In the previous subject, we saw how circuits applying the Boole laws and of Morgan imagined to obtain circuits that responded to the premises that we gave. In this case, we are going to establish a system more methodical and simplified to obtain the circuits. For it, we used the diagrams of Karnaugh.

On this method we were based on which:

1 becomes a system of squares so that each of them corresponds to a binary number

2 the squares are neighboring€ itself varies a bit

3 to obtain point 2, the variables in each coordinates are due to put so that only one variable changes.

4 the neighbor grouping becomes in group of 2, 4, 8, 16, etc

5 In the squares where the function must activate, puts a 1, and if the function does not activate, a 0 is put.

We happen to put an example of 4 variables to clarify a little but.

We suppose that we have a digital system that activates for combinations 0001, 0011, 0101, 0111 and 0110.

In the column, we put the two first variables and the row superior the other two variables.

We see in the table that has put a 1 in the combinations where the function activates

As we have a group of 4 squares, they are grouped and the variables are taken that, when moving from a square to another one, do not vary. Therefore in that group we have those variables are A (that is to 0) and D (that are to 1). In that case we have **A' D**

For 2 group we have a grouping of 2 terms. When moving us from a square to another one, we see that the variables that they do not change are **A'BC**

Therefore, our summarized function is:

We go in this first part, to solve problem 8 of the previous subject, where we must fill I deposit of water that have two sensors To and B and another one of security that we have it in the well from where we took the water.

In this case we have three variables of entrance to, b and c. So that the system is reasonable, the c is put in well so that if it generates a signal 0, the water pump cannot be put to work. The sensor to, that he is the one of spill of the deposit, when marks 1, must stop the pump. When the water arrives at the level inferior from the deposit (b =0), the motor must put to work to fill up of the deposit. We happen to make the logical table.

a | b | c | function (water Pump) |

0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 |

0 | 1 | 0 | 0 |

0 | 1 | 1 | * X |

1 | 0 | 0 | 0 |

1 | 0 | 1 | 0 |

1 | 1 | 0 | 0 |

1 | 1 | 1 | 0 |

*X we put an X because it can be 0 or 1. If I deposit is lowering, it would have to be 0 so that the pump does not work until it arrives at the minimum. If it is raising, it would have to work until arriving to the maximum. Soon we will see a solution, but so far we will put it as a 1. Therefore,

we make KarnaughThe solution would be A'C.

Here it would be all resolute, but, this has a disadvantage, because it is not just like the deposit is filling to that it is draining. The pump cannot be always working and, therefore, we are going to establish a second way to solve this.

a ) If it is raising, the motor is working and would have to raise until it is arrived at A

b) If it is lowering, the motor is stopped and would have to follow thus until it is arrived at the sensorial B

Dice this exposition, we needed a witness the motor, that we will call M. m will give a 1 if the motor is working and 0 if it is stopped.

We initiate the sequence.

1 the deposit is empty. Therefore we have it pump must start and F1 = a'b'c (to emptiness, full empty b and c)

2 When the water arrives at the minimum from I deposit, must continue working, therefore F2 = a' b c m (as m is activated the motor continues working)

3 When the water arrives above, activates and the motor is stopped. One is fulfilled both in previous cases, that is for F1 and F2

4 When the low water, is put To a 0. As m this unemployed, f2=0 and as b this to 1, F1 this to 0. Therefore no function to make work the motor.

5 When it is arrived at b, activates F1

Therefore, we reached the conclusion that the canonical function is F1 + F2, that is:

We mount the circuit to verify it in logycly. To mount it in the PC and to verify the operation

Now we are going to replace the witness of the motor (m) by the same exit f, and we have left this circuit.