Fault Tables and Boolean Algebra

Discussion in 'Homework Help' started by YRazor, May 9, 2011.

  1. YRazor

    Thread Starter New Member

    May 9, 2011
    hey forum, my first post and im sorry to be begging for advice already but i really need help understanding fault tables. I have an exam on wednesday and if i can understand them then im pretty sure i will pass. I understand how to generate a fault table and find the minimum test vectors and devise a fault tree and whatnot but I cannot get this far as I fall at the first hurdle. Finding the function of the circuit with a fault. Below is the circuit i am working with.


    Now the fault free circuit function is J = A.B + C + B'.D' i can understand that and how we got to that but what comes next is determining all the functions of the circuit with faults applied. This has been done for me below which is all well and good but I cant understand how my lecturer got the results shown i would just like some help on the working out. By the way according to my lecturer D/0 (D stuck at 0) and D/1 (D stuck at 1) are backwards on the worksheet.


    after getting annoyed with this question i tried one of the examples and now ive also been struggling to even work out the algebraic functions of some circuits.

    I wont paste a picture of the whole circuit/workings (unless required) but all i really want to know is how my lecturer got from this

    I = A' + B' + B' + C' + B'CD + C'CD + D

    to this

    I = A' + B' + C' + D

    thanks in advance everyone
  2. Georacer


    Nov 25, 2009
    The second image is quite simple:
    It shows how the output function behaves with various faults applied. For example, if A is stuck on 0, the output will become:

    Another example is to get the gate I stuck to 1. That will make the output equal to:

    About your second circuit, I can't know how the faults will affect it unless I have its schematic and the specific faults you want to examine.
  3. YRazor

    Thread Starter New Member

    May 9, 2011
    thanks for that mate it was very helpful, now another thing that stumped me which im sure is ridiculously simple. Boolean Identities.

    A + 0 = A
    A . 0 = 0
    A + 1 = 1
    A . 1 = A

    I can understand this but i think i am making a mistake somewhere when working with NOT inputs

    for example does A' + 0 still = A' ?

    this is the boolean equation i am working with H = A' + B' + C'D' + D and i am trying to find the function with the fault A stuck at 0 applied.

    this should make it 0 + B' + C'D' + D

    B' + 0 = B' therefore H = B' + C'D' + D.

    now i have the correct answers and know that this is wrong. the correct answer for A stuck at 0 is H = 1 but this is however the correct answer for A stuck at 1. Hopefully i am about to anwer my own question but is it because with A stuck at 0 the equation (H = A' + B' + C'D' + D) should actually be H = 1 + B + C'D' + D because A is stuck at 0 but the equation uses NOT A (A') rather than A. If that is the case then i understand

    H = 1 + B' + C'D' + D
    B' + 1 = 1
    H = 1 + C'D' + D
    C'D' + 1 = 1
    H = 1 + D
    D + 1 = 1
    H = 1

    please confirm or state if i have gone wrong anyone thanks.
  4. Georacer


    Nov 25, 2009
    You thought correctly. When A is 0 then A' is 1 and H=1. Well done.