Question

1)

We would like to design a bus system for 32 registers of 16 bits each. How many multiplexers are needed for the design?

Select one:

5

16

1

4

32

2)

The basic computer can be interrupted while another interrupt is being serviced.

Select one:

True

False

3)

If the Opcode bits of an instruction is 111, then the basic computer instruction type is either memory-reference or input-output.

Select one:

True

False

4)

The content of AC in the basic computer is hexadecimal C244 and the initial value of E is 1. The initial value of the PC is hexadecimal 1AF. Determine the contents of AC, E, PC, AR and IR in hexadecimal after the execution of the INC instruction.

Hint: Consider the size of the register when giving your answer. For example, if the size of a register is 12 bits (= 3 hexadecimal digits), your answer should be 0AA, not 00AA.

AC
E
PC
AR
IR

5)

If the Opcode bits of an instruction is 110, then the basic computer instruction type must be memory-reference.

Select one:

True

False

6)

The following figure shows the contents of several memory locations for the basic computer. Determine the efficient address for the instruction at location 14F.

Answer:

7)

If a circular shift right microoperation and later a circular shift left microoperation is applied to an 8-bit register, the register holds its initial value.  

Select one:

True

False

8)

An 8-bit register contains the binary value of 11110000. What is the register value after two arithmetic shift rights?

Hint: Write down only the value of the register, not anything else. For example write, 11110000, do not write the answer is 11110000

Answer:

9)

A computer uses a memory unit of size 256K x 24. A binary instruction code is stored in one word of memory. The instruction has 3 parts: an indirect bit, an operation code and an address part. How many bits are there in the operation code?

Hint: Just type in the numerical value of your answer. For example, write 15, do not write 15 bits.

Answer:

10)

We would like to design a bus system for 16 registers of 32 bits each. How many selection inputs do the multiplexers have?

Select one:

1

32

16

5

4

Answer 1

Ans-1 is 16.

Explanation:-

No. of multiplexers = bits of register

= 16

16 multiplexers are needed to design a bus system.

Ans-2:- True

Explanation:-

the basic computer can be interrupted while other interrupr is under serviced because cpu assign the interruption based on interrupt handling if interrupt has high priority then it can interrupt other lower priority interrupt.

Ans-3:- False.

Explanation:-

When the three operation code bits are equal to 111, control unit inspects the bit in position 15. If the bit is 0, the instruction is a register-reference type. Otherwise, the instruction is an input-output type having bit 1 at position 15.

The memory reference type is from 000 to 110.

Ans-5:- True

Explanation:

the opcode instruction are from 000 to 110 for memory reference type.

Ans-7:- True

Explanation:-

Ans-8:- 11111100

Explanation:-

Date Page No – Explanation - Date T First Arithmatic shift Right :- I Binasy Value MSB LSB Second Asithmatic shift Right :- Binary value Ilololo so. The register value after two Asithmatic shift Right is = |||||100

Ans-9:- 5

Explanation:-

Here, indirect bit = 1 bit

Address = 256 * 1024 (bytes)

= 2^8  * 2^10 = 2^18

= 18 bits

Op code = 24 - 18 - 1 = 5

Ans-10:- 4

Explanation:-

2^n = No. of registers,

where, n = selection input of multiplexer.

2^n = 16 = 2^4

So, n = 4.

4 selection input lines should be in each multiplexer.

No of multiplexer = 32

Total selection input lines = 4 * 32 = 128