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A switching function of four variable, f (w, x y, z) is to equal the product of two other function f_{1} and f_{2}, of the same variable f = f_{1}f_{2} . The function f and f_{1} are as follows :
f = ∑m(4,7,15)
f = ∑m(0,1,2, 3, 4,7, 8,9,10,11,15)
Que: The number of full specified function, that will satisfy the given condition, is
f = ∑m(4,7,15)
f_{1} = ∑m(0,1,2, 3, 4,7, 8,9,10,11,15)
f_{2} = ∑m(4,7,15) + ∑dc(5, 6, 12, 13, 14)
There are 5 don't care condition. So 2^{5} = 32 different functions f_{2}
A switching function of four variable, f (w, x y, z) is to equal the product of two other function f_{1} and f_{2}, of the same variable f = f_{1}f_{2} . The function f and f_{1} are as follows :
f = ∑m(4,7,15)
f_{1} = ∑m(0,1,2, 3, 4,7, 8,9,10,11,15)
Que: The simplest function for f_{2} is
A fourvariable switching function has minterms m_{6 }and m_{9}. If the literals in these minterms are complemented, the corresponding minterm numbers are
The minimum function that can detect a “divisible by 3’’ 8421 BCD code digit (representation D_{8} D_{4} D_{2} D_{1} ) is given by
0, 3, 6 and 9 are divisible by 3
f =
For a binary half subtractor having two input A and B, the correct set of logical expressions for the outputs D = (A  B) and X (borrow) are
What type of logic circuit is represented by the figure shown below?
After solving the circuit we get (A’B’)+AB as output, which is XNOR operation. Thus, it will produce 1 when inputs are even number of 1s or all 0s, and produce 0 when input is odd number of 1s.
The building block shown in fig. is a active high output decoder.
Que: The output X is
The building block shown in fig. is a active high output decoder.
Que: The output Y is
A logic circuit consist of two 2 x 4 decoder as shown in fig.
The output of decoder are as follow
The value of f ( x, y, z) is
Number of 2 × 1 Multiplexers are required to implement 64 × 1 Multiplexers
A 64 × 1 multiplexer has 64 inputs so if we use 2 × 1 multiplexers 32 are needed in the first stage for 64 inputs, the output of these 32 multiplexers are connected to inputs of 16 multiplexers in the second stage.
Similarly, in third stage, 8 (2 × 1) multiplexers are used, in fourth stage 4 are used and finally 2 (2 × 1) multiplexers in the fifth stage, 1 in the sixth stage.
Total 2 × 1 multiplexers needed are 32 + 16 + 8 + 4 + 2 + 1 = 63
How many 3lineto8line decoders are required for a 1of32 decoder?
Number of 2 × 1 Multiplexers are required to implement 64 × 1 Multiplexers
A 64 × 1 multiplexer has 64 inputs so if we use 2 × 1 multiplexers 32 are needed in the first stage for 64 inputs, the output of these 32 multiplexers are connected to inputs of 16 multiplexers in the second stage.
Similarly, in third stage, 8 (2 × 1) multiplexers are used, in fourth stage 4 are used and finally 2 (2 × 1) multiplexers in the fifth stage, 1 in the sixth stage.
Total 2 × 1 multiplexers needed are 32 + 16 + 8 + 4 + 2 + 1 = 63
The MUX shown in fig. P4.2.31 is 4 * 1 multiplexer. The output Z is
Correct Answer : c
Explanation : Z = (bar AB)C + (bar A)B + (bar B)A + AB
= (bar A)[(barB)C + B) + A[(bar B) + B]
= (bar A)[(B + C)] + A
= A + B + C
The output of the 4 x 1 multiplexer shown in fig. is
The MUX shown in fig. is a 4 x 1 multiplexer. The output Z is
The output from the upper first level multiplexer is f_{a} and from the lower first level multiplexer is f_{b}
For the logic circuit shown in fig.the output Y is
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