Difference between the use of 'bandwidth' in DSP and computer networking

Question

As far as I know, bandwidth is measured in herz or 'radian/second' in digital signal processing or in analog communications. But in computer networking it is measured in 'bits/second'. I don't get how these two are related. Can anybody please help?

https://en.wikipedia.org/wiki/Shannon-Hartley_theorem If you had a zero-noise system, you could transmit infinite bits per second by using different voltage levels for different strings of bits (for instance). In real life, noise in the channel prevents you from distinguishing between very similar levels. Bandwidth is the width of the channel, and SNR is the "height" of the channel. bit/s is the area — endolith, Jan 14 '15 at 17:22

Marcus Müller · Accepted Answer · 2015-01-14T17:09:29.123

2

They both define the amount of data you can get through a channel -- the channel bandwidth, together with the SNR, defines the channel capacity $C$, which has the unit of $\frac{\mathrm{bit}}{\mathrm{s}}$ (which, by the way, is simply $1$ binary decision per second), giving you the maximum amount of information you can get through that:

$ C_\mathit{Shannon} = f_\mathrm{bandwidth} \log_2 ( 1+\mathrm{SNR} )$

You see, bandwidth in the signal sense just says "we can see so and so many changes per second", and you can directly translate that to bits you get through that bandwidth, if you know how well your signal gets through.

edited Jan 14 '15 at 17:09

answered Jan 14 '15 at 16:54

Marcus Müller

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i dont understand,can you suggest me a website or some book with more information.? – spectre Jan 14 '15 at 16:57
Any of the books in the Digital Comm section of this page. – Marcus Müller Jan 14 '15 at 17:01
@spectre: better (see edit)? – Marcus Müller Jan 14 '15 at 17:03
i am a beginner in digital comm,and i am still confused – spectre Jan 14 '15 at 17:06
Then I recommend to rephrase your question. A bit/s is an event per second, and I've answered the wording of your question. – Marcus Müller Jan 14 '15 at 17:08
You're aware that $\textrm{Hz}= \frac{1}{\textrm{s}}$? – Marcus Müller Jan 14 '15 at 17:10
oh yeah,i am aware and i know basics of digg comm. – spectre Jan 14 '15 at 17:13
This is the absolute basic of digital communications -- the relationship between information and bandwidth is what digital communication always tries to optimize. If you haven't come across this, I'd strongly recommend going further into your literature (or where ever you got your basics from). – Marcus Müller Jan 14 '15 at 17:14
BTW, a slightly more general formula for channel capacity is $$C_\mathit{Shannon} = \int\limits_0^{f_\mathrm{bandwidth}} \log_2 \left( 1 + \frac{S(f)}{N(f)} \right) \ df $$ where $S(f)$ is the power spectrum of the signal and $N(f)$ is the power spectrum of the noise (so $S(f)/N(f)$ is the SNR at frequency $f$). – robert bristow-johnson Jan 14 '15 at 18:35
True, and covers the cases where $S(f)$ or $N(f)$ aren't ratio-constant. However, @robertbristow-johnson, I think the original poster is not yet somewhere where he has to consider that (which usually is the case for non-white noise or un-flat channels). – Marcus Müller Jan 15 '15 at 14:16
it was just a BTW. – robert bristow-johnson Jan 15 '15 at 18:33
:) and an interesting one, too, yielding a lot of interesting results about maximizing the throughput of a system! – Marcus Müller Jan 15 '15 at 23:52
@ marcus, So are you saying ,if we have to consider noise effect ,then we have to go for this 'bits/second'? – spectre Jan 16 '15 at 01:53
@MarcusMüller and spectre, i first came upon this reading a paper from Gerzon and Craven. and it sorta turns the Shannon capacity formula around to tell us, given a number of bits in our quantized word, what kind of noise spectrum we can expect to achieve. commonly called the "Gerzon-Craven limit". – robert bristow-johnson Jan 16 '15 at 18:08

score 1 · Answer 2 · answered Jan 14 '15 at 17:04

As far as I'm aware, the original use of 'bandwidth' comes from communications engineering and is measured in hertz, and its use predates digital communications. The data (or information) rate in a link is meaured in bits per second. Since the data rate in a link is related to the system's bandwidth, the term was also relevant to the computer networking community. Over time, and probably due to marketing, the two terms started to be used interchangably in casual conversation. I would hope that networking engineers are aware of the difference, though.

So: the proper unit of bandwidth is hertz. Using orthogonal signaling, the maximum achievable symbol rate (or baud rate), measured in symbols per second, is twice the bandwidth. The maximum bit rate is equal to the symbol rate times the number of bits per symbol.

In networking, the data rate (or link speed, or link rate) is usually measured in bits per second. This is what is commonly, but wrongly, called the network's or the link's "bandwidth".

so,are you saying 'bandwidth' and data rate are not equal.But commonly and wrongly interpreted as equal? — spectre, Jan 14 '15 at 17:34
I can confirm that that's actually what @MBaz is saying, yes. MBaz, would you confirm my confirmation? — Marcus Müller, Jan 15 '15 at 16:42

Fat32 · Answer 3 · 2015-01-15T20:14:34.800

bits/s is data rate of a signal in digital form. Bandwidth of a baseband signal refers to its highest frequency. Bandwidth of a lowpass channel refers to the frequency that passes with -3dB attenuation through the channel. (with respect to 0dB reference) Capacity of a channel refers to maximum rate of information that can be transmitted reliably over a noisy channel with a given bandwidth and SNR.

For a given transmission channel and a prescribed modulation technique to send those bits over its channel, then each bits/second will also be equivalent to a physical bandwidth.

Let's put solid examples: Consider an analog speech signal x(t) first it is lowpass filtered (by an analog RLC network) to 3khz, and then sampled at 8000 samples/second rate at 8-bits/sample resolution by a practical 8-bit ADC. Now you have x[n] signal. This digital signal has 8 bits/sample x 8k sample/second = 64 kbits/second = 64 kpbs data rate.

Now to transmit these bits over a network, will require different bandwidths depending on channel noise and modulation technique employed. The more efficient your modulating scheme, the less physical channel bandwidth you will require. Assuming that, your transmission hertz utilization is 4 bits/hz, then you will require a 16 kHz channel bandwith to transmit this speech signal at 64 kbps data rate.

Take another example as digital video broadcast: Now becoming old and obsolete (nevertheless still in use) digital STDV (Standard Definition TV) PAL system data rate is approximately 625 x 720 x 25 x 16 = 180 Mbps of which 165 Mbps is pure video signal. This huge data rate is reduced to fit into TV broadcast channel bandwidth of 6-Mhz. Greatest part of this reduction comes from video compression which might well reduce the data bitrate down below 30 Mbps, further which is channel encoded to fit into 6 Mhz bandwidth. Hence here, appr. 30 Mbps data rate is transmitted over 6 Mhz channel bandwidth.

hotpaw2 · Answer 4 · 2015-01-15T17:08:00.553

Bandwidth in the sense of Hz of radio channel or spectral frequency width does not take into account the signal-to-noise ratio.

The bits per second that can be reliably sent using a given frequency channel width does take into account the channel noise and signal power, and thus can be quite different (nearly zero BPS in high noise, or much higher bits/second than the Hz frequency width in near zero noise, relative to signal power, given a fancy enough modulation scheme).

Wider channels in Hz at the same S/N can usually carry a greater BPS of information, given suitable modulation schemes. But different S/N ratios or the use of different modulation schemes can kill that relationship.

Difference between the use of 'bandwidth' in DSP and computer networking

4 Answers4