Notice: This put up 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 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 manner as was potential to me, solid a light-weight 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 Rework, or FFT), once more with each conceptual/explanatory in addition to sensible, code-it-yourself elements.
Collectively, these cowl way more materials than might sensibly match right into a weblog put up; due to this fact, please contemplate what follows extra as a “teaser” than a totally fledged article.

Within the sciences, the Fourier Rework is nearly in all places. Said very usually, it converts information from one illustration to a different, with none lack of info (if accomplished appropriately, that’s.) In the event you use torch, it’s only a perform name away: torch_fft_fft() goes a method, torch_fft_ifft() the opposite. For the consumer, that’s handy – you “simply” must know find out how to interpret the outcomes. Right here, I wish 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 best way we set issues up, there might be sixty-four samples; the sampling interval thus equals N = 64. The content material of frequency(), the under helper perform used to assemble the sign, displays how we symbolize the cosine. Particularly:

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

Right here (x) values progress over time (or area), and (ok) is the frequency index. A cosine is periodic with interval (2 pi); so if we wish it to first return to its beginning state after sixty-four samples, and (x) runs between zero and sixty-three, 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 affirm this did what it was alleged 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 now we have 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-about will equal the variety of sampling factors: So (X) might 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 circumstances, you would name torch_fft_rfft() as a substitute, which yields “nicer” (within the sense of shorter) vectors to work with. Right here although, I wish to clarify the final case, since that’s what you’ll discover accomplished in most expositions on the subject.)

Even with the sign being actual, the Fourier coefficients are advanced 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 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, particularly, that at (ok = 1). This matches (and it higher needed to) the best way we constructed the sign: particularly, as undertaking a single revolution over the whole sampling interval.

Since, in idea, each coefficient might have non-zero 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 section, indicating a potential 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 elements are each precisely zero; however attributable to 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 all these “pretend non-zeroes” and divide it by one other, as occurs within the angle calculation, massive values may result. To stop this from taking place, our customized implementation rounds each inputs earlier than triggering the division.

section <- perform(Ft, threshold = 1e5) {
  torch_atan2(
    torch_abs(torch_round(Ft$imag * threshold)),
    torch_abs(torch_round(Ft$actual * threshold))
  )
}

as.numeric(section(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 section 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,
  section(Ft),
  "section"
)

p_real + p_imag + p_magnitude + p_phase

It’s honest to say that now we have no motive to doubt what torch_fft_fft() has accomplished. 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, 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 fluctuate broadly on a dimension of math and sciences schooling, my probabilities to fulfill your expectations, pricey reader, have to be very near zero. Nonetheless, I wish to take the danger. In the event you’re an professional on these items, you’ll anyway be simply scanning the textual content, looking for items of torch code. In the event you’re reasonably accustomed to the DFT, you should still like being reminded of its internal workings. And – most significantly – when you’re somewhat new, and even utterly new, to this subject, you’ll hopefully take away (not less than) one factor: that what looks like one of many best wonders of the universe (assuming there’s a actuality by some means akin to what goes on in our minds) could be a marvel, 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}^{(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 (ok) operating by the premise vectors, they are often written:

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

Like (ok), (n) runs from (0) to (N-1). To grasp what these foundation vectors are doing, it’s useful to quickly change 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 are able to characterize these 4 foundation vectors by way of their “velocity”: how briskly they transfer across the unit circle. To do that, we merely have 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 totally different closing dates. 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); another step, and it could 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 manner, it finally ends up finishing two revolutions general. 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 by way of the premise vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok n}), the internal product between the perform and every foundation vector might 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 advanced 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 instantly 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 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), now we have a contribution of (frac{1}{2}), leaving us with a last 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 have been capable of arrive on the similar consequence. And we’ve realized one thing alongside the best way.

So long as we stick with indicators composed of a number of foundation vectors, we will compute the DFT on this manner. On the finish of the chapter, we’ll develop code that can work for all indicators, however first, let’s see if we will dive even deeper into the workings of the DFT. Three issues we’ll wish to discover:

• What would occur if frequencies modified – say, a melody have been sung at the next pitch?

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

• What about section – e.g., there have been an offset earlier than the piece began?

In all circumstances, we’ll name torch_fft_fft() solely as soon as we’ve decided the consequence ourselves.

And eventually, we’ll see how advanced sinusoids, made up of various parts, can nonetheless be analyzed on this manner, offered they are often expressed by way of the frequencies that make up the premise.

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 will specific it like so:

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

We already see that non-zero coefficients might be obtained just for frequency indices (4) and (60). Selecting the previous, we get hold of

[
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 sure that our evaluation is appropriate. The next code snippet accommodates 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)  p_imag) /
    (p_magnitude 

plot_ft(x)

This does certainly affirm 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, akin 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 fluctuate amplitude. For instance, say the sign will get twice as loud. Now, there might be a multiplier of two that may be taken outdoors the internal product. In consequence, the one factor that adjustments 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)

To date, now we have not as soon as seen a coefficient with non-zero imaginary half. To alter this, we add in section.

Including section

Altering the section 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 – place to begin 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 (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 wish to get hold of 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 destructive section shift.

Now, we’re going to calculate the DFT for a shifted model of our instance sign. However when 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 observe our familiar-by-now technique. The sign now appears like this:

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

First, we specific 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 non-zero coefficients just for frequencies (4) and (60). However they’re advanced now, and each coefficients are now not an identical. As a substitute, one is the advanced 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 normal, we verify 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 section 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 have a look at a wave that has greater than a single sinusoidal element.

Superposition of sinusoids

The sign we assemble should be expressed by way of the premise 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 totally 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 number of issues we will say:

• For the reason that sign consists of two pure cosines and one pure sine, there might be 4 coefficients with non-zero actual elements, and two with non-zero imaginary elements. The latter might be advanced 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 elements obtained for the person sinusoids.

Let’s verify:

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 indicators?

Coding the DFT

Luckily, we already know what must be accomplished. We wish to undertaking the sign onto every of the premise vectors. In different phrases, we’ll be computing a bunch of internal merchandise. Logic-wise, nothing adjustments: The one distinction is that on the whole, it won’t be potential to symbolize the sign by way of 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 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 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, undertaking the sign onto the corresponding foundation vector.

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

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 (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 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 – when you look again – the outcomes are the identical.

Above, did I say “little code”? In truth, a loop just isn’t even wanted. As a substitute 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 might be (N) of them. The columns correspond to positions (0) to (N-1); there might 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 appears much more elegant:

dft_vec <- perform(x) {
  n_samples <- size(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 consequence is identical.

Thanks for studying!

Picture by Trac Vu on Unsplash