A Microprocessor Instruction Cycle Analysis
Microprocessor Instruction Cycle Analysis
a. Determine the length of the instruction cycle for the case of a string of 56 bytes:
b. What is the worst-case delay for acknowledging an interrupt if the instruction is non-interruptible?
c. Repeat part (b) assuming the instruction can be interrupted at the beginning of each byte transfer.
Final answer: The instruction cycle time for a string of 56 bytes is 113.5 nanoseconds. If the instruction is non-interruptible, the worst case delay is also 113.5 nanoseconds. However, if the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be 2 nanoseconds.
Explanation: The cycle length of the instruction for a string of 56 bytes can be calculated as the sum of the initial 15 clock cycle, 20 cycles for each byte, multiplied by the total number of bytes. Total cycles = 15 + (20 * 56) = 1135 cycles. Given that the processor clock speed is 10GHZ, The instruction cycle time = 1135 cycles / 10^9 (frequency in Hertz) = 113.5 nanoseconds. The worst-case delay for acknowledging an interrupt, if the instruction is non-interruptible, would be the total time of execution which is 113.5 nanoseconds. If the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be the time to transfer one byte, which is 20 clock cycles or 2 nanoseconds.
a. Determine the length of the instruction cycle for the case of a string of 56 bytes. b. What is the worst-case delay for acknowledging an interrupt if the instruction is non-interruptible? c. Repeat part (b) assuming the instruction can be interrupted at the beginning of each byte transfer. The length of the instruction cycle for a string of 56 bytes is 113.5 nanoseconds. If the instruction is non-interruptible, the worst-case delay is also 113.5 nanoseconds. However, if the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be 2 nanoseconds.