Digital circuits miscellaneous
- The minimum number of NAND gates required to implement A + AB + ABC is equal to
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ƒ = A + A B + A B C
= A + A B = A
Hence, number of NAND gates required is 0.Correct Option: A
ƒ = A + A B + A B C
= A + A B = A
Hence, number of NAND gates required is 0.
- Output of the circuit shown in the figure is equal to
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The gates are XNOR
F = (A ⊕ B) ⊕ (A ⊕ B) = (AB + AB) ⊕ (AB + AB)
= (AB + AB) ⊕ (A B + AB)
= (AB + AB) ⊕ (AB + AB)
= 0Correct Option: A
The gates are XNOR
F = (A ⊕ B) ⊕ (A ⊕ B) = (AB + AB) ⊕ (AB + AB)
= (AB + AB) ⊕ (A B + AB)
= (AB + AB) ⊕ (AB + AB)
= 0
- A carry look ahead adder is frequently used for addition because it
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Carry look ahead adder is after since the carry is generated in parallel at all stages of addition rather than sequentially as in ripple adders.
Correct Option: A
Carry look ahead adder is after since the carry is generated in parallel at all stages of addition rather than sequentially as in ripple adders.
- I n the given network of AND and OR gates, ƒ can be written as
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In terms of Boolean operations
Output of 1 is x0 x1
Output of 2 is (x0 x1 + x2)
Output of 3 is (x0 x1 + x2) x3 = x0 x1 x3 + x2 x3
Output of 4 is x0 x1 x3 + x2 x3 + x4
Output of 5 would be x0 x1 x3 x5 + x2 x3 x5 + x4 x5
Output of 6 would be x0 x1 x3 x5 + x2 x3 x5 + x4 x5 x6
Thus for n gates connected as shown, the output would be
x0 x1 x3 ...................... xn– 1
+ x2 x3 x5 ...................... xn– 1
+ x4 x5 x7 ...................... xn– 1
+ xh – 2 xn– 1
+ xn .Correct Option: D
In terms of Boolean operations
Output of 1 is x0 x1
Output of 2 is (x0 x1 + x2)
Output of 3 is (x0 x1 + x2) x3 = x0 x1 x3 + x2 x3
Output of 4 is x0 x1 x3 + x2 x3 + x4
Output of 5 would be x0 x1 x3 x5 + x2 x3 x5 + x4 x5
Output of 6 would be x0 x1 x3 x5 + x2 x3 x5 + x4 x5 x6
Thus for n gates connected as shown, the output would be
x0 x1 x3 ...................... xn– 1
+ x2 x3 x5 ...................... xn– 1
+ x4 x5 x7 ...................... xn– 1
+ xh – 2 xn– 1
+ xn .
- Output Y of the circuit shown in the figure is
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Y = (A + B)C. DE.
= (A + B) C + DECorrect Option: A
Y = (A + B)C. DE.
= (A + B) C + DE
