Observe: 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 partially 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 approach as was attainable to me, forged a lightweight on what’s behind the magic; it additionally reveals how, surprisingly, you’ll be able to code the DFT in merely half a dozen traces. The second focuses on quick implementation (the Quick Fourier Rework, or FFT), once more with each conceptual/explanatory in addition to sensible, codeityourself elements.
Collectively, these cowl way more materials than might sensibly match right into a weblog submit; subsequently, please take into account what follows extra as a “teaser” than a totally fledged article.
Within the sciences, the Fourier Rework is nearly in all places. Said very typically, it converts information from one illustration to a different, with none lack of data (if carried out accurately, that’s.) In the event 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 easy methods to interpret the outcomes. Right here, I need to assist with that. We begin with an instance operate name, taking part in 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 best attainable instance sign, a pure cosine that performs one revolution over the whole sampling interval.
Start line: A cosine of frequency 1
The way in which we set issues up, there can be sixtyfour samples; the sampling interval thus equals N = 64
. The content material of frequency()
, the beneath helper operate used to assemble the sign, displays how we signify the cosine. Particularly:
[
f(x) = cos(frac{2 pi}{N} k x)
]
Right here (x) values progress over time (or area), and (okay) 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 (okay) 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 rapidly verify this did what it was presupposed to:
df < information.body(x = sample_positions, y = as.numeric(x))
ggplot(df, aes(x = x, y = y)) +
geom_line() +
xlab("time") +
ylab("amplitude") +
theme_minimal()
Now that we’ve the enter sign, torch_fft_fft()
computes for us the Fourier coefficients, that’s, the significance of the assorted frequencies current within the sign. The variety of frequencies thought of will equal the variety of sampling factors: So (X) can be of size sixtyfour as nicely.
(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 would 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 final case, since that’s what you’ll discover carried out 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 true 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 trying 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 (okay = 1). This matches (and it higher needed to) the best way we constructed the sign: specifically, as carrying out a single revolution over the whole sampling interval.
Since, in concept, each coefficient might have nonzero actual and imaginary elements, typically what you’d report is the magnitude (the sq. root of the sum of squared actual and imaginary elements):
[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 elements.
Lastly, there’s the part, indicating a attainable shift of the sign (a pure cosine is unshifted). In torch
, we’ve 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 true and imaginary elements are each precisely zero; however because of finite precision in how numbers are offered in a pc, the precise values will typically not be zero. As a substitute, they’ll be very small. If we take one in every of these “pretend nonzeroes” and divide it by one other, as occurs within the angle calculation, massive values may result. To forestall this from occurring, 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)) %>% spherical(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 isn’t a part shift within the sign.
Let’s visualize what we discovered.
create_plot < operate(x, y, amount) {
df < information.body(
x_ = x,
y_ = as.numeric(y) %>% spherical(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 we’ve no purpose to doubt what torch_fft_fft()
has carried out. However with a pure sinusoid like this, we will perceive precisely what’s happening by computing the DFT ourselves, by hand. Doing this now will considerably assist us later, after 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 range extensively on a dimension of math and sciences training, my probabilities to satisfy your expectations, pricey reader, have to be very near zero. Nonetheless, I need to take the danger. In the event you’re an skilled on this stuff, you’ll anyway be simply scanning the textual content, looking for items of torch
code. In the event you’re reasonably acquainted with the DFT, you should still like being reminded of its inside workings. And – most significantly – should you’re quite new, and even fully new, to this matter, you’ll hopefully take away (a minimum of) one factor: that what looks like one of many best wonders of the universe (assuming there’s a actuality in some way comparable to what goes on in our minds) could 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 seems to be 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 (okay) working by the idea vectors, they are often written:
[
mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}k n}
] {#eqdft1}
Like (okay), (n) runs from (0) to (N1). To grasp what these foundation vectors are doing, it’s useful to briefly swap to a shorter sampling interval, (N = 4), say. If we achieve this, we’ve 4 foundation vectors: (mathbf{w}^{0n}_N), (mathbf{w}^{1n}_N), (mathbf{w}^{2n}_N), and (mathbf{w}^{3n}_N). The primary one seems to be 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 at last, 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 are able to characterize these 4 foundation vectors when it comes to 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 time limits. Because of this taking a look at a single “replace of place”, we will 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 could 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 approach, it finally ends up finishing two revolutions general. 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 when it comes to the idea vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}okay n}), the inside product between the operate and every foundation vector can 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 immediately outcomes from Euler’s method, 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 (okay)th Fourier coefficient is obtained by projecting the sign onto foundation vector (okay).
Because of the 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 inside product between the operate and the idea vector in query, that’s, by summing over (n). For every (n) ranging between (0) and (N1), we’ve a contribution of (frac{1}{2}), leaving us with a ultimate 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, trying again at what torch_fft_fft()
gave us, we see we have been in a position to arrive on the similar consequence. 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 will compute the DFT on this approach. On the finish of the chapter, we’ll develop code that may work for all alerts, however first, let’s see if we will 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 have been sung at a better pitch?

What about amplitude modifications – say, the music have 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 consequence ourselves.
And at last, we’ll see how complicated sinusoids, made up of various elements, can nonetheless be analyzed on this approach, supplied they are often expressed when it comes to the frequencies that make up the idea.
Various frequency
Assume we quadrupled the frequency, giving us a sign that seemed like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]
Following the identical logic as above, we will specific it like so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N)
]
We already see that nonzero coefficients can be obtained just for frequency indices (4) and (60). Choosing the previous, we acquire
[
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 similar consequence.
Now, let’s make certain our evaluation is right. The next code snippet comprises 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) p_phase)
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 to be so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{32n}_N + mathbf{w}^{32n}_N)
]
Consequently, we find yourself with a single coefficient, comparable to a frequency of 32 revolutions per pattern interval, of double the magnitude (64, thus). Listed here 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 after we range amplitude. For instance, say the sign will get twice as loud. Now, there can be a multiplier of two that may be taken outdoors the inside product. In consequence, the one factor that modifications 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, we’ve 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 have 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 (okay=1), in order that the instance is an easy cosine:
[
f(x) = cos(x – phi)
]
The minus signal might look unintuitive at first. But it surely does make sense: We now need to acquire 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 unfavourable part shift.
Now, we’re going to calculate the DFT for a shifted model of our instance sign. However should 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 comply with our familiarbynow technique. The sign now seems to be like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*x – frac{pi}{2})
]
First, we specific it when it comes to 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, we’ve nonzero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are not equivalent. 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 typical, 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 when it comes to the idea vectors, however it’s 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 elements, and assemble the outcomes. With none calculation, nevertheless, there’s fairly just a few issues we will say:
 For the reason that sign consists of two pure cosines and one pure sine, there can be 4 coefficients with nonzero actual elements, and two with nonzero imaginary elements. The latter can be complicated conjugates of one another.
 From the best way the sign is written, it’s simple to find the respective frequencies, as nicely: The allreal coefficients will correspond to frequency indices 2, 8, 56, and 62; the allimaginary ones to indices 4 and 60.
 Lastly, amplitudes will consequence from multiplying with (frac{64}{2}) the scaling components obtained for the person sinusoids.
Let’s test:
Now, how will we calculate the DFT for much less handy alerts?
Coding the DFT
Fortuitously, we already know what needs to be carried out. We need to venture the sign onto every of the idea vectors. In different phrases, we’ll be computing a bunch of inside merchandise. Logicwise, nothing modifications: The one distinction is that generally, it won’t be attainable to signify the sign when it comes to only a few foundation vectors, like we did earlier than. Thus, all projections will really need to be calculated. However isn’t automation of tedious duties one factor we’ve computer systems for?
Let’s begin by stating enter, output, and central logic of the algorithm to be applied. 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, venture the sign onto the corresponding foundation vector.
To implement that concept, we have to create the idea vectors, and for each, compute its inside product with the sign. This may be carried out in a loop. Surprisingly little code is required to perform the purpose:
dft < operate(x) {
n_samples < size(x)
n < torch_arange(0, n_samples  1)$unsqueeze(1)
Ft < torch_complex(
torch_zeros(n_samples), torch_zeros(n_samples)
)
for (okay in 0:(n_samples  1)) {
w_k < torch_exp(1i * 2 * pi / n_samples * okay * n)
dot < torch_matmul(w_k, x$to(dtype = torch_cfloat()))
Ft[k + 1] < dot
}
Ft
}
To check the implementation, we will take the final sign we analysed, and examine 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 – should you look again – the outcomes are the identical.
Above, did I say “little code”? In reality, a loop just isn’t even wanted. As a substitute of working with the idea vectors onebyone, we will stack them in a matrix. Then every row will maintain the conjugate of a foundation vector, and there can be (N) of them. The columns correspond to positions (0) to (N1); there can be (N) of them as nicely. 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 seems to be much more elegant:
dft_vec < operate(x) {
n_samples < size(x)
n < torch_arange(0, n_samples  1)$unsqueeze(1)
okay < torch_arange(0, n_samples  1)$unsqueeze(2)
mat_k_m < torch_exp(1i * 2 * pi / n_samples * okay * n)
torch_matmul(mat_k_m, x$to(dtype = torch_cfloat()))
}
As you’ll be able to simply confirm, the consequence is identical.
Thanks for studying!
Photograph by Trac Vu on Unsplash