I came across this problem and it has a solution and everything, but I don't understand it. Here's the illustration:

Crossroad illustration

So, the picture shows a crossroad which has sensors along the higher-priority DC road and the lower-priority AB road. They detect if there's a car on any of them. If there is a car in any of those roads, the sensor output is 1, otherwise it's 0. The crossroad is operated by a traffic light which functions like this:

1) W/E state of the traffic light is green if:

  • both C and D roads are busy
  • C or D is busy and neither A and B is busy
  • no cars on the crossroad

2) N/S state of the traffic light is green if:

  • both roads A and B are busy and neither D or C is busy
  • A or B is busy and neither D or C is busy Implement the circuit for controlling the traffic light using only MUX 4/1 multiplexers if the inputs of the circuit are the outputs of the sensor and outputs of the circuit are states of the traffic light (1-green state, 0-red state).



The part I don't understand are the information inputs of the multiplexers. What do they represent? Please explain in detail.


1 Answer 1


The multiplexers are being used as logic gates. In the case where just 1 or 0 is attached to the input, the multiplexers are essentially truth tables.

S1 and S0 are 2 bit binary selection inputs. For the case of the one on the top left,

C  |  D  |  Dec  |  Selected | Output
0  |  0  |   0   |    I0     |   0
0  |  1  |   1   |    I1     |   1
1  |  0  |   2   |    I2     |   1
1  |  1  |   3   |    I3     |   1

In that case, the mux is essentially acting like an OR gate.

The one below that will only pass a 1 if both S1 and S0 = 1, which makes it an AND gate.

The third multiplexer in that circuit is more complicated because its inputs can vary. You can still model it in terms of boolean logic, but it isn't as simple as the previous two.

Here's a logic-gate representation of the left hand multiplexer circuit (please feel free to point out any mistakes).


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ This is the answer I was typing up. It's easiest to just replace the muxes with "black box truth tables" and look at how it works from there. \$\endgroup\$
    – Greg d'Eon
    Jan 28, 2015 at 14:28

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