1
$\begingroup$

The Volterra series contains a number of terms of order 2 and above. Each of the addends is a generalization of the convolution integral of 2 and higher output signals with 2 and higher output signals, respectively. I read several articles and books - they are used to linearize nonlinear elements in control systems. How to use this series to linearize the parallel product of signals growing exponentially? In Mathcad i made a small calculation - but this is not at all what should turn out. The output of such a system should be the square of the input signal + dynamics of the selected link (if two identical links are multiplied together). Apparently, I do not quite understand the meaning of this series. Please help solve the issue.

https://ibb.co/t8kx604 https://ibb.co/6nxqCMh

$\endgroup$

closed as unclear what you're asking by Yanior Weg, Jean-Claude Arbaut, Lee David Chung Lin, Hayk, Adrian Keister May 30 at 13:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It is quite hard to understand what you are asking. Can you state your question in a more precise way? $\endgroup$ – fedja May 28 at 21:34
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$ – dantopa May 29 at 23:45
0
$\begingroup$

You read my scheme right. Suppose I give a stepped signal to both of these links = 3. At the output I have to get the output signal 9. The convolution integral is obtained correctly, but the dynamics ... it turns out just an exponential with a double degree. Shouldn't something like 1 / (s + 1) ^ 2 be obtained if the Laplace transform is done after convolution?

$\endgroup$
  • $\begingroup$ It is a non-linear mapping, so the Laplace transform of the output is not a fixed multiple of the Laplace transform(s) of the input(s) in general. Even if you fix one input to make the mapping linear in the other one, it won't be a pure convolution, so the transfer function will make little or no sense. $\endgroup$ – fedja May 31 at 21:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.