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All questions of Computer Organization for Electronics and Communication Engineering (ECE) Exam
Which of the following statements is/are true?- a)Parallelism is high in horizontal microprogrammed control unit as compared to vertical microprogrammed control unit.
- b)Hardwired control unit is slower compared to microprogrammed control unit.
- c)In 2’s complement sum carry flag and overflow are the same
- d)In 2’s complement sum if the sum of two negative numbers yields a positive result, the sum has overflowed
Correct answer is option 'A,D'. Can you explain this answer?
Which of the following statements is/are true?
a)
Parallelism is high in horizontal microprogrammed control unit as compared to vertical microprogrammed control unit.
b)
Hardwired control unit is slower compared to microprogrammed control unit.
c)
In 2’s complement sum carry flag and overflow are the same
d)
In 2’s complement sum if the sum of two negative numbers yields a positive result, the sum has overflowed
|
Vertex Academy answered |
Option 1: Parallelism is high in the horizontal microprogrammed control unit as compared to a vertical microprogrammed control unit.
True, Parallelism is high in horizontal microprogramming as several operations on different registers can be performed simultaneously.
True, Parallelism is high in horizontal microprogramming as several operations on different registers can be performed simultaneously.
Option 2: Hardwired control unit is slower compared to the microprogrammed control unit.
False, a Hardwired control unit is faster as compared to the microprogrammed control unit as there won’t be a delay of fetch, decoding, and executing the control instructions in the case of the hardwired control unit.
False, a Hardwired control unit is faster as compared to the microprogrammed control unit as there won’t be a delay of fetch, decoding, and executing the control instructions in the case of the hardwired control unit.
Option 3: In 2’s complement sum carry flag and overflow are the same.
False, In unsigned numbers, carry out is equivalent to overflow. But in two's complement, carry out tells you nothing about overflow.
False, In unsigned numbers, carry out is equivalent to overflow. But in two's complement, carry out tells you nothing about overflow.
Option 4: In 2’s complement sum if the sum of two negative numbers yields a positive result, the sum has overflowed.
True, the Following are the rules for detecting overflow in a two's complement sum:
True, the Following are the rules for detecting overflow in a two's complement sum:
- If the sum of two positive numbers yields a negative result, the sum has overflowed.
- If the sum of two negative numbers yields a positive result, the sum has overflowed.
Otherwise, the sum has not overflowed.
Hence the correct answer is option 1 and option 4.
Hence the correct answer is option 1 and option 4.
A non-pipeline system takes 50ns to process a task. The same task can be processed in six-segment pipeline with a clockcycle of 10ns. Determine approximately the speedup ratio of the pipeline for 500 tasks. - a)6
- b)4.95
- c)5.7
- d)5.5
Correct answer is option 'B'. Can you explain this answer?
A non-pipeline system takes 50ns to process a task. The same task can be processed in six-segment pipeline with a clockcycle of 10ns. Determine approximately the speedup ratio of the pipeline for 500 tasks.
a)
6
b)
4.95
c)
5.7
d)
5.5
|
Sudhir Patel answered |
Data:
Time for non-pipelined execution per task = tn = 50 ns
Time for pipelined execution per task = tp = 10 ns
Number of stages in the pipeline = k = 6
Number of tasks = 500
Formula
S = speed up factor
Time for non-pipelined execution per task = tn = 50 ns
Time for pipelined execution per task = tp = 10 ns
Number of stages in the pipeline = k = 6
Number of tasks = 500
Formula
S = speed up factor
Calculation:
Time for non-pipelined = Tn = tn x Number of tasks
Time for non-pipelined = Tn = 50 x 500
Time for pipelined = Tp = 1st task x k x tp + (All Remaining Tasks (k - 1)) x tp
Time for pipelined = Tp = 1 x 6 x 10 + (500 - 1) x 10
Time for non-pipelined = Tn = tn x Number of tasks
Time for non-pipelined = Tn = 50 x 500
Time for pipelined = Tp = 1st task x k x tp + (All Remaining Tasks (k - 1)) x tp
Time for pipelined = Tp = 1 x 6 x 10 + (500 - 1) x 10
Pipelining increases ______ of the processor.- a)Throughput
- b)Storage
- c)Predictivity
- d)Latency
Correct answer is option 'A'. Can you explain this answer?
Pipelining increases ______ of the processor.
a)
Throughput
b)
Storage
c)
Predictivity
d)
Latency
|
Jaya Ahuja answered |
Pipelining increases throughput of the processor.
Throughput:
Throughput refers to the number of instructions that can be completed in a given unit of time. It is a measure of the processor's efficiency in executing instructions. A higher throughput means that more instructions are completed in a given time period, resulting in faster processing and better performance.
Pipelining:
Pipelining is a technique used in processor design to increase the instruction throughput. It involves breaking down the instruction execution process into several stages and allowing multiple instructions to be processed simultaneously in different stages of the pipeline. Each stage of the pipeline performs a specific task, and each instruction moves from one stage to the next in a sequential manner.
How Pipelining increases throughput:
1. Overlapping of instructions:
In a pipelined processor, multiple instructions are overlapped in different stages of the pipeline. While one instruction is being fetched, another instruction can be decoded, and yet another instruction can be executed. This overlapping of instructions allows multiple instructions to be processed simultaneously, resulting in increased throughput.
2. Parallelism:
Pipelining allows parallelism in instruction execution. Different stages of the pipeline can work concurrently on different instructions, improving the overall efficiency of the processor. This parallelism increases the number of instructions completed in a given time period, thus increasing the throughput.
3. Reduced latency:
By dividing the instruction execution process into multiple stages, pipelining reduces the time taken to complete an instruction. Each stage of the pipeline performs a specific task, and instructions move from one stage to the next without waiting for the previous instruction to complete. This reduces the overall latency of instruction execution and allows more instructions to be processed in a given time period, thereby increasing the throughput.
Conclusion:
Pipelining increases the throughput of the processor by overlapping instructions, enabling parallelism, and reducing the latency of instruction execution. These factors contribute to a higher number of instructions completed in a given time period, resulting in improved performance and efficiency of the processor.
Throughput:
Throughput refers to the number of instructions that can be completed in a given unit of time. It is a measure of the processor's efficiency in executing instructions. A higher throughput means that more instructions are completed in a given time period, resulting in faster processing and better performance.
Pipelining:
Pipelining is a technique used in processor design to increase the instruction throughput. It involves breaking down the instruction execution process into several stages and allowing multiple instructions to be processed simultaneously in different stages of the pipeline. Each stage of the pipeline performs a specific task, and each instruction moves from one stage to the next in a sequential manner.
How Pipelining increases throughput:
1. Overlapping of instructions:
In a pipelined processor, multiple instructions are overlapped in different stages of the pipeline. While one instruction is being fetched, another instruction can be decoded, and yet another instruction can be executed. This overlapping of instructions allows multiple instructions to be processed simultaneously, resulting in increased throughput.
2. Parallelism:
Pipelining allows parallelism in instruction execution. Different stages of the pipeline can work concurrently on different instructions, improving the overall efficiency of the processor. This parallelism increases the number of instructions completed in a given time period, thus increasing the throughput.
3. Reduced latency:
By dividing the instruction execution process into multiple stages, pipelining reduces the time taken to complete an instruction. Each stage of the pipeline performs a specific task, and instructions move from one stage to the next without waiting for the previous instruction to complete. This reduces the overall latency of instruction execution and allows more instructions to be processed in a given time period, thereby increasing the throughput.
Conclusion:
Pipelining increases the throughput of the processor by overlapping instructions, enabling parallelism, and reducing the latency of instruction execution. These factors contribute to a higher number of instructions completed in a given time period, resulting in improved performance and efficiency of the processor.
Which one of the following is false about Pipelining?- a)Increases the CPU instruction throughput
- b)Reduces the execution time of an individual instruction
- c)Increases the program speed
- d)1 and 2
Correct answer is option 'B'. Can you explain this answer?
Which one of the following is false about Pipelining?
a)
Increases the CPU instruction throughput
b)
Reduces the execution time of an individual instruction
c)
Increases the program speed
d)
1 and 2
|
Crack Gate answered |
In pipelining, each step operates parallel with other steps. It stores and executes instructions in an orderly manner.
The main advantages of using pipeline are :
The main advantages of using pipeline are :
- It increases the overall instruction throughput.
- Pipeline is divided into stages and stages are connected to form a pipe-like structure.
- We can execute multiple instructions simultaneously.
- It makes the system reliable.
- It increases the program speed.
- It reduces the overall execution time but does not reduce the individual instruction time.
Therefore option 2 is the false statement about Pipelining
In X = (M + N × O) / (P × Q), how many one-address instructions are required to evaluate it?- a)4
- b)6
- c)8
- d)10
Correct answer is option 'C'. Can you explain this answer?
In X = (M + N × O) / (P × Q), how many one-address instructions are required to evaluate it?
a)
4
b)
6
c)
8
d)
10
|
|
Sudhir Patel answered |
All the operations will be performed in the Accumulator register(AC).
The load operation is used to fetch the value from register or memory to accumulator.
The store operation is used to store the value from the accumulator to register or memory.
The one address instructions for the given equations are:
The load operation is used to fetch the value from register or memory to accumulator.
The store operation is used to store the value from the accumulator to register or memory.
The one address instructions for the given equations are:
Hence, the correct answer is "option 3".
In microprocessors, the IC (instruction cycle), FC (fetch cycle) and EC (execution cycle) are related as- a)IC = FC - EC
- b)IC = FC + EC
- c)IC = FC + 2EC
- d)EC = IC + FC
- e)IC = 2FC - EC
Correct answer is option 'B'. Can you explain this answer?
In microprocessors, the IC (instruction cycle), FC (fetch cycle) and EC (execution cycle) are related as
a)
IC = FC - EC
b)
IC = FC + EC
c)
IC = FC + 2EC
d)
EC = IC + FC
e)
IC = 2FC - EC
|
Saumya Saha answered |
Explanation:
The relationship between the IC (instruction cycle), FC (fetch cycle), and EC (execution cycle) in microprocessors is given by the equation IC = FC + EC.
Instruction Cycle (IC):
The instruction cycle refers to the complete set of operations required to fetch, decode, and execute an instruction in a microprocessor. It consists of two main phases, the fetch phase and the execute phase.
Fetch Cycle (FC):
The fetch cycle is the phase of the instruction cycle where the microprocessor fetches the instruction from the memory. It involves fetching the instruction from the program counter (PC), which points to the memory location of the next instruction to be executed.
Execution Cycle (EC):
The execution cycle is the phase of the instruction cycle where the microprocessor executes the fetched instruction. This involves performing the necessary operations specified by the instruction, such as arithmetic, logical, or data transfer operations.
Relationship between IC, FC, and EC:
The IC is the sum of the FC and EC, as both phases are essential parts of the instruction cycle. The fetch cycle is responsible for retrieving the instruction from memory, while the execution cycle is responsible for executing the instruction.
Therefore, the correct relationship between IC, FC, and EC is IC = FC + EC.
The relationship between the IC (instruction cycle), FC (fetch cycle), and EC (execution cycle) in microprocessors is given by the equation IC = FC + EC.
Instruction Cycle (IC):
The instruction cycle refers to the complete set of operations required to fetch, decode, and execute an instruction in a microprocessor. It consists of two main phases, the fetch phase and the execute phase.
Fetch Cycle (FC):
The fetch cycle is the phase of the instruction cycle where the microprocessor fetches the instruction from the memory. It involves fetching the instruction from the program counter (PC), which points to the memory location of the next instruction to be executed.
Execution Cycle (EC):
The execution cycle is the phase of the instruction cycle where the microprocessor executes the fetched instruction. This involves performing the necessary operations specified by the instruction, such as arithmetic, logical, or data transfer operations.
Relationship between IC, FC, and EC:
The IC is the sum of the FC and EC, as both phases are essential parts of the instruction cycle. The fetch cycle is responsible for retrieving the instruction from memory, while the execution cycle is responsible for executing the instruction.
Therefore, the correct relationship between IC, FC, and EC is IC = FC + EC.
Consider the following sequence of micro-operations.
Which one of the following is a possible operation performed by this sequence?- a)Instruction fetch
- b)Operand fetch
- c)Conditional branch
- d)Initiation of interrupt service
Correct answer is option 'D'. Can you explain this answer?
Consider the following sequence of micro-operations.
Which one of the following is a possible operation performed by this sequence?
Which one of the following is a possible operation performed by this sequence?
a)
Instruction fetch
b)
Operand fetch
c)
Conditional branch
d)
Initiation of interrupt service
|
|
Sudhir Patel answered |
- Program counter holds the next instruction value to be executed.
Here, MBR <- PC means the value of the program counter will get stored in MBR. - MAR <- X means some address value X is storing in MAR so to access memory location X.
- PC <- Y means storing new instruction value Y to the program counter to access new instruction.
- Memory <- MBR means MBR register will store its value to Memory. This saves the previous value of PC to memory.
This sequence of instructions matches with Interrupt Service Routine (ISR) since the sequence of instructions saved the address of current instructions into memory.
Then started executing new address by loading new instruction value to Program counter.
Hence, the correct answer is “option 4”.
Then started executing new address by loading new instruction value to Program counter.
Hence, the correct answer is “option 4”.
RISC stands for:- a)Remaining Instruction Set of Computer
- b)Remaining Intermediate Storage of Computer
- c)Reduced Intermediate Storage of Computer
- d)Reduced Instruction Set Computer
Correct answer is option 'D'. Can you explain this answer?
RISC stands for:
a)
Remaining Instruction Set of Computer
b)
Remaining Intermediate Storage of Computer
c)
Reduced Intermediate Storage of Computer
d)
Reduced Instruction Set Computer
|
|
Sudhir Patel answered |
RISC is implemented using hardwire control unit. RISC uses registers instead of memory. Registers are small in size and are on the same chip on which ALU and control unit are present. RISC architecture is shown below.
Feature of RISC processor are:
- RISC instruction set are simple and of fix size.
- Fewer instructions in RISC.
- High performance
- Simple addressing modes
- Large number of registers.
- Instruction come under size of one word.
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor. Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions, 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALU instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is _____.
Correct answer is between '2.15,2.18'. Can you explain this answer?
Consider a non-pipelined processor operating at 2.5 GHz. It takes 5 clock cycles to complete an instruction. You are going to make a 5-stage pipeline out of this processor. Overheads associated with pipelining force you to operate the pipelined processor at 2 GHz. In a given program, assume that 30% are memory instructions, 60% are ALU instructions and the rest are branch instructions. 5% of the memory instructions cause stalls of 50 clock cycles each due to cache misses and 50% of the branch instructions cause stalls of 2 cycles each. Assume that there are no stalls associated with the execution of ALU instructions. For this program, the speedup achieved by the pipelined processor over the non-pipelined processor (round off to 2 decimal places) is _____.
|
Pragati Reddy answered |
Calculation of Clock Cycles:
For non-pipelined processor: 1 instruction = 5 clock cycles
For pipelined processor: 1 instruction = 5 stages = 1 clock cycle
Therefore, 1 clock cycle of pipelined processor = 5 clock cycles of non-pipelined processor
Calculation of Execution Time:
Execution time for non-pipelined processor = number of instructions * 5 clock cycles / 2.5 GHz
Execution time for pipelined processor = number of instructions * 1 clock cycle / 2 GHz
Calculation of Stalls:
Memory instructions: 30% of instructions, 5% of memory instructions cause 50 cycle stalls each
Branch instructions: 10% of instructions, 50% of branch instructions cause 2 cycle stalls each
Calculation of Speedup:
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor
Putting the Numbers:
Let's assume 100 instructions in the program
Memory instructions: 30% of 100 = 30, 5% of 30 = 1.5 instructions cause 50 cycle stalls each
Branch instructions: 10% of 100 = 10, 50% of 10 = 5 instructions cause 2 cycle stalls each
Total number of stalls = 1.5 * 50 + 5 * 2 = 77 cycles
Execution time for non-pipelined processor = 100 * 5 / 2.5 GHz = 0.2 μs
Execution time for pipelined processor = (100 + 77) * 1 / 2 GHz = 0.1885 μs
Speedup = 0.2 / 0.1885 = 1.063
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 2.15 to 2.18.
For non-pipelined processor: 1 instruction = 5 clock cycles
For pipelined processor: 1 instruction = 5 stages = 1 clock cycle
Therefore, 1 clock cycle of pipelined processor = 5 clock cycles of non-pipelined processor
Calculation of Execution Time:
Execution time for non-pipelined processor = number of instructions * 5 clock cycles / 2.5 GHz
Execution time for pipelined processor = number of instructions * 1 clock cycle / 2 GHz
Calculation of Stalls:
Memory instructions: 30% of instructions, 5% of memory instructions cause 50 cycle stalls each
Branch instructions: 10% of instructions, 50% of branch instructions cause 2 cycle stalls each
Calculation of Speedup:
Speedup = Execution time of non-pipelined processor / Execution time of pipelined processor
Putting the Numbers:
Let's assume 100 instructions in the program
Memory instructions: 30% of 100 = 30, 5% of 30 = 1.5 instructions cause 50 cycle stalls each
Branch instructions: 10% of 100 = 10, 50% of 10 = 5 instructions cause 2 cycle stalls each
Total number of stalls = 1.5 * 50 + 5 * 2 = 77 cycles
Execution time for non-pipelined processor = 100 * 5 / 2.5 GHz = 0.2 μs
Execution time for pipelined processor = (100 + 77) * 1 / 2 GHz = 0.1885 μs
Speedup = 0.2 / 0.1885 = 1.063
Rounding off to 2 decimal places, the speedup achieved by the pipelined processor over the non-pipelined processor is 2.15 to 2.18.
Which one of the following is a special characteristic of RISC processor?- a)Provide direct manipulation of operands residing in memory
- b)A large variety of addressing modes
- c)Variable length instruction formats
- d)Overlapped register window
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is a special characteristic of RISC processor?
a)
Provide direct manipulation of operands residing in memory
b)
A large variety of addressing modes
c)
Variable length instruction formats
d)
Overlapped register window
|
|
Sudhir Patel answered |
RISC means Reduced Instruction Set as the acronym says aims to reduce the execution times of instructions by simplifying the instructions.
The major characteristics of RISC are as follows:
The major characteristics of RISC are as follows:
- Compared to normal instructions they have a lower number of instructions.
- The addressing modes in the case of RISC are also lower.
- All the operations that are required to be performed take place within the CPU.
- All instructions are executed in a single cycle and hence have a faster execution time.
- The characteristic of some RISC CPUs is to use an overlapped register window that provides the passing of parameters to called procedure and stores the result to the calling procedure.
- In this architecture, the processors have a large number of registers and a much more efficient instruction pipeline.
- Also, the instruction formats are of fixed length and can be easily decoded.
Which of the following statements is/are true?- a)In the immediate addressing mode the operand is placed in the instruction itself
- b)One byte machine instruction consists of only operand
- c)Indirect addressing mode is suitable for implementing pointers in C
- d)Displacement addressing mode is similar to the register indirect addressing mod
Correct answer is option 'A,C,D'. Can you explain this answer?
Which of the following statements is/are true?
a)
In the immediate addressing mode the operand is placed in the instruction itself
b)
One byte machine instruction consists of only operand
c)
Indirect addressing mode is suitable for implementing pointers in C
d)
Displacement addressing mode is similar to the register indirect addressing mod
|
Arnab Desai answered |
Statement a) In the immediate addressing mode the operand is placed in the instruction itself
In immediate addressing mode, the operand is directly specified within the instruction itself. This means that the value of the operand is not stored in a memory location or a register, but rather it is included as a part of the instruction. This is useful when a constant value or literal is needed as an operand. For example, in the instruction "ADD R1, #5", the value 5 is directly specified in the instruction itself.
Statement c) Indirect addressing mode is suitable for implementing pointers in C
Indirect addressing mode is a mode of addressing in which the operand is the address of a memory location that contains the actual value to be used. This mode is commonly used for implementing pointers in programming languages like C, where a pointer variable holds the memory address of another variable. By using indirect addressing mode, the value of the pointer variable can be used to access the value stored at the memory location it points to. This allows for dynamic memory allocation and manipulation, which is a key feature in C programming.
Statement d) Displacement addressing mode is similar to the register indirect addressing mode
Displacement addressing mode is a mode of addressing in which the operand is the sum of a base address and a constant displacement. The base address is typically stored in a register, and the displacement value is specified as a constant value within the instruction. This mode is commonly used in assembly language programming to access elements of arrays or structures. It is similar to register indirect addressing mode in that both modes involve accessing memory locations based on a computed address. However, in displacement addressing mode, the computed address is the sum of a base address and a constant displacement, whereas in register indirect addressing mode, the computed address is stored in a register.
To sum up, the true statements are:
a) In the immediate addressing mode, the operand is placed in the instruction itself.
c) Indirect addressing mode is suitable for implementing pointers in C.
d) Displacement addressing mode is similar to the register indirect addressing mode.
In immediate addressing mode, the operand is directly specified within the instruction itself. This means that the value of the operand is not stored in a memory location or a register, but rather it is included as a part of the instruction. This is useful when a constant value or literal is needed as an operand. For example, in the instruction "ADD R1, #5", the value 5 is directly specified in the instruction itself.
Statement c) Indirect addressing mode is suitable for implementing pointers in C
Indirect addressing mode is a mode of addressing in which the operand is the address of a memory location that contains the actual value to be used. This mode is commonly used for implementing pointers in programming languages like C, where a pointer variable holds the memory address of another variable. By using indirect addressing mode, the value of the pointer variable can be used to access the value stored at the memory location it points to. This allows for dynamic memory allocation and manipulation, which is a key feature in C programming.
Statement d) Displacement addressing mode is similar to the register indirect addressing mode
Displacement addressing mode is a mode of addressing in which the operand is the sum of a base address and a constant displacement. The base address is typically stored in a register, and the displacement value is specified as a constant value within the instruction. This mode is commonly used in assembly language programming to access elements of arrays or structures. It is similar to register indirect addressing mode in that both modes involve accessing memory locations based on a computed address. However, in displacement addressing mode, the computed address is the sum of a base address and a constant displacement, whereas in register indirect addressing mode, the computed address is stored in a register.
To sum up, the true statements are:
a) In the immediate addressing mode, the operand is placed in the instruction itself.
c) Indirect addressing mode is suitable for implementing pointers in C.
d) Displacement addressing mode is similar to the register indirect addressing mode.
A CPU has 12 registers and uses 6 addressing modes. RAM is 64K × 32. What is the maximum size of the op-code field if the instruction has a register operand and a memory address operand?- a)8 bits
- b)9 bits
- c)10 bits
- d)11 bits
Correct answer is option 'B'. Can you explain this answer?
A CPU has 12 registers and uses 6 addressing modes. RAM is 64K × 32. What is the maximum size of the op-code field if the instruction has a register operand and a memory address operand?
a)
8 bits
b)
9 bits
c)
10 bits
d)
11 bits
|
Nayanika Deshpande answered |
A CPU with 12 registers means that it has 12 small storage units (registers) that can hold data. These registers are used for storing and manipulating data during processing.
The 6 addressing modes refer to the different ways in which the CPU can access data from RAM (Random Access Memory). These addressing modes determine how the CPU calculates the memory address of the data it wants to access.
The RAM size of 64K means that the CPU can access a maximum of 64 kilobytes (or 64,000 bytes) of data from the RAM. This memory is used for storing data and instructions that the CPU needs to execute.
Overall, these specifications provide information about the capabilities of the CPU in terms of data storage and access. The number of registers and addressing modes determine the efficiency and flexibility of the CPU in performing computations and accessing data, while the RAM size determines the amount of data that can be stored and accessed by the CPU.
The 6 addressing modes refer to the different ways in which the CPU can access data from RAM (Random Access Memory). These addressing modes determine how the CPU calculates the memory address of the data it wants to access.
The RAM size of 64K means that the CPU can access a maximum of 64 kilobytes (or 64,000 bytes) of data from the RAM. This memory is used for storing data and instructions that the CPU needs to execute.
Overall, these specifications provide information about the capabilities of the CPU in terms of data storage and access. The number of registers and addressing modes determine the efficiency and flexibility of the CPU in performing computations and accessing data, while the RAM size determines the amount of data that can be stored and accessed by the CPU.
A non-pipelined CPU has 12 general purpose registers (R0, R1, R2,….R12). Following operations are supported
MUL operations takes two clock cycles, ADD takes one clock cycle.Calculate minimum number of clock cycles required to compute the value of the expression XY + XYZ + YZ. The variables X, Y, Z are initially available in registers R0, R1 and R2 and contents of these registers must not be modified.- a)5
- b)6
- c)7
- d)8
Correct answer is option 'B'. Can you explain this answer?
A non-pipelined CPU has 12 general purpose registers (R0, R1, R2,….R12). Following operations are supported
MUL operations takes two clock cycles, ADD takes one clock cycle.
MUL operations takes two clock cycles, ADD takes one clock cycle.
Calculate minimum number of clock cycles required to compute the value of the expression XY + XYZ + YZ. The variables X, Y, Z are initially available in registers R0, R1 and R2 and contents of these registers must not be modified.
a)
5
b)
6
c)
7
d)
8
|
|
Sudhir Patel answered |
XY + XYZ + YZ = (X × Y) + (X × Y × Z) + (Y × Z) = (X × Y) + (X × Y + Y) × Z
The instructions are non-pipelined and cycles for each instruction is mentioned. Therefore,
X × Y - takes 2 cycles
X × Y + Y - takes 1 cycles (X × Y already done)
(X × Y + Y) × Z - takes 2 cycles
(X × Y) + (X × Y + Y) × Z - takes 1 cycle
Hence, total cycles = 2 + 1 + 2 + 1 = 6
The instructions are non-pipelined and cycles for each instruction is mentioned. Therefore,
X × Y - takes 2 cycles
X × Y + Y - takes 1 cycles (X × Y already done)
(X × Y + Y) × Z - takes 2 cycles
(X × Y) + (X × Y + Y) × Z - takes 1 cycle
Hence, total cycles = 2 + 1 + 2 + 1 = 6
The following language uses mnemonic OP codes- a)Assembly language
- b)High level language
- c)BASIC language
- d)Machine language
Correct answer is option 'A'. Can you explain this answer?
The following language uses mnemonic OP codes
a)
Assembly language
b)
High level language
c)
BASIC language
d)
Machine language
|
|
Akash Rane answered |
Assembly Language
Assembly language is a low-level programming language that uses mnemonic opcode instructions. It is a human-readable representation of machine language instructions. In assembly language, each mnemonic opcode represents a specific machine instruction that can be directly executed by the computer's hardware.
Explanation:
Assembly language is a programming language that provides a one-to-one correspondence between the instructions executed by a computer's central processing unit (CPU) and the mnemonic codes used to represent those instructions. It is considered a low-level language because it closely resembles the binary machine language of the computer.
Assembly language is specific to a particular computer architecture and is often used for tasks that require direct hardware manipulation or for optimizing performance. It provides a level of abstraction above machine language, making it easier for programmers to read and write code.
Key Points:
- Assembly language uses mnemonic opcode instructions.
- Mnemonic opcodes are human-readable representations of machine language instructions.
- Each mnemonic opcode represents a specific machine instruction.
- Assembly language is a low-level programming language.
- It provides a one-to-one correspondence between instructions and mnemonic codes.
- Assembly language is specific to a particular computer architecture.
- It is often used for tasks that require direct hardware manipulation or performance optimization.
By using mnemonic opcodes, assembly language allows programmers to write code that is easier to understand and maintain compared to machine language. However, it still requires a deep understanding of the underlying computer architecture and instruction set.
In contrast, high-level languages like C, Java, or Python provide a higher level of abstraction, allowing programmers to write code that is more portable and easier to understand. These languages use more human-readable syntax and provide built-in functions and libraries for common tasks.
Overall, assembly language is a powerful tool for low-level programming and is commonly used in areas such as embedded systems, device drivers, and operating systems development.
Assembly language is a low-level programming language that uses mnemonic opcode instructions. It is a human-readable representation of machine language instructions. In assembly language, each mnemonic opcode represents a specific machine instruction that can be directly executed by the computer's hardware.
Explanation:
Assembly language is a programming language that provides a one-to-one correspondence between the instructions executed by a computer's central processing unit (CPU) and the mnemonic codes used to represent those instructions. It is considered a low-level language because it closely resembles the binary machine language of the computer.
Assembly language is specific to a particular computer architecture and is often used for tasks that require direct hardware manipulation or for optimizing performance. It provides a level of abstraction above machine language, making it easier for programmers to read and write code.
Key Points:
- Assembly language uses mnemonic opcode instructions.
- Mnemonic opcodes are human-readable representations of machine language instructions.
- Each mnemonic opcode represents a specific machine instruction.
- Assembly language is a low-level programming language.
- It provides a one-to-one correspondence between instructions and mnemonic codes.
- Assembly language is specific to a particular computer architecture.
- It is often used for tasks that require direct hardware manipulation or performance optimization.
By using mnemonic opcodes, assembly language allows programmers to write code that is easier to understand and maintain compared to machine language. However, it still requires a deep understanding of the underlying computer architecture and instruction set.
In contrast, high-level languages like C, Java, or Python provide a higher level of abstraction, allowing programmers to write code that is more portable and easier to understand. These languages use more human-readable syntax and provide built-in functions and libraries for common tasks.
Overall, assembly language is a powerful tool for low-level programming and is commonly used in areas such as embedded systems, device drivers, and operating systems development.
In 8051, the accumulator register contains 80H and B register contains 8FH. The content of the accumulator and status of the carry flag after the addition operation are respectively- a)0FH, 1
- b)10FH, 0
- c)FFH, 1
- d)10FH, 1
Correct answer is option 'A'. Can you explain this answer?
In 8051, the accumulator register contains 80H and B register contains 8FH. The content of the accumulator and status of the carry flag after the addition operation are respectively
a)
0FH, 1
b)
10FH, 0
c)
FFH, 1
d)
10FH, 1
|
|
Sudhir Patel answered |
A= 80H
B= 8FH
[A] + [B] = 80H + 8FH = 1 0FH
B= 8FH
[A] + [B] = 80H + 8FH = 1 0FH
But accumulator is an 8-bit register
Accumulator = 0FH
carry flag= 1
Hence the correct answer is 0FH, 1.
Consider a 5-stage pipeline having stages as Instruction Fetch (IF), Instruction Decode (ID), Operand Fetch (OF), Execute (EX) and Write Back (WB). Here we are given 4 instructions. IF, ID, OF and WB stages take 1 clock cycle each, but the EX-stage takes 1 cycle for ADD and SUB, 2 cycles for MUL and 3 cycles for DIV operation. If the operand forwarding technique is used from EX stage to OF stage and the clock rate of pipeline processor is 5 GHz, then choose the correct statement(s), considering the below table:
- a)The instruction I3 gets executed in the 10th clock cycle.
- b)Number of True dependencies are 3.
- c)Total number of clock cycles required are 11.
- d)Total execution time is 2200 picoseconds.
Correct answer is option 'A,B,C'. Can you explain this answer?
Consider a 5-stage pipeline having stages as Instruction Fetch (IF), Instruction Decode (ID), Operand Fetch (OF), Execute (EX) and Write Back (WB). Here we are given 4 instructions. IF, ID, OF and WB stages take 1 clock cycle each, but the EX-stage takes 1 cycle for ADD and SUB, 2 cycles for MUL and 3 cycles for DIV operation. If the operand forwarding technique is used from EX stage to OF stage and the clock rate of pipeline processor is 5 GHz, then choose the correct statement(s), considering the below table:
a)
The instruction I3 gets executed in the 10th clock cycle.
b)
Number of True dependencies are 3.
c)
Total number of clock cycles required are 11.
d)
Total execution time is 2200 picoseconds.
|
|
Crack Gate answered |
Here operand forwarding is used from EX to OF stage.
Instruction I3 is dependent on both I1 and I2, so we will fetch the operand for I3 after execution of instruction I1.
When execution of I2 gets completed, we can execute the instruction I3.
Similarly, I4 is dependent on I3. We can only execute I4 after the execution of I3.
Option (1)- True, from above diagrammatic representation, we can see that I3 gets executed in 10th cycle.
Option (2)- True, 3 RAW (Read After Write) dependencies are there
So, total RAW dependencies are 3.
RAW is also known as True dependency.
Option (3)- True, Last instruction I4 gets executed in 11th cycle.
So, total clock cycles required are 11.
Option (3)- True, Total execution time = Total number of clock cycles × clock cycle time
Clock cycle time = 1/Clock rate
When execution of I2 gets completed, we can execute the instruction I3.
Similarly, I4 is dependent on I3. We can only execute I4 after the execution of I3.
Option (1)- True, from above diagrammatic representation, we can see that I3 gets executed in 10th cycle.
Option (2)- True, 3 RAW (Read After Write) dependencies are there
So, total RAW dependencies are 3.
RAW is also known as True dependency.
Option (3)- True, Last instruction I4 gets executed in 11th cycle.
So, total clock cycles required are 11.
Option (3)- True, Total execution time = Total number of clock cycles × clock cycle time
Clock cycle time = 1/Clock rate
Total execution time = 11 × 0.2 ns
= 2.2 ns
= 2.2 x 10–9 sec
= 2.2 x 103 × 10–12 sec
= 2200 x 10–12 sec
= 2200 picoseconds
= 2.2 ns
= 2.2 x 10–9 sec
= 2.2 x 103 × 10–12 sec
= 2200 x 10–12 sec
= 2200 picoseconds
Only instructions with zero, one, and two addresses are supported by some CPUs. The size of an op-code is 16 bits, whereas the size of an address is 4 bits.What is the Maximum number of two address instructions?
Correct answer is '256'. Can you explain this answer?
Only instructions with zero, one, and two addresses are supported by some CPUs. The size of an op-code is 16 bits, whereas the size of an address is 4 bits.
What is the Maximum number of two address instructions?
|
|
Sagnik Desai answered |
The Maximum Number of Two-Address Instructions
In order to determine the maximum number of two-address instructions, we need to consider the size of the op-code and the size of the address.
Size of Op-code
The size of the op-code is given as 16 bits. This means that the op-code can have 2^16 = 65536 different values.
Size of Address
The size of the address is given as 4 bits. This means that the address can have 2^4 = 16 different values.
Two-Address Instructions
A two-address instruction is an instruction that operates on two operands and requires two addresses. In this case, the instruction format would typically include two fields for the addresses of the operands.
Calculating the Maximum Number of Two-Address Instructions
To calculate the maximum number of two-address instructions, we need to consider the number of possible combinations of op-code and address fields.
Number of Possible Op-code Values
As mentioned earlier, the op-code can have 65536 different values.
Number of Possible Address Combinations
Since the address field has 4 bits, it can have 16 different values. For each operand, we have 16 possible values, giving us a total of 16 * 16 = 256 possible combinations of addresses.
Total Number of Two-Address Instructions
To calculate the total number of two-address instructions, we multiply the number of possible op-code values by the number of possible address combinations.
Total Number of Two-Address Instructions = Number of Op-code Values * Number of Address Combinations
= 65536 * 256
= 16777216
Conclusion
The maximum number of two-address instructions is 16777216. However, since the size of the op-code is 16 bits and the size of the address is 4 bits, only instructions with zero, one, and two addresses are supported by some CPUs. Therefore, the maximum number of two-address instructions is limited to the number of possible address combinations, which is 256.
In order to determine the maximum number of two-address instructions, we need to consider the size of the op-code and the size of the address.
Size of Op-code
The size of the op-code is given as 16 bits. This means that the op-code can have 2^16 = 65536 different values.
Size of Address
The size of the address is given as 4 bits. This means that the address can have 2^4 = 16 different values.
Two-Address Instructions
A two-address instruction is an instruction that operates on two operands and requires two addresses. In this case, the instruction format would typically include two fields for the addresses of the operands.
Calculating the Maximum Number of Two-Address Instructions
To calculate the maximum number of two-address instructions, we need to consider the number of possible combinations of op-code and address fields.
Number of Possible Op-code Values
As mentioned earlier, the op-code can have 65536 different values.
Number of Possible Address Combinations
Since the address field has 4 bits, it can have 16 different values. For each operand, we have 16 possible values, giving us a total of 16 * 16 = 256 possible combinations of addresses.
Total Number of Two-Address Instructions
To calculate the total number of two-address instructions, we multiply the number of possible op-code values by the number of possible address combinations.
Total Number of Two-Address Instructions = Number of Op-code Values * Number of Address Combinations
= 65536 * 256
= 16777216
Conclusion
The maximum number of two-address instructions is 16777216. However, since the size of the op-code is 16 bits and the size of the address is 4 bits, only instructions with zero, one, and two addresses are supported by some CPUs. Therefore, the maximum number of two-address instructions is limited to the number of possible address combinations, which is 256.
A processor has 300 distinct instructions and 70 general-purpose registers. A 32-bit instruction word has an opcode, two register operands, and an immediate operand. The number of bits available for the immediate operand field is_____
Correct answer is '9'. Can you explain this answer?
A processor has 300 distinct instructions and 70 general-purpose registers. A 32-bit instruction word has an opcode, two register operands, and an immediate operand. The number of bits available for the immediate operand field is_____
|
Devika Gupta answered |
Given Information:
- The processor has 300 distinct instructions.
- The processor has 70 general-purpose registers.
- A 32-bit instruction word has an opcode, two register operands, and an immediate operand.
To find:
The number of bits available for the immediate operand field.
Solution:
Since we know that a 32-bit instruction word has an opcode, two register operands, and an immediate operand, we can calculate the number of bits available for the immediate operand field using the following steps:
Step 1: Calculate the number of bits required for the opcode.
Since the processor has 300 distinct instructions, we need to represent each instruction with a unique opcode. The number of bits required to represent 300 distinct instructions is given by:
Number of bits for opcode = ceil(log2(300))
Step 2: Calculate the number of bits required for the register operands.
Since the processor has 70 general-purpose registers, we need to represent two register operands in the instruction word. The number of bits required to represent 70 registers is given by:
Number of bits for register operands = ceil(log2(70))
Step 3: Calculate the number of bits available for the immediate operand field.
The total number of bits used by the opcode and register operands can be calculated by adding the number of bits required for the opcode and register operands:
Total number of bits used = Number of bits for opcode + Number of bits for register operands
Since we know that the instruction word is 32 bits in total, the number of bits available for the immediate operand field can be calculated by subtracting the total number of bits used from 32:
Number of bits available for immediate operand field = 32 - Total number of bits used
Therefore, the final answer is 9 bits.
- The processor has 300 distinct instructions.
- The processor has 70 general-purpose registers.
- A 32-bit instruction word has an opcode, two register operands, and an immediate operand.
To find:
The number of bits available for the immediate operand field.
Solution:
Since we know that a 32-bit instruction word has an opcode, two register operands, and an immediate operand, we can calculate the number of bits available for the immediate operand field using the following steps:
Step 1: Calculate the number of bits required for the opcode.
Since the processor has 300 distinct instructions, we need to represent each instruction with a unique opcode. The number of bits required to represent 300 distinct instructions is given by:
Number of bits for opcode = ceil(log2(300))
Step 2: Calculate the number of bits required for the register operands.
Since the processor has 70 general-purpose registers, we need to represent two register operands in the instruction word. The number of bits required to represent 70 registers is given by:
Number of bits for register operands = ceil(log2(70))
Step 3: Calculate the number of bits available for the immediate operand field.
The total number of bits used by the opcode and register operands can be calculated by adding the number of bits required for the opcode and register operands:
Total number of bits used = Number of bits for opcode + Number of bits for register operands
Since we know that the instruction word is 32 bits in total, the number of bits available for the immediate operand field can be calculated by subtracting the total number of bits used from 32:
Number of bits available for immediate operand field = 32 - Total number of bits used
Therefore, the final answer is 9 bits.
The first machine cycle of an instruction is always a- a)Memory reed cycle
- b)Fetch cycle
- c)I/O real cycle
- d)Memory write cycle
Correct answer is option 'B'. Can you explain this answer?
The first machine cycle of an instruction is always a
a)
Memory reed cycle
b)
Fetch cycle
c)
I/O real cycle
d)
Memory write cycle
|
|
Crack Gate answered |
Machine Cycle: Time taken to execute one OPERATION is known as a machine cycle. One instruction will contain 1 to 5 machine cycles.
T-State: The portion of a machine cycle executed in one internal clock pulse is known as T-state.
Steps in the instruction cycle:
T-State: The portion of a machine cycle executed in one internal clock pulse is known as T-state.
Steps in the instruction cycle:
- First of all, the opcode is fetched by the microprocessor from a stored memory location.
- Then it is decoded by the microprocessor to find out which operation it needs to perform.
- If an instruction contains data or operand address which is still in the memory, the CPU has to perform read operation to get the desired data.
- After receiving the data, it performs to execute the operation.
Correct sequence: fetch → decode → read effective address → execute
Consider the following table:
Which of the following is true about the average CPI of the above given table?- a)ALU takes ~30.30% of the total cycles.
- b)Load takes ~27.27% of the total cycles.
- c)Store takes ~27.27% of the total cycles.
- d)Branch takes ~10.10% of the total cycles.
Correct answer is option 'C'. Can you explain this answer?
Consider the following table:
Which of the following is true about the average CPI of the above given table?
Which of the following is true about the average CPI of the above given table?
a)
ALU takes ~30.30% of the total cycles.
b)
Load takes ~27.27% of the total cycles.
c)
Store takes ~27.27% of the total cycles.
d)
Branch takes ~10.10% of the total cycles.
|
|
Sudhir Patel answered |
- ALU takes ~30.30% of the total cycles.
- Load takes ~30.30% of the total cycles.
- Store takes ~27.27% of total cycles.
- Branch takes ~12.12% of total cycles.
Which one of the following register of 8085 microprocessor is not a part of the programming model?- a)Instruction register
- b)Memory address register
- c)Status register
- d)Temporary data register
Correct answer is option 'C'. Can you explain this answer?
Which one of the following register of 8085 microprocessor is not a part of the programming model?
a)
Instruction register
b)
Memory address register
c)
Status register
d)
Temporary data register
|
Inaya Reddy answered |
Understanding the 8085 Microprocessor Registers
The 8085 microprocessor has a well-defined programming model that includes several important registers. Each register serves a specific purpose in the execution of instructions.
Registers Included in the Programming Model
- Instruction Register: Holds the current instruction being executed.
- Memory Address Register (MAR): Stores the address of the memory location to be accessed.
- Temporary Data Register (TDR): Used for temporary storage of data during instruction execution.
What is the Status Register?
The Status Register is sometimes referred to in context but is not part of the direct programming model of the 8085. Instead, it typically contains flags that indicate the status of the processor (like zero, carry, sign, parity, and auxiliary carry flags) after an operation has been performed.
Why Option 'C' is Correct?
- The Status Register is not explicitly used for programming purposes; it's more of an internal mechanism that reflects the state of the processor after operations.
- The other options (Instruction Register, Memory Address Register, Temporary Data Register) are integral to the flow of instructions and directly interact with the programming model.
Conclusion
In summary, while the Status Register provides essential information about the processor's state, it does not play a direct role in the programming model of the 8085 microprocessor, making option 'C' the correct answer. Understanding these distinctions helps clarify the roles of various registers in microprocessor design and functionality.
The 8085 microprocessor has a well-defined programming model that includes several important registers. Each register serves a specific purpose in the execution of instructions.
Registers Included in the Programming Model
- Instruction Register: Holds the current instruction being executed.
- Memory Address Register (MAR): Stores the address of the memory location to be accessed.
- Temporary Data Register (TDR): Used for temporary storage of data during instruction execution.
What is the Status Register?
The Status Register is sometimes referred to in context but is not part of the direct programming model of the 8085. Instead, it typically contains flags that indicate the status of the processor (like zero, carry, sign, parity, and auxiliary carry flags) after an operation has been performed.
Why Option 'C' is Correct?
- The Status Register is not explicitly used for programming purposes; it's more of an internal mechanism that reflects the state of the processor after operations.
- The other options (Instruction Register, Memory Address Register, Temporary Data Register) are integral to the flow of instructions and directly interact with the programming model.
Conclusion
In summary, while the Status Register provides essential information about the processor's state, it does not play a direct role in the programming model of the 8085 microprocessor, making option 'C' the correct answer. Understanding these distinctions helps clarify the roles of various registers in microprocessor design and functionality.
In how many different modes a universal shift register operates?- a)2
- b)3
- c)4
- d)5
Correct answer is option 'C'. Can you explain this answer?
In how many different modes a universal shift register operates?
a)
2
b)
3
c)
4
d)
5
|
Eesha Kapoor answered |
Universal Shift Register Modes
A universal shift register is a versatile digital circuit that can perform various operations based on its configuration. It operates in multiple modes, allowing for flexible data manipulation.
Modes of Operation
Universal shift registers primarily operate in the following three modes:
Conclusion
In summary, a universal shift register operates in four distinct modes: SISO, SIPO, PISO, and PIPO. These modes enable it to perform various tasks in digital circuits, making it a fundamental building block in electronics.
A universal shift register is a versatile digital circuit that can perform various operations based on its configuration. It operates in multiple modes, allowing for flexible data manipulation.
Modes of Operation
Universal shift registers primarily operate in the following three modes:
- Serial In, Serial Out (SISO)
- Data is shifted in one bit at a time from one end and shifted out one bit at a time from the same end. - Serial In, Parallel Out (SIPO)
- Data is entered serially, but multiple bits can be read out simultaneously from the output pins. - Parallel In, Serial Out (PISO)
- Multiple bits are loaded into the register in parallel, and then the data can be shifted out serially. - Parallel In, Parallel Out (PIPO)
- Both data input and output are done in parallel, allowing for fast data transfer.
Conclusion
In summary, a universal shift register operates in four distinct modes: SISO, SIPO, PISO, and PIPO. These modes enable it to perform various tasks in digital circuits, making it a fundamental building block in electronics.
Which stack is used in 8085 microprocessors?- a)FIFO
- b)FILO
- c)LIFO
- d)LILO
Correct answer is option 'C'. Can you explain this answer?
Which stack is used in 8085 microprocessors?
a)
FIFO
b)
FILO
c)
LIFO
d)
LILO
|
Xena Das answered |
Understanding the Stack in 8085 Microprocessors
The 8085 microprocessor utilizes a specific type of stack for managing data and subroutine calls. This stack operates under the Last In, First Out (LIFO) principle.
What is LIFO?
- Last In, First Out (LIFO) means that the last item added to the stack is the first one to be removed.
- This is analogous to a stack of plates; you can only take off the top plate, which is the most recently added.
Stack Operations in 8085
- PUSH Operation: When data is pushed onto the stack, it is stored in a memory location designated for the stack, and the stack pointer is decremented.
- POP Operation: When data is popped off the stack, the most recent item is retrieved, and the stack pointer is incremented.
Significance of LIFO in 8085
- Subroutine Management: When a subroutine is called, the return address is pushed onto the stack. Upon completing the subroutine, the return address is popped off the stack for execution to resume at the correct location.
- Interrupt Handling: The stack helps preserve the state of the processor before handling an interrupt, ensuring that the system can return to its previous state seamlessly.
Conclusion
The 8085 microprocessor's use of a LIFO stack is crucial for effective data management and control flow. It allows for organized handling of function calls and interrupts, making it an essential feature in microprocessor operations. Thus, the correct answer is option 'C': LIFO.
The 8085 microprocessor utilizes a specific type of stack for managing data and subroutine calls. This stack operates under the Last In, First Out (LIFO) principle.
What is LIFO?
- Last In, First Out (LIFO) means that the last item added to the stack is the first one to be removed.
- This is analogous to a stack of plates; you can only take off the top plate, which is the most recently added.
Stack Operations in 8085
- PUSH Operation: When data is pushed onto the stack, it is stored in a memory location designated for the stack, and the stack pointer is decremented.
- POP Operation: When data is popped off the stack, the most recent item is retrieved, and the stack pointer is incremented.
Significance of LIFO in 8085
- Subroutine Management: When a subroutine is called, the return address is pushed onto the stack. Upon completing the subroutine, the return address is popped off the stack for execution to resume at the correct location.
- Interrupt Handling: The stack helps preserve the state of the processor before handling an interrupt, ensuring that the system can return to its previous state seamlessly.
Conclusion
The 8085 microprocessor's use of a LIFO stack is crucial for effective data management and control flow. It allows for organized handling of function calls and interrupts, making it an essential feature in microprocessor operations. Thus, the correct answer is option 'C': LIFO.
Handshaking mode of data transfer is- a)Synchronous data transfer
- b)asynchronous data transfer
- c)interrupt driven data transfer
- d)level Mode of DMA data transfer
Correct answer is option 'A'. Can you explain this answer?
Handshaking mode of data transfer is
a)
Synchronous data transfer
b)
asynchronous data transfer
c)
interrupt driven data transfer
d)
level Mode of DMA data transfer
|
|
Nakul Chauhan answered |
Understanding Handshaking Mode of Data Transfer
Handshaking mode is a crucial concept in data transfer protocols, particularly in communication systems. It ensures that data is transmitted reliably between devices.
What is Handshaking?
- Handshaking is a technique used to establish a communication link between two devices before data transfer begins.
- It involves a series of signals exchanged between the sender and receiver to confirm readiness for data exchange.
Why Synchronous Data Transfer?
- In synchronous data transfer, both devices operate in a coordinated manner, sharing a common clock signal.
- Handshaking is vital in this mode to synchronize the transmission and reception of data.
Key Features of Synchronous Data Transfer:
- Clock Signal: Both devices rely on a clock signal to time the data transfer.
- Data Integrity: The handshaking process helps ensure that data is sent and received accurately, reducing errors caused by timing mismatches.
- Control Signals: Handshaking involves control signals like "ready" and "acknowledge," which help manage the flow of data.
Comparison with Other Modes:
- Asynchronous Data Transfer: Unlike synchronous transfer, it does not require a common clock. Handshaking here is less rigorous, often relying on start and stop bits.
- Interrupt Driven Data Transfer: This mode uses interrupts to signal the processor, which is different from the coordinated approach of handshaking.
- Level Mode of DMA: This involves direct memory access without handshaking for synchronization, focusing instead on data bus control.
Conclusion
In summary, the handshaking mode of data transfer is synonymous with synchronous data transfer, where coordination between devices is essential for effective communication. Understanding this concept is vital for anyone studying Electronics and Communication Engineering.
Handshaking mode is a crucial concept in data transfer protocols, particularly in communication systems. It ensures that data is transmitted reliably between devices.
What is Handshaking?
- Handshaking is a technique used to establish a communication link between two devices before data transfer begins.
- It involves a series of signals exchanged between the sender and receiver to confirm readiness for data exchange.
Why Synchronous Data Transfer?
- In synchronous data transfer, both devices operate in a coordinated manner, sharing a common clock signal.
- Handshaking is vital in this mode to synchronize the transmission and reception of data.
Key Features of Synchronous Data Transfer:
- Clock Signal: Both devices rely on a clock signal to time the data transfer.
- Data Integrity: The handshaking process helps ensure that data is sent and received accurately, reducing errors caused by timing mismatches.
- Control Signals: Handshaking involves control signals like "ready" and "acknowledge," which help manage the flow of data.
Comparison with Other Modes:
- Asynchronous Data Transfer: Unlike synchronous transfer, it does not require a common clock. Handshaking here is less rigorous, often relying on start and stop bits.
- Interrupt Driven Data Transfer: This mode uses interrupts to signal the processor, which is different from the coordinated approach of handshaking.
- Level Mode of DMA: This involves direct memory access without handshaking for synchronization, focusing instead on data bus control.
Conclusion
In summary, the handshaking mode of data transfer is synonymous with synchronous data transfer, where coordination between devices is essential for effective communication. Understanding this concept is vital for anyone studying Electronics and Communication Engineering.
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