I have an EEG signal of length 2 sec. sampled at a rate of 200 Hz. If $\vec{x}$ denotes my signal in $\mathbb{R}^{400}$, then I can obtain the DFT by computing $W\vec{x}$, where $W$ denotes the 400-by-400 DFT Matrix. (I use the convention that columns are normalized.) The signal and its "two-sided ?" amplitude spectrum are the top two graphs shown below

Suppose I focus on bin #7 of 400, which corresponds to 3.5 Hz. My understanding is that, from a linear algebra perspective, I can compute the spectral component of my signal at 3.5 Hz by projecting my signal onto columns 7 and 393 (=400-7) of $W$ and adding the two projections together. More specifically,if $\vec{w}_0$ and $\vec{w}_1$ denote columns 7 and 393 of $W$, respectively, then the spectral component of $\vec{x}$ at frequency $f=3.5$ Hz is merely

\begin{equation*} \vec{x}_f=(\vec{x} \bullet \vec{w}_0)\vec{w}_0+(\vec{x} \bullet \vec{w}_1)\vec{w}_1 \end{equation*}

(I tested this principle on some simple linear combinations of sinusoids and it worked--please tell me if I'm completely off base here.) The graph of the signal component $\vec{x}_f$ is the sinusoid shown in the bottom graph below.

Now suppose I had a second EEG signal of same time length and sampling frequency, $\vec{y}$, whose spectral component at $f=3.5$ is denoted as $\vec{y}_f$. It's graph would also be a sinusoid having the same frequency as $\vec{x}_f$, but likely differing in amplitude and phase.

In theory is the coherence at frequency $f$ merely a measure describing the extent to which I could obtain the graph of $\vec{y}_f$ from $\vec{x}_f$ using a linear mapping of sorts?

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1 Answer 1


Nice write-up!

The coherence for two signals $x$ and $y$ is defined as:

$$\texttt{coh}_{xy}(f) = \frac{|P_{xy}(f)|^2}{P_x(f)P_y(f)}$$ where $P_{xy}(f)$ is the Cross Power Spectral Density between $x$ and $y$, and $P_x(f), P_y(f)$ the Auto Power Spectral Densities of $x$ and $y$ respectively. Note both $P_x(f)$ and $P_y(f)$ are real functions.

It indeed provides an estimate measure of linearity between two signals $x$ and $y$ at every frequency, through an input-output lense, and is widely used in audio measurements for example, to rate the linearity of transfer function measurements.

Let's call $X(f)$ and $Y(f)$ the Fourier transforms of input $x$ and output $y$, and consider an ideal linear system $H(f)$, such that: $$Y(f) = H(f)X(f)$$

Then $P_{xy} = H(f)P_x(f)$ and $P_y = |H(f)|^2 P_x(f)$, and $$\texttt{coh}_{xy}(f) = \frac{|H(f)P_x(f)|^2}{P_x(f)|H(f)|^2 P_x(f)} = 1\frac{}{}$$

In plain English, if $x$ and $y$ are ideally linearly related, then the coherence at every frequency $f$ is $1$.

Of course ideal systems are un-realizable and $\texttt{coh}$ is merely an estimate (mainly because of measurement noise, but also less mentioned because of the finite nature of measurements).

In practice, $\texttt{coh}_{xy}(f) \leq 1$

  • $\begingroup$ So, I'm familiar with the equation for $Coh_{xy}(f)$ in terms of power- and cross-power spectral densities, but let me ask about the equation $Y(f)=H(f)X(f)$. Is $H$ what's referred to as a "transfer function?" I don't know really much about those but I know about convolution and just learned that the preceding equation is the same as saying $y=h\ast x$ in the time domain, where $h$ is an impulse response function. So, does the ideal situation correspond to the existence of such an impulse response function $h$? $\endgroup$
    – fishbacp
    Commented Dec 22, 2022 at 19:52
  • 1
    $\begingroup$ Yes, $H$ is the transfer function. To be exact, $H(z)$ ($\mathbb{Z}$-transform of $h$) is the transfer function, and $H(f) = H(z = e^{j2\pi f})$ is the frequency response. The ideal situation corresponds to $H(f)$ being an ideal linear frequency response, which in practice is never the case. For example, a common estimate of $H(f)$ is $H_1 = P_{xy}(f) / P_x(f)$. Another one is $H_2 = P_{y}(f) / P_{xy}(f)$. Notice $\texttt{coh} = H1/H2$. If $H(f)$ was ideal, then of course $H_1 = H_2 = H(f)$ and you get back to $\texttt{coh} = 1$. $\endgroup$
    – Jdip
    Commented Dec 22, 2022 at 21:24
  • $\begingroup$ I should correct myself here: $H_2 = P_y(f)/P_{yx}(f)$ $\endgroup$
    – Jdip
    Commented Dec 22, 2022 at 21:32
  • $\begingroup$ For clarification, when you say "linearly related," is this in the sense of superposition, $y=mx$, as opposed to the sense of affine transformation, $y=mx+b$? Would a coherence of one at a particular frequency then imply that the spectral components of $x$ and $y$ at that frequency are simply multiples of other, i.e. no phase difference? $\endgroup$
    – fishbacp
    Commented Dec 23, 2022 at 18:15
  • 1
    $\begingroup$ As soon as you take the squared magnitude of $P_{xy}$, all phase information is lost. $\endgroup$
    – Jdip
    Commented Dec 23, 2022 at 19:10

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