NEED help with Reducing switching expressions and Converting SEs into SoM and PoS

Thread Starter

pghaffari

Joined Oct 1, 2011
13
Please check if I did the following correctly. Thanks. Basically, #1 and #4 I need help with (those are primary) and #2,#3 i need you to check if i did it correctly.

Reduce the following SE's to number of literals specified.
1) abc'd + ab'c + bc'd + ab'c' + acd + a'bcd (4 Literals)

bc'd(a+1) + ab'(c+c') + cd(a+a'b)
bc'd + ab' + cd(a+b)
bc'd + ab' + acd + bcd
bd(c+c') + ab' + acd

bd + ab' + acd (From here, I have 7 literals, and I don't know what to do. I've tried the following but i still end up with 6 literals and can't get 4)

ab'(d+d') + bd(a+a') + acd
ab'd + ab'd' + abd + a'bd + acd
ad(b'+b) + ab'd' + a'bd + acd
ad + ab'd' + a'bd + acd
ad(1+c) + ab'd' + a'bd
ad + ab'd' + a'bd
a(d+d'b) + a'bd
a(d+b) + a'bd
ad + ab + a'bd
ad + b(a+a'd)
ad + ab + bd ... Stuck here.. What can i do? I've played with it for like 2 hours straight and can't figure it out. Any help would be greatly appreciated.

Problem 2 (I finished this, please check if its correct)
2) acb + ac'd + bc'd' + a'b'c' + ab'c'd' + bc'd (3 Literals)

bc'(d'+d) + b'c'(ad' + a') + abc + ac'd
bc' + b'c'(a'+d') + abc + ac'd
bc' + a'b'c' + b'c'd' + abc + ac'd
b(c' + ca) + a'b'c' + b'c'd' + ac'd
b(c' + a) + a'b'c' + b'c'd' + ac'd
bc' + ab + a'b'c' + b'c'd' + ac'd
c'(b+b'a') + ab + b'c'd' + ac'd
c'(b+a') + ab + b'c'd' + ac'd
bc' + a'c' + ab + b'c'd' + ac'd
c'(b + b'd') + a'c' + ab + ac'd
c'(b + d') + a'c' + ab + ac'd
bc' + c'd' + a'c' + ab + ac'd
bc' + c'd' + c'(a'+ad) + ab
bc' + c'd' + c'(a'+d) + ab
a'c' + c'd + c'd' + bc' + ab
a'c' + c'(d+d') + bc' + ab
a'c' + c' + bc' + ab
a'c' + c'(b+1) + ab
a'c' + c' + ab
c'(a'+1) + ab
ab + c'

Problem 3: Convert following SEs into sum of minterms
{[(a+b+a'c')c+d]' + ab'}
too much to type, but basically i got m(0,1,2,3,4,5,8,9,11,12) let me know if you agree with this if it disagrees ill type out what i have.

Problem 4: Convert following SEs into product of maxterms without obtaining sum of minterms first

xyz + yw + x'z' + xy'

So basically to do this we just keep expanding until we get a product of maxterms? like this..:

x(yz + y') + yw + x'z'
x(y' + z) + yw + x'z'
xy' + xz + yw + x'z'
(xy' + x)(xy' + z) + (yw+x')(yw+z')
x(xy'+z) + (yw+x')(yw+z')
(xy' +xz) + (yw+x')(yw+z')
((xy'+xz)) + (yw+x'))((xy'+xz) + (yw+z'))

I dont get what i do.. if i keep expanding it gets really complicated.. any suggestions??
 
Last edited:

Georacer

Joined Nov 25, 2009
5,181
I find creating the PoS form of a Boolean expression through Boolean operations extremely difficult, mainly because it involves factorizing the expression. Factorization is a very difficult mathematical problem as is. Unless you already have the solution, it's very hard to imagine the correct steps towards the solution.
I have roamed the internet looking for the solution to that problem several times, and each time I returned empty handed.

I suggest forming the truth table of the expression and extracting the Maxterms from it. It violates the exercise guidelines indirectly, but I can't think of any other way.
 
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