0
$\begingroup$

In electronics, analog signal processing/computation suffers from parasitic capacitance which leads to slew rate—which is the maximum change of voltage allowed, so handling super-high frequency becomes impossible.

Is this still true for optical signal processing/computation, such as optical (frequency) mixer or optical frequency multiplier? If so, what would be the physics behind?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

Your question includes both the conversion (since you speak of processing) and light propagation. Conversion involves electronics, as @Nasha mentions, and thus is directly impacted by the slew rate.

Light propagation speed is reduced (with respect to that in vacuum) by the refractive index of the material. The physics causing the finite slew rate is also also causing the fact that the refractive index of the material is always greater than 1.

Improve this answer
$\endgroup$
3
  • $\begingroup$ OK. Let us forget about processing. Let me stick to optical mixer and/or optical frequency multipliers. I do not think they necessarily involve electronics, if the original signal is already optical. In such a case, would there be slew rate issues? According to your answer, it seems that only light propagation speed problem exists.. $\endgroup$
    – NCL
    Commented Nov 14, 2015 at 16:27
  • $\begingroup$ I am raising this question, because it is often said that optical signal processing allows for frequencies higher than allowed by electronics signal processing. Even adding and multiplying signals, which may count as processing or not, is affected by slew rate in electronics domain. Thus my question. $\endgroup$
    – NCL
    Commented Nov 14, 2015 at 16:29
  • $\begingroup$ @MBK, Light propagation is involved as long as you are in optical domain: unlike electroncs, photons cannot be stationary. Hence the same logic applies to 'any' light-related process: you will always have light in some material, this material will consist of atoms, refractive index will be greater than 1. $\endgroup$
    – texnic
    Commented Nov 14, 2015 at 16:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.