Digital Electronics for Engineering Class Notes

3rd Feb 2025

DEE: Digital Electronics for Engineering

• Course Objectives:
  • To provide the basic understanding of properties and theorems of Boolean Algebra.
  • To provide knowledge on logic gates, universal gates, and their classifications.
  • To teach the techniques to reduce Boolean expressions by using K-map (Karnaugh Map).
  • To give an introduction to logic families and different types of Integrated Circuits (IC 741, operational amplifier Op-Amp).

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Unit 1: Number Systems and Boolean Algebra

1.1: Number Systems:

• Number systems, complements of Number systems, and binary codes.

1.2: Boolean Algebra:

• Properties and theorems of Boolean Algebra, Logic gates and universal gates, Multilevel NAND, NOR gates.

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Unit 2: Minimization of Boolean Functions and Combinational Circuits

2.1: Minimization Techniques:

• K-Map (up to four variables) and Don’t Care conditions, Tabular method.

2.2: Combinational Circuits:

• Design and analysis of adders, subtractors, comparators, multiplexers (MUX), demultiplexers (DMUX), decoders, encoders, code converters, and hazard-free realizations.

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4th Feb 2025

Unit 3: Sequential Circuits

• Study of sequential circuits, including:
  • Flip-Flops (SR, JK, JK Master-Slave, D, T types).
  • Registers (SISO, SIPO, PISO, PIPO).
  • Counters (Ripple counters, Decade counters).

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Unit 4: Logic Families

• Overview of logic families such as RTL, DTL, TTL, CMOS, ECL etc.
• AND, OR, NOT gates using diodes and transistors.
• TTL Characteristics
  • Standard TTL.
  • Open Collector.
  • Tri-State.

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Unit 5: Integrated Circuits (ICs)

• IC Interface:
  • IC families/Classification.
• Operational Amplifiers:
  • Ideal (IC 741 Op-Amp) and practical characteristics.
  • DC and AC characteristics of Op-Amps.
  • Modes: Inverting, Non-Inverting.

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Unit 1

Number Systems

Number systems are useful in digital computer technology which give reliable and efficient arithmetic operations. It is broadly classified into two types:

1. Non-Positional Number Systems (e.g., Roman Numerals).
2. Positional Number Systems.

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Positional Number Systems

A positional number system must contain a Radix/Base. The weight of a digit depends on its relative position within the number.

Radix/Base:
It is a term used to describe the positional number system. or It specifies the number of symbols used for corresponding number system.

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S.No	Number System	Base/Radix	Essential Numbers
1	Binary	2	0, 1
2	Octal	8	0, 1, 2, 3, 4, 5, 6, 7
3	Decimal	10	0, 1, 2, 3, 4, 5, 6, 7, 8, 9
4	Hexadecimal	16	0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A(10), B(11), C(12), D(13), E(14), F(15)

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Representation of Numbers

Each binary digit is called a BIT (Binary Digit).

• 1 BIT = Single Binary Digit.

3-Bit Binary Representation

Octal	Binary (3-bit) $2^2,2^1,2^0$
0	000
1	001
2	010
3	011
4	100
5	101
6	110
7	111

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4-Bit Binary Representation

Hexadecimal	Binary (4-bit) $2^3,2^2,2^1,2^0$
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001
A	1010
B	1011
C	1100
D	1101
E	1110
F	1111

2-Bit Binary Representation

Decimal	Binary (2-bit) $2^1,2^0$
0	00
1	01
2	10
3	11

5th Feb 2025

Number System Conversion

Any decimal system is converted into any base.

• Convert Integer part of any base number system by successive division.
• Convert Fractional part of any base number system by successive multiplication.
• This method is popularly called as double dabble Method.

Decimal to Binary:

Example:$(24.8{)}_{10}$

• Integer Part:$24$

  $$\begin{array}{rl}& 2|\underline{24}\text{Remainder: }0\\ & 2|\underline{12}\text{Remainder: }0\\ & 2|\underline{6}\text{Remainder: }0\\ & 2|\underline{3}\text{Remainder: }1\\ & 2|\underline{1}\text{Remainder: }1\end{array}$$

  Binary Integer:$(11000{)}_2$

• Fractional Part:$0.8$

  $$\begin{array}{rl}& 0.8\times 2=1.6\text{Carry: }1\\ & 0.6\times 2=1.2\text{Carry: }1\\ & 0.2\times 2=0.4\text{Carry: }0\\ & 0.4\times 2=0.8\text{Carry: }0\end{array}$$

  Binary Fraction:$(0.1100{)}_2$

Final Result:$(11000.1100{)}_2$

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Decimal to Octal:

Example:$(215.14{)}_{10}$

• Integer Part:$215$

  $$\begin{array}{rl}& 8|\underline{215}\text{Remainder: }7\\ & 8|\underline{26}\text{Remainder: }2\\ & 8|\underline{3}\text{Remainder: }3\end{array}$$

  Octal Integer:$(327{)}_8$

• Fractional Part:$0.14$

  $$\begin{array}{rl}& 0.14\times 8=1.12\text{Carry: }1\\ & 0.12\times 8=0.96\text{Carry: }0\\ & 0.96\times 8=7.68\text{Carry: }7\end{array}$$

  Octal Fraction:$(0.107{)}_8$

Final Result:$(327.107{)}_8$

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Decimal to Hexadecimal:

Example:$(1056{)}_{10}$

• Integer Part:$1056$ $$\begin{array}{rl}& 16|\underline{1056}\text{Remainder: }0\\ & 16|\underline{66}\text{Remainder: }2\\ & 16|\underline{4}\text{Remainder: }4\end{array}$$ Hexadecimal Integer:$(420{)}_{16}$

Final Result:$(420{)}_{16}$

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6th Feb 2025

Any Base to Decimal:

Example:$(6BE{)}_{16}$

$$\begin{array}{rl}(6BE{)}_{16}& =6\times 16^2+11\times 16^1+14\times 16^0\\ & =6\times 256+11\times 16+14\times 1\\ & =1536+176+14\\ & =(1726{)}_{10}\end{array}$$

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Problems:

1. $(101011{)}_2=(?{)}_{16}$

   • Group binary into sets of 4 bits:$00101011$
   • Convert each group to hexadecimal:$2B$
   • Final Result:$(2B{)}_{16}$

2. $(ZEBA9{)}_{16}=(?{)}_{10}$

   • Expand using positional values: $$ZEBA{9}_{16}=15\times 16^4+14\times 16^3+11\times 16^2+10\times 16^1+9\times 16^0$$

3. $(798{)}_{10}=(?{)}_8$

   • Perform successive division by 8: $$\begin{array}{rl}& 8|\underline{798}\text{Remainder: }6\\ & 8|\underline{99}\text{Remainder: }3\\ & 8|\underline{12}\text{Remainder: }4\\ & 8|\underline{1}\text{Remainder: }1\end{array}$$
   • Octal Result:$(1436{)}_8$

4. $(145{)}_8=(?{)}_2$

   • Convert each octal digit to 3-bit binary: $$1_8=001_2,4_8=100_2,5_8=101_2$$
   • Combine:$(001100101{)}_2$
   • Final Result:$(1100101{)}_2$

Here is the corrected and properly formatted version of your notes using single $ for mathematical notations:

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Octal to Binary:

• One octal digit is converted into binary; we require 3 binary bits.

Octal	Binary
0	000
1	001
2	010
3	011
4	100
5	101
6	110
7	111

Example: $(2357{)}_8$

• Convert each octal digit to 3-bit binary: $$2\to 010,3\to 011,5\to 101,7\to 111$$
• Combine:$(010011101111{)}_2$
• Final Result:$(010011101111{)}_2$

Binary to Octal:

• Group binary digits into sets of 3 bits (starting from the right).
• Convert each group to its corresponding octal digit.

Examples:

1. Binary:$1010111011$

   • Group into 3 bits:$001010111011$
   • Convert:$1273$
   • Final Result:$(1273{)}_8$

2. Binary:$1110111011011$

   • Group into 3 bits:$001110111011011$
   • Convert:$16733$
   • Final Result:$(16733{)}_8$

3. Binary:$10110101110$

   • Group into 3 bits:$010110101110$
   • Convert:$2656$
   • Final Result:$(2656{)}_8$

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Hexadecimal to Binary:

• One hexadecimal digit is converted into binary; we require 4 binary bits.

Hexadecimal	Binary
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001
A	1010
B	1011
C	1100
D	1101
E	1110
F	1111

Examples:

1. Hexadecimal:$F294B$

   • Convert each hexadecimal digit to 4-bit binary: $$F\to 1111,2\to 0010,9\to 1001,4\to 0100,B\to 1011$$
   • Combine:$(11110010100101001011{)}_2$
   • Final Result:$(11110010100101001011{)}_2$

2. Hexadecimal:$379BDF$

   • Convert each hexadecimal digit to 4-bit binary: $$3\to 0011,7\to 0111,9\to 1001,B\to 1011,D\to 1101,F\to 1111$$
   • Combine:$(001101111001101111011111{)}_2$
   • Final Result:$(001101111001101111011111{)}_2$

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Binary to Hexadecimal:

• Group binary digits into sets of 4 bits (starting from the right).
• Convert each group to its corresponding hexadecimal digit.

Examples:

3. Binary:$101011011011$

   • Group into 4 bits:$101011011011$
   • Convert:$ADB$
   • Final Result:$(ADB{)}_{16}$

4. Binary:$1011011011.111011$

   • Group into 4 bits:$001011011011.11101100$
   • Convert:$2DB.EC$
   • Final Result:$(2DB.EC{)}_{16}$

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Hexadecimal to Octal:

• Convert hexadecimal to binary first (4 bits per digit).
• Then convert binary to octal (3 bits per group).

Example: Hexadecimal:$79BC54$

• Convert to binary: $$7\to 0111,9\to 1001,B\to 1011,C\to 1100,5\to 0101,4\to 0100$$ Binary:$(011110011011110001010100{)}_2$
• Group into 3 bits for octal: $$011110011011110001010100$$
• Convert:$36336124$
• Final Result:$(36336124{)}_8$

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Octal to Hexadecimal:

• Convert octal to binary first (3 bits per digit).
• Then convert binary to hexadecimal (4 bits per group).

Example: Octal:$46321$

• Convert to binary: $$4\to 100,6\to 110,3\to 011,2\to 010,1\to 001$$ Binary:$(100110011010001{)}_2$
• Group into 4 bits for hexadecimal: $$0100110011010001$$
• Convert:$4CD1$
• Final Result:$(4CD1{)}_{16}$

7th Feb 2025

Question:$(FB39{)}_{16}=({)}_4$To convert FB39 (hexadecimal) to base 4:

Step 1: Hexadecimal to Decimal

Expand FB39:

$$FB39_{16}=15\cdot 16^3+11\cdot 16^2+3\cdot 16^1+9\cdot 16^0$$

Calculate:

$$FB39_{16}=61440+2816+48+9=64313_{10}$$

Step 2: Decimal to Base 4

Divide$64313$by$4$repeatedly, recording remainders:

• $64313÷4=16078$, Remainder =$1$
• $16078÷4=4019$, Remainder =$2$
• $4019÷4=1004$, Remainder =$3$
• $1004÷4=251$, Remainder =$0$
• $251÷4=62$, Remainder =$3$
• $62÷4=15$, Remainder =$2$
• $15÷4=3$, Remainder =$3$
• $3÷4=0$, Remainder =$3$

Read remainders from bottom to top:

$$64313_{10}=3320321_4$$ $$\boxed{3320321_4}$$

Binary Arithmetic

1. Binary Addition

Binary addition follows simple rules:

• $0+0=0$
• $0+1=1$
• $1+0=1$
• $1+1=0$ (carry over 1 to the next higher bit)

2. Binary Subtraction

Binary subtraction follows these rules:

• $0-0=0$
• $1-0=1$
• $1-1=0$
• $0-1=1$ (borrow 1 from the next higher bit)

3. Binary Multiplication

Binary multiplication is similar to decimal multiplication but simpler because it only involves$0$and$1$:

• $0\times 0=0$
• $0\times 1=0$
• $1\times 0=0$
• $1\times 1=1$

4. Binary Division

Operation	Result
$0÷0$	$(0)$Undefined
$0÷1$	$0$
$1÷0$	$(0)$Undefined
$1÷1$	$1$

Example: Add $26$ and $13$ in binary

• Decimal:$26+13=39$
• Binary equivalents: -$26=11010$ -$13=1101$

Perform binary addition:

   11010   (26)
+   1101   (13)
---------
  100111   (39)

Steps:

1. Start from the rightmost bit.
2. Add bits column by column, carrying over if necessary.
3. Final result:$100111_2$, which equals$39_{10}$.

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Example: Subtract $13$ from $26$ in binary

• Decimal:$26-13=13$
• Binary equivalents:
  • $26=11010$
  • $13=1101$

Perform binary subtraction:

   11010   (26)
-   1101   (13)
---------
    1101   (13)

Steps:

1. Start from the rightmost bit.
2. Borrow from the next higher bit if necessary.
3. Final result: $1101_2$, which equals$13_{10}$.

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• Decimal:$13-26=-13$
• Binary equivalents:
  • $26=11010$
  • $13=1101$

Perform binary subtraction:

    1101   (13)
-  11010   (26)
---------
  110011   (13's 2's complement)