Digital exchange and three-stage crosspoints

Working principle of a digital telephone exchange

A digital exchange switches speech as PCM samples (8-bit words, 8000 per second) using time-division techniques, under stored program control (SPC).

 Subscriber +-------+  +----------+  +-------+
 lines ---->| Line  |->| Digital  |->| Trunk |--> other
 (analog)   | units |<-| switching|<-| units |    exch.
            |BORSCHT|  | (TST)    |  |(E1)   |
            +-------+  +----------+  +-------+
                ^           ^            ^
            +-------------------------------+
            | Control processor (SPC)       |
            | + signaling                   |
            +-------------------------------+

1. Line unit: each analogue line has a line card doing the BORSCHT functions: Battery feed, Over-voltage protection, Ringing, Supervision, Coding (codec: A/D and D/A), Hybrid (2-wire to 4-wire), Testing.
2. Concentration and multiplexing: coded samples from many lines are multiplexed onto 32-channel (2.048 Mbps) PCM highways.
3. Digital switching network: time switches (TSIs) and space switches (usually in TST form) move each sample from its input slot and highway to the required output slot and highway.
4. Control: the processor detects off-hook, collects digits, finds a free path, writes the control memories of the T and S stages, sends ringing and tones, supervises and records charging data.
5. Trunk and signaling units: connect to other exchanges over E1 links; signaling is by CAS or CCS (SS7).

Crosspoints in a three-stage switch

Take $N$ inputs and $N$ outputs, divided into groups of $n$ lines, with $k$ middle-stage arrays.

• First stage: $N/n$ arrays, each $n \times k$
• Second stage: $k$ arrays, each $(N/n) \times (N/n)$
• Third stage: $N/n$ arrays, each $k \times n$

$$\begin{aligned} N_x &= \frac{N}{n}(nk) + k\left(\frac{N}{n}\right)^2 + \frac{N}{n}(kn) \\ &= 2Nk + k\left(\frac{N}{n}\right)^2 \end{aligned}$$

For a strictly non-blocking (Clos) network, $k = 2n - 1$:

$$N_x = (2n-1)\left[2N + \left(\frac{N}{n}\right)^2\right]$$

Example: $N = 100$, $n = 10$, $k = 2n-1 = 19$:

$$\begin{aligned} N_x &= 2(100)(19) + 19\left(\frac{100}{10}\right)^2 \\ &= 3800 + 1900 = 5700 \end{aligned}$$

A single-stage $100 \times 100$ switch needs $N^2 = 10\,000$ crosspoints, so the three-stage design saves 43% and is still non-blocking.

Answer: $N_x = 2Nk + k(N/n)^2$; for $N = 100$, $n = 10$, $k = 19$, $N_x = 5700$ crosspoints.