INSTRUCTIONS: Answer Any Five (5) Questions. i. Draw the block diagram of a typical microprocessor-based computing system including components of the central processing unit (CPU).

here are the solutions to your questions: 1. a) i) Draw the block diagram of a typical microprocessor-based computing system including the components of the central processing unit (CPU). A typical microprocessor-based computing system consists of a Central Processing Unit (CPU), Memory (RAM and ROM), and Input/Output (I/O) devices, all connected via buses (Address Bus, Data Bus, Control Bus). The CPU itself comprises three main components: • Arithmetic Logic Unit (ALU): Performs arithmetic and logical operations. • Control Unit (CU): Manages and coordinates all operations within the CPU and other components. • Registers: Small, high-speed storage locations within the CPU for temporary data storage during processing. 1. a) ii) Describe the function of these components that form the CPU. • The Arithmetic Logic Unit (ALU) is responsible for executing all arithmetic operations (like addition, subtraction) and logical operations (like AND, OR, NOT). • The Control Unit (CU) fetches instructions from memory, decodes them, and then generates control signals to execute the instructions, coordinating the flow of data between the CPU, memory, and I/O devices. • Registers are used by the CPU to quickly store data, instructions, and memory addresses that are actively being used. They provide fast access to data, improving CPU performance. 1. b) i) Differentiate between a programmable port and a dedicated port. • A programmable port is an I/O port whose function (input or output) can be configured by software. This offers flexibility in how the port interacts with external devices. • A dedicated port is an I/O port with a fixed, unchangeable function (either always input or always output), determined by its hardware design. 1. b) ii) Draw and label the registers that form the internal structure of the 8086/88 processor and the segment registers that work with each processor register. The 8086/88 processor has several general-purpose registers and segment registers. General-Purpose Registers (16-bit, can be accessed as two 8-bit halves): • AX (Accumulator Register): Used for arithmetic, logic, and data transfer operations. (AH for high byte, AL for low byte) • BX (Base Register): Primarily used as a base pointer for memory access. (BH for high byte, BL for low byte) • CX (Count Register): Used as a loop counter and for shift/rotate operations. (CH for high byte, CL for low byte) • DX (Data Register): Used for I/O operations and as an operand for multiplication/division. (DH for high byte, DL for low byte) Pointer and Index Registers (16-bit): • SP (Stack Pointer): Points to the top of the stack. • BP (Base Pointer): Used to access data on the stack. • SI (Source Index): Used as a source pointer for string operations. • DI (Destination Index): Used as a destination pointer for string operations. Segment Registers (16-bit): • CS (Code Segment): Stores the starting address of the code segment. • DS (Data Segment): Stores the starting address of the data segment. • SS (Stack Segment): Stores the starting address of the stack segment. • ES (Extra Segment): Stores the starting address of an extra data segment. Other Registers: • IP (Instruction Pointer): Stores the offset of the next instruction to be executed within the code segment. • FLAGS Register: Stores the status of the CPU and the results of arithmetic/logic operations. 2. a) i) Explain the importance of interrupt in computing and differentiate between the two types of interrupts. Interrupts are crucial in computing as they allow the CPU to respond to external events or internal conditions, enabling multitasking and efficient resource management. They temporarily halt the current program execution to handle a higher-priority event. The two main types of interrupts are: • Hardware Interrupts: Generated by external hardware devices (e.g., keyboard press, mouse movement, timer, disk controller) to signal the CPU that an event requires attention. • Software Interrupts: Generated by software instructions (e.g., system calls, division by zero error) to request a service from the operating system or to handle an exceptional condition. 2. a) ii) Differentiate between an Interrupt Value and an Interrupt; explain why some interrupts are dedicated. • An Interrupt is an event that causes the CPU to temporarily suspend its current program execution and transfer control to a special routine called an Interrupt Service Routine (ISR) to handle the event. • An Interrupt Value (or Interrupt Vector Number) is a unique identifier (typically an 8-bit number) associated with a specific interrupt. This value is used by the CPU to locate the address of the corresponding Interrupt Service Routine in the Interrupt Vector Table. Some interrupts are dedicated because they are assigned to critical, non-negotiable events or devices. For example, the Non-Maskable Interrupt (NMI) is dedicated to severe hardware errors (like memory parity errors) that must be handled immediately and cannot be ignored by the CPU. This ensures system stability and responsiveness to critical failures. 2. b) During an interrupt, the stack, the stack pointer and the interrupt return instruction work together ensuring that the CPU returns to the last instruction in the main Programme it was executing before the interrupt request being made. Explain how this happens. When an interrupt occurs, the CPU performs the following steps to ensure a proper return: 1. The CPU first saves the current state of the program, including the address of the next instruction to be executed (from the Instruction Pointer, IP) and the current status of the CPU (from the FLAGS register), onto the stack. 2. The stack pointer (SP) register is automatically decremented to point to the new top of the stack where this information is stored. 3. The CPU then loads the address of the appropriate Interrupt Service Routine (ISR) from the Interrupt Vector Table and begins executing the ISR. 4. After the ISR completes its task, it executes an Interrupt Return (IRET) instruction. This instruction tells the CPU to pop the saved IP and FLAGS values from the stack back into their respective registers. 5. The stack pointer (SP) is automatically incremented as the values are popped. 6. With the original IP and FLAGS restored, the CPU resumes execution of the main program exactly from where it was interrupted. 3. a) Briefly explain the following numbering systems and give example of each: i) Binary Numbering System The Binary Numbering System is a base-2 system that uses only two digits: 0 and 1. It is the fundamental language of computers, where each digit (bit) represents a state of "off" or "on." Example: The binary number 1011_2 is equivalent to 11_10 in decimal. ii) Hexadecimal Numbering System The Hexadecimal Numbering System is a base-16 system that uses 16 distinct symbols: 0-9 and A-F (where A represents 10, B represents 11, and so on, up to F for 15). It is often used in computing as a compact way to represent binary numbers, as each hexadecimal digit corresponds to four binary digits (bits). Example: The hexadecimal number 2A_16 is equivalent to 42_10 in decimal. 3. b) i) Convert the octal number 7654 to binary. Step 1: Convert each octal digit to its 3-bit binary equivalent. 7_8 = 111_2 6_8 = 110_2 5_8 = 101_2 4_8 = 100_2 Step 2: Combine the binary equivalents. 7654_8 = 111110101100_2 7654_8 = 111110101100_2 3. b) ii) Convert the hexadecimal number ABC to binary. Step 1: Convert each hexadecimal digit to its 4-bit binary equivalent. A_16 = 1010_2 B_16 = 1011_2 C_16 = 1100_2 Step 2: Combine the binary equivalents. ABC_16 = 101010111100_2 ABC_16 = 101010111100_2 3. c) The 24-bit binary number 01111100100101101010 is to be entered into a computer. State how it be entered using: i) Octal code Step 1: Group the 24-bit binary number into groups of 3 bits from right to left. 011 \ 111 \ 001 \ 001 \ 011 \ 010 \ 100 (added two leading zeros to make it 24 bits, 01111100100101101010_2 is 22 bits, assuming it should be 0001111100100101101010_2 for 24 bits, or the last group is 10_2 if it's exactly 22 bits. Let's assume the given string is exactly 24 bits, so the last group is 010_2 and the first group is 011_2. The provided number is 01111100100101101010. This is 22 bits. If it's 24 bits, it should be 0001111100100101101010_2. Let's assume the given string is the number, and we pad with leading zeros to make it divisible by 3 for octal, and by 4 for hexadecimal.) Given binary number: 01111100100101101010_2 (22 bits) To convert to octal, group in 3s from the right. Pad with leading zeros to make it divisible by 3. 00 \ 011 \ 111 \ 001 \ 001 \ 011 \ 010 \ 100 (padded to 24 bits for convenience, or 0001111100100101101010_2 if it's exactly 24 bits) Let's use the given 22 bits and pad to 24 bits for octal conversion: 0001111100100101101010_2 Group in 3s: 000 \ 111 \ 110 \ 010 \ 010 \ 110 \ 101 \ 0_2 (This is 22 bits, so the last group is 10_2, which is 010_2 with padding) Let's re-evaluate the grouping for 01111100100101101010_2: 011 \ 111 \ 001 \ 001 \ 011 \ 010 \ 10_2 Pad the leftmost group to 3 bits: 0001111100100101101010_2 Group in 3s: 000 \ 111 \ 110 \ 010 \ 010 \ 110 \ 101 \ 010_2 (This is 24 bits) 000_2 = 0_8 111_2 = 7_8 110_2 = 6_8 010_2 = 2_8 010_2 = 2_8 110_2 = 6_8 101_2 = 5_8 010_2 = 2_8 So, 01111100100101101010_2 (assuming it's 0001111100100101101010_2 for 24 bits) is 07622652_8. If the number is exactly 01111100100101101010_2 (22 bits), then: 011 \ 111 \ 001 \ 001 \ 011 \ 010 \ 10_2 011_2 = 3_8 111_2 = 7_8 001_2 = 1_8 001_2 = 1_8 011_2 = 3_8 010_2 = 2_8 010_2 = 2_8 (padded 10_2 to 010_2) Octal code: 3711322_8 ii) Hexadecimal code Step 1: Group the 24-bit binary number into groups of 4 bits from right to left. Given binary number: 01111100100101101010_2 (22 bits) Pad with leading zeros to make it divisible by 4. 0001111100100101101010_2 (padded to 24 bits) Group in 4s: 0001 \ 1111 \ 0010 \ 0101 \ 1010 \ 1010_2 0001_2 = 1_16 1111_2 = F_16 0010_2 = 2_16 0101_2 = 5_16 1010_2 = A_16 1010_2 = A_16 Hexadecimal code: 1F25AA_16 4. a) List four (4) functions of the Bus Interface Unit (BIU) of the 8086 microprocessors. The Bus Interface Unit (BIU) of the 8086 microprocessor is responsible for all external bus operations. Its four main functions are: • Instruction Fetching: Fetches instructions from memory. • Operand Fetching: Fetches operands (data) from memory. • Operand Storing: Writes operands (results) to memory. • Address Generation: Generates the 20-bit physical addresses for memory and I/O operations. • Bus Control: Manages the address, data, and control buses. • Queue Management: Maintains an instruction queue (prefetch buffer) to speed up instruction execution. 4. b) Briefly describe the following purpose registers of the 8086 microprocessors: i) AX Register The AX (Accumulator) Register is a 16-bit general-purpose register primarily used for arithmetic, logical, and data transfer operations. It is often the default operand for many instructions and can be accessed as two 8-bit registers, AH (high byte) and AL (low byte). ii) BX Register The BX (Base) Register is a 16-bit general-purpose register often used as a base pointer for memory addressing. It can hold the starting address of a data table in memory and can also be accessed as two 8-bit registers, BH (high byte) and BL (low byte). iii) CX Register The CX (Count) Register is a 16-bit general-purpose register primarily used as a counter in loop instructions and for shift/rotate operations. It can also be accessed as two 8-bit registers, CH (high byte) and CL (low byte). iv) DX Register The DX (Data) Register is a 16-bit general-purpose register used for I/O operations (e.g., specifying port addresses) and as an operand for multiplication and division instructions (e.g., holding the high-order word of a 32-bit product or dividend). It can also be accessed as two 8-bit registers, DH (high byte) and DL (low byte). 4. c) What is meant by the following terms? i) Software Software refers to the set of instructions, data, or programs used to operate computers and execute specific tasks. It is intangible and includes operating systems, applications, and programming languages. ii) Hardware Hardware refers to the physical components of a computer system that can be seen and touched. This includes the CPU, memory, storage devices, input/output devices, and all other electronic and mechanical parts. iii) Firmware Firmware is a specific type of software that is permanently stored in a hardware device's non-volatile memory (like ROM or flash memory). It provides low-level control for the device's specific hardware, acting as an interface between the hardware and higher-level software. Examples include BIOS in computers or embedded software in microcontrollers. Got more? Send 'em!