6
$\begingroup$

The Analytic representation of a signal has no negative frequencies.

Does this mean that everywhere, it's instantaneous frequency is positive?

$\endgroup$
2
  • $\begingroup$ It's difficult to say for sure. Any single sinusoidal component (which is really an exponential with a purely imaginary exponent), will have a positive frequency. But the instantaneous frequency is computed on the entire analytic signal as a whole. The complex $\arg\{\cdot\}$ of the analytic signal is the instantaneous phase and the derivative w.r.t. time of the unwrapped instantaneous phase is the instantaneous frequency. I am not sure that the unwrapped instantaneous phase would always be strictly increasing w.r.t. time. $\endgroup$ Sep 24, 2023 at 23:39
  • $\begingroup$ I guess in this answer I am implying that the phase increase per sample is always between $0$ and $+\pi$, which means it's never decreasing and that the instantaneous frequency is always positive. But that was for a single sinusoid (with noise), not a broadbanded baseband analytic signal. $\endgroup$ Sep 24, 2023 at 23:45

3 Answers 3

8
$\begingroup$

It is not the case that the instantaneous frequency of an analytic signal is always positive. In general, the instantaneous frequency can become negative, also for analytic signals.

I'll show this using the example of an analytic two-tone signal:

$$s(t)=A_1e^{j\omega_1t}+A_2e^{j\omega_2t}=A(t)e^{j\phi(t)}\tag{1}$$

with real-valued $A_i>0$, $\omega_i>0$, $i\in\{1,2\}$, and $\omega_1\neq \omega_2$.

From $(1)$, the derivative of $s(t)$ is given by

$$s'(t)=A'(t)e^{j\phi(t)}+A(t)e^{j\phi(t)}j\phi'(t)\tag{2}$$

Hence,

$$\frac{s'(t)}{s(t)}=\frac{A'(t)}{A(t)}+j\phi'(t)\tag{3}$$

and, consequently, the instantaneous (angular) frequency $\phi'(t)$ can be expressed as

$$\phi'(t)=\textrm{Im}\left\{\frac{s'(t)}{s(t)}\right\}=\frac{1}{|s(t)|^2}\textrm{Im}\left\{s'(t)s^*(t)\right\},\qquad |s(t)|>0\tag{4}$$

where $s^*(t)$ is the complex conjugate of $s(t)$.

With

$$s'(t)=jA_1\omega_1e^{j\omega_1t}+jA_2\omega_2e^{j\omega_2t}\tag{5}$$

and

$$|s(t)|^2=A_1^2+A_2^2+2A_1A_2\cos\big[(\omega_1-\omega_2)t\big]\tag{6}$$

a straightforward calculation gives

$$\phi'(t)=\frac{A_1^2\omega_1+A_2^2\omega_2+(\omega_1+\omega_2)A_1A_2\cos\big[(\omega_1-\omega_2)t\big]}{A_1^2+A_2^2+2A_1A_2\cos\big[(\omega_1-\omega_2)t\big]}\tag{7}$$

This result can be rewritten as

$$\phi'(t)=\frac{\omega_1+\omega_2}{2}+\frac{\omega_1-\omega_2}{2}\frac{A_1^2-A_2^2}{A_1^2+A_2^2+2A_1A_2\cos\big[(\omega_1-\omega_2)t\big]}\tag{8}$$

The representation $(8)$ shows that for $A_1=A_2$, the instantaneous frequency of $s(t)$ is the arithmetic average of $\omega_1$ and $\omega_2$, which is intuitively pleasing. However, it also shows that for $A_1\neq A_2$ it is always possible for $\phi'(t)$ to become negative for some $t$ if $\omega_{1,2}$ and $A_{1,2}$ are chosen appropriately.

The figure below shows the instantaneous phase and instantaneous (angular) frequency of the signal $(1)$ with $A_1=2$, $A_2=1$, $\omega_1=1$ and $\omega_2=5$. It can be seen that for $t\in[0,3]$, there are two time intervals inside which the derivative of the phase, and hence the instantaneous frequency, become negative.

enter image description here

$\endgroup$
5
  • 1
    $\begingroup$ Really nice Matt- (+1 from me) I wasn't thinking of different amplitudes, that's really interesting. I confirmed equation 8 with some examples where it does go negative and does match the derivative of the phase. $\endgroup$ Sep 25, 2023 at 12:19
  • $\begingroup$ I might suggest that we worry very little about the individual down voter. This is good answer. $\endgroup$ Sep 25, 2023 at 13:31
  • $\begingroup$ Matt, I'm not the down-voter. But may I suggest that you and I harmonize our symbols? Question: Should the analytic signal be "$x_\mathrm{a}(t)$" or "$s(t)$"? I know I'm discrete time and you're continuous time, and that's good. We're doing this from different angles. Should I change my symbols? $\endgroup$ Sep 25, 2023 at 13:50
  • $\begingroup$ @robertbristow-johnson: I don't think that it's necessary that you (or I) change our symbols. We're both being consistent in what we're doing, and we both adhere to useful standards, such as brackets for discrete-time signals. But there's no agreed standard on what to call an analytic signal. I do use $x_a(t)$ sometimes, but only if I also used $x(t)$ for the corresponding real-valued signal. In this answer I started out with the analytic signal, that's why I chose a simpler symbol without subscript. $\endgroup$
    – Matt L.
    Sep 25, 2023 at 14:36
  • $\begingroup$ yeah.. But you're also using brackets for something other than discrete-time indices. Like $$\cos\big[(\omega_1-\omega_2)t\big]$$ Once I started using brackets for $x[n]$, (and I like to use curly braces for operators on functions, like $\mathscr{H}\{\cdot\}$ and for $\Re e\{\cdot\}$ or $\arg\{\cdot\}$) I stopped using them as parenths. So I would say $$\cos\big((\omega_1-\omega_2)t\big)$$ $\endgroup$ Sep 25, 2023 at 16:28
5
$\begingroup$

Okay, so let's get a little specific about the math...

The Hilbert Transform:

$$\begin{align} \hat{x}[n] &= \mathscr{H}\big\{ x[n] \big\} \\ \\ &= \sum\limits_{i=-\infty}^{+\infty} \frac{1-(-1)^i}{\pi \ i} \ x[n-i] \end{align}$$

The Analytic signal:

$$ x_\mathrm{a}[n] \triangleq x[n] + j \hat{x}[n] $$

The instantaneous phase (wrapped):

$$\phi[n] \triangleq \arg\{x_\mathrm{a}[n]\}$$

The discrete-time instantaneous frequency:

$$ \omega[n] \triangleq \phi[n] - \phi[n-1] $$

Sometimes $2\pi$ has to be added to that to undo the effect of phase wrapping, but that goes away with this:

$$\begin{align} \omega[n] &\triangleq \phi[n] - \phi[n-1] \\ \\ &= \arg\{x_\mathrm{a}[n]\} - \arg\{x_\mathrm{a}[n-1]\} \\ \\ &= \arg \left\{ \frac{x_\mathrm{a}[n]}{x_\mathrm{a}[n-1]} \right\} \\ \\ &= \arg \left\{ \frac{x_\mathrm{a}[n] \, (x_\mathrm{a}[n-1])^*}{x_\mathrm{a}[n-1] \, (x_\mathrm{a}[n-1])^*} \right\} \qquad \qquad (\cdot)^* \text{ is complex conjugate}\\ \\ &= \arg \left\{ \frac{x_\mathrm{a}[n] \, (x_\mathrm{a}[n-1])^*}{\big| x_\mathrm{a}[n-1] \big|^2} \right\} \\ \\ &= \arg \Big\{ x_\mathrm{a}[n] \, (x_\mathrm{a}[n-1])^* \Big\} \\ \\ &= \arg \Big\{ (x[n] + j \hat{x}[n]) \, (x[n-1] - j \hat{x}[n-1]) \Big\} \\ \\ &= \arg \Big\{ (x[n]x[n-1] + \hat{x}[n]\hat{x}[n-1]) \, + \, j \big(\hat{x}[n]x[n-1] - x[n]\hat{x}[n-1] \big) \Big\} \\ \end{align}$$

The phase wrapping problem goes away if we compute the phase difference this way.

A complete definition for the complex $\arg\{\cdot \}$:

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

If the phase difference was always positive, we could use solely the second line of the above. This means that the imaginary part, $v>0$, is always positive.

$$ \omega[n] = \tfrac{\pi}{2} - \arctan\left(\frac{x[n]x[n-1] + \hat{x}[n]\hat{x}[n-1]}{\hat{x}[n]x[n-1] - x[n]\hat{x}[n-1]}\right) $$

But for that to be, then

$$ 0 < \hat{x}[n]x[n-1] - x[n]\hat{x}[n-1] $$

or

$$ \hat{x}[n]x[n-1] > x[n]\hat{x}[n-1] $$

The instantaneous frequency is positive only if the above holds.

$\endgroup$
1
  • $\begingroup$ Thank you for the discrete perspective. The data I am processing is of course discrete. There is probably a counter-example similar to Matt's answer in this case too. $\endgroup$ Sep 26, 2023 at 7:24
-2
$\begingroup$

How to classify this laconic question? If sprouted from homework, it is a no-effort candidate. Compassionate ones might comment with a counterexample of 100·exp(jωt) + exp(j(100ω)t) and encourage OP to do homework. Or is this question an invitation to write an essay on the subject? Then maybe the existing body of literature on the subject should suffice. For example, in 1992, Proceedings of the IEEE, vol. 80, no. 4, April 1992, published an article by Boualem Boashash (a free copy of this article is available by courtesy of https://www.math.ucdavis.edu/~saito/data/sonar/boashash1.pdf). The article relates the story of the application-oriented (telecommunications, seismics, electric circuit theory, radars) generalization of the concept of frequency, from pioneering work by Carson and Fry and Van der Pol (FM modulation) to Gabor (unique complex signal via Hilbert transform) to Ville (instantaneous frequency as the first momentum of Wigner-Ville distribution) to Mandel (physical interpretation of IF) to Priestly, who considered nonstationary processes.

Mandel's interpretation of IF can become an eye-opener for a student of IF concept in signal processing:

Mandel strongly promoted the idea that the IF and Fourier frequencies are different quantities, and that one source of their mutual confusion is the same name-frequency-attached to both of them. Finally, Mandel asks a question: Which of these two quantities is most closely related to measurements? He also provides the answer: It strongly depends “on the nature of the experiment.”

[T]he IF and Fourier frequencies are different quantities (emphasis mine), therefore, analytic signal's IF sign is not prescribed by its 'analyticity'; IF values can be positive, negative or zero.

"Frequency" is anything but unique among signal processing terms that have more than one definition, although not as many as some other words like bandwidth, spectrum, distribution. The problem with "frequency" is that the meaning of the other words is often clear from a context which can be parsed from the natural language, while the context for intermediate frequency has yet to be uncovered from equations, diagrams and codes.

What I like most in engineering science is that it never stops its reciprocal development, with engineering part stimulating generation of new ideas and concepts in theoretical part followed by consequent adoption of the novel instruments developed.

Mandel had clearly delineated the two concepts, IF and Fourier frequency. Now, analytic signal concept extended the range of signal functions to the complex plane; predictably, there appeared a need to extend the domain of signal functions to the complex plane. In this aspect, the analytic signal is considered to be a boundary value of analytic function. For example, an analytic function defined in the upper half plane is the progenitor of an analytic signal on the entire real axis like $-∞ < t < +∞$.

Similar to a real-valued signal decomposition that, if exists, has a unique Fourier integral representation, decomposition of the generalized analytic signal (domain is the upper half complex plane), can be expressed in the form $$ f(z) = \exp(iαz)B(z)S(z)G(z) $$ which is also unique (if exists). Here, B(z) is a Blaschke product, S(z) and G(z) are singular functions in the upper half plan (keywords Hardy space, Blaschke product).

This decomposition enables one to define the instantaneous frequency of a complex valued 'analytic' (cf., meromorphic) signal as the complex valued logarithm $$ ω_f(z) = {d \over dt}{\Im(\ln (f))} $$ which supersedes Gabor's definition. (Analytic signals with nonnegative instantaneous frequency by D.V.Vliet)

This dramatic progress of the IF concept that adds semantic overload to words 'frequency' and 'analytic signal' in signal processing, also makes a seeming U-turn as compared with Mandel's delineation of IF and FF concepts. Citing Vliet's paper:

... At first glance, this seems like a stark contrast between the IF and FF perspectives. A more careful read on the situation brings to light a surprising parallel: Outer functions (that are reasonably nice) have zero mean instantaneous frequency; the Fourier transform of a purely amplitude modulated (AM) signal has symmetric support. This leads to analogous procedures for making signals with positive frequency in both the IF and FF realms. In the FF realm, a band limited signal is made to have positive Fourier frequency by modulating it with a carrier frequency (a pure FF function) high enough to shift the Fourier frequencies above zero. (This is the well-known procedure for making an analytic signal out of an arbitrary band-limited signal.) In the IF realm, an analytic signal is made to have positive instantaneous frequency by modulating it with an inner function (a pure IF function) appropriately chosen to shift the instantaneous frequencies above zero. The analogy is quite clear: ASNIFs are to analytic signals as analytic signals are to “left band-limited signals.”

(ASNIF stands for Analytic Signals with Nonnegative Instantaneous Frequency)

On second thought, however, this citation sharpens the picture under this delineation, making analytic signal and IF versatile instruments of signal processing.

Also, that there exists an ASNIF tool does not prove that ALL analytic signals have nonnegative instantaneous frequency ;). Quite the contrary.

$\endgroup$
6
  • $\begingroup$ Hi V.V.T. I tried looking at the paper you referenced before asking my question. It says among other things, that the average IF is equal to the average frequency (in terms of energy), but not much more I could immediately use. I suppose the two concepts are quite different, unless the signals have same amplitude, as Matt showed. Thank you for the perspective. $\endgroup$ Sep 26, 2023 at 11:57
  • $\begingroup$ Hi Sebastian: IMO you are not under obligation to upvote the answer you do not find useful or do not understand (I believe the upvote is yours), especially when there are answers nearby that you sincerely like. Upvoting an answer is not an act of politeness but an indicator of answer's usefulness: and honesty rulez! Notice the prompts 'This answer is useful/This answer is not useful' that pop up when hovering at voting buttons. $\endgroup$
    – V.V.T
    Sep 26, 2023 at 14:12
  • $\begingroup$ Personally I never use "This answer is not useful" (from PoV of community, my personal likings/dislikings are not a big deal), but "This answer is useful" saves me typing my recommendations to the community when it is the only message that I want to share. I'm editing my answer so that you can undo your upvote pressing the upvote button one more time (see Help Center instructions). If you like, I can delete my answer. $\endgroup$
    – V.V.T
    Sep 26, 2023 at 14:12
  • $\begingroup$ Removed the vote 👍 $\endgroup$ Sep 27, 2023 at 18:08
  • $\begingroup$ "This leads to analogous procedures for making signals with positive frequency in both the IF and FF realms [...]": Does this mean that a real signal that is modulated by a carrier frequency (satisfying this) is ASNIF, after forming the analytic representation with a Hilbert transform? $\endgroup$ Sep 27, 2023 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.