Problem
N° 2 |
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Created it, 06/09/09
Update it, 06/09/16
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6. - SOLVED PROBLEMS
6. 1. - PROBLEM N° 1
Three pieces of equipment: a, b, c,
function together. If, at least two of this equipment break down, one
wishes to feed automatically an element of help X.
The logical level “1” will indicate the correct operation,
The logical level “0” will indicate the faulty operation.
To give the truth table,
To trace the table of Karnaugh,
After groupings, to give the equation minimum,
To draw the logic diagram,
To carry out this diagram with the integrated circuits reversers,
AND, OR … of which you lay out (figure 76),
To indicate on the diagram the stitching of the integrated circuits,
To carry out the same assembly while using as functions NAND.
You lay out of the three circuits represented figure 76.
6. 2. - SOLUTION OF the PROBLEM N° 1
We see that X must be to 1 each time at least two elements on a, b, c are to 0 from where the truth table of figure 77.
Let us defer in the table of Karnaugh of figure 78 levels 1 of X for the corresponding states of a, b and c.
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According to the table of Karnaugh, we can write :
Red grouping = 
Green grouping = 
Blue grouping =
From where the expression is drawn :
X = +
+
One can now build the logic diagram or logigramme figure 79.
We see that in practice one “OR” at three entries can be replaced by two “OR” at two entries (figure 80).
The diagram becomes that of figure 81 then.
Let us replace each switching function by the equivalents starting from NAND.
According to the theorem of MORGAN, one can replace one OR by three NAND (figure 82).
Indeed, 
= .
From where S
= a + b = 
One can thus transform the diagram of figure 83 so that it becomes that of figure 84 and by removing the useless reversers that of figure 85.
6. 3. - PROBLEM N° 2
Let us consider the device of figure 86.
Operation :
One has a button Marche / Arrêt K,
The electromagnet “M” can be excited
only when the contact “d” = 1 (closed
valve),
“c” is then released and becomes “c”
= 0,
Sand flows in the device of weighing B (a
is then to 1),
The spring of weighing is crushed, “a”
master key with 0, but “b”
still remains to 0 (intermediate state
between “a” and “b”),
“b” = 1, this causes the de-energizing
of the electromagnet M,
The contact “c” passes by again to 1,
“c” excites the electromagnet N
then and thus “d” passes to 0,
Sand flows in the tip truck, “b” = 0,
but “a” remainder still with 0,
«a» = 1, the electromagnet N
is de-energized, the valve is closed again,
The contact d is then actuated, “d”
= 1 and the cycle starts again (M
is excited, c is released…).
1°) - To give the truth table of the system for the relays M and N,
2°) - To draw up the tables of Karnaugh relating to M and N,
3°) - Logigrammes.
6. 4. - SOLUTION OF the PROBLEM N° 2
6. 4. 1. - TRUTH TABLE FOR THE EXITS M AND N (FIGURE 87)
We will examine the combinations of variables in the order of the course of the process, we will not take account of the combinations not used.
The equations draw from the truth table are :
Equation of M
M = acd
+ ad
+ d
Equation of N
N = bcd
+ bc
+ c
The functions M and N are incompletely defined (certain combinations of the variables are not used), we consider them to 0 in the table of Karnaugh.
6. 4. 2. - TABLE OF KARNAUGH FOR THE EXIT M (FIGURE 88)
One can carry out two groupings :
Red grouping = d
Green grouping = ad
The simplified equation is thus M
= ad
+ d
who can be written : M = d
(a + )
6. 4. 3. - TABLE OF KARNAUGH FOR THE EXIT N (FIGURE 89)
One can carry out two groupings :
Blue grouping = c
Green grouping = bc
The simplified equation is thus N
= c
+ bc
who can be written : N = c
(b + )
6. 4. 4. - LOGIGRAMME FOR M (FIGURE 90).
6. 4. 5. - LOGIGRAMME FOR N (FIGURE 91).
The examination of the combinatory logic is now finished, we will see sequential logic thereafter, which will enable us to return ahead in the electronic circuits by making many practical experiments.
The last problem which was proposed to you is besides in extreme cases of sequential because we see for the first time an order of closing of quite precise contacts but this problem can still be solved by the methods which we know.
This 3rd theoretical lesson finishes now, we advise you to carry out the tests which follow (on a site envisaged to this end) in order to re-examine the concepts a little that we exposed up to now.
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