---
title: "TST vs STS and three-stage design"
subject: "Telecommunications"
chapter: 4
key: "semester_8/telecommunications:tst-sts-design-crosspoints"
source: https://www.bctengineer.com/q/semester_8/telecommunications/tst-sts-design-crosspoints
site: BCT Engineering
---

# TST vs STS and three-stage design

Telecommunications · Chapter 4: Digital Switching

- Asked in: 2075 Chaitra, 3+6 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/tst-sts-design-crosspoints
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-4

## Question

Differentiate between TST and STS switch. Design a three stage switching system having 4 stage array of 5 input line and 6 second stage array. Also calculate the total number of cross points of the switching system.

## Short answer

TST uses time-space-time with TSIs outside and a space switch center; STS uses space-time-space with TSIs in the center. For 4 first-stage arrays of 5x6, 6 middle arrays of 4x4, and 4 third-stage arrays of 6x5, total crosspoints = 120 + 96 + 120 = 336.

## Answer

#### Difference between TST and STS switches

| Point | TST | STS |
|---|---|---|
| Structure | Time – Space – Time | Space – Time – Space |
| Outer stages | TSIs (memory) | Time-multiplexed space switches |
| Centre stage | Time-multiplexed space switch | Array of TSIs |
| Number of alternative paths | $l$ (internal time slots) | $k$ (centre TSIs) |
| Expansion | Time expansion (cheap: faster memory) | Space expansion (costly: more crosspoints and TSIs) |
| Blocking | Lower, $[1-(1-p')^2]^l$ | Higher, $[1-(1-p')^2]^k$ |
| Control | Path search over time slots | Path search over centre TSIs |
| Best for | Large exchanges, heavy traffic | Small exchanges, light traffic |

#### Three-stage design
**Reading of the data:** the first stage has 4 arrays with 5 input lines each, so $N = 4 \times 5 = 20$ lines, $n = 5$; the second stage has $k = 6$ arrays.

- Stage 1: 4 arrays of $5 \times 6$
- Stage 2: 6 arrays of $4 \times 4$ (each middle array has one link from each of the 4 first-stage arrays)
- Stage 3: 4 arrays of $6 \times 5$

```text
  5 in [5x6] \           / [6x5] 5 out
  5 in [5x6] --> 6 x [4x4] --> [6x5] 5 out
  5 in [5x6] -->  middle  --> [6x5] 5 out
  5 in [5x6] /           \ [6x5] 5 out
  (20 inputs)               (20 outputs)
```

#### Crosspoints
$$
\begin{aligned}
N_1 &= 4 \times (5 \times 6) = 120 \\
N_2 &= 6 \times (4 \times 4) = 96 \\
N_3 &= 4 \times (6 \times 5) = 120 \\
N_x &= 120 + 96 + 120 = 336
\end{aligned}
$$
Check: $N_x = 2Nk + k(N/n)^2 = 2(20)(6) + 6(4)^2 = 240 + 96 = 336$.

A single-stage $20 \times 20$ switch needs 400 crosspoints, so the saving is 64 crosspoints (16%). Since $k = 6 < 2n - 1 = 9$, this network is **blocking**; 9 middle arrays would be needed for strict non-blocking.

**Answer:** total crosspoints = **336**.

## Check yourself

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

1. Which statement correctly contrasts TST and STS?
   A. TST has TSIs at the outer stages and a space switch in the centre; STS the reverse
   B. TST has a space switch at the outer stages
   C. Both have identical stage order
   D. STS uses time expansion
   Answer: A. TST is Time-Space-Time and STS is Space-Time-Space. (Key point: TST vs STS structure)

2. For 4 first-stage arrays of 5 inputs and k = 6 middle arrays, the middle stage consists of
   A. 6 arrays of 4 x 4
   B. 4 arrays of 6 x 6
   C. 6 arrays of 5 x 5
   D. 5 arrays of 4 x 4
   Answer: A. Each middle array has one link from each of the 4 first-stage arrays, so it is 4 x 4, and there are 6 of them. (Key point: Stage sizes)

3. The total crosspoints for the design are
   A. 336
   B. 400
   C. 240
   D. 96
   Answer: A. 120 + 96 + 120 = 336, matching 2(20)(6) + 6(4)^2. (Key point: Total crosspoints)

4. Why is this network blocking?
   A. k = 6 is less than 2n - 1 = 9
   B. N is too small
   C. It has three stages
   D. The middle arrays are 4 x 4
   Answer: A. Strict non-blocking needs 2n - 1 = 9 middle arrays. (Key point: Blocking check)

## Related in Telecommunications

- [Multistage switching; TST vs STS](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Amultistage-tst-sts-compare) — Telecommunications, chapter 4
- [Design 200×200 three-stage switch](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Athree-stage-200x200-clos) — Telecommunications, chapter 4
- [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×
- [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×
- [PCM switching and three-stage crosspoints](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Apcm-switching-crosspoints-4x5) — Telecommunications, chapter 4
- [Space and time switches; STS vs TST](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aspace-time-switch-sts-tst) — Telecommunications, chapter 4

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