# Recent questions tagged min-sum-of-products-form 1
The sum of all the minterms of the boolean function of the n variable is 1. Prove the above statement with n = 3. give a Procedure for a general proof
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Express the complement of the following function in the sum of minterms. $(a) F(A,B,C,D) = \sum(0,2,6,11,13,14)$ $(b) F(x,y,z) = \prod(0,3,6,7)$
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What is the minimum SOP expression for the following f(a,b,c,d) = Σm(8, 10, 11, 12, 13, 15) + Σd(7, 9 , 14)? I’m getting only “a” as the answer, but the answer given is “a + bcd”. My doubt – the “Dont care term” generating the expression “bcd” is NOT necessary, and it is NOT considered as an Essential Prime Implicant. Please correct if I’m wrong
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Please verify my approach for calculation the minimal POS form for a function f. STEPS: Find out the f' in sum of minterms. Minimize f' and find out minimal SOP form. Using DeMorgan's law, find out f. The calcuated f will be minimal and in SOP form. ... this approach. However, I need a confirmation whether this method is actually correct and is bound to give me correct results all the time..!
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Consider the minterm list form of a Boolean function $F$ given below. $F(P, Q, R, S) = \Sigma m(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)$ Here, $m$ denotes a minterm and $d$ denotes a don't care term. The number of essential prime implicants of the function $F$ is ___
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A switching function of four variable, is equal to the product of two other functions f1and f2, of the same variable, i.e. f = f1 f2. The function f and f1 are as follows: The number of full specified function, that will satisfy the given condition, is
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From the following tables Two bits equality detector: x1 x2 y1 y2 f 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 ... the min-term expression of the f output of two bits equlity detector. Write down the max-term expression of the f output of the truth table below: x y f 0 0 0 0 1 1 1 0 0 1 1 1
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Number of boolean function with 3 boolean variable such that the function contain exactly 2 or 7 min term in their canonical SOP? Please explain the logic!
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Three switching functions $f_1, \: f_2 \:$ and $f_3$ are expressed below as sum of minterms. $f_1 (w, x, y, z) = \sum \: 0, 1, 2, 3, 5, 12$ $f_2 (w, x, y, z) = \sum \: 0, 1, 2, 10, 13, 14, 15$ $f_3 (w, x, y, z) = \sum \: 2, 4, 5, 8$ Express the function $f$ realised by the circuit shown in the below figure as the sum of minterms (in decimal notation).
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Find the minimum sum of products form of the logic function $f(A,B,C,D) = \Sigma m(0,2,8,10,15)+ \Sigma d(3,11,12,14)$ where $m$ and $d$ represent minterm and don't care term respectively.
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Let $f=(\bar{w} + y)(\bar{x} +y)(w+\bar{x}+z)(\bar{w}+z)(\bar{x}+z)$ Express $f$ as the minimal sum of products. Write only the answer. If the output line is stuck at $0$, for how many input combinations will the value of $f$ be correct?
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Consider the following Boolean function of four variables $f(A, B, C, D) = Σ(2, 3, 6, 7, 8, 9, 10, 11, 12, 13)$ The function is independent of one variable independent of two variables independent of three variable dependent on all the variables
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The simplified SOP (Sum of Product) from the Boolean expression $(P + \bar{Q} + \bar{R}) . (P + \bar{Q} + R) . (P + Q +\bar{R})$ is $(\bar{P}.Q+\bar{R})$ $(P+\bar{Q}.\bar{R})$ $(\bar{P}.Q+R)$ $(P.Q+R)$
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Consider the following minterm expression for $F$: $F(P,Q,R,S) = \sum 0,2,5,7,8,10,13,15$ The minterms $2$, $7$, $8$ and $13$ are 'do not care' terms. The minimal sum-of-products form for $F$ is $Q \bar S+ \bar QS$ $\bar Q \bar S+QS$ $\bar Q \bar R \bar S+ \bar QR \bar S+Q \bar R S+QRS$ $\bar P \bar Q \bar S+ \bar P QS+PQS+P \bar Q \bar S$
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Consider the $4\text{-to-1}$ multiplexer with two select lines $S_1$ and $S_0$ given below The minimal sum-of-products form of the Boolean expression for the output $F$ of the multiplexer is $\bar{P}Q + Q\bar{R} + P\bar{Q}R$ $\bar{P}Q + \bar{P}Q\bar{R} + PQ\bar{R} + P\bar{Q}R$ $\bar{P}QR + \bar{P}Q\bar{R} + Q\bar{R} + P\bar{Q}R$ $PQ\bar{R}$
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Consider a Boolean function $f(w,x,y,z)$. Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors $i_{1}=\left \langle w_{1}, x_{1}, y_{1},z_{1}\right \rangle$ ... $wx\overline{y} \overline{z}, xz, w\overline{x}yz$ $wx\overline{y}, wyz, wxz, \overline{w}xz, x\overline{y}z, xyz$
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Consider the following Boolean expression for F: $F(P,Q,R,S)= PQ + \bar{P}QR + \bar{P}Q\bar{R}S$ The minimal sum$-$of$-$products form of $F$ is $PQ+QR+QS$ $P+Q+R+S$ $\bar{P} + \bar{Q}+ \bar{R}+ \bar{S}$ $\bar{P}R + \bar{R} \bar{P}S+P$
The switching expression corresponding to $f(A,B,C,D)=\Sigma(1, 4, 5, 9, 11, 12)$ is: $BC’D’ + A’C’D + AB’D$ $ABC’ + ACD + B’C’D$ $ACD’ + A’BC’ + AC’D’$ $A’BD + ACD’ + BCD’$
Consider the following Boolean function of four variables: $f(w, x, y, z) = \Sigma(1, 3, 4, 6, 9, 11, 12, 14)$ The function is independent of one variables. independent of two variables. independent of three variables. dependent on all variables
Is the $3\text{-variable}$ function $f= \Sigma(0,1,2,4)$ its self-dual? Justify your answer. Give a minimal product-of-sum form of the $b$ output of the following $\text{excess-3}$ to $\text{BCD}$ converter.