Summer 2022
ECE 456: Wireless and Mobile Communications
Assignment #1 Solution
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1. Given the duplexing distance of 175 MHz, the downlink channels should occupy the spectrum of 1885 + 175 = 2060 MHz – 2200 MHz. The spectrum support 140/5 = 28 CDMA
channels. The corresponding uplink channels are over the frequency range of 1885 – 2025
MHz. As such, the maximum number of two-way user channels is 28*80 = 2240.
Note that the spectrum of 2025 - 2060 MHz can not be used with FDD implementation.
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2. (i) Given the duplexing distance of 30 MHz, the downlink channels should occupy the
spectrum of 1698 + 30 = 1728 MHz – 1746 MHz, which will support 18/2 = 9 one-way
channels. The corresponding uplink channels are over the frequency range of 1698 – 1716
MHz. Therefore, the spectrum support 9 two-way channels.
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(ii) With TDD implementation, the number of two-way channels is calculated as (17461698)/4.8 = 10. TDD implementation exploits the unused spectrum of 1716 – 1728 MHz
with FDD.
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3. Sample answer: a sample implementation of 2G GSM system with FDD/TDMA uses 890915 MHz for uplink transmission and 935-960 MHz for downlink transmission (duplexing
scheme). The uplink spectrum is divided into 125 frequency channels of 200 kHz. Each
frequency channel is shared by 8 users for uplink transmission in TDMA fashion (multiple
access scheme). Each user accesses a frequency channel repetitively every 4.6 ms for a
slot duration of 0.58 ms. The same division and sharing patterns are applied to downlink
spectrum.