# Normalization purpose in signal processing

What is the purpose of normalizing the signal?

If we have two signals on hand, how is it used when comparing these two signals?

• What signals are you processing? This would help answering the question. – AliceD Jan 5 '15 at 13:40

Normalization is basically bringing the two signals to the same range or a predefined range. A typical example of a predefined range is the statistical perception of the normalization, which is transforming the signal so that its mean is $0$ and standard deviation is $1$. After such transform the canonical form is obtained. In layman's terms, such the operation consists of subtracting the minimum value and dividing by the range (difference between the maximum and minimum values).

Transforming all signals to such canonical form eases and robustifies the process of comparisons as well as serving too different needs such as visualization, and analysis. If we are to speak in terms of image normalization the effect would be the following:

Un-normalized:

Normalized:

Note that, the distinction between the colors is clearer and more apparent.

when comparing two different signals that mean two different things (but equally important or with an equal significance regarding the information the signals carry), would you expect to be comparing two signals, one with values in the ballpark of $\pm$1 and the other with no samples with magnitude at least $2^{-8}$ or so?
This question is too vague. Normalization can be done in many ways and for many different purposes. For instance, in addition to the example of @tbirdal for image processing, I could think of other examples. A very simple one is one where where we know the signal is \begin{align} y_1(t) &= x_1(t) + n_1(t) \\ y_2(t) &= x_2(t) + n_2(t) \end{align} where $n_1(t)\sim\mathcal{N}(0,\sigma^2)$, $n(t)\sim\mathcal{N}(0,\sigma_2^2)$, and $x_1(t)$ and $x_2(t)$ are deterministic. In such case, in order to compare both signals, a logical normalization would be to compute \begin{gather} z_1(t) = \frac{y_1(t)}{\sigma_1} \\ z_2(t) = \frac{y_2(t)}{\sigma_2}. \end{gather} Now $z_1(t)$ and $z_2(t)$ have a noise with variance 1. The advantage is that now if $z_1(t)$ is larger than $z_2(t)$, you know $z_1(t)$ has a better signal-to-noise ratio (SNR).