I came across the following question at ResearchGate, but I fail to comprehend the answer to it.

The question states:

Why do we only study the DC gain in VLC channel modeling?

In all the papers and books that I read, only the DC channel gain of the VLC system is studied without any further justification.

I looked at many relevant sources in the literature, and I agree that there is no explanation provided as to why the channel DC gain is sufficient to modeling the optical channel.

The answer might be so obvious that it does not require an explanation, however it is not immediately straightforward to me.

Literature closest to a satisfactory explanation were the following:


The frequency responses of infrared channels are relatively flat near dc, so for most purposes, the single most important quantity characterizing a channel is the dc gain


For intensity-in intensity-out channels, the zero-frequency (DC) value of their frequency responses can be expressed as [52]. The expression in equation (2) is commonly referred to as channel DC gain, which is the fraction of power emitted from a continuous-wave transmitter that is detected by the receiver

My understanding from the above is that, since optical wireless communication uses Intensity Modulation with Direct Detection (IM/DD), there is no frequency component that one should be concerned with, but on the other hand I do not understand what "the zero-frequency (DC) value of their (referring to the channels) frequency responses" means?

  • \$\begingroup\$ When you say "channel" do you mean the path that conveys light from the transmitter to the receiver? (As opposed to, say, considering the receiver and decision circuit as parts of "the channel") \$\endgroup\$
    – The Photon
    May 25, 2022 at 16:53
  • \$\begingroup\$ which answer are you not comprehending? \$\endgroup\$
    – jsotola
    May 25, 2022 at 17:13
  • \$\begingroup\$ @ThePhoton That would be "the path that conveys light from the transmitter to the receiver". \$\endgroup\$ May 26, 2022 at 8:21

1 Answer 1


Say you're modulating a red LED with 10 Mb/s data.

Say the LED has 10 nm linewidth, with a spectrum spanning 645 to 655 nm. Put in frequency terms (\$f_o=c/\lambda\$), that means the LED spectrum spans roughly 458 - 465 THz. It has a bandwidth of 7 THz.

The modulation introduced by your 10 Mb/s data signal, will only require about 5-10 MHz of bandwidth.

10 MHz of modulation on a carrier with 7 THz spread is essentially insignificant. If we measure the channel loss for the unmodulated LED output, the loss for the modulated optical signal won't be measurably different.

(If you start modulating lasers, which have significantly narrower spectra, at higher data rates, for example in the 10's of Gb/s, then you can easily see the change in the spectrum due to the modulation and may need to be concerned about the optical path disturbing the modulation signal)

  • \$\begingroup\$ To make sure I understood it correctly: The fact that the modulation bandwidth << carrier bandwidth is sufficient for us to conclude that there will be no frequency dependent effects in the channel. Additional question: how does this relate to the IM/DD? If the modulation bandwidth ~= carrier bandwidth then the situation would be different, righ? \$\endgroup\$ May 26, 2022 at 8:54
  • \$\begingroup\$ @KristofTak, If modulation bandwidth > carrier bandwidth, I could design an optical filter that specifically picks off, for example, one side of the spectrum, or that cuts off the tails of the spectrum. You run into such things in DWDM systems where you're trying to pack as many channels as possible into a given band. For free-space VLC, I don't expect you'd run into many narrow filters like that in practice. Maybe if you used a really narrow-band input filter on your receiver? \$\endgroup\$
    – The Photon
    May 26, 2022 at 15:21

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