Observe: 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 means as was potential to me, forged a lightweight 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; subsequently, please think about what follows extra as a “teaser” than a completely fledged article.
Within the sciences, the Fourier Rework is nearly in all places. Said very typically, it converts knowledge from one illustration to a different, with none lack of data (if carried out accurately, that’s.) Should you use torch, it’s only a perform name away: torch_fft_fft() goes a technique, torch_fft_ifft() the opposite. For the consumer, that’s handy – you “simply” have to know interpret the outcomes. Right here, I need to assist with that. We begin with an instance perform name, enjoying round with its output, after which, attempt to get a grip on what’s going on behind the scenes.
Understanding the output of torch_fft_fft()
As we care about precise understanding, we begin from the 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 characterize the cosine. Specifically:
[
f(x) = cos(frac p_imag) /
(p_magnitude plot_spacer()) /
(p_real k x)
]
Right here (x) values progress over time (or area), and (okay) 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 (okay) to be equal to (1). Like that, we’ll attain the preliminary state once more at place (x = frac
df <- knowledge.body(x = sample_positions, y = as.numeric(x))
p_signal <- ggplot(df, aes(x = x, y = y)) +
geom_line() +
xlab(“time”) +
ylab(“amplitude”) +
theme_minimal()
# within the code, I am utilizing Ft as an alternative of X as a result of not
# all working techniques deal with variables as case-sensitive
Ft <- torch_fft_fft(x)
p_real <- create_plot(
sample_positions,
Ft$actual,
“actual half”
)
p_imag <- create_plot(
sample_positions,
Ft$imag,
“imaginary half”
)
p_magnitude <- create_plot(
sample_positions,
torch_abs(Ft),
“magnitude”
)
p_phase <- create_plot(
sample_positions,
part(Ft),
“part”
)
(p_signal
df <- knowledge.body(x = sample_positions, y = as.numeric(x))
p_signal <- ggplot(df, aes(x = x, y = y)) +
geom_line() +
xlab(“time”) +
ylab(“amplitude”) +
theme_minimal()
# within the code, I am utilizing Ft as an alternative of X as a result of not
# all working techniques deal with variables as case-sensitive
Ft <- torch_fft_fft(x)
p_real <- create_plot(
sample_positions,
Ft$actual,
“actual half”
)
p_imag <- create_plot(
sample_positions,
Ft$imag,
“imaginary half”
)
p_magnitude <- create_plot(
sample_positions,
torch_abs(Ft),
“magnitude”
)
p_phase <- create_plot(
sample_positions,
part(Ft),
“part”
)
(p_signal * 1 * 64).
Let’s rapidly affirm this did what it was presupposed 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-about will equal the variety of sampling factors: So (X) shall 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 instances, you would name torch_fft_rfft() as an alternative, which yields “nicer” (within the sense of shorter) vectors to work with. Right here although, I need to clarify the final case, since that’s what you’ll discover carried out 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 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 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 0At this level we all know that there’s only a single frequency current within the sign, specifically, that at (okay = 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 principle, 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 32Unsurprisingly, these values precisely mirror the respective actual components.
Lastly, there’s the part, 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 take into consideration roundoff error right here. We all know that in every however a single case, the true and imaginary components are each precisely zero; however on account 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 certainly one of these “faux non-zeroes” and divide it by one other, as occurs within the angle calculation, large values may result. To stop this from occurring, our customized implementation rounds each inputs earlier than triggering the division.
part <- perform(Ft, threshold = 1e5) {
torch_atan2(
torch_abs(torch_round(Ft$imag * threshold)),
torch_abs(torch_round(Ft$actual * threshold))
)
}
as.numeric(part(Ft)) %>% spherical(5)[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[29] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[57] 0 0 0 0 0 0 0 0As anticipated, there is no such thing as a part 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,
part(Ft),
"part"
)
p_real + p_imag + p_magnitude + p_phase
It’s truthful to say that we’ve no purpose to doubt what torch_fft_fft() has carried out. However with a pure sinusoid like this, we are able to perceive precisely what’s occurring by computing the DFT ourselves, by hand. Doing this now will considerably assist us later, after we’re writing the code.
Reconstructing the magic
One caveat about this part. With a subject as wealthy because the Fourier Rework, and an viewers who I think about to differ broadly on a dimension of math and sciences schooling, my probabilities to satisfy your expectations, pricey reader, have to be very near zero. Nonetheless, I need to take the chance. Should you’re an professional on these items, you’ll anyway be simply scanning the textual content, looking for items of torch code. Should you’re reasonably aware of the DFT, you should still like being reminded of its inside workings. And – most significantly – should you’re quite new, and even fully new, to this matter, you’ll hopefully take away (at the least) one factor: that what looks as if one of many biggest wonders of the universe (assuming there’s a actuality by some means equivalent 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 features – 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 (okay) operating via the premise vectors, they are often written:
[
mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}k n}
] {#eq-dft-1}
Like (okay), (n) runs from (0) to (N-1). To know what these foundation vectors are doing, it’s useful to quickly 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 at last, the fourth:
[
mathbf{w}^{3n}_N
=
begin{bmatrix}
e^{ifrac{2 pi}{4}* 3 * 0}
e^{ifrac{2 pi}{4}* 3 * 1}
e^{ifrac{2 pi}{4}* 3 * 2}
e^{ifrac{2 pi}{4}* 3 * 3}
end{bmatrix}
=
begin{bmatrix}
1
e^{ifrac{3 pi}{2}}
e^{i 3 pi}
e^{ifrac{9 pi}{2}}
end{bmatrix}
=
begin{bmatrix}
1
-i
-1
i
end{bmatrix}
]
We will characterize these 4 foundation vectors by way of their “pace”: how briskly they transfer across the unit circle. To do that, we merely take a look at the rightmost column vectors, the place the ultimate calculation outcomes seem. The values in that column correspond to positions pointed to by the revolving foundation vector at totally different cut-off dates. Which means 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); yet one more 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 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 characterize this perform by way of the premise vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}okay n}), the inside 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 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 (okay)th Fourier coefficient is obtained by projecting the sign onto foundation vector (okay).
Because of the orthogonality of the premise vectors, solely two coefficients is not going to be zero: these for (mathbf{w}^{1n}_N) and (mathbf{w}^{63n}_N). They’re obtained by computing the inside 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 remaining sum of (32) for each coefficients. For instance, for (mathbf{w}^{1n}_N):
[
begin{aligned}
X_1 &= langle mathbf{w}^{1n}_N, mathbf{x}_n rangle
&= langle mathbf{w}^{1n}_N, frac{1}{2} (mathbf{w}^{1n}_N + mathbf{w}^{63n}_N) rangle
&= frac{1}{2} * 64
&= 32
end{aligned}
]
And analogously for (X_{63}).
Now, trying again at what torch_fft_fft() gave us, we see we have been capable of arrive on the similar outcome. And we’ve realized one thing alongside the way in which.
So long as we stick with alerts composed of a number of foundation vectors, we are able to compute the DFT on this means. On the finish of the chapter, we’ll develop code that can 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 need to discover:
What would occur if frequencies modified – say, a melody have been sung at a better pitch?
What about amplitude modifications – say, the music have 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 outcome ourselves.
And at last, we’ll see how complicated sinusoids, made up of various parts, can nonetheless be analyzed on this means, 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 seemed like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]
Following the identical logic as above, we are able to 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). 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 outcome.
Now, let’s be sure that our evaluation is appropriate. 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 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 to be 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 below are the sign and its DFT:
x <- torch_cos(frequency(32, N) * sample_positions)
plot_ft(x)
Various amplitude
Now, let’s take into consideration what occurs after we differ amplitude. For instance, say the sign will get twice as loud. Now, there shall 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)
Thus far, we’ve not as soon as seen a coefficient with non-zero imaginary half. To vary this, we add in part.
Including part
Altering the part of a sign means shifting it in time. Our instance sign is a cosine, a perform whose worth is 1 at (t=0). (That additionally was the – arbitrarily chosen – 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 (okay=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 need 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 damaging part shift.
Now, we’re going to calculate the DFT for a shifted model of our instance sign. However should you like, take a peek on the phase-shifted model of the time-domain image now already. You’ll see {that a} cosine, delayed by (frac{pi}{2}), is nothing else than a sine beginning at 0.
To compute the DFT, we comply with our familiar-by-now technique. The sign now appears to be like like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*x – frac{pi}{2})
]
First, we categorical 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, we’ve non-zero coefficients just for frequencies (4) and (60). However they’re complicated now, and each coefficients are not similar. As an alternative, one is the complicated conjugate of the opposite. First, (X_4):
[
begin{aligned}
X_4 &= langle mathbf{w}^{4n}_N, mathbf{x}_n rangle
&=langle mathbf{w}^{4n}_N, frac{1}{2} (e^{-i frac{pi}{2}} mathbf{w}^{4n}_N + e^{i frac{pi}{2}} mathbf{w}^{60n}_N) rangle
&= 32 *e^{-i frac{pi}{2}}
&= -32i
end{aligned}
]
And right here, (X_{60}):
[
begin{aligned}
X_{60} &= langle mathbf{w}^{60n}_N, mathbf{x}_N rangle
&= 32 *e^{i frac{pi}{2}}
&= 32i
end{aligned}
]
As typical, we examine our calculation utilizing torch_fft_fft().
x <- torch_cos(frequency(4, N) * sample_positions - pi / 2)
plot_ft(x)
For a pure sine wave, the non-zero Fourier coefficients are imaginary. The part shift within the coefficients, reported as (frac{pi}{2}), displays the time delay we utilized to the sign.
Lastly – earlier than we write some code – let’s put all of it collectively, and take a look at a wave that has greater than a single sinusoidal element.
Superposition of sinusoids
The sign we assemble should still be expressed by way of 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 just a few issues we are able to 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 complicated conjugates of one another.
- From the way in which 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:
Now, how will we calculate the DFT for much less handy alerts?
Coding the DFT
Fortuitously, we already know what must be carried out. We need 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 normally, it is not going to be potential to characterize 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 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 every one, compute its inside product with the sign. This may be carried out 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)$unsqueeze(1)
Ft <- torch_complex(
torch_zeros(n_samples), torch_zeros(n_samples)
)
for (okay in 0:(n_samples - 1)) {
w_k <- torch_exp(-1i * 2 * pi / n_samples * okay * n)
dot <- torch_matmul(w_k, x$to(dtype = torch_cfloat()))
Ft[k + 1] <- dot
}
Ft
}To check the implementation, we are able to 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 – should you look again – the outcomes are the identical.
Above, did I say “little code”? In actual fact, a loop is just not even wanted. As an alternative 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 shall be (N) of them. The columns correspond to positions (0) to (N-1); there shall 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 to be like much more elegant:
dft_vec <- perform(x) {
n_samples <- size(x)
n <- torch_arange(0, n_samples - 1)$unsqueeze(1)
okay <- torch_arange(0, n_samples - 1)$unsqueeze(2)
mat_k_m <- torch_exp(-1i * 2 * pi / n_samples * okay * n)
torch_matmul(mat_k_m, x$to(dtype = torch_cfloat()))
}As you possibly can simply confirm, the outcome is identical.
Thanks for studying!