this fft code doesn't make sense

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune

BillJosephson skrev:
> Rune Allnor wrote:
> > BillJosephson skrev:
> > > % What do I fail to understand, when I expect this code to produce two
> > > % waveforms of the same frequency, and identical fft peaks?
> > ....
> > > % but the second one shows the peak at 121 Hz (which is 120).
> >
> > No, it doesn't. The peak shows up in bin 121 which still is at 12 Hz.
> > Look at the exponential term in your DFT. There you will find
> >
> > 2*pi*k*n/N
> >
> > These factors seem thrown together at random, but they are not:
> >
> > - The 2*pi factor is there because the DFT is defined in terms of
> > angular
> >   frequencies. The n is a running index over samples in the input data
> >   sequence.
> > - The N is the number of samples in the input data sequence.
> > - The k is the bin index, and the factor you have trouble with
> >   understanding.
> >
> > To see how the k fits in the big bicture, we need first to draw the
> > big picture.
> >
> > The DFT takes a signal that consist of N samples that are
> > regularly sampled in time as input, and transform them to
> > a signal that is regularly sampled in frequency. In time domain
> > we have the sampling interval T, and the signal lasts for
> > (N-1)T seconds. The signal duration changes as N changes.
> >
> > In frequency domain, we have the sampling frequency Fs = 1/T
> > and the bin width dF = Fs/N. The spectrum is (N-1)*dF wide,
> > or (N-1)/N*Fs.
> >
> > Here is the key point: The sampling frequency is fixed when
> > N changes. The bin width dF changes when N changes.
> >
> > The question now is: At what frequency does coefficient X_k
> > fall, given N and T?
> >
> > The k'th frequency occur at F_k = k*dF = k*Fs/N =k/(N*T).
> >
> > Looking at that expression, one might ask why the T factor
> > is important or even necessary; and the answer is that it is
> > neither.

********************************************************************

Forget about this for now!

One can get rid of it by normalizing the frequency
> > scale with respect to the sampling frequency:
> >
> > f_k = F_k/Fs = (k/N*T)/(1/T) = k/N
> >
> > And now we recognize the exonential term in the DFT:
> >
> > 2*pi*n*F_k -> 2*pi*n*k/N
> >
> > when the T factors are "distilled" away.

*******************************************************************

> > To find out in what bin a certain frequency falls, you need
> > to reverse this whole excercise:
> >
> > Given N and T, in what bin (indexed by k) contains the
> > frequency f?
> >
> > First, find Fs and dF:
> >
> > Fs = 1/T
> > dF = 1/(NT)
> >
> > Next, k is found by comparing ratios:
> >
> > f/Fs = k*dF/Fs
> > f*T = (k/N*T)/(1/T)
> > f*T = kT/NT
> > f*T = k/N
> >
> > which, at last, gives
> >
> > k = f*T*N
> > =======
> >
> > and that's it. Just remember that k has to be an integer
> > while the ratio f*T needs not be an integer.
> >
> > Rune
>
>
> Rune,
>
> If  you  knew how much this message is helping me you'd know how much I
> appreciate your most kind help. I've been struggling with this off and
> on for the last year, and gotten help here and there, but this is the
> first message that is clear and complete enough that I think I have a
> chance of thorougly "getting it."  Actually, I'm still working my way
> through it and implementing a function as I go. I may come back with a
> question, but at the moment am feeling like for once I have all the
> parts together with the plan all at the same time.
>
> Many, many thanks.

Y're welcome.

Seems like you would benefit from reading Rick Lyons'
"Undertsanding Digital Signal Processing", Prentice Hall,
2004.

After I posted yesterday I realized that one section does not
fit in the overall post. I have outlined it above. The T factor
needst to be preserved in the DFT all the way through,
it can not be "distilled away" by mere arithmetics.

The FFT is, however, a generic maths tool that is used
in many applications where similar factors need not
appear. Besides, leaving the T factor out also saves
some precious FLOPs in computation-expensive
applications. So by convention, everybody knows
that the T factor ought to be present in the exponential
term.

I'm sorry for confusing you, but these conventions are
second nature to me. I very seldom thing through
the details about how things work.

Rune