Chapter 4: Problem 6
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.
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.
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:
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.
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:
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.