knitr::opts_chunk$set( collapse = TRUE, comment = "#>", fig.width = 7 )
In this vignette we show how to detect quadratic phase coupling (QPC) of
one-dimensional or multi-dimensional real-valued time series by bicoherence
or
cross_bicoherence
of rhosa, respectively.
The first section gives an example applying bicoherence
to the data from a
simple model exhibiting QPC at each frequency pair.
In the second section we describe a generative model of three channels with
another kind of QPC revealed by cross_bicoherence
.
The third section summarizes when and why bicoherence
or cross_bicoherence
fails to recognize a certain type of QPC.
We begin with importing rhosa:
library(rhosa)
Our first mathematical model, adapted from [1], is a superposition of six cosine curves of unit amplitude with different frequencies, named $x_1$:
$$x_1(t) = \sum_{i=1}^{6} \cos(\lambda_i t + \varphi_i)$$
where, for each $i = 1,2,...,6$, $\lambda_i$ is given and fixed, namely
$$ \lambda_1 = 0.55; \lambda_2 = 0.75; \lambda_3 = \lambda_1 + \lambda_2; $$ $$ \lambda_4 = 0.6; \lambda_5 = 0.8; \lambda_6 = \lambda_4 + \lambda_5. $$
On the other hand, we choose $\varphi_i$ ($i = 1, ..., 5$) independently from the uniform variable of range $[0, 2\pi)$, and define
$$ \varphi_6 = \varphi_4 + \varphi_5. $$
Note that the trigonometric identities implies
$$\cos(\lambda_6 t + \varphi_6) = \cos((\lambda_4 + \lambda_5) t + (\varphi_4 + \varphi_5)) = \cos(\lambda_4 t + \varphi_4) \cos(\lambda_5 t + \varphi_5) - \sin(\lambda_4 t + \varphi_4) \sin(\lambda_5 t + \varphi_5),$$
so $\cos(\lambda_6 t + \varphi_6)$ is positively correlated with the product of $\cos(\lambda_4 t + \varphi_4)$ and $\cos(\lambda_5 t + \varphi_5)$. But $\cos(\lambda_3 t + \varphi_3)$ is not correlated with the product of $\cos(\lambda_1 t + \varphi_1)$ and $\cos(\lambda_2 t + \varphi_2)$ in general as the phase $\varphi_3$ is randomly assigned.
Once $\varphi_i$s are chosen, $x_1(t)$ is a periodic function of $t$.
So it turns out that $x_1$ is a (strictly) stationary stochastic process.
The wave length of any frequency component of $x_1$ is shorter than $4\pi$.
Now consider sampling a realization of $x_1$ repeatedly during a fixed length of
time, say 256. The sampling rate is 1. That is, the interval of consecutive
samples is 1. The following R code effectively simulates $x_1$ as x1
.
triple_lambda <- function(a, b) c(a, b, a + b) lambda <- c(triple_lambda(0.55, 0.75), triple_lambda(0.6, 0.8)) x1 <- function(k) { set.seed(k) init_phi <- runif(5, min = 0, max = 2*pi) phi <- c(init_phi, init_phi[4] + init_phi[5]) function(t) do.call(sum, Map(function(l, p) cos(l * t + p), lambda, phi)) } observe <- function(f) { sapply(seq_len(256), f) } N1 <- 100 m1 <- do.call(cbind, Map(observe, Map(x1, seq_len(N1))))
Each column of matrix m1
in the last line represents a realization of x1
.
That is, we have taken 100 realizations of $x_1$ as m1
.
Let's plot them:
ith_sample <- function(i) { data.frame(i = i, t = seq_len(256), v = m1[,i]) } r1 <- do.call(rbind, Map(ith_sample, seq_len(100))) library(ggplot2) ggplot(r1) + geom_line(aes(t, v)) + facet_wrap(vars(i)) + theme(axis.title.y = element_blank(), axis.text.y = element_blank(), strip.background = element_blank(), strip.text.x = element_blank())
A realization of $x_1$ looks like:
plot(m1[,1], type = "l", ylim = c(-5, 5), xlab = "t", ylab = "")
Its spectrum estimation shows peaks at six frequencies as expected:
spectrum(m1)
Now we approximate $x_1$'s bicoherence by bicoherence
:
bc1 <- bicoherence(m1)
... and define an R function to plot the heat map of the estimated bicoherence:
heatmap_bicoherence <- function(bc) { ggplot(bc) + geom_raster(aes(f1, f2, fill = value)) + coord_fixed() + scale_alpha(guide = "none") } heatmap_bicoherence(bc1)
Note that the highest peak is at the bifrequency $(f_1, f_2) = (\frac{\lambda_5}{2 \pi}, \frac{\lambda_4}{2 \pi}) \approx (0.127, 0.095)$.
Another example of QPC consists of three channels, which accept series of periodic input signals and suffer from Gaussian noises. From a couple of the channels, say $C_1$ and $C_2$, we observe the summation of input and noises as their output. On the other hand the last channel, called $C_3$, adds $C_1$'s output multiplied by $C_2$'s to its own input and noise.
The following block diagram shows the skeleton of our three-channel model.
Here, assume that $C_1$'s input is a triangle wave of a fixed frequency with varying initial phases. A rectangle wave of another frequency for $C_2$'s input. A cosinusoidal curve of yet another frequency for $C_3$'s input. Running the following code simulates the model.
Fcoef1 <- 1.2 Fcoef2 <- 0.7 Fcoef3 <- 0.8 i1 <- function(x, p) {2 * asin(sin(Fcoef1 * x + p))} i2 <- function(x, p) {ifelse(cos(Fcoef2 * x + p) >= 0, -1, 1)} i3 <- function(x, p) {cos(Fcoef3 * x + p)} Qcoef <- 0.3 tc <- function(k) { set.seed(k) ps <- runif(3, min = 0, max = 2*pi) function(x) { c1 <- i1(x, ps[1]) + rnorm(length(x), mean = 0, sd = 1) c2 <- i2(x, ps[2]) + rnorm(length(x), mean = 0, sd = 1) c3 <- Qcoef * c1 * c2 + i3(x, ps[3]) + rnorm(length(x), mean = 0, sd = 1) data.frame(c1, c2, c3) } } N2 <- 100 sample_tc <- function() { Map(function(f) {f(seq_len(256))}, Map(tc, seq_len(N2))) } c1_data_frame <- function(y) { do.call(cbind, Map(function(k) {y[[k]]$c1}, seq_len(N2))) } c2_data_frame <- function(y) { do.call(cbind, Map(function(k) {y[[k]]$c2}, seq_len(N2))) } c3_data_frame <- function(y) { do.call(cbind, Map(function(k) {y[[k]]$c3}, seq_len(N2))) } y1 <- sample_tc() d1 <- c1_data_frame(y1) d2 <- c2_data_frame(y1) d3 <- c3_data_frame(y1)
That is, we obtain 100 series of data with 256 points. To be specific, $C_1$'s triangle wave has cycle $\frac{2 \pi}{1.2}$, the cycle of $C_2$'s rectangle wave is $\frac{2 \pi}{0.7}$, and the one of $C_3$'s cosinusoidal wave is $\frac{2 \pi}{0.8}$.
$C_1$'s output looks like:
plot(d1[,1], type = "l", ylim = c(-5, 5), xlab = "t", ylab = "")
And $C_2$'s output:
plot(d2[,1], type = "l", ylim = c(-5, 5), xlab = "t", ylab = "")
$C_3$'s output:
plot(d3[,1], type = "l", ylim = c(-5, 5), xlab = "t", ylab = "")
Their spectrum estimation show significant peaks at expected frequencies:
spectrum(d1)
spectrum(d2)
spectrum(d3)
It is expected that $C_3$'s bicoherence shows no outstanding pair of frequencies as $C_3$ does not have any big component of frequency $\frac{1.2}{2 \pi}$ nor $\frac{0.7}{2 \pi}$, but of frequency $\frac{0.8}{2 \pi}$ and $\frac{1.2+0.7}{2 \pi}$ only:
bc3 <- bicoherence(d3) heatmap_bicoherence(bc3)
Now we calculate the estimated cross-bicoherence between $C_1$, $C_2$, and $C_3$
by cross_bicoherence
, and plot the heat map of the results:
cb123 <- cross_bicoherence(d1, d2, d3) heatmap_cross_bicoherence <- function(cb) { ggplot(cb) + geom_raster(aes(f1, f2, fill = value)) + coord_fixed() + scale_alpha(guide = "none") } heatmap_cross_bicoherence(cb123)
Notice the high values around bifrequencies $(f_1, f_2) = (\frac{1.2}{2 \pi}, \frac{0.7}{2 \pi}) \approx (0.191, 0.111)$ and $(f_1, f_2) = (\frac{1.2}{2 \pi}, -\frac{0.7}{2 \pi}) \approx (0.191, -0.111)$. It means that the QPC in the model is correctly identified.
How about changing the order of cross-bicoherence's arguments? Let's estimate the one between $C_3$, $C_1$, and $C_2$:
cb312 <- cross_bicoherence(d3, d1, d2) heatmap_cross_bicoherence(cb312)
Now the pairs of frequencies $(f_1, f_2) = (\frac{1.2-0.7}{2 \pi}, -\frac{1.2}{2 \pi}) \approx (0.080, -0.191)$ and $(f_1, f_2) = (\frac{1.2+0.7}{2 \pi}, -\frac{1.2}{2 \pi}) \approx (0.302, -0.191)$ are in highlights.
Using the three-channel model in the previous section, let's see some examples of QPC which does not exhibit a higher value of the coupling bifrequency in problem.
The most straightforward reason of invisible QPC is the weakness of the
coupling strength. It is reproducible by taking a small Qcoef
in our model.
Qcoef <- 0.01 y2 <- sample_tc() cb2 <- cross_bicoherence(c1_data_frame(y2), c2_data_frame(y2), c3_data_frame(y2)) heatmap_cross_bicoherence(cb2)
Another reason of undetectable QPC is that the coupling bifrequency is in the
outside of the region sufficiently sampled. The following example demonstrates
that the f1
of the bifrequency almost equals to the Nyquist frequency. If
Fcoef1
exceeds $\pi$, it will be no longer identifiable.
Fcoef1 <- pi - 0.1 Fcoef2 <- 2.3 Fcoef3 <- 1.5 Qcoef <- 0.3 y3 <- sample_tc() cb3 <- cross_bicoherence(c1_data_frame(y3), c2_data_frame(y3), c3_data_frame(y3)) heatmap_cross_bicoherence(cb3)
Last but not least, undersampling also ends up with missing QPC. If the number of
samples N2
is too small, we cannot distinguish the coupling bifrequency from
the others.
Fcoef1 <- 1.2 Fcoef2 <- 0.7 Fcoef3 <- 0.8 Qcoef <- 0.3 N2 <- 10 y4 <- sample_tc() cb4 <- cross_bicoherence(c1_data_frame(y4), c2_data_frame(y4), c3_data_frame(y4)) heatmap_cross_bicoherence(cb4)
Using numerical simulation of time series of stationary stochastic processes,
we have demonstrated that rhosa's bicoherence
and cross_bicoherence
can
suggest the bifrequencies of potential QPC. Although some assumptions on the
time series, e.g. stationarity, must be satisfied for accurate estimation,
there is no additional cost except for the computing time if the first-order
spectral analysis has been adopted already. The API lowers the barrier preventing
from applying the bispectral analysis to time series for an exploratory
purpose. Considering appropriate generative models helps interpretation of the
result obtained from the analysis.
[1] Petropulu, A.P., 1994. Higher-Order Spectra in Biomedical Signal Processing. IFAC Proceedings Volumes, IFAC Symposium on Modelling and Control in Biomedical Systems, Galveston, TX, USA, 27-30 March 1994 27, 47–52. https://doi.org/10.1016/S1474-6670%2817%2946158-1
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