Andor wrote:> Andor wrote:...> Is recording onto a recorder and playing back at a different speed a > "continuous process"? How does it make a difference if the recorder is > analog or digital?A device that plays back faster than it records will shortly predict future inputs. One that plays back more slowly must eventually choke. Such a device can at best be intermittently continuous. :-) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Analog modulation
Started by ●January 13, 2006
Reply by ●January 14, 20062006-01-14
Reply by ●January 15, 20062006-01-15
Jerry Avins wrote:> Andor wrote: > > Andor wrote: > > ... > > > Is recording onto a recorder and playing back at a different speed a > > "continuous process"? How does it make a difference if the recorder is > > analog or digital? > > A device that plays back faster than it records will shortly predict > future inputs. One that plays back more slowly must eventually choke. > Such a device can at best be intermittently continuous. :-)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. In other words, if doubling the transmission time for any message is not an option, you need at least two channels (assuming the SNR stays constant) with half the bandwidth. One can think of a scheme with several recorders alternately recording and playing back two continuous signals on the two channels. For the general case, the number of recorders required depends only on the ratio of the two bandwidths. Such a transmission scheme can be implemented with arbitrary small, but finite, delay. A second option would be to separate the signal with bandwidth B into two signals with bandwidth B/2 - this would be even simpler using analog technology (filters and mixers) - and recombine the original signal at the receiving end. 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). Regards, Andor
Reply by ●January 15, 20062006-01-15
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? 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? Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●January 15, 20062006-01-15
Andor wrote:> 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). >A nonlinear operator, like FM, for example, converts the finite bandwidth signal to the infinite bandwidth. Why there could not be an operator with the opposite property of shrinking of the bandwidth for any signal? You can map a finite piece to an infinite line. Why there is no such maping in the frq domain? Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●January 15, 20062006-01-15
Andor wrote:> Jerry Avins wrote: > > >>Andor wrote: >> >>>Andor wrote: >> >> ... >> >> >>>Is recording onto a recorder and playing back at a different speed a >>>"continuous process"? How does it make a difference if the recorder is >>>analog or digital? >> >>A device that plays back faster than it records will shortly predict >>future inputs. One that plays back more slowly must eventually choke. >>Such a device can at best be intermittently continuous. :-) > > > 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. > > In other words, if doubling the transmission time for any message is > not an option, you need at least two channels (assuming the SNR stays > constant) with half the bandwidth. One can think of a scheme with > several recorders alternately recording and playing back two continuous > signals on the two channels. > > For the general case, the number of recorders required depends only on > the ratio of the two bandwidths. Such a transmission scheme can be > implemented with arbitrary small, but finite, delay. > > A second option would be to separate the signal with bandwidth B into > two signals with bandwidth B/2 - this would be even simpler using > analog technology (filters and mixers) - and recombine the original > signal at the receiving end. > > 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. -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 15, 20062006-01-15
Vladimir Vassilevsky wrote: ...> 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? > 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?A proof probably lies in the meaning of bandwidth. We have a pile of oranges to big to fit into a particular box. We want to transport those oranges, all at the same time and all in that box, without making orange juice. Can it be done? I think not. Is there a proof? I think the same proof applies your question and mine. 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:>> 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? >> 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?> A proof probably lies in the meaning of bandwidth. > > We have a pile of oranges to big to fit into a particular box. We want > to transport those oranges, all at the same time and all in that box, > without making orange juice. Can it be done? I think not. Is there a > proof? I think the same proof applies your question and mine.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? Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●January 15, 20062006-01-15
Jerry Avins wrote:> Andor wrote: > > Jerry Avins wrote: > > > > > >>Andor wrote: > >> > >>>Andor wrote: > >> > >> ... > >> > >> > >>>Is recording onto a recorder and playing back at a different speed a > >>>"continuous process"? How does it make a difference if the recorder is > >>>analog or digital? > >> > >>A device that plays back faster than it records will shortly predict > >>future inputs. One that plays back more slowly must eventually choke. > >>Such a device can at best be intermittently continuous. :-) > > > > > > 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. > > > > In other words, if doubling the transmission time for any message is > > not an option, you need at least two channels (assuming the SNR stays > > constant) with half the bandwidth. One can think of a scheme with > > several recorders alternately recording and playing back two continuous > > signals on the two channels. > > > > For the general case, the number of recorders required depends only on > > the ratio of the two bandwidths. Such a transmission scheme can be > > implemented with arbitrary small, but finite, delay. > > > > A second option would be to separate the signal with bandwidth B into > > two signals with bandwidth B/2 - this would be even simpler using > > analog technology (filters and mixers) - and recombine the original > > signal at the receiving end. > > > > 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.No, the data can be a continuous stream. Re-read what I wrote.
Reply by ●January 15, 20062006-01-15
Vladimir Vassilevsky wrote:> > > Jerry Avins wrote: > >>> 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? >>> 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? > > >> A proof probably lies in the meaning of bandwidth. >> >> We have a pile of oranges to big to fit into a particular box. We want >> to transport those oranges, all at the same time and all in that box, >> without making orange juice. Can it be done? I think not. Is there a >> proof? I think the same proof applies your question and mine. > > > 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. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 15, 20062006-01-15
Andor wrote:> Jerry Avins wrote:...>>Yes, of course, but the message must have finite length, so it is not a >>continuous data stream in a strict sense. > > > No, the data can be a continuous stream. Re-read what I wrote.Do you mean that two channels of B/2 can replace one of B? Sure. Anyway, signals sent in binary form on low-noise channels rarely use the channel's full capacity. We get 56K bits/second worth of information on a telephone line, but we that is achieved with a lower rate of symbols each packed with more information. There is no magic, but there is inspiring sleight of hand. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
