Dr. Nermin Hamza

x · y = y · x x + y = y + x x · (y · z) = (x · y) · z x + (y + z) = (x + y) + z x · (y + z) = (x · y) + (x · z) x + (y · z) = (x + y) · (x + z) x · x = x x + x = x

x · (x + y) = x x + (x · y) = x x · x ‘ = 0 x + x’ = 1 (x’) ‘ = x (x · y)’ = x’ + y’ (x + y)’ = x’.y’

Exercise Proof that : X+ XY = X X+X’Y = X+Y X(X’+Y)= XY

Exercise Simplify : 1- X + X’.Y 2- X. Y +X’. Z + Y. Z 3- A + ((A. B’) ‘.C

Exercise Simplify : 1- X + X’.Y = X +Y 2- X. Y +X ‘ Z + Y. Z = X.Y + X’.Z 3- A + ((A. B’) ‘.C = B+C

Exercise Simplify : 1- A + ((B+C)’. A 2- ((A. B’)’ +B’). B

Exercise Simplify : 1- A + ((B+C)’. A = A 2- ((A. B’)’ +B’). B = A’.B

Exercise Get the function as sum of product, product of sum, proof that both cases are equal: XYF

Exercise F= ∑ m(0,1) = X’.Y’ + X.Y F= ∏ M(1,2) = (X’+Y).(X+Y’) PROOF: (X’+Y).(X+Y’) = (X’.X)+(X’.Y’)+(Y. X)+(Y.Y’) = 0+ X’Y’ + X.Y + 0= MIDTERMS

Exercise GET sum of product and product of sum : ABCF

Exercise Solution is : ∑ m=(3,4,5,6,7) = A’BC + AB’C’ + AB’C + ABC’ + ABC ∏ M (0,1,2) = (A+B+C).(A+B+C’). (A+ B’ + C) PROOF ??== A+BC

Exercise GET sum of product and product of sum : ABCF

Exercise ∑m(0,1,2)= Proof ? A’ (B’+C)

Other Digital Logic Operation 15 Basic Combinational Logic, NAND and NOR gates

16 Combinational logic How would your describe the output of this combinational logic circuit?

17 NAND Gate The NAND gate is the combination of an NOT gate with an AND gate. The Bubble in front of the gate is an inverter.

18 Combinational logic How would your describe the output of this combinational logic circuit?

19 NOR gate The NOR gate is the combination of the NOT gate with the OR gate. The Bubble in front of the gate is an inverter.

20 NAND and NOR gates The NAND and NOR gates are very popular as they can be connected in more ways that the simple AND and OR gates. NAND : F= (XY)’ NOR : F= (X+Y)’

21 Truth Table Complete the Truth Table for the NAND and NOR Gates InputOutput InputOutput NAND NOR

XOR F= X’Y +X Y’ = (X XOR Y)

XNOR F= XY +X’Y’ = (X XOR Y) ‘

Exercise

Map Method m0m1 m2m3

Map Method

27 Map Representation A two-variable function has four possible minterms. We can re- arrange these minterms into a Karnaugh map (K-map). Now we can easily see which minterms contain common literals. Minterms on the left and right sides contain y’ and y respectively. Minterms in the top and bottom rows contain x’ and x respectively.

Map Method F(a,b) = Σm(0,3) F(a,b) =A’B’ + AB

Let f= m1 +m2 + m3 … present in map

Exercise