Question

Solving both questions would be great but question 1 is the priority right now.

Question 1: Data representation (16 points) a) Given the following signed integer values, what are the corresponding represen- tation in different bases on 16 bits? (4 points) 1) 2's complement AE86(16): Binary= Decimal= 2) 2020(10): Binary= Hexadecimal= b) Given the following signed integer values, what are the corresponding represen- tation in different bases on 32 bits? (4 points) 1) 2's complement BF88(16): Binary= Decimal= 2) -2601(10): Binary: Hexadecimal=

Answer 1

Q1

Part 1)

(AE86) equivalent binary is = 1010 1110 1000 0110

2's complement of AE86 in binary = (0101 0001 0111 1010) base 2

2) 2's complement of AE86 in decimal = (20858) in base 10

2)

a)

2020 in binary of 16bits = 0000 0111 1110 0100

b)

2020 In Hexadecimal = 07E4

Part 2)

a)

BF88 into binary = 1011 1111 1000 1000

2's Complement of BF88 in binary = 0100 0000 0111 1000

b)

2's Complement of BF88 in decimal = 16504

B)

-2601

First do binary of 2601 = 0000 1010 0010 1001

Now do 2's complement for

-2601 in binary = 1111 0101 1101 0111

-2601 in hexadecimal = F5D7

Part C

IEEE single precison format is

Sign Bit (S)	Exponent (E)	Mantisa (M)
1bit	8 Bits	23 bits

We represent number as

FORMAT = +/- 1.M * 2 ^ E-Bias

Here in single precision bias = 127

a)

-687.625

As number is negative so

Sign bit = 1

Now covert 687.625 into binary

678.625= 1010100110.101

Now we convert into format given above =

- 1.010100110101 * 2^9

Now M =0101 0011 0101 0000 0000000 i.e 23 bits

So to calculate E

Do 9 + Bias = 0000 1001 + 0111 1111 = 1000 1000

So to

-678.625 in IEEE single-precision

Sign	exponent	Mantisa
1	10001000	0101 0011 0101 0000 0000000

b)

0.105

As number is positive

S= 0

Now convert 0.105 in binary

0.105=0.000110101

Now convert into given format

1.10101 * 2^-4

So now M = 1010 1000 0000 0000 0000000 i.e 23 bits

To calculate E

Do -4+ Bias = 1111 1100 + 0111 1111

= 0111 1011

Sign	exponent	Mantisa
0	0111 1011	1010 1000 0000 0000 0000000

Part D )

a)

1 1000 0011 1010 1110 rest 0's

Now Follow above and divide it into S ,E and M

S= 1

E=1000 0011

M = 1010 1110 rest 0's

So acc to format

- 1.1010 1110 * 2^E- bias

E- Bias = 1000 0011 - 0111 1111

= 1111 1100 so as E >Bias so E is +ve but it is in 2's complement form s0

2's complement is = 0000 0100 = +4

So number

- 1.1010 1110 * 2^+4
= -11010.1110

=- 26.875

b)

0 0000 0000 1010100000 rest 0's

Agn Divide into S ,E and M

S= 0

E= 0000 0000

M = 1010100000rest 0's

So Acc to fomat number is

1. 101010 * 2^0 = 1.101010

So number = +1.65625

Here all the parts are solved

Q2 )

Part a

I think it is already in simplified form as no boolean algebra property is present among given expression

Ans = AB + B'C + AC

Part b)

Date... - AB + B C + AB - Ving de- morgan's Rule (AB). ( Bc ). (A i (Ā + ).( 5 + 7).(Ā +B) » à à t Ă C + Š + B ī ) Cà +B) 2) Ā Ē + à łt Ā Š Ā Šit À CB 4 2 Ā Š + À č + Ā B C + À CB a) Ā Š + à c t Ă Ĉ C B+ ß ] 2) Ā B + à čo o à [ B + c ] . . » Ā [ B ] 2) A Ā B C

Part C)

Dale... AB + BC & A + AB Using De-morgan's Rule (AB). CBC )+ AC + AB (Ā + Š ) ( B to ) + AC + AB À Ř t à c + B + B łtałt AB combining B TIHA - [lt à ] + A [ [ à + B + + AB Ř t ł[it B J +AB 2 B + č + AB Ans] - AB+ B + c

ThanK You

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