## Uncertainty principle in time frequency analysis

If you like playing music or listening to music, you would probably know how different instruments playing their rthyum. You can hear how a jazz drum was druming  |B S  BB S|B S  BB S| … (B- Bass drum, S– Snare drum). You may hear how the guitar (fill with chords) and the bass guitar ( making backup rthyum)  playing their note. You may hear wonderful keyboard  filling in their note.  All are done instantaneously, and experienced musicians can even write them immediately.

Can we work it out by a computer? Can we plot an “﻿﻿﻿﻿﻿﻿instantaneous﻿﻿﻿” spectrum of frequency. Let’s think mathematically. Let $f\in L^2({\Bbb R}^n)$, its fourier-Plancherel transform  defined by

$\widehat{f}(\omega) = \int_{-\infty}^{\infty} f(t)e^{-2\pi i \omega t}dt.$

We think $f(t)$ as a incoming wave function as a musical signal, its fourier transform ${\widehat{f}}(\omega)$ as the overall frequency spectrum. ﻿﻿﻿﻿﻿ Our problem is asking whether at a particular time, we know the frequency spectrum or not.  This turns out the unertainty principle makes all these instantaeous frequency impossible. Heuristically,

a function $f$ and its fourier transform $\widehat{f}$ cannot be supported on arbitrarily small sets together.

There are many variant forms.

Theorem 1. (Naive uncertainty principle) If $f$ is of compact support on $\mathbb {R}^n$, then $\widehat{f}$ cannot be of compact support.

Proof. If so, then we assume the support in on $[0,1/2]^n$, then develop its fourier series on $[0,1]^n$, by assumption, the series must be a trigonmetric polynomial. This is impossible, since the trigonmetric polynomial cannot vanish on a set of posifive measure. Q.E.D.

We work on more specfically, for two linear operators (may not be bounded) on a Hilbert space, we define the commutator

$[A,B] = AB-BA$

Theorem 2. (Hesienberg uncertainty principle) Let $A, B$ be selfadjoint operators on the a Hilbert space. Then for all $a,b\in {\mathbb Z}$

$\|(A-a)f\|\|(B-b)f\|\geq \frac{1}{2}|\langle[A,B]f,f\rangle|.$

In particular on $L^2({\Bbb R}^n)$,

$(\int_{-\infty}^{\infty}(x-a)^2)|f(x)|^2dx)^{1/2}\cdot(\int_{-\infty}^{\infty}(\omega-b)^2|\widehat{f}(\omega)|^2d\omega)^{1/2}\geq \frac{1}{4\pi}\|f\|_2^2.$

Proof. (Sketch due to limited space) By selfadjointness

$\langle[A,B]f,f\rangle = \langle{(A-a)(B-b)-(B-b)(A-a)}f,f\rangle = 2i Im\langle(B-b)f,(A-a)f\rangle$.

We then apply Cauchy-Schwarz inequality.

For the second one, we consider

$Af(x) = xf(x)$,

$Bf(x)=\frac{1}{2 \pi i}f'(x)$

defined on the Schwartz space of rapidly decreasing functions. Then apply the abstract inequality. Q.E.D.

More to come, we define a function $f\in L^2({\mathbb R}^n)$ is $\epsilon$concentrated on a measurable set $T$ if

$(\int_{T^c}|f(x)|^2dx)^{1/2}\leq \epsilon \|f\|_2.$

Theorem 3. (Donoho and Stark uncertainty principle) Let $f\in L^2({\mathbb R}^n)$, $f\neq 0$, is $\epsilon_T$ concentrated on $T$, and ${\widehat{f}}$ is $\epsilon_{\Omega}$-concntrated. Then

$|T||\Omega|\geq (1-\epsilon_T-\epsilon_{\Omega})^2.$

Proof. Consider the Landau projection operators,

$P_Tf = \chi_T f$

$Q_{\Omega}f = \int_{\Omega}{\widehat{f}}(\omega)e^{2\pi i \omega x}d\omega$

Then we have by concentratedness,

$\|f-P_Tf\|_2\leq\epsilon_T\|f\|_2$.

$\|f-Q_{\Omega}f\|_2 = \|\chi_{\Omega^c}{\widehat{f}}\|_2\leq\epsilon_{\Omega}\|f\|_2$.

We have by triangle inequality,

$\|f-Q_{\Omega}P_{T}f\|_2\leq \|f-Q_{\Omega}f\|_2+\|Q_{\Omega}(f-P_Tf)\|_2\leq (\epsilon_T+\epsilon_{\Omega}\|f\|_2)$,

Hence,

$\|Q_{\Omega}P_Tf\|_2\geq\|f\|_2-\|f-Q_{\Omega}P_Tf\|_2\geq (1-\epsilon_T-\epsilon_{\Omega})\|f\|_2.$

But $Q_{\Omega}P_T$ is a Hilbert-Schmidt operator, with $\|Q_{\Omega}P_T\|_{H.S.} = |T||\Omega|$ (by direct computation), hence the result follows from this. Q.E.D.

We may say do  a short time windowed fourier transform. Let $g$ be a bounded function of compact support, we define

$V_gf(x,\omega) = \int f(t)g(t-x)e^{-2\pi i t\omega}dt$.

Can we define some intstantaneous frequency conecpt, the answer is no. We state the following theorem without proof.

Theorem 4.(Liebs uncertainty principle) Suppose that $\|f\|_2 = \|g\|_2=1$.  If $U\subset{\mathbb R}^{2n}$ and $\epsilon>0$ are such that

$\int\int_{U}|V_gf(x,\omega)|^2dxd\omega\geq 1-\epsilon$,

then

$|U|\geq \sup_{p>2}(1-\epsilon)^{\frac{p}{p-2}}(\frac{p}{2})^{\frac{2n}{p-2}}\geq (1-\epsilon)^22^n$.

Uncertainty principle also appears in quantum mechanics where $x$ is the position and $\omega$ is the momentum.

Back to our time frequency analysis, we must overcome uncertainty principle in order to make precise analysis of musical signal. However, our ear is competent to it. Why? May be our brain never interpret music as fourier transform, it never does fourier transform, but understand the music  in a mixture of experience, feeling and understanding ^^ We must appreciate our brain being  far better than the computer.

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### 2 Responses to Uncertainty principle in time frequency analysis

1. KKK says:

Is there a physical way to interpret the mathematical results? For example, in your “native uncertainty principle”, does it mean if we can quite accurately determine its position, it’s hard to determine its pitch? Or something like that?

• tejonbiker says:

Yes, first I am not mathematician, I am engineer, the Fourier transform is orthogonal (time and frequency are not related). Let’s go to discrete world, in discrete world you need sampling a signal (conver to a number representation). If you want know with very high precision the frequency analisis you need much samples, samples require time to get, in other words, if you want more frequency precision you need much time (and samples),but remember, in the frecuency domain you lost the information of time, that means that you lost the exact position in time of the frequency information.

Other example is as follow, you sample one second of time, let’s get 1000 samples over the second, we have sampling a sine waveform that appear in the time interval of 100 mili seconds to 400 ms, if you apply Fourier Transform you get the frequency representation.

let’s take other escenario, the same situtation except that the waveform appears in the time interval of 500 ms to 900 ms, if you apply Fourier transform, you get the same frequency representation!!., that means you lost in wich time interval appear the waveform.

With 1000 points at 1 (kilo sample)/second you get a resolution of 500 bands, separated with 1 Hz between each band.

500 bands, not bad, but you lost in wich moment of the second the frecuency appear, let’s take other aproximation.

let’s apply the Fourier transform every 100 samples, this means 10 Fourier transform in one second, with this aproximation we divided the time in ten intervals, this get a coarse aproximation, but get some time information (this is called Short Time Fourier Transform), the cost, well, let’s see:

With 100 points at 1 ks/s you get 50 bands separated with 10 Hz between each band, yes the resolution in frequency is lost, but, you get some information in time!!!!, (Uncertainty principle).

Sorry is the answer is to long, but I think is sustancial.

Greetings from Mexico :)