# Convert Prod of Sums to Sum of Prod

1. Jan 28, 2015

### kukumaluboy

Question: Convert to Sum of Products

(p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)

= ((p+q’+r+s’)’ + (p’+q+r+s’)’ + (p’+q’+r+s’)’)’

(Demorgans each Sum-Term)

= ((p’.q.r’.s) + (p.q’.r’.s) + (p.q.r’.s))’ (Factor p.r’.s)

= ((p’.q.r’.s) + (p.r’.s).(q’ + q))’ (Inverse Law)

= ((p’.q.r’.s) + (p.r’.s))’ (Factor r’.s)

= ((r’.s).(p’.q + p))’ (Absorbtion Variant)

= ((r’.s). (q + p))’ (Distributive Law)

= (q.r’.s + p.r’.s)’ (Demorgans outer expression)

= ((q.r’.s)’. (p.r’.s)’) (Demorgans inner expression)

= .(q’+r+s’) . (p’+r+s’)

Still not in SOP

2. Jan 28, 2015

### electronicsguy

There are many ways to solve this.
The BASIC WAY is to simplify the expression. Lets name the Expression F
F = (p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)
F = (p + q')(p'+q)(p'+q') + (r+s') ---By factoring out (r+s')
F = (pp' + pq + p'q' + qq')(p'+q') + (r+s') ---Multiplying (p + q')(p'+q)
(pq+p'q')(p'+q') + (r+s') --- Some terms are canceled pp' = 0, qq' = 0
(pp'q + pqq' + p'q' + p'q') + (r+s') --Multiplying (pq+p'q')(p'+q')
(p'q') + r + s' --- Some terms are canceled pp'q = 0, pqq' = 0 and p'q' + p'q' = p'q'

F = (p'q') + r + s'

THE OTHER METHOD
I think the easiest and most efficient way to solve this is to use a K-map especially in really long expressions.
Computing for complement of F
F' = ((p+q’+r+s’) + (p’+q+r+s’)’ + (p’+q’+r+s’))'= (p'qr's)(pq'r's)(pqr's)
Then plot it on the K-MAP
(SEE BELOW)

Then group the remaining ones (1s)
Which will yield:

F = s'+(p'q') + r.

Which is the same as the above expression.

#### Attached Files:

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Last edited: Jan 28, 2015
3. Feb 5, 2015

Thanks!

4. Feb 7, 2015

### Korisnik

In order to get sum of minterms you should multiply what you have gotten so far, then simplify. Also, you could've done that at the very beginning. What electronicsguy suggested is preferred; when you complement the maxterms given in the original function, you get positions of zeros in your K-table, then minimize the ones as to get sum of minterms.