Observe: This publish is an excerpt from the forthcoming ebook, Deep Studying and Scientific Computing with R torch. The chapter in query is on the Discrete Fourier Rework (DFT), and is situated 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 method as was doable 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 components.
Collectively, these cowl much more materials than may sensibly match right into a weblog publish; due to this fact, please take into account what follows extra as a “teaser” than a totally fledged article.
Within the sciences, the Fourier Rework is nearly in all places. Acknowledged very usually, it converts knowledge from one illustration to a different, with none lack of info (if achieved appropriately, that’s.) For those who use torch, it’s only a operate name away: torch_fft_fft() goes a technique, torch_fft_ifft() the opposite. For the consumer, that’s handy – you “simply” must know 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 doable instance sign, a pure cosine that performs one revolution over the entire 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 beneath helper operate 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 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 <- 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 have now 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) might be of size sixty-four as properly.
(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 wish to clarify the final 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 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 best way we constructed the sign: specifically, as engaging in a single revolution over the entire sampling interval.
Since, in principle, each coefficient may have non-zero actual and imaginary components, usually 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 replicate the respective actual components.
Lastly, there’s the part, indicating a doable shift of the sign (a pure cosine is unshifted). In torch, we have now 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 components are each precisely zero; however because of finite precision in how numbers are offered in a pc, the precise values will usually 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, large values may end up. To forestall this from occurring, our customized implementation rounds each inputs earlier than triggering the division.
part <- operate(Ft, threshold = 1e5) {
torch_atan2(
torch_abs(torch_round(Ft$imag * threshold)),
torch_abs(torch_round(Ft$actual * threshold))
)
}
as.numeric(part(Ft)) %>% spherical(5)[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 0As anticipated, there isn’t any part shift within the sign.
Let’s visualize what we discovered.
create_plot <- operate(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,
part(Ft),
"part"
)
p_real + p_imag + p_magnitude + p_phase
It’s truthful to say that we have now no cause 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, 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 broadly on a dimension of math and sciences training, my possibilities to satisfy your expectations, expensive reader, should be very near zero. Nonetheless, I wish to take the chance. For those who’re an professional on these items, you’ll anyway be simply scanning the textual content, searching for items of torch code. For those who’re reasonably acquainted with the DFT, you should still like being reminded of its inside workings. And – most significantly – in the event you’re reasonably new, and even fully new, to this matter, you’ll hopefully take away (a minimum of) one factor: that what looks like one of many best wonders of the universe (assuming there’s a actuality by some means equivalent to what goes on in our minds) could be a surprise, however neither “magic” nor a factor reserved to the initiated.
In a nutshell, the Fourier Rework is a foundation transformation. Within the case of the DFT – the Discrete Fourier Rework, the place time and frequency representations each are finite vectors, not features – the brand new foundation seems 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 way of the idea 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 know 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 have now 4 foundation vectors: (mathbf{w}^{0n}_N), (mathbf{w}^{1n}_N), (mathbf{w}^{2n}_N), and (mathbf{w}^{3n}_N). The primary one seems 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 time limits. Which means that taking a look at a single “replace of place”, we will see how briskly the vector is transferring in a single time step.
Wanting first at (mathbf{w}^{0n}_N), we see that it doesn’t transfer in any respect. (mathbf{w}^{1n}_N) goes from (1) to (i) to (-1) to (-i); another step, and it will be again the place it began. That’s one revolution in 4 steps, or a step dimension of (frac{pi}{2}). Then (mathbf{w}^{2n}_N) goes at double that tempo, transferring a distance of (pi) alongside the circle. That method, it finally ends up finishing two revolutions total. Lastly, (mathbf{w}^{3n}_N) achieves three full loops, for a step dimension of (frac{3 pi}{2}).
The factor that makes these foundation vectors so helpful is that they’re mutually orthogonal. That’s, their dot product is zero:
[
langle mathbf{w}^{kn}_N, mathbf{w}^{ln}_N rangle = sum_{n=0}^{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 signify this operate when it comes to the idea vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok n}), the inside product between the operate and every foundation vector might be both zero (the “default”) or a a number of of 1 (in case the operate has a element matching the idea vector in query). Fortunately, sines and cosines can simply be transformed into complicated exponentials. In our instance, that is how that goes:
[
begin{aligned}
mathbf{x}_n &= cos(frac{2 pi}{64} n)
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{-ifrac{2 pi}{64} n})
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{ifrac{2 pi}{64} 63n})
&= frac{1}{2} (mathbf{w}^{1n}_N + mathbf{w}^{63n}_N)
end{aligned}
]
Right here step one straight 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).
Because of the orthogonality of the idea vectors, solely two coefficients won’t be zero: these for (mathbf{w}^{1n}_N) and (mathbf{w}^{63n}_N). They’re obtained by computing the inside product between the operate and the idea vector in query, that’s, by summing over (n). For every (n) ranging between (0) and (N-1), we have now 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 in a position to arrive on the similar end result. And we’ve realized one thing alongside the best way.
So long as we stick with alerts composed of a number of foundation vectors, we will compute the DFT on this 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 wish to discover:
-
What would occur if frequencies modified – say, a melody had been sung at a better 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 end result ourselves.
And eventually, we’ll see how complicated sinusoids, made up of various parts, can nonetheless be analyzed on this method, supplied they are often expressed when it comes to the frequencies that make up the idea.
Various frequency
Assume we quadrupled the frequency, giving us a sign that seemed like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]
Following the identical logic as above, we will 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 might be obtained just for frequency indices (4) and (60). Selecting 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 end result.
Now, let’s make certain our evaluation is appropriate. 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 <- operate(x) p_imag) /
(p_magnitude
plot_ft(x)
This does certainly verify our calculations.
A particular case arises when sign frequency rises to the very best one “allowed”, within the sense of being detectable with out aliasing. That would be the case at one half of the variety of sampling factors. Then, the sign will appear like so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{32n}_N + mathbf{w}^{32n}_N)
]
Consequently, we find yourself with a single coefficient, equivalent to a frequency of 32 revolutions per pattern interval, of double the magnitude (64, thus). Listed here are the sign and its DFT:
x <- torch_cos(frequency(32, N) * sample_positions)
plot_ft(x)
Various amplitude
Now, let’s take into consideration what occurs once we range 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 inside 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)
To date, we have now 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 operate 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 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. But it surely does make sense: We now wish to receive a worth of 1 at (x=frac{pi}{2}), so (x – phi) ought to consider to zero. (Or to any a number of of (pi).) Summing up, a delay in time will seem as a unfavorable part 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 seems 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 have now non-zero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are now not similar. 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 normal, 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 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 idea 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 just a few issues we will say:
- Because the sign consists of two pure cosines and one pure sine, there might be 4 coefficients with non-zero actual components, and two with non-zero imaginary components. The latter might be complicated conjugates of one another.
- From the best way the sign is written, it’s straightforward to find the respective frequencies, as properly: 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 end result from multiplying with (frac{64}{2}) the scaling elements obtained for the person sinusoids.
Let’s examine:
Now, how will we calculate the DFT for much less handy alerts?
Coding the DFT
Happily, we already know what needs to be achieved. We wish to challenge the sign onto every of the idea vectors. In different phrases, we’ll be computing a bunch of inside merchandise. Logic-wise, nothing modifications: The one distinction is that on the whole, it won’t be doable to signify the sign when it comes to just some 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 have now 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, challenge the sign onto the corresponding foundation vector.
To implement that concept, we have to create the idea 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 objective:
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 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 0Reassuringly – in the event you look again – the outcomes are the identical.
Above, did I say “little code”? The truth is, a loop isn’t even wanted. As a substitute of working with the idea 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 properly. 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 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 end result is identical.
Thanks for studying!
Picture by Trac Vu on Unsplash
