A (8 points) For each part, do or answer the following Show how you would add th

Question

A (8 points) For each part, do or answer the following Show how you would add the two 8-bit binary numbers by filling in each of the blank boxes with a 8 or 1. NOTE1: Addition of 2's complement numbers is done just like the addition of unsigned whole numbers, so you only need to show each addition once NOTE2: There are9 blank boxes for the Carry bits and 8 blank boxes for the Sum bits CAUION: Every blank box should be filled with a or 1.f you leave any boxes blank, your answer will be taken as incomplete. - Circle one observation from the list below that is applicable and useful to help determine whether overflow has occurred, if the numbers are unsigned whole numbers and only 8 bits are available for holding the result CAUTION: There wll be penalty for additional observations circled that are not applicable helpful Circle Y or N to indicate vour inference on whether overnow has occurred or not (if the numbers are znwyec hole numbers and only 8 bits are available for holding the result) based on the circled observation(s). - Circle two observations from the list below either of which is applicable and se to help determine whetherlow has occurred, if the numbers are 2's complement numbers and only 8 bits are available for holding the result. CAUTION: There will be penalty for observations circled that are not applicablehelpful. - Circle Y or N to indicate your inferencc on whether overflow has occurred or not (if the numbers arc 2's complement umbers and only 8 bits are available for holding the result) based on the circled observalion(s). List of observations (note that each is in terms of the state(s) of certain bit(s) in or 2 positions something that can be casily made in hardwarc): (A) Carry-out is 0 when bils at mosl-signilicant-bil position are added (B) Carry-out is I when bits at most-significant-bit position arc added. (C) Addends have opposile signs. (D) Addends are both positive, sum is positive. (E) Addends are bolh positive, sum is negative (F) Addends are both negative, sum is positive. (G) Addends are bolh negalive, sum is negative (H) Carry-in and carry-out at most-significant-bit position are the same. (I) Carry-in and carry-out at mosl-signilicanl-bil position are nol the same CAUTION You will carn NO CREDITS if you simple give answers for the overflow part without doing the binary addition part. (a)Carry |observation ifunsigned: ( A / BIC/D1E/F/6/H/1) Overflow if unsigned?YN) 0101011 0 0 11 0 0 1 1 1 2 observations if 2's complement. ( A / B / C / D / E / F f G / H / 1 ) Overflow if 2's complement? ( Y N) Sum: l observation if unsigned. ( A / B / / D I E / F / G } H / 1 ) Overflow if unsigned?Y N) Carry: 0 10 0 0 0 1 0 2 observations if 2's complement: (A BCIDIE/FIGIH I) Overflow if 2's complement? ( Y / N ) Sum:

Explanation / Answer

A)

a)

1. observation if unsigned: A,D,I

overflow if unsigned: N

2. observation if 2's complement:A,E,I

overflow if 2's complement:N

b)

1. observation if unsigned: B,D,H

overflow if unsigned: Y

2. observation if 2's complement:B,C,H

overflow if 2's complement:Y

c)

1. observation if unsigned: B,D,H

overflow if unsigned: Y

2. observation if 2's complement:B,G,H

overflow if 2's complement:Y

d)

1. observation if unsigned: A,D,I

overflow if unsigned: N

2. observation if 2's complement:A,E,I

overflow if 2's complement:N

B)

number = -314.9

converting it into binary number,

-314.910 = 100111010.11100110011001100110011001100110011001100112

Now, in IEEE 754-1985, double precision has,

1 sign bit

11 exponent bit and

52 fraction bit

Now, -314.910 = 100111010.11100110011001100110011001100110011001100112 = 1.00111010111001100110011001100110011001100110011001102 * 28 (scientific notation)

So, sign bit = 1 (because -314.9 is negative)

exponent = 8 + bias. for double precision, bias is 1023. So, exponent = 8 + 1023 = 103110 = 100000001112

fraction = .00111010111001100110011001100110011001100110011001102 . taking 52 bit, fraction = 0011101011100110011001100110011001100110011001100110

So, final, in IEEE 754-1985 double notation,

-314.9 =

C)

number = 423A600016 = 111112192010 = 10000100011101001100000000000002

Now, in IEEE 754-1985, single precision has,

1 sign bit

8 exponent bit and

23 fraction bit

Now, 111112192010 = 10000100011101001100000000000002 = 1.0000100011101001100000000000002 * 230 (scientific notation)

So, sign bit = 0 (because 1111121920 is positive)

exponent = 30 + bias. for double precision, bias is 127. So, exponent = 30 + 127= 15710 = 100111012

fraction = .0000100011101001100000000000002 . taking 23 bit, fraction = 00001000111010011000000

So, final, in IEEE 754-1985 single notation,

423A600016 = 111112192010 =  

Carry 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 Sum 1 0 1 1 1 1 0 1

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