Why is it that in older textbooks the DFT or FFT is defined as having scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer of 1 makes more sense as it is the averagevalue. of 4 ones.
I know people say - "Oh we can scale it later", but it doesn't make mathematical sense. The other point is that the Fourier transform doesn't apply to a step function anyway so why does the DFT?
Am 02.09.20 um 07:00 schrieb Tom Killwhang:
Why is it that in older textbooks the DFT or FFT is defined as having
scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If
we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer of
1 makes more sense as it is the average value. of 4 ones.
I know people say - "Oh we can scale it later", but it doesn't make
mathematical sense. The other point is that the Fourier transform
doesn't apply to a step function anyway so why does the DFT?
I think this for technical reasons. If you program FFT with twiddle
factors taken from the unit circle you get a total scale factor of N for
a full turn around. Typically it is less work to compensate for this
factor at another place.
Take also into account the scale factor of the inverse FFT. If you scale
by 1/N you must *not* scale the inverse FFT to get the original values
back. If you want to operate symmetrically you need to scale both by 1/sqrt(N) which is even more non intuitive. But this keeps the RMS
energy constant.
Why is it that in older textbooks the DFT or FFT is defined as having
scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If
we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer
of 1 makes more sense as it is the average value. of 4 ones.
I know people say - "Oh we can scale it later", but it doesn't make mathematical sense. The other point is that the Fourier transform
doesn't apply to a step function anyway so why does the DFT?
The question of how to normalize FFTs goes back way before computers
were even invented, on account of that vexing factor of 2*pi.
The physics and math contingents tend to use omega = 2 pi f as the independent frequency variable, so that the 2*pi winds up as a
multiplicative constant outside the integral, either asymmetrically as
1/(2 pi) on the inverse transform, or symmetrically, as
in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the statement of FT theorems.
IMNSHO putting the 2*pi in the exponent is a huge win, in both the
continuous and discrete cases
On Thursday, September 3, 2020 at 3:47:41 PM UTC-7, Phil Hobbs wrote:
(snip)
The question of how to normalize FFTs goes back way before computers
were even invented, on account of that vexing factor of 2*pi.
(snip of EE contingent)
The physics and math contingents tend to use omega = 2 pi f as the
independent frequency variable, so that the 2*pi winds up as a
multiplicative constant outside the integral, either asymmetrically as
1/(2 pi) on the inverse transform, or symmetrically, as
in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the
statement of FT theorems.
IMNSHO putting the 2*pi in the exponent is a huge win, in both the
continuous and discrete cases
Not only in the Fourier transform, but in all the rest of the equations, omega works out much better than f. It is only convenient to use f
when you are actually counting frequencies. (That is, no-one makes
frequency counters to output in omega.) There are some who use
wavelength over 2pi, but that is rare. Using omega and k, you get
exp(i.k.x - i.omega.t) for propagating waves. (Where k is the
wave number or wave vector, magnitude 2 pi over wavelength.)
On Thursday, September 3, 2020 at 3:47:41 PM UTC-7, Phil Hobbs wrote:
(snip)
The question of how to normalize FFTs goes back way before computers(snip of EE contingent)
were even invented, on account of that vexing factor of 2*pi.
The physics and math contingents tend to use omega = 2 pi f as the independent frequency variable, so that the 2*pi winds up as a multiplicative constant outside the integral, either asymmetrically as 1/(2 pi) on the inverse transform, or symmetrically, as
in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the statement of FT theorems.
IMNSHO putting the 2*pi in the exponent is a huge win, in both the continuous and discrete casesNot only in the Fourier transform, but in all the rest of the equations, omega works out much better than f. It is only convenient to use f
when you are actually counting frequencies. (That is, no-one makes
frequency counters to output in omega.) There are some who use
wavelength over 2pi, but that is rare. Using omega and k, you get
exp(i.k.x - i.omega.t) for propagating waves. (Where k is the
wave number or wave vector, magnitude 2 pi over wavelength.)
It is, then, convenient for the Fourier transform to put the 1/(2pi)
along with the d omega.
In the case of FFT in fixed point, best is to do all with enough bits so
as not to overflow and not divide by N until the end, if it is needed.
In floating point with radix other than 2, it is a little complicated where to put the divide, but note that bits can be lost when dividing by two. (Specifically, IBM used radix 16 for many years, and still supports it, along with 2 a little recently, and 10 on newer machines. Yes,
some machines support all three.)
Step functions in continuous Fourier transforms are not rare, but
a little complicated. Delta functions are not unusual. But for DFT
(defined for periodic band-limited signals) a step function is
not all that special. Especially for small N, where it isn't all
that steep. And note that this all applies for DST or DCT, too.
I noticed another problem that arises from this. The units don't match up when you
use Parseval's theorem and the FFT. Look ta the equation near the bottom that uses the FFT for Parseval's theorem
https://en.wikipedia.org/wiki/Parseval%27s_theorem
I did a simulation with white noise through a bandpass filter.
Now I found the periodogram and this gives me power vs frequency. when you sum them
up you do indeed get the same answer - but the expression on the left of the equation
is the sum of squares of a random variable and that is not power.
The power is variance after you divide by N. Hence both sides need to be divided by N
otherwise we are equating energies and not power.
I always understood variance (with no dc) is average power and not energy.
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