This set of Optical Communications Multiple Choice Questions & Answers (MCQs) focuses on “Digital System Planning Considerations”.

1. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine number of bits in a frame.

a) 64

b) 128

c) 32

d) 256

View Answer

Explanation: Number of bits in a frame can be calculated as follows:

Bits in a frame = No. of channels * Sampling rate for each channel.

2. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the transmission rate for system with 256 bits in a frame.

a) 2.96 Mbits/s

b) 2.048 Mbits/s

c) 3.92 Mbits/s

d) 4 Mbits/s

View Answer

Explanation: Transmission rate can be determined by-

Transmission rate = Sampling rate * No. of bits in a frame.

3. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the bit duration with transmission rate of 2.048 M bits/s.

a) 388 ns

b) 490 ns

c) 488 ns

d) 540 ns

View Answer

Explanation: Bit duration is the reciprocal of the transmission rate. Thus, it is given by-

Bit duration = 1/transmission rate.

4. The bit duration is 488 ns. Sampling rate for each channel on 32-channel PCM is 8 KHz encoded into 8 bits. Determine the time slot duration.

a) 3.2 μs

b) 3.1 μs

c) 7 μs

d) 3.9 μs

View Answer

Explanation: Time slot duration is given by –

Time slot duration = Encoded bits * bit duration.

5. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine duration of frame with time slot duration of 3.9μs.

a) 125 μs

b) 130 μs

c) 132 μs

d) 133 μs

View Answer

Explanation: Duration of a frame is determined by –

Duration of a frame = 32 * time slot duration.

6. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the duration of multi-frame if duration of a frame is 125μs.

a) 2ms

b) 3ms

c) 4ms

d) 10ms

View Answer

Explanation: Multi-frame duration can be determined by –

Multi-frame duration = 16 * Duration of a single frame.

7. Determine excess avalanche noise factor F(M) if APD has multiplication factor of 100, carrier ionization rate of 0.02.

a) 3.99

b) 3.95

c) 4.3

d) 4

View Answer

Explanation: Excess avalanche noise factor is computed by –

F (M) = k*M + (2-1/M) (1-k), where k is ionization rate and M is the multiplication factor.

8. Compute average number of photons incident at receiver in APD if quantum efficiency is 80%, F (M) = 4, SNR = 144.

a) 866

b) 865

c) 864

d) 867

View Answer

Explanation: Average number of photons arez

_{m}=[2β

_{ς}F(M)]*[S/N*η]

Here, η = quantum efficiency, S/N = signal to noise ratio.

9. Determine incident optical power if z_{m}=864, wavelength = 1μm.

a) -85 dBm

b) -80 dBm

c) -69.7 dBm

d) -60.7 dBm

View Answer

Explanation: Incident optical power is P0=z

_{m}h

_{c}B

_{T}/2λ. Here z

_{m}=average number of photons, h

_{c}=Planck’s constant.

10. Determine wavelength of incident optical power if z_{m}=864, incident optical power is -60.7 dB, B_{T}=1 * 10^{7}.

a) 1 μs

b) 2 μs

c) 3 μs

d) 4 μs

View Answer

Explanation: Wavelength is determined by λ=z

_{m}h

_{c}B

_{T}/2P

_{0}. Here z

_{m}=average number of photons, h

_{c}=Planck’s constant, P

_{0}=incident optical power.

11. Determine total channel loss if connector loss at source and detector is 3.5 and 2.5 dB and attenuation of 5 dB/km.

a) 34 dB

b) 35 dB

c) 36 dB

d) 38 dB

View Answer

Explanation: The total channel loss is C

_{L}=(α

_{fc}+α

_{j})L + α

_{cr}. Here α

_{cr}=loss at detector and source combined, α

_{fc}= attenuation in dB/km.

12. Determine length of the fiber if attenuation is 5dB/km, splice loss is 2 dB/km, connector loss at source and detector is 3.5 and 2.5.

a) 5 km

b) 4 km

c) 3 km

d) 8 km

View Answer

Explanation: Length of the fiber is L = C

_{L}/(α

_{fc}+α

_{j}) – α

_{cr}. Here α

_{cr}= loss at detector and source combined, α

_{fc}= attenuation in dB/km.

13. Determine total RMS pulse broadening over 8 km if RMS pulse broadening is 0.6ns/km.

a) 3.6 ns

b) 4 ns

c) 4.8 ns

d) 3 ns

View Answer

Explanation: Total RMS pulse broadening is given by –

σ

_{T}= σ*L Where σ = rms pulse broadening and L = length of the fiber.

14. Determine RMS pulse broadening over 8 km if total RMS pulse broadening is 5.8ns/km.

a) 0.2ns/km

b) 0.1ns/km

c) 0.4ns/km

d) 0.72ns/km

View Answer

Explanation: RMS pulse broadening is given by –

σ = σ

_{T}/L where σ = rms pulse broadening and L = length of the fiber.

**Sanfoundry Global Education & Learning Series – Optical Communications.**

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