# How to convert my transfer function to the frequency domain

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?

• 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 $Y(\omega) = H(\omega)X(\omega)$. – jithin Apr 22 at 12:32
• Meaning that working in the frequency domain isn't an option? – Chandran Goodchild Apr 22 at 12:33
• You can but you will not be able to get $H(\omega)$ dependent only on $\theta_1(\omega)$ – jithin Apr 22 at 12:34
• 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 $H\left(\omega\right)$... – Chandran Goodchild Apr 22 at 12:37
• Okay so based on the information given I have answered it. – jithin Apr 22 at 12:51

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.