Unformatted text preview: EE200 DIGITAL LOGIC CIRCUIT DESIGN
Class Notes CLASS 61 The material covered in this class will be as follows: ⇒ ⇒ ⇒ ⇒ Prime implicants Five variables Karnaugh map Product of sums (POS) simplification Don’t care conditions Prime Implicants We must insure that all minterms of the function are covered using the minimum number of groups. Also redundant terms should be avoided. Sometimes there could be two or more expressions that satisfy the simplification criteria. Prime implicants and essential prime implicants help in organizing the procedure for simplification. Prime Implicant A product term obtained by combining the maximum possible number of adjacent squares in the map. Essential Prime Implicant If a minterm is covered only by a prime implicant, then this prime implicant is called an essential prime implicant. EE200( EE200 EE200(class 61) Prof. M.M. Dawoud 43 Example Consider the function :
F ( A, B , C , D ) = ∑ (0, 2, 3,5,7,8,9,10,11,13,15)
C 1 BD 1 1 A 1 B’D’ D 1 AB’ D B’C 1 B 1 A 1 1 1 1 1 1 1 CD 1 AD 1 1 B 1 C 1 The two essential prime implicants are shown on the left and are: BD and B’D’ There are also four prime implicants shown on the right. Only two of them can be used to cover the remaining three minterms 3, 9, and 11. These prime implicants are: Also, B’C and CD AB’ and AD The simplified function can take one of four different forms: F = BD + B′D′ + CD + AD F = BD + B′D′ + CD + AB′
EE200( EE200 EE200(class 61) Prof. M.M. Dawoud 44 F = BD + B′D′ + B′C + AD F = BD + B′D′ + B′C + AB′
Five Variable Map The map of five variables is shown below. It consists of two four variable maps.
A D m0 m4 m12 B m8 m9 E m11 m10 m1 m5 m13 m3 m7 m15 m2 m6 C m14 B m24 m25 E m27 m28 m16 m20 m17 m21 m29 m19 m23 m31 D m18 m22 C m30 m26 Example Simplify the Boolean function
F ( A, B , C , D , E ) = ∑ (0, 2,4,6,9,13, 21, 23, 25, 29, 31)
A D 1 1 1 B 1 E B 1 E 1 C 1 1 1 1 C D EE200( EE200 EE200(class 61) Prof. M.M. Dawoud 45 F ( A, B , C , D, E ) = A′ B′E ′ + BD′E+ACE
Product of Sums (POS) Simplification We combine the zeros of the function to obtain a simplified expression in a POS form. This follows from the fact that the zeros of the function are the ones of the complement of the function. Example Simplify the Boolean function F ( A, B , C , D ) = Σ(0,1, 2,5,8,9,10) a. In SOP form b. In POS form.
C 1 BD’ 0 0 A 1 AB 1 D 0 1 1 1 0 0 0 0 1 0 B 0 CD Combining the ones of the function, we get: F = B′C ′ + B′D′ + A′ C ′D
The complement of the function can be expressed in SOP by combing the zeros of the function: EE200( EE200 EE200(class 61) Prof. M.M. Dawoud 46 F ′ = AB + CD + BD′
We complement the last espression to get the function F in POS form, F = ( AB + CD + BD′ )′ = ( A′ + B′ )(C ′ + D′ )( B′ + D )
Don’t care conditions When we design combinational logic circuits, we sometimes encounter situations where combinations of the input variables will never occur. In such cases, we can assume that these conditions can take on the value 1 or 0, whichever goins to give us a simpler expression. These conditions are indicated by letter X or D. Example Simplify the function F ( w , x , y , z ) = Σ(1, 3,7,11,15) , which has the don’t care conditions d ( w , x , y , z ) = Σ(0, 2,5)
y X W’x’ W’z w 1 z 1 X 1 1 x 1 X yz or F = w ′x′ + yz F = w ′z + yz
Prof. M.M. Dawoud EE200( EE200 EE200(class 61) 47 ...
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 Spring '09
 prof.ahmed
 Boolean Algebra, Logic gate, Karnaugh map, Prof. M.M. Dawoud

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