Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 1)
1.

Convert the following SOP expression to an equivalent POS expression.

Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
37 comments Page 1 of 4.

Aditya Warnulkar said:   1 decade ago
Minterms are = 7, 4, 5, 6, 1.

Hence corresponding maxterms will be 0, 2, 3.

Therefore option B.

Rajasekhar said:   1 decade ago
How to find maxterms?

Ganesh Rathod said:   1 decade ago
SOP is complement of POS and vice-versa. So option B is the correct answer.

Maha said:   1 decade ago
Draw a three variable k-map and plot the given SOP forms. The remaining boxes which are not filled will give the answer for POS form. But you have to inverse each variable for example A means Abar.

Jahoor alam said:   1 decade ago
Please draw the three variable k map & give explanation.

Mounika said:   1 decade ago
If sop is complement of pos then please explain that.

Venkatesh said:   1 decade ago
Max terms:0 2 3
a b c->000:0,
a bbar c->010:2,
a bbar cbar->011:3.

VINOD said:   1 decade ago
SHORT CUT :NO.OF COMBINATION POSSIBLE WITH 3 VARIABLE IS(01234567)
min terms of the abv eq is: 7,4,5,6,1
IN POS IS TAKING MAX TERMS :REMAINING NUMBERS IS =0,2,3
MIN TERM IS :1 IF U GET 0 IS COMPLIMENT USING METHOD IS :SOP
MAX TERM IS :0 IF U GET 1 IS COMPLEMENT USING METHOD IS :POS

Dheeraj said:   1 decade ago
Ya Vinod is correct.

Shipra said:   1 decade ago
Nice answer vinod.


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