Important Short Questions and Answers: Digital Logic Circuits: Number Systems and Digital Logic Families

NUMBER SYSTEM AND DIGITAL LOGIC FAMILIES

1. Convert the hexadecimal number E3FA to binary.

Solution:

E3FA16 – Hexadecimal E 3 F A

11102 00112 11112 10102

So the equivalent binary value is 11100011111110102

2. Perform the following conversion (1029)10 to gray

Solution:

1          0 2 9 ----- Decimal

0001 0000 0010 1001 ----- BCD

0001 0000 0011 1101 ----- Gray

Thus the Gray code of 102910 is 00010000001111012

3. Add 1A816 and 67B16

Solution:

1 A 816

6 7 B16

--------------------------

8 2 316

--------------------------

= 82316

4. Show the Karnaugh map with the encircled groups for the Boolean function,

F =C’ +A’D’ + A’B’D’ .

Solution:

1          1 1

1          1 1

1          1

1          1 1

5. Perform 2s complement subtraction of 010110-100101.

Solution:

1’s complement of minuend100101 = 011010

2’s complement of 011010 = 011011

Addition of 010110 + 011011 = 110001

There is no end carry.

Therefore, the answer is –(2’s complement of 110001)

Answer = - 001111

6. Apply Demorgan’s theorems to simplify (A+BC') '.

Solution:

(A +BC') ' = A'. (BC') ' = A' . (B'+ C)

7. Plot the expression on K-Map F (w, x, y) =Σ (0,1,3,5,6) + d(2,4).

Solution:

W XY

1 1 1 X

X 1 1

F = x' +x y' + w’

8. If A and B are Boolean variables and if A=1 and (A+B) ' = 0, find B.

Solution:

If B=0 (A+B)’=0

If B=1 (A+B)’=0

Hence the value of B= 0 (or) 1 i.e Don’t care

9. What is the feature of gray code? What are its applications

Solution:

The advantage of gray code also called reflected code over pure binary numbers is that a number in gray code changes by only one bit as it proceeds from one number to the next. A typical application of the reflected code occurs when the analog data are represented by a continuous change of a shaft position. The shaft is portioned into segments and each segment is assigned a number. If adjacent segment are made to correspond to adjacent reflected-code numbers, ambiguity is reduced when detection is sensed in the line that separates any two segments.

So in 3-bit code, error may occur due to one bit position, other two bit positions of

adjacent sectors are always same and hence there is no possibility of error. Thus in 3-bit code, probability of error is reduced to 66 % and in 4-bit code it is reduced upto 25%.

10.   Convert the gray code number 11011 to binary.

Solution: gray code

binary code =10010

11.   What is even parity?

Solution:

A parity bit is an extra bit included with a message to make the total number of 1’s either odd or even. If the total number of 1’s is even then it is called even parity.

12.   Find the 2’s complement and 1’s complement of 101101.

Solution:

1’s complement of 101101 = 010010 2’s complement of 101101 = 010010

1           

----------

010011

----------

13.   Simplify X1 +X1 X2.

Solution: x1 + x1x2

=          x1(1+x2)

=          x1

14.   Find the standard sum for the following function.

f = x1 x2 x3 + x1 x3 x4 + x1 x2 x4.

Solution:

f = x1 x2 x3 + x1 x3 x4 + x1 x2 x4

= x1 x2 x3(x4+x4’) + x1(x2+x2’) x3x4+x1x2(x3+x3’) x4 = x1x2x3x4 +x1x2x3x4’+x1x2’x3x4 + x1x2x3’x4

15.   Convert binary number 11011110 into its decimal equivalent.

Solution:

1          1 0 1 1 1 1 0

----------- 0 * 20 =0

----------- 1 * 21=2

----------- 1 * 22=4

----------- 1 * 23=8

----------- 1 * 24=16

----------- 0 * 25=0

----------- 1 * 26=64

----------- 1 * 27=128