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Boolean Equations Conversion Tutorial

Learn to convert English sentences to Boolean equations using minterms and maxterms. Understand minterm expansions, algebraic expressions, and truth tables. Illustrative examples provided.

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Boolean Equations Conversion Tutorial

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  1. Lecture 4 Minterm and Maxterm • Conversion of English sentences to Boolean equations. • Example • The alarm will ring iff the alarm switch is on and the door is not closed or it is after 6 PM and the window is not closed. • Boolean equation • Z = AB’ + CD’ • If Z = 1 , the alarm will ring. • Draw the network. Z will drive the alarm. Chap 4

  2. Combinational Network Specified in a Truth Table • Problem statement • Input 3 bits A, B, C = Binary number N. Output f. Output f = 1 if N >= 011 and f = 0 if N < 011. A B C f 000 001 010 011 100 101 110 111 0 0 0 1 1 1 1 1 Chap 4

  3. Derive Algebraic Expression from Truth Table • Using f = 1gives the SOP form. f = A’BC +AB’C’ + AB’C + ABC’ + ABC = A’BC + AB’ +AB = A’BC + A = A + BC. • Using f = 0 gives the POS form. Maxterms are multiplied together so that if any one of them is 0, f will be 0. See what happens if using OR. f = (A+B+C)(A+B+C’)(A+B’+C) = (A+B)(A+B’+C) = A+BC Chap 4

  4. Minterm and Maxterm • Minterm • A minterm of n variables = product of n literals in which each variable appears exactly once either in T or F form, but not in both. (Also known as a standard product term) • Each minterm has value 1 for exactly one combination of values of variables. E.g. ABC, 111 => m7 • A function can be written as a sum of minterms, which is referred to as a minterm expansion or a standard sum of products. Chap 4

  5. Minterm/Maxterm • Three variables Chap 4

  6. Minterm Notation • f = A’BC + AB’C’ + AB’C + ABC’ +ABC; The other way to represent f is: f (A,B,C) = m3 + m4 + m5 + m6 + m7 or f (A,B,C) = m(3,4,5,6,7) Another view, f (A,B,C) =0.m0+ 0.m1 +0.m2+ 1. m3 +1. m4 +1. m5 +1. m6 +1. m7 • Minterms present in f correspond with the 1’s of f in the truth table. Chap 4

  7. Maxterm • Maxterm • A maxterm of n variables = sum of n literals in which each variable appears exactly once in T or F from, but not in both. • Each maxterm has a value of 0 for exactly one combination of values of variables. E. g. A + B + C’, 001 => M1 (the value is 0). • Therefore Mi = m’i. • A function can be written as a product of maxterms, which is referred to as a maxterm expansion or a standard product of sums. Chap 4

  8. Maxterm Notation f = (A+B+C)(A+B+C’)(A+B’+C) f (A,B,C) = M0M1M2 or f (A,B,C) =  M (0,1,2) • Maxterms present in f correspond with the 0’s of f in the truth table. Chap 4

  9. M and m Relationship • If the minterm expansion for f (A,B,C) = m3 + m4 + m5 + m6 + m7 , what is the maxterm expansion for f(A,B,C)? • Choose those not present in the minterms. • So the Maxterm expansion for f(A,B,C) = M0M1M2. Chap 4

  10. Complement of minterm • Complement of a minterm is the corresponding maxterm. • Example if f = f (A,B,C) = m3 + m4 + m5 + m6 + m7 • f’ = (m3 + m4 + m5 + m6 + m7)’ = m’3 m’4 m’5 m’6 m’7 = M3 M4 M5M6M7 Chap 4

  11. Find the Minterm Expansion • f(a,b,c,d) = a’(b’+d) + acd’. = a’b’ +a’d + acd’ = a’b’(c+c’)(d+d’) + a’d(b+b’)(c+c’) + acd’(b+b’) = a’b’c’d’ + a’b’c’d + a’b’cd’ +a’b’cd + a’bc’d + a’bcd + abcd’ +ab’cd’ = Σm(0,1,2,3,5,7,10,14) What is the maxterm expansion for f? Chap 4

  12. Find the Maxterm Expansion f(a,b,c,d) = a’(b’+d) + acd’. = (a’+cd’)(a +b’+d);Use (x+y)(x’+z)=xz +x’y. = (a’+c)(a’+d’)(a+b’+d); Use (x+y)(x+z) = x+yz. = (a’+bb’+c+dd’)(a’+bb’+cc’+d’)(a+b’+cc’+d) = (a’+bb’+c+d)(a’+bb’+c+d’)(a’+bb’+c+d’)(a’+bb’+c’+d’)(a+b’+cc’+d) = (a’+b+c+d)(a’+b’+c+d)(a’+b+c+d’)(a’+b’+c+d’)(a’+b+c’+d’)(a’+b’+c’+d’)(a+b’+c+d)(a+b’+c’+d) =ΠM(4,6,8,9,11,12,13,15); primed = 1, unprimed = 0. Note that maxterm = 0. Chap 4

  13. General Expressions • n variables (i = 0 to 2n-1values) Minterm : F =Σaimi If ai = 1, then minterm mi exists. Maxterm : F =Π(ai+Mi); If ai = 1, then the maxterm does not exist. • Note that Σaimi = Π(ai+Mi) • F’ = [Π(ai+Mi)]’= Σai’Mi’ = Σai’mi = Π(ai’+Mi) Chap 4

  14. Incompletely Specified Functions • Don’t care terms. • A’B’C and ABC’ are “don’t care” term. We don’t care the value of these terms,whether it is 1 or 0. • Example F= A’B’C’+A’BC +ABC = A’B’C’ + BC (assign 0 to both X’s) F = A’B’C’+A’B’C+A’BC+ABC = A’B’+BC (assign 1 to first X and 0 to the second) F = A’B’+BC+AB (assign 1 to both X’s). Chap 4

  15. Minterm Expansion for Don’t Care • Example • Minterm • F = Σm(0,3,7) + Σd(1,6) • Maxterm • F = ΠM(2,4,5) . ΠD(1,6) Chap 4

  16. Don’t Care What if A B C f x x 0 1 f = c’ f = 000 010 100 110 Chap 4

  17. Examples • 1’s complement adder Chap 4

  18. Full Adder • One bit Chap 4

  19. 2’s Complement Adder • Form 2’s complement for minus operand for subtraction Chap 4

  20. Full Adder Using Gates module fulladd (Cin, x, y, s, Cout); input Cin, x, y; output s, Cout; xor (s, x, y, Cin); and (a, x, y); and (b, x, Cin); and (c, y, Cin); or (Cout, a, b, c); endmodule module fulladd (Cin, x, y, s, Cout); input Cin, x, y; output s, Cout; xor (s, x, y, Cin); and (a, x, y), (b, x, Cin), //omit and (c, y, Cin); or (Cout, a, b, c); endmodule

  21. Full Adder Using Functional Expression module fulladd (Cin, x, y, s, Cout); input Cin, x, y; output s, Cout; assign s = x ^ y ^ Cin, Cout = (x & y) | (x & Cin) | (y & Cin); endmodule

  22. 3-bit Ripple Adders module adder3 (c0, x2, x1, x0, y2, y1, y0, s2, s1, s0, carryout); input c0, x2, x1, x0, y2, y1, y0; output s2, s1, s0, carryout; fulladd b0 (c0, x0, y0, s0, c1); fulladd b1 (c1, x1, y1, s1, c2); fulladd b2 (c2, x2, y2, s2, carryout); endmodule modulefulladd (Cin, x, y, s, Cout); input Cin, x, y; output s, Cout; assign s = x ^ y ^ Cin, assign Cout = (x & y) | (x & Cin) | (y & Cin); endmodule Instantiate the fulladd module

  23. 3-bit Ripple Adders Using Vectored Signals module adder3 (c0, x2, x1, x0, y2, y1, y0, s2, s1, s0, carryout); input c0, x2, x1, x0, y2, y1, y0; output s2, s1, s0, carryout; fulladd b0 (c0, x0, y0, s0, c1); fulladd b1 (c1, x1, y1, s1, c2); fulladd b2 (c2, x2, y2, s2, carryout); endmodule modulefulladd (Cin, x, y, s, Cout); input Cin, x, y; output s, Cout; assign s = x ^ y ^ Cin, assign Cout = (x & y) | (x & Cin) | (y & Cin); endmodule Vectored signals used module adder3 (c0, X, Y, S, carryout); input c0; input [2:0] X, Y; output [2:0] S; output carryout; wire [2:1] C; fulladd b0 (c0, X[0], Y[0], S[0], C[1]); fulladd b1 (C[1], X[1], Y[1], S[1], C[2]); fulladd b2 (C[2], X[2], Y[2], S[2], carryout); endmodule

  24. Overflow and Carry-Out detection For unsigned number carry out from n-1 bit position: If both xn-1 and yn-1 are 1 or If either xn-1 or yn-1 is 1 and sn-1 is 0. Hence, carryout = xn-1 yn-1 + ~sn-1 xn-1 + ~sn-1 yn-1 n-bit signed number: -2n-1 to 2n-1 -1 Detect overflow for signed number: Overflow = Cn-1⊕ Cn Overflow = Xn-1 Yn-1 ~Sn-1 (110) + ~Xn-1 ~Yn-1 Sn-1 (001) (summation of two same signs produce different sign) where X and Y represent the 2’s complement numbers, S = X+Y. (sign bits 0, 0 ≠ 1 ) 0111 0111 1111

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