Solution

Two binary variables can have 2x2= 2^2= 4 input combinations.

Neutral functions need to have half of the Input combinations mapped to 1. In other words, a neutral function contains exactly half of the minterms(or maxterms).

Selection of two minterms out of four possible minterms can be done in 4C₂ = 6 ways.

So, the number of neutral functions=6.

Solution

XOR obeys commutative law.

A⊕B= B⊕A

XOR obeys associative law.

A⊕(B⊕C)=(A⊕B)⊕C.

Solution

Johnson’s counter with 2k states i.e, or Mod 2k Johnson’s counter can be constructed with k flip-flops.

2k= 16

Hence, k=8

Solution

XOR is neither distributive over AND nor distributive over OR.

A⊕(B+C) ≠ (A⊕B)+(A⊕C)

A⊕BC ≠(A⊕B).(A⊕C)

Solution

Gray code is a reflexive code. Gray codes can be written easily by using mirror image technique.

Gray codes of 0 and 15 are mirror images of each other except MSB.

Gray Code for decimal 0 is 0 000.

Gray code for decimal 15 is 1 000.

Solution

The function F assumes the value 1 whenever G does.

G= 1 whenever

w=x=1, y=0 and z= 0 or 1.

F= 1 whenever

w=x=1, y=0 or 1, z=0 or 1

or

w= 0 or 1, x =0 or 1, y=1 and z=1.

Solution

Let n is the length of the information.

Let k is the number of parity bits.

Then, k must satisfy the inequality 2^k - k -1 >= n.

If n=4, the minimum value of k which satisfies above inequality is 3.

Solution

Time T = 1/F

= 1/(0.1 GHz)

= (1/0.1) ns

= 10 ns

Solution

Solution

{OR,NOT} and {AND, NOT} are functionally complete sets which can implement any function.

{XNOR, 1} can’t derive NOT. So, {XNOR, 1, OR} can’t derive {NOT,OR}

Let f(A,B)) = A xnor B = AB + A’B’

f(1,B)= B and f(A,1)= A.

Solution

Sum = A ⊕ B.

Carry = AB

Total 5 NAND Gates are needed to implement Half Adder

Solution

P: Y = Z ⊗ X

RHS = Z ⊗ X

= (X⊗Y) ⊗ X

= (X⊗ X) ⊗ Y

= 1 ⊗ Y

= Y

= LHS

Q: X⊗ Y⊗ Z = 0

LHS = X⊗ Y⊗ Z

= (X⊗Y) ⊗ (X⊗ Y)

= 1

≠ RHS

R: X = Z ⊗ Y

RHS = Z ⊗ Y

= (X⊗ Y) ⊗ Y

= X ⊗ (Y⊗ Y)

= X ⊗ 1

= X

Solution

Solution

xyz’ + xyz + x’z + x’yz

= xyz’ + xyz + x’z + x’yz + xyz

=xy(z’+z) + x’z + (x’+x)yz

=xy + x’z + yz

By consensus theorem

= xy + x’z

Solution

A function can have more than one minimal expression.

Irredundant expression need not be minimal.

Solution

(x ⊕ y)’ = xy + x’y’

= 0 + x’y’

= (x+y)’

Complement on both sides.

(x ⊕ y)= x+y

Solution

Output of XNOR gate is 1 whenever its inputs are equal.

AND gate produces output 1 when both of its inputs are 1.

So, the output z= 1 whenever the two binary numbers x₁x₀ and y₁y₀ are equal. That is x₁=y₁ and x₀=y₀ which implies x₁x₀ = y₁y₀.

Solution

Solution

The counting loop has 8 states. So, the counter reaches the same state after 8 clock cycles.

1023 = 127x8 +7= 1016 +7.

The counter will be in state 15 after 1016 +7 clocks.

It reaches the state 14 after 7 more clocks.

Solution

Solution

f(a,b,c)= Σ(0,1,2,4).

f(a,b,c)= a’b’+ b’c’ + a’c’

Dual of f = (a’+b’)(b’+c’)(a’+c’)

= a’b’ + b’c’ + c’a’

= f.

Hence, f is self dual.

Solution

Solution

F(P,Q,R) = ( (P+Q)(Q+R) + (P+R)(Q+R) )’

= (PQ+Q+QR+PQ+PR+QR+R)’

= (Q(P+1+R+P+R)+(P+1)R )’

= (Q+R)’

Solution

S=0; A is a positive number.

Exponent e= E-16 = 19-16= 3

Mantissa M= 11100..0

Value of A= (-1)^S x 1.M x 2^e.

= +( 1.111 x 2^3 ).

= + (1111.0)

= 15

Solution

Solution

A = 1001 0101,

B= 1000 1010

B’= 0111 0101

B’ + 1= 0111 0110

A-B = A+ B’ + 1

= 1001 0101 + 0111 0110

= 0000 1011

= (11)₁₀

Solution

Solution

Essential Prime Implicants are A’C and AC’.

Minterm AB’C can be covered by either B’C or AB’.

So, A’C + AC’ + B’C and A’C + AC’ + AB’ are minimal expressions.

Solution

Given circuit is a synchronous circuit.

Initial state (P, Q, R) = (0, 0, 0).

In a D flip-flop, next state Q_n= D.

P_n = R= 0

Q_n = (P + R)’ = 1

R_n = QR’ = 0

(P_n,Q_n,R_n) = (0, 1, 0)

Solution

The bit pattern of floating point implicit normalized numbers in IEEE 754 single precision format is

So, contents of Z are

Solution

Q_0N= 1, Q_1N=,0 Q_2N=0.

Solution

Solution