Word: This submit is an excerpt from the forthcoming guide, 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 means as was potential to me, solid a light-weight on what’s behind the magic; it additionally exhibits 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, code-it-yourself components.
Collectively, these cowl way more materials than may sensibly match right into a weblog submit; due to this fact, please take into account what follows extra as a “teaser” than a completely fledged article.
Within the sciences, the Fourier Rework is nearly in every single place. Said very typically, it converts knowledge from one illustration to a different, with none lack of info (if executed accurately, that’s.) If you happen to 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” have to know the way 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 only potential instance sign, a pure cosine that performs one revolution over the entire sampling interval.
Start line: A cosine of frequency 1
The best way we set issues up, there shall 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 signify the cosine. Specifically:
[
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 rapidly affirm this did what it was speculated to:
df <- knowledge.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) shall 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 may 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 executed 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 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 32Solely 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 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 0At this level we all know that there’s only a single frequency current within the sign, specifically, that at (ok = 1). This matches (and it higher needed to) the way in which we constructed the sign: specifically, as conducting a single revolution over the entire sampling interval.
Since, in idea, each coefficient may 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 32Unsurprisingly, these values precisely mirror the respective actual components.
Lastly, there’s the section, indicating a potential shift of the sign (a pure cosine is unshifted). In torch, we’ve 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 an alternative, 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 end up. To forestall 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 0As anticipated, there is no such thing as a section shift within the sign.
Let’s visualize what we discovered.
create_plot <- perform(x, y, amount) {
df <- knowledge.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 truthful to say that we’ve no cause to doubt what torch_fft_fft() has executed. 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, should be very near zero. Nonetheless, I wish to take the danger. If you happen to’re an professional on this stuff, you’ll anyway be simply scanning the textual content, looking for items of torch code. If you happen to’re reasonably aware of the DFT, you should still like being reminded of its interior workings. And – most significantly – in the event you’re moderately new, and even fully new, to this subject, you’ll hopefully take away (at the very least) one factor: that what looks like one of many best wonders of the universe (assuming there’s a actuality in some way similar to what goes on in our minds) might be a surprise, however neither “magic” nor a factor reserved to the initiated.
In a nutshell, the Fourier Rework is a foundation transformation. Within the case of the DFT – the Discrete Fourier Rework, the place time and frequency representations each are finite vectors, not capabilities – the brand new foundation appears to be like 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) working 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 briefly change 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 appears to be like like this:
[
mathbf{w}^{0n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 0 * 0}
e^{ifrac{2 pi}{4}* 0 * 1}
e^{ifrac{2 pi}{4}* 0 * 2}
e^{ifrac{2 pi}{4}* 0 * 3}
end{bmatrix}
=
begin{bmatrix}
1
1
1
1
end{bmatrix}
]
The second, like so:
[
mathbf{w}^{1n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 1 * 0}
e^{ifrac{2 pi}{4}* 1 * 1}
e^{ifrac{2 pi}{4}* 1 * 2}
e^{ifrac{2 pi}{4}* 1 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ifrac{pi}{2}}
e^{i pi}
e^{ifrac{3 pi}{4}}
end{bmatrix}
=
begin{bmatrix}
1
i
-1
-i
end{bmatrix}
]
That is the third:
[
mathbf{w}^{2n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 2 * 0}
e^{ifrac{2 pi}{4}* 2 * 1}
e^{ifrac{2 pi}{4}* 2 * 2}
e^{ifrac{2 pi}{4}* 2 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ipi}
e^{i 2 pi}
e^{ifrac{3 pi}{2}}
end{bmatrix}
=
begin{bmatrix}
1
-1
1
-1
end{bmatrix}
]
And eventually, the fourth:
[
mathbf{w}^{3n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 3 * 0}
e^{ifrac{2 pi}{4}* 3 * 1}
e^{ifrac{2 pi}{4}* 3 * 2}
e^{ifrac{2 pi}{4}* 3 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ifrac{3 pi}{2}}
e^{i 3 pi}
e^{ifrac{9 pi}{2}}
end{bmatrix}
=
begin{bmatrix}
1
-i
-1
i
end{bmatrix}
]
We will characterize these 4 foundation vectors when it comes to their “velocity”: 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 totally different cut-off dates. Because of this a single “replace of place”, we will see how briskly the vector is shifting 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 another 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, shifting a distance of (pi) alongside the circle. That means, it finally ends up finishing two revolutions total. Lastly, (mathbf{w}^{3n}_N) achieves three full loops, for a step 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 way in which 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 perform when it comes to the premise vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok n}), the interior product between the perform and every foundation vector shall 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 immediately 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).
Because of the orthogonality of the premise vectors, solely two coefficients is not going to be zero: these for (mathbf{w}^{1n}_N) and (mathbf{w}^{63n}_N). They’re obtained by computing the interior 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 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 had been capable of arrive on the similar consequence. And we’ve realized one thing alongside the way in which.
So long as we stick with indicators composed of a number of foundation vectors, we will compute the DFT on this means. 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 had been sung at a better pitch?
-
What about amplitude adjustments – 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 consequence ourselves.
And eventually, we’ll see how advanced sinusoids, made up of various parts, can nonetheless be analyzed on this means, supplied 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 appeared 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 shall be obtained just for frequency indices (4) and (60). Choosing the previous, we receive
[
begin{aligned}
X_4 &= langle mathbf{w}^{4n}_N, mathbf{x}_n rangle
&= langle mathbf{w}^{4n}_N, frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N) rangle
&= 32
end{aligned}
]
For the latter, we’d arrive on the similar consequence.
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 <- 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 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 after we range amplitude. For instance, say the sign will get twice as loud. Now, there shall be a multiplier of two that may be taken exterior the interior product. In consequence, the one factor that adjustments is the magnitude of the coefficients.
Let’s confirm this. The modification relies on the instance we had earlier than the final one, with 4 revolutions over the sampling interval:
x <- 2 * torch_cos(frequency(4, N) * sample_positions)
plot_ft(x)
To date, we’ve 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 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 (ok=1), in order that the instance is an easy cosine:
[
f(x) = cos(x – phi)
]
The minus signal could look unintuitive at first. However it does make sense: We now wish to receive 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 in the event 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 to be like 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 non-zero coefficients just for frequencies (4) and (60). However they’re advanced now, and each coefficients are not similar. As an alternative, 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 standard, 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 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 totally different from the earlier ones. You compute the DFT for every of the three parts, and assemble the outcomes. With none calculation, nevertheless, there’s fairly a couple of issues we will say:
- For the reason that sign consists of two pure cosines and one pure sine, there shall be 4 coefficients with non-zero actual components, and two with non-zero imaginary components. The latter shall be advanced conjugates of one another.
- From the way in which the sign is written, it’s straightforward 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:
Now, how can we calculate the DFT for much less handy indicators?
Coding the DFT
Thankfully, we already know what must be executed. We wish to venture the sign onto every of the premise vectors. In different phrases, we’ll be computing a bunch of interior merchandise. Logic-wise, nothing adjustments: The one distinction is that on the whole, it is not going to be potential to signify the sign when it comes to only a few 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 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 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 each, compute its interior product with the sign. This may be executed 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)$unsqueeze(1)
Ft <- torch_complex(
torch_zeros(n_samples), torch_zeros(n_samples)
)
for (ok in 0:(n_samples - 1)) {
w_k <- torch_exp(-1i * 2 * pi / n_samples * ok * n)
dot <- torch_matmul(w_k, x$to(dtype = torch_cfloat()))
Ft[k + 1] <- dot
}
Ft
}To check the implementation, we will take the final sign we analysed, and 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 0Reassuringly – in the event you look again – the outcomes are the identical.
Above, did I say “little code”? In truth, a loop shouldn’t 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 shall be (N) of them. The columns correspond to positions (0) to (N-1); there shall 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 to be like 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’ll be able to simply confirm, the consequence is identical.
Thanks for studying!
Picture by Trac Vu on Unsplash