Convert Prod of Sums to Sum of Prod

In summary: Once you get the sum of minterms, you could get the maxterms by complementing the minterms you've gotten.
  • #1
kukumaluboy
61
1
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
 
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  • #2
There are many ways to solve this.
The BASIC WAY is to simplify the expression. Let's 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.
lets name your function F
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.
 

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  • #3
Thanks!
 
  • #4
kukumaluboy said:
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
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.
 

1. How do you convert a product of sums to a sum of products?

To convert a product of sums to a sum of products, you need to use the distributive property. This means multiplying each term in the first set of parentheses by each term in the second set of parentheses, and then combining like terms.

2. Why is it useful to convert a product of sums to a sum of products?

Converting a product of sums to a sum of products can make it easier to simplify and solve algebraic expressions. It can also help to identify common factors and make it easier to factor polynomials.

3. Are there any specific steps to follow when converting a product of sums to a sum of products?

Yes, there are specific steps to follow when converting a product of sums to a sum of products. First, use the distributive property to multiply each term in the first set of parentheses by each term in the second set. Then, combine like terms to simplify the expression.

4. Can you provide an example of converting a product of sums to a sum of products?

Yes, for example, to convert (x + 2)(x + 3) to a sum of products, we would use the distributive property to get x^2 + 3x + 2x + 6. Then, we would combine like terms to get x^2 + 5x + 6.

5. Is it possible to convert a sum of products back to a product of sums?

Yes, it is possible to convert a sum of products back to a product of sums. This process is called factoring, and it involves finding common factors and rewriting the expression in a different form. However, it may not always be possible to factor completely, especially if the expression is not easily factorable or if there are no common factors.

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