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’

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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.

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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.

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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

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Simplify Y (A,B,C,D)=

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m (7,9,10,11,12,13,14,15)

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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)=

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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)

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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

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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)

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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

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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)

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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

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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

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Y= C(B+D)