Thank you for that, it has helped me think through my question.

I used convert(arctanh(-2),polar) to see it in the form you suggest and I drew an Argand diagram.

My question is then why for eg x<-1 does the imaginary part appear to remain a constant value (Pi/2) even though the argument and modulus change values?

Is there a graphical explanation to this?

Thank you once again

Thank you for that, it has helped me think through my question.

I used convert(arctanh(-2),polar) to see it in the form you suggest and I drew an Argand diagram.

My question is then why for eg x<-1 does the imaginary part appear to remain a constant value (Pi/2) even though the argument and modulus change values?

Is there a graphical explanation to this?

Thank you once again

No change for me.

Can I check if the second plot was printed and can you repeat the problem?

Thanks for looking at this.  I have made a 'submit' as suggested.

No change for me.

Can I check if the second plot was printed and can you repeat the problem?

Thanks for looking at this.  I have made a 'submit' as suggested.

I too am using XP SP3, Standard Interface and 14.01

Any other options anyone can think of

Thank you for that: I have worked through your sheet and my understanding increases.  I know it is mathematics but it also has, for me, a sort of magic about it!

Regarding the HarmonicAnalyses package which I do not need as I can now, using the method above, do what I want and understand what is going on which is even better, I did never the less down load it: unpacking the zip reported that the file was corrupt.  Is this still a 'pay for' package?  Is that why it does not unpack?

A further question if I may; given that it is so easy to zero the frequencies in the DFT in order to filter what is achieved by dedicated filters (I think there is one called Butterworth) which is not achieved by this method?

Thank you for that: I have worked through your sheet and my understanding increases.  I know it is mathematics but it also has, for me, a sort of magic about it!

Regarding the HarmonicAnalyses package which I do not need as I can now, using the method above, do what I want and understand what is going on which is even better, I did never the less down load it: unpacking the zip reported that the file was corrupt.  Is this still a 'pay for' package?  Is that why it does not unpack?

A further question if I may; given that it is so easy to zero the frequencies in the DFT in order to filter what is achieved by dedicated filters (I think there is one called Butterworth) which is not achieved by this method?

Thank you, I have read and found the link helpful.

I have plotted the data multiplied by the window function you provided and it progresses from zero and back to zero as expected. 

However, when I pass the data multiplied by the window function through the filter the results at the start and end data show much greater deviation from the signal than occurs without using the window function.   Is this possible or have I made an error in implementing your code?

On the other hand when I choose my window start and end so that the signal has a similar magnitude at the start and end, the filtered signal follows the unfiltered signal much more closely as the article suggest.

Thank you, I have read and found the link helpful.

I have plotted the data multiplied by the window function you provided and it progresses from zero and back to zero as expected. 

However, when I pass the data multiplied by the window function through the filter the results at the start and end data show much greater deviation from the signal than occurs without using the window function.   Is this possible or have I made an error in implementing your code?

On the other hand when I choose my window start and end so that the signal has a similar magnitude at the start and end, the filtered signal follows the unfiltered signal much more closely as the article suggest.

Thank you very much.

I can understand why ‘zeroing the result of the FFT’ produces the result: I think that is the first time in signal-processing that the how-to seems so straight forward.

I have your code up and running with my data (collected at 50 Hz) and it works with range 0..5 and so on.

If I might take advantage of your expertise with a question:  

The lower I set the second frequency (high pass?) the more the filtered signal deviates from the unfiltered signal at the signal start and end.

Am I correct in thinking that this is related to the fact that the signal I pass to the DFT is actually the combination of a unit impulse for the duration of the signal and the signal itself?  

If I am correct why does the filtered signal start and end deviation from the unfiltered signal decrease as higher frequencies are introduced?

Many thanks once again.    

Thank you very much.

I can understand why ‘zeroing the result of the FFT’ produces the result: I think that is the first time in signal-processing that the how-to seems so straight forward.

I have your code up and running with my data (collected at 50 Hz) and it works with range 0..5 and so on.

If I might take advantage of your expertise with a question:  

The lower I set the second frequency (high pass?) the more the filtered signal deviates from the unfiltered signal at the signal start and end.

Am I correct in thinking that this is related to the fact that the signal I pass to the DFT is actually the combination of a unit impulse for the duration of the signal and the signal itself?  

If I am correct why does the filtered signal start and end deviation from the unfiltered signal decrease as higher frequencies are introduced?

Many thanks once again.    

Thank you for that; I will order Maple 14.

Thank you for that; I will order Maple 14.