Microprogrammed-Control-Unit
Question 1 |
Consider a CPU where all the instructions require 7 clock cycles to complete execution. There are 140 instructions in the instruction set. It is found that 125 control signals are needed to be generated by the control unit. While designing the horizontal microprogrammed control unit, single address field format is used for branch control logic. What is the minimum size of the control word and control address register?
125, 7 | |
125, 10 | |
135, 7 | |
135, 10 |
Question 1 Explanation:
Each instruction takes 7 cycles,
i.e., 140 instruction takes = 140 * 7 =980 cycles.
So, size of control address register = ⌈log2 980⌉
= 10 bit
Since horizontal microprogramming is used, so 125 control signals will require 125 bits.
Hence, size of control word = 125 + 10 = 135 bits
i.e., 140 instruction takes = 140 * 7 =980 cycles.
So, size of control address register = ⌈log2 980⌉
= 10 bit
Since horizontal microprogramming is used, so 125 control signals will require 125 bits.
Hence, size of control word = 125 + 10 = 135 bits
Question 2 |
A microprogrammed control unit
Is faster than a hard-wired control unit. | |
Facilitates easy implementation of new instruction. | |
Is useful when very small programs are to be run. | |
Usually refers to the control unit of a microprocessor. |
Question 2 Explanation:
In micro-programmed control unit we can add new instruction by changing the content of control memory.
Question 3 |
Consider a 32-bit processor which supports 70 instructions. Each instruction is 32 bit long and has 4 fields namely opcode, two register identifiers and an immediate operand of unsigned integer type. Maximum value of the immediate operand that can be supported by the processor is 8191. How many registers the processor has?
32 | |
64 | |
128 | |
16 |
Question 3 Explanation:
Given that each instruction is 32 bit long and it has 4 fields.
As there are 70 instructions, each opcode needs 7 bits as ceil(log70) = 7.
It is given that maximum value of the immediate operand is 8191.
Number of bits needed for the immediate operand = ceil(log8191) = 13.
Out of the total 32 bits, remaining bits for the two register identifiers = 32 - 7 - 13 = 32 - 20 = 12
These 12 bits are equally divided to identify the two registers. So to identify each register we need 6 bits. So there are a total of 2^6 = 64 registers.
As there are 70 instructions, each opcode needs 7 bits as ceil(log70) = 7.
It is given that maximum value of the immediate operand is 8191.
Number of bits needed for the immediate operand = ceil(log8191) = 13.
Out of the total 32 bits, remaining bits for the two register identifiers = 32 - 7 - 13 = 32 - 20 = 12
These 12 bits are equally divided to identify the two registers. So to identify each register we need 6 bits. So there are a total of 2^6 = 64 registers.
There are 3 questions to complete.