T1 signaling rate of 1.544 Mbps

In the T1 (DS1) TDM-PCM system, 24 voice channels are sampled, quantised, encoded and interleaved in time into one frame.

Given standard values

• Voice band limited to about 3.4 kHz, so the sampling rate used is $f_s = 8$ kHz (above the Nyquist rate of 6.8 kHz).
• Each sample is coded with $n = 8$ bits (256 levels, µ-law).
• Number of channels $N = 24$.
• One framing bit is added per frame for synchronisation.

Frame structure

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)

Proof

One frame holds one sample from every channel, so frames repeat at the sampling rate:

$$T_{frame} = \frac{1}{f_s} = \frac{1}{8000} = 125\ \mu\text{s}$$

Bits in one frame:

$$\begin{aligned} \text{Information bits} &= N \times n = 24 \times 8 = 192\ \text{bits} \\ \text{Framing bit} &= 1\ \text{bit} \\ \text{Total} &= 192 + 1 = 193\ \text{bits/frame} \end{aligned}$$

Signalling rate:

$$\begin{aligned} R_b &= \frac{\text{bits per frame}}{T_{frame}} = \frac{193}{125 \times 10^{-6}} \\ &= 193 \times 8000 = 1\,544\,000\ \text{b/s} \end{aligned}$$

Without the framing bit, the rate would be $24 \times 64\ \text{kb/s} = 1.536$ Mb/s. The framing bit adds 8 kb/s of overhead.

Each bit lasts $T_b = 1/1.544\ \text{MHz} \approx 0.648$ µs, and the minimum (Nyquist) transmission bandwidth is $R_b/2 = 772$ kHz.

Answer: Signalling rate $R_b = 193 \times 8000 = 1.544$ Mb/s (proved).