Switching functions and three-stage crosspoints

Basic switching functions

1. Attending: detect a call request (off-hook) and identify the calling line.
2. Information receiving: send dial tone and receive the dialled digits.
3. Information processing: analyse the digits to find the called party and route.
4. Busy testing: check that the called line or a trunk is free.
5. Interconnection: set up a path through the switching network.
6. Alerting: ring the called party and send ring-back tone to the caller.
7. Supervision: detect answer and clear-down; release the path.
8. Information sending: send signals to other exchanges for outgoing calls.
9. Charging: record call details for billing.

Crosspoints in the three-stage switch

Reading of the data: 3 first-stage arrays with 4 input lines each, so $N = 3 \times 4 = 12$, $n = 4$; 4 second-stage arrays, $k = 4$.

• Stage 1: 3 arrays of $4 \times 4$
• Stage 2: 4 arrays of $3 \times 3$
• Stage 3: 3 arrays of $4 \times 4$

 4 in [4x4] \             / [4x4] 4 out
 4 in [4x4] --> 4 x [3x3] --> [4x4] 4 out
 4 in [4x4] /    middle   \ [4x4] 4 out
 (12 inputs)                 (12 outputs)
 each outer array has one link to each
 of the 4 middle arrays

$$\begin{aligned} N_1 &= 3 \times (4 \times 4) = 48 \\ N_2 &= 4 \times (3 \times 3) = 36 \\ N_3 &= 3 \times (4 \times 4) = 48 \\ N_x &= 48 + 36 + 48 = 132 \end{aligned}$$

Check with the formula: $N_x = 2Nk + k(N/n)^2 = 2(12)(4) + 4(3)^2 = 96 + 36 = 132$.

A single-stage $12 \times 12$ switch needs $12^2 = 144$ crosspoints, so the saving is 12 crosspoints. Since $k = 4 < 2n - 1 = 7$, the network is blocking; 7 middle arrays would make it strictly non-blocking.

Answer: total crosspoints = 132.