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I am wondering how the electrical interface for a frequency multiplexed telephone line works. If two senders are connected to a single line, how is it that the final output stage amplifiers don't clobber each other's signals, possibly even burning out the amplifiers? Are there analog filters after the amplifiers to ensure that the impedance the amplifier sees for signals outside the passband is very high? Does this mean that each device (presumably the modem) can only transmit on one frequency, or can the filters be tuned to a desired passband or selected from a bank of filters in the device? Or, alternately, is the output impedance of each amplifier high enough such that the loading effects are tolerated? Would this imply that the signal level per channel drops with every connected device?

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Frequency Division Multiplexing (FDM) was first demonstrated by the Bell System in 1918, but was not put into wide commercial use until the 1930's. It works by translating a single analog voice telephone channel to one of several higher frequency bands and then combining them into one signal.

Telephone voice channels have a frequency range of 300 to 3400 Hz. Each voice channel (corresponding to a single voice conversation) is modulated by a carrier frequency, stating with 60 kHz, then 64 kHz etc.

This shows the multiplexing/modulation scheme for three channels out of a 12 channel group:

enter image description here

Two sidebands are created in this way, but only the upper sideband is needed since they are redundant. The carrier is not transmitted either, and instead it is added back at the other end. So the transmission scheme is called "single sideband suppressed carrier".

By using channel spacing of 4 kHz, this provided a "guard band" between channels. The channels were filtered through a low-pass filter to remove any high frequency components.

At the other end, the carrier was added back in, and precise (and expensive) band-pass filters were used to separate out each channel. These filters used quartz crystal elements to provide a very steep rolloff:

enter image description here

This shows the response for a filter design for the 64-68 kHz band:

enter image description here

When sending over long distance circuits, telephone channels are split up into signals, one going to the far end, and the other coming back from the far end (4-wire circuit). This splitting is done by a circuit call a hybrid. Thus each channel over the FDM system actually required two sets of equipment, one in each direction:

enter image description here

Twelve sets of channels were combined into a "Group". These were combined into higher and higher levels as the capacity of telephone cable (particularly coaxial) increased. This table shows groups defined by the CCITT (now ITU-T), a international telecommunications standards group,m the highest being a 60 MHz system with over 10,000 channels.

enter image description here

To answer your specific questions:

"If two senders are connected to a single line, how is it that the final output stage amplifiers don't clobber each others signals, possibly even burning out the amplifiers?"

The only place where signals are mixed is following the modulators, to create one signal to be sent over the line. The inputs are high impedance, there is just one output stage. At the other end, there is one output stage per channel.

"Are there analog filters after the amplifiers to ensure that the impedance the amplifier sees for signals outside the passband is very high?"

The passband filters are before the amplifiers.

"Does this mean that each device (presumably the modem) can only transmit on one frequency, or can the filters be tuned to a desired passband or selected from a bank of filters in the device? Or, alternately, is the output impedance of each amplifier high enough such that the loading effects are tolerated?"

Each modulator is the combination of an audio channel and carrier frequency.

"Would this imply that the signal level per channel drops with every connected device?"

No, they are all independent.

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  • \$\begingroup\$ I am familiar with mixing and guard bands, etc. I approach signal processing from my experience with DSP in hobby projects and two summer internships at a test and measurement company, so I tend to have some gaps in analog. There's a lot of useful information here about the specific implementation and I especially enjoy the table at the end, but my specific question has not been sufficiently answered. More information about the bandpass filters in the muxing equipment would be useful. The most useful answers will even demonstrate how a filter may be designed. \$\endgroup\$ – Void Star Sep 13 '15 at 15:15
  • \$\begingroup\$ @BigEndian I was trying to answer the title of your question, "How does frequency multiplexing on telephone lines work?" I have added answers to each of your questions to my answer. I have also included some more information of the bandpass filters used to my answer. You seem to be concerned a lot about loading and impedances. All though they didn't have them back then, think of opamps. They have very high input impedances, and low output impedance. So you can sum a lot of channels together very easily without signals interfering with each other. \$\endgroup\$ – tcrosley Sep 13 '15 at 20:15
  • \$\begingroup\$ Your answer implies that subscribers on a telephone network have their own line for the "first mile", i.e. until you get to a piece of telecom infrastructure, which then does a SSB modulation, filters out of band noise, sums with the other channels, and then goes on to the next piece of infrastructure. What I was imagining was a single transmission line to which subscribers connect to "in parallel". Given that the latter is possible, and uses much less cabling, I do not understand why the former would be used. Can you update your answer with references explaining the choice of technologies? \$\endgroup\$ – Void Star Sep 14 '15 at 4:58
  • \$\begingroup\$ As to the title of the question, that was horribly misleading and I will edit it. \$\endgroup\$ – Void Star Sep 14 '15 at 4:59
  • \$\begingroup\$ Can you also comment on the purpose of the arrows connecting the crystals and inductors? \$\endgroup\$ – Void Star Sep 14 '15 at 5:04
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Think about a much bigger system - commercial broadcast radio - there are literally hundreds of high quality transmissions occuring simultaneously but the system works just fine because a radio receiver can tune into one part of the spectrum and ignore the closeby transmissions that are not desired.

As for the transmit antenna it may be handling several different transmissions in one particular band and electrically, the different power amps (that feed the antenna) will use "tight" filters to prevent leakage back from other power amps.

This means the filters do as you suggest in your question - they block by providing a high impedance but are fairly transparent to the frequency they wish to pass.

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  • \$\begingroup\$ I upvote this because it actually answers my question more thoroughly than tcrosley's answer. I would still like to see more detail before I accept an answer. \$\endgroup\$ – Void Star Sep 13 '15 at 15:21
  • \$\begingroup\$ @BigEndian the basic LC series resonant circuit can be used for each band - it acts like a low impedance for the frequencies it wishes to pass and a high impedance for those it wishes to block. This allows several PAs (each with their own tuned filter) to output to the same node simultaneously. However, I suspect that pro equipment doing this will take the low level modulated signals from each modulator and combine with a summer and then feed to a PA for transmission onto the line (or antenna). \$\endgroup\$ – Andy aka Sep 13 '15 at 17:17
  • \$\begingroup\$ An LC circuit wouldn't have a very flat response in the passband or very fast roll off. Is this really how it is done? \$\endgroup\$ – Void Star Sep 13 '15 at 18:19
  • \$\begingroup\$ You can make LC circuits as complex as you need to achieve the passband character whilst remaining high impedance to bands co-located. It's all a matter of how complex you want it to be. \$\endgroup\$ – Andy aka Sep 14 '15 at 7:10
  • \$\begingroup\$ Ah, I took you to mean 1 L and 1 C, but yes, complex filters can give more desirable responses. \$\endgroup\$ – Void Star Sep 14 '15 at 7:20

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