Observe: This submit 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 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 attainable to me, forged a lightweight 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 much 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 totally fledged article.
Within the sciences, the Fourier Rework is nearly in all places. Said very usually, it converts information from one illustration to a different, with none lack of data (if carried out 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” have to know tips on how to interpret the outcomes. Right here, I wish to assist with that. We begin with an instance operate name, taking part in 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 best way we set issues up, there will likely be sixty-four samples; the sampling interval thus equals N = 64. The content material of frequency(), the under helper operate used to assemble the sign, displays how we characterize 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 shortly verify this did what it was presupposed to:
df <- information.body(x = sample_positions, y = as.numeric(x))
ggplot(df, aes(x = x, y = y)) +
geom_line() +
xlab("time") +
ylab("amplitude") +
theme_minimal()
Now that now we have the enter sign, torch_fft_fft() computes for us the Fourier coefficients, that’s, the significance of the varied frequencies current within the sign. The variety of frequencies thought of will equal the variety of sampling factors: So (X) will likely 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 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 carried out 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 should 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, particularly, that at (ok = 1). This matches (and it higher needed to) the way in which we constructed the sign: particularly, as engaging in a single revolution over the entire sampling interval.
Since, in concept, 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 part, indicating a attainable shift of the sign (a pure cosine is unshifted). In torch, now we have 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 as a result of finite precision in how numbers are offered in a pc, the precise values will typically not be zero. As a substitute, they’ll be very small. If we take one in every of 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 taking place, 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 a part 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,
part(Ft),
"part"
)
p_real + p_imag + p_magnitude + p_phase
It’s honest to say that now we have no purpose to doubt what torch_fft_fft() has carried out. 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, 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 fluctuate broadly on a dimension of math and sciences schooling, my possibilities to satisfy your expectations, expensive reader, should be very near zero. Nonetheless, I wish to take the danger. For those who’re an skilled on this stuff, you’ll anyway be simply scanning the textual content, looking for items of torch code. For those who’re reasonably aware of the DFT, you should still like being reminded of its internal workings. And – most significantly – should you’re somewhat new, and even fully new, to this subject, you’ll hopefully take away (a minimum of) one factor: that what looks like one of many biggest wonders of the universe (assuming there’s a actuality in some way akin to what goes on in our minds) could be a marvel, however neither “magic” nor a factor reserved to the initiated.
In a nutshell, the Fourier Rework is a foundation transformation. Within the case of the DFT – the Discrete Fourier Rework, the place time and frequency representations each are finite vectors, not features – the brand new foundation 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 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 quickly swap to a shorter sampling interval, (N = 4), say. If we accomplish that, now we have 4 foundation vectors: (mathbf{w}^{0n}_N), (mathbf{w}^{1n}_N), (mathbf{w}^{2n}_N), and (mathbf{w}^{3n}_N). The primary one 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 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 closing 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 will 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 method, 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 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 operate by way of the idea vectors (mathbf{w}^{kn}_N = e^{ifrac{2 pi}{N}ok n}), the internal product between the operate and every foundation vector will likely 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 advanced exponentials. In our instance, that is how that goes:
[
begin{aligned}
mathbf{x}_n &= cos(frac{2 pi}{64} n)
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{-ifrac{2 pi}{64} n})
&= frac{1}{2} (e^{ifrac{2 pi}{64} n} + e^{ifrac{2 pi}{64} 63n})
&= frac{1}{2} (mathbf{w}^{1n}_N + mathbf{w}^{63n}_N)
end{aligned}
]
Right here step one instantly outcomes from Euler’s components, and the second displays the truth that the Fourier coefficients are periodic, with frequency -1 being the identical as 63, -2 equaling 62, and so forth.
Now, the (ok)th Fourier coefficient is obtained by projecting the sign onto foundation vector (ok).
As a result of orthogonality of the 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 internal 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), now we have 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 in a position to 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 method. On the finish of the chapter, we’ll develop code that can work for all indicators, however first, let’s see if we will dive even deeper into the workings of the DFT. Three issues we’ll wish to discover:
-
What would occur if frequencies modified – say, a melody have been sung at the next pitch?
-
What about amplitude adjustments – say, the music have been performed twice as loud?
-
What about 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 consequence ourselves.
And eventually, we’ll see how advanced sinusoids, made up of various parts, can nonetheless be analyzed on this method, supplied they are often expressed by way of the frequencies that make up the idea.
Various frequency
Assume we quadrupled the frequency, giving us a sign that appeared like this:
[
mathbf{x}_n = cos(frac{2 pi}{N}*4*n)
]
Following the identical logic as above, we will specific it like so:
[
mathbf{x}_n = frac{1}{2} (mathbf{w}^{4n}_N + mathbf{w}^{60n}_N)
]
We already see that non-zero coefficients will likely be obtained just for frequency indices (4) and (60). Choosing the previous, we acquire
[
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 be certain 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 <- operate(x) p_phase)
plot_ft(x)
This does certainly verify 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, akin to a frequency of 32 revolutions per pattern interval, of double the magnitude (64, thus). Listed below are the sign and its DFT:
x <- torch_cos(frequency(32, N) * sample_positions)
plot_ft(x)
Various amplitude
Now, let’s take into consideration what occurs after we fluctuate amplitude. For instance, say the sign will get twice as loud. Now, there will likely be a multiplier of two that may be taken outdoors the internal product. In consequence, the one factor that adjustments is the magnitude of the coefficients.
Let’s confirm this. The modification 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)
Thus far, now we have not as soon as seen a coefficient with non-zero imaginary half. To alter 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 have been seeing at zero strikes over to (frac{pi}{2}); and if we nonetheless begin “recording” at zero, we should discover a worth of zero there. An equation describing that is the next. For comfort, we assume a sampling interval of (2 pi) and (ok=1), in order that the instance is a straightforward cosine:
[
f(x) = cos(x – phi)
]
The minus signal might look unintuitive at first. However it does make sense: We now wish to acquire 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 adverse 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 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 specific it by way of foundation vectors:
[
begin{aligned}
mathbf{x}_n &= cos(frac{2 pi}{64} 4 n – frac{pi}{2})
&= frac{1}{2} (e^{ifrac{2 pi}{64} 4n – frac{pi}{2}} + e^{ifrac{2 pi}{64} 60n – frac{pi}{2}})
&= frac{1}{2} (e^{ifrac{2 pi}{64} 4n} e^{-i frac{pi}{2}} + e^{ifrac{2 pi}{64} 60n} e^{ifrac{pi}{2}})
&= frac{1}{2} (e^{-i frac{pi}{2}} mathbf{w}^{4n}_N + e^{i frac{pi}{2}} mathbf{w}^{60n}_N)
end{aligned}
]
Once more, now we have non-zero coefficients just for frequencies (4) and (60). However they’re advanced now, and each coefficients are not equivalent. As a substitute, one is the advanced conjugate of the opposite. First, (X_4):
[
begin{aligned}
X_4 &= langle mathbf{w}^{4n}_N, mathbf{x}_n rangle
&=langle mathbf{w}^{4n}_N, frac{1}{2} (e^{-i frac{pi}{2}} mathbf{w}^{4n}_N + e^{i frac{pi}{2}} mathbf{w}^{60n}_N) rangle
&= 32 *e^{-i frac{pi}{2}}
&= -32i
end{aligned}
]
And right here, (X_{60}):
[
begin{aligned}
X_{60} &= langle mathbf{w}^{60n}_N, mathbf{x}_N rangle
&= 32 *e^{i frac{pi}{2}}
&= 32i
end{aligned}
]
As 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 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 be expressed by way of the idea vectors, however it’s not a pure sinusoid. As a substitute, it’s a linear mixture of such:
[
begin{aligned}
mathbf{x}_n &= 3 sin(frac{2 pi}{64} 4n) + 6 cos(frac{2 pi}{64} 2n) +2cos(frac{2 pi}{64} 8n)
end{aligned}
]
I received’t undergo the calculation intimately, however it’s no totally different from the earlier ones. You compute the DFT for every of the three parts, and assemble the outcomes. With none calculation, nonetheless, there’s fairly a number of issues we will say:
- Because the sign consists of two pure cosines and one pure sine, there will likely be 4 coefficients with non-zero actual components, and two with non-zero imaginary components. The latter will likely 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 components obtained for the person sinusoids.
Let’s examine:
Now, how can we calculate the DFT for much less handy indicators?
Coding the DFT
Happily, we already know what needs to be carried out. We wish to challenge the sign onto every of the idea vectors. In different phrases, we’ll be computing a bunch of internal merchandise. Logic-wise, nothing adjustments: The one distinction is that basically, it won’t be attainable to characterize the sign by way of just some foundation vectors, like we did earlier than. Thus, all projections will really must be calculated. However isn’t automation of tedious duties one factor now we have computer systems for?
Let’s begin by stating enter, output, and central logic of the algorithm to be 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, 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 internal product with the sign. This may be carried out 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 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 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 will likely be (N) of them. The columns correspond to positions (0) to (N-1); there will likely 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 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’ll be able to simply confirm, the consequence is similar.
Thanks for studying!
Picture by Trac Vu on Unsplash
