Fundamentals of Digital Systems and Arithmetic: Two Marks Important Questions and Answers

Digital Principles and Computer Organization:

Chapter 1: Fundamentals of Digital Systems and Arithmetic

Two Marks Questions with Answers

 

1. Define an analog system with one example.

 Answer:

An analog system represents information using continuous signals that vary smoothly over time.

Example : A mercury thermometer, where the height of mercury continuously varies with temperature.

2. State two characteristics of analog systems.

 Answer:

1. Continuous : Signal varies smoothly over time and can have infinite values.

2. Noise Sensitive : Analog signals are easily affected by noise and distortion.

3. What is a digital system ?

 Answer:

 A digital system represents information using discrete signals, most commonly the binary system (0 and 1). It is the foundation of modern computers and electronics.

4. State two characteristics of digital systems.

 Answer:

1. Discrete : Signals exist only at specific levels (0 or 1).

2. Noise–Immune : Small fluctuations due to noise do not affect interpretation.

5. What is a binary signal ?

 Answer:

 A binary signal is the most common digital signal with two levels : logic 0 (OFF/low) and logic 1 (ON/high).

6. Write two differences between analog and digital systems.

 Answer:

• Analog : Continuous signals, low noise immunity.

• Digital : Discrete signals, high noise immunity.

7. Why are digital systems widely used? (Any two reasons)

 Answer:

1. They provide higher accuracy and are less affected by noise.

2. They are easy to design, flexible, and can store and process large data using memory.

8. What is the binary number system?

 Answer:

The binary system uses only two digits, 0 and 1, with base 2. Each digit is called a bit.

9. Convert (1011)₂ into decimal.

 Answer:

 (1011)₂ = (1×23+0×22+1×21+1×20) = 8+0+2+1 = (11)₁₀ (1011)₂

= (1×2³ +0×2² +1×2¹ +1×2⁰) = 8+0+2+1 = (11)₁₀ (1011)₂.

= (1×23 + 0×22 + 1×21 + 1×20) = 8+0+2+1 = (11)₁₀

10. What is the octal number system ?

 Answer:

The octal system uses eight digits (0–7) with base 8.

11. Convert (57)₈ into decimal.

 Answer:

(57)₈ = (5×81+7×80) = 40+7 = (47)₁₀ (57)₈ = (5×8¹+7×8⁰) = 40+7 =(47)₁₀ (57)₈

= (5×81+7×80) = 40+7 = (47)₁₀

12. What is the hexadecimal number system ?

 Answer:  

The hexadecimal system has base 16 and uses digits 0–9 and letters A–F (A=10, … F=15).

11. Convert (2F)₁₆ into decimal.

 Answer: 

(2F)₁₆ = (2×161+15×160) = 32+15=(47)₁₀ (2F)₁₆ = (2×16¹+15×16⁰)

= 32+15 = (47)₁₀ (2F)₁₆ = (2×161+15×160) = 32+15=(47)₁₀

Complements

14. What is 1's complement of a binary number?

 Answer:

The 1's complement is obtained by changing all 1's to 0's and all 0's to 1's in a binary number.

15. Find the 1's complement of (11010100) ?

 Answer:

(11010100)₂ → (00101011)₂ (11010100)₂ → (00101011)₂ (11010100)₂ → (00101011)₂

16. What is 2's complement of a binary number?

 Answer:  

The 2's complement is obtained by adding 1 to the 1's complement of a binary number.

17.  Find the 2's complement of (11000100)₂

 Answer:

1's complement of (11000100)₂ =(00111011)₂

Add 1 → (00111011)₂ + (1)₂ = (00111100)₂

18. Why is 2's complement preferred over 1's complement ?

 Answer:

Because the 2's complement system has only one zero representation and allows simpler arithmetic operations compared to 1's complement.

19. Represent +9 and –9 using 8–bit 2's complement representation.

 Answer:

+9 = (00001001)₂

– 9 = Take 2's complement of 00001001 = (11110111)₂

20. State the rules of binary addition.

 Answer:

• 0 + 0 = 0

• 0 + 1 = 1

• 1 + 0 = 1

• 1 + 1 = 10 (sum = 0, carry = 1)

21. State the rules of binary subtraction.

 Answer:

• 0 –  0  = 0

• 1 – 0  = 1

• 1 – 1 = 0

• 0 – 1 = 1 (with borrow = 1)

22. Perform subtraction using 1's complement (11010)₂ − (10000)₂.

 Answer:

23. Find the octal equivalent of hexadecimal number AB.CD.

 Answer:

(231.4065)₈

24. Perform the following code conversions :

(1010.10)₁₆ → (?)₂ → (?)₂ → (?)₁₀

 Answer:

 (1000000010000.0001)₂, (10020.02)₈, (4112.0625)₁₀

25. Subtract 11001 from 01101 using 2's complement.

 Answer:

1100.

26. Convert (2.B2)₁₆ to binary and octal numbers.

  Answer:

 (2.B2)₁₆ = (2.544)₈ = (10.1011001)₂

27. What is meant by weighted and non–weighted coding ?

 Answer:

Weighted codes : In weighted codes, each digit position of the number represents a specific weight. For example, in decimal code, if number is 567 then weight of 5 is 100, weight of 6 is 10 and weight of 7 is 1. In weighted binary codes each digit has a weight 8, 4, 2 or 1.

The codes 8421, 2421 and 5211 are all weighted codes. (Refer Tables 1.6.1, 1.6.3 and 1.6.5)

Non–weighted codes : Non–weighted codes are not assigned with any weight to each digit position, i.e., each digit position within the number is not assigned fixed value. Excess–3 and gray codes are the non–weighted codes.

28. Encode the ten decimal digits in the 2 out of 5 code.

 Answer:

Table 1.6.9 shows some 5–bit BCD codes having special characteristics. These special characteristics of the code are useful for error detection. As shown in Table 1.6.9, 63210 and two–out–of–five codes have two 1s in every code group. The 63210 code is a weighted code (except for decimal digit 0) and it is used for storing data on magnetic tape. On the other hand two–out–of–five code is unweighted code and is used in the telephone and communications industries. Shift counter code, also known as Johnson so known code and 51111 code has specific sequences of 1s which are useful for error detection.

29. Show that the excess–3 code is self–complementing.

 Answer:

In excess–3 code we get 9's complement of a number by just complementing each bit. Due to this excess–3 code is called self–complementing code or reflective code.

30. What is the advantage of gray codes over the binary number sequence?

 Answer:

• If we use gray code to represent disk position then error due to improper brush alignment can be reduced. This is because the gray code assures that only one bit will change each time the decimal number is incremented.

• So in 3–bit code, error may occur due to one bit position. Other two bits positions two bits of two adjacent sectors are always same and hence there is no possibility of error.

• Therefore in 3–bit code probability of error is reduced upto 66 %. In case of 4–bit code it is reduced upto 75 %. This is an important advantage of gray code.

31. What are error detecting codes ?

 Answer:

• When the digital information in the binary form is transmitted from one circuit or system to another circuit or system an error may occur. This means a signal corresponding to 0 may change to 1 or vice versa due to presence of noise.

• To maintain the data integrity between transmitter and receiver, extra bit or more than one bit are added in the data.

• These extra bits allow the detection and sometimes correction of error in the data.

• The data along with the extra bit/bits forms the code. Codes which allow only error detection are called error detecting codes and codes which allow error detection and correction are called error detecting and correcting codes.