3
$\begingroup$

One of the most common techniques used for functional neuroimaging nowadays is functional near infra-red spectroscopy (fun fact: IIRC Natalie Portman worked on a research paper involving fNIRS as the modality), which shines near infrared light into the brain from a source to a detector (both called optodes) in a "banana" shape.

It's not uncommon to read that most of these devices, be they continuous wave (CW) or one of the two kinds that involve fast modulation, frequency domain (FD) or time domain (TD), require two separate frequencies to be emitted. For instance, NIRx explains it as follows on their website:

"For neuro-imaging applications it is by far most common to illuminate with two discrete wavelength, which is the minimum requirement to assess relative variations of both oxygenation states of the hemoglobin molecule independently."

Why is that the case?

I haven't delved into the intricacies of it, but no reason immediately jumps out at me. For instance, in the case of CW, the relative difference in intensity is all that matters, so why do we need two frequencies?

$\endgroup$
3
  • $\begingroup$ By "frequency" I assume you refer to (inverse of) different wavelengths of light? $\endgroup$
    – Bryan Krause
    Dec 6, 2023 at 13:15
  • $\begingroup$ @BryanKrause yes, that is correct $\endgroup$
    – David Cian
    Dec 7, 2023 at 9:38
  • $\begingroup$ @BryanKrause I added a source from the NIRx website $\endgroup$
    – David Cian
    Dec 7, 2023 at 9:40

1 Answer 1

1
$\begingroup$

Basically, you have two unknowns, so you need two equations.

From Wikipedia:

Using a dual wavelength system, measurements for HbO2 and Hb can be solved from the matrix equation

fnirs matrix equation

With just one measurement, you could still describe relative changes in the signal over time, but you wouldn't be able to measure the ratio of the two. That might be sufficient within a given experiment, but wouldn't result in numbers that you could compare to other recordings. Each recording is going to have a distinct signal-to-noise ratio based on things like the amount of tissue you're imaging through: most of the light is going to be lost no matter what the signal is.

$\endgroup$

Your Answer

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

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