is it possible to apply CSD to bough wooley multiplier
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000000 is the lowest number in a 6 bit unsigned binary number (meaning the high order bit is not the sign bit). If it is a signed number then the lowest number would be represented by 100000 which is equivalent to -32 in decimal. Highest unsigned number in 6 bits is decimal 63. Highest signed number in 6 bits is decimal 31.
A nibble is 4 bits, so the largest unsigned number is 1111, or 15. Also, the largest signed number is 0111, or 7.
The highest unsigned integer is 255; The highest signed integer is 127.
A 32 binary number is a number stored by a computer in 32 bits. it can represent: 1) An unsigned number in the range 0 to 4,294,967,295 2) A signed number in the range -2,147,483,648 to 2,147,483,647 3) A single precision IEEE floating point number with 1 sign bit, 8 exponent bits and 23 mantissa bits give an accuracy of about 7.2 decimal digits and a range of ± 10^-38 to 10^38
It depends. If you are using unsigned numbers, then the following assumption is made: 0b11 = 0b00000011, in which case the answer is; 2^1 + 2^0 = 2 + 1 = 3 If you are using signed numbers, than a binary number in the form 0b11 would be interpreted as negative because the leading bit is equal to 1. For signed numbers, the '1' in the leading bit is extended, thus: 0b11 = 0b11111111 In order to interpret this number, negate the number by flipping the bits and adding 1: 0b11111111 0b00000000 (bits flipped) 0b00000001 (added one) The positive representation of 0b11111111 is equal to 0b00000001, which is equal to 1, thus 0b11 = 0b11111111 = -1