Could you explain this signed fixed point number equation?

Question

How do I interpret this equation?

Decimal value of signed fixed point number:

$$V=(-1)^{b_{N-1}}\times 2^{N-P-1}+\sum_{i=0}^{N-2}(b_i\times 2^i)\times 2^{-P}$$

Original picture

$N$ is number of bits. The point is on a fixed position, $P$, from the right.

Answer 1

You have a bit string with a point, $(b_{N-1}b_{N-2}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}$, with $P$ bits to the right of the point (which is easy to miss, so I’ve colored it red). Let’s ignore the leftmost bit for a moment. The remaining $N-1$ bits are first interpreted in the usual way. The binary number $(b_{N-2}b_{N-3}\ldots b_0)_{\text{two}}$ (without any binary point) is

$$b_{N-2}\cdot 2^{N-2}+b_{N-3}\cdot 2^{N-3}+\ldots+b_1\cdot 2^1+b_0\cdot 2^0=\sum_{i=0}^{N-2}b_i\cdot2^i\;.$$

For example,

$$(11001)_{\text{two}}=1\cdot2^4+1\cdot2^3+0\cdot2^2+0\cdot2^1+1\cdot2^0=25\;.$$

Multiplying by $2^{-P}$ or, equivalently, dividing by $2^P$, shifts the binary point $P$ places to the left, just as dividing a number in ordinary base ten notation by $10^P$ shifts the decimal point $P$ places to the left. If I take $P=2$ in my example, for instance, I get

$$(110\color{red}.01)_{\text{two}}=\frac{25}{2^2}=6.25\;,$$

which is correct: $(110)_{\text{two}}=6$, and $(0.01)_{\text{two}}=0\cdot2^{-1}+1\cdot2^{-2}=0.25$.

Thus, the term

$$\sum_{i=0}^{N-2}b_i\cdot2^i\cdot2^{-P}$$

is the correct value of $(b_{N-2}b_{N-3}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}$ interpreted as a binary number.

The string $b_{N-2}b_{N-3}\ldots b_P\color{red}.b_{P-1}\ldots b_0$ has $N-1$ bits, $P$ of which are to the right of the point, so $N-1-P=N-P-1$ are to the left of the point. The binary representation of $2^{N-P-1}$ is

$$1\underbrace{00\ldots00}_{N-P-1}\;,$$

so

$$2^{N-P-1}+\sum_{i=0}^{N-2}b_i\cdot2^i\cdot2^{-P}$$

has the binary representation $(1b_{N-2}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}$. This is the value of $V$ when $b_{N-1}=0$, so that $(-1)^{b_{N-1}}=1$. In other words, the string $$0b_{N-2}\ldots b_0$$ represents the binary number $$(1b_{N-2}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}\;,$$ whose value is

$$2^{N-P-1}+\sum_{i=0}^{N-2}b_i\cdot2^i\cdot2^{-P}\;.$$

When $b_{N-1}=1$, however, $(-1)^{b_{N-1}}=-1$, and the string

$$1b_{N-2}\ldots b_0$$

represents the number

$$-2^{N-P-1}+\sum_{i=0}^{N-2}b_i\cdot2^i\cdot2^{-P}\;.$$

In binary notation this is

$$(\underbrace{b_{N-2}b_{N-3}\ldots b_P}_{N-P-1}\color{red}.\underbrace{b_{P-1}\ldots b_0}_P)_{\text{two}}-(1\underbrace{00\ldots00}_{N-P-1}\color{red}.\underbrace{00\ldots00}_P)_{\text{two}}\;,\tag{1}$$

where I’ve padded the second term to show the same number of places to the right of the point as in the first term.

For $b\in\{0,1\}$ let $\bar b=1-b$, so that $\bar 0=1$ and $\bar 1=0$. It’s not hard to see that

$$(b_{N-2}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}+(\bar b_{N-2}\ldots \bar b_P\color{red}.\bar b_{P-1}\ldots \bar b_0)_{\text{two}}=(\underbrace{11\ldots11}_{N-P-1}\color{red}.\underbrace{11\ldots11}_P)_{\text{two}}$$

and that

$$(\underbrace{11\ldots11}_{N-P-1}\color{red}.\underbrace{11\ldots11}_P)_{\text{two}}+(0\color{red}.\underbrace{00\ldots00}_{P-1}1)_{\text{two}}=(1\underbrace{00\ldots00}_{N-P-1}\color{red}.\underbrace{00\ldots00}_P)_{\text{two}}\;.$$

Thus, in this case the value of the expression is the negative of the binary number

$$\begin{align*} &(1\underbrace{00\ldots00}_{N-P-1}\color{red}.\underbrace{00\ldots00}_P)_{\text{two}}-(\underbrace{b_{N-2}b_{N-3}\ldots b_P}_{N-P-1}\color{red}.\underbrace{b_{P-1}\ldots b_0}_P)_{\text{two}}\\\\ &\qquad=(\underbrace{11\ldots11}_{N-P-1}\color{red}.\underbrace{11\ldots11}_P)_{\text{two}}+(0\color{red}.\underbrace{00\ldots00}_{P-1}1)_{\text{two}}-(\underbrace{b_{N-2}b_{N-3}\ldots b_P}_{N-P-1}\color{red}.\underbrace{b_{P-1}\ldots b_0}_P)_{\text{two}}\\\\ &\qquad=(\bar b_{N-2}\ldots \bar b_P\color{red}.\bar b_{P-1}\ldots \bar b_0)_{\text{two}}+(0\color{red}.\underbrace{00\ldots00}_{P-1}1)_{\text{two}}\;. \end{align*}$$

You can find this easily from the string $1b_{N-2}\ldots b_0$: drop off the leading $1$, replace each $b_i$ by $\bar b_i$, add

$$2^{-P}=(0\color{red}.\underbrace{00\ldots00}_{P-1}1)_{\text{two}}\;,$$

and stick a minus sign in front.