Notice: This submit is an excerpt from the forthcoming e-book, Deep Studying and Scientific Computing with R torch. The chapter in query is on the Discrete Fourier Remodel (DFT), and is positioned partially three. Half three is devoted to scientific computation past deep studying.
There are two chapters on the Fourier Remodel. The primary strives to, in as “verbal” and lucid a method as was potential to me, forged a lightweight on what’s behind the magic; it additionally exhibits how, surprisingly, you possibly can code the DFT in merely half a dozen traces. The second focuses on quick implementation (the Quick Fourier Remodel, or FFT), once more with each conceptual/explanatory in addition to sensible, code-it-yourself elements.
Collectively, these cowl way more materials than may sensibly match right into a weblog submit; due to this fact, please contemplate what follows extra as a “teaser” than a completely fledged article.

Within the sciences, the Fourier Remodel is nearly in all places. Said very usually, it converts information from one illustration to a different, with none lack of info (if achieved appropriately, that’s.) Should you use torch, it’s only a perform name away: torch_fft_fft() goes a technique, torch_fft_ifft() the opposite. For the person, that’s handy – you “simply” must know how one can interpret the outcomes. Right here, I need to assist with that. We begin with an instance perform 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 best potential instance sign, a pure cosine that performs one revolution over the whole sampling interval.

Place to begin: A cosine of frequency 1

The way in which we set issues up, there will probably be sixty-four samples; the sampling interval thus equals N = 64. The content material of frequency(), the beneath helper perform used to assemble the sign, displays how we symbolize the cosine. Specifically:

[
f(x) = cos(frac{2 pi}{N} k x)
]

Right here (x) values progress over time (or house), 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 sixty-four samples, and (x) runs between zero and sixty-three, 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 shortly affirm this did what it was imagined 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 got 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 thought of will equal the variety of sampling factors: So (X) will probably be of size sixty-four 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 real-valued 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 achieved 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 non-zero, the one at place 1. (We begin counting from zero, and should 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, particularly, that at (okay = 1). This matches (and it higher needed to) the best way we constructed the sign: particularly, as conducting a single revolution over the whole sampling interval.

Since, in concept, each coefficient may have non-zero actual and imaginary elements, usually 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 mirror the respective actual elements.

Lastly, there’s the part, indicating a potential shift of the sign (a pure cosine is unshifted). In torch, we’ve got torch_angle() complementing torch_abs(), however we have to keep in mind roundoff error right here. We all know that in every however a single case, the actual and imaginary elements are each precisely zero; however on account of finite precision in how numbers are introduced in a pc, the precise values will usually not be zero. As an alternative, they’ll be very small. If we take one among these “faux non-zeroes” and divide it by one other, as occurs within the angle calculation, large values may result. To forestall this from taking place, our customized implementation rounds each inputs earlier than triggering the division.

part <- perform(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 any part shift within the sign.

Let’s visualize what we discovered.

create_plot <- perform(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 got no purpose to doubt what torch_fft_fft() has achieved. However with a pure sinusoid like this, we will perceive precisely what’s occurring 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 Remodel, and an viewers who I think about to differ extensively on a dimension of math and sciences schooling, my possibilities to satisfy your expectations, pricey reader, should be very near zero. Nonetheless, I need to take the danger. Should you’re an professional on this stuff, you’ll anyway be simply scanning the textual content, looking for items of torch code. Should you’re reasonably accustomed to the DFT, you should still like being reminded of its internal workings. And – most significantly – in case you’re slightly new, and even fully new, to this subject, you’ll hopefully take away (no less than) one factor: that what looks as if one of many best wonders of the universe (assuming there’s a actuality one way or the other comparable 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 Remodel is a foundation transformation. Within the case of the DFT – the Discrete Fourier Remodel, the place time and frequency representations each are finite vectors, not capabilities – 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}^{(N-1)n}_N = e^{ifrac{2 pi}{N}* (N-1) * n} = e^{ifrac{2 pi}{N}(N-1)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 way of the premise vectors, they are often written:

[
mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}k n}
] {#eq-dft-1}

Like (okay), (n) runs from (0) to (N-1). To know what these foundation vectors are doing, it’s useful to briefly change to a shorter sampling interval, (N = 4), say. If we accomplish that, we’ve got 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 deadlines. Which means a single “replace of place”, we will see how briskly the vector is transferring in a single time step.

Trying 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 one more step, and it might be again the place it began. That’s one revolution in 4 steps, or a step measurement of (frac{pi}{2}). Then (mathbf{w}^{2n}_N) goes at double that tempo, transferring a distance of (pi) alongside the circle. That method, it finally ends up finishing two revolutions total. Lastly, (mathbf{w}^{3n}_N) achieves three full loops, for a step measurement 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}^{N-1} ({e^{ifrac{2 pi}{N}k n}})^* e^{ifrac{2 pi}{N}l n} = sum_{n=0}^{N-1} ({e^{-ifrac{2 pi}{N}k n}})e^{ifrac{2 pi}{N}l n} = 0
] {#eq-dft-2}

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 symbolize this perform when it comes to the premise vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}okay n}), the internal product between the perform and every foundation vector will probably be both zero (the “default”) or a a number of of 1 (in case the perform has a element matching the premise 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 formulation, 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 premise 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 internal product between the perform and the premise vector in query, that’s, by summing over (n). For every (n) ranging between (0) and (N-1), we’ve got 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, trying again at what torch_fft_fft() gave us, we see we had been in a position to arrive on the identical consequence. And we’ve discovered 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 method. 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 had been sung at the next pitch?

• What about amplitude modifications – 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 consequence ourselves.

And at last, we’ll see how complicated sinusoids, made up of various parts, can nonetheless be analyzed on this method, offered they are often expressed when it comes to the frequencies that make up the premise.

Various frequency

Assume we quadrupled the frequency, giving us a sign that regarded like this:

[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]

Following the identical logic as above, we will categorical it like so:

[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N)
]

We already see that non-zero coefficients will probably be obtained just for frequency indices (4) and (60). Selecting 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 identical consequence.

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 <- perform(x)  plot_spacer()) /
    (p_real 

plot_ft(x)

This does certainly affirm our calculations.

A particular case arises when sign frequency rises to the 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, comparable 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 after 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 internal product. In consequence, the one factor that modifications is the magnitude of the coefficients.

Let’s confirm this. The modification is predicated 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)

Up to now, we’ve got not as soon as seen a coefficient with non-zero 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 perform 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 (okay=1), in order that the instance is an easy cosine:

[
f(x) = cos(x – phi)
]

The minus signal might look unintuitive at first. However it does make sense: We now need to acquire a price 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 phase-shifted model of the time-domain 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 familiar-by-now technique. The sign now seems to be like this:

[
mathbf{x}_n = cos(frac{2 pi}{N}*4*x – frac{pi}{2})
]

First, we categorical 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 got non-zero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are not similar. As an alternative, 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 examine 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 non-zero 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 element.

Superposition of sinusoids

The sign we assemble should be expressed when it comes to the premise vectors, however it’s not a pure sinusoid. As an alternative, 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 gained’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, nonetheless, there’s fairly a couple of issues we will say:

• Because the sign consists of two pure cosines and one pure sine, there will probably be 4 coefficients with non-zero actual elements, and two with non-zero imaginary elements. 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 nicely: The all-real coefficients will correspond to frequency indices 2, 8, 56, and 62; the all-imaginary 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 examine:

x <- 3 * torch_sin(frequency(4, N) * sample_positions) +
  6 * torch_cos(frequency(2, N) * sample_positions) +
  2 * torch_cos(frequency(8, N) * sample_positions)

plot_ft(x)

Now, how will we calculate the DFT for much less handy alerts?

Coding the DFT

Happily, we already know what must be achieved. We need to venture the sign onto every of the premise vectors. In different phrases, we’ll be computing a bunch of internal merchandise. Logic-wise, nothing modifications: The one distinction is that usually, it won’t be potential to symbolize the sign when it comes to just some foundation vectors, like we did earlier than. Thus, all projections will truly should be calculated. However isn’t automation of tedious duties one factor we’ve got 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 one-dimensional tensor, encoding a sign. The output is a one-dimensional vector of Fourier coefficients, of the identical size because the enter, every holding details about a frequency. The central thought is: To acquire a coefficient, venture the sign onto the corresponding foundation vector.

To implement that concept, we have to create the premise vectors, and for each, compute its internal product with the sign. This may be achieved in a loop. Surprisingly little code is required to perform the purpose:

dft <- perform(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 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”? In reality, a loop will not be even wanted. As an alternative of working with the premise vectors one-by-one, we will 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 (N-1); there will probably 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}
] {#eq-dft-3}

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 <- perform(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 possibly can simply confirm, the consequence is identical.

Thanks for studying!

Photograph by Trac Vu on Unsplash