Posit AI Weblog: Discrete Fourier Rework

Be aware: This publish 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 situated 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 approach as was attainable to me, forged a light-weight on what’s behind the magic; it additionally reveals 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, code-it-yourself components.
Collectively, these cowl much more materials than might sensibly match right into a weblog publish; due to this fact, please contemplate what follows extra as a “teaser” than a completely fledged article.

Within the sciences, the Fourier Rework is nearly in all places. Acknowledged very typically, it converts information from one illustration to a different, with none lack of data (if achieved appropriately, that’s.) In the event you use torch, it’s only a operate name away: torch_fft_fft() goes a method, torch_fft_ifft() the opposite. For the consumer, that’s handy – you “simply” have to know easy methods to interpret the outcomes. Right here, I wish 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 best attainable instance sign, a pure cosine that performs one revolution over the entire sampling interval.

Start line: A cosine of frequency 1

The way in which we set issues up, there can be sixty-four 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 symbolize 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 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 rapidly verify this did what it was purported 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 now 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 thought-about will equal the variety of sampling factors: So (X) can be of size sixty-four 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 real-valued sign. In such circumstances, you possibly can 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 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 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 non-zero, 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, particularly, that at (ok = 1). This matches (and it higher needed to) the best way we constructed the sign: particularly, as conducting a single revolution over the entire sampling interval.

Since, in concept, each coefficient might have non-zero actual and imaginary components, typically 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 mirror the respective actual components.

Lastly, there’s the section, indicating a attainable shift of the sign (a pure cosine is unshifted). In torch, we now have torch_angle() complementing torch_abs(), however we have to have in mind roundoff error right here. We all know that in every however a single case, the true and imaginary components are each precisely zero; however as a consequence of finite precision in how numbers are introduced in a pc, the precise values will typically not be zero. As a substitute, they’ll be very small. If we take considered one of these “pretend non-zeroes” and divide it by one other, as occurs within the angle calculation, huge values may result. To stop this from occurring, our customized implementation rounds each inputs earlier than triggering the division.

section <- operate(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 any section 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,
  section(Ft),
  "section"
)

p_real + p_imag + p_magnitude + p_phase

It’s honest to say that we now have no motive to doubt what torch_fft_fft() has achieved. 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 range extensively on a dimension of math and sciences training, my possibilities to satisfy your expectations, pricey reader, have to be very near zero. Nonetheless, I wish to take the danger. In the event you’re an knowledgeable on this stuff, you’ll anyway be simply scanning the textual content, looking for items of torch code. In the event you’re reasonably conversant in the DFT, you should still like being reminded of its inside workings. And – most significantly – if you happen to’re slightly new, and even fully new, to this matter, you’ll hopefully take away (no less than) one factor: that what looks like one of many best wonders of the universe (assuming there’s a actuality by some means similar to what goes on in our minds) might 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 via 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 briefly change to a shorter sampling interval, (N = 4), say. If we achieve this, we now 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 when it comes to their “pace”: 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. Because of this 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.

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 approach, 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 operate when it comes to the premise vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok 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 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 instantly outcomes from Euler’s system, 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 inside product between the operate and the premise vector in query, that’s, by summing over (n). For every (n) ranging between (0) and (N-1), we now 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, trying again at what torch_fft_fft() gave us, we see we had been in a position to arrive on the identical outcome. 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 are able to 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 are able to 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 had been sung at the next pitch?

• What about amplitude modifications – say, the music had 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 outcome ourselves.

And eventually, we’ll see how complicated sinusoids, made up of various elements, can nonetheless be analyzed on this approach, 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 seemed like this:

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

Following the identical logic as above, we are able to 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 can 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 identical outcome.

Now, let’s make sure that 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 seem like so:

[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{32n}_N + mathbf{w}^{32n}_N)
]

Consequently, we find yourself with a single coefficient, similar 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 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 now have not as soon as seen a coefficient with non-zero imaginary half. To vary 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 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 an easy cosine:

[
f(x) = cos(x – phi)
]

The minus signal could look unintuitive at first. Nevertheless it 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 detrimental section shift.

Now, we’re going to calculate the DFT for a shifted model of our instance sign. However if you happen to 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 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 now have non-zero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are now not an identical. 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 regular, 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 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 still be expressed when it comes to 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 elements, and assemble the outcomes. With none calculation, nevertheless, there’s fairly a number of issues we are able to say:

• Because the sign consists of two pure cosines and one pure sine, there can be 4 coefficients with non-zero actual components, and two with non-zero imaginary components. 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 effectively: 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 outcome 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 can we calculate the DFT for much less handy alerts?

Coding the DFT

Fortuitously, we already know what needs to be achieved. We wish to venture the sign onto every of the premise vectors. In different phrases, we’ll be computing a bunch of inside merchandise. Logic-wise, nothing modifications: The one distinction is that basically, it won’t be attainable to symbolize the sign when it comes to 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 we now have 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 concept 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 every one, compute its inside product with the sign. This may be achieved in a loop. Surprisingly little code is required to perform the aim:

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 (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 – if you happen to look again – the outcomes are the identical.

Above, did I say “little code”? Actually, a loop is just not even wanted. As a substitute of working with the premise vectors one-by-one, we are able to 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 (N-1); there can 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}
] {#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 <- operate(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 outcome is identical.

Thanks for studying!

Picture by Trac Vu on Unsplash