Question

Design a combinational circuit which has six inputs y2,y1,y0,x2,x1,x0 and four

outputs Z3, Z2, Z1, Z0.

a) Z0 is high whenever vector Y is equal to vector X

b) Z1 is high whenever vector Y is greater than vector X

c) Z2 is high whenever vector Y is less than vector X

d) Z3 is high whenever parity of the all six individual inputs y2, y1, y0, x2,

x1, x0 is even.

Note that Y represents y2,y1,y0 and X represents x2,x1,x0. (msb to lsb

respectively)

Answer 1

Answer:

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Explanation:

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Answer 2

Six inputs : y₂ , y₁ , y₀ , x₂ , x₁ , x₀

Four outputs : z₃ , z₂ , z₁ , z₀

a) z₀ is high whenever vector Y is equal to vector X

Y is equal to X if each of the individual bits are equal.

i.e., EX-NOR of each pair of bits is 1

Therefore, z₀ = (y₂ EX-NOR x₂)(y₁ EX-NOR x₁)(y₀ EX-NOR x₀)

b) z₁ is high whenever vector Y is greater than vector X

y₂>x₂   = y₂x₂' = 1

y₂=x₂ then y₁>x1 = (y₂ EX-NOR x₂)(y₁x₁')

y₂=x₂ , y₁=x₁ then y₀>x₀ = (y₂ EX-NOR x₂)(y₁ EX-NOR x₁)(y₀x₀')

Therefore, z₁ = (y₂x₂') + (y₂ EX-NOR x₂)(y₁x₁') + (y₂ EX-NOR x₂)(y₁ EX-NOR x₁ )(y₀x₀')

c) z₂ is high whenever vector Y is less than vector X

y₂<x₂   = y₂'x₂ = 1

y₂=x₂ then y₁<x1 = (y₂ EX-NOR x₂)(y₁'x₁)

y₂=x₂ , y₁=x₁ then y₀<x₀ = (y₂ EX-NOR x₂)(y₁ EX-NOR x₁)(y₀'x₀)

Therefore, z₂ = (y₂'x₂) + (y₂ EX-NOR x₂)(y₁'x₁) + (y₂ EX-NOR x₂)(y₁ EX-NOR x₁ )(y₀'x₀)

d) z₃ is high whenever parity of the all six individual inputs is even.

The parity is even if it has even no. of 1's. But as these all are single bits, thus if it is 0 then even parity as no. 1'z and if the bit is 1 , then even parity=0.

Thus if all the individual bits are zero then z₃ is high

Therefore, z₃ = (y₂y₁y₀x₂x₁x₀)'