250+ TOP MCQs on Noise and Answers

Optical Communications Multiple Choice Questions on “Noise”.

1. _____________ refers to any spurious or undesired disturbances that mask the received signal in a communication system.
a) Attenuation
b) Noise
c) Dispersion
d) Bandwidth

Answer: b
Explanation: Noise is an unwanted and undesirable quantity. It affects the received signal in a communication system. In optical fiber communication systems, noise is due to the spontaneous fluctuations rather than erratic disturbances.

2. How many types of noise are observed because of the spontaneous fluctuations in optical fiber communication systems?
a) One
b) Four
c) Two
d) Three

Answer: d
Explanation: There are three types of noise because of the spontaneous fluctuations in optical fiber communication systems. These are thermal noise, the dark current noise and quantum noise. These noise types are not caused by the electronic interference.

3. ______________ is caused due to thermal interaction between the free electrons and the vibrating ions in the conduction medium.
a) Thermal noise
b) Dark noise
c) Quantum noise
d) Gaussian noise

Answer: a
Explanation: Thermal noise is basically a spontaneous fluctuation caused due to thermal interaction of electrons and ions. It is especially prevalent in resistors at room temperature. Thermal noise is measured in the form of current and is called as thermal noise current.

4. A small leakage current still flows from the device terminals even if there is no optical power incident on the photo detector.
a) True
b) False

Answer: a
Explanation: A reverse leakage current that flows from the device terminals is called as dark current. This dark current contributes to the total system noise. This gives random fluctuations about the average particle flow of the photocurrent.

5. ___________ distribution provides the description the random statistics of light emitted in black body radiation.
a) Poisson
b) Cumulative
c) Probability
d) Bose-Einstein

Answer: d
Explanation: Incoherent light is emitted by independent atoms and therefore there is no phase relationship between the emitted photons. The property dictates an exponential intensity distribution which is identical to Bose-Einstein distribution.

6. The probability of zero pairs being generated when a light pulse is present is given by which of the following equation?
a) P(0/1) = exp(-Zm)
b) P(x) = exp (Zm)
c) P(y) = x (0) + x(1)
d) P(z) = P(-Zm)

Answer: a
Explanation: The probability of zero pairs being generated when a light pulse is present is given by equation –
P (0/1) = exp(-Zm)
Where, P (0/1) represents the system error probability p(e) and Zm is variance of the probability distribution.

7. The minimum pulse energy needed to maintain a given bit-error-rate (BER) which any practical receiver must satisfy is known as ___________
a) Minimal energy
b) Quantum limit
c) Point of reversed
d) Binary signaling

Answer: b
Explanation: A perfect photo detector emits no electron-hole pairs in the absence of illumination. The error probability determines a standardized fundamental limit in digital optical communications. This limit is termed as quantum limit.

8. A digital optical fiber communication system requires a maximum bit-error-rate of 10-9. Find the average number of photons detected in a time period for a given BER.
a) 19.7
b) 21.2
c) 20.7
d) 26.2

Answer: c
Explanation: The probability of error is given by-
P(e) = exp(-Zm)
Where, Zm = No. of photons
Here P(e) = 10-9, therefore Zm is calculated from above relation.

9. For a given optical fiber communication system, P(e) = 10-9, Zm = 20.7, f = 2.9×1014, η = 1. Find the minimum pulse energy or quantum limit.
a) 3.9×10-18
b) 4.2×10-18
c) 6.2×10-14
d) 7.2×10-14

Answer: a
Explanation: The minimum pulse energy or quantum limit is given by –
Emin = Zmhf/η
Where, Zm = Number of photons
h = Planck’s constant
f = frequency
η = Quantum efficiency.

10. An analog optical fiber system operating at wavelength 1μmhas a post-detection bandwidth of 5MHz. Assuming an ideal detector and incident power of 198 nW, calculate the SNR (f = 2.99×1014Hz).
a) 46
b) 40
c) 50
d) 52

Answer: c
Explanation: The SNR is given by –
S/N = ηP0/2hfB
Where, η = 1 (for ideal detector)
P0 = incident power
h = Planck’s constant
B = Bandwidth.

11. The incident optical power required to achieve a desirable SNR is 168.2nW. What is the value of incident power in dBm?
a) -37.7 dBm
b) -37 dBm
c) – 34 dBm
d) -38.2 dBm

Answer: a
Explanation: Incident power in denoted by P0. It is given by –
P0 = 10log10(P0(watts))
Where P0(watts) = incident power in Watts/milliWatt.

12. In the equation given below, what does τstands for?

Zm = ηP0τ/hf

a) Velocity
b) Time
c) Reflection
d) Refractive index

Answer: b
Explanation: In the given equation, Zm is the variance of the probability distribution. The number of electrons generated in time τis equal to the average of the number of photons detected over this time period Zm. Hence, τ is the time and P0 is the incident power, ηis the quantum efficiency and f is the frequency.