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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

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?

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?

Calculate the latency (from first bit sent to last bit received) for the following: (a) 10-Mbps Ethernet with a single store-and-forward switch in the path, and a packet size of 5000 bits. Assume that each link introduces a propagation delay of \(10 \mu \mathrm{s}\) and that the switch begins retransmitting immediately after it has finished receiving the packet. (b) Same as (a) but with three switches. (c) Same as (a) but assume the switch implements "cut-through" switching: It is able to begin retransmitting the packet after the first 200 bits have been received.

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 \(88.2 \mathrm{KHz}\).

Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress \(50 \%\), or 2 seconds to compress \(60 \%\). (a) Calculate the bandwidth at which each compression option takes the same total compression + transmission time. (b) Explain why latency does not affect your answer.

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