Digital Electronics (BOE-310) - Complete Unit 1 Notes[cite: 2]

University: Dr. A.P.J. Abdul Kalam Technical University (AKTU)[cite: 2]

Lecture 1: Number System and Binary Codes[cite: 2]

1. Common Number Systems[cite: 2]

System	Base	Symbols Used	Used by Humans?	Used in Computers?
Decimal	10	0, 1, 2, 3, 4, 5, 6, 7, 8, 9	Yes	No
Binary	2	0, 1	No	Yes
Octal	8	0, 1, 2, 3, 4, 5, 6, 7	No	No
Hexadecimal	16	0-9, A, B, C, D, E, F	No	No

Example of Equivalence: (25)₁₀ = (11001)₂ = (31)₈ = (19)₁₆[cite: 2]

2. Conversion Techniques[cite: 2]

A. Any Base to Decimal[cite: 2]

Technique: Multiply each digit by Baseⁿ (where n is the weight, starting from 0 at the right/LSB).[cite: 2]

• Binary to Dec: (101011)₂ = 1×2⁵ + 0×2⁴ + 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = (43)₁₀[cite: 2]
• Octal to Dec: (724)₈ = 7×8² + 2×8¹ + 4×8⁰ = 448 + 16 + 4 = (468)₁₀[cite: 2]
• Hex to Dec: (ABC)₁₆ = A(10)×16² + B(11)×16¹ + C(12)×16⁰ = 2560 + 176 + 12 = (2748)₁₀[cite: 2]

B. Decimal to Any Base (Successive Division)[cite: 2]

Technique: Divide the number by the target base (2, 8, or 16) and track the remainder. Read remainders from bottom to top (MSB to LSB).[cite: 2]

• Dec to Binary: 125 ÷ 2 continuously gives (1111101)₂.[cite: 2]
• Dec to Octal: 1234 ÷ 8 continuously gives (2322)₈.[cite: 2]
• Dec to Hex: 1234 ÷ 16 continuously gives (4D2)₁₆.[cite: 2]

3. Binary Codes[cite: 2]

• Gray Code: Also known as Cyclic Code. Only one bit changes between two successive numbers.[cite: 2] Conversion: EX-OR consecutive binary bits.[cite: 2]
• BCD Code (8421): Each decimal digit is replaced by a 4-bit binary equivalent.[cite: 2] Valid range is 0 to 9. 10 to 15 are invalid.[cite: 2]
• Excess-3 Code: Obtained by adding 3 (0011 in binary) to the BCD code.[cite: 2] It is a self-complementing code.[cite: 2]

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Lecture 2: Number System Arithmetic & Hamming Code[cite: 2]

1. Negative Number Representation & Subtraction[cite: 2]

• 1's Complement Method: Change all 1s to 0s and 0s to 1s.[cite: 2]
  Subtraction: Add 1's complement of subtrahend to minuend. If carry is generated, add it to the LSB (Result is positive). If no carry, take 1's complement of result (Result is negative).[cite: 2]
• 2's Complement Shortcut: Scan from right (LSB) to left. Retain bits up to the first '1', then invert the remaining bits.[cite: 2]
  Subtraction: Add 2's complement of subtrahend. If carry is generated, discard it (Result is positive). If no carry, take 2's complement of result (Result is negative).[cite: 2]

2. Hamming Code (Error Detection & Correction)[cite: 2]

Hamming code uses additional parity bits for error detection and correction.[cite: 2]

• Condition: 2^p ≥ m + p + 1 (where m = message bits, p = parity bits).[cite: 2]
• Example: For message "1101001" (m=7), we need p=4. Total bits = 11.[cite: 2]
• Parity bits are placed at positions 1, 2, 4, 8 (powers of 2).[cite: 2]
• Error Checking: If there is a single conflict (parity failure), it indicates an error. The sum of the failing parity bit positions gives the exact location of the error.[cite: 2]

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Lecture 3: Logic Gates and Boolean Algebra[cite: 2]

1. Logic Gates with Symbols[cite: 2]

There is a total of 7 Logic Gates (Basic, Universal, and Exclusive).[cite: 2]

NOT Gate[cite: 2]

AND Gate[cite: 2]

OR Gate[cite: 2]

NAND Gate[cite: 2]

NOR Gate[cite: 2]

XOR Gate[cite: 2]

2. Basic Theorems of Boolean Algebra[cite: 2]

• Commutative Law: A + B = B + A, A · B = B · A[cite: 2]
• Distributive Law: A(B + C) = AB + AC, A + (BC) = (A + B)(A + C)[cite: 2]
• Absorption Law: A + AB = A, A(A + B) = A[cite: 2]
• De-Morgan's Theorem: (X + Y)' = X' · Y' and (X · Y)' = X' + Y' (Rule: Break the line & change the sign)[cite: 2]

3. Canonical Forms (SOP & POS)[cite: 2]

In standard SOP and POS, each term must contain all the variables (literals) used in the expression.[cite: 2]

💡 Solved Example: Converting to Standard SOP (Extra Practice)

Question: Convert F = AB + AC to Canonical SOP form (Variables: A, B, C).

Solution:
F = AB(C + C') + AC(B + B')
F = ABC + ABC' + ABC + AB'C
F = ABC + ABC' + AB'C (Duplicate ABC is removed)

💡 Solved Example: Converting to Standard POS (Extra Practice)

Question: Convert F = (A + B) to Canonical POS form (Variables: A, B, C).

Solution:
Add missing literal C as C.C' (since C.C' = 0).
F = (A + B + C.C')
F = (A + B + C)(A + B + C') (Using distributive law: X + YZ = (X+Y)(X+Z))

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Lecture 4: Karnaugh Maps & Tabulation Method[cite: 2]

1. Karnaugh Map (K-Map) Rules[cite: 2]

K-Maps provide a systematic method to minimize Boolean expressions. Cells are labeled using Gray Code counting (00, 01, 11, 10).[cite: 2]

• Pair: Groups of 2 adjacent 1s. Eliminates 1 variable.[cite: 2]
• Quad: Groups of 4 adjacent 1s. Eliminates 2 variables.[cite: 2]
• Octet: Groups of 8 adjacent 1s. Eliminates 3 variables.[cite: 2]

Standard 4-Variable K-Map Structure[cite: 2]

AB \ CD	00 (C'D')	01 (C'D)	11 (CD)	10 (CD')
00 (A'B')	m0	m1	m3	m2
01 (A'B)	m4	m5	m7	m6
11 (AB)	m12	m13	m15	m14
10 (AB')	m8	m9	m11	m10

💡 Solved Example: 3-Variable K-Map (Extra Practice)

Minimize: F(A, B, C) = Σm(0, 1, 2, 3)

Solution: We place 1s in cells 0, 1, 2, and 3. All these four cells form a single Quad in the top row (A' row). Since the columns span all combinations of B and C, B and C are eliminated.

Result: F = A'

💡 Solved Example: 4-Variable K-Map (Extra Practice)

Minimize: F(A, B, C, D) = Σm(0, 2, 4, 6, 8, 10, 12, 14)

Solution: By placing 1s in the corresponding cells, we see that all the corner cells (0, 2, 8, 10) and middle edge cells (4, 6, 12, 14) are filled. These 8 cells wrap around to form an Octet.

Result: F = D'

2. Tabulation Method (Quine Mc-Clusky Method)[cite: 2]

Used for 5 or 6 variable problems where K-maps become too complex.[cite: 2]

Example Question: Minimize F(A,B,C,D) = Σm(0, 1, 2, 3, 5, 7, 8, 10, 12, 13, 15)[cite: 2]

1. Step 1: Write all terms in binary form.[cite: 2]
2. Step 2: Arrange terms in increasing order of their Index (number of 1s in the binary representation).[cite: 2]
3. Step 3: Compare terms and form groups, placing a dash '-' where a bit differs.[cite: 2]
4. Step 4: Create a Prime Implicant (PI) chart to find essential prime implicants.[cite: 2]

Final Minimized Expression for above function:
F = A'B' + B'D' + A'D + BD + AC'D' + ABC'[cite: 2]