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In the below table why can't we form minterms where the function takes value 0 , say for the first combination why can't we write a'b'c' as one of the minterms ?

What's the issue in considering this minterm in the function ? | 66 views
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I think we make minterm only for f=1.
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Or, a'b'c' is minterm but it's not corresponds to f. It'll not include in SOP for f (here).
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Yes we can write it as $a'b'c'$. Lets say you are writing entire thing in canonical SOP then all the entries will be there in the expression. But all those entries for which output is $0$ will get removed and we will be remaining with only those entries which has output as $1$. Thats why we take minterm only for those entries which outputs $1$
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if you want to associate with zero's ====> those are maxterms but not minterms.

0 0 0 ===> (a+b+c)
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@Karan , I am not getting why will we remove a'b'c' when it can give Op 1 since we complemented all the variables a ,b and C .

To Consider as Minterm, we have to take input combinations for which the function (Digital Circuit) gives us '1' as output.

if you consider a'b'c' as minterm, For input combination (0,0,0) it will give output as '1' (1.1.1 = 1) for which our function produce '1' as shown in the above truth table.

Consider only terms which produce output 1  or 0 but not both.

if you consider terms with output '1' they are written as sum of the product terms i so it is called SOP expression.

otherwise they are called as POS terms.

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