5G Wireless Standard Design Week 1 NPTEL Assignment Answers 2025

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NPTEL 5G Wireless Standard Design Week 1 Assignment Answers 2025

1. In an OFDM system, inter-symbol interference can be avoided using

• Orthogonal sub-carriers
• Cyclic prefix
• IFFT
• Encoding

Answer : For Answers Click Here 

2. Calculate the sampling rate for an 5G NR System with 30kHz subcarrier spacing and 50MHz bandwidth.

• 245.76 MHz
• 122.88 MHz
• 61.44 MHz
• 30.72 MHz

Answer :

3. Consider a 5G system with a bandwidth of 100MHz. If the signal-to-noise ratio (SNR) of the received signal is 9dB, what is the maximum rate (in Mbps) at which the receiver can decode the signal successfully without any error?

• 180Mbps
• 280Mbps
• 316Mbps
• 459Mbps

Answer :

4. The number of slots/ subframe if the system uses a sub-carrier spacing of 60kHz.

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5. Calculate the IFFT size required for a 50MHz 5G NR system with subcarrier spacing ∆f = 15kHz.

Answer :

6. The 5G system which uses incremental redundancy retransmission mode with an effective code rate of 1/4 did an initial transmission. But the receiver failed to successfully decode the data in the first attempt and requested for 2 more retransmissions. Calculate the code rate at which the receiver was able to successfully decode the data.

• 1/8
• 1/4
• 1/12
• 3/7

Answer :

7. The minimum SNR (in dB) required to achieve the capacity of 5 bps/Hz is approximately

• 5 dB
• 15 dB
• 21 dB
• 31 dB

Answer :

8. What is the best approach to maintain a link when UE experiences sudden drop in SNR?

• Decrease the code-rate
• Increase the code rate
• Increase transport block CRC length
• Modify CRC polynomial with same degree as before

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NPTEL 5G Wireless Standard Design Week 1 Assignment Answers 2024

Q1. What is the minimum SNR (in dB) required to achieve a channel capacity of 10 bps/Hz using the Shannon formula?

a) 20 dB
b) 10 dB
c) 30 dB
d) 40 dB

✅ Answer: c) 30 dB
Short Explanation:
Shannon’s formula: C=log⁡2(1+SNR)C = \log_2(1 + \text{SNR})C=log2​(1+SNR)
Solving: 10=log⁡2(1+SNR)⇒SNR=1023⇒10log⁡10(1023)≈30 dB10 = \log_2(1 + \text{SNR}) \Rightarrow \text{SNR} = 1023 \Rightarrow 10 \log_{10}(1023) ≈ 30 \text{ dB}10=log2​(1+SNR)⇒SNR=1023⇒10log10​(1023)≈30 dB

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Q2. How many slots are there in a subframe if the subcarrier spacing is 60 kHz?

a) 2
b) 3
c) 4
d) 5

✅ Answer: c) 4
Short Explanation:
At 60 kHz, numerology index μ = 2 → Slots = 2μ=42^μ = 42μ=4

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Q3. What is the effective code rate after two retransmissions if the original code rate is 3/4?

a) 1/4
b) 1/3
c) 2/3
d) 3/4

✅ Answer: a) 1/4
Short Explanation:
Effective code rate = (Original rate) ÷ (Number of transmissions) = 3/4 ÷ 3 = 1/4

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Q4. What is Ceil(Sampling Rate / 10⁶) when subcarrier spacing = 15kHz and FFT size = 4096?

a) 60
b) 61
c) 62
d) 63

✅ Answer: c) 62
Short Explanation:
Sampling Rate = 15,000 × 4096 = 61.44 MHz → Ceil(61.44) = 62

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Q5. What is the value of (Slot Duration in ms) × (Subframe Duration in ms)?

a) 10
b) 100
c) 1
d) 5

✅ Answer: a) 10
Short Explanation:
Slot duration = 1 ms, Subframe = 10 ms → 1 × 10 = 10

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Q6. If CP length of each symbol is 288 samples, what is the CP length of the first symbol?

a) 300
b) 320
c) 352
d) 384

✅ Answer: c) 352
Short Explanation:
First OFDM symbol in LTE/NR has longer CP → Usually 352 samples

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Q7. What is the spectral efficiency if modulation order is 2 and code rate is 894/1024?

a) 1.0
b) 1.7461
c) 2.0
d) 1.5

✅ Answer: b) 1.7461
Short Explanation:
Spectral Efficiency = Modulation Order × Code Rate = 2 × (894 / 1024) = 1.7461