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Microwave, fiber optics and satellite: board exam questions with solutions

On the ECE exam under Electronic Systems and Technologies (EST). Try each question before you open its solution, then practise with fresh numbers below: 2 computed question types on this topic.

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Questions

1. What is the TE10 cutoff frequency of an air-filled rectangular waveguide whose broad dimension is a = 2.286 cm?

  1. 13.11 GHz
  2. 3.28 GHz
  3. 10.00 GHz
  4. 6.56 GHz
Show the answer and solution

Answer: D. 6.56 GHz

fc = c/(2a) = 3 × 10⁸/(2 × 0.02286) = 6.56 GHz.

2. In that same waveguide (a = 2.286 cm, TE10), what is the guide wavelength at 10 GHz?

  1. 3.00 cm
  2. 2.26 cm
  3. 3.97 cm
  4. 4.57 cm
Show the answer and solution

Answer: C. 3.97 cm

λ0 = 3.00 cm; λg = λ0/√(1 − (fc/f)²) = 3.00/√(1 − (6.557/10)²) = 3.97 cm.

3. A step-index fiber has core index 1.48 and cladding index 1.46. What is its acceptance half-angle in air?

  1. 80.6°
  2. 14.0°
  3. 28.1°
  4. 1.1°
Show the answer and solution

Answer: B. 14.0°

NA = √(1.48² − 1.46²) = 0.2425; θa = sin⁻¹(NA) = 14.0°.

4. A fiber link launches 0 dBm into 40 km of fiber at 0.25 dB/km, with two connectors at 0.5 dB each and eight splices at 0.1 dB each. What is the received power?

  1. −11.8 dBm
  2. −10.0 dBm
  3. −11.6 dBm
  4. −12.8 dBm
Show the answer and solution

Answer: A. −11.8 dBm

Loss = 40 × 0.25 + 2 × 0.5 + 8 × 0.1 = 10 + 1 + 0.8 = 11.8 dB, so Pr = 0 − 11.8 = −11.8 dBm.

5. What is the main reason single-mode fiber supports far higher bandwidth-distance products than multimode fiber?

  1. It has a larger core that accepts more light
  2. It has zero attenuation
  3. It uses LEDs instead of lasers
  4. It eliminates intermodal (modal) dispersion
Show the answer and solution

Answer: D. It eliminates intermodal (modal) dispersion

With only one guided mode there are no differing path delays between modes, leaving only much smaller chromatic and polarization dispersion.

6. Approximately how high above the equator is a geostationary satellite?

  1. 20,200 km
  2. 42,200 km
  3. 35,800 km
  4. 1,200 km
Show the answer and solution

Answer: C. 35,800 km

Orbital radius r = (GM·T²/4π²)^(1/3) with T = 1 sidereal day gives r ≈ 42,164 km; minus Earth's equatorial radius (6378 km) gives about 35,786 km. (20 200 km is GPS; 42 200 km is the radius, not the altitude.)

7. What is the free-space path loss from a satellite to an earth station 36,000 km away at 4 GHz?

  1. 175.6 dB
  2. 195.6 dB
  3. 215.6 dB
  4. 185.6 dB
Show the answer and solution

Answer: B. 195.6 dB

FSPL = 32.45 + 20 log 4000 + 20 log 36 000 = 32.45 + 72.04 + 91.13 = 195.6 dB.

8. What is the critical angle (measured from the normal) at the core-cladding boundary of a fiber with core index 1.50 and cladding index 1.45?

  1. 75.2°
  2. 14.8°
  3. 22.6°
  4. 45.0°
Show the answer and solution

Answer: A. 75.2°

θc = sin⁻¹(n2/n1) = sin⁻¹(1.45/1.50) = 75.2°. Rays striking the boundary at more than this angle are totally internally reflected.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. What is the numerical aperture of a fiber with core index 1.5 and cladding index 1.495?

  1. 0.122
  2. 0.997
  3. 0.005
  4. 0.0707
Show the answer and solution

Answer: A. 0.122

NA = √(n₁² − n₂²) = √(1.5² − 1.495²) = 0.122.

10. What is the TE₁₀ cutoff frequency of an air-filled rectangular waveguide whose broad wall is 10.67 mm?

  1. 28.1 GHz
  2. 14.1 GHz
  3. 9.37 GHz
  4. 7.03 GHz
Show the answer and solution

Answer: B. 14.1 GHz

TE₁₀ cutoff: fc = c/(2a) = 3×10⁸/(2 × 10.67 mm) = 14.1 GHz.

Other topics in the same subjects

Communications fundamentals (dB, noise) · AM, FM and PM · Digital communications · Transmission lines · Antennas and wave propagation · Data communications and networks · Broadcasting and acoustics

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