bandwidthledoptical-fibre
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.
Here is the open-loop bandwidth of a certain op-amp shown in red: -
The blue line is when certain closed-loop components are applied to the op-amp.
Bandwidth is normally measured at the 3dB point of the frequency response and in the case of an op-amp (open-loop) this will be at 24Hz in the diagram.
If closed loop components were present, the gain would be reduced to (say) 20dB (blue line) but the bandwidth would increase to 1MHz.
The above example is for simple resistors "closing" the loop with negative feedback and the resulting bandwidth (3dB point) is always greater.
However, if an op-amp filter circuit was required that cut-off frequencies above 10Hz, the filter would have a bandwidth of 10Hz. In this example the closed-loop bandwidth is less than the open-loop bandwidth.
Does this help?
In some cases it works out to be more convenient. Some things (like human perception of pitch, and filter responses) respond to frequency logarithmically. That is, higher frequency octaves have more bandwidth than low frequency octaves.
It's the same reason low-pass or high-pass filter topologies have their roll-off described in dB/octave rather than dB/Hz. For example, we know that all first-order filters have a roll-off of 6dB/octave. This holds if we are designing a filter for 1kHz or 1GHz.
The same sort of issue applies to band-pass and band-reject filters. If we took a band-pass filter 100Hz wide at 1kHz, then scaled all the component values to move it up to 1GHz, then its bandwidth would be 100MHz. However, it would have the same fractional bandwidth.
Thus, it's convenient to specify the bandwidth of a particular topology as a fractional bandwidth because it does not change as the passband is moved by scaling all the component values.
Best Answer
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:
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.