--- title: "PCM switching and three-stage crosspoints" subject: "Telecommunications" chapter: 4 key: "semester_8/telecommunications:pcm-switching-crosspoints-4x5" source: https://www.bctengineer.com/q/semester_8/telecommunications/pcm-switching-crosspoints-4x5 site: BCT Engineering --- # PCM switching and three-stage crosspoints Telecommunications · Chapter 4: Digital Switching - Asked in: 2073 Shrawan, 3+5 marks - Video: [Switching Engineering - Cross-points and Number of Switches - Calculations](https://www.youtube.com/watch?v=Bbm3lKI7O5Y) — KINEMARC - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/telecommunications/pcm-switching-crosspoints-4x5 - Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-4 ## Question What are advantages and issues of PCM switching when it compared with analog switching? Calculate and draw, how many cross points are found in three stages switching system, whereas 3 stages array of 4 input lines and 5 second stages array. ## Short answer PCM switching advantages: no noise accumulation, time-shared hardware, integration with PCM transmission, low cost VLSI, data/ISDN. Issues: synchronization/slips, codec per line, delay/echo, memory speed, software complexity. For 3 first-stage 4x5 arrays, 5 middle 3x3 arrays, 3 last 5x4 arrays, total crosspoints = 60+45+60 = 165. ## Answer #### Advantages of PCM switching over analog switching - **No accumulation of noise:** digital signals are regenerated; quality does not depend on the number of switching stages. - **Time sharing of hardware:** one crosspoint or memory location serves many calls, so fewer components are needed. - **Integration with PCM transmission:** digital trunks are switched without A/D conversion at each exchange (integrated digital network). - **Low cost, size and power** using VLSI; easy maintenance and SPC features. - **Data/ISDN services** and fast common channel signaling. #### Issues (problems) of PCM switching - **Synchronisation:** all exchanges must run on a common clock, otherwise frame **slips** occur. - **Codec and hybrid per line:** each analogue subscriber line needs BORSCHT functions, which costs more per line. - **Delay and echo:** buffering adds delay (up to a frame per T stage), and 2-wire/4-wire hybrids cause echo that may need cancellers. - **Memory speed** limits the size of a single time switch. - **Software complexity** and dependence on processors. #### Crosspoints in the three-stage switch **Reading of the data:** 3 first-stage arrays, each with 4 input lines, so $N = 3 \times 4 = 12$, $n = 4$; 5 second-stage arrays, $k = 5$. - Stage 1: 3 arrays of $4 \times 5$ - Stage 2: 5 arrays of $3 \times 3$ - Stage 3: 3 arrays of $5 \times 4$ ```text 4 in [4x5] \ / [5x4] 4 out 4 in [4x5] --> 5 x [3x3] --> [5x4] 4 out 4 in [4x5] / middle \ [5x4] 4 out (12 inputs) (12 outputs) each outer array has 1 link to each of the 5 middle arrays ``` $$ \begin{aligned} N_1 &= 3 \times (4 \times 5) = 60 \\ N_2 &= 5 \times (3 \times 3) = 45 \\ N_3 &= 3 \times (5 \times 4) = 60 \\ N_x &= 60 + 45 + 60 = 165 \end{aligned} $$ Check: $N_x = 2Nk + k(N/n)^2 = 2(12)(5) + 5(3)^2 = 120 + 45 = 165$. **Comment:** a single-stage $12 \times 12$ switch needs only 144 crosspoints, so for such a small $N$ the three-stage form gives no saving; it pays off only for large $N$. Also $k = 5 < 2n - 1 = 7$, so the network is blocking. **Answer:** total crosspoints = **165**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens if PCM exchanges do not share a common clock? A. Echo increases B. Frame slips occur C. Crosstalk increases D. BORSCHT fails Answer: B. Lack of synchronisation causes slips. (Key point: Synchronisation issue) 2. With 3 first-stage arrays of 4 inputs, and 5 second-stage arrays, what is the size of each first-stage array? A. 4 x 4 B. 3 x 3 C. 4 x 5 D. 5 x 4 Answer: C. Each first-stage array has n = 4 inputs and one link to each of the k = 5 middle arrays: 4 x 5. (Key point: Stage sizes) 3. What size is each second-stage array here (N = 12, n = 4)? A. 3 x 3 B. 4 x 5 C. 5 x 5 D. 12 x 12 Answer: A. There are N/n = 3 first-stage arrays, so each middle array is 3 x 3. (Key point: Middle stage size) 4. What is the total number of crosspoints? A. 144 B. 120 C. 165 D. 180 Answer: C. 60 + 45 + 60 = 165, equal to 2Nk + k(N/n)^2 = 120 + 45. (Key point: Total crosspoints) ## Related in Telecommunications - [Switching functions and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aswitching-functions-crosspoints-4x4) — Telecommunications, chapter 4, asked 2× - [Crosspoints in three-stage switch (4 inputs, 5 arrays)](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Acrosspoints-4-input-5-second) — Telecommunications, chapter 4, asked 2× - [Crosspoints and three-stage switching](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Acrosspoint-switching-array) — Telecommunications, chapter 4 - [Digital exchange and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adigital-exchange-crosspoints) — Telecommunications, chapter 4 - [TST vs STS and three-stage design](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atst-sts-design-crosspoints) — Telecommunications, chapter 4 - [Conventional exchange functions; PCM switching advantages](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aconventional-exchange-pcm-advantages) — Telecommunications, chapter 4