Revision day two - logic design, single cycle, multi cycle, MIPS and C

Question 1

Data path

Answer 1

performs the data operations as commanded by the program instructions

Question 2

control

Answer 2

controls the data path, memory and I/O according to program instructions

Question 3

disadvantages of single cycle

Answer 3

speed - the cycle time is the Time for the most complex instruction but the average instruction needs less time

cost - functional units cannot be resused within one instructions execution

Question 4

execution time

Answer 4

instruction count x number of cycles per instruction x cycle time

Question 5

fetch step

Answer 5

instruction is fetched from memory
increment pc to the next instruction (pc +4)

Question 6

instruction decode

Answer 6

retrieve the appropriate values from register files
decode the instruction
create control signals that will accompany the execution
calculate the target address - sign extend the immediate, shift left by 2 (word align) and add to (pc +4)

Question 7

R-format instruction

Answer 7

3. perform arithmetic and store output in ALUOut
4. write result to correct register

Question 8

Store instruction

Answer 8

3. calculate address - 16 bit offset (sign extended) + first register source operand
4. write to memory

Question 9

Branch instructions

Answer 9

3. compare operands using ALU subtraction

if 0 - address target is changed to one calculated in step two

else pc register will remain at PC+4

Question 10

RISC

Answer 10

reduced instruction set computer

Question 11

CISC

Answer 11

complex instruction set computer

Question 12

instruction set architecture

Answer 12

the language of the computer

Question 13

features of CISC

Answer 13

very few registers
very little memory

Question 14

features of RISC

Answer 14

more memory + faster clock frequency
fixed length and fixed format instructions (for easy, fast decoding)

Question 15

big endian

Answer 15

the msbyte of the data is placed at the location with the lowest address

Question 16

little endian

Answer 16

the least significant byte of the data is placed at the byte with the lowest address

Question 17

how can a word be stored in memory

Answer 17

1. unaligned - placed anywhere
2. aligned - at an address which is a multiple of the word size

Question 18

memory regions and management in c

Answer 18

heap - dynamically allocated memory (managed by the programmer)
stack - for function / method calls (managed by compiler)
static - for data that lives during the whole lifetime of the program (initialised when the process starts

Question 19

extern

Answer 19

extends the scope before the definition and to other files

Question 20

static

Answer 20

lets it hide from other files

Question 21

what does xor instruction do

Answer 21

inverts the bits

Question 22

in a processor with caches, an index is used to

Answer 22

locate the data in a direct mapped cache

Question 23

data representation affects what two things

Answer 23

circuit complexity
circuit speed

Question 24

name one thing that will not trigger a processor exception

Answer 24

cache miss

Question 25

combinational logic

Answer 25

logical system whose blocks do NOT contain memory computes some output given some input

Question 26

sequential logic

Answer 26

logical system whose blocks contain memory hence the output depends on the input and current values in the memory

Question 27

three ways all logical circuits can be built

Answer 27

and, or, not nand nor

Question 28

decoder

Answer 28

a logic block that has n-bits input and 2^n bits output where each output is asserted for each input combination

Question 29

multiplexer (MUX)

Answer 29

output is one of the inputs selected by a controller

Question 30

what does a multiplexer consist of

Answer 30

1. detector that generates n different input values 2. array of n AND gates 3. one OR gate (that combines the output of the AND gates)

Question 31

32-bit adder

Answer 31

a circuit made by connecting 32 adders

Question 32

propagation delay

Answer 32

time between input signal change and output signal change at the other end

Question 33

three things that the propagation delay depends on

Answer 33

1. technology 2. delay through each gate 3. number of gates driven by gates output

Question 34

asynchronous

Answer 34

state of circuit changes whenever input changes (there is no clock)

Question 35

synchrouous

Answer 35

state can only change at times synchronised to an external signal (or clock)

Question 36

edge triggered clock

Answer 36

clock scheme in which all changes occur on the clock edge

Question 37

D flip flop

Answer 37

a memory element for which an output is equal to the value of the stored state inside the element for which the internal state is changed only on the clocks edge

Question 38

D latch

Answer 38

A memory element for which an output is equal to the value of the stored state inside the element for which the internal state changes whenever the appropriate input changes and clock is asserted

Question 39

setup time

Answer 39

the minimum time that the input to the memory device must be valid before the clock edge

Question 40

hold time

Answer 40

a minimum time during which the input must be valid after a clock edge

Question 41

Two details of the FSM

Answer 41

synchronised, every change happens in a clock cycle sequential logic can be described via a FSM

Question 42

glitch

Answer 42

when different inputs have different delays

Question 43

Moores law

Answer 43

the number of transistors doubles every 18 - 24 months. at the same time the cost of the transistor reduces

Question 44

compiler

Answer 44

a program that translates a high level language (C, java) into aa sequence of assembly language statements

Question 45

assembly language

Answer 45

a symbolic representation of machine instructions

Question 46

operating system

Answer 46

a supervising program that manages all programs in a computer such as 1. basic I/O interaction 2. allocation and memory 3. protection of sharing the computer

Question 47

load instruction

Answer 47

3. calculate address - 16 bit offset (sign extended) + first register source operand 4. read from memory 5. store in destination register