Optical Communications Multiple Choice Questions 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
Answer: d
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
Answer: b
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
Answer: c
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
Answer: d
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
Answer: a
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
Answer: a
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
Answer: b
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
Answer: c
Explanation: Average number of photons arezm=[2βςF(M)]*[S/N*η]
Here, η = quantum efficiency, S/N = signal to noise ratio.
9. Determine incident optical power if zm=864, wavelength = 1μm.
a) -85 dBm
b) -80 dBm
c) -69.7 dBm
d) -60.7 dBm
Answer: d
Explanation: Incident optical power is P0=zmhcBT/2λ. Here zm=average number of photons, hc=Planck’s constant.
10. Determine wavelength of incident optical power if zm=864, incident optical power is -60.7 dB, BT=1 * 107.
a) 1 μs
b) 2 μs
c) 3 μs
d) 4 μs
Answer: a
Explanation: Wavelength is determined by λ=zmhcBT/2P0. Here zm=average number of photons, hc=Planck’s constant, P0=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
Answer: a
Explanation: The total channel loss is CL=(αfc+αj)L + αcr. Here αcr=loss at detector and source combined, αfc = 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
Answer: b
Explanation: Length of the fiber is L = CL/(αfc+αj) – αcr. Here αcr = loss at detector and source combined, αfc = 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
Answer: c
Explanation: Total RMS pulse broadening is given by –
σT = 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
Answer: d
Explanation: RMS pulse broadening is given by –
σ = σT/L where σ = rms pulse broadening and L = length of the fiber.
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