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Communications fundamentals (dB, noise): 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: 4 computed question types on this topic.

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Questions

1. An amplifier's output power is 1000 times its input power. What is its gain in decibels?

  1. 60 dB
  2. 30 dB
  3. 3 dB
  4. 1000 dB
Show the answer and solution

Answer: B. 30 dB

G = 10 log(1000) = 30 dB. (60 dB would be the voltage-ratio formula misapplied to power.)

2. What power corresponds to +20 dBm?

  1. 100 mW
  2. 20 mW
  3. 10 mW
  4. 1 W
Show the answer and solution

Answer: A. 100 mW

P = 1 mW × 10^(20/10) = 100 mW.

3. What is the thermal noise power, in dBm, in a 1 MHz bandwidth at 290 K?

  1. −174 dBm
  2. −144 dBm
  3. −84 dBm
  4. −114 dBm
Show the answer and solution

Answer: D. −114 dBm

N = kTB = 1.38 × 10⁻²³ × 290 × 10⁶ = 4.00 × 10⁻¹⁵ W → 10 log(N/1 mW) = −114.0 dBm (−174 dBm/Hz + 60 dB).

4. What rms noise voltage does a 10 kΩ resistor at 290 K produce across a 100 kHz bandwidth?

  1. 2.00 µV
  2. 16.02 µV
  3. 4.00 µV
  4. 4002 µV
Show the answer and solution

Answer: C. 4.00 µV

Vn = √(4kTRB) = √(4 × 1.38 × 10⁻²³ × 290 × 10⁴ × 10⁵) ≈ 4.00 µV.

5. A receiver front end has a first stage with 20 dB gain and 2 dB noise figure, followed by a stage with 10 dB noise figure. What is the overall noise figure?

  1. 12.00 dB
  2. 2.24 dB
  3. 2.00 dB
  4. 10.00 dB
Show the answer and solution

Answer: B. 2.24 dB

Friis: F = F1 + (F2 − 1)/G1 = 1.585 + (10 − 1)/100 = 1.675 → NF = 10 log(1.675) = 2.24 dB.

6. What is the equivalent noise temperature of an amplifier with a 3 dB noise figure (reference 290 K)?

  1. 289 K
  2. 870 K
  3. 580 K
  4. 87 K
Show the answer and solution

Answer: A. 289 K

Te = 290(F − 1) = 290(10^0.3 − 1) = 290 × 0.995 = 289 K.

7. How much information, in bits, is conveyed by a message whose probability of occurrence is 1/8?

  1. 8 bits
  2. 0.125 bits
  3. 0.375 bits
  4. 3 bits
Show the answer and solution

Answer: D. 3 bits

I = log₂(1/p) = log₂ 8 = 3 bits.

8. A source emits three symbols with probabilities 0.5, 0.25 and 0.25. What is its entropy?

  1. 1.0 bits/symbol
  2. 1.58 bits/symbol
  3. 1.5 bits/symbol
  4. 2.0 bits/symbol
Show the answer and solution

Answer: C. 1.5 bits/symbol

H = Σ p log₂(1/p) = 0.5(1) + 0.25(2) + 0.25(2) = 1.5 bits/symbol (1.58 would require equal probabilities).

Worked examples with computed numbers

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

9. An amplifier takes 1 mW in and delivers 5 mW. What is its power gain in dB?

  1. 14 dB
  2. 0.5 dB
  3. 0.699 dB
  4. 6.99 dB
Show the answer and solution

Answer: D. 6.99 dB

Power gain in dB = 10 log(Po/Pi) = 10 log 5 = 6.99 dB.

10. Express 2 mW in dBm.

  1. 6.02 dBm
  2. 3.01 dBm
  3. 33 dBm
  4. -27 dBm
Show the answer and solution

Answer: B. 3.01 dBm

dBm = 10 log(P/1 mW) = 10 log(2 mW/1 mW) = 3.01 dBm.

11. What is the thermal noise power in a 10 kHz bandwidth at 290 K?

  1. -174 dBm
  2. -134 dBm
  3. -164 dBm
  4. -104 dBm
Show the answer and solution

Answer: B. -134 dBm

N = kTB = 1.38×10⁻²³ × 290 × 10 kHz = 0.00004 pW = -134 dBm (−174 dBm/Hz + 10 log B).

12. A preamp (NF 3 dB, gain 20 dB) drives a receiver with NF 8 dB. What is the overall noise figure?

  1. 3.4 dB
  2. 3.11 dB
  3. 11 dB
  4. 9.19 dB
Show the answer and solution

Answer: B. 3.11 dB

Friis: F = F₁ + (F₂ − 1)/G₁ = 1.995 + (6.31 − 1)/100 = 2.048, NF = 3.11 dB.

Other topics in the same subjects

AM, FM and PM · Digital communications · Transmission lines · Antennas and wave propagation · Microwave, fiber optics and satellite · Data communications and networks · Broadcasting and acoustics

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