Digital System and Binary Numbers

PART-1: Number System and its Arithmetic, Signed Binary Numbers

Que 1.1. Define number system. Give classification of number system. Also define signed and unsigned binary number.

A. Number system: It is a language of digital systems consisting of a set of symbols called digits with rules defined for their addition, multiplication and other mathematical operations.

B. Classification of number system: The classification of number system is as follows:

1. Decimal number system: It has 10 symbols, so the base or radix of this number system is 10. The 10 symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
2. Binary number system: It is a base 2 number system. The two binary digits are 1 and 0.
3. Octal number system: It has a base of 8, it has eight possible digits 0, 1, 2, 3, 4, 5, 6 and 7.
4. Hexadecimal number system: It is a base 16 number system. It has digits from 0 to 9, A, B, C, D, E and F.

C. Signed binary number: Binary number that carry identification as to their polarity is called signed binary number. Plus and minus signs for positive and negative numbers can be represented in a digital format. The three major signed binary notations are sign magnitude notation, 1's complement notation and 2's complement notation.

D. Unsigned binary number: In these type of numbers we do not consider the (+) or (-) sign and concentrate only on the magnitude (absolute value) of numbers.

Que 1.2. Convert the following.

1. (5162)₁₀ = ( )₂

2	5162
2	2581	- 0 ↑
2	1290	- 1 ↑
2	645	- 0 ↑
2	322	- 1 ↑
2	161	- 0 ↑
2	80	- 1 ↑
2	40	- 0 ↑
2	20	- 0 ↑
2	10	- 0 ↑
2	5	- 0 ↑
2	2	- 1 ↑
2	1	- 0 ↑
	0	- 1 ↑

Result: (5162)₁₀ = (1010000101010)₂

2. (11011001)₂ = ( )₁₀

1 × 2⁰ = 1
0 × 2¹ = 0
0 × 2² = 0
1 × 2³ = 8
1 × 2⁴ = 16
0 × 2⁵ = 0
1 × 2⁶ = 64
1 × 2⁷ = 128
------------------
Total = 217

Result: (11011001)₂ = (217)₁₀

3. (6273)₁₀ = ( )₈

8	6273
8	784	- 1 ↑
8	98	- 0 ↑
8	12	- 2 ↑
8	1	- 4 ↑
	0	- 1 ↑

Result: (6273)₁₀ = (14201)₈

4. (7860)₁₀ = ( )₁₆

16	7860
16	491	- 4 ↑
16	30	- 11 (=B) ↑
16	1	- 14 (=E) ↑
	0	- 1 ↑

Result: (7860)₁₀ = (1EB4)₁₆

5. (A23B8)₁₆ = ( )₁₀

8 × 16⁰ = 8
B(11) × 16¹ = 176
3 × 16² = 768
2 × 16³ = 8192
A(10) × 16⁴ = 655360
------------------
Total = 664504

Result: (A23B8)₁₆ = (664504)₁₀

Que 1.3. Convert the following: i. (62.7)₈ = ( )₁₆ = ( )₂ ii. (BC6)₁₆ = ( )₁₀ = ( )₂

i. (62.7)₈ = ( )₁₆ = ( )₂

Octal to Binary:
6 → 110, 2 → 010, 7 → 111
Binary = 110010.111

Binary to Hex: Group in 4s (0011 0010 . 1110) → 3 2 . E

Result: (62.7)₈ = (32.E)₁₆ = (00110010.1110)₂

ii. (BC6)₁₆ = ( )₁₀ = ( )₂

(BC6)₁₆ = 11 × 16² + 12 × 16¹ + 6 × 16⁰ = (3014)₁₀

Hex to Binary: B → 1011, C → 1100, 6 → 0110

Result: (BC6)₁₆ = (3014)₁₀ = (101111000110)₂

Que 1.4. Add the following numbers: i. (ABC)₁₆ + (CDE)₁₆ ii. (77)₈ + (107)₈

i. (ABC)₁₆ = 1010 1011 1100 (in binary)
(CDE)₁₆ = 1100 1101 1110
-----------------------------
Sum = 10111 1001 1010 = (179A)₁₆

ii. (77)₈ = 000 111 111 (in binary)
(107)₈ = 001 000 111
-----------------------------
Sum = 010 000 111 = (206)₈

Que 1.5. Perform the following subtraction using 2's complement method: i. 01000 - 01001 ii. 0011.1001 - 0001.1110

i. 01000 - 01001
Assuming X = 01000, Y = 01001
1's complement of Y = 10110
2's complement of Y = 10111
X + (2's comp of Y) = 01000 + 10111 = 11111
There is no end carry.
X - Y = -(2's complement of 11111) = -(00001)

ii. 0011.1001 - 0001.1110
Assuming X = 0011.1001, Y = 0001.1110
1's complement of Y = 1110.0001
2's complement of Y = 1110.0010
X + (2's comp of Y) = 0011.1001 + 1110.0010 = 1 0001.1011
Since carry is generated, discard carry. Answer is 0001.1011

PART-2: Logic Simplification and Combinational Logic Design

Que 1.6. Describe the binary codes. Show the classification of binary codes in tabular format.

1. Code is the representation of group of symbols, words, or letters. As the digital data is used as group of binary numbers so, we call it as the binary codes.

2. These binary codes are used for the designing and analysis of digital circuit, computer applications, in digital communication. The codes are classified into certain following categories: Weighted codes, Non-weighted codes, Reflective codes, Sequential codes, Alphanumeric codes, Error detecting & correcting codes.

• Codes
  • Weighted codes
    • Binary
    • BCD (8421, 2421)
  • Non-weighted codes
    • Excess-3
    • Gray
  • Reflective codes
    • 2421
    • Excess-3
  • Sequential codes
    • 8421
    • Excess-3
  • Alphanumeric codes
    • ASCII
    • EBCDIC
  • Error detecting codes
    • Hamming
    • Parity

5. Weighted binary codes are those which obey the positional weight for the number to represent.
6. In non-weighted codes, the positional weights are not assigned.
7. In reflective code, the reflectivity is desirable. For example, in complement subtraction, code for 9 is the complement for 0.
8. In sequential code, each succeeding code is one binary number greater than the preceding code.
9. The alphanumeric codes are designed to represent numbers as well as characters.
10. The error detecting and correcting codes are used to detect and correct the error.

Que 1.7. Represent the decimal number 6 in (i) excess-3 code, (ii) BCD code, (iii) Gray code, (iv) 8421 code and (v) 2421 codes.

i. Excess-3 code: 6 (in BCD) = 0110. Add 3 (0011) → 1001

ii. BCD code: (6)₁₀ = 0110

iii. Gray code: Binary = 0110 → Gray code = 0101

iv. 8421 code: (6)₂ = 0110

v. 2421 code: (6)_2,4,2,1 = 1100

Que 1.8. Design Binary code to Gray code converter.

Truth Table:

Binary Code				Gray Code
D	C	B	A	G3	G2	G1	G0
0	0	0	0	0	0	0	0
0	0	0	1	0	0	0	1
0	0	1	0	0	0	1	1
0	0	1	1	0	0	1	0
... (and so on up to 1111)

K-Map Simplification:

Expression for G0

DCBA	00	01	11	10
00	0	11	3	12
01	4	15	7	16
11	12	113	15	114
10	8	19	11	110

G0 = B ⊕ A

Expression for G1

DCBA	00	01	11	10
00	0	1	13	12
01	4	5	7	6
11	12	13	15	14
10	8	9	111	110

G1 = C ⊕ B

Simplified Boolean expressions: G3 = D, G2 = C ⊕ D, G1 = C ⊕ B, G0 = B ⊕ A

Logic Diagram:

PART-4: Review of Boolean Algebra and Demorgans Theorem

Que 1.11. Discuss about the concept of Boolean algebra.

Boolean algebra is a fundamental concept that deals with the manipulation of binary values, represented as 0 and 1.

Basic concepts: Boolean algebra operates on variables that can only take on two values: 0 (false) or 1 (true). These variables are combined using three basic operations:

• AND (·): The AND operation represents the logical conjunction. The result of A · B is 1 only if both A and B are 1. Otherwise, the result is 0.
• OR (+): The OR operation represents the logical disjunction. The result of A + B is 1 if either A or B is 1. Otherwise, the result is 0.
• NOT (¯): The NOT operation represents the logical negation. The result of ¯A is the opposite of A.

Que 1.12. Define the De Morgans theorem of logic simplification for SOP & POS forms.

First De Morgans theorem: It states, complement of two or more variables and then AND operation on these is equivalent to NOR operation on these variables.
¯(A · B) = ¯A + ¯B

Second De Morgan theorem: It states that complement of two or more variables and then OR operation on these is equivalent to a NAND operation on these variables.
¯(A + B) = ¯A · ¯B

Example: (AB)(CD)(EFG) = (¯A + ¯B) + (¯C + ¯D) + (¯E + ¯F + ¯G)

PART-5: SOP and POS Forms, Canonical Forms

Que 1.13. Explain sum of products (SOP) and product of sums (POS) with example.

Sum of products (SOP): SOP is a way to represent a logical expression using a sum of several product terms. SOP is also known as the "Minterm" representation because each product term represents a unique combination of input variables for which the expression evaluates to 1.
Example SOP expression: F(A, B, C) = A'B'C + A'BC' + AB'C + ABC

Product of sums (POS): POS is another way to represent a logical expression, but it uses a product of several sum terms. POS is also known as the "Maxterm" representation because each sum term represents a unique combination of input variables for which the expression evaluates to 0.
Example POS expression: F(A, B, C) = (A+B+C)(A+B+C')(A+B'+C)(A'+B+C)

PART-6: Karnaugh Maps Method up to Five Variables, Don't Care Conditions

Que 1.16. Minimize the given boolean function using K-map. F(A, B, C,D) = Σm(3, 4, 5, 7, 9, 13, 14, 15)

ABCD	00	01	11	10
00	0	1	13	2
01	14	15	17	6
11	12	113	115	114
10	8	19	11	10

F = ¯A B ¯C + ¯A C D + A ¯C D + A B C

Que 1.18. Simplify the following Boolean function using K-map and also draw the simplified logic circuit using basic gates. f(A, B, C,D) = Σm(0, 1, 5, 6, 12, 13, 14) + d(2, 4)

ABCD	00	01	11	10
00	10	11	3	X2
01	X4	15	7	16
11	112	113	15	114
10	8	9	11	10

F = B ¯C + ¯A ¯C + A B ¯D

Logic Diagram:

Que 1.22. Implement the Boolean function F(x,y,z) = (1, 2, 3, 4, 6, 7) using NAND gates.

Simplified function via K-map: F(x,y,z) = y + ¯x z + x ¯z

Using De Morgan's: F(x,y,z) = ¯( ¯y · ¯(¯x z) · ¯(x ¯z) )

Que 1.23. Design an XOR gate by using NAND gate implementation.

The output of 1st NAND gate is Y1 = ¯(AB)

Similarly outputs of second and third NAND gates are Y2 = ¯(A · Y1) and Y3 = ¯(B · Y1)

Output Y = ¯(Y2 · Y3) = A ⊕ B

PART-8: Quine McClusky Method (Tabular Method)

Que 1.26. Minimize the following using Quine-McCluskey method: F(A, B, C, D) = Σm(0, 1, 9, 15, 24, 29, 30) + Σd(8, 11, 31)

1. Arrange minterms according to categories of 1's:

No. of 1's	Minterms	Binary ABCDE	Minterms (2 cell)	Binary	Minterms (4 cell)	Binary
0	m0	00000	0,1	0000-	0,1,8,9	0--00
1	m1 m8	00001 01000	0,8 1,9 8,9 8,24	0-000 0-001 0100- 1-000

2. List of prime implicants:

Prime implicants	A B C D E
8, 24	1 - 0 0 0
11, 15	0 1 - 1 1
15, 31	- 1 1 1 1
0, 1, 8, 9	0 - 0 0 -

3. Select the minimum number of prime implicants which must cover the minterms except don't care minterms:

Resulting expression: Y = B C D E + A B C E + A B C D + A C D

Que 1.28. Use Quine-Mc Cluskey (QM) method to solve the following function: F(A,B,C,D) = Σm(5, 7, 8, 9, 10, 11, 14, 15)

Prime Implicant Chart:

Prime implicants	m5	m7	m8	m9	m10	m11	m14	m15
¯A B D (5,7)	✓	✓
B C D (7,15)		✓						✓
A ¯B (8,9,10,11)			✓	✓	✓	✓
A C (10,11,14,15)					✓	✓	✓	✓

Therefore, F(A, B, C, D) = A C + A ¯B + ¯A B D