--- title: "T1 and E1 bit rate and bandwidth; DPCM encoder" subject: "Communication Systems" chapter: 5 key: "semester_6/communication_systems:t1-e1-bitrate-dpcm" source: https://www.bctengineer.com/q/semester_6/communication_systems/t1-e1-bitrate-dpcm site: BCT Engineering --- # T1 and E1 bit rate and bandwidth; DPCM encoder Communication Systems · Chapter 5: Multiplexing Techniques - Asked in: 2070 Chaitra (CS II), 3+4 marks - Video: [T1 Digital Systems](https://www.youtube.com/watch?v=o4OOA9LkXPk) — Ashutosh Rasttogi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/t1-e1-bitrate-dpcm - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-5 ## Question What are the signalling (bit) rate and bandwidth requirement for the T1 and E1 digital carrier systems? Explain briefly about Differential Pulse Code Modulation (DPCM) encoder. ## Short answer T1 bit rate is 1.544 Mb/s, min BW 772 kHz; E1 bit rate is 2.048 Mb/s, min BW 1.024 MHz. DPCM encodes prediction error, needing fewer bits than PCM. ## Answer #### Bit rate and bandwidth of T1 and E1 Both systems sample speech at $f_s = 8$ kHz with 8 bits per sample. **T1:** 24 channels + 1 framing bit: $$ R_b = (24 \times 8 + 1) \times 8000 = 193 \times 8000 = 1.544\ \text{Mb/s} $$ **E1:** 32 slots (30 voice + 2 overhead): $$ R_b = 32 \times 8 \times 8000 = 256 \times 8000 = 2.048\ \text{Mb/s} $$ **Bandwidth.** The minimum (Nyquist) bandwidth for binary transmission is $B = R_b/2$. With practical NRZ/AMI pulses, the first-null bandwidth is about $R_b$. | System | Bit rate | Min. bandwidth $R_b/2$ | First-null BW ≈ $R_b$ | |---|---|---|---| | T1 | 1.544 Mb/s | 772 kHz | 1.544 MHz | | E1 | 2.048 Mb/s | 1.024 MHz | 2.048 MHz | #### DPCM encoder **Differential PCM** uses the fact that speech samples next to each other are strongly correlated. It does not quantise each sample directly. It quantises the **difference** between the sample and a **prediction** made from earlier samples. The difference is smaller than the sample, so fewer bits are needed for the same quality. ```text m[n] ->(+)--e[n]-->[Quantizer]--eq[n]--+-->[Encoder]-> bits ^- | | v | m^[n] ---> (+) | ^ | mq[n] +-------- m^[n] --[Predictor]<--+ ``` - Prediction error: $e[n] = m[n] - \hat m[n]$. - Quantised error: $e_q[n] = e[n] + q[n]$. - Predictor input: $m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n]$. The predictor (a linear filter on past values of $m_q$) works on the **quantised** samples. The receiver has the same predictor, so encoder and decoder stay in step and quantisation errors do not build up. The SNR improves by the **prediction gain** $G_p = \sigma_m^2 / \sigma_e^2$. DPCM saves about 1–2 bits per sample compared with PCM, e.g. 48–56 kb/s instead of 64 kb/s for speech. Adaptive DPCM (ADPCM) goes further, to 32 kb/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the minimum (Nyquist) bandwidth for binary transmission of the E1 signal at 2.048 Mb/s? A. 2.048 MHz B. 4.096 MHz C. 1.024 MHz D. 772 kHz Answer: C. B = Rb/2 = 2.048/2 = 1.024 MHz. (772 kHz is for T1.) (Key point: Minimum bandwidth) 2. What does a DPCM encoder quantise? A. Each sample directly B. The difference between the sample and a prediction from earlier samples C. The sum of two adjacent samples D. Only the sign of the sample Answer: B. Neighbouring speech samples are correlated, so the prediction error is smaller than the sample and needs fewer bits. (Key point: DPCM principle) 3. On which samples does the DPCM predictor operate, and why? A. On the original samples, because they are exact B. On the quantised samples, so the receiver predictor can match and errors do not build up C. On the noise only D. On the sampling clock Answer: B. The predictor works on mq[n] = m[n] + q[n], which the receiver also has, so encoder and decoder stay in step. (Key point: Predictor input) 4. How is the SNR improvement of DPCM over PCM described? A. Gp = sigma_e^2 / sigma_m^2 B. Gp = sigma_m^2 / sigma_e^2, the ratio of message variance to prediction error variance C. Gp = 2^n D. Gp = Rb/2 Answer: B. The prediction gain is the ratio of message power to prediction-error power; it allows 1-2 fewer bits than PCM. (Key point: Prediction gain) ## Related in Communication Systems - [DPCM vs delta modulation; E1 hierarchy](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adpcm-vs-dm-e1) — Communication Systems, chapter 4 - [T1 digital carrier system; adaptive delta modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3At1-hierarchy-adm) — Communication Systems, chapter 5 - [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 7-bit at 50 Mbps: maximum message bandwidth](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-7bit-50mbps-bandwidth) — Communication Systems, chapter 4 - [Differences between T1 and E1 hierarchy](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3At1-vs-e1) — Communication Systems, chapter 5 - [Error detection vs correction; rate ½ convolutional encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-detect-correct-conv-half-rate) — Communication Systems, chapter 8 ## Related in other subjects - [TDM and T1/E1 carrier standards](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atdm-t1-e1-carriers) — Telecommunications, chapter 3 - [T1 and E1 frame structures and bit rates](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3At1-e1-frame-structures) — Telecommunications, chapter 3 - [Trellis decoding of 1/2 rate conv. code (3-bit errors)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aconv-trellis-3bit) — Data Communication, chapter 8, asked 2× - [Trellis decoding of 1/2 rate conv. code (2-bit errors)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aconv-trellis-2bit) — Data Communication, chapter 8, asked 2×