[FieldTrip] spectral resolution & interpretation of power estimates

Lam, N.H.L. (Nietzsche) n.lam at fcdonders.ru.nl
Tue Nov 25 15:33:33 CET 2014


Hi Helene,

The idea in general is to design one's experiment in a way that allows one to observe effects at the frequency of interest.  Otherwise, what you might end up with is an experiment that doesn't allow you to estimate power at the frequencies of interest. 

For wavelets, given your specifications, you would get a spectral bandwidth of  8/7*2 = 2.28 Hz.  
So  8 +/-  2.28 Hz  ( 5.72 - 10.28) i.e. the bandwidth calculated refers to the bandwidth on one end, not both. Therefore, you shouldn't divide by two.  Similarly with multitapers, if you have a smoothing for 8 Hz, that's -8Hz and +8Hz. 

For the Hanning taper,  1/0.5s  = 2Hz, means you get estimates of power that are multiples of 2 Hz, so 10, 12, 14 Hz ..etc.   With regards to what the spectral bandwidth is, the short answer is that it is  plus and minus half the frequency resolution.  So, for 10 Hz, it is 9 - 11Hz.   The reasons behind this answer are to do with the spectral profile of the taper (e.g., a Hanning window) that you apply to your data.  I would suggest that you take a look at one of our lecture video's here by Robert: https://www.youtube.com/watch?feature=player_detailpage&v=QLvsa1r1Voc#t=741   where he provides a much more detailed explanation.

cfg.pad parameter in ft_freqanalysis:  This parameter is helpful if your trials are of different length.  If you set cfg.pad = 7, that means any trial shorter than 7s long will be padded to become 7s worth of data.  The type of data that you use to pad is specified in cfg.padtype.    The frequency resolution of your data is indeed 1/cfg.pad.    There is a suggested limit for how much you should pad given your data, but I'm not sure what the limit is, maybe someone else knows.

Now with your code, 
  cfg.toi = -1.75:0.05:1.75;
  cfg.taper = 'hanning';
  cfg.foi = 3:35;
  cfg.t_ftimwin = ones(length(cfg.foi),1) .* 0.4;  

You have a 0.4 second window.  So 1/0.4 = 2.5Hz frequency resolution, as you figured out. 
This means you can only estimate in steps of 2.5Hz so doing power estimates at 1 Hz steps does *not* make sense.  To get any sensible estimate, you would need something like this cfg.foi = 2.5:2.5:35;

Finally, no, the frequency resolution does not have to be an integer. We don't know what frequency(ies) the brain oscillates, but integers allow for us to deal with the data more easily. 

I hope this helps, and I hope that if I've not been entirely accurate with my explanations that someone will jump in to correct me.

Best,
Nietzsche 

________________________________________
From: fieldtrip-bounces at science.ru.nl [fieldtrip-bounces at science.ru.nl] on behalf of Helene Gudi [helene.gudi at uni-hamburg.de]
Sent: 25 November 2014 13:15
To: fieldtrip at science.ru.nl
Subject: [FieldTrip] spectral resolution & interpretation of power estimates

Dear FieldTrip List,

Since  a while I am struggeling with a question regarding the spectral
resolution and its consequences for interpreting the power vaules. I
would appreciate any comments or hints which help me understand the issue.

As described in the tutorial on TFR analysis when using the Morlet
wavelets the spectral bandwidth at a given frequency is determined by
the formula: F/width*2, meaning that when I define cfg.width=7 and look
into the estimated power values of e.g. 8Hz, what I get is the power
values not for 8 Hz but for a spectral band of 8Hz+/- (8/7*2)/2.
Am I correct?
If yes, is this also true when estimating the power values using a
Hanning window? As described in the tutorial when using the Hanning
taper the frequency resolution is defined by: 1/length of the sliding
window. Let's say my sliding window = .5s, resulting in a frequency
resolution of 2 Hz. Now, if i look into the power estimate for 10 Hz, do
I get the spectral band of 2Hz at a given frequency eg. 9-11Hz, for 10Hz
etc.?

My next question refers to the 'cfg.pad' parameter, in the help it says:
"the padding determines the spectral resolution". I unfortunately could
not find any further explanations. What exactly does it mean? How do I
compute the exact frequency resolution once cfg.pad has been used? Is it
1/cfg.pad?
Does the padding allow to estimate power values for frequencies other
than given by '1/length of time window'-resolution? As in the code below
power estimates are calculated in 1Hz steps. Do the power estimates at
3Hz,4Hz etc. make sense at all, given the frequency resolution of 2.5Hz
(as defined by 1/cfg.t_ftimwin)? What spectral band is actually included
in the single 1Hz bins? Is it a problem if the frequency resolution ist
not an integer?

                 cfg = [];
                 cfg.output = 'pow';
                 cfg.channel = 'all';
                 cfg.keeptapers  = 'no';
                 cfg.pad = 7;
                 cfg.method = 'mtmconvol';
                 cfg.toi = -1.75:0.05:1.75;
                 cfg.taper = 'hanning';
                 cfg.keeptrials  = 'yes';
                 cfg.foi = 3:35;
                 cfg.t_ftimwin = ones(length(cfg.foi),1) .* 0.4;
                 ft_freqanalysis(cfg, data);

I would be very thankful for any help!
Lena

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