Hi-
Sorry, things are slow around holiday time.
… and this is going to be a loooong message, because there are a lot of considerations here-- my apologies.
Before addressing the mechanics of bandpassing, I want to comment on some consequences of bandpassing, in terms of your GLM and allowed degrees of freedom. If you set up your problem to bandpass to keep only frequencies between 0-0.01 Hz, that will be quite problematic if you are including other things in the model, because you will run out of degrees of freedom (DFs).
If you have N time points in your time series, you start with N DFs. Every time point you censor, you lose 1 DF, and every regressor you put into your model, you lose 1 DF.
Bandpassing can be done by including frequencies in the regression model-- which you are doing. In a uniformly sample, finite time series (=FMRI data going into the model), there are frequencies ranging from 0 to the Nyquist frequency, where the Nyquist frequency f_N depends on your TR; essentially, f_N = 1/(2*TR). So, if your TR = 2 s, then f_N = 0.25 Hz. The frequencies are discrete, too. To censor a frequency in the GLM, you have 1 cos and 1 sin of that frequency. Going back to our counting of DFs, when you “bandpass” out a frequency in the GLM, you lose 2 DFs.
Now, to your case: you can approximate the number of DFs you lose with a bandpass in the following:
- Say you are keeping frequencies between fbot (bottom freq) and ftop (top freq), and these are in the range [0, f_N].
- Then, the fraction of DFs you keep is approx: (ftop - fbot)/f_N.
- And the fraction of DFs you lose is approx: 1 - [(ftop - fbot)/f_N].
- If ftop >> fbot, then you can approximate the loss as: 1 - ftop/f_N = 1 - (2 * ftop * TR).
- Note: from this last relation (though it is true in general, too), if you bandpass through a given range and decrease your TR, then you purge more DFs.
In standard RS-FMRI, people might have TR=2s, and bandpass to keep 0.01-0.1 Hz, then you are losing:
1 - (0.1 - 0.01)/0.25 ~ 1 - 0.1/0.25 = 1 - 0.4 = 0.6
… which means you are losing 60% of your DFs, just by bandpassing. (And you will almost certainly lose more with baseline regressors, motion regressors, and motion censoring.) That is a big loss.
Now, in your case, if you have TR = 2s. then bandpassing to keep freq between fbot=0 and ftop=0.01 Hz, leads to losing:
1 - (0.01 - 0)/0.25 = 1 - 0.01/0.25 = 1 - 0.04 = 0.96
… means you are losing 96% of your data (!!!). That is why you are getting the errors you are seeing in many cases: you are using up too many DFs in total. The modeling won’t let you set up a model that asks to over-use the available DFs (which is N, to start).
To do “high pass”, which means getting rid of low frequencies and keeping everything above a certain value, then you specify your “fbot”, and make your “ftop” the f_N or higher. For example, you could have your fbot = 0.01 and ftop = 999 (and the program should chop of ftop at the Nyquist frequency).
Note that often the reason for “high pass” filtering like this is to put a regressor in the model for low frequency, slow baseline drift. There are multiple ways to do this: I think SPM processing tends to use this “high pass filtering”, which is probably fine. In AFNI, we tend to use low order polynomials in the model. I don’t know that it makes that much difference, but there you have it.
OK, to your examples:
It seems you are basically using 1dBPort to make the set of regressors to purge from your data.
Ex 1:
1dBport -nodata 525 0.8 -band 0.01 0.1 -invert -nozero
… You have 525 time points, TR = 0.8 s (so f_N = 0.625 Hz), and you are making regressors <0.1 Hz and >0.1 Hz to put in your model, for bandpassing. According to the above rule, I think you are removing:
~ 1 - 20.10.8 = 1 - 0.16 = 0.84
… so you are purging 84% of your DFs right off the bat. You see that in the message:
++ 525 retained time points MINUS 462 regressors ==> 63 D.O.F. left
… where you are actually losing 88% of your DFs in total, I assume because you have other regressors in your model, too.
Ex. 2:
1dBport -nodata 525 0.8 -band 0.01 5100000 -invert -nozero > bandpass_rall.1D
Here, you are making regressors of a different range, just <0.01 Hz, which are very few (though your ftop is reaalllly unnecessarily large; it just needs to be > f_N). Within the program, your fbot = 0.01 and ftop = f_N, so we can see the loss is:
1 - (f_N - 0.01)/f_N = 1 - 1 + (0.01 / 0.625) = 0.016,
—> so you are only purging about 1.6% of your data from bandpassing (= about 8 DFs), which is consistent with this message:
++ 525 retained time points MINUS 23 regressors ==> 502 D.O.F. left
Ex 3:
1dBport -nodata 525 0.8 -band 0.01 5300000 -invert -nozero > bandpass_rall.1D
… should, in theory, be the exact same as Ex 2, but the “ftop” number here is so big (5300000), that it appears to be breaking the input reading somehow-- it appears to be being treated as if ftop=fbot; that is, no regressors are made, just 1s. Why are you using such a large number? I don’t think you should be using that large-- ftop just needs to be bigger than f_N to have it be “maximal”.
Ex 4:
1dBport -nodata 525 0.8 -band 0 0.01 > bandpass_rall.1D
… which means make regressors just within fbot=0 and ftop=0.01, with no inversion. This should be approximately the same as Example 2, with the difference coming from the value of freq = 0.01 Hz being included in the “made” regressors here, but the value freq = 0.01 Hz not being included within Ex 2, the way the inversion is done, basically whether 0.01 is within the range of bandpassing (case in Ex 4) or whether it is not (case of inversion in Ex 2).
–pt
