Answer:
a. Unsigned integer 107
b. Signed magnitude integer -43
c. One’s complement integer -20
d. Two’s complement integer 21
e. ASCII character k
Explanation:
a) For unsigned integer,
We put this value in representation of binary and put binary number in it.
we will place 1, 2, 4, 8, 16, 32, 64, 128 ...(powers of two)
64 32 16 8 4 2 1
<u>1 1 0 1 0 1 1</u>
The positions at 1 is present,we will add those numbers.In this (64+32+8+2+1) =107 is there.
So,107 will be the representation.
b)For signed magnitude integer,
The representation is just the same,but as signed integers the first bit represent the negative number.
64 32 16 8 4 2 1
<u>1 1 0 1 0 1 1</u>
The first bit is for Negative(-),then we will add other number where 1 is present.In this (32+8+2+1)=43.We will add (-) due to signed integers.
So,-43 will be the representation.
c) For One's complement integer,
We will compliment the bits of binary number.At the place of 1 ,place 0 and at the place of 0,place 1.
<u>1 1 0 1 0 1 1</u>
Compliment <u> 0 0 1 0 1 0 0</u>
Then,We put this value in representation of binary and put binary number in it, we will place 1, 2, 4, 8, 16, 32, 64, 128 ...(powers of two)
64 32 16 8 4 2 1
<u> 0 0 1 0 1 0 0</u>
The positions at 1 is present,we will add those numbers.In this (16+8)=20 we will put negative at the starting because of the compliment
So, -20 will be the representation.
d)For Two's complement integer,
After compliment of bits At the place of 1 ,place 0 and at the place of 0,place 1.Then,we add 1 bit to the Least significant bit(Lsb).
<u>1 1 0 1 0 1 1</u>
Compliment <u> 0 0 1 0 1 0 0</u>
Add 1 to Lsb <u> 0 0 1 0 1 0 0</u>
+ <u> 1</u>
Number <u> 0 0 1 0 1 0 1</u>
Then,We put this value in representation of binary and put binary number in it, we will place 1, 2, 4, 8, 16, 32, 64, 128 ...(powers of two)
64 32 16 8 4 2 1
<u> 0 0 1 0 1 0 1</u>
The positions at 1 is present,we will add those numbers.In this (16+8+1)=21
So,21 will be the representation.
e. For ASCII character,
First,convert it into decimal
We multiply bits with 2^n,from ascending numbers to 0 to (n-1),and add them
= 1 * 2^6 + 1*2^5 + 0* 2^4 + 1*2^3 + 0* 2^2 + 1*2^1 + 1* 2^0
= 64+32+8+2+1
= 107
Then,we check in Ascii table 107 decimal number's position ,k is there.
So,k will be the representation.
