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If a binary signal is sent over a \(3-\mathrm{kHz}\) channel whose signal-to- noise ratio is \(20 \mathrm{~dB}\), what is the maximum achievable data rate?

Short Answer

Expert verified
Maximum achievable data rate is approximately 19,974 bits per second.

Step by step solution

01

Understanding the Problem

We need to find the maximum achievable data rate of a binary signal transmitted over a given channel. To do that, we utilize the Shannon Capacity Formula, which relates the bandwidth, the signal-to-noise ratio (SNR), and channel capacity.
02

Identifying the Given Values

We have been given a channel bandwidth ( ) of 3 kHz and a signal-to-noise ratio (SNR) of 20 dB. We need to convert the SNR from decibels to a linear scale.
03

Converting SNR from decibels to linear scale

SNR in decibels can be converted to a linear scale using the formula: \[ SNR_{ ext{linear}} = 10^{ rac{SNR_{ ext{dB}}}{10}} \]Substituting the given SNR (20 dB), we have:\[ SNR_{ ext{linear}} = 10^{ rac{20}{10}} = 10^2 = 100 \]
04

Applying the Shannon Capacity Formula

The Shannon Capacity formula is given by:\[ C = B imes ext{log}_2(1 + SNR_{ ext{linear}}) \]where:- \( C \) is the channel capacity in bits per second,- \( B \) is the bandwidth in Hz.Substituting the known values:\[ C = 3000 imes ext{log}_2(1 + 100) \]
05

Calculating the Maximum Achievable Data Rate

Using the formula above, we first calculate \( \text{log}_2(101) \):\[ \text{log}_2(101) \approx 6.658 \]Thus, the channel capacity is:\[ C = 3000 imes 6.658 \approx 19974 ext{ bits per second} \]
06

Final Answer

Hence, the maximum achievable data rate is approximately 19,974 bits per second.

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Key Concepts

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

Understanding Channel Bandwidth
The concept of channel bandwidth is fundamental in understanding how data is transmitted over communication channels. Bandwidth refers to the difference between the highest and lowest frequencies that a channel can carry. In simpler terms, it is the range of frequencies the channel can manage without distortion. For example, a channel with a bandwidth of 3 kHz can support frequencies up to 3,000 Hz.

The channel bandwidth directly influences how much data can be sent over a communication channel within a given period. So, when we're discussing data rates and transmission capacity, we'll typically refer to Shannon's Capacity Formula, where bandwidth plays a vital role. This formula helps us see how bandwidth, along with other factors, determines the maximum data rate achievable in any channel, thereby shaping efficient communication systems. Keeping this in mind will help you grasp more complex concepts easily!
  • Higher bandwidth means more data per second
  • It determines the amount of information transmitted
  • Measured in hertz (Hz)
Signal-to-Noise Ratio (SNR) in Communication
Signal-to-Noise Ratio, or SNR, is another key factor affecting data transmission. It describes the power ratio between the signal's strength and the noise level in the channel. Essentially, SNR tells us how "loud" the signal is compared to the background noise.

SNR is measured in decibels (dB), but for many calculations, it needs to be converted to a linear scale. This conversion is crucial as it simplifies the application of the Shannon Capacity Formula. In more practical terms, a high SNR means a clearer and stronger signal, leading to more accurate data transmission. Conversely, a low SNR indicates the presence of noise overpowering the signal, resulting in potential errors.

Understanding SNR is vital because it directly influences the quality and speed of communication:
  • Higher SNR means less noise interference
  • Indicates signal clarity
  • Inverse relationship with potential data errors
Data Rate Calculation with Shannon's Formula
Calculating the data rate using Shannon's Capacity Formula is a fundamental aspect of understanding communications and networks. The formula essentially provides a theoretical maximum data rate for a channel given its bandwidth and SNR. This calculation is crucial when designing systems that need high data throughput.

The equation used is: \[ C = B \times \log_2(1 + SNR_{\text{linear}}) \]where \( C \) is the capacity, \( B \) is the bandwidth, and \( SNR_{\text{linear}} \) represents the linear scale SNR. Through this calculation, we determine the number of bits per second that can be effectively transmitted over the channel.

Employing this formula not only helps in determining current capacities but also in optimizing them:
  • Produces maximum data rates
  • Helps in channel efficiency evaluation
  • Useful in system design and improvement
Binary Signal Transmission Basics
Binary signal transmission is the simplicity behind the screens, a fundamental concept in digital communication. In this process, complex data is broken down into binary code, consisting of 1s and 0s. These binary numbers are then transmitted as electrical signals over channels.

Whenever we talk about high data rates and efficient communications, we're also discussing how well binary signals are transmitted. The speed and quality can be greatly improved by minimizing noise, maximizing bandwidth, and applying thorough calculations like the Shannon Capacity.

This understanding is key in fields like data communications and information technology:
  • Translates complex data into binary form
  • Ensures data integrity during transmission
  • Forms the backbone of digital communication systems

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Most popular questions from this chapter

Make a rough estimate of the number of PCS microcells \(100 \mathrm{~m}\) in diameter it would take to cover San Francisco (120 square \(\mathrm{km}\) ).

A regional telephone company has 15 million subscribers. Each of their telephones is connected to a central office by a copper twisted pair. The average length of these twisted pairs is \(10 \mathrm{~km}\). How much is the copper in the local loops worth? Assume that the cross section of each strand is a circle \(1 \mathrm{~mm}\) in diameter, the density of copper is \(9.0 \mathrm{grams} / \mathrm{cm}^{3}\), and that copper sells for \(\$ 6\) per kilogram.

At the low end, the telephone system is star shaped, with all the local loops in a neighborhood converging on an end office. In contrast, cable television consists of a single long cable snaking its way past all the houses in the same neighborhood. Suppose that a future TV cable were 10-Gbps fiber instead of copper. Could it be used to simulate the telephone model of everybody having their own private line to the end office? If so, how many one-telephone houses could be hooked up to a single fiber?

It is desired to send a sequence of computer screen images over an optical fiber. The screen is \(3840 \times 2160\) pixels, each pixel being 24 bits. There are 50 screen images per second. What data rate is needed is needed?

A cable company decides to provide Internet access over cable in a neighborhood consisting of 5000 houses. The company uses a coaxial cable and spectrum allocation allowing \(100 \mathrm{Mbps}\) downstream bandwidth per cable. To attract customers, the company decides to guarantee at least 2 Mbps downstream bandwidth to each house at any time. Describe what the cable company needs to do to provide this guarantee.

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