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Recent questions tagged canonicalnormalform
+1
vote
2
answers
1
Don't Care term in POS
Should we use Don't care terms while calculating POS expression?
asked
Mar 21
in
Digital Logic
by
Jason
Active
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1.5k
points)

249
views
digitallogic
canonicalnormalform
kmap
+1
vote
1
answer
2
Canonical Normal Form, Boolean Expression, Minimization
asked
Oct 17, 2017
in
Digital Logic
by
rishi71662data4
Active
(
2.4k
points)

188
views
digitallogic
canonicalnormalform
+13
votes
3
answers
3
GATE19905a
Find the minimum product of sums of the following expression $f=ABC + \bar{A}\bar{B}\bar{C}$
asked
Nov 24, 2016
in
Digital Logic
by
makhdoom ghaya
Boss
(
40.5k
points)

731
views
gate1990
descriptive
digitallogic
canonicalnormalform
+9
votes
6
answers
4
ISRO201616
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
by
Arjun
Veteran
(
369k
points)

3.7k
views
digitallogic
canonicalnormalform
isro2016
+5
votes
3
answers
5
ISRO201328
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
by
makhdoom ghaya
Boss
(
40.5k
points)

1.9k
views
isro2013
digitallogic
canonicalnormalform
+15
votes
2
answers
6
TIFR2015B9
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 ... . 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
by
makhdoom ghaya
Boss
(
40.5k
points)

513
views
tifr2015
canonicalnormalform
+2
votes
3
answers
7
F = {X > YZ, Y > XZ, Z > X} How many no. of minimal and canonical covers are possible?
asked
Jul 22, 2015
in
Databases
by
Shefali
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(
1.2k
points)

583
views
functionaldependencies
databases
canonicalnormalform
+22
votes
6
answers
8
GATE2015344
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. $(S1) F = \sum(4, 5, 6)$ $(S2) F = \sum(0, 1, 2, 3, 7)$ $(S3) F = \Pi (4, 5, 6)$ ... (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
by
jothee
Veteran
(
112k
points)

2k
views
gate20153
digitallogic
canonicalnormalform
normal
+23
votes
7
answers
9
GATE2015343
The total number of prime implicants of the function $f(w, x, y, z) = \sum (0, 2, 4, 5, 6, 10)$ is __________
asked
Feb 15, 2015
in
Digital Logic
by
jothee
Veteran
(
112k
points)

2.9k
views
gate20153
digitallogic
canonicalnormalform
normal
numericalanswers
+2
votes
4
answers
10
Kindly have a try .... IN canonical POS form following equation is written as (ABC)=AB+BC+AC (A)πM(0,1,2,4) (B)πM(3,5,6,7) (A)πM(0,1,2,3) (A)πM(4,5,6,7)
asked
Oct 11, 2014
in
Digital Logic
by
Gobind
(
37
points)

493
views
digitallogic
canonicalnormalform
+13
votes
3
answers
11
GATE20106
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
by
jothee
Veteran
(
112k
points)

1.8k
views
gate2010
digitallogic
canonicalnormalform
normal
+30
votes
2
answers
12
GATE200221
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
by
Kathleen
Veteran
(
59.7k
points)

2.9k
views
gate2002
digitallogic
normal
canonicalnormalform
circuitoutput
+19
votes
3
answers
13
GATE20088
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
by
Kathleen
Veteran
(
59.7k
points)

2.3k
views
gate2008
digitallogic
canonicalnormalform
easy
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