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Kmap Simplifications: Y(A,B,C,D)=∑m(1,2,3,6,8,9,10,1213,14)
Y(A,B,C,D)=AC’ + A’B’D+CD’
14 September 2020
Overlapping Groups• You are allowed to use the same 1 more than once. The 1 representing the
fundamental product ABC D is part of the pair and part of the octet. The simplified equation for the overlapping groups is Y=A + BCD It is valid to encircle the ls as shown in Fig.
• use the 1 s more than once to get the largest groups you can.
14 September 2020
Rolling of Maps Visualize picking up the Kamaugh map and
rolling it so that the left side touches the right side.
If you are visualizing correctly, you will realize
the two pairs actually form a quad.
14 September 2020
14 September 2020
Simplify using K Map method
m( 0,1,4,5,6,8,9,12,13,14) m(0,1,2,4,5,6,8,9,10,12,13)Y(A,B,C,D)=Y=
Eliminating Unnescessary Groups
14 September 2020
Simplify Y (A,B,C,D)=
14 September 2020
m (7,9,10,11,12,13,14,15)
14 September 2020
14 September 2020
Simplify using K Map method: Y(A,B,C,D)= m( 9) + d(10,11,12,13,14,15)
Y(A,B,C,D)=AD
Y(A,B,C,D)=
14 September 2020
m( 0) + D( 10,11,12,13,14,15)
Give the simplest logic circuit for following logic equation where d represents don't-care condition for following locations.F(A, B, C, D) = m(7) + d(l0, 11, 12, 13, 14, 15)
14 September 2020
PRODUCT-Of-SUMS METHOD (POS)
A B C Maxterm
0 0 0 A+B+C M0
0 0 1 A+B+C’ M1
0 1 0 A+B’+C M2
0 1 1 A+B’+C’ M3
1 0 0 A’+B+C M4
1 0 1 A’+B+C’ M5
1 1 0 A’+B’+C M6
1 1 1 A’+B’+C’ M7
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Obtain POS equation : Y(A,B,C)=∏M(0,3,6)
• PRODUCT-Of-SUMS METHOD
14 September 2020
A B C Y Maxterm
0 0 0 0 M0 (A+B+C)
0 0 1 1 M1
0 1 0 1 M2
0 1 1 0 M3 (A+B’+C’)
1 0 0 1 M4
1 0 1 1 M5
1 1 0 0 M6 (A’+B’+C)
1 1 1 1 M7
Y(A,B,C,D)=(A+B+C) (A+B’+C’)(A’+B’+C)
Conversion between SOP and POS
Thus SOP and POS occupy complementary locations in a truth table and one representation can be obtained from the other by
(i) identifying complementary locations,
(ii) changing minterm to maxterm or reverse, and finally
(iii) changing summation by product or reverse.
Ex1. Y = F(A, B, C) = ∏M(O, 3, 6) = ∑m(l, 2, 4, 5, 7)
Ex 2. Y = F(A, B, C) = ∑m(3, 5, 6, 7) = ∏M(O, 1, 2, 4)
14 September 2020
Suppose a truth table has a low output for the first three input conditions: 000, 001, and 010. If all other outputs are high, what is the simplified product-of-sums equation?
C’ C
A’B’ 0 0
A’B 0 1
AB 1 1
AB’ 1 1
14 September 2020
A B C Y Maxterm
0 0 0 0 M0
0 0 1 0 M1
0 1 0 0 M2
0 1 1 1 M3
1 0 0 1 M4
1 0 1 1 M5
1 1 0 1 M6
1 1 1 1 M7
Y=∏M(0,1,2)
Y=(A+B)( A+C ) Simplified POS
Y=(A+B+C)(A+B+C’)(A+B’+C) POS equation
Obtain POS equation : (A,B,C,D)=∏M(4,5,6,7,8,9)
14 September 2020
Y=(A’+B+C)( A’+B)
C’D’ C’D CD CD’
A’B’ 1 1 1 1
AB’ 0 0 0 0
AB 1 1 1 1
AB’ 0 0 1 1
Give simplest POS form of Karnaugh map shown:
C’D’ C’D CD CD’
A’B’ 0 0 0 0
A’B 0 0 0 1
AB 1 1 1 1
AB’ 1 1 1 1
14 September 2020
Y=(A+C)(A+D’)(A+B)
Ex: Y(A,B,C,D)=∑m(3,6,7,14)+d(8,9,11,12,13,15)Obtain simplified POS equation
C’D’ C’D CD CD’
A’B’ 0 0 1 0
A’B 0 0 1 1
AB X X X 1
AB’ X X X 0
14 September 2020
Y= C(B+D)