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Recent questions tagged canonical-normal-form

6 votes
4 answers
1
Consider the Boolean function $z(a,b,c)$. Which one of the following minterm lists represents the circuit given above? $z=\sum (0,1,3,7)$ $z=\sum (1,4,5,6,7)$ $z=\sum (2,4,5,6,7)$ $z=\sum (2,3,5)$
asked Feb 12, 2020 in Digital Logic Arjun 3.2k views
20 votes
11 answers
2
What is the minimum number of $2$-input NOR gates required to implement a $4$ -variable function expressed in sum-of-minterms form as $f=\Sigma(0,2,5,7, 8, 10, 13, 15)?$ Assume that all the inputs and their complements are available. Answer: _______
asked Feb 7, 2019 in Digital Logic Arjun 16.8k views
1 vote
2 answers
3
Should we use Don't care terms while calculating POS expression?
asked Mar 21, 2018 in Digital Logic Jason 1.1k views
2 votes
1 answer
4
17 votes
3 answers
5
12 votes
6 answers
6
The simplified SOP (Sum of Product) from the Boolean expression $(P + \bar{Q} + \bar{R}) . (P + {Q} + R) . (P + Q +\bar{R})$ is $(\bar{P}.Q+\bar{R})$ $(P+{Q}.\bar{R})$ $({P}.\bar{Q}+R)$ $(P.Q+R)$
asked Jul 4, 2016 in Digital Logic Arjun 6.5k views
7 votes
4 answers
7
The most simplified form of the Boolean function $x (A, B, C, D) = \sum (7, 8, 9, 10, 11, 12, 13, 14, 15)$ (expressed in sum of minterms) is? A + A'BCD AB + CD A + BCD ABC + D
asked Apr 27, 2016 in Digital Logic makhdoom ghaya 2.9k views
26 votes
2 answers
8
A Boolean expression is an expression made out of propositional letters (such as $p, q, r$) and operators $\wedge$, $\vee$ and $\neg$; e.g. $p\wedge \neg (q \vee \neg r)$. An expression is said to be in sum of product form ... operator. Every Boolean expression is equivalent to an expression without $\wedge$ operator. Every Boolean expression is equivalent to an expression without $\neg$ operator.
asked Dec 8, 2015 in Digital Logic makhdoom ghaya 1.3k views
2 votes
3 answers
9
Answer is 2 minimal and 2 canonical covers. Please give full explanation of how to solve.
asked Jul 22, 2015 in Databases Shefali 1k views
30 votes
6 answers
10
Given the function $F = P' +QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P'=!P$, consider the following statements. $(S1) F = \sum(4, 5, 6)$ $(S2) F = \sum(0, 1, 2, 3, 7)$ $(S3) F = \Pi (4, 5, 6)$ ... )-False (S1)-True, (S2)-False, (S3)-False, (S4)-True (S1)-False, (S2)-False, (S3)-True, (S4)-True (S1)-True, (S2)-True, (S3)-False, (S4)-False
asked Feb 15, 2015 in Digital Logic jothee 4.4k views
35 votes
8 answers
11
2 votes
4 answers
12
Kindly have a try .... IN canonical POS form following equation is written as F(A,B,C)=AB+BC+AC (A)πM(0,1,2,4) (B)πM(3,5,6,7) (C)πM(0,1,2,3) (D)πM(4,5,6,7)
asked Oct 11, 2014 in Digital Logic Gobind 1.3k views
19 votes
4 answers
13
The minterm expansion of $f(P,Q,R) = PQ +Q \bar{R}+P\bar{R}$ is $m_2+m_4+m_6+m_7$ $m_0+m_1+m_3+m_5$ $m_0+m_1+m_6+m_7$ $m_2+m_3+m_4+m_5$
asked Sep 29, 2014 in Digital Logic jothee 4.5k views
47 votes
3 answers
14
Consider the following logic circuit whose inputs are functions $f_1, f_2, f_3$ and output is $f$ Given that $f_1(x,y,z) = \Sigma (0,1,3,5)$ $f_2(x,y,z) = \Sigma (6,7),$ and $f(x,y,z) = \Sigma (1,4,5).$ $f_3$ is $\Sigma (1,4,5)$ $\Sigma (6,7)$ $\Sigma (0,1,3,5)$ None of the above
asked Sep 16, 2014 in Digital Logic Kathleen 7.7k views
30 votes
5 answers
15
Given $f_1$, $f_3$ and $f$ in canonical sum of products form (in decimal) for the circuit $f_1 = \Sigma m(4, 5, 6, 7, 8)$ $f_3 = \Sigma m(1, 6, 15)$ $f = \Sigma m(1, 6, 8, 15)$ then $f_2$ is $\Sigma m(4, 6)$ $\Sigma m(4, 8)$ $\Sigma m(6, 8)$ $\Sigma m(4, 6, 8)$
asked Sep 11, 2014 in Digital Logic Kathleen 5.6k views
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