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.

  • \$\begingroup\$ 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. \$\endgroup\$ May 19, 2015 at 22:41
  • \$\begingroup\$ 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... \$\endgroup\$ May 19, 2015 at 22:44
  • \$\begingroup\$ 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! \$\endgroup\$ May 19, 2015 at 23:05

1 Answer 1


Your system is a modulator.

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\$):

\begin{align*} y(t) = \cos(2\pi t)\, x(t) = \dfrac{e^{\,j2\pi f_0 t} + e^{\,-j2\pi f_0 t}}{2} x(t) = \dfrac 1 2 x(t) e^{\,j2\pi f_0 t} + \dfrac 1 2 x(t) e^{\,-j2\pi f_0 t} \end{align*}

Given the following property of the Fourier transform:

\begin{align*} F\{x(t)e^{\,j2\pi f_0 t}\} = X(f-f_0) \end{align*}

Transforming with Fourier y(t) gives:

\begin{align*} Y(f) = \dfrac 1 2 X(f-f_0) + \dfrac 1 2 X(f + f_0) = \dfrac 1 2 X(f-1) + \dfrac 1 2 X(f + 1) \end{align*}


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