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Olutions To Assignment #4

This document contains the solutions to Assignment #4 for the NPTEL MOOC "Principles of Modern CDMA/ MIMO/ OFDM Wireless Communications" taught by Associate Professor Aditya Jagannatham from IIT Kanpur. It includes explanations and calculations for questions related to multipath propagation, delay spread, coherence bandwidth, Doppler shift, and coherence time in wireless communications.

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0% found this document useful (0 votes)
402 views1 page

Olutions To Assignment #4

This document contains the solutions to Assignment #4 for the NPTEL MOOC "Principles of Modern CDMA/ MIMO/ OFDM Wireless Communications" taught by Associate Professor Aditya Jagannatham from IIT Kanpur. It includes explanations and calculations for questions related to multipath propagation, delay spread, coherence bandwidth, Doppler shift, and coherence time in wireless communications.

Uploaded by

Lakshmi Kumari
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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NPTEL MOOC on “Principles of Modern CDMA/ MIMO/ OFDM Wireless Communications”

[NOC15 EC05]

Aditya Jagannatham
Associate Professor
Dept. of Electrical Engineering
IIT Kanpur
adityaj@iitk.ac.in

Solutions to Assignment #4

1. Delay spread of a wireless channel arises due to Multipath Propagation. Ans b


2. Delay spread of the wireless channel can be calculated from the Multipath Power Profile. Ans d
3. Given an L = 4 component multipath wireless channel with components arriving at 0 s, 2 s, 3 s and 4 s. The maximum
delay spread of the wireless channel is the difference in the time of arrival between the last arriving component and the first,
i.e. 4 s – 0 s = 4 s. Ans d
4. Given an L = 4 component multipath wireless channel with components arriving at 0 s, 2 s, 3 s and 4 s and respective
powers of components as -10 dB, -10 dB, 0 dB and -20 dB. The average delay of the wireless channel can be calculated as
(0.1  0 + 0.1  2 + 1  3 + 0.01  4)/(0.1+0.1+1+0.01) = 2.68 s

Ans c

5. Given an L = 4 component multipath wireless channel with components arriving at 0 s, 2 s, 3 s and 4 s and respective
powers of components as -10 dB, -10 dB, 0 dB and -20 dB, the RMS delay spread of the wireless channel can be calculated as
follows. From solution to problem 4 above, the average delay is given as 2.68 s. Hence, the RMS delay spread is

((0.1  (0-2.68)2 + 0.1  (2-2.68)2 + 1  (3-2.68)2 + 0.01  (4-2.68)2)/(0.1+0.1+1+0.01))


~ 0.86 s.
Ans a
6. As described in the lectures, a large delay spread leads to Inter Symbol Interference. Ans b
7. Condition for Inter Symbol Interference in wireless communication is Delay spread greater than symbol time. Ans a
8. Typical outdoor delay spread of the wireless channel is ~ 2 s = 0.002 ms. Ans d
9. As described in the lectures, coherence bandwidth of wireless channel is the Bandwidth over which frequency response of the
channel is approximately flat. Hence there is no distortion. Ans c.
10. As defined in the lectures, Coherence Bandwidth is approximately equal to 1/τ, where τ is the delay spread. Ans a
11. Coherence bandwidth of typical outdoor channel is 500 KHz. Ans b
12. Frequency selective distortion occurs when signal bandwidth is larger than the coherence bandwidth or in other words,
coherence bandwidth is lower than signal bandwidth. Ans d
13. As described in the lectures, frequency selective fading leads to Intersymbol Interference in the time domain. Ans b
14. Doppler effect is caused by Relative motion between wireless transmitter and receiver. Ans c
15. Doppler shift does NOT depend on which of the quantities Symbol rate. Ans a
16. Given velocity = 50 Kmph = 50  5/18 m/s and carrier frequency = 3 GHz. Angle  = 60 degrees = π/3 radians. The Doppler
shift is v cos  c/fc = 50  5/18  cos(π/3)  3  109 /(3  108) ~ 69 Hz. Ans d
17. Doppler shift leads to Time Varying Channel. Ans b
18. Coherence time is the Time interval over which channel is approximately constant. Ans c
19. As demonstrated in the lecture, for Doppler shift fd, coherence time is 1/(4fd). Ans c
20. For Doppler shift fd = 80 Hz, coherence time = 1/(4fd) = 1/320 = 0.003125 s = 3.125 ms. Ans a

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