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KTU B.Tech Solved ModelQuestion – Computer Organisation

KTU B.Tech Solved ModelQuestion paper

Computer Organisation



KTU B.Tech Solved ModelQuestion paper – Computer Organisation






Time:3 hrs                                                                                      Max. Marks:100



(Answer all questions. Each carries 3 marks)


  1. Differentiate between big endian and little endian byte ordering.


Big-endian and little-endian are terms that describe the order in which a sequence of bytes are stored in computer memory. Big-endian is an order in which the “big end” (most significant value in the sequence) is stored first (at the lowest storage address). Little-endian is an order in which the “little end” (least significant value in the sequence) is stored first. For example, in a big-endian computer, the two bytes required for the hexadecimal number 4F52 would be stored as 4F52 in storage (if 4F is stored at storage address 1000, for example, 52 will be at address 1001). In a little-endian system, it would be stored as 524F (52 at address 1000, 4F at 1001).

Big Endian

In big endian, you store the most significant byte in the smallest address. Here’s how it would look:

Address Value
1000 90
1001 AB
1002 12
1003 CD

Little Endian

In little endian, you store the least significant byte in the smallest address. Here’s how it would look:

Address Value
1000 CD
1001 12
1002 AB
1003 90


  1. Describe the basic instruction types.





  1. Give the control sequence for execution of instruction Add[R3],R1




  1. Design a 2×2 array multiplier


The multiplicand bits are b1 and b0, the multiplier bits are a1 and a0, and the product is c3 c2 c1 c0• The first partial product is formed by multiplying a0 by b1 b0• The multiplication of two bits such as a0 and b0 produces a 1 if both bits are 1; otherwise, it produces a 0. This is identical to an AND operation and can be implemented with an AND gate. As shown in the diagram, the first partial product is formed by means of two AND gates. The second partial product is formed by multiplying a1 by b1 b0 and is shifted one position to the left. The two partial products are added with two half-adder (HA) circuits. Usually, there are more bits in the partial products and it will be necessary to use full-adders to produce the sum. Note that the least significant bit of the product does not have to go through an adder since it is formed by the output of the first AND gate.




(Answer any two. Each carries 9 marks)

  1. a) Describe the different addressing modes. (5)


  1. Register Addressing
  2. Immediate Addressing 
  3. PC-Relative Addressing
  4. Base Addressing
  5. Pseudo-Direct Addressing.
  • Register Addressing is considered the simplest addressing mode.
  • This is because both operands are in a register. Which allow instructions to be executed much more faster in comparison with other addressing modes because they does not involves with memory access.
  • The number of registers is limited since only a few bits are reserved to select a register.
  • Register Addressing is a form of direct addressing , this is because we are only interested in the number in the register , rather than using that number as a memory address.

Here’s an example of Register Addressing :
add $s1 , $s2 , $s3 also means that $s1 ←$s2 + $s3

where ; $s1 = rd

$s2 = rs

$s3 = rt
That’s about it for Register Addressing.

Immediate Addressing is a numeric value embedded in the instruction in the actual operand.

  • In immediate addressing , the operand is a constant within the encoded instruction.
  • Immediate addressing has the advantage of not requiring an extra memory access to fetch the operand , hence will be executed faster. However , the size of operand is limited to 16 bits.
  • The jump instruction format also falls under immediate addressing , where the destination is held in the instruction.

example of Immediate Addressing :

addi $t1 , $zero , 1 means $t1 ← 0 + 7

(add immediate , uses the I-type format)

where ; $t1 = rd

$zero = r1

1 = immediate value


PC-Relative Addressing also known as Program Counter Addressing is a data or instruction memory location is specified as an offset relative to the incremented PC.

  • PC-relative addressing is usually used in conditional branches. PC refers to special purpose register , Program Counter that stores the address of next instruction to be fetched.
  • In PC-relative addressing , the offset value can be an immediate value or an interpreted label value.
  • The effective address is the sum of the Program Counter and offset value in the instruction. The effective address determines the branch target.
  • PC-relative addressing implements position-independent codes. Only a small offset is adequate for shorter loops.
  • Branch instructions can only move 32768 above or below the program counter because the offset is a 16-bit two’s complement number.

Another word of saying to explain PC-Relative Addressing :

The operand address = PC + an offset

Implements position-independent codes. A small

offset is adequate for short loops.

Example: beqz $t0 , strEnd

where ; $t0 = rs

100 = offset

Thus ; if ($t1 == 0) goto PC + 4 + (4*2)

In this instruction , beqz is a conditional instruction that branches to label in the code if the content of $t0 is equal to zero. If the current address for branch instruction in execution is 0x4000000C , the effective address will be 40000018.


Base Addressing is a data or instruction memory location is specified as a signed offset from a register.

  • Base addressing is also known as indirect addressing , where a register act as a pointer to an operand located at the memory location whose address is in the register.
  • The register is called base that may point to a structure or some other collection of data and immediate value is loaded at a constant offset from the beginning of the structure. The offset specifies how far the location of the operand data from the memory location pointed by the base.
  • The address of the operand is the sum of the offset value and the base value(rs). However, the size of operand is limited to 16 bits because each MIPS instruction fits into a word.
  • The offset value is a signed number which is represented in a two’s complement format. Therefore , offset value can also be a negative value.

Here’s an example for Base Addressing :

Instruction : lw $t1 , 4 ($t2)

where $t1 = rs

$t2 = base (memory address)

4 = offset value

Thus ; $t1 = Memory [$t2 +4]

In the example above , $t2 pointed to the base of a memory structure. The instruction the load register $t1 with the contents of the memory location four words onward from the location pointed by register $t2.

Pseudo-direct Addressing is the memory address which (mostly) embedded in the instructions.

  • Pseudo-Direct addressing is specifically used for J-type instructions , j and jal. The instruction format is 6 bits of opcode and 26 bits for the immediate value (target).
  • In Pseudo-Direct addressing , the effective address is calculated by taking the upper 4 bits of the Program Counter(PC) , concatenated to the 26 bit immediate value , and the lower two bits are 00.
  • Therefore , the new effective address will always be a word-aligned and we can never have a target address of a jump instruction with the two bits anything other than 0 0 and creates a complete 32-bit address. Since the upper 4 bits of the PC are used, this constrains the jump target to anywhere within the current 256 MB block of code (1/16 of the total 4 GB address space). To jump anywhere within the 4 GB space, the R-type instructions jr and jalr are used , where the complete 32 – bit target address is specified in a register.

*Note :

Address in Pseudo-Direct must be a multiple of four.

b) Give the flow chart for Booth’s Multiplication. (4)





  1. Explain how nested subroutines are processed internally.


subroutine nesting, is to have one subroutine call another. In this case, the return address of the second call is also stored in the link register, destroying its previous contents. Hence, it is essential to save the contents of the link register in some other location before calling another subroutine. Otherwise, the return address of the first subroutine will be lost. Subroutine nesting can be carried out to any depth. Eventually, the last subroutine called completes its computations and returns to the subroutine that called it. The return address needed for this first return is the last one generated in the nested call Memory location Subroutine SUB 1000 first instruction …. …. Return 204 Return sequence. That is, return addresses are generated and used in a last-in-first-out order. This suggests that the return addresses associated with subroutine calls should be pushed onto a stack. A particular register is designated as the stack pointer, SP, to be used in this operation. The stack pointer points to a stack called the processor stack. The Call instruction pushes the contents of the PC onto the processor stack and loads the subroutine address into the PC. The Return instruction pops the return address from the processor stack into the PC.


Since PC saving and recovery is automatic by the use of stack, it is possible to execute nested subroutines as shown in figure 2. In other words, it is possible for a subroutine to call another subroutine.

As in figure 2, first the subroutine 1 is called from part ‘a’ of main program. At this point, the returning address i.e. 1001 should pe pushed on the top of the return stack. From this subroutine, subrotuine2 is called and jump is made to address 0923 i.e. jump 2. At this point, the returning address in subroutine 1 should be pushed onto stack. Thus the return stack should contain the returning addresses in the reverse order of the calling order. First ‘a’ is called, and then ‘b’ is called; so returning address of ‘b’ should be on top of stack and below it should be the returning address of ‘a’. 


  1. Explain restoring method of division with an example.


  • Assume ─ that there is an accumulator and MQ register, each of k-bits • MQ 0, (lsb of MQ) bit gives the quotient, which is saved after a subtraction or addition
  • Total number of additions or subtractions are k-only and total number of shifts = k plus one addition for restoring remainder if needed
  • Assume ─ that X register has (2 k−1) bit for dividend and Y has the k-bit divisor • Assume ─ a sign-bit S shows the sign


  1. Load (upper half k −1 bits of the dividend X) into accumulator k-bit A and load dividend X (lower half bits into the lower k bits at quotient register MQ • Reset sign S = 0 • Subtract the k bits divisor Y from S-A (1 plus k bits) and assign MQ 0 as per S
  2. . If sign of A, S = 0, shift S plus 2 k-bit register pair A-MQ left and subtract the k bits divisor Y from S-A (1 plus k bits); else if sign of A, S = 1, shift S plus 2 k-bit register pair A – MQ left and add the divisor Y into S-A (1 plus k bits) • Assign MQ 0 as per S
  3. Repeat step 2 again till the total number of operations = k.
  4. . If at the last step, the sign of A in S = 1, then add Y into S -A to leave the correct remainder into A and also assign MQ 0 as per S, else do nothing.
  5. . A has the remainder and MQ has the quotient





(Answer all questions. Each carries 3 marks)


  1. Write notes on vectored interrupts.


In a computer, a vectored interrupt is an I/O interrupt that tells the part of the computer that handles I/O interrupts at the hardware level that a request for attention from an I/O device has been received and and also identifies the device that sent the request.

A vectored interrupt is an alternative to a polled interrupt , which requires that the interrupt handler poll or send a signal to each device in turn in order to find out which one sent the interrupt request

  1. Differentiate between synchronous and asynchronous buses.

 Synchronous bus:

  • Transmitter and receivers are synchronized of clock.
  • Data bits are transmitted with synchronization of clock.
  • Character is received at constant Rate.
  • Data transfer takes place in block.
  • Start and stop bit are required to establish communication of each character.
  • Used in high – speed transmission.

Asynchronous bus:

  • Transmitters and receivers are not synchronized by clock.
  • Bit’s of data are transmitted at constant rate.
  • Character may arrive at any rate at receiver.
  • Data transfer is character oriented.
  • Start and stop bits are required to establish communication of each character.
  • Used in low – speed transmission.


  1. Briefly explain static memory.

SRAM (static RAM) is random access memory (RAM) that retains data bits in its memory as long as power is being supplied. Unlike dynamic RAM (DRAM), which stores bits in cells consisting of a capacitor and a transistor, SRAM does not have to be periodically refreshed.


  • Low power consumption
  • Simplicity – a refresh circuit is not needed
  • Reliability


  • Price
  • Capacity


  1. Describe the LRU algorithm for cache replacement.


Discards the least recently used items first. This algorithm requires keeping track of what was used when, which is expensive if one wants to make sure the algorithm always discards the least recently used item. General implementations of this technique require keeping “age bits” for cache-lines and track the “Least Recently Used” cache-line based on age-bits. In such an implementation, every time a cache-line is used, the age of all other cache-lines changes.

The access sequence for the below example is A B C D E D F.


In the above example once A B C D gets installed in the blocks with sequence numbers (Increment 1 for each new Access) and when E is accessed, it is a miss and it needs to be installed in one of the blocks. According LRU Algorithm, since A has the lowest Rank(A(0)), E will replace A.



(Answer any two. Each carries 9 marks)


  1. a) Which are the different bus arbitration schemes? (5)
  • There are two approaches to bus arbitration: Centralized and distributed.
  1. Centralized Arbitration
    • In centralized bus arbitration, a single bus arbiter performs the required arbitration. The bus arbiter may be the processor or a separate controller connected to the bus.
    • There are three different arbitration schemes that use the centralized bus arbitration approach. There schemes are:
  1. Daisy chaining
  2. Polling method
  3. Independent request
  4. a) Daisy chaining
    • The system connections for Daisy chaining method are shown in fig below.
  • It is simple and cheaper method. All masters make use of the same line for bus request.
  • In response to the bus request the controller sends a bus grant if the bus is free.
  • The bus grant signal serially propagates through each master until it encounters the first one that is requesting access to the bus. This master blocks the propagation of the bus grant signal, activities the busy line and gains control of the bus.
  • Therefore any other requesting module will not receive the grant signal and hence cannot get the bus access.
  1. b) Polling method
  • The system connections for polling method are shown in figure above.
  • In this the controller is used to generate the addresses for the master. Number of address line required depends on the number of master connected in the system.
  • For example, if there are 8 masters connected in the system, at least three address lines are required.
  • In response to the bus request controller generates a sequence of master address. When the requesting master recognizes its address, it activated the busy line ad begins to use the bus.
  1. c) Independent request
  • The figure below shows the system connections for the independent request scheme.
  • In this scheme each master has a separate pair of bus request and bus grant lines and each pair has a priority assigned to it.
  • The built in priority decoder within the controller selects the highest priority request and asserts the corresponding bus grant signal.
  1. Distributed Arbitration
  • In distributed arbitration, all devices participate in the selection of the next bus master.
  • In this scheme each device on the bus is assigned a4-bit identification number.
  • The number of devices connected on the bus when one or more devices request for the control of bus, they assert the start-arbitration signal and place their 4-bit ID numbers on arbitration lines, ARB0 through ARB3.
  • These four arbitration lines are all open-collector. Therefore, more than one device can place their 4-bit ID number to indicate that they need to control of bus. If one device puts 1 on the bus line and another device puts 0 on the same bus line, the bus line status will be 0. Device reads the status of all lines through inverters buffers so device reads bus status 0as logic 1. Scheme the device having highest ID number has highest priority.
  • When two or more devices place their ID number on bus lines then it is necessary to identify the highest ID number on bus lines then it is necessary to identify the highest ID number from the status of bus line. Consider that two devices A and B, having ID number 1 and 6, respectively are requesting the use of the bus.
  • Device A puts the bit pattern 0001, and device B puts the bit pattern 0110. With this combination the status of bus-line will be 1000; however because of inverter buffers code seen by both devices is 0111.
  • Each device compares the code formed on the arbitration line to its own ID, starting from the most significant bit. If it finds the difference at any bit position, it disables its drives at that bit position and for all lower-order bits.
  • It does so by placing a 0 at the input of their drive. In our example, device detects a different on line ARB2 and hence it disables its drives on line ARB2, ARB1 and ARB0. This causes the code on the arbitration lines to change to 0110. This means that device B has won the race.
  • The decentralized arbitration offers high reliability because operation of the bus is not dependent on any single device.


  1. b) Write notes on flash memory (4)


Flash Memory (sometimes called “Flash RAM“) is a type of RAM that, like a ROM, retains its contents when the power supply is removed, but whose contents can be easily erased by applying a short pulse of higher voltage. This is called flash erasure, hence the name. Flash memory is currently both too expensive and too slow to serve as MAIN MEMORY, but is used as removable storage cards for digital cameras and pocket computers.

It is a variation of electrically erasable programmable read-only memory (EEPROM) which, unlike flash memory, is erased and rewritten at the byte level, which is slower than flash memory updating. Flash memory is often used to hold control code such as the basic input/output system (BIOS) in a personal computer. When BIOS needs to be changed (rewritten), the flash memory can be written to in block (rather than byte) sizes, making it easy to update. On the other hand, flash memory is not useful as random access memory (RAM) because RAM needs to be addressable at the byte (not the block) level.

Flash Memory gets its name because the microchip is so organized that a section of memory cells are erased in a single action or “flash.” The erasure is caused by tunneling in which electrons pierce through a thin dielectric material to remove an electronic charge from a floating gate associated with each memory cell. Intel offers a form of flash memory that holds two bits (rather than one) in each memory cell, thus doubling the capacity of memory without a corresponding increase in price. Flash memory is used in digital cellular phones, digital cameras, LAN switches, PC Cards for notebook computers, digital set-up boxes, embedded controllers, and other devices.

13. Explain the working of Universal Serial Bus (USB).

USB, short for Universal Serial Bus, is a standard type of connection for many different kinds of devices.

Generally, USB refers to the types of cables and connectors used to connect these many types of external devices to computers.

The Universal Serial Bus standard has been extremely successful. USB ports and cables are used to connect hardware such as printers, scanners, keyboards, mice, flash drives, external hard drives, joysticks, cameras, and more to computers of all kinds, including desktops, tablets, laptops, netbooks, etc.

In fact, USB has become so common that you’ll find the connection available on nearly any computer-like device such as video game consoles, home audio/visual equipment, and even in many automobiles.

Many portable devices, like smartphones, ebook readers, and small tablets, use USB primarily for charging. USB charging has become so common that it’s now easy to find replacement electrical outlets at home improvement stores with USB ports built it, negating the need for a USB power adapter.

USB Versions

There have been three major USB standards, 3.1 being the newest:

  • USB 3.1: Called Superspeed+, USB 3.1 compliant devices are able to transfer data at 10 Gbps (10,240 Mpbs).
  • USB 3.0: Called SuperSpeed USB, USB 3.0 compliant hardware can reach a maximum transmission rate of 5 Gbps (5,120 Mbps).
  • USB 2.0: Called High-Speed USB, USB 2.0 compliant devices can reach a maximum transmission rate of 480 Mbps.
  • USB 1.1: Called Full Speed USB, USB 1.1 devices can reach a maximum transmission rate of 12 Mbps.

How USB Works

When a computer is powered up and USB devices are connected to a hub, the system will query and request from them the information on how much bandwidth is needed. Enumeration process will then occur where each device is assigned with a unique address. After that, the system will determine what kind of data the USB devices wish to transfer. There will be 4 different modes of transfers:

  • Interrupt: used for devices which transfer little amount of data but need fast response (E.g. mouse, keyboard)
  • Bulk: used for devices which receive big packet of data (E.g. printer)
  • Isochronous: used for devices which requires streaming process (E.g. speaker, webcam)
  • Control: short, simple commands to the device, and a status response.


After all connected devices are enumerated, the computer system will take care of the overall bandwidth and allocate it to different devices according to their transfer mode. Most of the bandwidth will be used for interrupt and isochronous transfer to ensure their requests are guaranteed. Once 90% of bandwidth is taken, the computer will refuse any other transfer from these two modes. Bulk or control transfer (if available) will then take up the remaining bandwidth amount, which is up to 10%.

USB Main Features

  • A maximum of 127 peripherals can be connected to a single USB host controller.
  • USB device has a maximum speed up to 480 Mbps (for USB 2.0).
  • Length of individual USB cable can reach up to 5 meters without a hub and 40 meters with hub.
  • USB acts as “plug and play” device.
  • USB can draw power by its own supply or from a computer. USB devices use power up to 5 voltages and deliver up to up to 500 mA.
  • If a computer turns into power-saving mode, some USB devices will automatically convert themselves into “sleep” mode
  1. a) Describe the different types of DRAMS. (5)


DRAM types

The different types of DRAM are used for different applications as a result of their slightly varying properties. The different types are summarised below:

  • Asynchronous DRAM:   Asynchronous DRAM is the basic type of DRAM on which all other types are based. Asynchronous DRAMs have connections for power, address inputs, and bidirectional data lines.

    Although this type of DRAM is asynchronous, the system is run by a memory controller which is clocked, and this limits the speed of the system to multiples of the clock rate. Nevertheless the operation of the DRAM itself is not synchronous.

    There are various types of asynchronous DRAM within the overall family:

    • RAS only Refresh, ROR:   This is a classic asynchronous DRAM type and it is refreshed by opening each row in turn. The refresh cycles are spread across the overall refresh interval. An external counter is required to refresh the rows sequentially.
    • CAS before RAS refresh, CBR:   To reduce the level of external circuitry the counter required for the refresh was incorporated into the main chip. This became the standard format for refresh of an asynchronous DRAM. (It is also the only form generally used with SDRAM).
  • FPM DRAM:   FPM DRAM or Fast Page Mode DRAM was designed to be faster than conventional types of DRAM. As such it was the main type of DRAM used in PCs, although it is now well out of date as it was only able to support memory bus speeds up to about 66 MHz.
  • EDO DRAM:   Extended Data Out DRAM was a form of DRAM that provided a performance increase over FPM DRAM. Yet this type of DRAM was still only able to operate at speeds of up to about 66 MHz.

    EDO DRAM is sometimes referred to as Hyper Page Mode enabled DRAM because it is a development of FPM type of DRAM to which it bears many similarities. The EDO DRAM type has the additional feature that a new access cycle could be started while the data output from the previous cycle was still present. This type of DRAM began its data output on the falling edge of /CAS line. However it did not inhibit the output when /CAS line rises. Instead, it held the output valid until either /RAS was dis-asserted, or a new /CAS falling edge selected a different column address. In some instances it was possible to carry out a memory transaction in one clock cycle, or provide an improvement from using three clock cycles to two dependent upon the scenario and memory used.

    This provided the opportunity to considerably increase the level of memory performance while also reducing costs.

  • BEDO DRAM:   The Burst EDO DRAM was a type of DRAM that gave improved performance of the straight EDO DRAM. The advantage of the BEDO DRAM type is that it could process four memory addresses in one burst saving three clock cycles when compared to EDO memory. This was done by adding an on-chip address counter count the next address.

    BEDO DRAM also added a pipelined to enable the page-access cycle to be devided in to two components

  1. the first component accessed the data from the memory array to the output stage
  2. the second component drove the data bus from this latch at the appropriate logic level

Since the data was already in the output buffer, a faster access time is achieved – up to 50% improvement when compared to conventional EDO DRAM.

BEDO DRAM provided a significant improvement over previous types of DRAM, but by the time it was introduced, SDRAM had been launched and took the market. Therefore BEDO DRAM was little used.

  • SDRAM:   Synchronous DRAM is a type of DRAM that is much faster than previous, conventional forms of RAM and DRAM. It operates in a synchronous mode, synchronising with the bus within the CPU.
  • RDRAM:   This is Rambus DRAM – a type of DRAM that was developed by Rambus Inc, obviously taking its name from the company. It was a competitor to SDRAM and DDR SDRAM, and was able to operate at much faster speeds than previous versions of DRAM.
  1. b) Compare the speed, size and cost of different types of memories
Dynamic RAM (DRAM)

Each memory cell in a DRAM chip holds one bit of data and is composed of a transistor and a capacitor. The transistor functions as a switch that allows the control circuitry on the memory chip to read the capacitor or change its state, while the capacitor is responsible for holding the bit of data in the form of a 1 or 0.

In terms of function, a capacitor is like a container that stores electrons. When this container is full, it designates a 1, while a container empty of electrons designates a 0. However, capacitors have a leakage that causes them to lose this charge, and as a result, the “container” becomes empty after just a few milliseconds.

Thus, in order for a DRAM chip to work, the CPU or memory controller must recharge the capacitors that are filled with electrons (and therefore indicate a 1) before they discharge in order to retain the data. To do this, the memory controller reads the data and then rewrites it. This is called refreshing and occurs thousands of times per second in a DRAM chip. This is also where the “Dynamic” in Dynamic RAM originates, since it refers to the refreshing necessary to retain the data.

Because of the need to constantly refresh data, which takes time, DRAM is slower.

Static RAM (SRAM)

Static RAM, on the other hand, uses flip-flops, which can be in one of two stable states that the support circuitry can read as either a 1 or a 0. A flip-flop, while requiring six transistors, has the advantage of not needing to be refreshed. The lack of a need to constantly refresh makes SRAM faster than DRAM; however, because SRAM needs more parts and wiring, an SRAM cell takes up more space on a chip than a DRAM cell does. Thus, SRAM is more expensive, not only because there is less memory per chip (less dense) but also because they are harder to manufacture.


Because SRAM does not need to refresh, it is typically faster. The average access time of DRAM is about 60 nanoseconds, while SRAM can give access times as low as 10 nanoseconds.

Capacity and Density

Because of its structure, SRAM needs more transistors than DRAM to store a certain amount of data. While a DRAM module only requires one transistor and one capacitor to store every bit of data, SRAM needs 6 transistors. Since the number of transistors in a memory module determines its capacity, for a similar number of transistors, a DRAM module can have up to 6 times more capacity than an SRAM module.

Power Consumption

Typically, an SRAM module consumes less power than a DRAM module. This is because SRAM only requires a small steady current while DRAM requires bursts of power every few milliseconds to refresh. This refresh current is several orders of magnitude greater than the low SRAM standby current. Thus, SRAM is used in most portable and battery-operated equipment.

However, the power consumption of SRAM does depend on the frequency at which it is accessed. When SRAM is used at a slower pace, it draws nearly negligible power while idled. On the other hand, at higher frequencies, SRAM can consume as much power as DRAM.


SRAM is much more expensive than DRAM. A gigabyte of SRAM cache costs around $5000, while a gigabyte of DRAM costs $20-$75. Since SRAM uses flip-flops, which can be made of up to 6 transistors, SRAM needs more transistors to store 1 bit than DRAM does, which only uses a single transistor and capacitor. Thus, for the same amount of memory, SRAM requires a higher number of transistors, which increases the production cost.




(Answer any four. Each carries 10 marks)


  1. Which are the different methods of processor organization?

Processor Organization

There are several components inside a CPU, namely, ALU, control unit, general purpose register, Instruction registers etc. Now we will see how these components are organized inside CPU. There are several ways to place these components and inteconnect them. One such organization is shown in the Figure 5.6.

In this case, the arithmatic and logic unit (ALU), and all CPU registers are connected via a single common bus. This bus is internal to CPU and this internal bus is used to transfer the information between different components of the CPU. This organization is termed as single bus organization, since only one internal bus is used for transferring of information between different components of CPU. We have external bus or buses to CPU also to connect the CPU with the memory module and I/O devices. The external memory bus is also shown in the Figure 5.6 connected to the CPU via the memory data and address register MDR and MAR.

The number and function of registers R0 to R(n-1) vary considerably from one machine to another. They may be given for general-purpose for the use of the programmer. Alternatively, some of them may be dedicated as special-purpose registers, such as index register or stack pointers.

In this organization, two registers, namely Y and Z are used which are transperant to the user. Programmer can not directly access these two registers. These are used as input and output buffer to the ALU which will be used in ALU operations. They will be used by CPU as temporary storage for some instructions.


Figure 5.6 : Single bus organization of the data path inside the CPU

For the execution of an instruction, we need to perform an instruction cycle.  An instruction cycle consists of two phase,

  • Fetch  cycle  and
  • Execution   cycle.

Most of the operation of a CPU can be carried out by performing one or more of the following functions in some prespecified sequence:

  1. Fetch the contents of a given memory location and load them into a CPU register.
  2. Store a word of data from a CPU register into a given memory location.
  3. Transfer a word of data from one CPU register to another or to the ALU.
  4. Perform an arithmatic or logic operation, and store the result in a CPU register.

Now we will examine the way in which each of the above functions is implemented in a computer. Fetching a Word from Memory:

Information is stored in memory location indentified by their address. To fetch a word from memory, the CPU has to specify the address of the memory location where this information is stored and request a Read operation. The information may include both, the data for an operation or the instruction of a program which is available in main memory.

To perform a memory fetch operation, we need to complete the following tasks:

The CPU transfers the address of the required memory location to the Memory Address Register (MAR).

The MAR is connected to the memory address line of the memory bus, hence the address of the required word is transfered to the main memory.

Next, CPU uses the control lines of the memory bus to indicate that a Read operation is initiated. After issuing this request, the CPU waits until it receives an answer from the memory, indicating that the requested operation has been completed.

This is accomplished by another control signal of memory bus known as Memory-Function-Complete (MFC).

The memory set this signal to 1 to indicate that the contents of the specified memory location are available in memory data bus.

As soon as MFC signal is set to 1, the information available in the data bus is loaded into the Memory Data Register (MDR) and this is available for use inside the CPU.

Storing a word into memory

The procedure of writing a word into memory location is similar to that for reading one from memory. The only difference is that the data word to be written is first loaded into the MDR, the write command is issued.

As an example, assumes that the data word to be stored in the memory is in register R1 and that the memory address is in register R2. The memory write operation requires the following sequence:

  1. MAR  [R2]
  2. MDR [R1]
  3. Write
  4. Wait for  MFC

–  In this case step 1 and step 2 are independent and so they can be carried out in any order. In fact, step 1 and 2 can be carried out simultaneously, if this is allowed by the architecture, that is, if these two data transfers (memory address and data) do not use the same data path.

In case of both memory read and memory write operation, the total time duration depends on wait for the MFC signal, which depends on the speed of the memory module.

There is a scope to improve the performance of the CPU, if CPU is allowed to perform some other operation while waiting for MFC signal. During the period, CPU can perform some other instructions which do not require the use of MAR and MDR.

Register Transfer Operation

Register transfer operations enable data transfer between various blocks connected to the common bus of CPU. We have several registers inside CPU and it is needed to transfer information from one register another. As for example during memory write operation data from appropriate register must be moved to MDR.

Since the input output lines of all the register are connected to the common internal bus, we need appropriate input output gating. The input and output gates for register Ri are controlled by the signal Ri in and Ri out respectively.

Thus, when Ri in set to 1 the data available in the common bus is loaded into Ri . Similarly when, Ri out is set to 1, the contents of the register Ri are placed on the bus. To transfer data from one register to other register, we need to generate the appropriate register gating signal.

Performing the arithmetic or logic operation:

    • Generally ALU is used inside CPU to perform arithmetic and logic operation. ALU is a combinational logic circuit which does not have any internal storage.

Therefore, to perform any arithmetic or logic operation (say binary operation) both the input should be made available at the two inputs of the ALU simultaneously. Once both the inputs are available then appropriate signal is generated to perform the required operation.

We may have to use temporary storage (register) to carry out the operation in ALU .

The sequence of operations that have to carried out to perform one ALU operation depends on the organization of the CPU. Consider an organization in which one of the operand of ALU is stored in some temporary register Y and other operand is directly taken from CPU internal bus. The result of the ALU operation is stored in another temporary register Z. This organization is shown in the Figure 5.7.


Figure 5.7 : Organization for Arithmatic & Logic Operation.

  1. Explain the design of a 4bit Arithmetic unit with two selection variables, which performs the basic arithmetic functions.

An Arithmetic and Logic Unit (ALU) is a combinational circuit that performs logic and arithmetic micro-operations on a pair of n-bit operands (ex. A[3:0] and B[3:0]). The operations performed by an ALU are controlled by a set of function-select inputs. In this lab you will design a 4-bit ALU with 3 function-select inputs: Mode M, Select S1 and S0 inputs. The mode input M selects between a Logic (M=0) and Arithmetic (M=1) operation. The functions performed by the ALU are specified in Table I.

Table 1: Functions of ALU
M = 0 Logic
0 0 X Ai.Bi AND
0 1 X Ai + Bi OR
1 0 X Ai $ Bi XOR
1 1 X Ai !$ Bi XNOR
M = 1 Arithmetic
0 0 0 A Transfer A
0 0 1 A + 1 Increment A by 1
0 1 0 A + B Add A and B
0 1 1 A + B + 1 Increment the sum of A and B by 1
1 0 0 A + B’ A plus one’s complement of B
1 0 1 A – B Subtract B from A (i.e. B’ + A + 1)
1 1 0 A’ + B B plus one’s complement of A
1 1 1 B – A B minus A (or A’ + B + 1)

A block diagram is given in Figure 1.
Figure 1: Block diagram of the 4-bit ALU.

When doing arithmetic, we need to decide how to represent negative numbers. As is commonly done in digital systems, negative numbers are represented in  twos complement. This has a number of advantages over the sign and magnitude representation such as easy addition or subtraction of mixed positive and negative numbers. Also, the number zero has a unique representation in twos complement. The twos complement of a n-bit  number N is defined as,

2n – N = (2n – 1 – N) + 1

The last representation gives us an easy way to find twos complement: take the bit wise complement of  the number and add 1 to it. As an example, to represent the number -5, we take twos complement of 5 (=0101) as follows,

5   0 1 0 1   –>      1 0 1 0  (bit wise complement)

     + 1

1 0 1 1  (twos complement)
Numbers represented in twos complement lie within the range -(2n-1) to  +(2n-1 – 1). For a  4-bit number this means that the number is in  the range -8 to +7. There is a potential problem we still need to be aware of when working with two’s complement, i.e. over- and underflow as is illustrated in the example below,

0 1 0 0    (=carry Ci)

+5       0 1 0 1

+4   +   0 1 0 0

+9     0 1 0 0 1    = -7!

1 0 0 0    (=carry Ci)

-7        1 0 0 1

-2     +  1 1 1 0

-9      1 0 1 1 1    = +7!
Both calculations give the wrong results (-7 instead of +9 or +7 instead of -9) which is caused by the fact that the result +9 or -9 is out of the allowable range for a 4-bit twos complement number. Whenever the result is larger than +7 or smaller than -8 there is an overflow or underflow and the result of the addition or subtraction is wrong. Overflow and underflow can be easily detected when the carry out of the most significant stage (i.e. C4 ) is different from the carry out of the previous stage (i.e. C3).
You can assume that the inputs  A and B are in twos complement when they are presented to the input of the ALU.


  1. a) Explain the design of status register. (5)

The Status Register

Bit7 Bit6 Bit5 Bit4 Bit3 Bit2 Bit1 Bit0




The STATUS register is of most importance to programming the PIC, it contains the arithmetic status of the ALU (Arithmetic Logic Unit), the RESET status and the bank select bit for data memory. As with any register, the STATUS register can be the destination for any instruction. If the STATUS register is the destination for an instruction that affects the Z, DC or C bits, then the write to these three bits is disabled. These bits are set or cleared according to device logic. Furthermore, the TO and PD bits are not writable. Therefore, the result of an instruction with the STATUS register as destination may be different than intended. For example, CLRF STATUS will clear the upper-three bits and set the Z bit. This leaves the STATUS register as 000u u1uu (where u = unchanged).



The first three bits (STATUS<0> to STATUS<2>) are the carry (C), digit carry (DC) and zero (Z) flags of the ALU respectively. The values of these bits change depending on the results of arithmetic or logical operations performed during program execution. Bits 3 and 4 are the power down PD and watchdog timer timeout TO bits respectively and bits 5and6(RP0 and RP1) are the bank selection bits.


R/W-0 R/W-0 R/W-0 R-1 R-1 R/W-x R/W-x R/W-x




Bit7 Bit6 Bit5 Bit4 Bit3 Bit2 Bit1 Bit0
R = Readable bit

W= Writable bit

U = Unimplemented bit, read as ‘0’

-n= Value at POR reset


bit 7:

IRP: Register Bank Select bit (used for indirect addressing)
0 = Bank 0, 1 (00h – FFh)
1 = Bank 2, 3 (100h – 1FFh)
The IRP bit is not used by the PIC16F8X. IRP should be maintained clear.
bit 6-5:

RP1:RP0: Register Bank Select bits (used for direct addressing)
00 = Bank 0 (00h – 7Fh)
01 = Bank 1 (80h – FFh)
10 = Bank 2 (100h – 17Fh)
11 = Bank 3 (180h – 1FFh)
Each bank is 128 bytes. Only bit RP0 is used by the PIC16F8X. RP1 should be maintained clear.
bit 4:

TO: Time-out bit
1 = After power-up, CLRWDT instruction, or SLEEP instruction
0 = A WDT time-out occurred
bit 3:

PD: Power-down bit
1 = After power-up or by the CLRWDT instruction
0 = By execution of the SLEEP instruction
bit 2:

Z: Zero bit
1 = The result of an arithmetic or logic operation is zero
0 = The result of an arithmetic or logic operation is not zero
bit (for ADDWF and ADDLW instructions) (For borrow the polarity is reversed)
bit 1:

DC: Digit carry/borrow
1 = A carry-out from the 4th low order bit of the result occurred
0 = No carry-out from the 4th low order bit of the result
bit (for ADDWF and ADDLW instructions)
bit 0:

C: Carry/borrow
1 = A carry-out from the most significant bit of the result occurred
0 = No carry-out from the most significant bit of the result occurred
Note: For borrow the second operand the polarity is reversed. A subtraction is executed by adding the two’s complement of. For rotate (RRF, RLF) instructions, this bit is loaded with either the high or low order bit of the source register.



  1. b) Give the design of a 4 bit shifter. (5)

4-bit Serial-in to Parallel-out Shift Register


The operation is as follows. Lets assume that all the flip-flops ( FFA to FFD ) have just been RESET ( CLEAR input ) and that all the outputs QA to QD are at logic level “0” ie, no parallel data output.

If a logic “1” is connected to the DATA input pin of FFA then on the first clock pulse the output of FFA and therefore the resulting QA will be set HIGH to logic “1” with all the other outputs still remaining LOW at logic “0”. Assume now that the DATA input pin of FFA has returned LOW again to logic “0” giving us one data pulse or 0-1-0.

The second clock pulse will change the output of FFA to logic “0” and the output of FFB and QB HIGH to logic “1” as its input D has the logic “1” level on it from QA. The logic “1” has now moved or been “shifted” one place along the register to the right as it is now at QA.

When the third clock pulse arrives this logic “1” value moves to the output of FFC ( QC ) and so on until the arrival of the fifth clock pulse which sets all the outputs QA to QDback again to logic level “0” because the input to FFA has remained constant at logic level “0”.

The effect of each clock pulse is to shift the data contents of each stage one place to the right, and this is shown in the following table until the complete data value of  0-0-0-1 is stored in the register. This data value can now be read directly from the outputs of QAto QD.

Then the data has been converted from a serial data input signal to a parallel data output. The truth table and following waveforms show the propagation of the logic “1” through the register from left to right as follows.

Basic Data Movement Through A Shift Register

Clock Pulse No QA QB QC QD
0 0 0 0 0
1 1 0 0 0
2 0 1 0 0
3 0 0 1 0
4 0 0 0 1
5 0 0 0 0



  1. Explain the design of micro program sequencer with an example.


A micro-program sequencer works in a way to generate these control signals from the microprogram

by transitioning from one state to another in every clock cycle. A state is defined by the

micro-instruction that has to be run in that clock cycle.It has two main functions

  1. Control Function The micro-operations that need to be executed to perform a certain microinstruction

are to be defined and be known. The micro-operation(s) are dependent on parameters

like selected destination, operand etc.

  1. Sequencing Function The address of next micro-instruction to be executed is generated while

controlling test conditions etc.

Thus, to summarize, to execute an instruction, the microprogram sequencer executes a micro-instruction

in every clock cycle and determines which micro-instruction (state) to run next. It can be thought in terms

of a state diagram.


Work Flow

Figure 1: Flow of the Control Unit

The Instruction Register loads the opcode into the decoder which then translates the opcode into a control memory address. The control address register contains the address of the next micro-instruction to be read.

The micro-instructions are stored in the microprogram memory (also called control memory, can be used interchangeably). The address from the control address register is used to read from this microprogram memory.

When a micro-instruction is read from the microprogram memory, it is transferred to the control buffer register. This register activates the control signals.

Thus, reading a micro-instruction from the microprogram memory has the effect of executing that micro-instruction. The sequencing logic loads the control address register and activates the read signal. This read signal loads the next micro-instruction from the Instruction Register completing a cycle.


  1. Explain the procedure for designing a hardwired control, using an appropriate example.


For each instruction, the control unit causes the CPU to execute a sequence of steps correctly. In reality, there must be control signals to assert lines on various digital components to make things happen. For example, when we perform an Add instruction in assembly language, we assume the addition takes place because the control signals for the ALU are set to “add” and the result is put into the AC. The ALU has various control lines that determine which operation to perform. The question we need to answer is, “How do these control lines actually become asserted?” We can take one of two approaches to ensure control lines are set properly. The first approach is to physically connect all of the control lines to the actual machine instructions. The instructions are divided up into fields, and different bits in the instruction are combined through various digital logic components to drive the control lines. This is called hardwired control, and is illustrated in figure (1). The control unit is implemented using hardware (for example: NAND gates, flip-flops, and counters).We need a special digital circuit that uses , as inputs, the bits from the Opcode field in our instructions, bits from the flag (or status) register, signals from the bus, and signals from the clock. It should produce, as outputs, the control signals to drive the various components in the computer. The advantage of hardwired control is that is very fast. The disadvantage is that the instruction set and the control logic are directly tied together by special circuits that are complex and difficult to design or modify.


If someone designs a hardwired computer and later decides to extend the instruction set, the physical components in the computer must be changed. This is prohibitively expensive, because not only must new chips be fabricated but also the old ones must be located and replaced.


  1. a) Explain the different methods of control organization. (5)


  1. b) Explain micro programmed CPU organization with the help of a diagram. (5)

In hardwired control , we saw how all the control signals required inside the CPU can be generated using a state counter and a PLA circuit.

There is an alternative approach by which the control signals required inside the CPU can be generated . This alternative approach is known as microprogrammed control unit.

In microprogrammed control unit , the logic of the control unit is specified by a microprogram.A

microprogram consists of a sequence of instructions in a microprogramming language. These are very instructions that specify microoperations.  A microprogrammed control unit is a relatively simple logic circuit that is capable of (1) sequencing through microinstructions and (2) generating control signals to execute each microinstruction. The concept of microprogram is similar to computer program. In computer program the complete instructions of the program is stored in main memory and during execution it fetches the instructions from main memory one after another. The sequence of instruction fetch is controlled by program counter (PC) . Microprogram are stored in microprogram memory and the execution is controlled by microprogram counter (PC ) .Microprogram consists of microinstructions which are nothing but the strings of 0’s and 1’s . In a particular instance ,we read the contents of one location of microprogram memory , which is nothing but a microinstruction . Each output line ( data line )  of microprogram memory corresponds to one control signal. If the  contents of the memory cell is ) , it indicates that the signal is to generated and if the contents of memory cell is 1 , it indicates that generate that control signal at that instant of time.

Control Word (CW) :  Control word is defined as a word whose individual bits represent the various control signal. Therefore each of the control steps in the control sequence of an instruction defines a unique combination of 0s and 1s in the CW.

A sequence of control words ( CWs ) corresponding to the control sequence of a machine instruction constitutes the microprogram for that instruction.

The individual control words in this microprogram are referred to as microinstructions.The microprograms corresponding to the instruction set of a computer are stored ina aspecial memory which will be referred to as the microprogram memory. The control words related to an instructions are stored in microprogram memory.The control unit can generate the control signals for any instruction by sequencially reading the CWs of  the corresponding microprogram from the microprogram memory.

To read the control word sequentially from the microprogram memory a microprogram counter (PC ) is needed.

The basic organization of a microprogrammed control unit is shown in the figure.


Basic organization of a microprogrammed control
The “starting address generator “  block is responsible for loading the starting address of the microprogram into the PC everytime a new instruction is loaded in the IR.The PC is then automatically incremented the clock, and it reads the successive microinstruction from memory . Each microinstruction basically provides the required control signal at that time step. The microprogram counter ensures that the control signal will be delivered to the various parts of the CPU in correct sequence.We have some instructions whose execution depends on the status of condition codes and status flag , as for example , the branch instruction. During branch instruction execution it is required to take the decision between the alternative action.To handle such type of instructions with microprogrammed control , the design of control unit is based on the concept of  conditional branching  in the microprogram. For that it is required to include some conditional branch microinstructions. In conditional microinstructions , it is required to specify the address of the microprogram memory to which the control must direct. It is known as branch address. Apart from branch address , these microinstructions can specify which of the states flags ,  condition codes , or possibly , bits of the instruction register should be checked as a condition for branching to take place.


To support microprogram branching , the organization of control unit should be odified to accommodate the branching decision.To generate the branch address , it is required to know the status  of the condition codes and status flag .
To generate the starting  address , we need  the instruction which is present in IR. But for branch address generation we have to check the content of condition codes and status flag.The organization of control unit to enable conditional branching in the microprogram is shown in the figure.


The control bits of the microinstructions word which specify the branch conditions and address are fed to the “ Starting and branch address generator “ block.This block performs the function  of loading a new address into the PC when the condition of branch instruction is satisfied. In a computer program we have seen that execution of every instruction consists of two part – fetch phase and execution phase of the instruction. It is also observed that the fetch phase of all instruction is same. In microprogrammed controlled control unit , a common microprogram is used to fetch the instruction. This microprogram is stored in a specific location and execution of each instruction start from that memory location.


At the end of fetch microprogram , the starting address generator unit calculate the appropriate starting address of the microprogram for the instruction which is currently present in IR. After the PC controls the execution of microprogram which generates the appropriate control signal in proper sequence. During the execution of a microprogram , the PC is always incremented everytime a new microinstruction is fetched from the microprogram memory , except in the following situations :

  1. When an End instruction is encountered , the PC is loaded with the address of the first CW in the microprogram for the instruction fetch cycle.


  1. When a new instruction is loaded into the IR , the PC is loaded with the starting address of the microprogram for that instruction.


  1. When a branch microinstruction is encountered , and the branch condition is satisfied , the PC is loaded with the branch address.


Let us examine the contents of microprogram memory and how the microprogram of each instruction is stored or organized in microprogram memory.

Consider the two example First example is  the control sequence for execution of the instruction “ Add contents of memory location addressed in memory direct mode to register RI”
Step                       Action

  1. PCout , MARin Read , Clear Y, set carry_in to ALU, Add , Zin
  2. Zout , PCin , Wait for MFC
  3. MDRout , IRin
  4. Address_field_of_IRow , MARin , Read
  5.  RIout , Yin , Wait for MFC

    6.                MDRout , Add , Zin

    7.                Zout ,  RIin

    8.                End

Control sequence for Conditional Branch instruction (BRN) Branch on negative)


Step                             Action

1.                      PCout , MARin , Read , Clear Y , Set Carry_in_to ALU , Add , Zin

2.                      Zout , PCin , Wait for MFC

  1. MDRout , IRin

    4. IF    then End

    IF    N   then  PCout , Yin

    5.                      Address field of IRout , Add , Zin


First consider the control signal required for fetch instruction , which is same for all the instruction, we are listing them in a particular order.

KTU B.Tech Solved ModelQuestion paper – Computer Organisation


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