T1 signal is the output of a 24-to-1 digital multiplexer whose inputs are 64 kbps each (plus one framing bit).

But a multiplexer is simply an electronic device that selects between several inputs and connects it to a single output. Therefore the line rate of the output must be the line rate of which input it is connected to.

So if each input could only transmit at 64 kbps, how could the output line be at 1.544 Mbps? How could the multiplexer transmit at a rate higher than what could the input possibly do?

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3 Answers 3


A T1 is 8000 frames per second. Each frame is 193 bits: a frame bit, followed by 24 8bit channels.

  • 8000 * (1 + 8 * 24) gives a line rate of 1,544,000.

Remember, the T1 was designed for voice applications. This method of multiplexing delivers 5.2microsec of data (a single PCM code) every 125microsec. This makes buffering for analog phones unnecessary. (just a latch - PCM code equates to a voltage, one only needs to hold that voltage until the next code arrives. Vs. gathering 8000 codes in 1/24th of a second and playing them out over one second.)


Each of the 24 time slots gets 64,000 bits, giving you a data rate of 1,536,000. It is only if you count the framing bits do you get the 1.544 Mbps. You can get a fractional T1 where you use only certain time slots, or you can get a full T1 where you use all the time slots and get the full 1.536 Mbps.

The multiplexing is that each input gets one or more specific time slots. With a router that has a T1 line, the router could use all the time slots for the link. You could also have up to 24 routers, each using a single time slot, giving each a 64 Kbps link, or you could have some combination in between.

No matter how you arrange it, the T1 will transmit 1.536 Mbps of data in each second.


Yeah, the answer is "framing." It's funny because it's legacy technology. But I'm sure there are still people using T1s. Then you can get the old line code mismatch, AMI and B8ZS, pick one, not both...

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