Chapter 3 Simplification of Boolean functions


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1 3.1 Introduction Chapter 3 Simplification of Boolean functions In this chapter, we are going to discuss several methods for simplifying the Boolean function. What is the need for simplifying the Boolean functions? The reason is that by simplifying the Boolean function we can reduce the required number of gates and so we can reduce the size of the circuit and thereby reduces cost also. We will explain the need for simplification of Boolean function by taking an example. Consider another Boolean function F1 = x y z+x yz+xy. First obtain the required complement function. Note that for x and y only complemented variable needed, not for z. Figure 3.1
2 Figure 3.2 To implement this Boolean function, we need two NOT gates, two 3 inputs AND gates, one 2 inputs AND gate and one 3 inputs OR gate. Let us simplify this Boolean function using postulates and theorems, and then implement the Boolean function. F1 = x y z+x yz+xy =x z(y +y)+xy (by taking x z term common) = x z+xy (because y+y = y +y = 1) Now implement this simplified Boolean function. Figure 3.3
3 After simplification of Boolean function, now we need one NOT gate, one 2 inputs AND gate and one 2 inputs OR gate. We can see the difference between these two circuits and I hope this example will demonstrate you the need for simplifying the Boolean function. So in this chapter, we are going to learn the following methods to simplify the Boolean function. Simplification of Boolean function using postulates and theorems. Simplification of Boolean function using Karnaugh map. Simplification of Boolean function using tabulation method (Quine McClusky technique). 3.2 Simplification using theorems and postulates To simplify the Boolean function using postulates and theorems, there are no specific rules. The only way to simplify is that applying postulates basic theorems and many other familiar manipulation methods. Let us take some examples. Example1: x(x +y) Exampe2: x+x y Example3: (x+y)(x+y ) Example4: xy+x z+yz x(x +y) = xx +xy = 0+xy = xy. x+x y = (x+x )(x+y) (by using postulate 5) = 1.(x+y) = x+y (x+y)(x+y ) = x+yy (by using postulate 5) =x+0 =x xy+x z+yz = xy+x z+yz(x+x ) = xy+x z+xyz+x yz = xy(1+z)+x z(1+y) = xy+x z
4 Example5: (x+y)(x +z)(y+z) Example6: (A+B+C) xy+x z+yz = xy+x z (x+y)(x +z)(y+z) = (x+y)(x +z) (A+B+C) = (A+x) (Let B+C = x) = (A.x ) = (A.(B+C) ) = A.B C 3.3 Two and three variable map Simplifying Boolean expression using postulates and theorems will be difficult as there is no specific rule to apply. To simplify the Boolean expression we have to remember all the Boolean laws and theorems. The map method provides simple procedure to simplify the Boolean function. The map method is first proposed by Veitch and developed by Karnaugh. So this method is called as Karnaugh map method. In map method, the diagram can be seen as truth table which has been written in different format. A two variable map is shown in figure. Example: Table 3.1
5 Figure 3.4 The values inside the square are the output of the truth table. For two variable inputs we will have 2 2 = 4 outputs. So in the 2 variable Karnaugh map, we have four boxes. Edge of the Karnaugh map gives the inputs. A is down the left side. B is along the top. So for example top right hand corner of the map shows the output when the input A=0 and B=1. We can write canonical Boolean expression from the above truth table. As we have studied in the last chapter, each truth table can be mapped into two Boolean expressions namely, Sum of Product (SOP) expression using minterms and Product of Sum (POS) expression using Maxterms. We know that in the minterms 0 is complemented value whereas 1 is true value. In maxterms 1 is complemented value whereas 0 is true value. So to write the Boolean expression, consider the true output only. In SOP, 1 is the true output. So we get output 1 when A=0, B=1 OR A=1,B=1. Or we can say that output is true when the minterms are A B OR AB. The SOP expression is X = A B+AB. The same truth table can be mapped to POS also. In POS, 0 is the true output. So we get output 0 when A=0, B=0 AND A=1,B=0. Or we can say that output is true when the Maxterms are (A+B) AND (A +B). The POS expression is X = (A+B) (A +B). In other way, we can write SOP as X(A,B) = (1,3) and POS as X(A,B) = (0,2). For a three variable map, there are 2 3 =8 squares. Let us take three variable map with an example. Example: Consider this truth table. Now we have to represent this truth table in the Karnaugh map.
6 Table 3.2 We have three input variables A, B and C. So we need three variable map. Figure 3.5 Note that last two columns are interchanged from natural order of truth table. Because Karnaugh map uses minimum distance code i.e. gray code. So only one variable can be changed to obtain the next code. From the truth table we can write the canonical forms of Boolean expression as stated above. SOP expression is Y = A BC+AB C +ABC +ABC or Y(A,B,C) = (3, 4, 6, 7) and POS expression is Y = (A+B+C)(A+B+C )(A+B +C)(A +B+C ) or Y (A,B,C) = (0, 1, 2, 5).
7 Simplification rules for Karnaugh map Group 1 in case of SOP expression simplification or group 0 in case of POS expression simplification. Groups may be horizontal or vertical, but should not be diagonal. Groups may contain 2 n cells.i.e. 2 0 =1, 2 1 =2, 2 2 =4, 2 3 =8, 2 4 =16, 2 5 =32 etc. Each group should be as large as possible. All 1 should be covered in case of SOP expression simplification. All 0 should be covered in case of POS expression simplification. Groups may overlap. The leftmost cell in a row may be grouped with the rightmost cell and the top cell in a column may be grouped with the bottom cell. There should be as few groups as possible, as long as this does not contradict any of the previous rules. Example: Consider our previous example X(A,B) = (1,3). To simplify it using Karnaugh map, redraw the two variable map again for clarity. Figure 3.6 it. There are two adjacent 1 in the second column. So we can group it together and draw Figure 3.7
8 Considering the group, see the value of the input variable. The group contains the outputs when A=0 and A=1. So literal A will not be in the output.the group has both the output when B=1. So literal of the group is B. So simplified Boolean expression is X = B. Example: Let us take another example X = A B+AB +AB. Draw the truth table for this output. Table 3.3 Convert this truth table into K Map (Karnaugh Map). Figure 3.8 There are three 1 s in this map. We can make vertical two 1 s as one group as we have done in the previous example. Horizontal two 1 s can be formed as another group. Because as per rule groups may overlap when necessary.
9 Figure 3.9 Now find the literals for each group. First consider vertical group. Vertical group contains two 1 s. The corresponding inputs A=0,A=1. So A is not in the output of vertical group. But B=1 only for vertical group. So vertical group literal is B. Next consider horizontal group. In horizontal group A = 1 only. But B=0, B=1. So B will not be in the literal of horizontal group. So the Boolean expression is X = A+B. Example: Simplify the Boolean function Y(A, B, C) = (2, 3, 4, 5). Draw the truth table for this Boolean function. As this is Sum of Product (SOP) expression, the output is 1 when the input is 2, 3, 4, 5 i.e. 010, 011, 100, 101 in binary. Table 3.4
10 Now transform this truth table into Kmap. Figure 3.10 Group the 1 s. As there are two adjacent 1 s, we can make two groups. Figure 3.11 For the upper group, literal value A = 0, B=1 (both 3 rd and 4 th column) and C = 1(3 rd column) and C = 0 (4 th column). So for upper group the combined literal is A B (As A=0, complemented variable in SOP). For the lower group, literal value A = 1, B=0 (both 1 st and 2 nd column) and C = 0 (1 st column) and C = 1 (2 nd column). So for lower group the combined literal is AB (As B=0, complemented variable in SOP). So the simplified Boolean expression is Y = A B+AB. Example: Simplify the Boolean function Y(A, B, C) = (3, 4, 6, 7). Draw the truth table for this Boolean function. As this is Sum of Product (SOP) expression, the output is 1 when the input is 3, 4, 6, 7 i.e. 011, 100, 110, 111 in binary.
11 Table 3.5 Now transform this truth table into Kmap. Figure 3.12 Group the 1 s. As there are two adjacent 1 s in the vertical it can be one group. The 1 in the 1 st column and 4 th column can be considered as adjacent 1 s and can be grouped. Figure 3.13 For the vertical group, literal value A = 0 and A=1, so literal A will not be in the output for vertical group. But B = 1 C = 1. So for vertical group the combined literal is BC.
12 For the horizontal group, literal value A = 1. C=0(both 1 st column and 4 th column). But B = 0(1 st column) and B = 1 (4 th column). So for horizontal group the combined literal is AC (As C=0, complemented variable in SOP). So the simplified Boolean expression is Y = BC+AC. Example: Simplify the Boolean function F(x, y, z) = (0, 2, 4, 5, 6). Draw the truth table for this Boolean function. As this is Sum of Product (SOP) expression, the output is 1 when the input is 0, 2, 4, 5, 6 i.e. 000, 010, 100, 101, 110 in binary. Table 3.6 Now transform this truth table into Kmap. Figure 3.14 While grouping maximum possible 1 s should be considered. So in the above map, we can consider two 1 s in the 1 st column and two 1 s in the 4 th column as adjacent 1 s. So we can make four 1 s as a group. Remaining 1 can be grouped with adjacent one as overlapping is allowed when necessary.
13 Figure 3.15 For the four 1 s group, literal value x = 0 and x=1, so literal x will not be in the output. y = 0 (1 st column) and y = 1(4 th column). So y is also not in the output. z = 0 (both 1 st and 4 th column). So for four 1 s group the literal is z (because z=0 is complemented variable in SOP). For the horizontal group, literal value x = 1. y=0 (both 1 st column and 2nd column). But z = 0(1 st column) and z = 1 (2 nd column). So for horizontal group the combined literal is xy. (As y=0, complemented variable in SOP). So the simplified Boolean expression is F = z +xy. Example: Given the Boolean function F = A C+A B+AB C+BC. (a) Express it in sum of minterms (b) Simplify using K map. F = A C+A B+AB C+BC (given) = A C(B+B )+A B(C+C )+AB C+BC(A+A ) (missing literals are combined) = A BC + A B C + A BC + A BC + AB C + ABC + A BC = A BC + A B C + A BC + AB C + ABC (duplicate terms are cancelled) = m3 + m1 + m2 + m5 + m7 = (1, 2, 3, 5, 7). Draw the truth table for this Boolean function. As this is Sum of Product (SOP) expression, the output is 1 when the input is 1, 2, 3, 5, 7 i.e. 001, 010, 011, 101, 111 in binary.
14 Table 3.7 Now transform this truth table into Kmap. Figure 3.16 While grouping maximum possible 1 s should be considered. So in the above map, we four 1 s. So we can make four 1 s as a group. Remaining 1 can be grouped with adjacent one as overlapping is allowed when necessary. Figure 3.17
15 For the four 1 s group, literal value A = 0 and A=1, so literal A will not be in the output. B = 0 (2 nd column) and B = 1(3 rd column). So B is also not in the output. C = 1 (both 2 nd and 3 rd column). So for four 1 s group the literal is C. For the horizontal group, literal value A=0. B = 1(both 3 rd and 4 th column). But C = 1(3 rd column) and C = 0 (4 th column). So for horizontal group the combined literal is A B. So the simplified Boolean expression is F = C+A B. 3.4 Four variable map For four variable inputs we will have 2 4 = 16 outputs. So in the 4 variable Karnaugh map, we have 16 boxes. Edge of the Karnaugh map gives the inputs. Suppose A, B, C and D are considered as four inputs A and B is down the left side. C and D is along the top as show in figure. Figure 3.18 As we have seen in the three variable map, gray code is followed. Note that 3 rd and 4 th columns and 3 rd and 4 th rows are swapped. We have already mentioned that Kmap can be considered as truth table arranged in special order. The following map shows how to map the SOP expression into Kmap.
16 Figure 3.19 Example: Simplify the Boolean function F(w, x, y, z) = (0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14). As this is SOP expression, the output will be 1 for the inputs 0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14. So fill 1 in the places 0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14 of above graph and remaining places should be filled by 0. Figure 3.20 Now we have to group maximum possible 1 s. As we have eight 1 s in the 1 st and 2 nd column, they can be grouped. After that three 1 s will be left. The 1 in the position of m2, m6 can be combined with 1 in the position of m0, m4. There will be only one 1 left which can be combined with 1 in the position of m6, m4, m12.
17 Figure 3.21 So we have combined all 1 s in three groups. First consider eight 1 s group. In that group, w = 0 in first two rows and w = 1 in next two rows. x = 0 in 1 st and 4 th row whereas x = 1 in the 2 nd and 3 rd row and so literal w and x will be out. z = 0 in 1 st column and z=1 in 2 nd 2 nd column and so z is also out of this group. y = 0 in both 1 st and 2 nd column and only literal from this group is y. In the same way other two group literals can be found. So the simplified Boolean expression is F = y +w z +xz. Example: Simplify the Boolean function F = A B C +B CD +A BCD +AB C One way to simplify the Boolean function is converting this expression into canonical forms and applying map procedure. Now we are going to explore another tricky method to solve this example. To map this equation into Kmap, consider 1 st term i.e. A B C. The missing literal in this term is D. A B C is equivalent to 000 inputs respectively. The fourth input is missing. D may be either 0 or 1. So the input could be either 0000 or So place 1 in the m0 (0000) and m1 (0001) position. In the same way remaining terms should be considered.
18 Figure 3.22 After grouping the 1 s in the map, the map becomes as shown in figure. Figure 3.23 Note that corner 1 s could be considered as adjacent and can be grouped. After obtaining literals for each group, the simplified Boolean function is F = B C +A CD +B D.
19 Example: Simplify the SOP expression F(A, B, C, D) = Σ(0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15). Draw the Karnaugh map. Figure 3.24 Group the 1 s using rules for simplifying the Boolean function using Kmap. Figure 3.25 The Simplified Boolean function is F = BD+B D +AB +CD. Example: Simplify the Product of Sum (POS) expression using Kmap. F = (A+B+C+D )(A+B+C +D)(A+B +C+D )(A+B +C +D)(A +B +C +D)(A +B+C +D). We have considered only SOP simplification up to this problem because to avoid confusion over SOP and POS simplification and to become familiar with the Kmap procedure. Now we will take POS expression and simplify it. The given expression can be written as F = Π(1, 2, 5, 6, 14, 10).
20 In POS, 0 is the true variable and 1 is complemented variable. So the output is 0 for the inputs of 1, 2, 5, 6, 10, 14. Fill 0 s in these places and 1 s in the remaining places. Figure 3.26 Now group 0 s instead of grouping 1 s. Figure 3.27 Finding the literals for each group is in the same way as we have done for SOP simplification except here 0 is true variable and 1 is complemented variable. For instance, in four 0 s group C = 1, D = 0. So the literal for this group is (C +D) as C=1 is complemented variable. In two 0 s group, A = 0, C = 0 and D = 1. So the literal for this group is (A+C+D ). So the simplified Boolean function is F = (C +D)(A+C+D ).
21 3.5 Five variable map For five variables we will have 2 5 =32 squares. Suppose the inputs are considered as A, B, C, D, E, the 5 variable map can be shown as follow. Figure 3.28 Example: Consider the minimization of the Boolean function F(A, B, C, D, E) = Σ(0, 1, 2, 8, 9, 15, 17, 21, 24, 25, 27, 31). Draw the Kmap for the given SOP expression. Figure 3.29
22 Let us group the 1 s in the maps. Figure 3.30 The simplified Boolean function is F = A C D +BC D +ABDE+BCDE+AB D E+A B C E. 3.6 Don t care conditions In practical, there may be some input combinations for which the output will not be specified. For example while converting BCD to Excess3 code, the output for inputs 10, 11, 12, 13, 14, 15 are not specified. These unspecified outputs are called as don t care conditions which are denoted by X. When simplifying Boolean function using map the don t care conditions could be either consider as 1 if it is useful to simplify the function or 0 if it is not necessary. Example: Simplify the Boolean function F(w, x, y, z) = Σ(1, 3, 7, 11, 15) which has don t care conditions d(w, x, y, z) = Σ(0, 2, 5). In the map, we will put 1 for the inputs 1, 3, 7, 11, 15 as it is SOP expression. For remaining place we fill 0 s usually. But in this problem, don t care conditions are specified for the inputs 0, 2, 5. So for these inputs, the output unspecified. So we have to put X for these inputs.
23 Figure 3.31 Seeing the 1 st row of the map, if don t care conditions are considered as 1, we could combine four cells and thereby the output literals will be reduced. The 1 s in the 3 rd columns are grouped together to form another group. But we don t need the X in m5 position. So it can be considered as 0 and can be left without grouping it. Figure 3.32 The simplified Boolean function is F = yz+w x.
24 3.7 Tabulation method The map method can be used for small number of variables as we have discussed now. Four variables or five variable inputs can be simplified by using Kmap methods. Beyond that the grouping will be difficult and there may be a good chance of manual error. Also map method is trial and error method and difficult to implement using computer. The tabulation method overcomes these disadvantages. The tabulation method is suitable for large number of variables and suitable for computer manipulations. The tabulation method was formulated by Quine and McCluskey. So this method is also called as Quine McCluskey (QM) technique. Procedure to simplify the Boolean function: Find prime implicant of the function. (The term prime implicant will be explained in the example). Construct prime implicant table and find essential prime implicant. Include essential prime implicant in the minimal sum. If all minterms are covered, then minimal sum is the simplified Boolean function. If any minterm(s) are not covered, then delete all prime implicants from prime implicant table. Determine dominated rows and dominating columns and then delete all dominated rows and dominating columns from the prime implicant table. Determine secondary essential prime implicant. Continue the last three steps as long as all minterms included in the minimal sum. Example: Simplify the following sum of product expression F(a, b, c, d) = (0, 1, 2, 3, 4, 6, 7, 11, 12, 15). Determine prime implicants. In the I reduction column, write the given minterms in the problem. First write index0 in which include minterm with all the binary value is 0. Then write index1 in which include minterms with only one 1 in the binary value. Index2 includes the minterms with only two 1 s in binary values. Likewise continue up to index 4 for four variable minterms or continue up to index 5 for five variable minterms etc. In II reduction column, index0 terms are compared with index1 terms. If there is only one binary value differs that will be included in the II reduction column.
25 Table 3.8 Then construct prime implicant table to determine essential prime implicant. In the prime implicant table, check each minterm columnwise. If a minterm included in only one prime implicant then it is called as essential prime implicant. In this problem we have three essential prime implicants A, D, E.
26 Table 3.9 A includes the minterms 3, 7, 11, 15. D includes the minetrs 0, 1, 2, 3. E includes the minterms 4, 12. Check whether all the given minterms in the problem are covered by A, D, E certainly not. Because 6 is not covered by any of these prime implicants A, D, E. So minimal sum includes this essential prime implicants but we have to find secondary essential prime implicant. Delete all essential prime implicant from the table. i.e. row A, row D, row E. Also delete all corresponding minterms i.e. 3, 7, 11, 15 column can be deleted as A includes these minterms. 0, 1, 2, 3 columns can be deleted as D includes these minterms. 4, 12 columns can be deleted as E includes these minterms. After deleting redraw the prime implicant table. Table 3.10 If you look row wise row B is equivalent to row C. Or we can say that set B = {6} and C = {6}. So both sets are equal. We can consider any one row as dominated row and then can be deleted. So the minimal sum includes B or C. So the simplified Boolean function is F = A+D+E+B (or) A+D+E+C. F = cd+a b +bc d +a c (or) cd+a b +bc d +a d
27 Example: Simplify the SOP expression using QM technique. F(a, b, c, d) = Σ(0,4,8,1012,13,15)+d(1,2). Determine prime implicants. Note that while determining prime implicant, the don t care conditions should also be included. Table 3.11 Construct prime implicant table. While constructing prime implicant table don t care conditions need not be included.
28 Table 3.12 From the table, essential prime implicants are A, B, C. A includes the minterms 0, 4, 8, 12. B includes 0, 8, 10. C includes 13, 15. So all the minterms are included in the essential prime implicants. The simplified Boolean function is F = A+B+C. F = c d +b d +abd. Example: Simplify the following product of sum expression using tabulation method. F(a, b, c, d) = Π(1, 3, 5, 7, 13, 15). Determine prime implicants.
29 Table 3.13 Construct prime implicant table. Here we have only two prime implicants A and B. Table 3.14 From the prime implicant table, we have found that both A and B are essential prime implicants. So the minimal product includes A and B. A and B includes all the maxterms within it. So the simplified POS expression is F = (b +d )(a+d ).
30 3.8 Summary In this chapter, we have discussed the need for simplifying the Boolean function, the simplification of Boolean function reduces the number of components required to build the circuit and thereby cost and size of the circuit reduced. Then we have solved few examples to simplify the Boolean function using postulates and basic theorems. As postulates and theorems are not standard procedure, also the designer has to remember all theorems; map method developed by Karnaugh may be used to simplify the Boolean function. K Map for two variables, three variables, four variables and five variables were discussed with examples. Don t care conditions are unspecified output in some applications. We have discussed don t care conditions with examples. As map method is difficult to solve for large number of variables, tabulation method formulated by Quine and McCluskey can be used for large number of variables. Review Questions 1. Simplify the following Boolean expression to a minimum number of literals using basic theorems. (a) x y +xy+x y (b) (x+y)(x+y ) (c) x y+xy +xy+x y (d) x +xy+xz +xy z 2. Simplify the following Boolean expression to a minimum number of literals using basic theorems. (a) ABC+A B+ABC (b) x yz+xz (c) (BC +A D)(AB +CD ) 3. Find the complement of F = x+yz. 4. Find the complement of following expressions. (a) xy +x y (b) (AB +C)D +E 5. Given the following Boolean function: F = xy'z + x'y'z + w!xy + wx'y + wxy (a) Obtain the truth table of the function. (b) Draw the logic diagram using the original Boolean expression. (c) Simplify the function to a minimum number of literals using Boolean algebra. (d) Obtain the truth table of the function from the simplified expression and show that it is the same as the one in part (a). (e) Draw the logic diagram from the simplified expression and compare the total number of gates with the diagram of part (b).
31 6. Simplify the following Boolean functions using threevariable maps: (a) F(x, y, z) = Σ(0, 1,5,7) (b) F(x, y, z)=σ(1, 2, 3, 6, 7) (c) F(x, y, z)=σ(3,5,6,7) (d) F(A, B, C) =Σ(0,2, 3, 4, 6) 7. Simplify the following Boolean expressions using threevariable maps: (a) xy + x'y'z' + x'yz (b) x'y' + yz + x'yz' (c) A'B + BC' + B'C' 8. Simplify the following Boolean functions using fourvariable maps: (a) F(A, B, C, D)=Σ(4, 6, 7,15) (b) F(w, x, y, z)=σ (2,3, 12, 13, 14, 15) (c) F(A, B, C, D)=Σ(3, 7, II, 13, 14, 15) 9. Simplify the following Boolean expressions using fourvariable maps: (a) w'z + xz + x'y + wx'z (b) B'D + A'BC' + AB'C + ABC' (c) AB'C + B'C'D' + BCD + ACD' + A'B'C + A'BC D (d) wxy + yz + xy'z + x'y 10. Simplify the following Boolean functions in product of sums: (a) F(w,x,y, z)=σ(0, 2, 5, 6, 7, 8,10) (b) F(A, B, C, D=Π(I, 3, 5, 7, 13, 15) (c) F(x, y, z)=σ(2, 3, 6, 7) (d) F(A, B, C, D)=Π(o, 1,2,3,4,10,11) 11. Minimize the following Boolean function using Quine McCluskey method, f(x1,x2,x3,x4) = Σ(0, 5, 7, 8, 9, 10, 11, 14, 15). 12. Minimize the following switching function using tabulation method, f(x1,x2,x3,x4,x5) = Σ(0, 1, 2, 8, 9, 15, 17, 21, 24, 25, 27, 31). 13. Simplify the following POS expression using tabulation method. F(a, b, c, d) = Π(0, 8, 10, 12, 13, 15) + d(1, 2, 3). 14. Simplify the following SOP expression using tabulation method. F(A, B, C, D) = Σ(0, 1, 2, 9, 11) + d(8,10, 14, 15).
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