problem with a circuit design

Discussion in 'Homework Help' started by mapisto, Dec 25, 2009.

  1. mapisto

    Thread Starter Member

    Dec 25, 2009
    36
    0
    Hi,
    i'm new here and so thankful i've found this forum.
    i've desperately tried to figure out how to make my homework, so i've searched the net and ended here eventualy.
    i've tried to figure most of the problems by reading older posts, but this one is different.

    The problem:

    There's a need to design an electric circuit which turns on a LED in exit F dependens on the income of A.B.C with the following terms:

    1) The LED will turn on when A appears and B appears after.

    2) The LED won't turn on when B appears and A appears after.

    3) in the 2 cases above, the circuit gets locked until C appears. (C=Stop=zeroing).

    Implement the circuit using logic gates.

    (I'm sorry if it sounds a little funny, i've just translated it from another language).


    Thanks a lot!!!!!
     
  2. mapisto

    Thread Starter Member

    Dec 25, 2009
    36
    0
    I've got another question about something else, i've had a problem which looks real easy and i've solved it, but it was just too easy so i'm afraid i've done something wrong.

    the problem is :

    You have to following 2 functions:

    F1(a,b,c,d)= ∏(0) (0,2,8,10), ∏(don't care) (5,13)

    F2(a,b,c,d)= ∏(0) (5,7,13,15), ∏(don't care) (8,10)

    a) You need to make an AND action between the 2 functions in order to recieve F3.

    b) You need to impliment F3 with minimal number of gates.


    ok, so i've put them both into karnough maps and i've got :
    F1=A+C
    F2=A'+C'

    if i'm "anding" them i get F3= (A+C)(A'+C') which leads to A(xor)C
    and now i've only got to impliment it.
    do you think i've done something wrong or it's only my teacher trying to give me a relief ? :)


    Thanks again :D
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    mapisto,

    Since you specify maxterms 0,2,8,10 , that is equivalent to minterms 1,3,5,7,9,11,13,15 . Karnaugh maps like minterms. Mapping the minterms of F1 gives B + D . F2 gives B' + D' .

    Ratch
     
  4. mapisto

    Thread Starter Member

    Dec 25, 2009
    36
    0
    i mean, i've used A and C, and you've used B and D.
    is there any difference? since we've done the same thing with the karnough maps.
     
  5. mapisto

    Thread Starter Member

    Dec 25, 2009
    36
    0
    does any1 has any idea how to accomplish the first task ? i mean, the big problem there is making the circuit work only when B is on AFTER a was on already and not the other way around. i've tried to write it with waves diagram and thought about truth table but nothing helps with this one... how is it possible doing it?

    Thanks!!!
     
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    mapisto,

    No.

    Yes, one way is right. The other way is wrong.

    Code ( (Unknown Language)):
    1.      A B C D   F = B+D        A B C D   G = B'+D'      A B C D   FG = BxD
    2. 0  | 0 0 0 0 | 0         0  | 0 0 0 0 | 1         0  | 0 0 0 0 | 0
    3. 1  | 0 0 0 1 | 1         1  | 0 0 0 1 | 1         1  | 0 0 0 1 | 1
    4. 2  | 0 0 1 0 | 0         2  | 0 0 1 0 | 1         2  | 0 0 1 0 | 0
    5. 3  | 0 0 1 1 | 1         3  | 0 0 1 1 | 1         3  | 0 0 1 1 | 1
    6. 4  | 0 1 0 0 | 1         4  | 0 1 0 0 | 1         4  | 0 1 0 0 | 1
    7. 5  | 0 1 0 1 | 1         5  | 0 1 0 1 | 0         5  | 0 1 0 1 | 0
    8. 6  | 0 1 1 0 | 1         6  | 0 1 1 0 | 1         6  | 0 1 1 0 | 1
    9. 7  | 0 1 1 1 | 1         7  | 0 1 1 1 | 0         7  | 0 1 1 1 | 0
    10. 8  | 1 0 0 0 | 0         8  | 1 0 0 0 | 1         8  | 1 0 0 0 | 0
    11. 9  | 1 0 0 1 | 1         9  | 1 0 0 1 | 1         9  | 1 0 0 1 | 1
    12. 10 | 1 0 1 0 | 0         10 | 1 0 1 0 | 1         10 | 1 0 1 0 | 0
    13. 11 | 1 0 1 1 | 1         11 | 1 0 1 1 | 1         11 | 1 0 1 1 | 1
    14. 12 | 1 1 0 0 | 1         12 | 1 1 0 0 | 1         12 | 1 1 0 0 | 1
    15. 13 | 1 1 0 1 | 1         13 | 1 1 0 1 | 0         13 | 1 1 0 1 | 0
    16. 14 | 1 1 1 0 | 1         14 | 1 1 1 0 | 1         14 | 1 1 1 0 | 1
    17. 15 | 1 1 1 1 | 1         15 | 1 1 1 1 | 0         15 | 1 1 1 1 | 0
    Yes, you are relying on a previous state for your output. So you have to employ sequential logic. You cannot do it with combinatorial logic.

    Ratch

    P.S. In English, sentences begin with a capital letter, as does the personal pronoun "I" .
     
  7. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    You need to make a truth table having A, B, and C. The C appears condition results in 0 and the the C does not appear is the go ahead for the rest of the logic. So in the case of C just have it disable the whole circuit if C appears.

    Now you have A and B to consider and this should just be a truth table with A, not A, and B and not B as the variables.
     
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