Question on discrete signals and quantization

Question

Suppose we have 2 different levels of a quantized signal.

Let's say that the range is $[-1,1]$. The number of bits necessary to encode these levels is$\log_2(2) = 1$. However if the value of the analog signal $x(n)$ is exactly equal to $0$, then can we choose freely to encode it as $0$ or $1$?

Answer 1

I think I understand what you're asking, but it depends on the quantization type, and your question lacks the details necessary to a definite answer.

However, in general, if you have an analog signal $x(t)$ that takes on values in a certain range, and a $m$-bit quantizer, your quantized (and sampled!) signal $x[n]$ can take on $M = 2^m$ values to represent the analog values within that range. This is done through a mapping operation. Which mapping is used depends on the quantizer.

---

For example, for an analog signal $x(t)$ in the range $[-1, 1]$, with $m = 2$ bits, the discrete signal $x[n]$ can take on $M = 2^2 = 4$ discrete values that could map to the analog values in the following ways:

\begin{align} &\text{4-bit representation} &\qquad &\text{1st mapping} &\qquad &\text{2nd mapping}\\ &\tt{11} &\qquad &0.5 &\qquad &0.75\\ &\tt{10} &\qquad &0 &\qquad &0.25\\ &\tt{01} &\qquad -&0.5 &\qquad -&0.25\\ &\tt{00} &\qquad -&1 &\qquad -&0.75\\ \end{align}

---

Let's look at the result $x[n_0]$ of these mappings on a couple analog values $x(t_0)$ and $x(t_1)$:

1. Analog value $x(t_0) = 1$:

   • 1st mapping: $x(t_0)$ would be represented by $\tt{11}$ which maps to $0.5$ (round down).
   • 2nd mapping: $x(t_0)$ would be represented by $\tt{11}$ which maps to $0.75$ (round down).

2. Analog value $x(t_1) = 0$:

   • 1st mapping: $x(t_1)$ would be represented by $\tt{10}$ which maps to $0$ (round down).
   • 2nd mapping: $x(t_1)$ would be represented by $\tt{10}$ which maps to $0.25$ (round up) or by $\tt{01}$ which maps to $-0.25$ (round down).