What does fixed-point number range represents? Why we use formula $2^a - 2^{-b}$, why minus $2^{-b}$? Where $a$ is number of integer bits, and $b$ is number of fractional bits.

If we have for example $a = 8$ and $b = 2$, don't we have a possibility to represent $2^a + 2^{-b}$ number and so that will represent the range?

  • $\begingroup$ yes @OlliNiemitalo I am trying to understand some fundamentals about fixed-point numbers, but some terms and formulas are confusing me. $\endgroup$
    – user20705
    May 12 '16 at 8:24

Let's assume that we are dealing with unsigned number types. If you would use all $a+b$ bits for the integer part then the set of possible numbers would be: $$\left\{0, 1, 2, 3, \dots, 2^{a+b}-1\right\}.$$

These numbers can be divided by $2^b$ (or multiplied by $2^{-b}$) to take use of $b$ bits for the fractional part, resulting in this set of possible numbers:

$$\left\{0, 1\times2^{-b}, 2\times2^{-b}, 3\times2^{-b},\dots, \left(2^{a+b}-1\right)\times2^{-b}\right\} \\=\left\{0, 2^{-b}, 2\times2^{-b}, 3\times2^{-b},\dots, \underline{\underline{2^a-2^{-b}}}\right\}.$$

So your formula gives the largest number that can be represented (double underlined).

The formula can also be understood as going a step $2^{-b}$ or one least significant bit (LSB) worth backwards from $2^a$ which is the first number that has the same truncated binary string representation as 0, the first number in the system. In a similar way in the 8-bit unsigned integer system we take a LSB-sized step backwards from 256 (1 0000 0000 binary) to obtain 255 (1111 1111 binary) which is the largest representable number in that system.

I just found out that range has an established meaning in arithmetic: Largest value minus smallest value. The smallest value happens to be zero so in this case the largest value equals the range.


For an unsigned fixed-point, the representation for a $N$-bit binary number $x$ is

$$x=\frac{1}{2^b}\sum_{n=0}^{N-1}2^n x_n, \quad \text{where $x_n$ is the $n^{\rm th}$ bit value of $x$}$$

For this $N$-bit binary number you can get values from $0$ to $2^N -1$ since the smallest number is the $N$-bit all-zeros ($x_n =0\ \forall n$), and the largest is the $N$-bit all-ones ($x_n =1\ \forall n$). The representation for the minimum and maximum values are then given as follow:

\begin{align} x_{\rm min}&=\frac{1}{2^b}\left(1\cdot 0+ 2^1\cdot 0+\ldots+2^{N-1}\cdot 0\right)=\frac{1}{2^b} \cdot 0=0\\ x_{\rm max}&=\frac{1}{2^b}\left(1\cdot 1+ 2^1\cdot 1+\ldots+2^{N-1}\cdot 1\right)=\frac{1}{2^b}\left[1\cdot\left(\frac{1-2^N}{1-2}\right)\right]=\frac{2^N - 1}{2^{b}}=2^{N-b} - 2^{-b}\\ \end{align} With $a=N-b$, you get the range.

You can read this great paper by Randy Yates on Fixed-Point Arithmetic. There you find the derivation of the range for the signed case and much more.


The formula expresses the difference between the largest, and smallest numbers we can represent with an integer/fractional representation.

Consider a simple example, such as 2.2 (where there are 2 integer bits and 2 fractional bits in our fixed point representation).

If we're using two's complement (signed) binary, the largest positive number we can represent is:


This number is equivalent to $2 - \frac{1}{4}$.

The largest negative number we can represent is:


This number is equivalent to $-2$.

The total range here is from the largest positive to largest negative, or $-2$ -> $2 - \frac{1}{4}$. Note that this is equal to:

$4 - \frac{1}{4}$, or $2^a - 2^{-b}$ where both $a$ and $b$ are equal to 2, according to our 2.2 representation.

  • $\begingroup$ why the largest positive is 3 and 3/4? Why not 4? $\endgroup$
    – user20705
    May 14 '16 at 17:47
  • $\begingroup$ Because we consider the range to include the value 0. If there are 4 bits, we can represent 16 values. If `zero' is one of these values, and each of the others represents 1/4, we can only add a further $15 * \frac{1}{4}$ to the range. This is why the largest range is 3 and 3/4. To represent 4 (in the binary representation used in this example) you would need to add an extra integer bit. $\endgroup$
    – Speedy
    May 16 '16 at 8:48

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