equations for circuit modeling

[talking]

my questions is about the relationship between polynomial modeling and transfer function for circuit.

As well known, the transfer function h(t) for LTI system has the relationship between input x(t) and output y(t):
y(t) = x(t) * h(t) ----- eq(1)
where * is convolution operation.

Then, a circuit is modeled using a polynomial series as
y(t) = Σ a(i) [x(t)]^i ----- eq(2)
where i is an index for the polynomial series, and a(i) is the polynomial coefficient.

Here is a question.
The polynomial model itself " Σ a(i) [x(t)]^i " in eq(2) corresponds to
" x(t) * h(t) " in eq(1)
OR
" h(t) " in eq(1)?

 

[dillikumar406]

by sustituting eq2 in eq1
Σ a(i) [x(t)]^i = x(t) * h(t)
let i=2 and since a(i) is a constant term, there is no problem with it
on expanding the above equation, then
a(1) x(t) + a(2) x(t)^2 = x(t) * h(t)
taking laplace transform to the above equation
a(1) X(s) + a(2) X(s) = X(s) H(s)
then you will get
H(s) = a(1) + a(2) X(s)
by taking inverse laplace transform, you will get h(t) in terms of x(t)
h(t) = a(1) del(t) + a(2) x(t)

 

[talking]

dillikumar406,

I don't fully understand the following calculation.

> a(1) x(t) + a(2) x(t)^2 = x(t) * h(t)
> taking laplace transform to the above equation
> a(1) X(s) + a(2) X(s) = X(s) H(s)

Isn't this as follows?
a(1) X(s) + a(2) [X(s)*X(s)] = X(s) H(s)

 

[dillikumar406]

X(s)*X(s) indicates convolution of two X(s) signals. but it is multiplication not convolution
i.e., X(s)^2 = X(s).X(s)
and convolution in time domain is equal to multiplication in frequency domain

 

[talking]

I apologize for unclear notation in my previous question; '*' in my question meant a convolution operator.
Basically, my original question was ...

In your previous equations, the laplace transform of the following equation (a) has been taken, and it results in equation (b).
> a(1) x(t) + a(2) x(t)^2 = x(t) * h(t) ----- (a)
> a(1) X(s) + a(2) X(s) = X(s) H(s) -----(b)

The laplace transform of "a(2) x(t)^2" in equation (a) is "a(2) X(s)" in equation (b)?

From the laplace transform theory, L{f1(t) X f2(t)} = F1(S) * F2(S), where L{} represents the laplace operator, X indicates time-domain multiplication, and * means a convolution operator.
So I though the laplace transform of "a(2) x(t)^2" in equation (a) needs to be "a(2) [X(s) * X(s)]" in equation (b), instead of being "a(2) X(s)". Am I wrong?

 

[Kevin Aylward]

Difficult problem. Have a look at
**broken link removed**