Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Number system
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Number system
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Subject: Basic Electronics
Class: Senior Secondary 3
Term: 1st Term
Week: 8
Theme: Digital Basics
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Identify different number systems Explain the basis of for mation of different number systems. Perform some additions and subtractions in binary numbers. Convert from one base number to another
$11000111_2$
Commentary: This demonstrates the efficiency of the grouping method for octal/hexadecimal to binary conversions, and vice-versa. Stress the importance of using the correct number of bits (3 for octal, 4 for hexadecimal).
Question 5 (Conversion: Binary to Octal, Binary to Hexadecimal): Convert the following numbers: a) $11100101_2$ to Octal b) $110110011110_2$ to Hexadecimal Solution 5: a) $11100101_2$ to Octal: Group in threes from the right: $011 \ 100 \ 101$ (added a leading zero)
Convert each group: $011_2 = 3_8$, $100_2 = 4_8$, $101_2 = 5_8$ Combine: $345_8$ b) $110110011110_2$ to Hexadecimal: Group in fours from the right: $1101 \ 1001 \ 1110$ Convert each group: $1101_2 = D_{16}$, $1001_2 = 9_{16}$, $1110_2 = E_{16}$ Combine: $D9E_{16}$
Commentary: This reinforces the grouping method for binary to octal/hexadecimal conversions. Remind students to add leading zeros to complete groups if necessary. The teacher should provide these questions and guide students through solving them, encouraging peer discussion before revealing solutions. Question 1 (Identifying and Explaining Bases): Identify the base and the set of unique digits for the following number systems: a) Decimal b)
Binary c)
Octal d)
Hexadecimal Solution 1: a)
Decimal: Base 10, Digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} b)
Binary: Base 2, Digits: {0, 1} c)
Octal: Base 8, Digits: {0, 1, 2, 3, 4, 5, 6, 7} d)
Hexadecimal: Base 16, Digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}
Commentary: This reinforces the basic definitions and characteristics of each number system, which is fundamental for further operations.
Question 2 (Binary Addition): Perform the following binary addition: $1011_2 + 110_2$.
Solution 2: ``` 1 0 1 1 (11 base 10) + 1 1 0 ( 6 base 10) --------- 1 0 0 0 1 (17 base 10) ``` Step 1 (Rightmost): $1 + 0 = 1$.
Step 2: $1 + 1 = 0$, carry
1. Step 3: $0 + 1 + (\text{carry } 1) = 0$, carry
1. Step 4 (Leftmost): $1 + (\text{carry } 1) = 10_2$. So, 0 here and carry 1 to the next position.
Step 5: Write down the final carry
1. Commentary: Students should pay close attention to the carry-overs, which are critical in binary addition. Verification using decimal conversion can help build confidence.
Question 3 (Binary Subtraction): Perform the following binary subtraction: $11010_2 - 1011_2$.
Solution 3: ``` 1 1^0 0^1 1^0 0^1 0 (26 base 10) - 1 0 1 1 (11 base 10) ---------------- 0 1 1 1 1 (15 base 10) ``` Step 1 (Rightmost): $0 - 1$. Borrow 1 from the next position. $10_2 - 1_2 = 1_2$. The '1' in the second position becomes '0'.
Step 2: The original '1' in the second position became '0'. So $0 - 1$. Borrow 1 from the next position. The '0' in the third position cannot give '1', so it borrows from the '1' in the fourth position. The '1' in the fourth position becomes '0'. The '0' in the third position becomes '2' ($10_2$). Now, the '0' in the second position borrows from the '2' in the third position, making the third position '1' ($01_2$) and the second position '2' ($10_2$). So, $10_2 - 1_2 = 1_2$.
Step 3: The '0' (originally '1') in the third position became '1' due to a previous borrow. So $1 - 0 = 1$.
Step 4: The '1' in the fourth position became '0' due to a borrow. So $0 - 1$. Borrow 1 from the leftmost '1'. This '1' becomes '0', and this '0' becomes '10_2' ($2_{10}$). So $10_2 - 1_2 = 1_2$.
Step 5 (Leftmost): The leftmost '1' became '0' due to a borrow. So $0 - 0 = 0$.
Final Result: $01111_2$ or simply $1111_2$.
Commentary: Binary subtraction, especially with multiple borrows, can be tricky. Emphasize showing the borrow process clearly. Two's complement could be mentioned as an alternative for practical digital systems.
Question 4 (Conversion: Octal to Binary, Hexadecimal to Binary): Convert the following numbers: a) $62_8$ to Binary b) $C7_{16}$ to Binary Solution 4: a) $62_8$ to Binary: Convert each octal digit to its 3-bit binary equivalent: $6_8 = 110_2$ $2_8 = 010_2$ Combine: $110010_2$ b) $C7_{16}$ to Binary: Convert each hexadecimal digit to its 4-bit binary equivalent: $C_{16} = 1100_2$ ($C_{16} = 12_{10}$) $7_{16} = 0111_2$ Combine: $11000111_2$
Commentary: This demonstrates the efficiency of the grouping method for octal/hexadecimal to binary conversions, and vice-versa. Stress the importance of using the correct number of bits (3 for octal, 4 for hexadecimal).
Question 5 (Conversion: Binary to Octal, Binary to Hexadecimal): Convert the following numbers: a) $11100101_2$ to Octal b) $110110011110_2$ to Hexadecimal Solution 5: a) $11100101_2$ to Octal: Group in threes from the right: $011 \ 100 \ 101$ (added a leading zero)
Convert each group: $011_2 = 3_8$, $100_2 = 4_8$, $101_2 = 5_8$ * Combine: $345_8$ Decimal Number System (Base 10)
Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 unique digits).
Basis of Formation: Each position represents a power of
1
0. The rightmost digit is the $10^0$ (units) place, the next is $10^1$ (tens), then $10^2$ (hundreds), and so on.
Example: $257_{10} = (2 \times 10^2) + (5 \times 10^1) + (7 \times 10^0) = 200 + 50 + 7 = 257$. This is the everyday number system used for counting market produce, currency, and time in Nigeria.
Binary Number System (Base 2)
Digits: 0, 1 (2 unique digits). These are often referred to as "bits" (binary digits).
Basis of Formation: Each position represents a power of
2. It is the fundamental system for digital computers and electronics because it directly corresponds to the ON (1) or OFF (0) states of electronic switches (transistors).
Example: $1101_2 = (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 8 + 4 + 0 + 1 = 13_{10}$.
Octal Number System (Base 8)
Digits: 0, 1, 2, 3, 4, 5, 6, 7 (8 unique digits).
Basis of Formation: Each position represents a power of
8. It is used as a compact way to represent binary numbers, as three binary digits can represent one octal digit ($2^3 = 8$).
Example: $37_8 = (3 \times 8^1) + (7 \times 8^0) = 24 + 7 = 31_{10}$. Hexadecimal Number System (Base 16)
Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F (16 unique digits). A = $10_{10}$, B = $11_{10}$, C = $12_{10}$, D = $13_{10}$, E = $14_{10}$, F = $15_{10}$.
Basis of Formation: Each position represents a power of
1
6. It is widely used in computing for memory addresses, colour codes (e.g., #FF0000 for red), and data representation because four binary digits (a nibble) can represent one hexadecimal digit ($2^4 = 16$).
Example: $2A_{16} = (2 \times 16^1) + (A \times 16^0) = (2 \times 16) + (10 \times 1) = 32 + 10 = 42_{10}$.
Binary Addition Rules: $0 + 0 = 0$ $0 + 1 = 1$ $1 + 0 = 1$ $1 + 1 = 0$ (carry 1 to the next position) $1 + 1 + 1 = 1$ (carry 1 to the next position) Worked
Example: Add $1101_2$ and $1011_2$. ``` 1 1 0 1 (13 base 10) + 1 0 1 1 (11 base 10) ``` Rightmost column: $1 + 1 = 0$, carry
1. Second column: $0 + 1 + (\text{carry } 1) = 0$, carry
1. Third column: $1 + 0 + (\text{carry } 1) = 0$, carry
1. Leftmost column: $1 + 1 + (\text{carry } 1) = 1$, carry 1. ``` 1 1 0 1 + 1 0 1 1 1 1 0 0 0 (24 base 10) ``` Binary Subtraction Rules: $0 - 0 = 0$ $1 - 0 = 1$ $1 - 1 = 0$ $0 - 1 = 1$ (borrow 1 from the next higher position, which makes the borrowed position $2_2$) Worked
Example: Subtract $101_2$ from $1110_2$. ``` 1 1 1 0 (14 base 10) 1 0 1 ( 5 base 10) ``` Rightmost column: $0 - 1$. Cannot do. Borrow 1 from the next position ($1_2$ becomes $0_2$). The $0_2$ becomes $10_2$ ($2_{10}$). So, $10_2 - 1_2 = 1_2$.
Second column: $0 - 0 = 0$. (Original $1_2$ became $0_2$ due to borrow).
Third column: $1 - 1 = 0$.
Leftmost column: $1 - 0 = 1$. ``` 1^1 1^0 1^0 0 <-- Showing borrows 1 0 1 1 0 0 1 (9 base 10) ```
Note: For more complex subtractions, the "two's complement" method is often used in digital systems, but for basic understanding, direct subtraction with borrowing is sufficient.
Digital Communications and Mobile Technology: In Nigeria, mobile phones are ubiquitous. Understanding binary, octal, and hexadecimal systems is crucial for comprehending how data (text messages, voice calls, images) is encoded, transmitted, and decoded by these devices. For example, network engineers often analyze data packets represented in hexadecimal. Students who grasp these concepts can better understand how their phones work and potentially pursue careers in telecommunications maintenance or software development for mobile platforms. Computer Hardware and Software Development: Every computer, from a desktop in an internet café in Lagos to a server managing government data, operates using binary logic. Memory addresses, processor instructions, and data storage are all fundamentally binary, often represented in hexadecimal for human readability. Nigerian students interested in computer science, software engineering, or hardware repair (e.g., diagnosing issues with a motherboard in a computer village in Ikeja) will find this knowledge indispensable for interpreting system logs, understanding assembly language, or working with microcontrollers used in local innovations like automated farming systems or smart home devices.
Electronic Systems and Automation: Number systems are integral to embedded systems and automation found in various Nigerian industries. This includes industrial control systems in manufacturing plants, traffic light controllers in Abuja, security systems in banks, or even simple electronic gadgets assembled by local technicians. Knowledge of binary arithmetic allows for the design and troubleshooting of these digital circuits, enabling students to contribute to local technological advancements and maintenance of critical infrastructure.