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How "wide" is a bit on a 1-Gbps link? How long is a bit in copper wire, where the speed of propagation is \(2.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
0.23 meters.

Step by step solution

01

- Understand the Problem

Determine how long a single bit takes to travel on a 1-Gbps link with a propagation speed of \(2.3 \times 10^8 \text{ meters/second}\).
02

- Calculate Bit Time

The bit rate is 1 Gbps (\(10^9 \text{ bits/second}\)). Therefore, the time to transmit one bit is the inverse of the bit rate: \[ \text{Bit Time} = \frac{1}{\text{Bit Rate}} = \frac{1}{10^9 \text{ bits/second}} \].
03

- Time Calculation

Solving the above equation: \[ \text{Bit Time} = 10^{-9} \text{ seconds (or nanoseconds)} \].
04

- Calculate Bit Length

To find the length of a bit in the wire, multiply the propagation speed by the bit time: \[ \text{Bit Length} = \text{Propagation Speed} \times \text{Bit Time} \].
05

- Length Calculation

Using the given propagation speed \(2.3 \times 10^8 \text{ meters/second}\) and bit time \(10^{-9} \text{ seconds}\): \[ \text{Bit Length} = 2.3 \times 10^8 \text{ meters/second} \times 10^{-9} \text{ seconds} \]. This simplifies to: \[ \text{Bit Length} = 0.23 \text{ meters} \].

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Key Concepts

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

bit rate
Bit rate is the number of bits transmitted per second in a communication channel. In this exercise, we deal with a 1-Gbps (Gigabit per second) link. This means that the channel transmits 1 billion bits every second. Understanding bit rate is vital because it directly influences how much data can be transferred in a given time frame. The formula for bit rate is simple: it's the total number of bits divided by the time in seconds. So, for our 1-Gbps link, the bit rate is:
  • 1 Gbps = \(1 \times 10^9\) bits per second.

Knowing the bit rate helps in determining other key parameters, such as bit time, which is critical for calculating how
propagation speed
Propagation speed refers to the speed at which a signal travels through a medium, such as copper wire or fiber optic cable. In the context of this exercise, the propagation speed is given as \(2.3 \times 10^8\) meters per second. This is the speed at which an electrical signal moves through the copper wire. To put it into perspective, light travels at a speed of approximately \(3 \times 10^8\) meters per second, so our signal is traveling at a speed that is quite fast but somewhat slower than light. Understanding propagation speed is crucial for calculating how long it takes for a signal to travel a certain distance. It helps in determining the bit length, which is the physical distance a single bit occupies in the medium.
bit time
Bit time is the duration it takes for one bit to be transmitted and it can be calculated using the bit rate. For a 1-Gbps link, the bit time is the inverse of the bit rate. The formula used is:
  • \text{Bit Time} = \frac{1}{\text{Bit Rate}}

In our exercise, the bit rate is 1 Gbps (\(10^9\) bits per second), so the bit time is:
  • \text{Bit Time} = \frac{1}{10^9 \text{ bits/second}} = 10^{-9}$ or 1 nanosecond.

To determine how

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Most popular questions from this chapter

Assume you wish to transfer an \(n\)-byte file along a path composed of the source, destination, seven point-to-point links, and five switches. Suppose each link has a propagation delay of \(2 \mathrm{~ms}\), bandwidth of \(4 \mathrm{Mbps}\), and that the switches support both circuit and packet switching. Thus you can either break the file up into 1-KB packets, or set up a circuit through the switches and send the file as one contiguous bit stream. Suppose that packets have 24 bytes of packet header information and 1000 bytes of payload, that store-and-forward packet processing at each switch incurs a 1 -ms delay after the packet has been completely received, that packets may be sent continuously without waiting for acknowledgments, and that circuit setup requires a 1-KB message to make one round-trip on the path incurring a 1-ms delay at each switch after the message has been completely received. Assume switches introduce no delay to data traversing a circuit. You may also assume that file size is a multiple of 1000 bytes. (a) For what file size \(n\) bytes is the total number of bytes sent across the network less for circuits than for packets? (b) For what file size \(n\) bytes is the total latency incurred before the entire file arrives at the destination less for circuits than for packets? (c) How sensitive are these results to the number of switches along the path? To the bandwidth of the links? To the ratio of packet size to packet header size? (d) How accurate do you think this model of the relative merits of circuits and packets is? Does it ignore important considerations that discredit one or the other approach? If so, what are they?

For the following, as in the previous problem, assume that no data compression is done. Calculate the bandwidth necessary for transmitting in real time: (a) HDTV high-definition video at a resolution of \(1920 \times 1080,24\) bits/pixel, 30 frames/second. (b) POTS (plain old telephone service) voice audio of 8-bit samples at \(8 \mathrm{KHz}\). (c) GSM mobile voice audio of 260 -bit samples at \(50 \mathrm{~Hz}\). (d) HDCD high-definition audio of 24-bit samples at \(88.2 \mathrm{KHz}\).

The Unix utility traceroute, or its Windows equivalent tracert, can be used to find the sequence of routers through which a message is routed. Use this to find the path from your site to some others. How well does the number of hops correlate with the RTT times from ping? How well does the number of hops correlate with geographical distance?

Consider a point-to-point link \(2 \mathrm{~km}\) in length. At what bandwidth would propagation delay (at a speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) equal transmit delay for 100 -byte packets? What about 512 -byte packets?

Suppose a 100-Mbps point-to-point link is being set up between Earth and a new lunar colony. The distance from the moon to Earth is approximately \(385,000 \mathrm{~km}\), and data travels over the link at the speed of light-3 \(\times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) Calculate the minimum RTT for the link. (b) Using the RTT as the delay, calculate the delay \(\times\) bandwidth product for the link. (c) What is the significance of the delay \(\times\) bandwidth product computed in (b)? (d) A camera on the lunar base takes pictures of Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the most current image, which is \(25 \mathrm{MB}\). What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?

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