# Determining Existence of Transfer Function H(s) and Fourier Transform H(w) for Non-LTI Systems?

Given that $$y(t) = \cos(2\pi t)x(t)$$ where $x(t)$ is a system input and $y(t)$ is the system's output, I need to determine whether an $H(s)/H(w)$ relation exists. Since this system is not LTI, I really don't know how to approach it. I'm not able to apply a Laplace Transform even when converted to polar form, and I'm at a loss.

In another case of a non-LTI system causing headaches, I've succeeded in determining that $$h(t) = \frac{\sin(10\pi t)}{\pi t}$$ (or $10\operatorname{sinc}(10\pi t)$ if you prefer), and thus I believe its Fourier Transform $H(w)$ should be a unit step centered about $10\pi$, but I don't know how to determine whether or not this form has a Laplace Transform (i.e. $H(s)$). I certainly don't see it as a standard form on any Laplace Transform tables I've come across.

Any pointers in the right direction would be greatly appreciated.

• Is this some kind of problem given to you by a professor or is it your own curiosity? As far as I know the simple I/O relationship afforded by L-transforms is obtainable only for LTI systems. That's because Y(s)=H(s)X(s) derives from applying the LT to a convolution integral in the time domain, i.e. the convolution between x(t) and h(t), where h is the impulse response of the LTI. In other words this kind of relationship is a consequence of the system being linear. I'm not aware of a similar technique for non-linear systems. – Lorenzo Donati supports Monica May 19 '15 at 22:41
• BTW, your system is a modulator, so you could apply the properties of Fourier transform to solve the issue. Decompose the cos function as two complex exponential functions... – Lorenzo Donati supports Monica May 19 '15 at 22:44
• It's a review problem for a refresher on material I covered earlier in my undergrad. I recognize the form as a simple coherent modulation, but wasn't sure how to relate the input/output beyond the trivial. Your answer below is very helpful, thanks! – Rome_Leader May 19 '15 at 23:05

Since $\cos(2\pi f_0 t)= \dfrac{e^{\,j2\pi f_0 t} + e^{\,-j2\pi f_0 t}}{2}$ it follows ($f_0=1$):