Question 1:

a. Perform the following computation using binary 2’s complement notation, assuming the register size to be of 8 bits. Also check for occurrence of overflow :
i. – 63 + 74
ii. – 128 + 39
iii. + 86 + 42

Answer:

i. -63 in 8-bit 2’s complement notation is: 11000011 74 in 8-bit binary is: 01001010

Adding these values: 11000011 + 01001010 = 100001001

The answer is -63 + 74 = 11 in 8-bit 2’s complement notation. No overflow occurs.

ii. -128 in 8-bit 2’s complement notation is: 10000000 39 in 8-bit binary is: 00100111

Adding these values: 10000000 + 00100111 = 10101111

The answer is -128 + 39 = -89 in 8-bit 2’s complement notation. An overflow occurs.

iii. +86 in 8-bit binary is: 01010110 42 in 8-bit binary is: 00101010

Adding these values: 01010110 + 00101010 = 01111100

The answer is +86 + 42 = 128 in 8-bit 2’s complement notation. No overflow occurs.

b. Explain the meaning of ‘minterm’ in the context of digital logic circuits. Make the truth table and simplify the following Boolean function in SOP form using K-maps. Also draw the logic diagram:

F (A, B, C) = Σ (0, 1, 4, 6, 7)

Answer:

In the context of digital logic circuits, a minterm is a Boolean expression in which a variable or its complement appears once and only once in each product term. Minterms are used in the Sum-of-Products (SOP) form of Boolean algebra, which is a way to represent a Boolean function using the OR of a set of ANDed variables or their complements.

The truth table for the Boolean function F(A, B, C) is:

To simplify the Boolean function using a K-map, we can identify the minterms that correspond to the output being 1 (0, 1, 4, 6, 7) and group them into adjacent squares.

The resulting simplified Boolean function in SOP form is : F(A,B,C) = (A’+ B’+ C) (A’+ B+ C’) + (A’+ B’+ C’) (A+ B+ C)

The logic diagram for the function can be represented by a 3-input AND gate and a 3-input OR gate.

Where the AND gate will have the minterm (A’+ B’+ C) and (A’+ B+ C’) as its input and the OR gate will have the output of AND gate and the minterm (A+ B+ C) as its input. The output will be the final output of the circuit

c. The main memory of a computer is of 64 K words size having a word size of 16 bits. The cache of this computer also has a block size of 16 bits having 256 blocks. Answer the following questions if direct mapping scheme has been followed :
i. Size of tag and index fields of cache address.
ii. In which address of cache a main memory address (AFBA) can be found ?
iii. What will be the action of memory management system if the stated memory address is not found in cache location ?

Answer:

i. The size of the tag field in the cache address would be 14 bits (since 64K words = 2^16 words and 256 blocks = 2^8 blocks, so the index field would be 8 bits and the remaining bits would be the tag field).

ii. The main memory address AFBA can be found in cache address (AFBA mod 256).

iii. If the stated memory address is not found in the cache location, the memory management system would perform a cache miss and would need to retrieve the data from the main memory and store it in the corresponding cache location.

d. What is an Interrupt ? Explain any one technique that can be used to determine which device has issued the interrupt.

Answer: An interrupt is a signal sent to the CPU by a device indicating that it needs attention or service. Interrupts are used to allow the CPU to perform other tasks while waiting for the device to complete its operation.

One technique that can be used to determine which device has issued the interrupt is the use of Interrupt Request (IRQ) lines. Each device that can generate an interrupt is assigned a unique IRQ line. When a device generates an interrupt, it sends a signal on its assigned IRQ line to the CPU. The CPU can then read the IRQ lines and determine which device has issued the interrupt. It then sends the corresponding interrupt vector to the device driver for the device that generated the interrupt. This allows the CPU to handle the interrupt and service the device without having to constantly poll all devices to see if they need attention.

e. Assume that an instruction has been fetched in Instruction Register (IR) of a computer, and has been decoded. R register DR is to be used for fetching the operand and AC register is to be used for calculation. Write and explain the various micro-operations for the purpose of execution of the instruction :

Add AC, A

where A is memory location which has the operand and the address of A is presently stored in MAR.

Answer:

The horizontal microinstruction format is a method of encoding microinstructions where the bit patterns for each microinstruction field (such as the opcode, operand, etc.) are arranged horizontally, with each field taking up a fixed number of bits. A diagram of this format might look something like this:

The vertical microinstruction format is a method of encoding microinstructions where the bit patterns for each microinstruction field are arranged vertically, with each field taking up one bit position in a fixed number of columns. A diagram of this format might look something like this:

It is generally believed that the horizontal microinstruction format is faster than the vertical microinstruction format because it requires fewer memory accesses to fetch a single microinstruction, and it also allows for faster and simpler decoding of microinstructions.

In horizontal microinstruction format, the instruction is stored in a single memory location and the entire instruction can be fetched in one memory access cycle. In vertical microinstruction format, the instruction is stored in multiple memory locations and it takes multiple memory access cycles to fetch the entire instruction. As a result, horizontal format is faster than vertical format.

f. Explain the horizontal and vertical micro-instruction format with the help of a
diagram each. Which of the two micro-instructions is faster ? Give reason in
support of your answer.

Answer: In a micro-programmed control unit, the control memory stores a sequence of micro-instructions that dictate the actions of the control unit. These micro-instructions are executed sequentially by the control unit and specify the operations to be performed on the data and the flow of control in the CPU.

The control memory is organized as a series of memory locations, each of which holds one micro-instruction. The control unit fetches the micro-instruction from the control memory and decodes it to determine the actions to be taken.

The organization of control memory can be represented as follows:

Control Memory |Location 0|Location 1|Location 2|…|Location n|

Each location in the control memory holds a micro-instruction, which is a binary word that contains the control signals for the CPU.

A horizontal micro-instruction is a micro-instruction where the control signals are encoded horizontally across the bits of the instruction. This is in contrast to a vertical micro-instruction where the control signals are encoded vertically in different fields or groups of bits.

A horizontal micro-instruction is organized as a bit-vector, where each bit corresponds to a specific control signal in the CPU. This organization allows for more compact representation of micro-instructions and is generally easier to implement in hardware.

An example of a horizontal micro-instruction format could be: [Opcode | ALU control signal | Register Select | Memory control signal | Next address]

Where each field of the instruction specifies a specific control signal in the CPU. Opcode field specifies the operation to be performed, ALU control signal field specifies the operation to be performed on ALU, Register Select field specifies the register to be selected, Memory control signal field specifies the memory operation and Next address field specifies the next micro-instruction address to be executed.

In summary, the control memory in a micro-programmed control unit is a memory that stores a sequence of micro-instructions that dictate the actions of the control unit. The micro-instructions are executed sequentially by the control unit and specify the operations to be performed on the data and the flow of control in the CPU. A horizontal micro-instruction is a micro-instruction where the control signals are encoded horizontally across the bits of the instruction. This format allows for a more efficient use of memory space and reduces the complexity of decoding the micro-instructions. It is a compact and easy to implement way of specifying the control signals in the CPU.

g. Write a program in 8086 assembly language that stores (FEDC)h in AX register and (BA98)h in BX register. It then stores the values of AL, AH, BL and BH in four consecutive byte locations in the memory. Make suitable assumptions.

Answer: Here is an example program in 8086 assembly language that stores the values (FEDC)h in the AX register and (BA98)h in the BX register, and then stores the values of AL, AH, BL, and BH in four consecutive byte locations in memory:

mov ax, 0FEDCh   ; Move the value FEDC hex into the AX register
mov bx, 0BA98h   ; Move the value BA98 hex into the BX register

mov [memory_location], al    ; Move the value of the AL register into memory_location
mov [memory_location+1], ah  ; Move the value of the AH register into memory_location+1
mov [memory_location+2], bl  ; Move the value of the BL register into memory_location+2
mov [memory_location+3], bh  ; Move the value of the BH register into memory_location+3

Note: You should define the memory_location variable before the code and make sure it’s a valid memory address that you can access. Also, it’s assumed that the memory is in real mode and the address of the memory_location is a valid address.

h. What is Memory Interleaving ? Discuss its advantages.

Answer: Memory interleaving is a technique used in computer memory architecture to increase the access speed of memory. It involves dividing the physical memory into several smaller banks, and then arranging these banks in such a way that when the processor requests a memory location, it can access multiple banks simultaneously, rather than having to access each bank one at a time.

Advantages of Memory Interleaving:

1. Increased memory bandwidth: With memory interleaving, the processor can access multiple banks of memory at the same time, increasing the overall memory bandwidth and improving performance.
2. Improved system performance: By increasing the memory bandwidth, memory interleaving can improve the overall performance of the system, leading to faster program execution and better responsiveness.
3. Better utilization of memory resources: Memory interleaving allows the processor to access multiple banks of memory at the same time, which can help to better utilize the memory resources in the system.
4. Better scalability: Memory interleaving allows the system to be easily scaled up by adding more memory banks, which can help to improve performance as the number of processors and memory requirements grow.
5. High availability: If one memory bank fails, the system can still operate with other memory banks, which increases the availability of the system.
6. Reduced memory access time: Memory interleaving can reduce the memory access time, which is the time required to read or write data to memory.

It’s important to note that memory interleaving can be applied to different types of memory such as DRAM, SRAM, and also on different memory channels, it also can be applied in different levels such as channel interleaving, bank interleaving, and even bit interleaving.

Question 2:

a. Explain the concept of S-R flip-flop with the help of logic diagram and characteristic
table. Make and explain the excitation table of S-R flip-flop.

Answer: An S-R (Set-Reset) flip-flop is a type of bistable multivibrator that has two inputs, S (Set) and R (Reset), and two outputs, Q and Q’ (the complement of Q). The S-R flip-flop can be in one of two states, either set (Q = 1) or reset (Q = 0).

The Logic Diagram of S-R Flip-flop :

The Characteristic Table of S-R Flip-flop:

The Excitation table of S-R Flip-flop:

The excitation table is very similar to the characteristic table, but it differs in the last case where S and R are both high (1) which is also known as forbidden or illegal state, in the excitation table the output Q(t+1) and Q'(t+1) are not defined and the output remains the same as previous state.

It’s important to note that the S-R Flip-flop is a basic component in digital circuits and it can be used in different applications such as counters, registers, and memory elements. Also, it can be implemented using different technologies such as transistors, gates, and also using different integrated circuits such as 74LS73 and 74LS74.

b. How normalization and biasing are used for representation of floating point numbers ? Explain using a suitable example.

Answer: Normalization and biasing are techniques used in the representation of floating-point numbers.

Normalization is a process where a number is scaled such that the most significant digit (MSD) is non-zero and is placed in the leftmost position of the number. This is done to represent the number in a more compact form. For example, the number 500 would be normalized to 5.00.

Biasing is a process where a fixed value is added to the exponent of a floating-point number. This is done to represent the exponent in a more compact form. The most common bias used is the excess-n bias where a value of n is added to the exponent. For example, if the exponent has a range of -3 to +3, an excess-3 bias would be used, and the exponent value of -3 would be represented as 0, and the exponent value of +3 would be represented as 6.

Example: Consider a floating-point number with a sign bit, 8-bit exponent field and a 23-bit mantissa field. Consider the number -1.2345 * 2^5.

First, the number is normalized by shifting the decimal point to the leftmost position of the mantissa, obtaining 12.345 * 2^2.

Next, the exponent is biased by adding a bias of 127, obtaining 12.345 * 2^(2+127)

The final representation of the number will be: 1 (sign bit) | 1000 0010 (biased exponent) | 00110011 01010001 00011000 (mantissa)

It’s important to note that normalization and biasing are important steps in the representation of floating point numbers, they allow the numbers to be represented in a more compact form which saves memory and computation resources. Also, the IEEE 754 standard for floating-point arithmetic defines the format and behavior of floating-point numbers and it’s widely used in computer systems and programming languages.

c. Briefly explain the following :
i. RAID
ii. Charge Coupled Devices
iii. Seek Time of a Disk

Answer:

i. RAID (Redundant Array of Independent Disks) is a technology that combines multiple physical disks into a single logical storage unit in order to provide data redundancy and increased performance. There are multiple levels of RAID, each with its own advantages and disadvantages. For example, RAID 0 provides improved performance by striping data across multiple disks, RAID 1 provides data redundancy by mirroring data across multiple disks, and RAID 5 provides both data redundancy and improved performance by using parity data across multiple disks.

ii. Charge-Coupled Devices (CCD) are a type of image sensor used in digital cameras and other imaging devices. CCDs are made up of a matrix of light-sensitive cells called photosites, which convert light into electrical charges. The charges are then transferred to a readout register, where they are converted into a digital signal that can be read by a computer. CCDs are known for their high sensitivity, low noise, and high resolution.

iii. Seek Time of a Disk refers to the time required for the read/write head of a disk drive to move to the location on the disk where the data is stored. It’s one of the main factors affecting the performance of a disk drive, especially for random access operations. The seek time is measured in milliseconds (ms) and it’s composed of the mechanical seek time, which is the time required for the head to move to the desired track, and the rotational latency, which is the additional time required for the disk to rotate to the desired sector. In general, lower seek time is preferable as it means faster access to data stored on the disk.

d. Describe the concept of address space and memory space in virtual memory with the help of an example.

Answer: In virtual memory, the address space is the set of virtual memory addresses that a process can use to access memory. Each process has its own unique address space, which is separate from the address spaces of other processes. The memory space is the physical memory (RAM) that is available to the system.

For example, consider a system with 4GB of physical memory and a single process that is running. The address space for that process would be 4GB, even though the process may not be using all of that memory. The memory space would also be 4GB, representing the total amount of physical memory available to the system. As the process runs, it may request additional memory, which would be temporarily stored on the hard drive in a file called the swap file. This allows the process to access more memory than is physically available, by temporarily swapping out parts of its own address space to make room for new data.

Question 3:

a. Explain the following addressing schemes with the help of an example of each :
i. Indexed Addressing
ii. Base Register Addressing
iii. Relative Addressing Scheme

Answer:

i. Indexed Addressing: In indexed addressing, an instruction includes an index value, which is added to a base address to form the final memory address. For example, consider a program with an array of integers called “numbers” starting at memory address 1000. To access the third element of the array, the instruction would use an indexed addressing mode with a base address of 1000 and an index value of 2 (since the array is 0-indexed). The final memory address would be 1000 + 2 = 1002, which is the memory location of the third element of the array.

ii. Base Register Addressing: In base register addressing, an instruction includes a register that contains the base address, and an offset value that is added to the value in the register to form the final memory address. For example, consider a program that uses a register called “base_register” to store the base address of an array of integers called “numbers”. To access the third element of the array, the instruction would use a base register addressing mode with an offset value of 2 (since the array is 0-indexed). The final memory address would be the contents of the base_register + 2.

iii. Relative Addressing: In relative addressing, an instruction includes a relative offset value that is added to the current instruction pointer to form the final memory address. For example, consider a program that has a instruction pointer register (IP) pointing to the current instruction being executed and a instruction that jumps to the next instruction + 10. The instruction would use a relative addressing mode with a relative offset value of 10. The final memory address would be the contents of the IP + 10.

b. Explain the concept of instruction pipelining with the help of a diagram.

Answer: Instruction pipelining is a technique used in computer architecture to improve the performance of processors by allowing multiple instructions to be in different stages of execution at the same time.

A pipeline is a series of stages, each of which performs a specific task on a instruction. The instruction is passed through the pipeline, with each stage performing its task before passing the instruction on to the next stage.

A simple diagram of an instruction pipeline might look something like this:

Fetch -> Decode -> Execute -> Memory -> Writeback

1. Fetch: In the fetch stage, the instruction is retrieved from memory and placed into the instruction register.
2. Decode: In the decode stage, the instruction is decoded and the necessary operands are identified.
3. Execute: In the execute stage, the instruction is executed, performing the operation specified by the instruction.
4. Memory: In the memory stage, the instruction may access memory to read or write data.
5. Writeback: In the writeback stage, the result of the instruction is written back to a register or memory location.

While one instruction is completing a stage, another instruction is entering another stage. This allows multiple instructions to be in various stages of execution at the same time, improving the overall performance of the processor.

This diagram is a simple representation of a pipeline, actual pipelines can be much more complex and have more stages.

c. Explain the following instructions of 8086 microprocessor :
i. CMP
ii. JMP
iii. RCL
iv. SHR

Answer:

i. CMP: CMP stands for Compare. The CMP instruction compares the value of a register or memory location with another value. It sets the flags in the flag register based on the result of the comparison. For example, CMP AX, BX compares the contents of the AX register with the contents of the BX register, and sets the flags in the flag register accordingly.

ii. JMP: JMP stands for Jump. The JMP instruction transfers control to a different location in the program. It can be used to jump to a specific memory address or to a label in the program. For example, JMP 100h transfers control to the instruction located at memory address 100h, while JMP start jumps to a label called start in the program.

iii. RCL: RCL stands for Rotate through Carry Left. The RCL instruction rotates the contents of a register or memory location to the left by a specified number of bits. It also takes the carry flag into account, so the leftmost bit of the result is the value of the carry flag before the rotate, and the carry flag after the rotate is the value of the bit that is shifted out. For example, RCL AX, 1 rotates the contents of the AX register to the left by 1 bit, taking the carry flag into account.

iv. SHR: SHR stands for Shift Right. The SHR instruction shifts the bits in a register or memory location to the right by a specified number of bits. The rightmost bits are filled with 0s. For example, SHR AX, 1 shifts the contents of the AX register to the right by 1 bit, filling the rightmost bit with 0.

These instructions are a subset of the instruction set of the 8086 microprocessor.

d. Explain the advantages of using segments in 8086 microprocessor.

Answer: Segments in the 8086 microprocessor are used to divide memory into logical groups to manage memory more efficiently. The advantages of using segments in the 8086 microprocessor include:

1. Protection: Segments allow the operating system to protect different parts of memory from unauthorized access. This is important for security and stability of the system.
2. Memory Management: Segments allow the operating system to manage memory more efficiently by allocating and deallocating memory as needed. This can help prevent fragmentation and improve overall system performance.
3. Relocation: Segments allow the operating system to move parts of memory around without affecting the programs that are using that memory. This is important for virtual memory systems, where memory may be swapped in and out of disk storage.
4. Sharing: Segments allow different programs to share the same memory space without interfering with each other. This is useful for running multiple programs at the same time.
5. Large Memory Space: The 8086 microprocessor supports up to 1MB of memory space by using segments, which is much larger than the 64KB of memory space that can be accessed without using segments. This allows for more memory and larger programs to be run on the system.
6. Portability: Programs written for 8086 microprocessor can be easily ported to other systems that support segments, as the memory management techniques are consistent across systems.

Overall, using segments in the 8086 microprocessor allows for more efficient use of memory and better protection of the system, which results in improved performance and stability of the system.

Question 4:

a. Draw the truth table of a 8 × 3 encoder. Also, write the expressions for the outputs in terms of inputs.

Answer: A 8×3 encoder is a digital circuit that has 8 input lines and 3 output lines. The truth table for an 8×3 encoder is shown below.

The output of the 8×3 encoder can be represented by the following Boolean expressions

OUT0 = IN0’IN1’IN2’IN3’IN4’IN5’IN6’IN7 OUT1 = IN0’IN1’IN2’IN3’IN4’IN5’IN6 + IN0’IN1’IN2’IN3’IN4’IN5’IN7 + IN0’IN1’IN2’IN3’IN4’IN6’IN7 + IN0’IN1’IN2’IN3’IN5’IN6’IN7 + IN0’IN1’IN2’IN4’IN

b. Explain the advantages of having densely packed integrated circuits.

Answer: Densely packed integrated circuits (ICs) refer to ICs that have a high number of transistors and other components in a small area. The advantages of having densely packed ICs include:

1. Increased Performance: Densely packed ICs can perform more complex operations and can process data at a faster rate than less densely packed ICs. This is because more transistors and other components can be packed into a smaller area, resulting in more processing power.
2. Reduced Size: Densely packed ICs are smaller in size than less densely packed ICs. This is important for applications where space is limited, such as mobile devices and other portable electronics.
3. Reduced Power Consumption: Densely packed ICs consume less power than less densely packed ICs. This is because the transistors are closer together, reducing the distance that the electrical signals need to travel. This results in less power required to operate the IC.
4. Increased Reliability: Densely packed ICs are more reliable than less densely packed ICs. This is because the transistors and other components are closer together, reducing the chances of damage caused by environmental factors such as heat and radiation.
5. Increased Integration: Densely packed ICs allows to integrate more functionality in a single IC than less densely packed ICs, which results in less space, less power consumption and lower cost.
6. Lower Cost: Densely packed ICs can be manufactured at a lower cost than less densely packed ICs. This is because more transistors and other components can be produced on a single wafer, which reduces the cost of production.

Overall, densely packed ICs offer a number of advantages, including increased performance, reduced size, reduced power consumption, increased reliability and lower cost. These advantages make them ideal for a wide range of applications, from mobile devices to high-performance computing.

c. What is an I/O interface in a computer ? List the functions of I/O interfaces.

Answer: An I/O interface (Input/Output interface) in a computer is a communication channel between the computer’s central processing unit (CPU) and the peripheral devices that it uses to input and output data. These peripheral devices include input devices (such as a keyboard, mouse, or scanner), output devices (such as a monitor or printer), and storage devices (such as a hard drive or flash drive).

The functions of an I/O interface are:

1. Data transfer: The I/O interface enables data transfer between the CPU and peripheral devices. The data can be transferred in both directions, from the CPU to the peripheral device (output) or from the peripheral device to the CPU (input).
2. Device control: The I/O interface allows the CPU to control the peripheral devices, such as setting the input/output modes, reading/writing data, and managing the device status.
3. Error detection: The I/O interface includes error detection mechanisms to identify and report any errors that occur during the data transfer or device control process.
4. Buffering: The I/O interface often includes buffers, which are temporary storage areas that can hold data while it is being transferred between the CPU and peripheral devices. Buffering can improve the performance and stability of the system.
5. Interrupt handling: The I/O interface supports the handling of interrupts, which are signals from peripheral devices that indicate that they need the CPU’s attention. Interrupts allow the CPU to respond quickly to input from peripheral devices, such as a keyboard or mouse.
6. Plug and Play: The I/O interface supports the plug and play capability, which allows the system to automatically detect and configure new peripheral devices as they are connected to the system.

In summary, the I/O interface plays a crucial role in the communication between the CPU and peripheral devices, enabling data transfer, device control, error detection, buffering and interrupt handling, and Plug and Play support.

d. Explain the features and uses of the following I/O devices :
i. DVD-ROM
ii. Printer
iii. Scanner

Answer:

i. DVD-ROM: A DVD-ROM (Digital Versatile Disc – Read Only Memory) is an optical disc storage device that is used to read data from DVDs. It is similar to a CD-ROM, but can store more data and has higher data transfer rates. Some of the features of a DVD-ROM include:

• High storage capacity: DVD-ROMs can store up to 17GB of data, which is much more than a CD-ROM.
• High data transfer rate: DVD-ROMs have a higher data transfer rate than CD-ROMs, which allows for faster data access.
• Versatility: DVD-ROMs can be used to store a variety of types of data, such as video, audio, and software.
• Multi-session support: DVD-ROMs support multiple sessions, which allows for multiple data sets to be recorded on the same disc.

Uses:

• DVD-ROMs are commonly used to store and distribute movies, music, and software.
• They are also used as backup storage media.
• In the educational field, it can be used to store video lectures, educational software, and multimedia materials.

ii. Printer: A printer is an output device that is used to print text and images on paper or other materials. Some of the features of a printer include:

• High-quality output: Printers can produce high-quality text and images that are similar to those produced by professional print shops.
• Speed: Some printers can print very quickly, allowing for large numbers of pages to be printed in a short amount of time.
• Connectivity: Many printers can be connected to computers and other devices via USB, Ethernet, or wireless connections.

Uses:

• Printers are commonly used to print documents, reports, and other text-based materials.
• They are also used to print photographs and other types of images.
• In many offices and workplaces, printers are used to print letters, memos, reports, and other documents that are needed on a daily basis.

iii. Scanner: A scanner is an input device that is used to convert analog images, such as photographs, into digital images. Some of the features of a scanner include:

• High resolution: Scanners can produce high-resolution digital images that are similar to the original.
• Color depth: Scanners can produce digital images with a wide range of colors, which allows for accurate reproductions of the original.
• Connectivity: Many scanners can be connected to computers and other devices via USB or other types of connections.

Uses:

• Scanners are commonly used to digitize photographs, documents, and other types of images.
• They are also used to create digital copies of drawings, maps, and other types of graphics.
• In many offices and workplaces, scanners are used to create

Question 5:
a. What is an Interrupt Vector Table (IVT) for a 8086 microprocessor ? Explain with the help of a diagram, how interrupts are processed using IVT.

The Interrupt Vector Table (IVT) is a memory area in the 8086 microprocessor that contains a list of memory addresses for handling different types of interrupts. When an interrupt occurs, the microprocessor looks up the appropriate memory address in the IVT and transfers control to the corresponding interrupt service routine (ISR) to handle the interrupt.

The IVT is a fixed memory area, typically located at the beginning of memory, that contains 256 memory addresses, each corresponding to a different interrupt type. The first address in the IVT is the address for interrupt 0, the second address is for interrupt 1, and so on, up to interrupt 255.

The following diagram illustrates how interrupts are processed using the IVT in a 8086 microprocessor:

1. Interrupt occurs: An interrupt is triggered by a peripheral device, such as a keyboard or timer.
2. Interrupt signal: The peripheral device sends an interrupt signal to the 8086 microprocessor, which signals that an interrupt has occurred.
3. Interrupt vector: The microprocessor looks up the appropriate memory address in the IVT based on the interrupt number received from the peripheral device.
4. Interrupt service routine: The microprocessor transfers control to the corresponding ISR located at the memory address retrieved from the IVT.
5. ISR execution: The ISR handles the interrupt and performs the appropriate actions, such as reading input data from the peripheral device or sending output data to the device.
6. Interrupt return: After the ISR has finished executing, the microprocessor returns control to the main program, and continues execution where it left off before the interrupt occurred.

In summary, the Interrupt Vector Table (IVT) is a memory area in the 8086 microprocessor that contains a list of memory addresses for handling different types of interrupts. When an interrupt occurs, the microprocessor looks up the appropriate memory address in the IVT and transfers control to the corresponding interrupt service routine (ISR) to handle the interrupt.

b. What is the role of flag register in 8086 microprocessor ? Explain the use of
i. overflow flag

ii. string direction flag,
iii. parity flag in 8086 microprocessor.

Answer: The flag register in the 8086 microprocessor is a special register that is used to store the current status of the microprocessor. It contains a set of flags that indicate the results of various operations and can be used to make decisions in the program.

i. Overflow flag: The overflow flag is used to indicate when an arithmetic operation produces a result that is too large or too small to be represented in the destination register. For example, when an addition operation results in a carry out from the most significant bit, the overflow flag is set.

ii. String direction flag: The string direction flag (also known as the direction flag) is used to indicate the direction of string operations (such as MOVS, CMPS, etc). When the direction flag is set, string operations will be performed from high memory addresses to low memory addresses, while when it is clear, operations will be performed from low memory addresses to high memory addresses.

iii. Parity flag: The parity flag is used to indicate the parity of a result of an operation. Parity is a method of error checking where the number of 1s in a binary number is counted. If the number of 1s is even, the parity is set to 0, and if the number of 1s is odd, the parity is set to 1.

In summary, the flag register in the 8086 microprocessor is a special register that is used to store the current status of the microprocessor. It contains a set of flags.

c. Explain the working of Wilkes control unit with the help of a diagram.

Answer: The Wilkes control unit is a type of control unit design proposed by Maurice Wilkes in the 1950s, it is considered one of the earliest designs of control units. The control unit is responsible for fetching instructions from memory, decoding them, and executing them.

The Wilkes control unit consists of several components, including a program counter (PC), an instruction register (IR), and a control memory. The following is a diagram of the working of the Wilkes control unit:

1. Program Counter: The program counter (PC) holds the memory address of the next instruction to be executed.
2. Fetch: The control unit fetches the instruction from memory using the address stored in the PC, and loads it into the instruction register (IR).
3. Decode: The instruction in the IR is then decoded by the control unit. This involves determining the operation to be performed, and the operands (data) required for the operation.
4. Execute: Once the instruction has been decoded, the control unit sends the necessary control signals to the appropriate functional units to execute the instruction.
5. Update PC: After the instruction has been executed, the control unit updates the PC to point to the next instruction to be executed.
6. Repeat: The control unit continues to fetch, decode, and execute instructions in a loop until the program completes or a termination condition is met.

The Wilkes control unit is a simple, elegant design that is easy to understand and implement. However, it has several limitations such as the lack of microprogramming and the need for a separate control memory, which makes it less flexible and less efficient

d. List any five characteristics of RISC machines.

Answer: RISC (Reduced Instruction Set Computer) is a type of computer architecture that emphasizes simplicity, speed, and efficiency. Some of the characteristics of RISC machines include:

1. Reduced instruction set: RISC machines have a smaller number of instructions compared to CISC (Complex Instruction Set Computer) machines. This simplifies the instruction set and makes it easier to implement.
2. Large register file: RISC machines have a large register file, which allows for more data to be stored and processed within the processor, reducing the need to access memory.
3. Simple instruction format: RISC instructions are typically encoded in a simple format, with all operands specified in registers, which eliminates the need for memory-to-memory operations.
4. Load-store architecture: RISC machines use a load-store architecture, where instructions that operate on memory are separated from instructions that operate on registers. This simplifies the instruction decoding and execution process.
5. Pipelining: RISC machines often use pipelining, which allows multiple instructions to be executed simultaneously. This improves the overall performance of the processor.
6. RISC machines are designed to execute a smaller number of simple instructions at a faster rate than CISC machines. This design allows for a more efficient use of transistors and die space which results in less power consumption.

In summary, RISC machines have a reduced instruction set, large register file, simple instruction.