University:
College
Course:
College Engineering
Academic year:

2023
Views:

349

Pages:

2
Author:

D. Krishnan

Question: The Doppler spread of a channel is 𝐵𝐷 = 80 Hz, if we need independent samples, what is the time separation needed for the samples in the received signal? Answer: The coherence time of the channel 𝑇𝑐 ≈ 1 1 = 𝐵𝐷 80 𝑇𝑐 = 12.5 ms (refer mobile wireless communication notes-5 for theory) the samples spaced 12.5 ms apart are uncorrelated, So the uncorrelated samples will be independent of each other. Question: A user moving with velocity 60 miles per hour, the frequency of transmission is 1850MHz. What is the Doppler shift, Doppler spread and coherence time? 𝜃 = 900 . Answer The Doppler frequency shift can be found by using the equation: 𝑓𝑑 = 𝑣 𝑐𝑜𝑠𝜃 𝜆 𝑐𝑜𝑠𝜃 = 1 𝑠𝑖𝑛𝑐𝑒 𝜃 = 900 𝑣 = 60 𝑚𝑖𝑙𝑒𝑠 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟 = 26.822 𝑚𝑒𝑡𝑒𝑟 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 (divide the speed value by 2.237) 𝜆= 𝑙𝑖𝑔ℎ𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 3 × 108 = = 0.1622 𝑚 𝑓𝑐 1850MHz 𝑓𝑑 = 𝑣 26.822 = 𝜆 0.1622 𝑓𝑑 = 165.47 Hz So the Doppler shift is 165.47Hz Doppler spread 𝐵𝐷 = 2 × 𝑓𝑑 = 330𝐻𝑧. coherence time 1 1 𝑇𝑐 ≈ = = 𝐵𝐷 330 𝑇𝑐 = 3 ms If the vehicle moving toward the base station the shifted frequency can be calculated as 𝑓𝑐 ′ = 𝑓𝑐 + 𝑓𝑑 𝑓𝑐 ′ = 1850MHz + 165.47 Hz