ANSWERS A YOUR KNOWLEDGE IN DIGITAL ELECTRONICS

Created it, 06/02/20

Last update it, 06/02/20

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Answers 1         Answers 2        Answers 3 

Answers to the test N° 1

1st series :

2nd series :

1 - a)      X =        Þ         X = +

1 - b)      X =        Þ          X = .

2 - a)      L = (a + b) . (c + d)

2 - b)       L = . b . c

2 - c)         L = ( . a) + ( . c) + (a . c)

It is noted that this diagram is simplified :

1)  . a = 0

2) In the two branches of the lower part, the contact “c” is common. Only one contact “c” would thus be enough, provided that it is put in series with + a

That is to say : L = c . ( + a)

The diagram becomes :

3rd series :

1)  X = a . . + a . b . + a . . c + a . b . c

Let us put “a” in factor

X = a ( . + b . + . c + b . c)

Let us put “c” and “” in factor

X = [ ( + b) + c ( + b) ]

We know that + b = 1

from where   X = a [ . 1 + c . 1]

                  X = a [ + c ]

However    + c = 1

                X = a

2) 

Let us apply the theorem of Morgan :

Again let us apply the theorem of Morgan to each member :

On reconnaît l'équation du OU Exclusif à deux entrées : a b

Answers to the test N° 2

1°) A sequential system can have for the same combination of the variables of entry two states different from its or its exits.

2°) To pass from a level L to a level H, one spends a certain time called time boarding which is translated on the chronograms by an oblique face.

3°) A secondary variable or intern is an additional variable function of the exit which makes it possible for an identical combination of the variables of entry of origin to discriminate two states different from the exit in a sequential system.

4°) Primitive Matrix of the states.

5°) Polygon of fusion.

6°) Contracted Matrix.

7°) Table of Karnaugh.

8°) Research of the logigramme.

Note : It is deduced that S = a + .

We see that in this example, there is no secondary variable. The problem was not a sequential problem but a combinative problem.

What we could see from the beginning: indeed, the primitive matrix does not present two stable states giving two different levels of exits.

The logigramme will be :

The electric diagram would be :

You can check that this assembly functions well as envisaged by deferring you to paragraph 4 of the questionnaire.

In conclusion, you can see that a combinative system is only one particular case of a sequential system for which there is no secondary variable.

We can note that here the contracted matrix, although there is no secondary variable, comprises two lines and that the combinative solution is due only to the grouping in the table of Karnaugh.

Resolution by the combinative method of the same problem.

a  - Truth table.

b - Table of Karnaugh.

S = a + : result identical to that found previously.

9°) The chronogram is represented below.

10°) A bistable rocker constitutes an elementary memory of 1 bit. It has two stable states.

11° - a)  To obtain a rocker R.S.C., one starts from a rocker .

One adds two doors NAND on and , which gives us the entries R and S.

Lastly, one connects together an entry of each one of his doors as represented below to obtain the order C.

11° - b) To obtain a rocker D, one gathers R and S in only one entry such as R = by means of a reverser as represented below.

12°) supplemented Chronogram relating to the rocker D of the latch type.

Answers to the test N° 3

1°) Chronogram supplemented for the exit Q :

2°) Chronogram supplemented for the exits Q1 and S.

T' = 4T. This assembly is a divider of frequency by 4.

3°) the role of entry PRESET is to position the exit Q of the rocker to state 1 independently of the state of the other entries of this rocker. The role of entry CLEAR is to position the exit Q of the rocker to state 0 independently of the state of the other entries of this rocker.

4°) Chronogram relating to rocker JK.

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Daniel