Quick Binary Representation Help

[ravenprp]

Homework Statement

What is the range of a 11bit code, using
c. Unsigned fixed-point with 8 binary places

Homework Equations

Base conversions
2^n where n is bits, for range/precision

The Attempt at a Solution

I have two ideas:

1)

The highest value that can be represented is: 111.11111111, which represents this ... 7.99609375 ... then I subtract one from it ... – 1 = 6.99609375
0 to 6.99609375

Or

2) Since it is a 11bit code, you have 2^11 possible combinations to store in memory. Since the binary pt is fictional, until it is programmed by the programmer, wouldn't you still need 2^11 binary points to represent the 11 bit code xxxxxxxxxxx, regardless of what value it is?

That's my reasoning.

 

[KataKoniK]

Hello,

Thank you for your question. The range of an 11bit code using unsigned fixed-point with 8 binary places can be calculated using the following formula:

Range = 2^(n-1)

Where n is the number of bits used, in this case n = 11. So the range would be:

Range = 2^(11-1) = 2^10 = 1024

This means that the range of values that can be represented using an 11bit code with 8 binary places is from 0 to 1023. This is because the first bit is used to represent the sign, leaving 10 bits for the actual number. Also, since we are using unsigned fixed-point, the range starts from 0 rather than -1023.

I hope this helps to clarify any confusion. Let me know if you have any further questions.

 

[piercas]

Your first idea is correct. The range of an 11-bit code, using unsigned fixed-point with 8 binary places, is 0 to 6.99609375. This is because the highest possible value that can be represented is 111.11111111, which is equivalent to 7.99609375 in decimal. Subtracting 1 from this gives us the lowest possible value of 0.

Your second idea is not entirely correct. While it is true that 2^11 possible combinations can be stored in memory, this does not necessarily mean that all 11 bits will be used to represent every value. In this case, only 8 bits are used to represent the integer part, and the remaining 3 bits are used to represent the fractional part. Therefore, the range is limited to 0 to 6.99609375.