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Let's do a simple case on the left with X and A user telephones. enter image description here

So the way an analog signal (user's speech X) is handled, is that it is sampled at 8000 bits / sec.

The high-speed C-channel has 128 kilo bits/sec of capacity.

X only transmits at 64kbps

A transmits at 64 kbps too.

However, if the switch 1 only "listens" to X for some time constant, and then switches to A, wouldn't it make the speech of X seem slury/ with pauses? Let's imagine user X is talking continuously for 5 minutes, if switch 1 is switching between X and A, then there will be short pauses in what user X is saying, isn't it?

So why is circuit switching even considered as a viable option in telephone communication? It's not.

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    \$\begingroup\$ "Circuit switching" is what wired telephones have done since the stone age, so clearly your assertion that it's not a viable option is false. TDMA was used by earlier generations of GSM mobile phones - and clearly that also worked just fine despite your assertion that it can't ... You're forgetting that the channel bandwidth far exceeds that necessary for a single call, that the audio is buffered at each end, and that for mobile phones, the 'circuit' was switched fast enough that the delay isn't noticeable. \$\endgroup\$ – brhans Oct 8 '18 at 1:20
  • \$\begingroup\$ @brhans how can the sw1 switch "fast enough" if both users X and A are talking at the same time? Data must be transmitted from both X and A to the high-speed channel "C". For the duration of time that sw1 only listens to A, it ignores whatever data X is sending at this time. \$\endgroup\$ – Jack Oct 8 '18 at 1:24
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    \$\begingroup\$ When X isn't sending, he's buffering. When X gets a turn to send, he sends his buffered data. When Y receives X's data, he also buffers. When there's no data coming in, Y is still playing buffered data. \$\endgroup\$ – brhans Oct 8 '18 at 1:27
  • \$\begingroup\$ @brhans well, if it buffers, then at the other end ("Y") the recipient will be hearing short pauses in between words/sentences of user X, since X has to "buffer" (wait) till sw1 comes back to X. \$\endgroup\$ – Jack Oct 8 '18 at 1:30
  • \$\begingroup\$ You write "buffer" (wait) as if they mean the same thing. They don't. Buffering means storing data. When it's your turn to talk you send all of that stored data in a much shorter timer period than it took to store the data. There are no pauses, there is only a consistent small delay. \$\endgroup\$ – brhans Oct 8 '18 at 1:33
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The key thing to notice is that the wire can send information twice as fast as X or A or Y or B can actually speak or listen. So the switch on one side speeds up the speech and the switch on the other side slows it down again.

Okay, so X is reciting the alphabet, and A's ordering a pizza. And let's say the time slot is 4 words long, which means they actually only say 2 words in a time slot. And let's say A goes first.

The switch takes two slots' worth of speech and sends it in one time slot. Then the switch on the other end takes one slots' worth of sped up speech and plays it back in two time slots.

       | slot 1    | slot 2   | slot 3        | slot 4  | slot 5          | slot 6  | slot 7  |
 X:    | A B       | C D      | E F           | G H     | I J             | K L     |
 A:    | Hi my     | name  is | A  and        | I would | like to         | order a |

 TDMA: |     Hi my | A B C D  | name is A and | E F G H | I would like to | I J K L |

 Y:    | <silence> | A B      | C D           | E F     | G H             | I J     | K L     |
 B:    | <silence> | Hi my    | name is       | A and   | I would         | like to | order a |

So yes there's a delay, no there isn't any stuttering.

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  • \$\begingroup\$ it'd be nice to put numbers. If we know the capacity of channel is 128 *10^3 bits/sec = C. Both X and Y have capacities of C_x = 64 *10^3 bits / sec. One slot duration in X would be 1/C_x, right? Anyhow, in your example. How can C-channel "grab" data from A faster than A could possibly transmit? If it takes 1 sec for A to say "Hi my" then that 1 second is how long it's gonna take for C-channel to "grab" the data "Hi my" from A, no? \$\endgroup\$ – Jack Oct 8 '18 at 5:23
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    \$\begingroup\$ @Jack if you look closely, I made it insert silence at the beginning. \$\endgroup\$ – user253751 Oct 8 '18 at 7:11
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The high-speed channel C can transmit two bits in the same time that X and A each produce one bit, so it can simply transmit both bits, one after the other. No bits get lost. Actually, in the PSTN, samples (bytes), rather than bits, are interleaved. In this way, a 1.544 Mbps T1 line can handle 24 conversations simultaneously. 8000 times a second, a complete 193-bit frame of data is transmitted, which contains one 8-bit audio sample for each of the 24 circuits (plus one bit of "overhead").

In wireless TDMA, FIFO buffers are used to bunch the bits from X and A into larger groups called "packets", and the packets are interleaved in the high-speed radio channel. Each packet is transmitted in much less time than it took to create it. At the other end, another buffer captures the packet and then the data is transferred to the DAC at the original sample rate.

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  • \$\begingroup\$ ">No bits get lost" I kinda get what you're trying to say, but unless TDMA delays the other bit from user X, it simply cannot accept data from X when sw1 is listening to A. Let's say both X and A transmit a 4 bit data at time T1, X transmits 1101 and A transmits 0011, at T1 when sw1 is on for A, it can take 0011, but it won't be taking 1101 as there is no physical connection for X at T1. At T2 when sw1 decides to switch to X, the 1101 is already gone from X and X is transmitting different data. So 1101 gets lost. \$\endgroup\$ – Jack Oct 8 '18 at 1:28
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    \$\begingroup\$ You're still ignoring the fact that the channel is twice as fast as each of the two data sources. It doesn't have to ignore one while sending the other; it just needs to save the bits somewhere temporarily. This interleaving can be done at the bit, byte or packet level. \$\endgroup\$ – Dave Tweed Oct 8 '18 at 1:30
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    \$\begingroup\$ Yes, there's a slight delay. This makes no difference in voice communication. \$\endgroup\$ – Dave Tweed Oct 8 '18 at 1:34
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    \$\begingroup\$ @Jack - now you're just flying off into hypothetical nonsense. If there are 1000 users all wanting to use their old mobile phones they're not all going to be on the same channel. And if there are too many users to be supported by the available channels then some of them will get a "network busy" signal. This was all figured out by smart engineers decades ago, even though you still think it's impossible. \$\endgroup\$ – brhans Oct 8 '18 at 1:39
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    \$\begingroup\$ @Jack If you have 1000 users then each sample has to wait for 999 other samples, but 8,000,000 samples are going through every second so it STILL only has to wait for 125 microseconds. \$\endgroup\$ – user253751 Oct 8 '18 at 3:12
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The key reason TDMA works without any apparent delay or pauses is the frequency at which the switching occurs. Audio sampling is best thought of as a stream of individual samples occurring at a certain rate. So for an example this might be 8000Hz, or one every 1/8000 of a second. Meaning we have 1/8000 of a second between transmitting samples. If a sample can be transmitted in 1/16000 of a second we can transmit 1 sample from each of 2 telephones in the same interval. And for each of those telephones, a sample will arrive every 1/8000th of a second. So the signal is transferred without any increased delay over a non-multiplexed line.

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    \$\begingroup\$ It would be 1/16000 of a second if it was twice as fast. \$\endgroup\$ – user253751 Oct 8 '18 at 3:13
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In the duration of time that sw1 only listens to A, A can't do anything meaningful because to do so would mean it would be faster than 8kbaud.

The two signals can be interleaved only because they are bandwidth-limited. On the receiving end, the "cut-up" signal can be filtered to recover the original bandwidth-limited signal.

enter image description here

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