I understand the concepts of electrical and optical bandwidths, but I encountered an example problem that used the following formula to calculate the electrical bandwidth:

$$electrical bandwidth = 1/\sqrt{2} * optical bandwidth$$

Can any one explain to me why the equation holds? (I couldn't derive it based on the definitions) Thanks!

The source is: Example 4.4 page 104 "Biophotonics, Concepts to applications" by Gerd Keiser, 1st ed. 2016 Edition.

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  • 2
    \$\begingroup\$ I believe the \$\sqrt{2}^{-1}\$ factor arrives because power is \$\propto\$ the square of voltage and the time average (integral) of the square of a sine or cosine is \$\frac12\$. But I don't have your book and can't get the context from what little you've written here. \$\endgroup\$
    – jonk
    Dec 19, 2018 at 15:35

1 Answer 1


Without more context we can't be 100% certain, but most likely this is because conversion from optical to electrical signals is very often a square-law process. That means that in an O/E converter the power in the electrical output generally varies as the square of the power in the optical input. This is because the current in the detector is generally proportional to the incident optical power and the power in the electrical signal is proportional to the square of the current.

So, if you have some bandwidth restriction in your optical path and it reduces the optical power by 3 dB, after passing through the photodetector the electrical signal power will be reduced by 6 dB. If you assume a single pole filter effect, you'd have to reduce the modulation frequency by a factor of \$\sqrt{2}\$ to find the frequency where the electrical signal is reduced by only 3 dB.

So your book's statement depends on a number of preconditions that should be stated somewhere in the text:

  • You're detecting the optical signal with a square law detector
  • The modulation response of the channel has a single-pole characteristic (this is pretty likely to be a reasonable assumption for the given case of measuring the modulation bandwidth of an LED. If we were talking about the bandwidth of an optical fiber transmission channel it might not be a very good assumption)
  • We define the bandwidth of the system by the frequencies with 3-dB attenuation relative to the peak response
  • etc.

In some other context you might, for example, define the optical bandwidth by the frequency with 6-dB attenuated response, while still defining the electrical bandwidth by the frequency with 3-dB attenuated response. Then you'd have the same bandwidth figure measured either way.

  • \$\begingroup\$ Very impressive. Especially not given any context, you nailed it. I added the context to the question. Glad you answered before I did. \$\endgroup\$ Dec 19, 2018 at 17:09
  • \$\begingroup\$ Thank you for the thorough response. I understand better now \$\endgroup\$
    – GolAzar
    Dec 20, 2018 at 18:38

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