It's not specific to wavelets and for wavelets there might be some
details I'm missing but fundamentally time and frequency resolution
are inversely related for any spectral estimation method. So to get 1
Hz resolution you need to work with 1 sec windows (that you can slide
in small increments, but they are still 1 sec windows). For 5 Hz
resolution 200 ms window would suffice etc. So by increasing Mfactor
you gain frequency resolution and lose time resolution. With other
methods these can be parametrised differently and the trade-off can be
better controlled but the fundamental limitation cannot be
circumvented.
Best,
Vladimir
On Sat, Nov 21, 2009 at 7:40 PM, Amir H Javadi <[log in to unmask]> wrote:
> Dear Vladimir
> Very many thanks for your detailed email. One more question. Your
> description for Mfactor was clear for time resolution. What does it mean in
> terms of frequency? How it contributes in frequency resolution?
> Have a good weekend :-)
> Amir
>
>
>
>
> 2009/11/21 Vladimir Litvak <[log in to unmask]>
>>
>> Dear Amir
>>
>> On Sat, Nov 21, 2009 at 12:19 PM, Amir H Javadi <[log in to unmask]>
>> wrote:
>>
>> > Sorry that I take your time a lot. I'm a little bit confused with the
>> > time-frequency analysis of SPM. I use the function "spm_eeg_tf" for
>> > time-frequency analysis. What is the output unit of "power" matrix of
>> > this
>> > function? Is it uV? The input is in uV. I use BioSemi to record my data.
>> > Thanks.
>>
>> If you don't do any rescaling then the units of power are the
>> original units squared so in your case uV^2. If you do rescaling then
>> you can get different units such as uV (sqrt), dB (LogR) or % (Rel).
>>
>> > One other question. What is the suggested value for "Mfactor" for
>> > different
>> > frequency bands? My intuition is that for high-frequencies it is better
>> > to
>> > use high Mfactor and for low-frequencies it is better to use low
>> > Mfactor.
>> > Currently I use 27 for gamma band and 11 for theta band. What is your
>> > idea?
>>
>> Mfactor determines time vs. frequency resolution trade-off. The
>> higher it is the better frequency resolution you will have at the
>> expense of the time resolution. So for instance if by theta you mean 5
>> Hz then one wavelet cycle is 200ms and 11 cycles is therefore 2.2
>> sec. That sounds like quite a long time to me but I'm not familiar
>> with your paradigm. The idea is then that power estimate for each
>> point is effectively an average of what's happening in +-1.1 sec sec
>> around this point.
>>
>> Similarly if by gamma your mean 40Hz then one cycle is 25ms and 27
>> cycles is 675 ms and you are averaging in +- 337.5 ms around your
>> point of interest.
>>
>> Best,
>>
>> Vladimir
>>
>>
>> > Thanks again for your attention.
>> > Have a good weekend :-)
>> > Amir
>> >
>> >
>
>
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