Does analytic signal have positive instantaneous frequency?

Question

The Analytic representation of a signal has no negative frequencies.

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

Answer 1

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.

Answer 2

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.

Answer 3

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.