CS314 Assignment 3 Solution and Discussion
Assignment No. 03
Semester: Fall 2019
CS314: Introduction to Cellular Networks Total Marks: 15
Due Date: 20-01-2020
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The objective of this assignment is to increase the learning capabilities of the students about:
• WCDMA Modulation
• Channel Capacity in WCDMA
• Direct Sequence Spread Spectrum (DSSS)
• Spread Code
Question – 01
Consider a CDMA network which has a chip rate of 384 MCPS (Megachips Per Second). The network has a downlink spreading factor (SF) of 8. The channel control overhead is 5%. By considering the given scenario, answer the following questions:
- What will be the symbol rate for downlink channels?
- What will be channel bit rate if bit rate per symbol is 2?
- What will be the maximum user available data rate?
Note: Provide complete steps of calculation otherwise marks will be deducted.
Question – 02
Consider a WCDMA network which uses direct sequence spread spectrum (DSSS) as a modulation technique to encode the transmitted data in a way to reduce the noise interference. Now, suppose you want to transmit a 16-bit data 8B 85 (given in hexadecimal form) over the network. What will be the output spread data if data is modulated by using the 8-bit spread code AB?
Note: The data values are given in hexadecimal for the sake of increased readability.
Best of Luck!
bc170403108 MUHAMMAD WALEED last edited by
What will be the maximum user available data rate?
What is the maximum load the cell can support while providing 2% call blocking?
From the Erlang B table with c= 56 channels and 2% call blocking, the maximum load = 45.9 Erlangs
What is the maximum number of users supported by the cell during the busy hour?
Load per active user = (1 call/3600 s) ×(120 s/call) = 33.3 m Erlangs
Number of active users = 45.9/(0.0333) = 1377
Total number of users = 4/3 number active users = 1836
What will be channel bit rate if bit rate per symbol is 2?
where 𝑅𝑝 is the number of pulses per second transmitted, 𝐵 is the available bandwidth, and the units are Bauds or symbols/sec (Bd). Here it is assumed that the pulses being used are sinc pulses; otherwise, the pulse rate will decrease. In this sense, 2𝐵
is an upper bound on the pulses per second that can be transmitted. It is correct to say that you can transmit two pulses for every Hz of bandwidth available.
Now, each pulse can carry several bits, by means of its amplitude. If you allow, say, two amplitudes (for example, -1 and 1), then each pulse carries one bit and 𝑅𝑏=𝑅𝑝
, where 𝑅𝑏 is the bit rate. In general, if you allow 𝑀=2𝑘 amplitudes, you’ll be able to transmit at a bit rate 𝑅𝑏=𝑘𝑅𝑝
Shannon capacity does constrain the number of bits that you can transmit per second; furthermore, the usual formulation assumes sinc pulses.