exercise 1 (sys)¶

EXERCISE 1:basic conversion and range of two's complement¶

convert the decimal number -105 into an 8-bit two's complement binary repretation.¶

ANS :$105=1*2^6+1*2^5+1*2^2+1*2^1+1*2^0$
$\therefore 105_{(10)}=01100111_{(2)}$
so the complete number is 01100111

convert the 8-bit two's complement binary number 10011100 into its decimal equivalent¶

the origin code is 10011011->11100100
$\therefore -(2^6+2^5+2^2)=-100$

what is the range of integers(in demical)that can be represented in an 8-bit two's complement system?¶

I think we should divide this question into to 3 situation.
Situation 1:there is no sign number and it is original code . The range of the number is 00000000 ~ 11111111. Therefore the integer range is 0~255.
situation 2: there is a sign number and 7 machine number . Also it is original code. The range of the number is 11111111 ~ 01111111. Therefore the range of integer number is -127 ~ 127.
situation 3: it adopts complete code:the code range is 10000000 ~ 01111111. Therefore the integer number range is -128～127

addition/subtraction in two's complement¶

Perform the following operations using 8-bit two's complement arithmetic. Determine whether overflow occurs in each case.¶

$(83)_{10}+(-45)_{10}$:

\[ \begin{gathered} (82)_{10}=01010010_2\\ -45=10011101_2\\ complement number: 01010010 11100011;\\ \therefore ans=00110101=53\\ \end{gathered} \]

it is overflown.

$(-98)_{10}+(-67)_{10}$

$$ \begin{gathered} (-98)_{10}=11100010_2\

-67=11000011_2\

complement number: 10011110 10111101\

\therefore ans=01000011_(2)=67_{10}\ \end{gathered} $$

IEEE 754 single-precision conversion¶

converting the demical number -13.625 into the IEEE 754 standard single -precision (32-bit)floating -point format.¶

sign:1

$13.625_{10}=1101.101$

Exponent:3+127=130

$\therefore 10000010$ value:10110100000000000000000

ANS:1100 0001 0101 1010 0000 0000 0000 0000 =0xC1 5A 00 00

interpreting IEEE754¶

Given the hexadecimal representation of an IEEE 754 single-precision floating-point number: 0xC2E8 0000.¶

1. Write down its binary form and break it down into the sign bit, exponent field, and mantissa (significand) field. - binary form: 1100 0010 1110 1000 0000 0000 0000 0000 - sign bit: 1 - exponent bit: 10000101 - signidicand bit:1101000 0000 0000 0000 0000

2. Calculate the decimal value it represents. - exp:$128+4+1-127=6$ - sign:11101000 0000 0000 0000 0000: $-1.1101_2=-1.8125_{10}$ $$ \begin{gathered} \therefore answer is -1.8125*2^6\ = -116 \end{gathered} $$