I have the following transfer function in the time domain:

$h\left(t\right) = B x\left(t\right) + a x^2\left(t\right) + b x^3\left(t\right)$.

My simulation is in the frequency domain, so I would like to have a transfer function of the form $H\left(\omega\right)$.

I didn't find any suitable Fourier/Laplace identities so I tried substituting as follows $\theta_1\left(t\right) = x\left(t\right)$, $\theta_2\left(t\right) = x^2\left(t\right)$ and $\theta_3\left(t\right) = x^3\left(t\right)$. The resultant frequency domain function

$H\left(\omega\right) = B \Theta_1 \left(\omega\right) + a \Theta_2 \left(\omega\right) + b \Theta_3 \left(\omega\right)$,

wasn't of much use to me because I only have $\Theta_1\left(\omega\right)$ and I don't know how $\Theta_2$ and $\Theta_3$ are related to $\Theta_1$.

I would like my transfer function to depend only on $\Theta_1$, which is the variable that I know.

How can I go about solving this problem?

  • 1
    $\begingroup$ This is not an LTI system, hence $e^{j\omega}$ is not eigen value. So if your input is $X(\omega)$, you will not get output $Y(\omega) = H(\omega)X(\omega)$. $\endgroup$
    – jithin
    Commented Apr 22, 2020 at 12:32
  • $\begingroup$ Meaning that working in the frequency domain isn't an option? $\endgroup$ Commented Apr 22, 2020 at 12:33
  • $\begingroup$ You can but you will not be able to get $H(\omega)$ dependent only on $\theta_1(\omega)$ $\endgroup$
    – jithin
    Commented Apr 22, 2020 at 12:34
  • $\begingroup$ I don't mind having $\Theta_2$ and $\Theta_3$ as dependencies as long as I know their values. Currently they are unknown so I can't use $H\left(\omega\right)$... $\endgroup$ Commented Apr 22, 2020 at 12:37
  • $\begingroup$ Okay so based on the information given I have answered it. $\endgroup$
    – jithin
    Commented Apr 22, 2020 at 12:51

1 Answer 1


Since this is not an LTI system, we cannot have $H(\omega) = K\Theta_1(\omega)$, where $K$ is a scalar complex number. But as OP mentioned in the comment he has knowledge of $\Theta_1(\omega)$, we can compute $\Theta_2(\omega)$ and $\Theta_3(\omega)$. As $x^2(t)=x(t)\times x(t)$, $$ \Theta_2(\omega)= \Theta_1(\omega)*\Theta_1(\omega) =\int_{-\infty}^{+\infty} \Theta_1(\alpha)\Theta_1(\omega-\alpha)d\alpha $$ where $*$ is the convolution operation. Similarly, $x^3(t)=x(t)\times x(t)\times x(t)$ $$ \Theta_3(\omega)= \Theta_1(\omega)*\Theta_1(\omega)*\Theta_1(\omega)=\Theta_1(\omega)*\Theta_2(\omega) $$ So $H(\omega)=B\Theta_1(\omega)+a\Theta_2(\omega)+b\Theta_3(\omega)$, where $\Theta_2$ and $\Theta_3$ are computed from $\Theta_1$ above.


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