1
\$\begingroup\$

I would like to use AC analysis (e.g. in spice) on a non-ideal op-amp integrator where the reset switch is open, and then apply a shaping function to the result of the AC simulation (in the frequency domain), so the final frequency domain output represents both the integration, gating and sampling function. The goal is to find the power spectral density (PSD) of a signal at the sampling moment \$t=t_{int}\$.

Idealized the system looks like this:

Switched integrator

The integrator on it's own has the transfer function:

$$\frac{Vo(\omega)}{I(\omega)} = - \frac{1}{j\omega C} \stackrel{\omega \to 2 \pi f}{=}- \frac{1}{j 2 \pi f C}$$

I think I can multiply the input current with the Fourier transform of the gate or rect function, and get the integrator output, e.g. like this:

$$ I_{gated}(\omega) = I(\omega) \cdot t_{int} \frac{\sin(\omega t_{int} / 2)}{w t_{int} / 2} $$

I assumed I could get the Vo result with something like:

$$ Vo(\omega) = I_{gated}(\omega) \cdot \frac{-1}{j\omega C}$$

However, this does not seem to be quite enough to find the PSD at the sampling moment which occurs at \$t=t_{int}\$. I feel like I am missing a step, like the sampling itself perhaps?

I don't really have a lot of experience with this type of problem, and I would normally try to go around it. However, knowing the solution to this would really help my analysis so I would like to know if anyone knows how to do this? A couple of hints might go a long way.

If it is of interest to know, the sampling is repeated periodically, but rather seldom ( \$Ts >> t_{int}\$ ).

\$\endgroup\$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.