Paper: A novel channel equalizer for chaotic digital communication system (DOI, RS link).

I am having problem in understanding their claim which is that a nonlinear system, considered here to be a chaotic system given by Eq (1) $x[n] = f(x[n-1])$ is represented as a symbolic system $x[n] = 0.b_nb_{n+1}...$ where the $b$'s are {0,1} and so a time series in terms of the binary expansion of a number $x[n]$ describes the dynamics of the system. This is explained in Section 2. Then they say that a linear filter of the form Eq(5)

$x[n] = \sum b[n-k]h[k]$

where the $b$'s are the information bit sequence and the output $x[n]$, is chaotic. So, this Equation according to them is a way of linearizing the nonlinear dynamics of Eq(1). If this is a linear representation, then how come it is chaotic?

I am unable to understand how the bits-- $b$'s that they are talking about is generated and how the binary bit sequence can be used as the input to the filter of Eq. (5) to get a chaotic data sequence i.e. why is this representation chaotic? One of the ways to produce symbolic dynamics of a trajectory of a map is by thresholding explained as:

b[n] = 1, if x[n] >= threshold, otherwise b[n] = 0

Is this how the $b$'s are generated from the same map or these bits are coming from some other source?

  • 2
    $\begingroup$ SKM: Daniels suggestion looks good to me, but we don't generally migrate to beta sites without the poster's permission. Odds are you'll get a better answer on a specialist site, while this question is hovering is the hazy boundary of off-topicalitiy on Physics. Just flag it and request that we migrate. $\endgroup$ Jun 28, 2015 at 23:13
  • $\begingroup$ @DanielSank: Please migrate the Question to dsp.stackexchange if you think that it will be on topic there. Thank you $\endgroup$
    – SKM
    Jun 29, 2015 at 3:13
  • $\begingroup$ I can't. I'm not a moderator. @dmckee looks like OP wants this migrated. $\endgroup$
    – DanielSank
    Jun 29, 2015 at 4:16

1 Answer 1


A typical communications system is composed of the following parts:

Message (Information Source) --> Encoding --> Modulation --> Channel --> Demodulation --> Decoding --> Recovered Message

A "chaotic modulator" is just another way of doing the Modulation building block. This is depicted in Figure 1 of the paper that is cited in the question.

The input to the Modulator is a binary sequence and the output of it is a digital (or directly analog, depending on the system) signal.

Therefore, to clarify (rather than answer) the first issue: As per Figure 1, the bs are essentially your Message (or Message after it has been encoded for whatever purpose (e.g. error correction).

A typical Modulator's job is to map binary symbols to carrier symbols with the carrier signal being one that is suitable for transmission effectively through the Channel. In a typical system, we can literally see this mapping in the Constellation Diagram.

In the case of a chaotic modulator though, binary symbols are mapped to different states of the chaotic modulator. This is mentioned in the introduction of the paper as:

Rather than using structured signals, such as rectangular pulses or sinusoids, to denote ‘0’s and ‘l’s, these communications systems embed the information in the time evolution, or dynamics, of the transmitted signal.

A simpler way to think about this is to take some difference equation (such as f) and produce a few values with some initial condition x0 and a few values for some initial condition x1. If the two time series are distinct enough, then perhaps it is possible to recover which of the two x0,x1 might have generated this time series. In this case, of going from time series to some initial condition, we are looking at some sort of Demodulator.

Now, chaotic of not, a difference equation provides a model. A model imposes a set of constraints and some sort of structure to the signal. This is what is exploited here (please keep in mind that we are talking about a pseudo-chaotic system). So, what if I take look at the difference equation and realise that I can express it as a typical digital filter?

Again, to clarify, rather than answer, this is what Drake & Williams did with "Pseudo-chaos for direct sequence spread spectrum communications". From the paper linked in the question:

More importantly, Drake and Williams [7] have shown that the sawtooth map's output is equivalent to the response of a linear filter to a binary sequence.

Having clarified this, it is also important to highlight Figure 2 (and 3 to some extent) from the original paper which shows that by inversing the filtering operation it is effectively possible to do Modulation and Demodulation.

This, hopefully, clarifies the "representation" missunderstanding.

Finally, I would like to stress the pseudo-chaotic nature of the generated signal which comes as a result of the modifications that are also explained in section 2, from "In practice this is not a possible because this filter..." (sic) onwards.

I would also like to recommend that you take a brief look at Pseudo-random number generators, the Linear congruential generator and finally the linear feedback shift register (LSFR) for examples of seemingly linear "representations" that still produce rather complex sequences and then taking a brief look at the Dyadic Mapping where it is also mentioned that:

The name bit shift map arises because, if the value of an iterate is written in binary notation, the next iterate is obtained by shifting the binary point one bit to the right, and if the bit to the left of the new binary point is a "one", replacing it with a zero.

Hope this helps.


In view of additional comment:

If the chaotic map is denoted by the state variable $x$, then these $b$ are the bits. Further, are the $b$′s obtained from the binary expansion of some numeric value of the iterate of chaotic map $x$?

Here is an alternative way to go about this:

Let's say that this is not a modulator. Ultimately, $x$ is the output from the mapping. We start with some initial value $x_0$ and produce $x_1$ which we send back into the map and get back $x_2$ and again through the map to get $x_3$ and back through it once more to get a $x_4$ and so on to get $x_n$. Since we are operating on a computer with numbers of finite wordlength, $x$ contains DIGITAL values (in the decimal system) with a well defined minimum and maximum. So, if $x$ is a byte, then the min value is 0 and the maximum value is 255.

Each one of these $x_n$ decimal values has a binary representation as well. Therefore:

$x_0$ is equivalent to some binary number $0100101$

$x_1$ is equivalent to some binary number $1001010$

$x_2$ is equivalent to some binary number $0010100$

And eventually...

$x_n$ will be equivalent to some $b_m b_{m+1} b_{m+2} ...$ where some $b \in {0,1}$

Therefore, it is possible to have a binary string of digits (0010001101...) for a pseudo-chaotic sequence $x$.

Let us now ask the reverse question: What if all we have is the binary sequence. Can we go from the binary sequence to the output it would produce?

To do this, you would have to establish a relationship between the $b$ (binary sequence) and $x$ (decimal sequence / digital signal). It turns out that this is possible via the work of reference 7 as mentioned above, through $h$.

To recap and put it in the context of communications:

$b$ is your Message. If the message was "ABC", its ASC representation would be 65,66,67 and the binary representation of that would be "010000010100001001000011". This is your $b$.

$x$ is the output from the Modulator with pseudo-chaotic dynamics.

$h$ is now the tool by which the map is applied to the data.

(Please see figures 2,3 in the original paper).

How can the convolution of $b$ with impulse give rise to a chaotic map, $x$?

Again...The input to the modulator is a binary sequence ($b$). Each $b$ is either $0$ or $1$. The output from the modulator is a digital signal ($x$) (decimal values of some wordlength).

If we take a look at:

$x[n] = \sum_{k=-\infty}^{\infty}{b[n-k] \times h[k]}$

We realise that:

  1. $x$ is a weighted sum of $b$.

  2. To produce one $x_n$, we need to sum many bits from $b$. How many? $k$ many.

  3. Obviously, $k$ cannot remain infinity in a practical system. That's where truncation of $h$ comes in.

I hope this helps.

  • $\begingroup$ THank you for the explanation. I had studied Dyadic map also called as the Bernoulli Shift Map. The area of confusion arises, which is still vague, due to the Equation $x[n] = \sum b[n-k] h[k]$ which is a linear filter mentioned in the paper. If the chaotic map is denoted by the state variable $x$, then these $b$ are the bits. How can the convolution of $b$ with impulse give rise to a chaotic map, $x$? Further, are the $b's$ obtained from the binary expansion of some numeric value of the iterate of chaotic map $x$ ? $\endgroup$
    – SKM
    Jun 29, 2015 at 19:05
  • $\begingroup$ Thank you for your comments. I have amended my response above and hope that you find it more helpful. $\endgroup$
    – A_A
    Jun 30, 2015 at 8:42
  • $\begingroup$ Thank you for such detailed explanation and appreciate your effort to go through the paper :) $\endgroup$
    – SKM
    Jun 30, 2015 at 21:20
  • $\begingroup$ (1) Last answer to the Question : How can the convolution of b with impulse give rise to a chaotic map, x ? is a bit unclear stiil. Is the output $x[n]$ of the modulator a chaotic or just random? It appears to be linear random and $h$ can be any impulse response? (2) And, just to clarify if my understanding on this part is right --the $b's$ that are convolved with the impulse response are just some random bits and not the bits generated from a chaotic signal. $\endgroup$
    – SKM
    Jun 30, 2015 at 21:25
  • $\begingroup$ I think that the current response as it stands contains the answer to your question. $x$ is the output of the modulator. It is pseudo-chaotic. $h$ is not ANY impulse response. $h$ is now the map that maps a bit sequence to a pseudo-chaotic waveform. The $b$s are your message. They are what you want to transmit, for example "Hello World". The $b$s are not random. Hope this helps. $\endgroup$
    – A_A
    Jul 1, 2015 at 10:32

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