Hello Forum, For a noiseless channels the maximum number of bits/sec= B* log V where V is the number of discrete levels. In digital communications, given a certain fixed bandwidth, say 6 MHz like in the TV channels, it seems that we can jam as many bits-per-second we want into that fixed bandwidth B.... That makes me conclude that if the noise is small enough, infinite data rate is possible. From Fourier theory I have learned that the more complex a signal is the more Frequency components (high frequency) it must contain, the bigger the bandwidth. A bandwidth of 6 MHz means that we have Fourier frequencies in that range. By playing with the amplitude and phase of those frequency components we can create different signals of different complexity. But isn't a signal representing a high data rate more complicated than one representing a lower data rate? And doesn't that extra complexity require a larger frequency bandwidth? How can two signals representing different data rates involve the same range of Fourier frequencies? I would naively think that the more bits in an interval T, the more frequencies we need to represent the signal.... Even using more than 2 discrete levels, the signal would turn out to be more complex and apparently need more bandwidth... what am I missing that is so fundamental? Thanks fisico32
Variable data rate with fixed bandwidth
Started by ●June 18, 2011
Reply by ●June 18, 20112011-06-18
On Jun 18, 9:15�am, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote:> Hello Forum, > > �For a noiseless channels the maximum number of > �bits/sec= B* log V where V is the number of discrete levels. > > In digital communications, given a certain fixed bandwidth, say 6 MHz like > in the TV channels, �it seems that we can jam as many bits-per-second we > want into that �fixed bandwidth B.... �That makes me conclude that if the > noise is small enough, infinite data rate is possible. > > From Fourier theory I have learned that the more complex a signal is the > more Frequency components (high frequency) it must contain, the bigger the > bandwidth. > > A bandwidth of 6 MHz means that we have Fourier frequencies in that range. > By playing with the amplitude and phase of those frequency components we > can create different signals of different complexity. > But isn't a signal representing a high data rate more complicated than one > representing a lower data rate? > > And doesn't that extra complexity require a larger frequency bandwidth? > How can two signals representing different data rates involve the same > range of Fourier frequencies? I would naively think that the more bits in > an interval T, the more frequencies we need to represent the signal.... > > Even using more than 2 discrete levels, the signal would turn out to be > more complex and apparently need more bandwidth... > > what am I missing that is so fundamental?I won't even try to answer in technical terms. The absence of noise and the ability to distinguish infinitesimal differences in level is simply unrealistic. Implement an unabridged dictionary in Tex; the result is an ASCII file. Concatenate the characters to make a single number, then put a binary point at the left end. The result is some number between zero and one. Cut a stick to that length in inches. An unabridged dictionary doesn't impress you? Then do the same with the entire Library of Congress. Imagine! All that information encoded onto a stick you can put on your key chain! Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●June 18, 20112011-06-18
On 06/18/2011 06:15 AM, fisico32 wrote:> Hello Forum, > > For a noiseless channels the maximum number of > bits/sec= B* log V where V is the number of discrete levels. > > In digital communications, given a certain fixed bandwidth, say 6 MHz like > in the TV channels, it seems that we can jam as many bits-per-second we > want into that fixed bandwidth B.... That makes me conclude that if the > noise is small enough, infinite data rate is possible.Claude Shannon thought so too, and came up with some mathematics to formalize this concept. It's called the "Channel Capacity", which is a part of information theory. It's new enough and esoteric enough that you may not have heard of it, though -- it's only been around since the mid 1940's, and is the foundation of nearly all modern communications theory.> From Fourier theory I have learned that the more complex a signal is the > more Frequency components (high frequency) it must contain, the bigger the > bandwidth. > > A bandwidth of 6 MHz means that we have Fourier frequencies in that range. > By playing with the amplitude and phase of those frequency components we > can create different signals of different complexity. > But isn't a signal representing a high data rate more complicated than one > representing a lower data rate?Not necessarily. The shape of a particular symbol determines it's spectrum, but you're free to set the amplitude of the symbol to anything you want (within bounds of the available power). See QAM.> And doesn't that extra complexity require a larger frequency bandwidth? > How can two signals representing different data rates involve the same > range of Fourier frequencies? I would naively think that the more bits in > an interval T, the more frequencies we need to represent the signal.... > > Even using more than 2 discrete levels, the signal would turn out to be > more complex and apparently need more bandwidth... > > what am I missing that is so fundamental?Very roughly, as a signal gets sharper corners it needs more spectrum. But there are ways to add 'complexity' (mostly by increasing the gradations of amplitude that you allow and can discern) that don't affect spectrum. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●June 18, 20112011-06-18
On Sat, 18 Jun 2011 08:15:51 -0500, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote:>Hello Forum, > > For a noiseless channels the maximum number of > bits/sec= B* log V where V is the number of discrete levels. > >In digital communications, given a certain fixed bandwidth, say 6 MHz like >in the TV channels, it seems that we can jam as many bits-per-second we >want into that fixed bandwidth B.... That makes me conclude that if the >noise is small enough, infinite data rate is possible. > >From Fourier theory I have learned that the more complex a signal is the >more Frequency components (high frequency) it must contain, the bigger the >bandwidth. > >A bandwidth of 6 MHz means that we have Fourier frequencies in that range. >By playing with the amplitude and phase of those frequency components we >can create different signals of different complexity. >But isn't a signal representing a high data rate more complicated than one >representing a lower data rate? > >And doesn't that extra complexity require a larger frequency bandwidth? >How can two signals representing different data rates involve the same >range of Fourier frequencies? I would naively think that the more bits in >an interval T, the more frequencies we need to represent the signal.... > >Even using more than 2 discrete levels, the signal would turn out to be >more complex and apparently need more bandwidth... > >what am I missing that is so fundamental? >Thanks > fisico32I'll add a little bit to what's been said: I'm restricting my comments to single-carrier signals, since pretty much all the relevant points can be made and it's the simplest case to understand. Generally the bandwidth of a single-carrier signal depends only on the symbol rate of the signal. The pulse filtering will determine the shape of the spectrum, but the 3dB points of the spectrum are pretty much generally taken as the bandwidth and are separated by the symbol rate. 6Msymbols/second takes up about 6MHz of BW between the 3dB points, and the slope of the rolloff is determined solely by the filtering and doesn't directly affect the information rate. So the information rate is essentially the symbol rate times the number of bits per symbol. As you say, as the noise level drops one can generally transmit more bits per symbol, or, as is often said, one can use a higher modulation order. e.g., QPSK works at low SNR but only carries two (potentially coded) bits per symbol, and 1024-QAM needs very high SNR and high fidelity components but can carry 10 (potentially coded) bits per symbol. Also as has been mentioned, Information Theory reveals that there is a theoretical (and proven practical) bound, i.e., Shannon's Channel Capacity, of how well one can expect to do in a typical channel. You can't expect to just keep jamming in more bits per symbol, because channels and electrical components are not perfect. There is another key contributor, channel coding, which can provide benefits and yield more reliable information bits/symbol than could be expected otherwise when the channels get noisy. Some overhead is required for the code, but the net gains are generally large when noise is present, which it always is, so the code rate also affects the information rate. Your curiosity is good. A basic comm engineering class would cover this sort of thing. Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●June 18, 20112011-06-18
Hey fizico, Think about this: if there was a way to convert a stupidity into an energy, we could probably build a huge power plant, do we? fisico32 wrote:> Hello Forum, > > For a noiseless channels the maximum number of > bits/sec= B* log V where V is the number of discrete levels. > > In digital communications, given a certain fixed bandwidth, say 6 MHz like > in the TV channels, it seems that we can jam as many bits-per-second we > want into that fixed bandwidth B.... That makes me conclude that if the > noise is small enough, infinite data rate is possible. > > From Fourier theory I have learned that the more complex a signal is the > more Frequency components (high frequency) it must contain, the bigger the > bandwidth. > > A bandwidth of 6 MHz means that we have Fourier frequencies in that range. > By playing with the amplitude and phase of those frequency components we > can create different signals of different complexity. > But isn't a signal representing a high data rate more complicated than one > representing a lower data rate? > > And doesn't that extra complexity require a larger frequency bandwidth? > How can two signals representing different data rates involve the same > range of Fourier frequencies? I would naively think that the more bits in > an interval T, the more frequencies we need to represent the signal.... > > Even using more than 2 discrete levels, the signal would turn out to be > more complex and apparently need more bandwidth... > > what am I missing that is so fundamental? > Thanks > fisico32
Reply by ●June 18, 20112011-06-18
My comments are interspersed with "***" below: On 6/18/2011 6:15 AM, fisico32 wrote:> Hello Forum, > > For a noiseless channels the maximum number of > bits/sec= B* log V where V is the number of discrete levels. > > In digital communications, given a certain fixed bandwidth, say 6 MHz like > in the TV channels, it seems that we can jam as many bits-per-second we > want into that fixed bandwidth B.... That makes me conclude that if the > noise is small enough, infinite data rate is possible. > > From Fourier theory I have learned that the more complex a signal is the > more Frequency components (high frequency) it must contain, the bigger the > bandwidth.***You need to be very careful with definitions here. What does "complex" mean? I would not accept this definition because I don't know what you mean by "complex". And, I'd say it isn't a commonly used term in this context.> > A bandwidth of 6 MHz means that we have Fourier frequencies in that range. > By playing with the amplitude and phase of those frequency components we > can create different signals of different complexity.***Not according to your definition of complexity above!! By playing with the amplitude and phase of those frequency components we can create different composites but don't add any new frequencies. Think of a Fourier Series that is strictly lowpass in nature.> But isn't a signal representing a high data rate more complicated than one > representing a lower data rate?***Heck, I don't know! What is "complicated"?> > And doesn't that extra complexity require a larger frequency bandwidth?***I don't know. What is "complexity"?> How can two signals representing different data rates involve the same > range of Fourier frequencies?***As commented on by others ... the data rate and the Fourier coefficients relative to bandwidth occupied are independent. I would naively think that the more bits in> an interval T, the more frequencies we need to represent the signal....***Yes. That would be somewhat naive. Consider PAM with varying numbers of bits per symbol / levels per pulse.> > Even using more than 2 discrete levels, the signal would turn out to be > more complex and apparently need more bandwidth...***Complex???> > what am I missing that is so fundamental?***It appears that you are misleading yourself by using the terms "complex" and "complexity" with no definition of what that means. ***Consider a very simple case: High SNR. - One signal goes between two levels in steps at a given rate. - Another signal goes between four levels in steps at the same given rate. Both "stepped waveforms" are passed through the same lowpass filter., Now any differences there might have been in their spectra are limited to the same set of Fourier Coefficients - but with perhaps different amplitude and phase between the two. ... which is really not much to the point here. With such a small number of levels, the filter is designed to let the amplitudes be consistently reached to something like 95% of the input level and with low intersymbol interference. So, there is no difference in the signal bandwidths at all. BUT One signal carries one bit per symbol The other signal carries two bits per symbol in the same bandwidth. The latter has double the data rate in the same bandwidth. Eventually intersymbol interference (the tails of one pulse overlap the peak of another) and noise limit one's ability to distinguish between levels / thus bits. Maybe this will help: Assume the 1-bit system transmits -2 volts for one level and +2 volts for the other level. The waveform out of the presumed filter has to reach 95% of 4 volts at each change. So that's what it's designed to do. Now increase the number of levels at the input by adding levels at +1 volt and -1 volt. Using the same filter: Surely if the output can traverse 4 volts in a symbol period then it can traverse 1 or 2 or 3 volts in the same symbol period, eh? Well, the percentage will generally be the same because the forcing function magnitude is changing - if you're used to solving transient responses of linear systems. In this manner you an keep adding levels until the system is no longer capable of detecting them. But the bandwidth stays the same. Fred
Reply by ●June 18, 20112011-06-18
fisico32 <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote: (snip)> A bandwidth of 6 MHz means that we have Fourier frequencies in that range. > By playing with the amplitude and phase of those frequency components we > can create different signals of different complexity. > But isn't a signal representing a high data rate more complicated than one > representing a lower data rate?> And doesn't that extra complexity require a larger frequency bandwidth? > How can two signals representing different data rates involve the same > range of Fourier frequencies? I would naively think that the more bits in > an interval T, the more frequencies we need to represent the signal....No, the extra complexity is carefully designed to reduce bandwidth, though it makes the decoding harder. Simpler modulation methods have a larger bandwidth for the data rate. 10 megabit ethernet is Manchester coded, with a 20MHz bandwidth. It trades bandwidth for easier decoding. 100baseTX use MLT3 coding to transmit 100 Mb/s in much less than 100 MHz. You have to consider that the higher frequency components are lost along the way, but the decoder can still recover the bits. The symbol rate is 125M symbols/second, with the fundamental (and peak for RFI) at about 32MHz.> Even using more than 2 discrete levels, the signal would turn > out to be more complex and apparently need more bandwidth...NO, the more complex signals need less bandwidth (per bit). -- glen
Reply by ●June 18, 20112011-06-18
On 06/18/2011 01:27 PM, Eric Jacobsen wrote:> I'll add a little bit to what's been said:Oh Eric, you're such a comedian! -- Randy Yates % "Watching all the days go by... Digital Signal Labs % Who are you and who am I?" mailto://yates@ieee.org % 'Mission (A World Record)', http://www.digitalsignallabs.com % *A New World Record*, ELO
Reply by ●June 19, 20112011-06-19
On 06/18/2011 01:27 PM, Eric Jacobsen wrote:> [...] > Generally the bandwidth of a single-carrier signal depends only on the > symbol rate of the signal.How do you explain, then, the difference in spectral efficiency between 4-PSK and 2-PSK here: http://http://www.digitalsignallabs.com/powervsbandwidth.png ? For the same Eb/No, we get identical BER but using half the bandwidth. Seems to me that the bandwidth of a single-carrier signal depends on the modulation type as well as the symbol rate. -- Randy Yates % "Watching all the days go by... Digital Signal Labs % Who are you and who am I?" mailto://yates@ieee.org % 'Mission (A World Record)', http://www.digitalsignallabs.com % *A New World Record*, ELO
Reply by ●June 19, 20112011-06-19
On 06/18/2011 11:20 PM, Randy Yates wrote:> [...] > http://http://www.digitalsignallabs.com/powervsbandwidth.pngCorrection: http://www.digitalsignallabs.com/powervsbandwidth.png -- Randy Yates % "Watching all the days go by... Digital Signal Labs % Who are you and who am I?" mailto://yates@ieee.org % 'Mission (A World Record)', http://www.digitalsignallabs.com % *A New World Record*, ELO
