What is the 8-bit binary representation of -49 in sign magnitude form

1. What is the 8-bit binary representation of -49 in sign magnitude form?
2. What is the 8-bit binary representation of -31 in excess 127 form?
3. What is the decimal value of 11001010 when viewed as an unsigned 8-bit integer?
4. What is the decimal value of 11001010 when viewed as a sign magnitude 8-bit integer?
5. What is the 32 bit floating point representation of 1.5 in binary and hex form?
6. How many bits would a complete load instruction need to have to hold a main memory address of decimal 65536

The Correct Answer and Explanation is :

Let’s break down each of your questions step by step:

1. 8-bit binary representation of -49 in sign magnitude form:

• Sign magnitude representation works by encoding the sign separately. In an 8-bit system:
  • First, convert 49 to binary: 49 in decimal = 00110001 in binary.
  • Since it is negative, the first bit is set to 1 to represent the sign.
  • Thus, the sign magnitude representation of -49 is: 10110001.

1. 8-bit binary representation of -31 in excess 127 form:

• Excess 127 is a form of bias representation where the number is offset by a constant value (127 in this case).
• First, convert 31 to binary: 31 in decimal = 00011111 in binary.
• Add the bias of 127 to it: 31 + 127 = 158.
• Convert 158 to binary: 158 in decimal = 10011110 in binary.
• Therefore, the excess 127 representation of -31 is 10011110.

1. Decimal value of 11001010 when viewed as an unsigned 8-bit integer:

• Treat the binary number as an unsigned 8-bit integer, where the value is simply the binary converted to decimal:
  • 11001010 in binary = 202 in decimal.

1. Decimal value of 11001010 when viewed as a sign magnitude 8-bit integer:

• In sign magnitude form, the most significant bit represents the sign, and the remaining bits represent the magnitude:
  • The first bit is 1, which indicates a negative number.
  • The magnitude part is 1001010, which is 74 in decimal.
  • Therefore, the decimal value is -74.

1. 32-bit floating point representation of 1.5 in binary and hex form:
   A 32-bit IEEE 754 floating point number is divided into three parts:

• 1 bit for the sign.
• 8 bits for the exponent (with a bias of 127).
• 23 bits for the fraction (mantissa).
• For 1.5, the normalized binary representation is 1.1 (which is 1.5 in binary).
• The exponent is 0 (since 1.5 = 1.1 * 2^0), so the exponent with bias 127 is 127 + 0 = 127.
• Convert 127 to binary: 01111111.
• The mantissa is the fractional part after the leading 1, which is .1, so the mantissa is 10000000000000000000000 (padding with zeros to 23 bits). The final IEEE 754 representation of 1.5 is:
• Binary: 0 01111111 10000000000000000000000
• Hex: 0x3F800000

1. How many bits would a complete load instruction need to have to hold a main memory address of decimal 65536?

• To store a memory address of 65536, we need to find how many bits are required to represent this value.
• First, convert 65536 to binary:
  • 65536 in decimal = 10000000000000000 in binary (17 bits).
• So, the instruction that holds a memory address must have at least 17 bits to accommodate the address 65536.

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