this fft code doesn't make sense

% What do I fail to understand, when I expect this code to produce two
% waveforms of the same frequency, and identical fft peaks?

t1 = (0:0.00048828:1);
y1  = sin(2*pi*12*t1);
t10 = (0:0.00048828:10);
y10  = sin(2*pi*12*t10);

% now I have two 12hz sine waves; y10 is 10 times as long as y1, and
has 120
% oscillations instead of the 12 for y1.....otherwise the waves are
% identical (I plotted and checked)

f1=abs(fft(y1));
f10=abs(fft(y10));

% now I have two ffts, f10 is 10 times as long as f1 (and both the same
as
% y1 and y10)

figure; plot(f1); figure; plot(f10);

% now I have two fft results for a 12 hz sine wave. The first one
correctly
% shows the peak at 13 Hz (which is 12).
% but the second one shows the peak at 121 Hz (which is 120).

% thanks in advance for any further help (this is my second, and
% hopefully better, question on this...)

Bill,

I think that what you don't understand is that one FFT is 10X longer
than the other but covers the same frequency range, so the index
spacing (in frequency) for the second case is 1/10 the index spacing of
the first case.

So adjusting for the MATLAB index starting at one and not 0, you would
expect the second case peak to be at the same frequency as the first
case peak, which it is, but this corresponds to an index value that is
10X the first case peak index, which it is.

(121-1)=10*(13-1)

Dirk Bell
DSP Consultant

BillJosephson wrote:
> % What do I fail to understand, when I expect this code to produce two
> % waveforms of the same frequency, and identical fft peaks?
>
> t1 = (0:0.00048828:1);
> y1  = sin(2*pi*12*t1);
> t10 = (0:0.00048828:10);
> y10  = sin(2*pi*12*t10);
>
> % now I have two 12hz sine waves; y10 is 10 times as long as y1, and
> has 120
> % oscillations instead of the 12 for y1.....otherwise the waves are
> % identical (I plotted and checked)
>
> f1=abs(fft(y1));
> f10=abs(fft(y10));
>
> % now I have two ffts, f10 is 10 times as long as f1 (and both the same
> as
> % y1 and y10)
>
> figure; plot(f1); figure; plot(f10);
>
> % now I have two fft results for a 12 hz sine wave. The first one
> correctly
> % shows the peak at 13 Hz (which is 12).
> % but the second one shows the peak at 121 Hz (which is 120).
>
> % thanks in advance for any further help (this is my second, and
> % hopefully better, question on this...)

On 2006-11-26, BillJosephson <billjosephson@hotmail.com> wrote:
> % now I have two fft results for a 12 hz sine wave. The first one
> correctly
> % shows the peak at 13 Hz (which is 12).
> % but the second one shows the peak at 121 Hz (which is 120).

The first is a coincidence.  Since your input sample was 1 second long,
the sample rate Fs and the size of the fft cancelled out.  When you
increased the size of the sample and thus the fft they no longer cancel.
You'll see the same discrepancy in the vertical scale.  Google fftplot()
(not sure if there's one in Matlab, there's not in Octave or Octave-forge
by default).  It is a wrapper which should give you a plot in the correct
units (Hz vs power).

-- 
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/

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. 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

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. 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

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. 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

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. 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

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. 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

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. 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

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. 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