5G Wireless Standard Design Nptel Week 4 Answers

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  1. 5G Wireless Standard Design Nptel Week 2 Answers (July-Dec 2025)

5G Wireless Standard Design Nptel Week 4 Answers
5G Wireless Standard Design Nptel Week 4 Answers

5G Wireless Standard Design Nptel Week 4 Answers (July-Dec 2025)

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Question 1. For the given channel y=x+ny = x + n the pdf of noise is given by … Calculate the MAP threshold value α\alpha if the transmitter is transmitting using BPSK symbol such that 0 is mapped to −1-1 and 1 is mapped to +1+1 with P(x=0)=13P(x=0)=\tfrac{1}{3} and P(x=1)=23P(x=1)=\tfrac{2}{3}.
a) 0
b) 1/2
c) -1/3
d) 1/6

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Question 2. Find the new threshold value in Question 1 if the transmitter is transmitting using BPSK symbol such that 0 is mapped to 0 and 1 is mapped to 1 with the rest of the parameters remaining same.
a) 0
b) -1/2
c) 1/3
d) -1/6

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Question 3. The scrambiling sequence will affect the LLR values which will be fed to the soft-decoder.
a) True
b) False

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Question 4. Consider a user for which 125 PRBs are allocated across 12 OFDM symbols. Calculate the total number of REs available for the user to transmit the PDSCH.
Note: Reserve the desired number of subcarriers for pilots (as taught in the class).

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Question 5. In the above question, if the user uses a 64‑QAM modulation with the code rate of 600/1000, calculate the number of intermediate information bits available to the user across 3 layers. Give NInfo/100N_{\text{Info}}/100 as the final answer.

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These are 5G Wireless Standard Design Nptel Week 4 Answers


Question 6. In Question 4, if the user is allocated 13 symbols with the rest of the conditions unchanged, calculate the gain in number of REs that the user will get to transmit the PDSCH.
a) 10%
b) 8.7%
c) 17.39%
d) 20%

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Question 7. The low pass filter is used in the baseband processing at the receiver to:
a) to remove the effect of noise
b) to obtain the transmitted IQ samples
c) to nullify the effect of carrier frequency offset
d) to estimate the channel

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Question 8. If the received RF signal r(t)=s(t)+nw(t)=2 I(t)cos⁡(ω0t)−2 Q(t)sin⁡(ω0t)+n(t)r(t)=s(t)+n_w(t)=\sqrt{2}\,I(t)\cos(\omega_0 t)-\sqrt{2}\,Q(t)\sin(\omega_0 t)+n(t).
Due to the phase‑offset of the local oscillator the receiver decodes the baseband signal after low pass filtering to be r~(t)=s(t)×(12−j32)+n~(t)\tilde r(t)= s(t)\times\left(\tfrac{1}{2}-j\tfrac{\sqrt{3}}{2}\right)+\tilde n(t).
Then the phase offset ϕ\phi of the local oscillator (2 I(t)cos⁡(ω0t−ϕ))(\sqrt{2}\,I(t)\cos(\omega_0 t-\phi)) w.r.t. carrier is:

a) −π/6-\pi/6
b) π/6\pi/6
c) −π/3-\pi/3
d) π/3\pi/3

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These are 5G Wireless Standard Design Nptel Week 4 Answers


Question 9. What is the purpose of using the passband filter at the receiver while recovering the baseband signal?
a) to supress the effect of the carrier frequency offset across passband
b) to supress the effect of the carrier phase offset inside passband
c) to limit the passband noise
d) to increase the gain of the system in the required band

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These are 5G Wireless Standard Design Nptel Week 4 Answers

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