EE DSP problem regarding Hilbert transform, analytic signals, and instantaneous frequency

Question

Sorry to import another mathematical problem from the Digital Signal Processing domain. I will try to pose the problem using notation mostly conventional for general mathematicians, not just electrical engineers. This is a discrete-time problem (but continuous-frequency).

So we have a discrete-time sequence of real numbers:

$$ x_n \in \mathbb{R} \qquad \qquad n \in \mathbb{Z} $$

The Discrete-Time Fourier Transform (or "DTFT") is

$$X(e^{i\omega}) = \mathscr{F} \Big\{ x_n \Big\} = \sum\limits_{n=-\infty}^{\infty} x_n \, e^{-i \omega n} \qquad \qquad \omega \in \mathbb{R}$$

It's essentially just the Z-transform with $z = e^{i\omega}$. Because $x_n$ is discrete time,

$$X(e^{i(\omega+ 2\pi)} ) = X(e^{i\omega}) \qquad \qquad \forall \omega \in \mathbb{R}$$

and because $x_n$ is real,

$$X(e^{i(-\omega)}) = \overline{X(e^{i\omega})}$$

where $\overline{z}$ is the complex conjugate of $z$. So because of the periodicity in $\omega$, we will generally only worry about $-\pi < \omega \le \pi$. And the description of $X(e^{i\omega})$ for $0 \le \omega \le \pi$ is sufficient to describe the real $x_n$.

The DTFT maps functions of discrete time (or sequences) to functions of periodic continuous frequency. The inverse DTFT maps it back.

Now, the Discrete-time Hilbert Transform maps a function of discrete time to another function of discrete time:

$$\begin{align} \hat{x}_n &= \mathscr{H} \Big\{ x_n \Big\} \\ \\ &= \sum\limits_{m=-\infty}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ x_{n-m} \\ \\ &= \sum\limits_{m=-\infty}^{-1} \frac{1-(-1)^m}{\pi \ m} \ x_{n-m} + \sum\limits_{m=1}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ x_{n-m} \\ \\ &= \sum\limits_{m=1}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ \big( x_{n-m} - \ x_{n+m} \big) \\ \end{align}$$

So we're resolving the $\frac{1-(-1)^m}{\pi \ m} \bigg|_{m=0}$ term to be zero for $m=0$. There are good reasons for this including that all other even $m$ terms in the summation are zero. So we could also say:

$$\begin{align} \hat{x}_n &= \mathscr{H}\Big\{ x_n \Big\} \\ \\ &= \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \\ \end{align}$$

I dunno if that's helpful or not.

Now the DTFT of the Hilbert transform of $x_n$ turns out to be:

$$\begin{align} \hat{X}(e^{i\omega}) &= \mathscr{F} \Big\{ \hat{x}_n \Big\} \\ \\ &= \sum\limits_{n=-\infty}^{\infty} \hat{x}_n \, e^{-i \omega n} \\ \\ &= -i \operatorname{sgn}(\omega) X(e^{i\omega}) \\ \\ &= \begin{cases} i X(e^{i\omega}) \qquad & -\pi < \omega < 0 \\ 0 \qquad & \omega = 0 \\ -i X(e^{i\omega}) \qquad & 0 < \omega < \pi \\ \end{cases} \\ \end{align}$$

The Hilbert transform essentially kills DC and delays every sinusoid by 90°, preserving their amplitude:

$$ \mathscr{H} \Big\{A \cos(\Omega n + \theta) \Big\} = A \sin(\Omega n + \theta) $$

Now there is one last preliminary definition. Given a real, discrete-time signal $x_n$, the complex Analytic signal is:

$$ {}^\mathrm{a}x_n = x_n + i\, \hat{x}_n $$

And the DTFT of the complex analytic signal is

$$\begin{align} {}^\mathrm{a}X(e^{i\omega}) = \mathscr{F} \Big\{ {}^\mathrm{a}x_n \Big\} &= \big(1 + \operatorname{sgn}(\omega) \big) X(e^{i\omega}) \\ \\ &= \begin{cases} 0 \qquad \qquad & -\pi < \omega < 0 \\ X(e^{i\omega}) \qquad \qquad & \omega = 0 \\ 2 X(e^{i\omega}) \qquad \qquad & 0 < \omega < \pi \\ \end{cases} \end{align}$$

These are all the preliminary definitions and relationships.

---

So the specific question I have to ask stems from this question and solution in the DSP Stack Exchange regarding deriving the Instantaneous frequency of an arbitrary real discrete-time signal $x_n$.

We know that when $x_n$ is a general sinusoid (having $0<\Omega<\pi$):

$$\begin{align} x_n &= A \cos(\Omega n + \theta) \\ \\ &= \frac{A}{2}\big( e^{i(\Omega n + \theta)} + e^{-i(\Omega n + \theta)} \big) \\ \end{align}$$

that the Hilbert transform will be:

$$\begin{align} \hat{x}_n &= A \sin(\Omega n + \theta) \\ \\ &= \frac{A}{2i}\big( e^{i(\Omega n + \theta)} - e^{-i(\Omega n + \theta)} \big) \\ \end{align}$$

and the analytic signal will be:

$$ {}^\mathrm{a}x_n = A \, e^{i(\Omega n + \theta)} $$

The instantaneous phase is defined to be the angle ($\arg\{\cdot\}$) of the complex analytic function:

$$ \phi_n = \arg \big\{ {}^\mathrm{a}x_n \big\}$$

which for the general sinusoid above is $ \phi_n = \Omega n + \theta$.

And the instantaneous frequency (a function of discrete-time $n$) is defined to be the rate of increment of the properly unwrapped phase:

$$\begin{align} \Omega_n &= \phi_n - \phi_{n-1} \\ \\ &= \arg\big\{ {}^\mathrm{a}x_n \big\} - \arg\big\{ {}^\mathrm{a}x_{n-1} \big\} \\ \\ &= \arg\left\{ \frac{ {}^\mathrm{a}x_n }{ {}^\mathrm{a}x_{n-1} } \right\} \\ \\ &= \arg\left\{ \frac{ {}^\mathrm{a}x_n \, \overline{{}^\mathrm{a}x_{n-1}} }{ {}^\mathrm{a}x_{n-1} \, \overline{{}^\mathrm{a}x_{n-1}}} \right\} \\ \\ &= \arg\left\{ \frac{ {}^\mathrm{a}x_n \, \overline{{}^\mathrm{a}x_{n-1}} }{ \big| {}^\mathrm{a}x_{n-1} \big|^2 } \right\} \\ \\ &= \arg\left\{ {}^\mathrm{a}x_n \, \overline{{}^\mathrm{a}x_{n-1}} \right\} \\ \\ &= \arg\left\{ \big(x_n + i\, \hat{x}_n\big) \big(x_{n-1} - i\, \hat{x}_{n-1}\big) \right\} \\ \\ &= \arg\left\{ x_n x_{n-1} + \hat{x}_n \hat{x}_{n-1} + i\big(\hat{x}_n x_{n-1} - x_n \hat{x}_{n-1}\big) \right\} \\ \end{align} $$

Now we really expect $0 < \Omega_n < \pi$ to be positive because there are no frequency components in the analytic signal ${}^\mathrm{a}x_n$ that are negative frequency (at least no components of ${}^\mathrm{a}X(e^{i\omega})=\mathscr{F} \big\{ {}^\mathrm{a}x_n \big\}$ in the interval $-\pi < \omega < 0$).

If we can know that $0 < \Omega_n < \pi$, then we need only the second case (quadrants I and II) for the $\arg\{\cdot\}$ operator. For both $u, v \in \mathbb{R}$,

$$ \arg\big\{ w \big\} = \arg \big\{ u \, + \, i \, v \big\} = \begin{cases} \arctan\left(\frac{v}{u}\right) &\text{if } u > 0, \\ \frac{\pi}{2} - \arctan\left(\frac{u}{v}\right) &\text{if } v > 0, \\ -\frac{\pi}{2} - \arctan\left(\frac{u}{v}\right) &\text{if } v < 0, \\ \arctan\left(\frac{v}{u}\right) \pm \pi &\text{if } u < 0, \\ \text{undefined} &\text{if } u = 0 \text{ and } v = 0 \end{cases} $$

Then, if we know that $v = \hat{x}_n x_{n-1} - x_n \hat{x}_{n-1} > 0$, then

$$ \boxed{ \Omega_n = \frac{\pi}{2} - \arctan\left( \frac{x_n x_{n-1} + \hat{x}_n \hat{x}_{n-1}}{\hat{x}_n x_{n-1} - x_n \hat{x}_{n-1}} \right) } $$

and we know that $0 < \Omega_n < \pi$. (!)

But do we know that?

So given any real, discrete-time sequence $x_n$ (not entirely zero nor any constant function and not necessarily a single sinusoid) and its Hilbert transform

$$\begin{align} \hat{x}_n &= \mathscr{H} \Big\{ x_n \Big\} \\ \\ &= \sum\limits_{m=-\infty}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ x_{n-m} \\ \\ &= \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \\ \end{align}$$

How can we show that $\hat{x}_n x_{n-1} - x_n \hat{x}_{n-1} > 0$? Or that

$$\begin{align} x_n \hat{x}_{n-1} &< \hat{x}_n x_{n-1} \qquad \qquad \forall n \in \mathbb{Z} \\ \\ x_n \sum\limits_{m=-\infty}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ x_{n-1-m} &< x_{n-1} \sum\limits_{m=-\infty}^{\infty} \frac{1-(-1)^m}{\pi \ m} \ x_{n-m} \\ \\ x_n \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-2} - \ x_{n+2m} \big) &< x_{n-1} \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \\ \\ x_n \sum\limits_{m=0}^{\infty} \frac{1}{2m+1} \ \big( x_{n-2m-2} - \ x_{n+2m} \big) &< x_{n-1} \sum\limits_{m=0}^{\infty} \frac{1}{2m+1} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \\ \\ \sum\limits_{m=0}^{\infty} \frac{x_n x_{n-2m-2} - \ x_n x_{n+2m}}{2m+1} &< \sum\limits_{m=0}^{\infty} \frac{x_{n-1} x_{n-2m-1} - \ x_{n-1} x_{n+2m+1}}{2m+1} \qquad \forall n \in \mathbb{Z} \\ \end{align}$$

or $$0 < \sum\limits_{m=0}^{\infty} \frac{x_{n-1} x_{n-2m-1} - \ x_{n-1} x_{n+2m+1} - x_n x_{n-2m-2} + \ x_n x_{n+2m}}{2m+1} \qquad \forall n \in \mathbb{Z} $$

How can I prove that this is true for a nearly arbitrary (not constant) real infinite sequence $x_n$? Or a good counter example to disprove it?

I can show it's true for a general sinusoid $x_n = A \cos(\Omega n + \theta)$ when $0<\Omega<\pi$. I know it's not true for $x_n=\mathrm{constant}$ (that's $\Omega=0$) or $x_n=(-1)^n$ (that's $\Omega=\pi$) or any linear combination because the Hilbert transform $\hat{x}_n=0$ (signals at DC or Nyquist are known to be funky), but I dunno how to deal with this for a virtually general real $x_n$. What conditions on $x_n$ guarantee this to be true? Or to be false?

---

Okay, I have started a bounty and I will attempt again to state the question/problem unambiguously.

Given any sequence of finite real numbers:

$$ x_n \in \mathbb{R} \qquad \qquad n \in \mathbb{Z} $$

except those sequences having this form:

$$ x_n = a + b(-1)^n \qquad \qquad a,b \in \mathbb{R} $$

what conditions on $x_n$ are necessary so that this inequality is satisfied:

$$ x_n \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-2} - \ x_{n+2m} \big) < x_{n-1} \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \qquad \forall n \in \mathbb{Z} $$

?

Answer 1

You wrote

$$0 < \sum\limits_{m=0}^{\infty} \frac{x_{n-1} x_{n-2m-1} - \ x_{n-1} x_{n+2m+1} - x_n x_{n-2m-2} + \ x_n x_{n+2m}}{2m+1} \qquad \forall n \in \mathbb{Z} $$

and

How can I prove that this is true for a nearly arbitrary (not constant) real infinite sequence $x_n$ ? Or a good counter example to disprove it?

I think there are infinitely many counterexamples.

If $\{x_n\}$ satisfies the followings :

• $x_1\gt 0$

• If $n\ge 0$ is even, $x_n\gt 0$

• If $n\ge 3$ is odd, $x_n\lt 0$

• If $n\le -1$ is odd, $x_n\gt 0$

• If $n\le -2$ is even, $x_n\lt 0$

• $x_0=x_1$

• $2x_{0}-x_2-x_{-1}\lt 0$

then, for $n=1$, the inequality does not hold.

This is because for $n=1$, $$\sum\limits_{m=0}^{\infty} \frac{x_{n-1} x_{n-2m-1} - \ x_{n-1} x_{n+2m+1} - x_n x_{n-2m-2} + \ x_n x_{n+2m}}{2m+1} $$ can be written as $$\underbrace{x_0(2x_{0} - x_{2} - x_{-1})}_{\text{negative}}+\sum\limits_{m=\color{red}1}^{\infty} \frac{\overbrace{x_0(x_{-2m} - x_{2m+2})}^{\text{negative}}+ \overbrace{x_0(- x_{-2m-1} + x_{2m+1})}^{\text{negative}}}{2m+1}$$ which is negative.

---

Added :

what conditions on $x_n$ are necessary so that this inequality is satisfied:$$ x_n \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-2} - \ x_{n+2m} \big) < x_{n-1} \sum\limits_{m=0}^{\infty} \frac{2}{\pi \ (2m+1)} \ \big( x_{n-2m-1} - \ x_{n+2m+1} \big) \qquad \forall n \in \mathbb{Z} $$

I'm going to write my observation which may help.

The inequality is equivalent to $$x_n\sum\limits_{m=0}^{\infty} \frac{x_{n-2m-2} - x_{n+2m}}{2m+1} < x_{n-1}\sum\limits_{m=0}^{\infty} \frac{x_{n-2m-1} - x_{n+2m+1}}{2m+1}$$

For $n=1$, we have to have $$x_1\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1} < x_0\color{red}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m} - x_{2m+2}}{2m+1}}\tag1$$

For $n=2$, we have to have $$x_2\color{red}{\sum\limits_{m=0}^{\infty}\frac{x_{-2m} - x_{2m+2}}{2m+1}} < x_1\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}\tag2$$

So, if both $x_0$ and $x_2$ are positive, then we have

$$\frac{x_1}{x_0}\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1} < \color{red}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m} - x_{2m+2}}{2m+1}} < \frac{x_1}{x_2}\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}$$ from which we get $$\frac{x_1}{x_0}\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1} < \frac{x_1}{x_2}\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}$$ So, if $x_1\gt 0$, then $$x_2\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1} < x_0\color{blue}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}}$$

If $x_1\lt 0$, then $$x_2\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1} \color{red}{\gt} x_0\color{blue}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}}$$

For $n=3$, we have to have $$x_3\color{blue}{\sum\limits_{m=0}^{\infty} \color{blue}{\frac{x_{-2m+1} - x_{2m+3}}{2m+1}}} < x_{2}\sum\limits_{m=0}^{\infty} \frac{x_{-2m+2} - x_{2m+4}}{2m+1}\tag3$$

So, if $x_1x_0x_3\gt 0$, then we get another inequality on the sum in blue.

(comments : Even when both $x_0$ and $x_2$ are negative, similar inequalities can be obtained.)

Of course, if $x_0\color{red}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m} - x_{2m+2}}{2m+1}}\lt x_2\color{red}{\sum\limits_{m=0}^{\infty}\frac{x_{-2m} - x_{2m+2}}{2m+1}}$, i.e. we have either

• $\color{red}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m} - x_{2m+2}}{2m+1}}\gt 0$ and $x_0\lt x_2$

• $\color{red}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m} - x_{2m+2}}{2m+1}}\lt 0$ and $x_0\gt x_2$

then we get $$x_1\sum\limits_{m=0}^{\infty} \frac{x_{-2m-1} - x_{2m+1}}{2m+1}\lt x_1\color{blue}{\sum\limits_{m=0}^{\infty} \frac{x_{-2m+1} - x_{2m+3}}{2m+1}}$$ You can divide the both sides by $x_1$, and with $(3)$, we get another inequality on the sum in blue.

These mean that we can get many similar necessary conditions (inequalities) because of the form of our inequality.

Answer 2

User mathlove presented one counter example. A simple search in any AI search gives many more counter examples. Such as $x[n]=2\cos[n]+\cos[5n]$ (the instantaneous frequency of the analytical version becomes negative for $n=2,3$, which I didn't rigorously verify).

In my opinion, the key point of OP's question to clarify for EE DSP engineers is: $\color{red}{\text{Is the instantaneous frequency of a signal mandatorily captured in the spectrum of the signal?}}$ The sign of spectrum range, sign of instantaneous frequency and as it applies to DTFT etc are all by-products of the answer to this question.

The Fourier Transform (CFT, DTFT etc) are infinite time averages of the frequency components (given by their amplitude and phase components). Therefore they kill the instantaneous frequency (rate of change of instantaneous phase) information. Therefore the key point to attack is your sentence "we really expect $\Omega_{n}$ to be positive because there are no frequency components in the analytic signal that are negative frequency.."

Hope I captured the essence of what the counter argument is. I am happy to rigorously provide counter examples that demonstrate this, in the spirit of mathlove's answer, but skipping it for a $10000$ feet zoomed out philosophical view.

Answer 3

$$0 < \sum\limits_{m=0}^{\infty} \frac{x_{n-1} x_{n-2m-1} - \ x_{n-1} x_{n+2m+1} - x_n x_{n-2m-2} + \ x_n x_{n+2m}}{2m+1} \qquad \forall n \in \mathbb{Z} $$

Suppose, for $m$ even, we could show that the numerator of the $m$th term is at least as large as the negative of the numerator of the $(m+1)$th term?

$$-\big(x_{n-1} x_{n-2(m+1)-1} \ - \ x_{n-1} x_{n+2(m+1)+1} \ - \ x_n x_{n-2(m+1)-2} \ + \ x_n x_{n+2(m+1)} \big) \\ \le \\ x_{n-1} x_{n-2m-1} \ - \ x_{n-1} x_{n+2m+1} \ - \ x_n x_{n-2m-2} \ + \ x_n x_{n+2m}$$

If we can show that for a general sinusoid where $0<\Omega<\pi$, would it not also be true for any sum of sinusoids?

Then every pair of terms, beginning with the first pair ($m=0,1$), would add to a positive value.

This is the beginning of an exploration. I dunno where it will lead. When the question is eligible for bounty, I think I'll put a bounty on it. As Srini suggested, the purpose is to prove for a general analytic signal (that has no negative frequency components) that the instantaneous frequency will always be positive. Maybe it's not true. But I would like to know either how it's true or how it's not.