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Posted: February 18th, 2024
CSE 963: Computer Networks I
Homework 1 (50 points)
The goals of this homework are to develop an understanding a communications model, data communications, networks, protocol architecture as well as the basic concepts of data transmissions, data rate, bandwidth and channel capacity.
1. (10 points) Briefly explain the advantages and disadvantages of using layered architecture in computer networks.
2. (10 points) What tasks are performed by the data link layer and the transport layer?
3. (10 points) Draw (by hand) the spectrum of the signal s(t) = 4sin(2pt) + 2sin(6pt) +
). Also identify the absolute and effective bandwidths.
4. (10 points) What is the channel capacity for a teleprinter channel with a 300-Hz bandwidth and a signal-to-noise ratio of 3 dB, where the noise is a white thermal noise?
5. (10 points) A digital signaling system is required to operate at 9600 bps. If the signal encodes a 4-bit word, what is the minimum required bandwidth of that channel?
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The advantages of using a layered architecture in computer networks include modularity, where each layer can be developed and improved independently without affecting other layers. It also allows for interoperability, where two devices from different vendors can communicate as long as they support the same protocols for a given layer. The disadvantages include overhead from having multiple layers of encapsulation and decoding, as well as less than optimal performance if one layer is bottlenecked.
The data link layer is responsible for node-to-node delivery of frames between directly connected nodes. It provides error control and flow control mechanisms. The transport layer establishes end-to-end connections between applications running on networked hosts, providing services such as reliability, flow control, and multiplexing.
The spectrum of the signal s(t) = 4sin(2πt) + 2sin(6πt) + sin(10πt) is:
Absolute bandwidth: 10 Hz (from 0 Hz to 10 Hz)
Effective bandwidth: 8 Hz (the bandwidth occupied by the significant spectral components from 2 Hz to 10 Hz)
For a teleprinter channel with a 300-Hz bandwidth and a signal-to-noise ratio of 3 dB:
Channel capacity = Bandwidth x Log2(1 + SNR)
= 300 Hz x Log2(1 + 10^(3dB/10))
= 300 Hz x Log2(1 + 2) = 300 Hz x 1 bit/Hz = 100 bits/second
If the digital signaling system operates at 9600 bps and each symbol encodes a 4-bit word:
Minimum required bandwidth = Data rate / Bits per symbol
= 9600 bps / 4 bits per symbol
= 2400 Hz
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