Instruction execution cycle _

_

1. COMPUTER ORGANIZATION &
ARCHITECTURE
(BCS-DS-402)
1.2 Instruction execution cycle
Ms. Swati Hans,
Assistant Professor
Department of Computer Science &
Engineering-SPL
School of Engineering & Technology
Manav Rachna International Institute of
Research and Studies (Deemed to be
University),Faridabad

2. Instruction Codes
A process is controlled by a program
A program is a set ofinstructionsthat
specify the operations, data, and the
control sequence
An instruction is stored in binary code that
specifies a sequence of microoperations
Instruction codes together with data are
stored in memory (Stored Program
Concept)

3. Program statements and
computer instructions
Computer instruction
Field specifying the
operation to be executed
Field specifying the data
To be operated on

4. Instruction code format
Instruction code format with two parts :
Opcode + Operand
Opcode : specify 16 possible operations(4 bits)
Operand : specify the address of an operand/data(12
bits)
If an operation in an instruction code does not need
an operand from memory, the rest of the bits in the
instruction(address field) can be used for other
purpose
Opcode Operand
15 12 11 0
instruction
data
15 12 11 0
Not an instruction

5. Instruction Format
The instruction format also defines the layout of the bits for an
instruction. It can be of variable lengths with multiple numbers of
addresses. The typical components of an instruction format include:
 15 14 12 11 0
Opcode: The operation to be performed (e.g., addition,
subtraction).
Operands: The data or addresses involved in the operation.
Addressing Mode:It defines how the operands are accessed
(e.g., direct, indirect).
Depending on the number of address fields in the instruction,
instruction is categorized as follows:
Three address instruction
Two address instruction
One address instruction
Zero address instruction
Mode Opcode Operand

6. Zero Address instructions
It operates without specifying any operands explicitly. Typically used
in stack-based architectures, these instructions rely on the stack to
store operands. The operations are performed on the top of the
stack, and the results are also stored on the stack.
Example
Uses Postfix Notation of expression
Push A
Push B

7. One Address Instructions
It specifies a single operand, typically held in an accumulator. The
accumulator is an implicit register used in the operation. The result can
either be stored in the accumulator or another location.
Example
Opcode Operand

8. Two Address Instructions
This instruction is most commonly used in commercial computers.
This address instruction format has three operand fields. The two
address fields can either be memory addresses or registers.

9. ThreeAddress Instructions
The format of a three address instruction requires three operand
fields. These three fields can be either memory addresses or
registers.

10. CPU organizations
A computer can have instructions of
different lengths containing varying
numbers of addresses. The number of
address fields of a computer depends on
the internal design of its registers. Most of
the computers fall into one of three types
of CPU organizations:
Single accumulator organization.
General register organization.
Stack organization.

11. Single Accumulator Organization
All the operations on a system are performed with an
implied accumulator register. The instruction format in
this type of computer uses one address field.
For example, the instruction for arithmetic addition is
defined by an assembly language instruction ‘ADD.’
Where X is the operand’s address, the ADD instruction
results in the operation.
AC ← AC + M[X].
AC is the accumulator register, M[X] symbolizes the
memory word located at address X.

12. General Register Organization
The general register type computers employ two or
three address fields in their instruction format. Each
address field specifies a processor register or a memory.
An instruction symbolized by ADD R1, X specifies the
operation R1 ← R + M [X].
This instruction has two address fields: register R1 and
memory address X.

13. Stack Organization
A computer with a stack organization has PUSH
and POP instructions that require an address
field.
Hence, the instruction PUSH X pushes the word
at address X to the top of the stack.
The stack pointer updates automatically. In
stack-organized computers, the operation type
instructions don’t require an address field as the
operation is performed on the two items on the
top of the stack

14. Common Bus System
Thebasiccomputerhaseightregisters,a
memoryunit,andacontrolunit.
Pathsmustbeprovidedtotransferinformation
fromoneregistertoanotherandbetween
memoryandregisters
Amoreefficientschemefortransferring
informationinasystemwithmanyregistersisto
useacommonbus.

15. Bus
s
1
s
2
s
0
16-bit common bus
Clock
LD
LD
LD
INR
OUTR
IR
INPR
LD INR CLR
LD INR CLR
LD INRCLR
LD INRCLR
WRITE
Address
Adder
& Logic
E
DR
PC
AR
CLR
7
1
2
3
4
5
6
Computer System Architecture, Mano, Copyright (C) 1993 Prentice-Hall, Inc.
AC
READ
Memory Unit
4096x16
TR

16. Common Bus System
The connection of the registers and memory of the
basic computer to a common bus system :
The outputs of seven registers and memory are connected to
the common bus
The specific output is selected by mux(S0, S1, S2) :
Memory(7), AR(1), PC(2), DR(3), AC(4), IR(5), TR(6)
When LD(Load Input) is enable, the particular register
receives the data from the bus
Control Input : LD, INC, CLR, Write, Read

17. COMMON BUS SYSTEM
Control variables: the bus is controlled
by
1-Selection switches for selecting the
source of information and
2-Enable switches at the destination
device to accept the information.

18. Selection variables
Selection variables: select a register or the
memory whose output is used as an input to the
common bus.
To select one device out of 8, we need 3 select
variables.
For example, if S2S1S0 = 011, the output of DR
is selected as an output of the common bus.

19. Load input
Load input (LD): Enables the input of a
register to download bits form the
common bus. When LD = 1 for a register,
the data on the common bus is read into
the register during the next clock pulse
transition.
> Increment input (INR): Increments the content of a
register.
> Clear input (CLR): Clear the content of a register to zero.

20. CONTROL UNIT HARDWARE (Hardwired)
• Inputs to the control unit come from IR where an instruction is stored.
• A hardwired control is implemented in the example computer using:
> A 3x8 decoder to decode opcode bits 12-14 into signals D0, ..., D7;
A flip-flop (I) to store the addressing mode bit in IR

21. A 4-bit binary
sequence counter
(SC) to count
from 0 to 15 to
achieve time
sequencing;
> A 4x16 decoder
to decode the
output of the
counter into 16
timing signals,
T0, ..., T15
A digital circuit
with inputs
D0, ..., D7, T0,
..., T15, I,
and address
bits in IR (11-0)
to generate
control outputs
supplied to
control inputs
and select signals
of registers , bus.

22. Instruction Cycle
A computer goes through the following
instruction cycle repeatedly:
do
1. Fetch an instruction from memory
2. Decode the instruction
3. Read the effective address from memory if the
instruction has an indirect address
4. Execute the instruction until a HALT instruction is
encountered

23. Instruction and Interrupt cycles
Fetch, decode
Next
Instruction
Execute
Instruction
START
HALT
Instruction cycle
Interrupt
cycle
Interrupt Cycle
Interrupts Enabled
Interrupts Disabled

24. Instruction Cycle Continued

25. Sequencing the Operations
Step 0: (Initialize)
•Initially, the program counter (PC)
is loaded with the address of the first
instructionin the program.
•The sequence counter (SC)is
initialized to 0, providing a decoded
timing signal T0.
SC ← 0
•After each clock pulse, SC is
incremented by one, so that the
timing signals go through a sequence
T0, T1, T2, and so on.

26. Fetch Cycle: (Fetch the Instruction)
Step 1:
During the first time pulse T0, the
instruction addressis transferred from the
PCto AR.
Step 2:
During the time pulse T1, the instruction read
from memory is then transferred to the
instruction register IR, and PCis incremented
by 1 for the address of the next instruction in
the program.
The micro operations for fetch cycle:
•T0: AR ← PC
•T1: IR ← M[AR], PC ← PC + 1

27. Decode Cycle: (Decode the Instruction)
Step 3:
At time T2,
•The operation codein IRis decoded.
•The address partof the instruction is
transferred to AR.
•The mode bit (direct/indirect bit)is
transferred to flip-flop I.
The microoperations for decode cycle:
•T2: D₀, D₁, ... D₇ ← Decode IR (12-14)
•AR ← IR (0-11)
•I ← IR (15) D₀ : 000
D₁ : 001
D₂ : 010
D₃ : 011
D₄ : 100
D₅ : 101
D₆ : 110
D₇ : 111
IOpcode Address
15 14 12 11 0

28. Determine the Type of Instruction:
(Decision)
Step 4: During time T₃, at this time the type of instruction is determined so
the data preparation and execution of the instruction takes place.
The three possible computer instruction types available in the basic
computers are:
•Memory-reference instructions
•Register-reference instructions, or
•Input-output instructions
Instruction Formats
Memory-Reference Instructions (OP-code = 000 ~ 110)
Register-Reference Instructions (OP-code = 111, I = 0)
Input-Output Instructions (OP-code = 111, I = 1)

29. Determine the Type of Instruction:
•Decoder output D₇is equal to 1, if the operation code is
equal to 111.
•Decoder output D₇is equal to 0, if the operation code is one
of the other seven values 000 through 110.
•If D₇ = 1, the instruction must be a register-reference or
input-output type.
•If D₇ = 0, the instruction must be a memory-reference.
Control then inspectsthe value of the first bitof the
instruction, which is in flip-flop I.
•If D₇ = 0 and I = 1, it is a memory reference
instructionwith an indirect address.
•It is then necessary to read the effective address
from memory. (Operand Fetch)
•T₃ : AR ← M[AR]
•If D₇ = 0 and I = 0, it is a memory reference
instructionwith a direct address.
•It is not necessary to do anything since the effective
address is already in AR.
•T₃ : Nothing
•If D₇ = 1 and I = 1, it is an input-output instruction.
•If D₇ = 1 and I = 0, it is a register-reference
instruction.

30. Execute the Instruction:
•A register-referenceor input-output instructioncan
be executed with the clock timing signal T₃.
•T₃ : Execute a register-reference instruction
•T₃ : Execute an input-output instruction
•After the instruction is executed, SC is cleared to 0and
control returns to the fetch phase with T₀ = 1.
•A memory reference instructioncan be executed with
the clock timing signal T₄.
•T₄ : Execute a memory-reference instruction
•After the instruction is executed, SC is cleared to 0and
control returns to the fetch phase with T₀ = 1.

31. Types of Instructions
Register Reference instructions
Memory Reference Instructions

32. REGISTER-REFERENCE INSTRUCTIONS
• The 12 register-reference instructions are recognized by I = 0
and D7 = 1 (IR(12-14) = 111). Each operation is designated by
the presence of 1 in one of the bits in IR(0-11). Therefore D7I`T3
r = 1 is common to all register-transfer instructions.

33. Memory Reference
Instructions
Since the data stored in memory cannot be
processed directly (the memory unit is not
connected to the ALU), the actual execution in
the bus system require a sequence of
microoperations.
(Note that T0-T2 for fetch an instruction; T3 for
AR M[AR]if indirect memory addressing.

34. Direct Addressing
Address AOpcode
Instruction
Memory
Operand

35. Direct Addressing
The value of address field is
the addressof the operand
Notation:
If X is an address then (X)
denotes the value contained
in the memory cell with
address X
EA= Effective Address in
memory of the operand
EA = A
e.g. ADD 457
Look in memory at address
457 for operand
Add contents to accumulator:
AC+M[457]AC
Or AC+(457)->AC

36. Indirect Addressing Diagram
Address AOpcode
Instruction
Memory
Operand
Pointer to operand

37. Indirect Addressing (1)
The memory cell referenced by the
address field contains the address of
(i.e., the pointer to) the operand
If A is the value of the address field,
then EA=(A)
e.g.
Above control word signifies 1 as indirect
addressing and operand part is containing
address and tells to look at address 300,
then go to address 1350 and look there for
operand
Add to accumulator the content of the
cell pointed to by the content of
Acc+M[1350]Acc
1 Add 300