Posts tagged: digital design prep

Que 6: What is the range of n bit 2’s complement and 1’s complement representation?

Ans  6: 2’s complement represents an integer in the range [−2ⁿ⁻¹, 2ⁿ⁻¹ − 1].
        1’s complement represents an integer in the range [−2ⁿ⁻¹ + 1, 2ⁿ⁻¹ − 1].

Que 7: In which two input gate, If we toggle any one of the inputs toggles the output?

Ans  7: In gray code, only one bit toggles in the next number, for every next number output should toggle, there could be two possibilities.
Refer to Table 1 for Possible solutions at Outputs 1 and 2.
This can be concluded that Output 1 is XOR and Output 2 is XNOR.

Que 8: In 7 input XOR gate, If 4 inputs are 1 and 3 inputs are 0, what will be the output?

Ans  8: In the XOR gate, When Input has even number of 1’s gives output 0 and if Odd number of inputs are 1 then output is 1.
If 4 inputs are 1 and 3 inputs are 0, then output will be 0.

Que 9: Which gates are universal gates? Design XOR gates using NAND gate.

Ans  9: A universal gate is a gate that can implement any Boolean function without the need to use any other gate type. The NAND and NOR gates are universal gates.
Refer to the Figure, for XOR gate implementation from the NAND gate.

Que 10: Implement OR gate using NAND gate.

Ans  10: OR gate can be converted to NAND gate using De Morgan’s theorem.
                A + B  = ((A + B )’)’
                = (A’ . B’)’
                Refer to the Figure, For OR gate using NAND gate implementation.

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Que 1: Prove that A+A'B = B+B'A = A+B.

Ans 1: A+A'B = (AB' + AB) + A'B
    = AB' + (AB + A'B)
    = AB' + B
A+A'B = (AB' + AB) + A'B
    = AB' + AB + AB + A'B
    = (AB' + AB ) + (AB + A'B)
    = A + B

Que 2: Convert 25.625 fraction decimal numbers into binary number system.

Ans  2: The weights in binary numbers are 2⁵(32), 2⁴(16), 2³(8), 2²(4), 2¹(2), 2⁰(1), 2^-1(.5), 2^-2(.25), 2^-3(.125) and so on.
25.625 can be written as 16+8+1+0.5+0.125, If the weight value exists it will be mentioned as 1 else 0. So, 25.625 can be written as 11001.101.

Que 3: Convert 1011001 in 2’s complement and 1’s complement.

Ans 3: To convert binary numbers into 2’s complement, bits of binary numbers are inverted. So, 1011001 can be represented as 0100110 in 2’s complement.
To convert binary numbers into 1’s complement, Bits of the binary numbers are inverted except the last 1 in the binary number. So, 1011001 can be represented as 0100111 in 1’s complement.  

Que 4: What is gray code representation and how to convert from binary number to gray code?

Ans  4: Gray code, is an ordering of the binary numeral system such that two successive values differ in only one bit. eg. 00, 01, 11, 10.
This representation can be used to reduce the combinational logic.
Refer the Fig. 1 for Gray code to binary code converter.

Que 5: Convert 10/3 into Base 3 system.

Ans 5: In base 3 system, weights of the number are 3⁴(81), 3³(27), 3²(9), 3¹(3), 3⁰(1), 3^-1 (1/3), 3^-2(1/9) and so on.
10/3 can be written as 9/3 +1/3 = 3 + 1/3, So the Base 3 representation of 10/3 would be 10.1.

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