The Poynting vector P = E × H represents:
Show hintHide hint
Check the units of E × H.
Show answerHide answer
Answer: A. The instantaneous power flow per unit area, in W/m²
E × H has units (V/m)(A/m) = W/m² and gives the direction and density of electromagnetic power flow.
51. A plane wave in air is normally incident on a lossless non-magnetic dielectric with εr = 4. The reflection coefficient for E is:
Show hintHide hint
Γ = (η2 − η1)/(η2 + η1).
Show answerHide answer
Answer: B. −1/3
η2 = 377/2; Γ = (η2 − η1)/(η2 + η1) = (0.5 − 1)/(0.5 + 1) = −1/3. (The transmission coefficient is 1 + Γ = 2/3.)
52. When a plane wave is normally incident on a perfect conductor:
Show hintHide hint
Tangential E must vanish at a perfect conductor.
Show answerHide answer
Answer: D. Γ = −1, the wave is totally reflected and a standing wave with an E-field null at the surface forms
For a perfect conductor η2 = 0, so Γ = −1: total reflection, tangential E = 0 at the surface, and a pure standing wave in front of it.
53. A plane wave in air is obliquely incident on a non-magnetic dielectric with εr = 3. The Brewster angle is:
Show hintHide hint
tan θB = √(εr2/εr1).
Show answerHide answer
Answer: C. 60°
tan θB = √(ε2/ε1) = √3, so θB = 60°.
54. A wave travels from a non-magnetic dielectric with εr = 4 toward air. The critical angle for total internal reflection is:
Show hintHide hint
sin θc = n2/n1.
Show answerHide answer
Answer: C. 30°
sin θc = √(ε2/ε1) = √(1/4) = 0.5, so θc = 30°.
55. The dominant mode of a rectangular waveguide with broad dimension a greater than narrow dimension b is:
Show hintHide hint
The mode with the lowest cutoff frequency.
Show answerHide answer
Answer: C. TE10
TE10 has the lowest cutoff frequency, fc = c/(2a), so it is the dominant mode when a > b.
56. The lowest-order TM mode that can exist in a rectangular waveguide is:
Show hintHide hint
What happens to Ez if m or n is zero?
Show answerHide answer
Answer: B. TM11
For TM modes Ez ∝ sin(mπx/a) sin(nπy/b); if m or n is zero all fields vanish, so the lowest TM mode is TM11.
57. A TEM wave cannot propagate inside a hollow rectangular waveguide because:
Show hintHide hint
Compare with a coaxial cable.
Show answerHide answer
Answer: D. A hollow guide has only one conductor, and TEM waves need at least two conductors
A TEM mode needs a static-like transverse potential difference between two separate conductors, which a single hollow conductor cannot support.
58. In a transverse electric (TE) mode of a waveguide:
Show hintHide hint
'Transverse electric' says which field has no axial component.
Show answerHide answer
Answer: A. Ez = 0 and Hz ≠ 0
TE means the electric field is entirely transverse (no component along the guide axis), while the magnetic field has a longitudinal component.
59. When a waveguide is operated below the cutoff frequency of a mode, that mode:
Show hintHide hint
A waveguide behaves like a high-pass filter.
Show answerHide answer
Answer: D. Is evanescent: its fields decay exponentially along the guide and carry no real power
Below cutoff the propagation constant is purely real (attenuation), so the waveguide behaves as a high-pass filter for that mode.
60. An air-filled rectangular waveguide has a = 2.286 cm and b = 1.016 cm. The cutoff frequency of the TE10 mode is about:
Show hintHide hint
TE10 cutoff depends only on the broad dimension.
Show answerHide answer
Answer: C. 6.56 GHz
fc(TE10) = c/(2a) = 3×10⁸/(2 × 0.02286) ≈ 6.56 GHz.
61. Electromagnetic radiation from an antenna is produced by:
Show hintHide hint
Static sources produce static fields.
Show answerHide answer
Answer: B. Time-varying (accelerating or decelerating) currents or charges
Static charges produce only electrostatic fields and steady currents only magnetostatic fields; radiation requires time-varying current, i.e. accelerated charge.
62. The reciprocity theorem applied to antennas implies that:
Show hintHide hint
Can the same dish be used to transmit and receive?
Show answerHide answer
Answer: B. An antenna's radiation pattern and impedance are the same whether it transmits or receives
Because Maxwell's equations are reciprocal in linear, isotropic media, an antenna's transmit and receive properties (pattern, gain, impedance) are identical.
63. The directivity of an antenna is the ratio of:
Show hintHide hint
Compare with an isotropic radiator of equal total power.
Show answerHide answer
Answer: A. Its maximum radiation intensity to the average radiation intensity over all directions
D = Umax/Uavg = 4πUmax/Prad, i.e. compared with an isotropic source radiating the same total power. Losses do not enter directivity.
64. An antenna has directivity 10 and radiation efficiency 80%. Its gain is about:
Show hintHide hint
Multiply first, then convert to dB.
Show answerHide answer
Answer: C. 9.0 dBi
G = ηD = 0.8 × 10 = 8, and 10 log10(8) ≈ 9.03 dBi.
65. A Hertzian (short) dipole has length λ/20. Its radiation resistance is about:
Show hintHide hint
Rr = 80π²(dl/λ)².
Show answerHide answer
Answer: A. 1.97 Ω
Rr = 80π²(dl/λ)² = 80 × 9.87 × (1/20)² ≈ 1.97 Ω.
66. The directivity of a thin half-wave dipole is about:
Show hintHide hint
Slightly more directive than a short dipole.
Show answerHide answer
Answer: D. 1.64 (2.15 dBi)
A half-wave dipole has D ≈ 1.64 (2.15 dBi) and Rr ≈ 73 Ω; 1.5 is the Hertzian dipole's directivity.
67. The half-power beamwidth (HPBW) of an antenna is the angular width of the main lobe between the points where:
Show hintHide hint
Half power, not half field.
Show answerHide answer
Answer: D. The radiated power density falls to half (−3 dB) of its maximum
HPBW is measured between the −3 dB (half-power, 0.707 field) points; the angle between the first nulls is FNBW.
68. An isotropic antenna is:
Show hintHide hint
It is the reference for dBi.
Show answerHide answer
Answer: A. A hypothetical lossless point source that radiates equally in all directions
The isotropic radiator cannot be built; its spherical pattern serves as the 0 dBi reference for gain and directivity.
69. An omnidirectional antenna, such as a vertical dipole, has a radiation pattern that is:
Show hintHide hint
Think of the doughnut shape.
Show answerHide answer
Answer: D. Uniform in one plane (e.g. azimuth) but directional in the perpendicular plane
An omnidirectional antenna radiates equally in all directions of one plane (a circle in the H-plane) while its E-plane pattern has nulls, giving a doughnut-shaped 3-D pattern.
70. The radiation pattern of a thin vertical half-wave dipole has:
Show hintHide hint
A dipole does not radiate off its ends.
Show answerHide answer
Answer: D. Maximum radiation broadside (perpendicular) to the wire and nulls along its axis
The dipole pattern varies roughly as sin θ (more precisely cos(π/2 cosθ)/sinθ), maximum at θ = 90° and zero along the axis.
71. A travelling-wave (non-resonant) antenna such as a long wire terminated in a matched load:
Show hintHide hint
What does the matched termination remove?
Show answerHide answer
Answer: A. Has no standing wave on it, a unidirectional pattern and a wide bandwidth
The matched termination absorbs the forward wave, so there is no reflected wave; the current is a travelling wave giving a unidirectional pattern toward the load and broad bandwidth.
72. A rhombic antenna is mainly used for:
Show hintHide hint
It needs a large area and wire lengths of several wavelengths.
Show answerHide answer
Answer: B. Long-distance HF point-to-point sky-wave links
The terminated rhombic is a large, broadband, directive travelling-wave antenna made of long wires, suited to HF point-to-point communication via the ionosphere.
73. The key geometric property of a paraboloidal reflector is that:
Show hintHide hint
This is why dish feeds sit at the focus.
Show answerHide answer
Answer: A. Rays from a source at the focus are reflected parallel to the axis, with equal path lengths to the aperture plane
All paths from the focus to the aperture plane via the parabola are equal, so a spherical wave from the focal feed becomes a plane wave, producing a narrow pencil beam.
74. In a Yagi-Uda antenna, compared with the driven element:
Show hintHide hint
The beam points toward the directors.
Show answerHide answer
Answer: D. The reflector is slightly longer and the directors are slightly shorter
A longer (inductive) reflector and shorter (capacitive) directors set the currents' phases so the radiation is concentrated toward the directors.
6.3 Communication System
34 questions · AEiE0603
75. In the basic block diagram of a communication system, which block lies between the transmitter and the receiver?
Show hintHide hint
Think of the medium the signal travels through.
Show answerHide answer
Answer: A. Channel
The channel is the physical medium (wire, fibre, free space) that carries the transmitted signal from transmitter to receiver.
76. The main purpose of the source encoder in a digital communication system is to:
Show hintHide hint
Compare source coding with channel coding.
Show answerHide answer
Answer: D. Remove redundancy so the message is represented with fewer bits
Source coding (compression) removes redundancy to represent the source efficiently; adding controlled redundancy is the job of the channel encoder.
77. The channel encoder in a digital communication system deliberately:
Show hintHide hint
It fights the effects of the channel, not the source.
Show answerHide answer
Answer: B. Adds controlled redundancy to detect or correct errors
Channel coding inserts extra (parity) bits in a controlled way so the receiver can detect or correct transmission errors.
78. Which block of a communication system converts a non-electrical message (such as sound) into an electrical signal?
Show hintHide hint
A microphone is an example.
Show answerHide answer
Answer: C. Input transducer
A transducer such as a microphone converts the physical message into an electrical signal before transmission.
79. A major advantage of digital communication over analog communication is that:
Show hintHide hint
Think of what a repeater can do with a 0/1 pulse.
Show answerHide answer
Answer: D. Regenerative repeaters can remove accumulated noise along the link
Digital pulses can be detected and regenerated afresh at each repeater, so noise does not accumulate as it does with analog amplifiers.
80. Shot noise in electronic devices is mainly due to:
Show hintHide hint
It exists only when DC current flows through a device.
Show answerHide answer
Answer: B. The random arrival of discrete charge carriers across a junction
Shot noise arises because current consists of discrete charges crossing a junction or gap at random instants; its mean-square current is 2qIB.
81. Which of the following is an example of external (not device-generated) noise?
Show hintHide hint
Where does it originate: inside or outside the receiver?
Show answerHide answer
Answer: B. Atmospheric noise from lightning discharges
Atmospheric, extraterrestrial (solar, cosmic) and man-made noise originate outside the receiver; the others are internal noise sources.
82. Noise that is simply added to the transmitted signal in the channel, independent of the signal, is modelled as:
Show hintHide hint
Recall the 'A' in AWGN.
Show answerHide answer
Answer: C. Additive noise
The AWGN channel model is r(t) = s(t) + n(t); the noise is added and does not depend on the signal.
83. A signal power of 10 mW is received along with a noise power of 1 µW. The signal-to-noise ratio is:
Show hintHide hint
Use 10 log for a power ratio.
Show answerHide answer
Answer: D. 40 dB
SNR = 10 log₁₀(10×10⁻³ / 1×10⁻⁶) = 10 log₁₀(10⁴) = 40 dB.
84. The noise figure of an amplifier is defined as:
Show hintHide hint
A noisy amplifier makes the SNR worse, so the ratio is at least 1.
Show answerHide answer
Answer: C. The ratio of input SNR to output SNR, expressed in dB
Noise factor F = (S/N)in / (S/N)out ≥ 1; noise figure NF = 10 log₁₀ F dB. A noiseless amplifier has NF = 0 dB.
85. A low-pass (baseband) signal is one whose spectrum:
Show hintHide hint
Where on the frequency axis is its energy?
Show answerHide answer
Answer: B. Is concentrated around zero frequency, from 0 up to some maximum frequency
A low-pass signal has significant spectral content from DC up to a highest frequency W; speech and video baseband signals are examples.
86. A band-pass signal has its spectral content:
Show hintHide hint
Modulated signals are typical examples.
Show answerHide answer
Answer: C. Confined to a band around a centre (carrier) frequency fc, away from zero
A band-pass signal occupies fc − B/2 to fc + B/2, with no significant content near DC; modulated signals are band-pass.
87. A band-pass signal can be written as x(t) = xI(t) cos 2πfct − xQ(t) sin 2πfct. Here xI(t) and xQ(t) are:
Show hintHide hint
These components vary slowly compared with the carrier.
Show answerHide answer
Answer: A. Low-pass in-phase and quadrature components
Any band-pass signal can be represented by two low-pass components, the in-phase xI(t) and quadrature xQ(t), which modulate quadrature carriers.
88. The 3-dB bandwidth of a system is the frequency range over which:
Show hintHide hint
−3 dB in power is a factor of about 0.5.
Show answerHide answer
Answer: B. The power gain is at least half of its maximum value
At the 3-dB points the power falls to 1/2 (voltage to 1/√2 ≈ 0.707) of its peak value.
89. An RC low-pass filter has R = 1 kΩ and C = 0.1 µF. Its 3-dB bandwidth is about:
Show hintHide hint
f = 1/(2πRC).
Show answerHide answer
Answer: D. 1.59 kHz
f3dB = 1/(2πRC) = 1/(2π × 10³ × 10⁻⁷) ≈ 1591 Hz ≈ 1.59 kHz.
90. A tuned band-pass circuit has a centre frequency of 10 MHz and a 3-dB bandwidth of 100 kHz. Its quality factor Q is:
Show hintHide hint
Q is centre frequency divided by bandwidth.
Show answerHide answer
Answer: A. 100
Q = f0/B = 10 MHz / 100 kHz = 100.
91. For distortionless transmission through a linear system, the transfer function must have:
Show hintHide hint
The output should be a scaled, delayed copy of the input.
Show answerHide answer
Answer: C. Constant magnitude and phase varying linearly with frequency over the signal band
H(f) = K e^(−j2πft₀) gives y(t) = K x(t − t₀): only scaling and delay, with no change of waveform shape.
92. A system has a constant magnitude response but a phase response that is not linear with frequency. The output will suffer from:
Show hintHide hint
Different components arrive at different times.
Show answerHide answer
Answer: A. Phase (delay) distortion
If phase is not linear, different frequency components are delayed by different amounts, which distorts the waveform even though their amplitudes are unchanged.
93. A distortionless channel has a linear phase response that reaches −π/2 rad at 1 kHz. The time delay introduced by the channel is:
Show hintHide hint
Linear phase θ = −2πft₀.
Show answerHide answer
Answer: C. 0.25 ms
θ(f) = −2πft₀, so t₀ = (π/2)/(2π × 1000) = 1/4000 s = 0.25 ms.
94. The Hilbert transformer is a linear filter whose transfer function is:
Show hintHide hint
It is a pure ±90° phase shifter.
Show answerHide answer
Answer: B. H(f) = −j sgn(f)
The Hilbert transformer shifts positive frequencies by −90° and negative frequencies by +90° with unit magnitude: H(f) = −j sgn(f).
95. The Hilbert transform of 3 cos(100t) is:
Show hintHide hint
Delay the phase of each component by 90°.
Show answerHide answer
Answer: D. 3 sin(100t)
The Hilbert transform shifts a cosine by −90°: cos(ωt − 90°) = sin(ωt), so 3 cos(100t) → 3 sin(100t).
96. The analytic signal (pre-envelope) x₊(t) = x(t) + j x̂(t) has a spectrum that is:
Show hintHide hint
1 + sgn(f) takes values 2 and 0.
Show answerHide answer
Answer: D. Zero for negative frequencies and twice X(f) for positive frequencies
X₊(f) = X(f)[1 + sgn(f)], which equals 2X(f) for f > 0, X(0) at f = 0 and 0 for f < 0.
97. A key application of the Hilbert transform in communication is:
Show hintHide hint
One sideband is cancelled using a 90°-shifted copy of the message.
Show answerHide answer
Answer: B. Generating SSB signals by the phase-shift method
The phase-shift SSB modulator combines m(t)cos ωct and m̂(t)sin ωct; the 90° shifted message m̂(t) is the Hilbert transform of m(t).
98. The signal s(t) = Ac[1 + ka m(t)] cos 2πfct represents:
Show hintHide hint
Expand the bracket: is a pure carrier term present?
Show answerHide answer
Answer: B. Conventional AM (DSB with full carrier)
The carrier term Ac cos 2πfct is present together with sidebands from Ac ka m(t) cos 2πfct, so this is standard DSB-FC AM.
99. An AM wave displayed on an oscilloscope has a maximum envelope amplitude of 15 V and a minimum of 5 V. The modulation index is:
Show hintHide hint
μ = (Vmax − Vmin)/(Vmax + Vmin).
Show answerHide answer
Answer: C. 0.5
μ = (Vmax − Vmin)/(Vmax + Vmin) = (15 − 5)/(15 + 5) = 0.5.
100. A 1 MHz carrier is amplitude modulated (DSB-FC) by a 5 kHz tone. The spectrum contains components at:
Show hintHide hint
Carrier plus upper and lower side frequencies.
Show answerHide answer
Answer: B. 995 kHz, 1000 kHz and 1005 kHz
Standard AM contains the carrier fc and the two side frequencies fc ± fm = 995 kHz and 1005 kHz; bandwidth 2fm = 10 kHz.
101. The transmission bandwidth of an SSB signal for a message of maximum frequency fm is:
Show hintHide hint
Only one of the two sidebands is sent.
Show answerHide answer
Answer: D. fm
SSB transmits only one sideband, so it needs half the bandwidth of DSB, i.e. fm.
102. Vestigial sideband (VSB) modulation was used for the video signal in analog television mainly because:
Show hintHide hint
Think of the spectrum of a video signal near 0 Hz.
Show answerHide answer
Answer: B. Video has significant low-frequency content, which makes pure SSB filtering impractical
Video extends almost to DC, so a sharp SSB filter is impossible; VSB keeps one sideband plus a vestige of the other, saving bandwidth compared with DSB.
103. Which AM variant cannot be demodulated by a simple envelope detector and requires coherent (synchronous) detection?
Show hintHide hint
Which one has no transmitted carrier?
Show answerHide answer
Answer: D. DSB-SC
In DSB-SC the envelope is |m(t)|, not m(t), so phase reversals are lost; a locally generated synchronous carrier is required.
104. A balanced modulator is used to generate:
Show hintHide hint
Its output has the carrier suppressed.
Show answerHide answer
Answer: B. DSB-SC signals
A balanced modulator cancels the carrier and outputs only the product m(t) cos ωct, i.e. DSB-SC.
105. An FM broadcast signal has a peak frequency deviation of 75 kHz and the highest audio frequency is 15 kHz. The modulation index is:
Show hintHide hint
β = Δf / fm.
Show answerHide answer
Answer: B. 5
β = Δf/fm = 75/15 = 5.
106. Using Carson's rule, the bandwidth of an FM signal with 75 kHz peak deviation and 15 kHz maximum modulating frequency is:
Show hintHide hint
Carson: B = 2(Δf + fm).
Show answerHide answer
Answer: A. 180 kHz
B = 2(Δf + fm) = 2(75 + 15) = 180 kHz.
107. The FM signal s(t) = 10 cos(2π×10⁶t + 5 sin 2π×10³t) has a peak frequency deviation of:
Show hintHide hint
Δf = β × fm.
Show answerHide answer
Answer: C. 5 kHz
The phase term is β sin 2πfmt with β = 5 and fm = 1 kHz, so Δf = βfm = 5 kHz.
108. An FM signal can be generated with a phase modulator by:
Show hintHide hint
Frequency is the derivative of phase.
Show answerHide answer
Answer: A. Integrating the message before applying it to the phase modulator
In FM the phase is proportional to the integral of m(t); so integrating m(t) and then phase-modulating gives FM (the basis of the Armstrong method).
6.4 Data communication and information theory
38 questions · AEiE0604
109. Noise generated by the random thermal motion of electrons in a resistor is called:
Show hintHide hint
It depends on temperature.
Show answerHide answer
Answer: D. Thermal (Johnson) noise
Thermal or Johnson-Nyquist noise arises from random thermal agitation of charge carriers in any conductor above 0 K.
110. The correct order of operations in converting an analog message into a PCM signal is:
Show hintHide hint
Time is discretised before amplitude.
Show answerHide answer
Answer: B. Sampling, quantising, encoding
The analog signal is first sampled in time, the sample values are then quantised to discrete levels, and each level is encoded into a binary code word.
111. According to the sampling theorem, a signal band-limited to W Hz can be recovered exactly from its samples if the sampling rate is:
Show hintHide hint
Twice the highest frequency.
Show answerHide answer
Answer: A. At least 2W samples per second
A band-limited signal is uniquely determined by samples taken at fs ≥ 2W; the minimum rate 2W is the Nyquist rate.
112. A 7 kHz sinusoid is sampled at 10 kHz and reconstructed with an ideal 5 kHz low-pass filter. The output is a tone at:
Show hintHide hint
Look for fs − f.
Show answerHide answer
Answer: A. 3 kHz
Sampling below the Nyquist rate causes aliasing: the 7 kHz tone appears at |10 − 7| = 3 kHz inside the 0–5 kHz band.
113. In which analog pulse modulation does the position (timing) of constant-width pulses vary with the message?
Show hintHide hint
The name says what varies.
Show answerHide answer
Answer: B. PPM
Pulse position modulation shifts the time of each pulse in proportion to the sample value; PAM varies amplitude and PWM varies width.
114. A telephone voice signal is sampled at 8 kHz and each sample is coded with 8 bits. The PCM bit rate is:
Show hintHide hint
Bits per sample × samples per second.
Show answerHide answer
Answer: A. 64 kbps
Bit rate = n fs = 8 × 8000 = 64 000 bps.
115. What is the minimum (Nyquist) transmission bandwidth needed for a binary PCM signal of 8 bits per sample sampled at 8 kHz?
Show hintHide hint
Minimum bandwidth is half the bit rate.
Show answerHide answer
Answer: D. 32 kHz
Rb = 64 kbps and the minimum baseband bandwidth for binary signalling is Rb/2 = 32 kHz.
116. For uniform quantisation with step size Δ, the mean-square quantisation noise (error uniformly distributed) is:
Show hintHide hint
Variance of a uniform distribution of width Δ.
Show answerHide answer
Answer: C. Δ²/12
The error is uniform over −Δ/2 to Δ/2, whose variance is Δ²/12.
117. In uniform PCM, increasing the number of bits per sample by one improves the SQNR by about:
Show hintHide hint
Halving Δ reduces Δ²/12 by a factor of four.
Show answerHide answer
Answer: D. 6 dB
Each extra bit doubles the number of levels and halves Δ, reducing noise power by 4, i.e. about 6 dB.
118. The main purpose of non-uniform quantisation (companding) in speech PCM is to:
Show hintHide hint
Speech has many low-amplitude samples.
Show answerHide answer
Answer: A. Give a more uniform SQNR for weak and strong signals
Companding uses small steps for small amplitudes and larger steps for large ones, so weak speech signals get adequate SQNR without more bits.
119. A mid-tread uniform quantiser is one that:
Show hintHide hint
Think of a staircase with a flat tread at the origin.
Show answerHide answer
Answer: C. Has zero as one of its output levels
In a mid-tread quantiser zero is an output (reconstruction) level; in a mid-rise quantiser zero is a decision threshold, so there is no zero output level.
120. Delta modulation (DM) transmits:
Show hintHide hint
It is the simplest form of differential coding.
Show answerHide answer
Answer: C. One bit per sample indicating whether the signal is above or below its staircase approximation
DM is 1-bit DPCM: each bit tells the receiver to step the staircase up or down by Δ.
121. In Huffman source coding, symbols with higher probability are assigned:
Show hintHide hint
Morse code uses the same idea.
Show answerHide answer
Answer: B. Shorter code words
Huffman coding is a variable-length prefix code that gives short code words to frequent symbols, minimising the average code length.
122. The Shannon-Hartley theorem gives the channel capacity of an AWGN channel of bandwidth B and signal-to-noise ratio S/N as:
Show hintHide hint
The logarithm is to base 2.
Show answerHide answer
Answer: B. C = B log₂(1 + S/N)
Shannon-Hartley: C = B log₂(1 + S/N) bits per second, with S/N as a power ratio (not in dB).
123. A telephone channel has a bandwidth of 3 kHz and an SNR of 30 dB. Its Shannon capacity is about:
Show hintHide hint
Convert 30 dB to a ratio first.
Show answerHide answer
Answer: C. 29.9 kbps
S/N = 10³; C = 3000 log₂(1001) ≈ 3000 × 9.97 ≈ 29.9 kbps.
124. A T1 frame carries 24 voice channels of 8 bits each plus 1 framing bit, at 8000 frames per second. The T1 line rate is:
Show hintHide hint
Count the bits in one frame first.
Show answerHide answer
Answer: D. 1.544 Mbps
(24 × 8 + 1) bits × 8000 frames/s = 193 × 8000 = 1 544 000 bps.
125. Guard bands are a feature of which multiplexing technique?
Show hintHide hint
Gaps between adjacent channel spectra.
Show answerHide answer
Answer: D. Frequency-division multiplexing
In FDM, unused frequency gaps (guard bands) are left between adjacent channels so practical filters can separate them without crosstalk.
126. The amount of information carried by a message is:
Show hintHide hint
A surprising message tells you more.
Show answerHide answer
Answer: B. Larger for less probable messages
Self-information I = log₂(1/p) increases as probability p decreases; a certain event (p = 1) carries no information.
127. A symbol occurs with probability 1/8. The information it conveys is:
Show hintHide hint
I = log₂(1/p).
Show answerHide answer
Answer: A. 3 bits
I = log₂(1/p) = log₂ 8 = 3 bits.
128. A source emits four symbols with probabilities 1/2, 1/4, 1/8 and 1/8. Its entropy is:
Show hintHide hint
H = Σ p log₂(1/p).
Show answerHide answer
Answer: A. 1.75 bits/symbol
H = ½(1) + ¼(2) + ⅛(3) + ⅛(3) = 0.5 + 0.5 + 0.375 + 0.375 = 1.75 bits/symbol.
129. A prefix (instantaneous) code is one in which:
Show hintHide hint
Think of decoding without waiting for the next bits.
Show answerHide answer
Answer: B. No code word is the beginning of any other code word
In a prefix code no code word is a prefix of another, so each code word can be decoded as soon as it ends; Huffman codes are prefix codes.
130. A disadvantage of unipolar NRZ line coding is that it:
Show hintHide hint
What is the average voltage of a 0/+V signal?
Show answerHide answer
Answer: B. Has a significant DC component and no guaranteed transitions for timing
Unipolar NRZ uses 0 and +V only, so the average is non-zero (DC), and long runs of 1s or 0s give no transitions for clock recovery.
131. In bipolar AMI line coding, binary 1s are represented by:
Show hintHide hint
'Alternate mark inversion' says it all.
Show answerHide answer
Answer: D. Pulses of alternating polarity, with 0s as zero voltage
AMI sends 0 as zero volts and successive 1s as +V, −V, +V …, which removes the DC component and allows detection of single errors as bipolar violations.
132. Manchester line coding is widely used because:
Show hintHide hint
Look at the middle of each bit.
Show answerHide answer
Answer: A. Every bit has a mid-bit transition, so it is self-clocking with no DC component
The guaranteed mid-bit transition carries timing and makes the average zero, at the cost of about twice the bandwidth of NRZ.
133. Inter-symbol interference (ISI) in a baseband system is caused mainly by:
Show hintHide hint
Neighbouring pulses overlap.
Show answerHide answer
Answer: C. Spreading of pulses beyond their symbol interval due to limited channel bandwidth
A band-limited channel spreads each pulse in time, so the tails of neighbouring pulses add to the sample of the current pulse.
134. According to the Nyquist criterion, the maximum symbol rate for zero-ISI transmission through an ideal channel of bandwidth 4 kHz is:
Show hintHide hint
Rs,max = 2B.
Show answerHide answer
Answer: C. 8000 symbols/s
An ideal low-pass channel of bandwidth B supports up to 2B = 8000 symbols/s without ISI (sinc pulses).
135. Binary data at 10 kbps is sent with raised-cosine pulse shaping of roll-off factor α = 0.5. The required baseband bandwidth is:
Show hintHide hint
B = (R/2)(1 + α).
Show answerHide answer
Answer: D. 7.5 kHz
B = (Rb/2)(1 + α) = 5 kHz × 1.5 = 7.5 kHz.
136. The Hamming distance between the code words 1011010 and 1001011 is:
Show hintHide hint
Count positions where the bits differ.
Show answerHide answer
Answer: B. 2
They differ in the 3rd and 7th positions, so the Hamming distance is 2.
137. A block code has a minimum Hamming distance of 5. The number of bit errors per code word it can always correct is:
Show hintHide hint
dmin ≥ 2t + 1.
Show answerHide answer
Answer: A. 2
Correction of t errors needs dmin ≥ 2t + 1, so t = ⌊(5 − 1)/2⌋ = 2.
138. The code rate of the (7, 4) Hamming code and the number of bit errors it can correct per code word are:
Show hintHide hint
Rate = k/n; dmin of Hamming codes is 3.
Show answerHide answer
Answer: B. About 0.57 and one
Code rate = k/n = 4/7 ≈ 0.57; with dmin = 3 it corrects one error per code word.
139. Which statement correctly distinguishes ARQ from FEC?
Show hintHide hint
Which one needs a feedback channel?
Show answerHide answer
Answer: C. ARQ detects errors and requests retransmission; FEC corrects errors at the receiver without a return channel
Automatic repeat request needs a feedback channel and retransmits erroneous frames; forward error correction adds enough redundancy to correct errors directly.
140. A random process is best described as:
Show hintHide hint
Fix the time and you get a random variable.
Show answerHide answer
Answer: D. An ensemble (collection) of time functions, one of which occurs in each trial
A random process X(t, s) assigns a sample function of time to each outcome s; at any fixed time it is a random variable.
141. A random process X(t) is wide-sense stationary (WSS) if:
Show hintHide hint
Only the mean and autocorrelation are checked.
Show answerHide answer
Answer: A. Its mean is constant and its autocorrelation depends only on the time difference τ
WSS needs only first- and second-order conditions: E[X(t)] = constant and R(t, t+τ) = R(τ). Full shift-invariance of all distributions is strict-sense stationarity.
142. The Wiener-Khinchin theorem states that the power spectral density of a WSS process is:
Show hintHide hint
Transform from the τ domain to the f domain.
Show answerHide answer
Answer: A. The Fourier transform of its autocorrelation function
S(f) = ∫ R(τ) e^(−j2πfτ) dτ; the PSD and autocorrelation form a Fourier transform pair.
143. White noise is defined as noise whose power spectral density:
Show hintHide hint
By analogy with white light.
Show answerHide answer
Answer: D. Is constant (flat) over all frequencies
White noise has a flat two-sided PSD of N0/2 W/Hz at all frequencies, by analogy with white light containing all colours equally.
144. The autocorrelation function of white noise with two-sided PSD N0/2 is:
Show hintHide hint
Inverse Fourier transform of a constant.
Show answerHide answer
Answer: A. (N0/2) δ(τ)
The inverse Fourier transform of a constant N0/2 is the impulse (N0/2)δ(τ).
145. The mean-square open-circuit thermal noise voltage of a resistor R at absolute temperature T over bandwidth B is:
Show hintHide hint
Don't confuse it with available power.
Show answerHide answer
Answer: A. 4kTRB
Johnson-Nyquist: v̄² = 4kTRB; kTB is the available noise power and 2qIB is shot noise.
146. The RMS thermal noise voltage of a 10 kΩ resistor at 300 K over a 10 kHz bandwidth is about (k = 1.38×10⁻²³ J/K):
Show hintHide hint
Take the square root of 4kTRB.
Show answerHide answer
Answer: D. 1.29 µV
v = √(4kTRB) = √(4 × 1.38×10⁻²³ × 300 × 10⁴ × 10⁴) = √(1.66×10⁻¹²) ≈ 1.29 µV.
6.5 Signal and system
32 questions · AEiE0605
147. A signal is best defined as:
Show hintHide hint
Think of what a microphone voltage or an image represents.
Show answerHide answer
Answer: A. A function of one or more independent variables that conveys information about a physical phenomenon
A signal is a function of independent variables (time, space, etc.) carrying information; a system is what processes it.
148. What is the total energy of the signal x(t) = e^(−2t)·u(t)?
Show hintHide hint
Square the signal before integrating.
Show answerHide answer
Answer: B. 0.25 J
E = ∫₀^∞ e^(−4t) dt = 1/4 = 0.25 J.
149. What is the average power of x(t) = 5 cos(100πt + π/3)?
Show hintHide hint
Average power of a sinusoid depends only on its amplitude.
Show answerHide answer
Answer: C. 12.5 W
For A cos(ωt + φ), the average power is A²/2 = 25/2 = 12.5 W; the phase does not matter.
150. What is the fundamental period of the discrete-time signal x[n] = cos(3πn/8)?
Show hintHide hint
N must be an integer, so find the smallest k that makes 16k/3 whole.
Show answerHide answer
Answer: D. 16 samples
N = 2πk/ω₀ = 16k/3; the smallest integer N occurs for k = 3, giving N = 16.
151. What is the value of ∫ (t² + 3)·δ(t − 2) dt taken over all time?
Show hintHide hint
The impulse picks out the value of the other factor at one instant.
Show answerHide answer
Answer: C. 7
By the sifting property the integral equals (t² + 3) at t = 2, i.e. 4 + 3 = 7.
152. The scaled impulse δ(2t) is equal to:
Show hintHide hint
Compare the areas under δ(2t) and δ(t).
Show answerHide answer
Answer: D. 0.5 δ(t)
δ(at) = δ(t)/|a|, so δ(2t) = δ(t)/2.
153. The unit step u(t) is related to the unit impulse δ(t) by:
Show hintHide hint
Think about the area accumulated by an impulse as time passes t = 0.
Show answerHide answer
Answer: B. u(t) is the running integral of δ(t)
u(t) = ∫ from −∞ to t of δ(τ) dτ, equivalently δ(t) = du(t)/dt.
154. The normalised sinc function sinc(x) = sin(πx)/(πx) has value 1 at x = 0 and is zero at:
Show hintHide hint
Where does sin(πx) vanish?
Show answerHide answer
Answer: D. All non-zero integer values of x
sin(πx) = 0 when x is an integer; at x = 0 the limit is 1, so the zeros are at x = ±1, ±2, ...
155. The signum function sgn(t) can be written in terms of the unit step as:
Show hintHide hint
Check the value of your expression for positive and for negative t.
Show answerHide answer
Answer: B. 2u(t) − 1
sgn(t) = +1 for t > 0 and −1 for t < 0, which equals 2u(t) − 1.
156. If x[n] = {1, 2, 1} (starting at n = 0) is applied to an LTI system with h[n] = {1, 1}, the output y[n] is:
Show hintHide hint
The output length is N₁ + N₂ − 1.
Show answerHide answer
Answer: C. {1, 3, 3, 1}
Linear convolution: y[0]=1, y[1]=2+1=3, y[2]=1+2=3, y[3]=1; the length is 3+2−1 = 4.
157. The convolution of the unit step with itself, u(t) * u(t), is:
Show hintHide hint
Convolving with u(t) is the same as integrating.
Show answerHide answer
Answer: C. t·u(t), the unit ramp
u(t)*u(t) = ∫₀^t 1 dτ = t for t ≥ 0, i.e. the unit ramp r(t) = t·u(t).
158. For an LTI system, the output for any input is completely determined by:
Show hintHide hint
Which single response lets you compute the output by convolution?
Show answerHide answer
Answer: B. Its impulse response
y(t) = x(t) * h(t); the impulse response fully characterises an LTI system.
159. Which of the following systems is linear and time-invariant?
Show hintHide hint
Test each one for additivity and for the effect of a shifted input.
Show answerHide answer
Answer: A. y(t) = 3 x(t − 2)
3x(t−2) is a scaled delay (LTI). x+2 fails homogeneity, t·x(t) and x(2t) are time-varying.
160. A real periodic signal has odd symmetry and half-wave symmetry (like a square wave centred on t = 0). Its trigonometric Fourier series contains:
Show hintHide hint
Apply the two symmetry rules one after the other.
Show answerHide answer
Answer: D. Only sine terms of odd harmonics
Odd symmetry removes a₀ and cosine terms; half-wave symmetry removes even harmonics, leaving odd-harmonic sines.
161. A periodic signal is x(t) = 3 + 4 cos(ω₀t) + 2 sin(3ω₀t). Its average power is:
Show hintHide hint
DC power is A², each sinusoid contributes A²/2.
Show answerHide answer
Answer: D. 19 W
By Parseval: P = 3² + 4²/2 + 2²/2 = 9 + 8 + 2 = 19 W.
162. Near a jump discontinuity, the partial sum of a Fourier series overshoots by about 9% of the jump no matter how many terms are taken. This is called:
Show hintHide hint
The effect is named after a physicist who explained it in 1899.
Show answerHide answer
Answer: B. Gibbs phenomenon
The persistent ≈9% overshoot near discontinuities is Gibbs phenomenon.
163. Which property holds for the exponential Fourier series coefficients cₖ of a real, even periodic signal?
Show hintHide hint
Combine conjugate symmetry with even symmetry.
Show answerHide answer
Answer: D. cₖ are real and even in k
A real signal gives c₋ₖ = cₖ*; even symmetry makes the coefficients real, so they are real and even.
164. The Fourier transform of x(t) = e^(−at)·u(t), a > 0, is:
Show hintHide hint
Combine the two exponents and integrate from 0 to ∞.
Show answerHide answer
Answer: C. 1/(a + jω)
X(ω) = ∫₀^∞ e^(−at) e^(−jωt) dt = 1/(a + jω).
165. The Fourier transform of the signum function sgn(t) is:
Show hintHide hint
The signum has zero average value, so no impulse at ω = 0.
Show answerHide answer
Answer: A. 2/(jω)
sgn(t) is the limit of e^(−a|t|)sgn(t); its transform tends to 2/(jω). Note u(t) = (1 + sgn t)/2 gives πδ(ω) + 1/(jω).
166. A rectangular pulse of width 2 ms is centred at t = 0. Its amplitude spectrum has the first zero crossing at:
Show hintHide hint
Nulls of the sinc occur at multiples of 1/T.
Show answerHide answer
Answer: A. 500 Hz
The transform is A·T·sinc(fT); the first null is at f = 1/T = 1/0.002 = 500 Hz.
167. A signal x(t) is band-limited to 4 kHz. The bandwidth of x(3t) is:
Show hintHide hint
Compression in time means expansion in frequency.
Show answerHide answer
Answer: B. 12 kHz
x(at) ↔ (1/|a|)X(ω/a); compressing time by 3 expands the spectrum by 3, giving 12 kHz.
168. Delaying a signal by t₀ seconds changes its Fourier transform X(ω) to:
Show hintHide hint
A delay should not change the magnitude spectrum.
Show answerHide answer
Answer: A. X(ω)·e^(−jωt₀)
The time-shift property: x(t − t₀) ↔ X(ω)e^(−jωt₀); only the phase changes.
169. According to the Wiener–Khinchin theorem, the power spectral density of a wide-sense stationary signal is:
Show hintHide hint
It links a time-domain correlation to a frequency-domain density.
Show answerHide answer
Answer: B. The Fourier transform of its autocorrelation function
Sₓ(f) = F{Rₓ(τ)} for a WSS process (Wiener–Khinchin).
170. The energy spectral density of x(t) = e^(−2t)·u(t) at ω = 0 is:
Show hintHide hint
ESD is the squared magnitude of the Fourier transform.
Show answerHide answer
Answer: B. 0.25
Ψ(ω) = |X(ω)|² = 1/(4 + ω²); at ω = 0 it is 1/4 = 0.25.
171. The DTFT X(e^jω) of any discrete-time sequence is always:
Show hintHide hint
Consider e^(−jωn) when ω increases by 2π.
Show answerHide answer
Answer: C. Periodic in ω with period 2π
Because e^(−j(ω+2π)n) = e^(−jωn) for integer n, X(e^jω) repeats every 2π.
172. For x[n] = (0.5)ⁿ·u[n], the value of its DTFT at ω = 0 is:
Show hintHide hint
At ω = 0 the DTFT is simply the sum of all samples.
Show answerHide answer
Answer: C. 2
X(e^jω) = 1/(1 − 0.5e^(−jω)); at ω = 0 it is 1/(1 − 0.5) = 2 (also the sum of the samples).
173. A periodic impulse train x[n] with period N = 4 has x[0] = 1 and x[1] = x[2] = x[3] = 0 in each period. Its DTFS coefficients aₖ (with the 1/N factor in the analysis equation) are:
Show hintHide hint
Only the n = 0 term survives in the sum.
Show answerHide answer
Answer: C. aₖ = 1/4 for every k
aₖ = (1/N) Σ x[n]e^(−j2πkn/N) = (1/4)·1 = 1/4 for k = 0,1,2,3.
174. A discrete-time periodic signal with fundamental period N has how many distinct Fourier series coefficients?
Show hintHide hint
Discrete-time complex exponentials repeat after N harmonics.
Show answerHide answer
Answer: A. N
The DTFS coefficients are themselves periodic with period N, so only N are distinct.
175. For a continuous-time LTI system, which condition guarantees BIBO stability?
Show hintHide hint
The test uses the absolute value of h(t).
Show answerHide answer
Answer: C. ∫|h(t)| dt over all t is finite
An LTI system is BIBO stable if and only if its impulse response is absolutely integrable.
176. Two LTI systems with impulse responses h₁(t) and h₂(t) are connected in cascade. The overall impulse response is:
Show hintHide hint
In frequency it is the product of H₁ and H₂.
Show answerHide answer
Answer: D. h₁(t) * h₂(t)
A cascade of LTI systems has the convolution of the impulse responses (product of frequency responses).
177. The complex exponential e^(jω₀t) applied to an LTI system produces the output H(jω₀)e^(jω₀t). For this reason complex exponentials are called:
Show hintHide hint
The input shape is preserved; only a complex scale factor appears.
Show answerHide answer
Answer: C. Eigenfunctions of LTI systems
The output is the same exponential scaled by the eigenvalue H(jω₀), so e^(jω₀t) is an eigenfunction.
178. For the system y[n] = 0.5(x[n] + x[n − 1]), what is |H(e^jω)| at ω = π/2?
Show hintHide hint
Factor out e^(−jω/2) to get a cosine.
Show answerHide answer
Answer: C. 0.707
H(e^jω) = 0.5(1 + e^(−jω)) ⇒ |H| = |cos(ω/2)| = cos(π/4) ≈ 0.707.
6.6 Digital Signal Processing
38 questions · AEiE0606
179. For stability of a causal LTI system, all poles of the transfer function must lie:
Show hintHide hint
The jω-axis maps onto the unit circle.
Show answerHide answer
Answer: D. In the left half of the s-plane for continuous time, and inside the unit circle of the z-plane for discrete time
Causal CT stability needs Re(poles) < 0; causal DT stability needs |poles| < 1.
180. The (bilateral) z-transform of a sequence x[n] is defined as:
Show hintHide hint
The power of z carries a negative sign.
Show answerHide answer
Answer: A. X(z) = Σ x[n]·z⁻ⁿ summed over all n
The bilateral z-transform is Σ from −∞ to ∞ of x[n]z⁻ⁿ; the DTFT is its value on the unit circle.
181. The z-transform of x[n] = aⁿ·u[n] and its region of convergence are:
Show hintHide hint
It is a geometric series in a z⁻¹.
Show answerHide answer
Answer: D. 1/(1 − a z⁻¹), |z| > |a|
Σ (a z⁻¹)ⁿ converges to 1/(1 − a z⁻¹) when |a z⁻¹| < 1, i.e. |z| > |a|.
182. Both aⁿu[n] and −aⁿu[−n−1] have the same algebraic z-transform 1/(1 − a z⁻¹). They are distinguished by:
Show hintHide hint
The expression is identical, so something else must differ.
Show answerHide answer
Answer: C. Their regions of convergence
The first has ROC |z| > |a| (right-sided), the second |z| < |a| (left-sided); X(z) alone is not unique without the ROC.
183. For a finite-duration causal sequence such as x[n] = {1, 2, 3} (n = 0, 1, 2), the ROC is:
Show hintHide hint
Where do terms in z⁻¹ become infinite?
Show answerHide answer
Answer: D. The entire z-plane except z = 0
X(z) = 1 + 2z⁻¹ + 3z⁻² is finite everywhere except at z = 0 where the negative powers blow up.
184. Which statement about the region of convergence of a z-transform is correct?
Show hintHide hint
What is the value of X(z) at a pole?
Show answerHide answer
Answer: B. The ROC cannot contain any pole
At a pole X(z) is infinite, so poles bound the ROC; a two-sided sequence has an annular ROC.
185. A system has H(z) with poles at z = 0.5 and z = 2. If the system is causal, then it is:
Show hintHide hint
Causality fixes the ROC; stability needs the unit circle inside it.
Show answerHide answer
Answer: C. Unstable, because its ROC |z| > 2 does not include the unit circle
Causal ⇒ ROC outside the outermost pole (|z| > 2), which excludes |z| = 1, so it is unstable.
186. The z-transform of x[n − k] (k > 0, causal x) is:
Show hintHide hint
z⁻¹ is the unit delay operator.
Show answerHide answer
Answer: D. z⁻ᵏ X(z)
Time-shift property: a delay of k samples multiplies X(z) by z⁻ᵏ.
187. The convolution y[n] = x[n] * h[n] corresponds in the z-domain to:
Show hintHide hint
Same idea as the Laplace and Fourier convolution properties.
Show answerHide answer
Answer: D. Y(z) = X(z)·H(z)
The convolution property of the z-transform: convolution in time becomes multiplication in z.
188. The inverse z-transform of X(z) = 1/(1 − 0.5z⁻¹), |z| > 0.5, evaluated at n = 2 is:
Show hintHide hint
Identify the standard pair aⁿu[n].
Show answerHide answer
Answer: A. 0.25
x[n] = (0.5)ⁿu[n], so x[2] = 0.25.
189. By Parseval's theorem, the energy of x[n] = (0.5)ⁿ·u[n] is:
Show hintHide hint
Square each sample and sum the geometric series.
Show answerHide answer
Answer: B. 4/3
Σ|x[n]|² = Σ(0.25)ⁿ = 1/(1 − 0.25) = 4/3; the same value results from (1/2π)∫|X(e^jω)|² dω.
190. A stable causal system H(z) = (1 + z⁻¹)/2 is driven by cos(πn/3). The steady-state output amplitude is:
Show hintHide hint
Evaluate H(z) on the unit circle at z = e^(jπ/3).
Show answerHide answer
Answer: C. 0.866
|H(e^jω)| = |cos(ω/2)|; at ω = π/3, cos(π/6) ≈ 0.866.
191. When a sinusoid is suddenly applied at n = 0 to a stable causal system H(z), the total response consists of:
Show hintHide hint
Separate the response into terms from the system poles and from the input poles.
Show answerHide answer
Answer: A. A transient part from the poles of H(z) that decays to zero, plus a steady-state sinusoid scaled by H(e^jω₀)
For a stable system the system-pole terms die out, leaving |H(e^jω₀)|cos(ω₀n + ∠H(e^jω₀)).
192. What is the DC gain of H(z) = 1/(1 − 0.8z⁻¹)?
Show hintHide hint
DC (ω = 0) maps to z = 1.
Show answerHide answer
Answer: B. 5
DC corresponds to z = 1: H(1) = 1/(1 − 0.8) = 5.
193. A filter has a single zero at z = −1 and a single pole at z = 0.9. It acts as a:
Show hintHide hint
z = 1 is ω = 0 and z = −1 is ω = π.
Show answerHide answer
Answer: B. Low-pass filter
The pole near z = 1 boosts low frequencies and the zero at z = −1 (ω = π) kills the highest frequency, giving a low-pass response.
194. The N-point DFT X[k] of a finite sequence is related to its z-transform by:
Show hintHide hint
The DFT samples lie on the unit circle.
Show answerHide answer
Answer: B. X[k] = X(z) evaluated at z = e^(j2πk/N)
The DFT samples the z-transform at N equally spaced points on the unit circle.
195. The 4-point DFT of x[n] = {1, 1, 1, 1} is:
Show hintHide hint
A constant has energy only at DC.
Show answerHide answer
Answer: D. {4, 0, 0, 0}
X[0] = sum of samples = 4; for k ≠ 0 the complex exponentials sum to zero.
196. For x[n] = {1, 2, 3, 4}, the 4-point DFT value X[1] is:
Show hintHide hint
For N = 4, W₄ = −j.
Show answerHide answer
Answer: C. −2 + 2j
X[1] = Σ x[n](−j)ⁿ = 1 − 2j − 3 + 4j = −2 + 2j.
197. A signal sampled at 8 kHz is analysed with a 256-point DFT. The frequency spacing between DFT bins is:
Show hintHide hint
Divide the sampling rate by the number of points.
Show answerHide answer
Answer: A. 31.25 Hz
Δf = fs/N = 8000/256 = 31.25 Hz.
198. The 2-point circular convolution of x[n] = {1, 2} and h[n] = {3, 4} is:
Show hintHide hint
Indices wrap modulo 2.
Show answerHide answer
Answer: C. {11, 10}
y[0] = 1·3 + 2·4 = 11 and y[1] = 1·4 + 2·3 = 10.
199. Two sequences of lengths 50 and 30 are to be linearly convolved using DFTs. The minimum DFT length that avoids wrap-around error is:
Show hintHide hint
Find the length of the linear convolution result.
Show answerHide answer
Answer: B. 79
Circular convolution equals linear convolution when N ≥ L + M − 1 = 50 + 30 − 1 = 79.
200. Multiplying the N-point DFTs of two sequences and taking the inverse DFT gives their:
Show hintHide hint
DFT treats sequences as one period of a periodic signal.
Show answerHide answer
Answer: A. N-point circular convolution
The DFT product property: X₁[k]X₂[k] ↔ x₁[n] ⊛ x₂[n] (circular convolution modulo N).
201. x[n] is real and its 8-point DFT has X[1] = 2 + 3j. Then X[7] equals:
Show hintHide hint
Real sequences have conjugate-symmetric DFTs.
Show answerHide answer
Answer: A. 2 − 3j
For real x[n], X[N − k] = X*[k], so X[7] = X*[1] = 2 − 3j.
202. A circular shift x[(n − m) mod N] in time corresponds in the DFT domain to:
Show hintHide hint
A shift in one domain is a phase factor in the other.
Show answerHide answer
Answer: C. Multiplying X[k] by e^(−j2πkm/N)
Circular time shift ↔ multiplication by the linear phase factor W_N^(km) = e^(−j2πkm/N).
203. For x[n] = {1, 2, 3, 4}, Σ|X[k]|² over its 4-point DFT equals:
Show hintHide hint
Mind the 1/N factor in the DFT form of Parseval's relation.
Show answerHide answer
Answer: C. 120
Parseval for the DFT: Σ|x[n]|² = (1/N)Σ|X[k]|², so Σ|X[k]|² = 4 × 30 = 120.
204. The N-point DFT of the unit impulse δ[n] is:
Show hintHide hint
Only the n = 0 term contributes.
Show answerHide answer
Answer: A. 1 for every k
X[k] = Σ δ[n]W^(kn) = W⁰ = 1 for all k.
205. In the classical approach to IIR filter design, a digital filter is obtained by:
Show hintHide hint
IIR design borrows from mature analog filter theory.
Show answerHide answer
Answer: D. Designing an analog prototype (e.g. Butterworth or Chebyshev) and mapping it to the z-domain
IIR design commonly converts a well-known analog filter to digital via impulse invariance or the bilinear transformation; the other options are FIR methods.
206. The magnitude response of a Butterworth low-pass filter is:
Show hintHide hint
No ripple anywhere.
Show answerHide answer
Answer: A. Maximally flat in the passband and monotonic everywhere
|H|² = 1/(1 + (Ω/Ωc)^(2N)) has all derivatives zero at Ω = 0 and decreases monotonically.
207. At its cutoff frequency Ωc, the gain of a Butterworth filter of any order is:
Show hintHide hint
Substitute Ω = Ωc in |H|².
Show answerHide answer
Answer: C. −3 dB
At Ω = Ωc, |H|² = 1/2, i.e. 10 log₁₀(0.5) ≈ −3 dB regardless of N.
208. What is the minimum order of a Butterworth low-pass filter that gives at least 30 dB attenuation at twice the cutoff (3 dB) frequency?
Show hintHide hint
Set 10 log₁₀(1 + 2^(2N)) ≥ 30 and solve for N.
Show answerHide answer
Answer: A. 5
Need 1 + 2^(2N) ≥ 10³ ⇒ N ≥ log₁₀(999)/(2 log₁₀2) ≈ 4.98, so N = 5.
209. In the impulse-invariance method, an analog pole at s = pₖ maps to a digital pole at:
Show hintHide hint
Sample the exponential e^(pₖt) at t = nT.
Show answerHide answer
Answer: B. z = e^(pₖT)
Sampling hₐ(t) = e^(pₖt) at t = nT gives (e^(pₖT))ⁿ, i.e. a pole at z = e^(pₖT).
210. Using impulse invariance with T = 0.1 s, the analog filter H(s) = 1/(s + 2) gives a digital pole at approximately:
Show hintHide hint
Use z = e^(pT) with p = −2.
Show answerHide answer
Answer: A. 0.819
z = e^(−2 × 0.1) = e^(−0.2) ≈ 0.819.
211. The impulse-invariance method is unsuitable for designing high-pass and band-stop filters because:
Show hintHide hint
Sampling a non-band-limited response causes what?
Show answerHide answer
Answer: A. The frequency response of the sampled impulse response suffers from aliasing
Impulse invariance samples hₐ(t); since high-pass responses are not band-limited, the spectral copies overlap (aliasing).
212. In FIR design by Fourier series (window) method, the ideal low-pass impulse response with cutoff ωc and delay α is hd[n] = sin(ωc(n − α))/(π(n − α)). For ωc = π/4, hd[α] equals:
Show hintHide hint
Use sin(x)/x → 1 as x → 0.
Show answerHide answer
Answer: D. 0.25
At n = α the limit is ωc/π = 0.25.
213. Truncating the ideal impulse response with a rectangular window causes:
Show hintHide hint
Abrupt truncation in time causes ripple in frequency.
Show answerHide answer
Answer: A. Gibbs oscillations, giving a minimum stopband attenuation of only about 21 dB
The rectangular window's large sidelobes produce ripple near the band edge; peak stopband attenuation is about 21 dB regardless of length.
214. Arrange the windows in increasing order of minimum stopband attenuation:
Show hintHide hint
The rectangular window is the worst.
Show answerHide answer
Answer: A. Rectangular, Hanning, Hamming, Blackman
Typical values: rectangular ≈ 21 dB, Hanning ≈ 44 dB, Hamming ≈ 53 dB, Blackman ≈ 74 dB.
215. The Hamming window w[n] = 0.54 − 0.46 cos(2πn/(M − 1)) has the value at its end points (n = 0 and n = M − 1) of:
Show hintHide hint
Put n = 0 into the formula.
Show answerHide answer
Answer: C. 0.08
At n = 0, cos(0) = 1 so w = 0.54 − 0.46 = 0.08 (unlike the Hanning window, which goes to zero).
216. Using the approximation that the transition width of a Hamming-window design equals its main-lobe width 8π/M, the length M needed for a transition width of 0.1π rad/sample is:
Show hintHide hint
Solve 8π/M = 0.1π for M.
Show answerHide answer
Answer: D. 80
M = 8π/Δω = 8π/(0.1π) = 80.