2. Data Representation

Question 1

Convert 0000 1011₂ to base 10.

Answer 1

11

Question 2

MIPS words are 32 bits in length. What is the range of unsigned integers representable in 1 word?

Answer 2

0 to 2³²-1

Question 3

Write -21 in 8-bit sign-magnitude form.

Answer 3

1 001 0101

Question 4

State two disadvantages of sign magnitude form over two’s complement.

Answer 4

• There is a positive and negative zero.
• Adders need an extra step to work out the sign.

Question 5

State an advantage of sign-magnitude form over two’s complement.

Answer 5

It is symmetric.

Question 6

Convert 1111 1100₂ (sign-magnitude) to base 10.

Answer 6

-124

Question 7

Convert 1111 1100₂ (2’s complement) to base 10.

Answer 7

-4

Question 8

Derive the ‘shortcut’ method for converting to two’s complement.

Answer 8

We can see that the inversion of a binary representation (x_i) satisfies:

x + x_i = -1

which implies:

-x = x_i + 1

Question 9

Describe one’s complement notation.

Answer 9

One’s complement is just a flip of bits to convert to negative. In other words, it is one less than the two’s complement.

Question 10

Convert eca8 6420 ₁₆ to binary.

Answer 10

1110 1100 1010 1000 0110 0100 0010 0000

Question 11

Convert 0001 0011 0101 0111 1001 1011 1101 1111 from binary to hexadecimal.

Answer 11

1357 9bdf

Question 12

State the range of signed integers in two’s complement.

Answer 12

-2^n-1 to -2^n-1 - 1

Question 13

Describe the process of sign extending a 16-bit binary number to a 32 bit binary number.

Answer 13

The msb (most significant bit) is copied into the 16 new bits at the start of the numbers.

Question 14

Explain the difference between logical right shift and arithmetic right shift. Why is there solely a left-shift?

Answer 14

Logical right shift fills in the spare bits with the msb, and arithmetic right shift fills it in with 0s. Left shift is always filled with 0s.

Question 15

Why can an overflow not occur when the operands have different signs?

Answer 15

Because the magnitude of the result is smaller than those of the operands, the result will fit in the size the operands are.

Question 16

When does an overflow occur in a binary addition?

Answer 16

When the (msb)s of the operands are the same, and both opposite to the msb of the result.

Question 17

Normalize 0.1 x 10^-8.

Answer 17

1 x 10^-9.

Question 18

In 1.xxxxxx x 2^yyyyyyy, what are xxxx and yyyy?

Answer 18

xxxx is the ‘fraction’ or ‘mantissa’, and yyyy is the ‘exponent’.

Question 19

In IEEE 754, why is the exponent placed before the mantissa?

Answer 19

To make sorting simpler.

Question 20

Explain the reasoning behind (exponent - bias) in IEEE 754.

Answer 20

Negative exponents cause sorting to be harder, and so we wish to have the most negative exponent as 0000 0000 and the most positive as 1111 1111. To do this, we add 127 to the exponent. (Double precision is 1023)

Question 21

What is -0.75 in IEEE 754 binary representation?

Answer 21

1 0111 1110 1000 0000 0000 0000 0000 000

Question 22

What is 1 1000 0001 0100 0000 0000 0000 0000 000

Answer 22

-5

Question 23

An exponent and mantissa of zero means….

An exponent of 255 and a mantissa of 0 means….

An exponent of 255 and a mantissa non-zero means….

Answer 23

a. 0

b. Infinity

c. NaN