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BC0046 Microprocessor Assignment Set 1

Q1.Convert the decimal number 123.145 to octal and

hexadecimal.Decimal to Octal conversion using Repeated Division Method

For absolute part For fractional part8 321 8 1458 40 1 LSB 8 18 1 LSB8 5 0 8 2 2

0 5 MSB 0 2 MSB

321 (10) = 501 (8) and 145 (10) = 221 (8)

So, 321.145 (10) = 501. 221 (8)

Decimal to Hexadecimal conversion using Repeated Division Method

For absolute part For fractional part16 321 16 14516 20 1 LSB 16 9 1 LSB1

6 1 4 MSB 0 9 MSB0 1

321 (10) = 141 (16) and 145 (10) = 91 (16)

So, 321.145 (10) = 141. 91 (16)

Q2. Write an assembly language program to find the smallestamong two numbers.

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Program

MVI B, 30H

MVI C, 40H

MOV A, B

CMP C

JZ EQU

JC GRT

OUT PORT1

HLT

EQU: MVI A, 01H

OUT PORT1

HLT

GRT: MOV A, C

OUT PORT1

HLT

Q3. Draw and explain the internal architecture of 8085.

System Bus

Typical system uses a number of busses, collection of wires, which transmit binarynumbers, one bit per wire. A typical microprocessor communicates with memory andother devices (input and output) using three busses: Address Bus, Data Bus andControl Bus.

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Fig 2.1: Internal Architecture of 8085

Address Bus

One wire for each bit, therefore 16 bits = 16 wires. Binary number carried alerts memoryto open the designated box. Data (binary) can then be put in or taken out. The AddressBus consists of 16 wires, therefore 16 bits. Its width is 16 bits. A 16 bit binary number allows 216 different numbers, or 32000 different numbers, ie 0000000000000000 up to1111111111111111. Because memory consists of boxes, each with a unique address,the size of the address bus determines the size of memory, which can be used. Tocommunicate with memory the microprocessor sends an address on the address bus,eg 0000000000000011 (3 in decimal), to the memory. The memory selects box number 3 for reading or writing data. Address bus is unidirectional, i.e numbers only sent frommicroprocessor to memory, not other way.

Data Bus

Data Bus: carries data, in binary form, between microprocessor and other externalunits, such as memory. Typical size is 8 or 16 bits. The Data Bus typically consists of 8wires. Therefore, 28 combinations of binary digits. Data bus used to transmit data, ieinformation, results of arithmetic, etc, between memory and the microprocessor. Bus isbi-directional. Size of the data bus determines what arithmetic can be done. If only 8 bitswide then largest number is 11111111 (255 in decimal). Therefore, larger number have

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to be broken down into chunks of 255. This slows microprocessor. Data Bus alsocarries instructions from memory to the microprocessor. Size of the bus therefore limitsthe number of possible instructions to 256, each specified by a separate number.

Control Bus

Control Bus are various lines which have specific functions for coordinating andcontrolling microprocessor operations. Eg: Read/NotWrite line, single binary digit.controls whether memory is being written to (data stored in mem) or read from (datataken out of mem) 1 = Read, 0 = Write. May also include clock line(s) for timing/synchronising, interrupts, reset etc. Typically microprocessor has 10 controllines. Cannot function correctly without these vital control signals.

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Q4. Draw and explain the internal architecture of 8086.

Figure 3.1 shows the internal architecture of the 8086. Except for the instructionregister, which is actually a 6-byte queue, the control unit and working registers aredivided into three groups according to their functions. There is a data group, which isessentially the set of arithmetic registers; the pointer group, which includes base andindex registers, but also contains the program counter and stack pointer; and thesegment group, which is a set of special purpose base registers. All of the registers are16 bits wide.

The data group consists of the AX, BX, CX and DX registers. These registers can be used tostore both operands and results and each of them can be accessed as a whole, or the upper andlower bytes can be accessed separately. For example, either the 2 bytes in BX can be usedtogether, or the upper byte BH or the lower byte BL can be used by itself by specifying BH or BL, respectively.

Fig 3.1: Internal Architecture of 8086

In addition to serving as arithmetic registers, the BX, CX and DX registers play specialaddressing, counting, and I/O roles:

BX may be used as a base register in address calculations.

CX is used as an implied counter by certain instructions.

DX is used to hold the I/O address during certain I/O operations.

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The pointer and index group consists of the IP, SP, BP, SI, and DI registers. The instruction pointer IP and SP registers are essentially the program counter and stack pointer registers, but thecomplete instruction and stack addresses are formed by adding the contents of these registers tothe contents of the code segment (CS) and stack segment (SS) registers. BP is a base register for accessing the stack and may be used with other registers and/or a displacement that is part of the

instruction. The SI and DI registers are for indexing. Although they may be used by themselves,they are often used with the BX or BP registers and/or a displacement. Except for the IP, a pointer can be used to hold an operand, but must be accessed as a whole.

To provide flexible base addressing and indexing, a data address may be formed byadding together a combination of the BX or BP register contents, SI or DI register contents, and a displacement. The result of such an address computation is called aneffective address (EA) or offset. The word displacement is used to indicate a quantitythat is added to the contents of a register(s) to form an EA. The final data address,however, is determined by the EA and the appropriate data segment (DS), extrasegment (ES), or stack segment (SS) register.

The segment group consists of the CS, SS, DS, and ES registers. As indicated above,the registers that can be used for addressing, the BX, IP, SP, BP, SI, and DI registers,are only 16 bits wide and, therefore, an effective address has only 16 bits. On the other hand, the address put on the address bus, called the physical address, must contain 20bits. The extra 4 bits are obtained by adding the effective address to the contents of oneof the segment registers as shown in Fig. 3.2. The addition is carried out by appendingfour 0 bits to the right of the number in the segment register before the addition is made;thus a 20-bit result is produced. As an example, if (CS) = 1234 and (IP) = 0002, then thenext instruction will be fetched from

0002 Effective address+ 12340 beginning segment address

12342 Physical address of instruction

[It is standard notation for parentheses around an entity to mean "contents of," e.g., (IP) meansthe contents of IP. Also, all addresses are given in hexadecimal.]

Fig. 3.2 Generation of physical Address

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The utilization of the segment registers essentially divide the memory space intooverlapping segments, with each segment being 64K bytes long and beginning at a 16-byte, or paragraph, boundary, i.e., beginning at an address that is divisible by 16. Wewill hereafter refer to the contents of a segment register as the segment address , andthe segment address multiplied by 16 as the beginning physical segment address , or

simply, the beginning segment address . An illustration of the example above is given inFig.3.3( a) and the overall segmentation of memory is shown in Fig. 3.3(b).

The advantages of using segment registers are that they:

1. Allow the memory capacity to be 1 MB even though the addresses associatedwith the individual instructions are only 16 bits wide.

2. Allows the instruction, data, or stack portion of a program to be more than 64Kbytes long by using more than one code, data, or stack segment.

3. Facilitate the use of separate memory areas for a program, its data, and thestack.Permit a program and/or its data to be put into different areas of memoryeach time the program is executed.

Fig 3.3Memory segmentation

The 8086s PSW contains 16 bits, but 7 of them are not used. Each bit in the PSW iscalled a flag . The 8086 flags are divided into the conditional flags,which reflect the result of the previous operation involving the ALU, and the controlflags, which control the execution of special functions. The flags are summarized in Fig.3.4.

The condition flags are:

SF (Sign FIag)- Is equal to the MSB of the result. Since in 2s complement negativenumbers have a 1 in the MSB and for nonnegative numbers this bit is 0, this flagindicates whether the previous result was negative or nonnegative.

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ZF (Zero Flag)- Is set to 1 if the result is zero and 0 if the result is nonzero.

PF (Parity Flag)- Is set to 1 if the low-order 8 bits of the result contains an even number of 1s; otherwise it is cleared.

CF (Carry Flag)- An addition causes this flag to be set if there is a carry out of the MSB,and a subtraction causes it to be set if a borrow is needed. Other instructions also affectthis flag and its value will be discussed when these instructions are defined.

AF (Auxiliary Carry Flag)- Is set if there is a carry out of bit 3 during an addition or aborrow by bit 3 during a subtraction. This flag is used exclusively for BCD arithmetic.

OF (Overflow Flag)- Is set if an overflow occurs, i.e., a result is out of range. Morespecifically, for addition this flag is set when there is a carry into the MSB and no carryout of the MSB or vice versa. For subtraction, it is set when the MSB needs a borrowand there is no borrow from the MSB or vice versa.

As an example, if the previous instruction performed the addition

0010 0011 0100 1101

+ 0011 0010 0001 0001

0101 0101 0101 1110

then following the instruction:

SF=0 ZF=0 PF=0 CF=0 AF=0 OF=0

Fig 3.4: PSW register of 8086

If the previous instruction performed the addition

0101 0100 0011 1001

+ 0100 0101 0110 1010

1001 1001 1010 0011

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then the flags would be:

SF = 1 ZF = 0 PF = 1 CF = 0 AF = 1 CF = 1

The control flags are:

DF (Direction Flag)-Used by string manipulation instructions. If zero, the string isprocessed from its beginning with the first element having the lowest address.Otherwise, the string is processed from the high address towards the low address.

IF (Interrupt Enable Flag)- If set, a certain type of interrupt (a maskable interrupt) can be recognized by the CPU;otherwise, these interrupts are ignored.

TF (Trap Flag) -If set, a trap is executed after each instruction.

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Q5. Write a sequence of instructions to reverse a two digithexadecimal number available in the register AX using shift and

rotate instructions?

Fig 4.15: Shift and rotate instructions

The shift and rotate instructions for the 8086 are defined in Fig. 4.15. These instructionsshift all of the bits in the operand to the left or right by the specified count, CNT. For the

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shift left instructions, zeroes are shifted into the right end of the operand and the MSBsare shifted out the left end and lost, except that the least significant bit of these MSBs isretained in the CF flag. The shift right instructions similarly shift bits to the right;however, SAR (shift arithmetic right) does not automatically insert zeroes from the left; itextends the sign of the operand by repeatedly inserting the MSB. The rotate instructions

differ from theshift instructions in that the operand is treated like a circle in which the bits shifted out of one end are shifted into the other end. The RCL and RCR instructions include the carryflag in the circle and the ROL and ROR do not, although the carry flag is affected in allcases.

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Q6. Explain the concept of Linking and Relocation.

Linking and Relocation

In constructing a program some program modules may be put in the same sourcemodule and assembled together; others may be in different source modules andassembled separately. If they are assembled separately, then the main module, whichhas the first instruction to be executed, must be terminated by an END statement withthe entry point specified, and each of the other modules must be terminated by an ENDstatement with no operand. In any event, the resulting object modules, some of whichmay be grouped into libraries, must be linked together to form a load module before theprogram can be executed. In addition to outputting the load module, normally the linker prints a memory map that indicates where the linked object modules will be loaded intomemory. After the load module has been created it is loaded into the memory of the

computer by the loader and execution begins. Although the I/O can be performed bymodules within the program, normally the I/O is done by I/O drivers that are part of theoperating system. All that appears in the users program are references to the I/Odrivers that cause the operating system to execute them.

The general process for creating and executing a program is illustrated in Fig 5.1. Theprocess for a particular system may not correspond exactly to the one diagrammed inthe figure, but the general concepts are the same. The arrowsindicate that corrections may be made after anyone of the major stages.

Fig. 5.1: Creation and Execution of a program

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Segment Combination

In addition to the linker commands, the ASM-86 assembler provides a means of regulating the way segments in different object modules are organized by the linker.Sometimes segments with the same name are concatenated and sometimes they are

overlaid. Just how the segments with the same name are joined together is determinedby modifiers attached to the SEGMENT directives. A SEGMENT directive may have theform

Segment name SEGMENT Combine-type

where the combine-type indicates how the segment is to be located within the loadmodule. Segments that have different names cannot be combined and segments withthe same name but no combine-type will cause a linker error. The possible combine-types are:

PUBLIC- If the segments in different object modules have the same name and thecombine-type PUBLIC, then they are concatenated into a single segment in the loadmodule. The ordering in the concatenation is specified by the linker command.

COMMON- If the segments in different object modules have the same name and thecombine-type is COMMON, then they are overlaid so that they have the samebeginning address. The length of the common segment is that of the longest segmentbeing overlaid.

STACK- If segments in different object modules have the same name and the combine-type

STACK, then they become one segment whose length is the sum of the lengths of theindividually specified segments. In effect, they are combined to form one large stack.

AT- The AT combine-type is followed by an expression that evaluates to a constantwhich is to be the segment address. It allows the user to specify the exact location of the segment in memory..

MEMORY- This combine-type causes the segment to be placed at the last of the loadmodule. If more than one segment with the MEMORY combine type is being linked, onlythe first one will be treated as having the MEMORY combine-type; the others will beoverlaid as if they had COMMON combine types.

Access to External Identifiers

Clearly, object modules that are being linked together must be able to refer to eachother. That is, there must be a way for a module to reference at least some of thevariables and/or labels in the other modules. If an identifier is defined in an objectmodule, then it is said to be a local (or internal) identifier relative to the module, and if it

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is not defined in the module but is defined in one of the other modules being linked, thenIt is referred to as an external (or global) identifier relative to the module.

For single-object module programs all identifiers that are referenced must be locally defined or an assembler error will occur. For multiple-module programs, the assembler must be informed in

advance of any externally defined identifiers that appear in a module so that it will not treat themas being undefined. Also, in order to permit other object modules to reference some of theidentifiers in a given module, the given module must include a list of the identifiers to which itwill allow access. Therefore, each module may contain two lists, one containing the externalidentifiers it references and one containing the locally defined identifiers that can be referred to

by other modules. These two lists are implemented by the EXTRN and PUBLIC directives,which have the forms:

EXTRN Identifier:Type, . . . , Identifier:Type

and

PUBLIC Identifier, . . ., Identifier

where the identifiers are the variables and labels being declared as external or as beingavailable to other modules. Because the assembler must know the type of all externalidentifiers before it can generate the proper machine code, a type specifier must beassociated with each identifier in an EXTRN statement. For a variable the type may beBYTE, WORD, or DWORD and for a label it may be NEAR or FAR. In the statement

INC VAR1

if VAR1 is external and is associated with a word, then the module containing thestatement must also contain a directive such as

EXTRN . . VAR1 :WORD.. . .

and the module in which VARl is defined must contain a statement of the form

PUBLIC .. ..VAR1.

One of the primary tasks of the linker is to verify that every identifier appearing in an EXTRNstatement is matched by one in a PUBLIC statement. If this is not the case, then there will be an

undefined external reference and a linker error will occur. Fig. 5.2 shows three modules and howthe matching is done by the linker while joining them together.

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Fig. 5.2: Illustration of the matching verified by the linker

As we have seen, there are two parts to every address, an offset and a segmentaddress. The offsets for the local identifiers can be and are inserted by the assembler,

but the offsets for the external identifiers and all segment addresses must be insertedby the linking process. The offsets associated with all external references can beassigned once all of the object modules have been found and their external symboltables have been examined. The assignment of the segment addresses is calledrelocation and is done after the king process has determined exactly where eachsegment is to be put in memory.

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Q7. Differentiate macros and procedures.

A macro is a group of repetitive instructions in a program which are codified only onceand can be used as many timesas necessary.

The main difference between a macro and a procedure is that in the macro the passageof parameters is possible and in theprocedure it is not, this is only applicable for the MASM - there are other programminglanguages which do allow it.

At the moment the macro is executed each parameter is substituted by the name or value specified at the time of the call.

We can say then that a procedure is an extension of a determined program, while themacro is a module with specificfunctions which can be used by different programs.

Another difference between a macro and a procedure is the way of calling each one, tocall a procedure the use of adirective is required, on the other hand the call of macros is done as if it were anassembler instruction.Example of procedure:For example, if we want a routine which adds two bytes stored in AH and AL each one,and keep the addition in the BX register:

Adding Proc Near ; Declaration of the procedure

Mov Bx, 0 ; Content of the procedureMov B1, AhMov Ah, 00

Add Bx, AxRet ; Return directive

Add Endp ; End of procedure declaration

and an example of Macro:

Position MACRO Row, ColumnPUSH AX

PUSH BXPUSH DXMOV AH, 02HMOV DH, RowMOV DL, ColumnMOV BH, 0INT 10HPOP DX

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POP BXPOP AXENDM

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Q9. Describe about each flag of a 8086 flag register.

8 086 CPU has 8 general purpose registers, each register has its own name:

AX - the accumulator register (divided into AH / AL):

1. Generates shortest machine code2. Arithmetic, logic and data transfer

3. One number must be in AL or AX

4. Multiplication & Division

5. Input & Output

BX - the base address register (divided into BH / BL ).

CX - the count register (divided into CH / CL ):

1. Iterative code segments using the LOOP instruction2. Repetitive operations on strings with the REP command

3. Count (in CL) of bits to shift and rotate

DX - the data register (divided into DH / DL):

1. DX:AX concatenated into 32-bit register for some MUL and DIV operations2. Specifying ports in some IN and OUT operations

SI - source index register:

1. Can be used for pointer addressing of data2. Used as source in some string processing instructions

3. Offset address relative to DS

DI - destination index register:

1. Can be used for pointer addressing of data2. Used as destination in some string processing instructions

3. Offset address relative to ES

BP - base pointer:

1. Primarily used to access parameters passed via the stack2. Offset address relative to SS

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SP - stack pointer:

1. Always points to top item on the stack2. Offset address relative to SS

3. Always points to word (byte at even address)

4. An empty stack will had SP = FFFEh

SEGMENT REGISTERS

CS - points at the segment containing the current program.

DS - generally points at segment where variables are defined.

ES - extra segment register, it's up to a coder to define its usage.

SS - points at the segment containing the stack.

Although it is possible to store any data in the segment registers, this is never a goodidea. The segment registers have a very special purpose - pointing at accessible blocksof memory.

Segment registers work together with general purpose register to access any memoryvalue. For example if we would like to access memory at the physical address 12345h(hexadecimal), we could set the DS = 1230h and SI = 0045h . This way we can accessmuch more memory than with a single register, which is limited to 16 bit values.The CPU makes a calculation of the physical address by multiplying the segment

register by 10h and adding the general purpose register to it (1230h * 10h + 45h =12345h):

The address formed with 2 registers is called an effective address .By default BX, SI and DI registers work with DS segment register;BP and SP work with SS segment register.Other general purpose registers cannot form an effective address.Also, although BX can form an effective address, BH and BL cannot.

SPECIAL PURPOSE REGISTERS

IP - the instruction pointer:

1. Always points to next instruction to be executed2. Offset address relative to CS

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IP register always works together with CS segment register and it points to currentlyexecuting instruction.

FLAGS REGISTER

Flags Register - determines the current state of the processor. They are modifiedautomatically by CPU after mathematical operations, this allows to determine the type of theresult, and to determine conditions to transfer control to other parts of the program.Generally you cannot access these registers directly.

1. Carry Flag (CF) - this flag is set to 1 when there is an unsigned overflow . For examplewhen you add bytes 255 + 1 (result is not in range 0...255). When there is no overflowthis flag is set to 0 .

2. Parity Flag (PF) - this flag is set to 1 when there is even number of one bits in result,and to 0 when there is odd number of one bits.

3. Auxiliary Flag (AF) - set to 1 when there is an unsigned overflow for low nibble (4bits).

4. Zero Flag (ZF) - set to 1 when result is zero . For non-zero result this flag is set to 0 .

5. Sign Flag (SF) - set to 1 when result is negative . When result is positive it is set to 0 .(This flag takes the value of the most significant bit.)

6. Trap Flag (TF) - Used for on-chip debugging.

7. Interrupt enable Flag (IF) - when this flag is set to 1 CPU reacts to interrupts fromexternal devices.

8. Direction Flag (DF) - this flag is used by some instructions to process data chains,when this flag is set to 0 - the processing is done forward, when this flag is set to 1 the processing is done backward.

9. Overflow Flag (OF) - set to 1 when there is a signed overflow . For example, whenyou add bytes 100 + 50 (result is not in range -128...127).

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Q10. Write an assembly program to add and display twonumbers.

Program

MVI D, 8BH

MVI C, 6FH

MOV A, C

ADD D

OUT PORT1

HLT

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BC0046 Microprocessor Assignment Set 2

Q2. When working with strings, what are the advantages of theMOVS and CMPS instructions over the MOV and CMPinstructions?

When working with strings, the advantages of the MOVS and CMPSinstructions over the MOV and CMP instructions are:

1. They are only 1 byte long.

2. Both operands are memory operands.

3. Their auto-indexing obviates the need for separate incrementing or decrementinginstructions, thus decreasing overall processing time.

As an example consider the problem of moving the contents of a block of memory to another area in memory. A solution that uses only the MOVinstruction, which cannot perform a memory-to-memory transfer, is shown inFig. 6.2(a).

Fig. 6.2: Program sequences for moving a block of data

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A solution that employs the MOVS instruction is given in Fig. 6.2(b). Note thatthe second program sequence may move either bytes or words, dependingon the type of STRING1 and STRING2.

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Q3. Explain the working of DMA.

If the data transfer rate to or from an I/O device is relatively low, then the communication can be performed using either programmed or interrupt I/O. Some devices, such as magnetic tape anddisk units and analog to-digital converters, may operate at data rates that are too high to behandled by byte or word transfers. Data rates for I/O and mass storage devices are oftendetermined by the devices, not the CPU, and the computer must be capable of executing I/Oaccording to the maximum speed of the device. For a disk unit the data rate is determined by thespeed with which data pass under the read/write head and quite often this rate exceeds 200,000

bytes per second. Thus, there is less than 5 microseconds to transfer each byte to or frommemory. For data rates of this magnitude, block transfers, which use DMA controllers tocommunicate directly with memory, are required.

During any given bus cycle, one of the system components connected to the system bus is given

control of the bus. This component is said to be the master during that cycle and the componentit is communicating with is said to be the slave. The CPU with its bus control logic is normallythe master, but other specially designed components can gain control of the bus by sending a busrequest to the CPU. After the current bus cycle is completed the CPU will return a bus grant signal and the component sending the request will become the master. Taking control of the busfor a bus cycle is called cycle stealing. Just like the bus control logic, a master must be capableof placing addresses on the address bus and directing the bus activity during a bus cycle. Thecomponents capable of becoming masters are processors (and their bus control logic) and DMAcontrollers. Sometimes a DMA controller is associated with a single interface, but they are oftendesigned to accommodate more than one interface.

The 8086 receives bus requests through its HOLD pin and issues grants fromits hold acknowledge (HLDA) pin. A request is made when a potential mastersends a 1 to the HOLD pin. Normally, after the current bus cycle is completethe 8086 will respond by putting a 1 on the HLDA pin. When the requestingdevice receives this grant signal it becomes the master. It will remain masteruntil it drops the signal to the HOLD pin, at which time the 8086 will drop thegrant on the HLDA pin. One exception to the normal sequence is that if aword which begins at an odd address is being accessed, then two bus cyclesare required to complete the transfer and a grant will not be issued untilafter the second bus cycle.

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Q4. Write short notes on (i) programmed I/O and (ii) Interrupt I/O

Programmed IO

Programmed I/O consists of continually examining the status of an interface andperforming an I/O operation with the interface when its status indicates that it has datato be input or its data-out buffer register is ready to receive data from the CPU. A typicalprogrammed input operation is flowcharted in Fig.7.3.

As a more complete example, suppose that a line of characters is to be input from aterminal to an 82-byte array beginning at BUFFER until a carriage return is encounteredor more than 80 characters are input. If a carriage return is not found in the first 81characters then the message BUFFER OVERFLOW is to be output to the terminal;otherwise, a line feed is to be automatically appended to the carriage return

Interrupt IO

Even though programmed I/O is conceptually simple, it can waste a considerableamount of time while waiting for ready bits to become active. In the above example, if the person typing on the terminal could type 10 characters per second and only 10 s isrequired for the computer to input each character, then approximately

99,990 x 100% = 99.99%100,000

of the time is not being utilized. This is all right if no other processing could be donewhile the input is taking place, but if other processing is unnecessarily delayed, then adifferent approach is needed.

Interruptis an event that causes the CPU to initiate a fixed sequence, known as an interruptsequence . Before an 8086 interrupt sequence can begin, the currently executinginstruction must be completed unless the current instruction is a HLT or WAITinstruction.

For a prefixed instruction, because the prefix is considered as part of the instruction, theinterrupt request is not recognized between the prefix and the instruction. In the case of the REP instruction, the interrupt request is recognized after the primitive operationfollowing the REP is completed, and the return address is the location of the REP prefix.For MOV and POP instructions in which the destination is a segment register, aninterrupt request is not recognized until after the instruction following the MOV or POPinstruction is executed. Provided that the segment register is filled first, this allows thecontents of both a segment register and a pointer to be changed without interruption. .

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Q5. Explain about the semaphore operations.

Semaphore Operations

In multiprogramming systems, processes are allowed to share common software anddata resources as well as hardware resources. In many situations, a common resourcemay be accessed and updated by only one process at a time and other processes mustwait until the one currently using the shared resource is finished with it. A resource of this type, which is commonly referred to as a serially reusable resource, must beprotected from being simultaneously accessed and modified by two or more processes.

A serially reusable resource may be a hardware resource (such as a printer, cardreader, or tape drive), a file of data, or a shared memory area.

For example, let us consider a personnel file that is shared by processes 1 and 2.Suppose that process 1 performs insertions, deletions, and changes, and process 2

puts the file in alphabetical order according to last names. If accessed sequentially, thisfile would either be updated by process 1 and then sorted by process 2, or vice versa.However, if both processes were allowed to access the file at the same time, the resultswould be unpredictable and almost certainly incorrect. The solution to this problem is toallow only one process at a time to enter its critical section of code, i.e., that section of code that accesses the serially reusable resource.

Preventing two or more processes from simultaneously entering their critical sections for accessing a shared resource is called mutual exclusion. One way to attain mutualexclusion is to use flags to indicate when the shared resource is already in use. Toexamine how this is done and some of the problems associated with this approach let

us consider the possibility of using only one flag.

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Q6. Explain what a virtual memory is?

Virtual Memory

The more sophisticated memory management scheme can be achieved by thehardware dynamic address translation design illustrated in Fig.8.6. For each memoryreference, the logical addressoutput by the CPU is translated into the physicaladdress , which is the address sent to the memory by the memory managementhardware. Logical addresses are the addresses that are generated by the instructionsaccording to the addressing modes.

Fig. 8.6: Dynamic Address Translation

Because the logical addresses may be different from the physical addresses, the user can design a program in a logical space, also called a virtual s pace, withoutconsideration for the limitations imposed by the real memory. This provides twomajor advantages. First, dividing a program into several pieces with each mapped intoan area in real memory, a program need not occupy a contiguous memory section.Therefore, memory fragmentation is reduced and less memory space is wasted.Second, it is not necessary to store the entire program in memory while it is executing.Whenever a portion of the program that is not currently in memory is referenced, theoperating system can suspend the program, load in the required section of code, andthen resume the programs execution. This allows a programs size to be larger than the

actual memory available. Inother words, a user virtually has more memory to work with than actually exists.

With address translation hardware, the program is divided into segments according tothe logical structure of the program and the resulting memory management scheme iscalled segmentation . When using segmentation, each logical address consists of twofields, the segment number and the offset within the segment. The number of bitsrepresenting the segment number governs the maximum number of segments allowed

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in a program, and the number of bits allocated to the offset specifies the maximumsegment size. For example, if the segment number and offset have mand nbits, respectively, a program may have up to 2 m segments, with each segment having amaximum size of 2 n

bytes, thus providing each user with a virtual storage of 2m+n

bytes.

Fig. 8.7 illustrates the mapping of logical addresses into physical addresses.

Fig. 8.7: Segmentation Scheme

The segment number S in a logical address is used as an index into the segment table,which returns the beginning physical address X of the referenced segment. Thisaddress added to the offset L to form the physical address of the memory location.Because each job may be assigned a separate area in the segment table, the baseaddress of that section of the segment table that is associated with the currentlyexecuting job is stored in a register called the segment table register. The index S ismade relative to the segment table register. When the system switches from one job toanother, the segment table register is updated to point to a new section of the segmenttable. Because the address translation must be performed for every memory reference,

either the entire segment table or that portion of the table containing the beginningaddresses of the segments that are currently in use must be stored in registers whichare part of the memory management hardware.

Each entry in the segment table is referred to as a segment descriptor. Depending on,the particular implementation, a segment descriptor may include attributes in addition tothe beginning segment address. The most common of these additional attributes arethe.:

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Status Field- Indicates whether or not the referenced segment is in the memory.

Segment Length Field- Indicates the size of the segment.

Protection Field- Provides protection against unauthorized reading, writing, or

execution.

Reference Field- Provides useful information in determining which segment is to bereplaced.

Change field : Indicates whether or not the segment has beenmodified since being brought into memory .

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Q8. Explain the 8288 Bus controller.

For reasons of simplicity, flexibility, and low cost, most microcomputers, including thoseinvolving multiprocessor configurations, are built around a primary system bus which

connects all of the major components in the system. In order to obtain a foundationwhile designing its products, a microcomputer manufacturer makes assumptions aboutthe bus that is to be used to connect its devices together Frequently, these assumptionsbecome formalized and constitute what is referred to as a bus standard

The Intel MULTIBUS has gained wide industrial acceptance and several manufacturersoffer MULTIBUS-compatible modules. This bus is designed to support both 8-bit and16-bit devices and can be used in multiprocessor systems in which several processorscan be masters. At any point in time, only two devices may communicate with eachother over the bus, one being the master and the other the slave. The master/slaverelationship is dynamic with bus allocation being accomplished through the bus

allocation (i.e., request/grant) control signals. The MULTIBUS has been physicallyimplemented on an etched backplane board which is connected to each module usingtwo edge connectors, denoted PI and P2, I shown in Fig. 9.13. The connector P1consists of 86 pins which provide-the major bus signals, and P2 is an optional connector consisting of 60 auxiliary lines, primarily used for power failure detection and handling.

Fig. 9.13: Illustration of a module being plugged into MULTIBUS

The P1 lines can be divided into the following groups according to their functions:

1. Address lines.

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2. Data lines.

3. Command and handshaking lines.

4. Bus access control lines.

5. Utility lines.

The MULTIBUS has 20 address lines, labeled through , where the numericsuffix represents the address bit in hexadecimal. The address lines are driven by thebus master to specify the memory location or I/O port being accessed.. The MULTIBUSstandard calls for all single bytes to be communicated over only the lower 8 bits of thebus; therefore, any 16-bit interface must include a swap byte buffer so that only thelower data lines are used for all byte transfers. (It should be pointed out that because an

8086 expects a byte to be put on the high-order byte of the bus when is active, one

may want to permit nonstandard MULTIBUS transfers between memory and an 8086.)

The two inhibit signals are provided for overlaying RAM, ROM, and auxiliary ROM in acommon address space. For example, a bootstrap loader may be stored in an auxiliary

ROM and a monitor in a ROM. Because the loader is needed only after a reset, and

could both be activated while the loader is executing.Then, when the monitor is in

control, could be raised while remains low. If control is passed to the user,

and could both be deactivated, thus allowing the RAM to fill the entire memory

space during normal operation.

There are 16 bidirectional data lines ( - ), only eight of which are used in an 8-bitsystem. Data transfers on the MULTIBUS bus are accomplished by handshakingsignals in a manner similar to that described in the preceding sections. The memory

read ( ), memory write ( ), I/O read ( ), and I/O write ( ) lines are definedto be the same as they were in the discussion of the 8288 bus controller. There is an

acknowledge ( ) signal which serves the same purpose as the READY signal in thediscussion of the bus control logic, i.e., to verify the end of a transfer. In a generalsetting it may be received by bus master. Because a master must wait to be notified of the completion transfer, the duration of a bus cycle varies depending on the speed of the bus master and the slave. This asynchronous nature enables the system to handleslow devices without penalizing fast devices

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Q9. Draw the block diagram of 8087.

The 8087 Numeric Data Processor

The 8087 numeric data processor (NDP) is specially designed to perform arithmeticoperations efficiently. It can operate on data of the integer, decimal, and real types,with lengths ranging from 2 to 10 bytes. The instruction set not only includes variousforms of addition, subtraction, multiplication, and division, but also provides many usefulfunctions, such as taking the square root, exponentiation, taking the tangent, and so on.

As an example of its computing power, the 8087 can multiply two 64-bit real numbers inabout 27 s and calculate a square root in about 36 s. If performed by the 8086through emulation, the same operations would require approximately 2 ms and 20 ms,respectively. The 8087 provides a simple andeffective way to enhance the performance of an 8086 based system, particular when anapplication is primarily computational in nature.

A pin diagram of the 8087 is shown in Fig. 10.3.

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Fig. 10.3: 8087 pin diagram

The address/data, status ready, reset, clock, ground, and power pins of the NDP havethe same pin positions as those assigned to the 8086/8088. Among the remaining eightpins, four of them are not used. The other pins are connected as follows: the BUSY pin

to the hosts pin; the / pin to the hosts / or / pin; the INT pin to theinterrupt

management logic (assuming the 8087 is enabled for interrupts); and the / couldbe connected to the bus request/grant pin of an independent processor such as an8089. This simple interface allows an existing maximum mode 8086-based system to beeasily upgraded by replacing the original CPU with a piggyback module, which consistsof an 8086/8088 and an 8087.

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Q10. Explain why the processor utilization rate can be improvedin a multiprocessor system by an instruction queue.

Although the 8086 is a powerful single-chip microprocessors, their instruction set is notsufficient to effectively satisfy some complex applications. For such applications, the8086 must be supplemented with coprocessors that extend the instruction set indirections that will allow the necessary special computations to be accomplished moreefficiently. For example, the 8086 has no instructions for performing floating pointarithmetic, but by using an Intel 8087 numeric data processor as a coprocessor, anapplication that heavily involves floating point calculations can be readily satisfied.

It will be seen that, except for the coprocessor itself, a coprocessor design does notrequire any extra logic other than that normally needed for a maximum mode system.Both the CPU and coprocessor execute their instructions from 1M same program, whichis written in a superset of the 8086 instruction set. Other than possibly fetching anoperand for the coprocessor, the CPU does not need to take any further action if theinstruction is executed by the coprocessor.

The interaction between the CPU and the coprocessor when an instruction is executedby the coprocessor is depicted in Fig.10.1. An instruction to be executed by thecoprocessor is indicated when an escape (ESC) instruction appears in the programsequence. Only the host CPU can fetch instructions, but the coprocessor also receivesall instructions and monitors the instruction sequencing of the host. An ESC instructioncontains an external operation code, indicating what the coprocessor is to do and issimultaneously decoded by both the coprocessor and the host. At this point the hostmay simply go on to the next instruction or it may fetch the first word of a memoryoperand for the coprocessor and then go on to the next instruction. If the CPU fetchesthe first word of an operand, the coprocessor will capture the data word and its 20-bitphysical address.

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Fig. 10.1: Synchronization between 8086 and its coprocessor

For a source operand that is longer than one word, the coprocessor can obtain theremaining words by stealing bus cycles. If the memory operand specified in the ESCinstruction is a destination, the coprocessor ignores the data word fetched by the hostand later the coprocessor will store the result into the captured address. In either case,

the coprocessor will send a busy (high) signal to the hosts pin and, as the hostcontinues processing the instruction stream, the coprocessor will perform the operationindicated by the code in the ESC instruction. This parallel operation continues until thehost needs the coprocessor to perform another operation or must have the results fromthe current operation. At that time the host should execute a WAIT instruction and wait

until its pin is activated by the coprocessor. The WAIT instruction repeatedly checks

the pin until it becomes activated and then executes the next instruction insequence.

An ESC assembler language instruction has two operands. The first of two operands

indicates the external opcode, which determines the action to be taken by thecoprocessor. If the second operand specifies a memory location, then as explainedabove, the 8086 will fetch a word from this location for the coprocessor and may passthe coprocessor an address for storing a result. If the second operand is a register, theregister address is treated as part of the external op code and the CPU does nothing.

The interfacing of a coprocessor to a host CPU is shown in Fig. 10.2.Both the host andthe coprocessor share the same clock generator and bus control logic. It is. possible to

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have two coprocessors connected to the same host CPU. When this is done, thecoprocessors must be assigned distinct subsets of the set of external opcodes and eachcoprocessor must be able to recognize and execute the members of its subset. For themost part parallel lines can be used to connect the host to its coprocessors. For two

coprocessors, one could use the / pin on the 8086 and the other could use the /

pin. The two coprocessors would be connected to separate 8259A interrupt requestpins.

In order for a coprocessor to determine when the host is executing an ESC instruction, it

must monitor the host CPUs status on the - lines and the ADl5-AD0 lines for fetched instructions. Because instructions are prefetched by the CPU, an ESCinstruction might not be executed immediately or, if it is preceded by a branchinstruction, it might not be executed at all.

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Fig. 10.2: Coprocessor configurarion

The coprocessor must track the instruction stream by monitoring the queue status bitsQS1 and QS0 and maintaining an instruction queue identical to that of the host. If thequeues status is 00, the coprocessor does nothing, but if it is 01, it will compare the fiveMSBs of the first byte in the queue to 11011. If there is a match, then an ESCinstruction is ready to be executed and, assuming that the coprocessor recognizes theexternal opcode, it will perform the indicated operation; otherwise, this byte is ignoredand is deleted from the queue. The queue status 10 indicates that the queue in the hostis being flushed and, therefore, causes the queue in the coprocessor to also beemptied. The 11 status combination indicates that the first byte in the queue is not the

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first byte of an instruction and this byte is looked at by the coprocessor only / if it isknown to be part of an ESC instruction. If not part of an ESC instruction, this byte isignored.

The coprocessor should be designed so that when an error occurs during the decodingor execution of an ESC instruction, it will send out an interrupt request (which isnormally sent to an 8259A). The coprocessor should also be designed so that it can

steal bus cycles by making bus requests through one of the hosts pins whenadditional data must be read from or stored in memory. Last, the coprocessor must be

able to apply a high signal to the hosts pin while it is busy.