---
title: "T1 signaling rate of 1.544 Mbps"
subject: "Communication Systems"
chapter: 5
key: "semester_6/communication_systems:tdm-24-channels-1544"
source: https://www.bctengineer.com/q/semester_6/communication_systems/tdm-24-channels-1544
site: BCT Engineering
---

# T1 signaling rate of 1.544 Mbps

Communication Systems · Chapter 5: Multiplexing Techniques

- Asked in: 2071 Magh (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/tdm-24-channels-1544
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-5

## Question

Prove that signaling rate for time division multiplexing of 24 voice channels is equal to 1.544 Mbps.

## Short answer

In T1 PCM, 24 channels sampled at 8 kHz with 8 bits/sample plus 1 framing bit per frame gives 193 bits/frame. The signaling rate is 193 × 8000 = 1.544 Mbps.

## Answer

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

```text
|<--------- 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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does a T1 frame last 125 microseconds?
   A. Frames repeat at the sampling rate of 8 kHz
   B. Because there are 193 bits
   C. Because voice is 3.4 kHz
   D. Because each bit lasts 0.648 us
   Answer: A. A frame holds one sample per channel, so T = 1/8000 = 125 us. (Key point: Frame duration)

2. How many bits are in a T1 frame?
   A. 192
   B. 193
   C. 200
   D. 256
   Answer: B. 24 x 8 = 192 information bits plus 1 framing bit = 193. (Key point: Bits per frame)

3. What is the signalling rate?
   A. 193 x 8000 = 1.544 Mb/s
   B. 192 x 8000 = 1.536 Mb/s
   C. 193 x 6800 = 1.312 Mb/s
   D. 24 x 64 kb/s = 1.544 Mb/s
   Answer: A. Rb = bits per frame / frame time = 193 x 8000 = 1 544 000 b/s. (Key point: Signalling rate)

4. How much overhead does the framing bit add?
   A. 8 kb/s
   B. 64 kb/s
   C. 1 kb/s
   D. 772 kb/s
   Answer: A. 1.544 - 1.536 Mb/s = 8 kb/s (1 bit x 8000 frames/s). (Key point: Framing overhead)

## Related in Communication Systems

- [TDM of 24 voice signals: PAM bandwidth and PCM sampling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-24-voice-pam-pcm) — Communication Systems, chapter 5
- [PCM block diagram; T1 signaling rate and frame](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-block-t1-rate-frame) — Communication Systems, chapter 5
- [TDM vs FDM for voice; E1 and T1 hierarchies](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-fdm-voice-e1-t1) — Communication Systems, chapter 5
- [Short note: Frequency division multiplexing](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashort-note-fdm) — Communication Systems, chapter 5, asked 2×
- [E1 frame, TDM hierarchy and signaling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-frame-hierarchy) — Communication Systems, chapter 5
- [T1 and E1 bit rate and bandwidth; DPCM encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3At1-e1-bitrate-dpcm) — Communication Systems, chapter 5

## Related in other subjects

- [Time division multiplexing explained](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Atdm-explain) — Data Communication, chapter 6, asked 3×
- [TDM frame, time slot and 24-channel system](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atdm-frame-24-channel) — Telecommunications, chapter 3
- [Analog vs digital multiplexing; 24-channel system](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aanalog-digital-mux-24-channel) — Telecommunications, chapter 3
- [FDM bandwidth for 10 voice channels](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afdm-10-voice-channels) — Data Communication, chapter 6