Jerry Avins wrote:>> >> What you are saying translates to "the perfect channels of the >> bandwidth B0 and B1 are not equivalent if B0 != B1". This does not >> seem to be correct. Any ideas? > > > Do you maintain that "Noiseless channels of different bandwidth are not > equivalent" is incorrect? I think there's a difference, but since the > carrying capacity of a noiseless channel is infinite, your aleph-nul is > as big as mine.OK, let's consider the noisy channels with the different bandwidth but with the same Shannon's capacity. What fundamental limitation does not allow for the existence of the operator which directly converts one channel into another? VLV
Analog modulation
Started by ●January 13, 2006
Reply by ●January 15, 20062006-01-15
Reply by ●January 15, 20062006-01-15
Vladimir Vassilevsky wrote:> > > Jerry Avins wrote: > > >>> >>> What you are saying translates to "the perfect channels of the >>> bandwidth B0 and B1 are not equivalent if B0 != B1". This does not >>> seem to be correct. Any ideas? >> >> >> >> Do you maintain that "Noiseless channels of different bandwidth are >> not equivalent" is incorrect? I think there's a difference, but since >> the carrying capacity of a noiseless channel is infinite, your >> aleph-nul is as big as mine. > > > OK, let's consider the noisy channels with the different bandwidth but > with the same Shannon's capacity. What fundamental limitation does not > allow for the existence of the operator which directly converts one > channel into another?Such an operator exists in certain cases. If each channel has the same capacity and they are used to the same fraction of that capacity, they can convey the same amount of information in the same time. Provided, of course, that a recoding exists which will allow it. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 15, 20062006-01-15
Jerry Avins wrote:>> OK, let's consider the noisy channels with the different bandwidth but >> with the same Shannon's capacity. What fundamental limitation does not >> allow for the existence of the operator which directly converts one >> channel into another? > > > Such an operator exists in certain cases. If each channel has the same > capacity and they are used to the same fraction of that capacity, they > can convey the same amount of information in the same time. Provided, of > course, that a recoding exists which will allow it.So, back to the original problem: it is possible to design such an operator which does NOT use the quantization in time and amplitude? VLV
Reply by ●January 15, 20062006-01-15
Vladimir Vassilevsky wrote:> > > Jerry Avins wrote: > > >>> OK, let's consider the noisy channels with the different bandwidth >>> but with the same Shannon's capacity. What fundamental limitation >>> does not allow for the existence of the operator which directly >>> converts one channel into another? >> >> >> >> Such an operator exists in certain cases. If each channel has the same >> capacity and they are used to the same fraction of that capacity, they >> can convey the same amount of information in the same time. Provided, >> of course, that a recoding exists which will allow it. > > > So, back to the original problem: it is possible to design such an > operator which does NOT use the quantization in time and amplitude?I don't see how, but someone clever might. One can in theory sample the signal with the LSB in the noise and encode in a way that suits the channel. At the far end, the signal can be reconstructed. If it can be done digitally, it can be probably done analog. I described analog signal in, channel you describe, and analog signal out. Are the means relevant? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 16, 20062006-01-16
Vladimir Vassilevsky wrote:> Given an analog bandlimited signal, is it possible to create a > continuous time reversible procedure to reduce the bandwidth without > losses? In the other words, is it possible to create a bandwidth > reducing analog modulation? > > Looks like the answer is no, however I can't make a mathematical proof > of this.A bandlimited signal can not be time limited, and when dealing with infinite anything (including infinite time signals), the results can look strange. An infinite number of filters with infinitely narrow bandwidths around center frequency f but driving oscillator outputs at f/2 might do the job, but each filter will have an infinite delay. The process is perfectly reversable. Does that long a start-up delay still count as continuous time? More realistically, one might try scaling the filter width by the inverse S/N ratio of the channel in exchange for imperfect reconstruction and delays slightly less than infinite. IMHO. YMMV. -- rhn A.T nicholson d.O.t C-o-M
Reply by ●January 17, 20062006-01-17
Vladimir Vassilevsky wrote:> Andor wrote: > > > > Let's assume you wanted to send a message of some given length M > > through a channel with some SNR and fixed bandwdith B, and the same > > message through a channel with equal SNR but bandwidth B/2. I think > > you'll agree that the time required needs to be 2M. > > We have an analog signal of the bandwidth B. We have a perfect channel > with the bandwidth less then B. Is it possible to transmit the signal > through this channel without any loss and at the same time scale?After sending any signal through a perfect channel with bandwidth B, it will be infinite in length - changing the time scale won't matter anymore.> Looks like any procedure which can accomplish that requires quantization > both in time and in amplitude, i.e. has to be digital and lossy. Is > there any proof?Well, if you have a perfect channel you have to assume non-quantized sampling. In that case, you don't lose information. Regards, Andor
Reply by ●January 17, 20062006-01-17
Jerry Avins wrote:> Andor wrote:<Snip>> > I don't see any problem once you have accepted the fact that the > > product of message time with the bandwdith has to stay constant > > (assuming equal SNR of all channels involved). > > Yes, of course, but the message must have finite length, so it is not a > continuous data stream in a strict sense. I call it a batch job, > "intermittently continuous". For two to one, one begins recording at > full speed and immediately transmitting at half speed*. That way, a > buffer only large enough to hold half the message is needed and the > delay is least. > > Jerry > ___________________________ > * Possible with digital storage, but that means sampling, again, not > truly continuous.Or a big bin for the tape :) However considering that most quality magnetic tape units used AC supersonic bias, maybe even audio tape recordings could be considered to be sampled. Regards Rocky
