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What is the length of a contention slot in CSMALCD for (a) a 2-km twin-lead cable (where signal propagation speed is \(82 \%\) of the signal propagation speed in vacuum)?, and (b) a \(40-k m\) multimode fiber optic cable (signal propagation speed is \(65 \%\) of the signal propagation speed in vacuum)?

Short Answer

Expert verified
(a) 16.3 microseconds, (b) 410 microseconds.

Step by step solution

01

Identify Information for Part (a)

For the 2-km twin-lead cable, we are given that the signal propagation speed is 82% of the speed of light in vacuum. The speed of light in vacuum is approximately \( c = 3 \times 10^8 \) meters per second.
02

Calculate Propagation Speed for Part (a)

The speed of signal propagation in the twin-lead cable is \( 0.82 \times 3 \times 10^8 \) m/s. Calculate this to get:\[ 0.82 \times 3 \times 10^8 = 2.46 \times 10^8 \text{ m/s}. \]
03

Calculate Contention Slot Length for Part (a)

The contention slot length is twice the time it takes for a signal to travel the length of the cable. Therefore, first calculate the travel time for 2 km:\[ \text{Travel time} = \frac{2000 \text{ m}}{2.46 \times 10^8 \text{ m/s}}. \]This gives a travel time of approximately \(8.13 \times 10^{-6} \text{ seconds}\). The contention slot length is twice this time:\[ 2 \times 8.13 \times 10^{-6} \approx 1.63 \times 10^{-5} \text{ seconds}. \]
04

Identify Information for Part (b)

For the 40-km multimode fiber optic cable, we are given that the signal propagation speed is 65% of the speed of light in vacuum. The speed of light in vacuum is \( c = 3 \times 10^8 \) m/s.
05

Calculate Propagation Speed for Part (b)

The speed of signal propagation in the fiber optic cable is \( 0.65 \times 3 \times 10^8 \) m/s. Calculate this to get:\[ 0.65 \times 3 \times 10^8 = 1.95 \times 10^8 \text{ m/s}. \]
06

Calculate Contention Slot Length for Part (b)

The contention slot length is twice the time it takes for a signal to travel the length of the cable. Therefore, first calculate the travel time for 40 km:\[ \text{Travel time} = \frac{40000 \text{ m}}{1.95 \times 10^8 \text{ m/s}}. \]This gives a travel time of approximately \(2.05 \times 10^{-4} \text{ seconds}\). The contention slot length is twice this time:\[ 2 \times 2.05 \times 10^{-4} = 4.10 \times 10^{-4} \text{ seconds}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Contention Slot
In the CSMA/CD protocol, a contention slot is a specific time period where a device on the network checks for collisions before beginning to transmit data. This is crucial as it helps manage data transmission over shared mediums, such as cable networks, to prevent multiple devices from talking at the same time.
To put it simply, when a device wants to send data, it listens to the cable to see if any other device is transmitting. If the cable is clear, the device will send the data. However, if multiple devices try to send data at the same time, a collision occurs and each device backs off for a random time period, known as the contention slot.
To calculate this slot, we need to consider the time it takes for a signal to travel the length of the cable and back, which entirely depends on the signal propagation speed and the cable length.
Signal Propagation Speed
Signal propagation speed is how fast a signal travels through a medium, such as air or cable. It's essential when calculating network aspects like contention slots since it directly impacts how quickly data can travel from one point to another within the network.
  • For vacuum: The speed of light is approximately \(3 \times 10^8\) m/s, which is the maximum speed possible for signal transmission.
  • For twin-lead cables: Usually signal speeds are around 82% of the speed of light, resulting in slower data travel compared to vacuum since the medium introduces resistance.
  • For fiber optic cables: These often carry signals at around 65% of the speed of light, which is slower than twin-lead but still highly efficient for data transfer over long distances due to low loss of signal strength.
Understanding propagation speed is vital when setting up networks, particularly to calculate time delays and determine maximum cable lengths to avoid data loss and ensure efficient transmission.
Fiber Optic Cable
Fiber optic cables are a common medium for network communication because they are capable of transmitting data over long distances with minimal signal loss. Made of thin strands of glass or plastic, these cables carry data via light pulses.
Fiber optics are known for:
  • High bandwidth: Capable of supporting vast amounts of data throughput.
  • Long-distance capabilities: Ideal for connections of up to 40-km as in our example with minimal quality degradation.
  • Low signal loss: Maintaining data integrity over various distances.
Typically, the signal propagation speed in a fiber optic cable is around 65% of the speed of light in a vacuum. This efficiency comes with a trade-off between speed and signal strength, making fiber optics a robust choice for wide area networks (WANs) or high-performance enterprise networks.
Twin-lead Cable
Twin-lead cables are one of the older types of cable used for signal transmission, often recognizable by their two parallel wires. They are common in RF (radio frequency) and TV antennas, especially those found in homes.
Here are some aspects of twin-lead cables:
  • Moderate speed: Signal propagation speed here is around 82% of the speed of light, making them decently fast for short-distance communication.
  • Low cost: These cables are generally cheaper than fiber optics, making them accessible for budget networks.
  • Susceptible to interference: Because of their exposed wiring, they are more prone to picking up noise and interference.
Twin-lead cables have seen less usage compared to modern alternatives like coaxial or fiber optic cables due to their limitations, but they still serve well in specific, less-demanding applications.

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Most popular questions from this chapter

Two devices connected to the same \(802.11\) network are both downloading a large file from the Intemet. Explain bow one device could obtain a higher data rate than the other by (ab)using a \(802.11\) mechanism intended to provide quality of service.

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