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# if binary channel has bandwidth 20mhz then what will be its channel capacity?

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## Maximum Data Rate (channel capacity) for Noiseless and Noisy channels

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## Maximum Data Rate (channel capacity) for Noiseless and Noisy channels

Difficulty Level : Easy

Last Updated : 29 Jun, 2021

As early as 1924, an AT&T engineer, Henry Nyquist, realized that even a perfect channel has a finite transmission capacity. He derived an equation expressing the maximum data rate for a finite-bandwidth noiseless channel. In 1948, Claude Shannon carried Nyquist’s work further and extended to it the case of a channel subject to random(that is, thermodynamic) noise (Shannon, 1948). This paper is the most important paper in all of the information theory.

Data rate governs the speed of data transmission. A very important consideration in data communication is how fast we can send data, in bits per second, over a channel. Data rate depends upon 3 factors:

The bandwidth available

Number of levels in digital signal

The quality of the channel – level of noise

Two theoretical formulas were developed to calculate the data rate: one by Nyquist for a noiseless channel, another by Shannon for a noisy channel.

1. Noiseless Channel: Nyquist Bit Rate –

For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate

proved that if an arbitrary signal has been run through a low-pass filter of bandwidth, the filtered signal can be completely reconstructed by making only 2*Bandwidth (exact) samples per second. Sampling the line faster than 2*Bandwidth times per second is pointless because the higher-frequency components that such sampling could recover have already been filtered out. If the signal consists of L discrete levels, Nyquist’s theorem states:

BitRate = 2 * Bandwidth * log2(L) bits/sec

In the above equation, bandwidth is the bandwidth of the channel, L is the number of signal levels used to represent data, and BitRate is the bit rate in bits per second.

Bandwidth is a fixed quantity, so it cannot be changed. Hence, the data rate is directly proportional to the number of signal levels.

Note –Increasing the levels of a signal may reduce the reliability of the system.Examples:Input1 : Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. What can be the maximum bit rate?Output1 : BitRate = 2 * 3000 * log2(2) = 6000bpsInput2 : We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?Output2 : 265000 = 2 * 20000 * log2(L)

log2(L) = 6.625

L = 26.625 = 98.7 levels

The amount of thermal noise present is measured by the ratio of the signal power to the noise power, called the SNR (Signal-to-Noise Ratio).

2. Noisy Channel : Shannon Capacity –

In reality, we cannot have a noiseless channel; the channel is always noisy. Shannon capacity is used, to determine the theoretical highest data rate for a noisy channel:

Capacity = bandwidth * log2(1 + SNR) bits/sec

In the above equation, bandwidth is the bandwidth of the channel, SNR is the signal-to-noise ratio, and capacity is the capacity of the channel in bits per second.

Bandwidth is a fixed quantity, so it cannot be changed. Hence, the channel capacity is directly proportional to the power of the signal, as SNR = (Power of signal) / (power of noise).

The signal-to-noise ratio (S/N) is usually expressed in decibels (dB) given by the formula:

10 * log10(S/N)

So for example a signal-to-noise ratio of 1000 is commonly expressed as:

10 * log10(1000) = 30 dB.

Examples:Input1 : A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communication. The SNR is usually 3162. What will be the capacity for this channel?Output1 : C = 3000 * log2(1 + SNR) = 3000 * 11.62 = 34860 bpsInput2 : The SNR is often given in decibels. Assume that SNR(dB) is 36 and the channel bandwidth is 2 MHz. Calculate the theoretical channel capacity.Output2 : SNR(dB) = 10 * log10(SNR)

SNR = 10(SNR(dB)/10)

SNR = 103.6 = 3981

Hence, C = 2 * 106 * log2(3982) = 24 MHz

Reference:Book – Computer Networks: A Top – Down Approach by FOROUZAN

स्रोत : www.geeksforgeeks.org

## Network Security Questions and Answers for Freshers

This set of Network Security Questions and Answers for freshers focuses on “Layers – I”. 1. If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? a) 2500 Hz b) 900 Hz c) 800 Hz d) can’t be determined with the ... Read more

## Networking Questions and Answers – Layers – I

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This set of Network Security Questions and Answers for freshers focuses on “Layers – I”.

1. If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth?

a) 2500 Hz b) 900 Hz c) 800 Hz

d) can’t be determined with the information given

2. Assume we need to download text documents at the rate of 100 pages per second. What is the required bit rate of the channel?

a) 1.846 Mbps b) 1.536 Mbps c) 2.4 Mbps

d) None of the Mentioned

3. A digitized voice channel is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?

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4. What is the bit rate for high-definition TV (HDTV)?

5. A device is sending out data at the rate of 1000 bps. How long does it take to send a file of 100,000 characters?

Check this: Cryptography and Network Security Books | Computer Science MCQs

6. A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel?

7. We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?

8. Calculate the theoretical highest bit rate of a regular telephone line. The signal-to-noise ratio is usually 3162.

a) 34.860 kbps b) 17.40 kbps c) 11.62 kbps

d) none of the Mentioned

9. Calculate the theoretical channel capacity. If SNR(dB) = 36 and the channel bandwidth is 2 MHz.

10. A channel has a 1-MHz bandwidth. The SNR for this channel is 63. What is the appropriate bit rate?

11. A channel has a 1-MHz bandwidth. The SNR for this channel is 63. What is the appropriate signel level?

What are the bits transmitted for the Unipolar system?

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To practice all areas of Network Security for freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.

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## If the channel has a capacity of 20 Mbps, the bandwidth of the channel is 5 MHz. What signal

Answer (1 of 3): C= B log(base2)(1+SNR) SNR=15 ; here C=20mbps and B=5Mhz

If the channel has a capacity of 20 Mbps, the bandwidth of the channel is 5 MHz. What signal-to-noise ratio is required to achieve this capacity?

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Sort Dwipayan

C= B log(base2)(1+SNR)

SNR=15 ; here C=20mbps and B=5Mhz

Related questions

Given a channel with an intended capacity of 20 Mbps. The bandwidth of the channel is 3MHz. What signal-to-noise ratio is required in order to achieve this capacity?

What is the channel capacity for a tele printer channel with a 300 Hz bandwidth and a signal-to-noise ratio of 3 dB?

Given a channel with an intended capacity of 20 Mbps, the bandwidth of the channel is 3MHz. Assuming white thermal noise, what signal-to-noise ratio is required to achieve this capacity?

What is a 160 MHz channel bandwidth?

How many channels should be used in a 5 GHz channel reuse pattern of 20 MHz channels?

Waldo Wheres 2y Related

Given a channel with an intended capacity of 20 Mbps. The bandwidth of the channel is 3MHz. What signal-to-noise ratio is required in order to achieve this capacity?

C = B log2(1+SNR)

You would then solve for SNR so…

2^(C/B) - 1= SNR

In this case C (capacity) = 20 and B (bandwidth) = 3

You can do the rest 😉

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Kurt Behnke

Former Telecom R&D and Manager Operations at Ericsson (company) (1997–2017)Author has 6K answers and 8.5M answer viewsFeb 4

Shannon-Hartley: Data rate D on a bandwidth B is less than or equal to

D≤B log 2 (1+S/N) D≤Blog2⁡(1+S/N) . which gives you 4≤ log 2 (1+S/N) 4≤log2⁡(1+S/N)

You should be able to extract S/N from this.

Richard Urwin

Related

How do I convert bandwidth MHz to Mbps?

Simplistically, 1MHz is 1Mbps.

Realistically, it’s nothing of the sort.

If we assume that the data is a sequence of 1’s and 0’s that are encoded in the signal amplitude, then it needs a bandwidth equal to its bit rate to encode the fastest-changing signal (10101010…).

However… this ignores that you can’t just send 1’s and 0’s out into the ether and expect them to be received at the other end. There will be errors, dropped bits, changed bits, and the clocks on either end will be a little out of sync with each other. Therefore we need to add a few other bits in order to frame them, and some more to

Rupert Baines

interersted in it. But primarily as an engineer not a theoreticianAuthor has 11.9K answers and 29.5M answer views5y

This is one of those questions that is Ruth very, very trivial and you should be embarrassed to ask — or can be rally complicated and interesting.

For a homework question, you really, really ought to know Shannon Hartley

In the real world, there are a lot of interesting things to discuss

What filter mask will you use to get your “5MHz”? Because you actually get less useful bandwidth. How much less…?

What is the nose like? Is it really AWGN? What difference would that make?

when you say 20Mbps what fits that mean? Information, payload, gross? What about signaling and overhead etc

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Paul Grimshaw

IT Architect in the Computer Industry (1983–present)Author has 3.2K answers and 10.1M answer views2y

Related

What does 20 MHz bandwidth mean?

What does 20 MHz bandwidth mean?

The word “bandwidth” is one of those words that has a different meaning in everyday and technical language. In everyday language “bandwidth” is pretty much interchangeable with “maximum bit rate”. So if you have a 20Mbps internet service for example then you might say that you have 20Mbps bandwidth. Some people might even get confused and say 20 MHz here.

But when you see 20MHz rather than 20Mbps then it’s rather more likely that we’re using the technical meaning of bandwidth. This is a very different concept to maximum data rate, although there is a relationship

Venkatraman S

Related

For a channel of bandwidth 4kHz, signal to noise ratio of 15dB, what is the channel capacity?

With the specified bandwidth and SNR, the channel capacity is about 20 kHz. The channel capacity is around 5 times more than the transmission bandwidth (4 kHz). This is due to the good ‘SNR’. The calculation is simple. Shannon’s formula for AWGN channels is to be used.

SNR should be converted to linear value and then calculations can be done as usual.