Be aware: This submit is an excerpt from the forthcoming ebook, Deep Studying and Scientific Computing with R torch. The chapter in query is on the Discrete Fourier Rework (DFT), and is positioned partly three. Half three is devoted to scientific computation past deep studying.
There are two chapters on the Fourier Rework. The primary strives to, in as “verbal” and lucid a means as was attainable to me, solid a lightweight on what’s behind the magic; it additionally exhibits how, surprisingly, you possibly can code the DFT in merely half a dozen strains. The second focuses on quick implementation (the Quick Fourier Rework, or FFT), once more with each conceptual/explanatory in addition to sensible, codeityourself components.
Collectively, these cowl way more materials than might sensibly match right into a weblog submit; subsequently, please contemplate what follows extra as a “teaser” than a totally fledged article.
Within the sciences, the Fourier Rework is nearly all over the place. Said very usually, it converts information from one illustration to a different, with none lack of data (if finished accurately, that’s.) In case you use torch
, it’s only a operate name away: torch_fft_fft()
goes a technique, torch_fft_ifft()
the opposite. For the person, that’s handy – you “simply” have to know find out how to interpret the outcomes. Right here, I need to assist with that. We begin with an instance operate name, enjoying round with its output, after which, attempt to get a grip on what’s going on behind the scenes.
Understanding the output of torch_fft_fft()
As we care about precise understanding, we begin from the only attainable instance sign, a pure cosine that performs one revolution over the whole sampling interval.
Place to begin: A cosine of frequency 1
The best way we set issues up, there will probably be sixtyfour samples; the sampling interval thus equals N = 64
. The content material of frequency()
, the under helper operate used to assemble the sign, displays how we signify the cosine. Specifically:
[
f(x) = cos(frac{2 pi}{N} k x)
]
Right here (x) values progress over time (or house), and (ok) is the frequency index. A cosine is periodic with interval (2 pi); so if we would like it to first return to its beginning state after sixtyfour samples, and (x) runs between zero and sixtythree, we’ll need (ok) to be equal to (1). Like that, we’ll attain the preliminary state once more at place (x = frac{2 pi}{64} * 1 * 64).
Let’s shortly verify this did what it was presupposed to:
df < data.frame(x = sample_positions, y = as.numeric(x))
ggplot(df, aes(x = x, y = y)) +
geom_line() +
xlab("time") +
ylab("amplitude") +
theme_minimal()
Now that now we have the enter sign, torch_fft_fft()
computes for us the Fourier coefficients, that’s, the significance of the varied frequencies current within the sign. The variety of frequencies thoughtabout will equal the variety of sampling factors: So (X) will probably be of size sixtyfour as effectively.
(In our instance, you’ll discover that the second half of coefficients will equal the primary in magnitude. That is the case for each realvalued sign. In such instances, you can name torch_fft_rfft()
as an alternative, which yields “nicer” (within the sense of shorter) vectors to work with. Right here although, I need to clarify the overall case, since that’s what you’ll discover finished in most expositions on the subject.)
Even with the sign being actual, the Fourier coefficients are complicated numbers. There are 4 methods to examine them. The primary is to extract the actual half:
[1] 0 32 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 32
Solely a single coefficient is nonzero, the one at place 1. (We begin counting from zero, and will discard the second half, as defined above.)
Now wanting on the imaginary half, we discover it’s zero all through:
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 0
At this level we all know that there’s only a single frequency current within the sign, specifically, that at (ok = 1). This matches (and it higher needed to) the best way we constructed the sign: specifically, as undertaking a single revolution over the whole sampling interval.
Since, in idea, each coefficient might have nonzero actual and imaginary components, usually what you’d report is the magnitude (the sq. root of the sum of squared actual and imaginary components):
[1] 0 32 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 32
Unsurprisingly, these values precisely replicate the respective actual components.
Lastly, there’s the part, indicating a attainable shift of the sign (a pure cosine is unshifted). In torch
, now we have torch_angle()
complementing torch_abs()
, however we have to bear in mind roundoff error right here. We all know that in every however a single case, the actual and imaginary components are each precisely zero; however as a result of finite precision in how numbers are introduced in a pc, the precise values will usually not be zero. As a substitute, they’ll be very small. If we take considered one of these “faux nonzeroes” and divide it by one other, as occurs within the angle calculation, large values may end up. To forestall this from taking place, our customized implementation rounds each inputs earlier than triggering the division.
part < operate(Ft, threshold = 1e5) {
torch_atan2(
torch_abs(torch_round(Ft$imag * threshold)),
torch_abs(torch_round(Ft$actual * threshold))
)
}
as.numeric(part(Ft)) %>% round(5)
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 0
As anticipated, there is no such thing as a part shift within the sign.
Let’s visualize what we discovered.
create_plot < operate(x, y, amount) {
df < data.frame(
x_ = x,
y_ = as.numeric(y) %>% round(5)
)
ggplot(df, aes(x = x_, y = y_)) +
geom_col() +
xlab("frequency") +
ylab(amount) +
theme_minimal()
}
p_real < create_plot(
sample_positions,
real_part,
"actual half"
)
p_imag < create_plot(
sample_positions,
imag_part,
"imaginary half"
)
p_magnitude < create_plot(
sample_positions,
magnitude,
"magnitude"
)
p_phase < create_plot(
sample_positions,
part(Ft),
"part"
)
p_real + p_imag + p_magnitude + p_phase
It’s truthful to say that now we have no motive to doubt what torch_fft_fft()
has finished. However with a pure sinusoid like this, we are able to perceive precisely what’s happening by computing the DFT ourselves, by hand. Doing this now will considerably assist us later, once we’re writing the code.
Reconstructing the magic
One caveat about this part. With a subject as wealthy because the Fourier Rework, and an viewers who I think about to differ broadly on a dimension of math and sciences schooling, my possibilities to fulfill your expectations, expensive reader, have to be very near zero. Nonetheless, I need to take the chance. In case you’re an knowledgeable on this stuff, you’ll anyway be simply scanning the textual content, searching for items of torch
code. In case you’re reasonably aware of the DFT, you should still like being reminded of its interior workings. And – most significantly – in case you’re slightly new, and even fully new, to this matter, you’ll hopefully take away (at the least) one factor: that what looks as if one of many biggest wonders of the universe (assuming there’s a actuality in some way equivalent to what goes on in our minds) might be a surprise, however neither “magic” nor a factor reserved to the initiated.
In a nutshell, the Fourier Rework is a foundation transformation. Within the case of the DFT – the Discrete Fourier Rework, the place time and frequency representations each are finite vectors, not features – the brand new foundation appears like this:
[
begin{aligned}
&mathbf{w}^{0n}_N = e^{ifrac{2 pi}{N}* 0 * n} = 1
&mathbf{w}^{1n}_N = e^{ifrac{2 pi}{N}* 1 * n} = e^{ifrac{2 pi}{N} n}
&mathbf{w}^{2n}_N = e^{ifrac{2 pi}{N}* 2 * n} = e^{ifrac{2 pi}{N}2n}& …
&mathbf{w}^{(N1)n}_N = e^{ifrac{2 pi}{N}* (N1) * n} = e^{ifrac{2 pi}{N}(N1)n}
end{aligned}
]
Right here (N), as earlier than, is the variety of samples (64, in our case); thus, there are (N) foundation vectors. With (ok) working by the idea vectors, they are often written:
[
mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}k n}
] {#eqdft1}
Like (ok), (n) runs from (0) to (N1). To know what these foundation vectors are doing, it’s useful to quickly swap to a shorter sampling interval, (N = 4), say. If we achieve this, now we have 4 foundation vectors: (mathbf{w}^{0n}_N), (mathbf{w}^{1n}_N), (mathbf{w}^{2n}_N), and (mathbf{w}^{3n}_N). The primary one appears like this:
[
mathbf{w}^{0n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 0 * 0}
e^{ifrac{2 pi}{4}* 0 * 1}
e^{ifrac{2 pi}{4}* 0 * 2}
e^{ifrac{2 pi}{4}* 0 * 3}
end{bmatrix}
=
begin{bmatrix}
1
1
1
1
end{bmatrix}
]
The second, like so:
[
mathbf{w}^{1n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 1 * 0}
e^{ifrac{2 pi}{4}* 1 * 1}
e^{ifrac{2 pi}{4}* 1 * 2}
e^{ifrac{2 pi}{4}* 1 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ifrac{pi}{2}}
e^{i pi}
e^{ifrac{3 pi}{4}}
end{bmatrix}
=
begin{bmatrix}
1
i
1
i
end{bmatrix}
]
That is the third:
[
mathbf{w}^{2n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 2 * 0}
e^{ifrac{2 pi}{4}* 2 * 1}
e^{ifrac{2 pi}{4}* 2 * 2}
e^{ifrac{2 pi}{4}* 2 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ipi}
e^{i 2 pi}
e^{ifrac{3 pi}{2}}
end{bmatrix}
=
begin{bmatrix}
1
1
1
1
end{bmatrix}
]
And eventually, the fourth:
[
mathbf{w}^{3n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 3 * 0}
e^{ifrac{2 pi}{4}* 3 * 1}
e^{ifrac{2 pi}{4}* 3 * 2}
e^{ifrac{2 pi}{4}* 3 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ifrac{3 pi}{2}}
e^{i 3 pi}
e^{ifrac{9 pi}{2}}
end{bmatrix}
=
begin{bmatrix}
1
i
1
i
end{bmatrix}
]
We will characterize these 4 foundation vectors by way of their “pace”: how briskly they transfer across the unit circle. To do that, we merely take a look at the rightmost column vectors, the place the ultimate calculation outcomes seem. The values in that column correspond to positions pointed to by the revolving foundation vector at completely different deadlines. Which means taking a look at a single “replace of place”, we are able to see how briskly the vector is transferring in a single time step.
Wanting first at (mathbf{w}^{0n}_N), we see that it doesn’t transfer in any respect. (mathbf{w}^{1n}_N) goes from (1) to (i) to (1) to (i); yet another step, and it might be again the place it began. That’s one revolution in 4 steps, or a step dimension of (frac{pi}{2}). Then (mathbf{w}^{2n}_N) goes at double that tempo, transferring a distance of (pi) alongside the circle. That means, it finally ends up finishing two revolutions total. Lastly, (mathbf{w}^{3n}_N) achieves three full loops, for a step dimension of (frac{3 pi}{2}).
The factor that makes these foundation vectors so helpful is that they’re mutually orthogonal. That’s, their dot product is zero:
[
langle mathbf{w}^{kn}_N, mathbf{w}^{ln}_N rangle = sum_{n=0}^{N1} ({e^{ifrac{2 pi}{N}k n}})^* e^{ifrac{2 pi}{N}l n} = sum_{n=0}^{N1} ({e^{ifrac{2 pi}{N}k n}})e^{ifrac{2 pi}{N}l n} = 0
] {#eqdft2}
Let’s take, for instance, (mathbf{w}^{2n}_N) and (mathbf{w}^{3n}_N). Certainly, their dot product evaluates to zero.
[
begin{bmatrix}
1 & 1 & 1 & 1
end{bmatrix}
begin{bmatrix}
1
i
1
i
end{bmatrix}
=
1 + i + (1) + (i) = 0
]
Now, we’re about to see how the orthogonality of the Fourier foundation considerably simplifies the calculation of the DFT. Did you discover the similarity between these foundation vectors and the best way we wrote the instance sign? Right here it’s once more:
[
f(x) = cos(frac{2 pi}{N} k x)
]
If we handle to signify this operate by way of the idea vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok n}), the interior product between the operate and every foundation vector will probably be both zero (the “default”) or a a number of of 1 (in case the operate has a part matching the idea vector in query). Fortunately, sines and cosines can simply be transformed into complicated exponentials. In our instance, that is how that goes:
[
begin{aligned}
mathbf{x}_n &= cos(frac{2 pi}{64} n)
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{ifrac{2 pi}{64} n})
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{ifrac{2 pi}{64} 63n})
&= frac{1}{2} (mathbf{w}^{1n}_N + mathbf{w}^{63n}_N)
end{aligned}
]
Right here step one straight outcomes from Euler’s components, and the second displays the truth that the Fourier coefficients are periodic, with frequency 1 being the identical as 63, 2 equaling 62, and so forth.
Now, the (ok)th Fourier coefficient is obtained by projecting the sign onto foundation vector (ok).
As a result of orthogonality of the idea vectors, solely two coefficients won’t be zero: these for (mathbf{w}^{1n}_N) and (mathbf{w}^{63n}_N). They’re obtained by computing the interior product between the operate and the idea vector in query, that’s, by summing over (n). For every (n) ranging between (0) and (N1), now we have a contribution of (frac{1}{2}), leaving us with a remaining sum of (32) for each coefficients. For instance, for (mathbf{w}^{1n}_N):
[
begin{aligned}
X_1 &= langle mathbf{w}^{1n}_N, mathbf{x}_n rangle
&= langle mathbf{w}^{1n}_N, frac{1}{2} (mathbf{w}^{1n}_N + mathbf{w}^{63n}_N) rangle
&= frac{1}{2} * 64
&= 32
end{aligned}
]
And analogously for (X_{63}).
Now, wanting again at what torch_fft_fft()
gave us, we see we had been capable of arrive on the identical outcome. And we’ve realized one thing alongside the best way.
So long as we stick with alerts composed of a number of foundation vectors, we are able to compute the DFT on this means. On the finish of the chapter, we’ll develop code that may work for all alerts, however first, let’s see if we are able to dive even deeper into the workings of the DFT. Three issues we’ll need to discover:

What would occur if frequencies modified – say, a melody had been sung at a better pitch?

What about amplitude adjustments – say, the music had been performed twice as loud?

What about part – e.g., there have been an offset earlier than the piece began?
In all instances, we’ll name torch_fft_fft()
solely as soon as we’ve decided the outcome ourselves.
And eventually, we’ll see how complicated sinusoids, made up of various parts, can nonetheless be analyzed on this means, supplied they are often expressed by way of the frequencies that make up the idea.
Various frequency
Assume we quadrupled the frequency, giving us a sign that appeared like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]
Following the identical logic as above, we are able to categorical it like so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N)
]
We already see that nonzero coefficients will probably be obtained just for frequency indices (4) and (60). Choosing the previous, we receive
[
begin{aligned}
X_4 &= langle mathbf{w}^{4n}_N, mathbf{x}_n rangle
&= langle mathbf{w}^{4n}_N, frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N) rangle
&= 32
end{aligned}
]
For the latter, we’d arrive on the identical outcome.
Now, let’s be sure that our evaluation is right. The next code snippet incorporates nothing new; it generates the sign, calculates the DFT, and plots them each.
x < torch_cos(frequency(4, N) * sample_positions)
plot_ft < operate(x) plot_spacer()) /
(p_real
plot_ft(x)
This does certainly verify our calculations.
A particular case arises when sign frequency rises to the very best one “allowed”, within the sense of being detectable with out aliasing. That would be the case at one half of the variety of sampling factors. Then, the sign will appear like so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{32n}_N + mathbf{w}^{32n}_N)
]
Consequently, we find yourself with a single coefficient, equivalent to a frequency of 32 revolutions per pattern interval, of double the magnitude (64, thus). Listed below are the sign and its DFT:
x < torch_cos(frequency(32, N) * sample_positions)
plot_ft(x)
Various amplitude
Now, let’s take into consideration what occurs once we differ amplitude. For instance, say the sign will get twice as loud. Now, there will probably be a multiplier of two that may be taken exterior the interior product. In consequence, the one factor that adjustments is the magnitude of the coefficients.
Let’s confirm this. The modification relies on the instance we had earlier than the final one, with 4 revolutions over the sampling interval:
x < 2 * torch_cos(frequency(4, N) * sample_positions)
plot_ft(x)
To this point, now we have not as soon as seen a coefficient with nonzero imaginary half. To vary this, we add in part.
Including part
Altering the part of a sign means shifting it in time. Our instance sign is a cosine, a operate whose worth is 1 at (t=0). (That additionally was the – arbitrarily chosen – start line of the sign.)
Now assume we shift the sign ahead by (frac{pi}{2}). Then the height we had been seeing at zero strikes over to (frac{pi}{2}); and if we nonetheless begin “recording” at zero, we should discover a worth of zero there. An equation describing that is the next. For comfort, we assume a sampling interval of (2 pi) and (ok=1), in order that the instance is a straightforward cosine:
[
f(x) = cos(x – phi)
]
The minus signal could look unintuitive at first. But it surely does make sense: We now need to receive a worth of 1 at (x=frac{pi}{2}), so (x – phi) ought to consider to zero. (Or to any a number of of (pi).) Summing up, a delay in time will seem as a damaging part shift.
Now, we’re going to calculate the DFT for a shifted model of our instance sign. However in case you like, take a peek on the phaseshifted model of the timedomain image now already. You’ll see {that a} cosine, delayed by (frac{pi}{2}), is nothing else than a sine beginning at 0.
To compute the DFT, we observe our familiarbynow technique. The sign now appears like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*x – frac{pi}{2})
]
First, we categorical it by way of foundation vectors:
[
begin{aligned}
mathbf{x}_n &= cos(frac{2 pi}{64} 4 n – frac{pi}{2})
&= frac{1}{2} (e^{ifrac{2 pi}{64} 4n – frac{pi}{2}} + e^{ifrac{2 pi}{64} 60n – frac{pi}{2}})
&= frac{1}{2} (e^{ifrac{2 pi}{64} 4n} e^{i frac{pi}{2}} + e^{ifrac{2 pi}{64} 60n} e^{ifrac{pi}{2}})
&= frac{1}{2} (e^{i frac{pi}{2}} mathbf{w}^{4n}_N + e^{i frac{pi}{2}} mathbf{w}^{60n}_N)
end{aligned}
]
Once more, now we have nonzero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are now not similar. As a substitute, one is the complicated conjugate of the opposite. First, (X_4):
[
begin{aligned}
X_4 &= langle mathbf{w}^{4n}_N, mathbf{x}_n rangle
&=langle mathbf{w}^{4n}_N, frac{1}{2} (e^{i frac{pi}{2}} mathbf{w}^{4n}_N + e^{i frac{pi}{2}} mathbf{w}^{60n}_N) rangle
&= 32 *e^{i frac{pi}{2}}
&= 32i
end{aligned}
]
And right here, (X_{60}):
[
begin{aligned}
X_{60} &= langle mathbf{w}^{60n}_N, mathbf{x}_N rangle
&= 32 *e^{i frac{pi}{2}}
&= 32i
end{aligned}
]
As standard, we test our calculation utilizing torch_fft_fft()
.
x < torch_cos(frequency(4, N) * sample_positions  pi / 2)
plot_ft(x)
For a pure sine wave, the nonzero Fourier coefficients are imaginary. The part shift within the coefficients, reported as (frac{pi}{2}), displays the time delay we utilized to the sign.
Lastly – earlier than we write some code – let’s put all of it collectively, and take a look at a wave that has greater than a single sinusoidal part.
Superposition of sinusoids
The sign we assemble should be expressed by way of the idea vectors, however it’s now not a pure sinusoid. As a substitute, it’s a linear mixture of such:
[
begin{aligned}
mathbf{x}_n &= 3 sin(frac{2 pi}{64} 4n) + 6 cos(frac{2 pi}{64} 2n) +2cos(frac{2 pi}{64} 8n)
end{aligned}
]
I received’t undergo the calculation intimately, however it’s no completely different from the earlier ones. You compute the DFT for every of the three parts, and assemble the outcomes. With none calculation, nevertheless, there’s fairly a number of issues we are able to say:
 For the reason that sign consists of two pure cosines and one pure sine, there will probably be 4 coefficients with nonzero actual components, and two with nonzero imaginary components. The latter will probably be complicated conjugates of one another.
 From the best way the sign is written, it’s simple to find the respective frequencies, as effectively: The allreal coefficients will correspond to frequency indices 2, 8, 56, and 62; the allimaginary ones to indices 4 and 60.
 Lastly, amplitudes will outcome from multiplying with (frac{64}{2}) the scaling elements obtained for the person sinusoids.
Let’s test:
Now, how can we calculate the DFT for much less handy alerts?
Coding the DFT
Happily, we already know what must be finished. We need to undertaking the sign onto every of the idea vectors. In different phrases, we’ll be computing a bunch of interior merchandise. Logicwise, nothing adjustments: The one distinction is that generally, it won’t be attainable to signify the sign by way of only a few foundation vectors, like we did earlier than. Thus, all projections will truly need to be calculated. However isn’t automation of tedious duties one factor now we have computer systems for?
Let’s begin by stating enter, output, and central logic of the algorithm to be carried out. As all through this chapter, we keep in a single dimension. The enter, thus, is a onedimensional tensor, encoding a sign. The output is a onedimensional vector of Fourier coefficients, of the identical size because the enter, every holding details about a frequency. The central concept is: To acquire a coefficient, undertaking the sign onto the corresponding foundation vector.
To implement that concept, we have to create the idea vectors, and for each, compute its interior product with the sign. This may be finished in a loop. Surprisingly little code is required to perform the objective:
dft < operate(x) {
n_samples < length(x)
n < torch_arange(0, n_samples  1)$unsqueeze(1)
Ft < torch_complex(
torch_zeros(n_samples), torch_zeros(n_samples)
)
for (ok in 0:(n_samples  1)) {
w_k < torch_exp(1i * 2 * pi / n_samples * ok * n)
dot < torch_matmul(w_k, x$to(dtype = torch_cfloat()))
Ft[k + 1] < dot
}
Ft
}
To check the implementation, we are able to take the final sign we analysed, and evaluate with the output of torch_fft_fft()
.
[1] 0 0 192 0 0 0 0 0 64 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 64 0 0 0 0 0 192 0
[1] 0 0 0 0 96 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 96 0 0 0
Reassuringly – in case you look again – the outcomes are the identical.
Above, did I say “little code”? The truth is, a loop isn’t even wanted. As a substitute of working with the idea vectors onebyone, we are able to stack them in a matrix. Then every row will maintain the conjugate of a foundation vector, and there will probably be (N) of them. The columns correspond to positions (0) to (N1); there will probably be (N) of them as effectively. For instance, that is how the matrix would search for (N=4):
[
mathbf{W}_4
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 0 * 0} & e^{ifrac{2 pi}{4}* 0 * 1} & e^{ifrac{2 pi}{4}* 0 * 2} & e^{ifrac{2 pi}{4}* 0 * 3}
e^{ifrac{2 pi}{4}* 1 * 0} & e^{ifrac{2 pi}{4}* 1 * 1} & e^{ifrac{2 pi}{4}* 1 * 2} & e^{ifrac{2 pi}{4}* 1 * 3}
e^{ifrac{2 pi}{4}* 2 * 0} & e^{ifrac{2 pi}{4}* 2 * 1} & e^{ifrac{2 pi}{4}* 2 * 2} & e^{ifrac{2 pi}{4}* 2 * 3}
e^{ifrac{2 pi}{4}* 3 * 0} & e^{ifrac{2 pi}{4}* 3 * 1} & e^{ifrac{2 pi}{4}* 3 * 2} & e^{ifrac{2 pi}{4}* 3 * 3}
end{bmatrix}
] {#eqdft3}
Or, evaluating the expressions:
[
mathbf{W}_4
=
begin{bmatrix}
1 & 1 & 1 & 1
1 & i & 1 & i
1 & 1 & 1 & 1
1 & i & 1 & i
end{bmatrix}
]
With that modification, the code appears much more elegant:
dft_vec < operate(x) {
n_samples < length(x)
n < torch_arange(0, n_samples  1)$unsqueeze(1)
ok < torch_arange(0, n_samples  1)$unsqueeze(2)
mat_k_m < torch_exp(1i * 2 * pi / n_samples * ok * n)
torch_matmul(mat_k_m, x$to(dtype = torch_cfloat()))
}
As you possibly can simply confirm, the outcome is identical.
Thanks for studying!
Photograph by Trac Vu on Unsplash