- 2073 Chaitra · 4 marks
4×4 omega switching network
Short answer
A 4×4 omega network has two stages of 2×2 switches, with perfect shuffle connections. To route input 3 (11) to output 1 (01), use destination bits MSB first: stage 1 switch B sets cross (lower input to upper output), stage 2 switch C sets straight (lower input to lower output) to reach Out1.
Design a 4×4 omega switching network and show the switch setting required to connect input 3 to output 1.
An omega network for N = 4 inputs/outputs has log₂4 = 2 stages, each with N/2 = 2 switches of size 2×2, so 4 switches in total. A perfect shuffle (rotate address bits left) connects the inputs to stage 1 and the stage 1 outputs to stage 2.
In0 -->+-----+--a0------------>+-----+--> Out0
In2 -->| A |--a1-------+ | C |--> Out1
+-----+ | +-->+-----+
In1 -->+-----+--b0-------|--+
In3 -->| B |--b1----+ | +-----+
+-----+ | +----->| D |--> Out2
+-------->| |--> Out3
+-----+
Connections: In0, In2 enter switch A; In1, In3 enter switch B. A's upper output (a0) goes to C's upper input, A's lower output (a1) to D's upper input, B's upper output (b0) to C's lower input, and B's lower output (b1) to D's lower input. Outputs 0, 1 come from C and 2, 3 from D.
Routing input 3 (11) to output 1 (01)
Use the destination address bits, MSB first: bit = 0 means the upper output of the switch, bit = 1 means the lower output.
| Stage | Switch | Input used | Destination bit | Output used | Setting |
|---|---|---|---|---|---|
| 1 | B | lower (In3) | 0 (MSB of 01) | upper (b0) | Lower input to upper output (cross) |
| 2 | C | lower (from b0) | 1 (LSB of 01) | lower (Out1) | Lower input to lower output (straight) |
Path: In3 → B (cross) → b0 → C (straight) → Out1. Switches A and D are not used for this connection and can be set arbitrarily.
✓ Answer checked by Claude Opus 5.5 (high effort)
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