Digital Electronics: Assignment

Q.1. (a) Convert (1101101)₂ to decimal number.
(b) Convert (11101.011)₂ to decimal number.
(c) Convert decimal number 1₂ to binary number.
(d) Convert decimal number 0.3125 to binary number.
(e) Convert octal number (2374)₈ to decimal number.
(f) Convert octal number (0.325)₈ to decimal number.
(g) Convert decimal number 359 to octal number.
(h) Convert each of the following octal numbers to binary numbers.
(i) 13₈ (ii) 25₈ (iii) 47₈ (iv) 170₈ (v) 752₈ (vi) 5276₈ (vii) 37.12₈
(i) Represent each of the following binary numbers by its octal equivalent:
(a) 110101₂ (b) 101111001₂ (c) 1011100110₂ (d) 1001101.1011₂
(j) Convert binary numbers to hexadecimal number.
(a) 1100101001010111₂ (b) 111111000101101001₂ (c) 1110011000.111₂
(k) Determine the binary numbers for the following hexadecimal numbers:
(a) 10A4₁₆ (b) CF83₁₆ (c) 9742₁₆ (d) D2E.8₁₆
(l) Convert the following hexadecimal numbers to decimal.
(a) 1C₁₆ (b) A85₁₆
(m) Convert (a) E 5₁₆ and (b) B 2F8₁₆ to decimal.
(n) Convert 650₁₀ to hexadecimal by repeated division by 16₁₀.
(o) Find 1's compliment and 2's compliments of binary number (10101), (10111) and (111100).

(a)

1 x 64 +1 x 32 + 0 x 16 +1 x 8 +1 x 4 + 0 x 2 +1 x 1 = 64 + 32 + 0 + 8 + 4 + 0 +1 = 109₁₀
(b) (16 + 8 + 4 + 1) . ( 0.25 + 0.125) = 29.375₁₀
(c)

(d) Step 1 (0.3125 x 2 = 0.625 → carry = 0 ( MSB))
Step 2 (0.625 x 2 = 1.25 → carry = 1
Step 3 (0.25 x 2 = 0.50 → carry = 0
Step 4 (0.50 x 2 = 1.00 → carry = 1 ( LSB) → (0.0101)₂)
(e)

2 x 512 + 3 x 64 + 7 x 8 + 4 x 1 = 1276₁₀
(f)

3 x 0.125 + 2 x 0.015625 + 5 x 0.001953 = 0.416015₁₀
(g)

(h)
 

(i)
(j)
 

(k)

(l)


(m) (a) E5₁₆ = E x 16 + 5 x 1 = 14 x 16 +5 x 1 = 224 +5 = 22910
(b) B2F8₁₆ = B x 4096 + 2 x 256 + F x16 + 8 x 1
= 11 x 4096 + 2 x 256 + 15 x 16 + 8 x 1
= 45056 + 512 + 240 + 8 = 45816₁₀
(n)

Therefore, 650₁₀ = 28A₁₆.
(o)

Q.2. For the logic circuit shown in figure, the input conditions (A, B, C) are given. In each case find output X.

(a) 1,0,1
(b) 0,0,1
(c) 1,1,1
(d) 0,1,1

XOR is inequality comparator and XNOR is equality comparator. In AND gate output will be high when all the input is 1.
(a) (A, B,C) = (1, 0,1) ; U1 = 1, U2 = 0 ⇒ U3 = 0
(b) (A, B,C) = (0,0,1) ; U1 = 0 , U2 = 0 ⇒ U3 = 0
(c) (A, B, C) = (1,1,1) ; U1 = 0 , U2 = 1 ⇒ U3 = 0
(d) (A, B, C) = ( 0,1,1) ; U1 = 1 , U2 = 1 ⇒ U3 = 1

Q.3. In each case find output Boolean expression.

Q.4. Find the output of combinational circuit shown in figure below.

Q.5. Consider the digital circuit shown below in which the input C is always low (0).

Write the truth table for the circuit shown in figure.

The truth table for the circuit can be written as

Q.6. Write the truth table for the circuit shown in figure.

Q.7. Using basic gates  implement the truth table?

Q.8. To implement the function f = (x + z)  how many two input NOR gate would require. 

f = ( x + z )

So, 4 NOR gate is required.

Q.9. Find the minimized expression for Boolean function
f(a, b, c, d) = ∑m (0, 2, 3, 5, 7, 8, 9,10,11) + ∑dc (4,15)

Q.10. A function with don't cares is as follows:
g (X , Y , Z) =∑m (5, 6)+∑dc (1, 2, 4)
For above function consider following expression
(1)  
(2)
(3)
(4)
Which of the above expressions are solution for g.

So expressions (1), (2) and (4) are solutions for g.