Paradoxes on Instantaneous Frequency a la Leon Cohen Time-Frequency Analysis, Prentice Hall, 1995...

Paradoxes on Instantaneous Frequency

a la Leon Cohen

Time-Frequency Analysis, Prentice Hall, 1995

Chapter 2: Instantaneous Frequency, P. 40

The Five Paradoxes

• 1. Instantaneous frequency of a signal may not be one of the frequencies in the spectrum.

• 2. For a signal with a line spectrum consisting of only a few sharp frequencies, the instantaneous frequency may be continuous and range over an infinite number of values.

• 3. Although the spectrum of analytic signal is zero for negative frequencies, the instantaneous frequency may be negative

• 4. For the band limited signal the instantaneous frequency may be outside the band.

• 5. The value of the Instantaneous frequency should depend only on the present time, but the analytic signal, from which the instantaneous frequency is computed, depends on the signal values for the whole time space.

Observations I

• By ‘spectrum’, Cohn is limiting the term to ‘Fourier spectrum’.

• By ‘instantaneous Frequency’, Cohn is limiting the terms to be the IF obtained through Hilbert Transform. In fact, as we see, IF could be determined through many other methods.

Observations II

• 1. Paradoxes 1, 2 and 4 are essentially the same: Instantaneous Frequency values may be different from the frequency in the spectrum.

• 2. The negative frequency in analytic signal seems to violate Gabor’s construction.

• 3. The analytic function, or the Hilbert Transform, involves the functional values over the whole time domain; therefore, it is not local.

Resolution for paradoxes 1, 2 and 4

Two Examples

The First Example

Sin A + c*Sin B

Data: Sin (πt/360) + Sin (πt/320) : t=0:23040

Hilbert Spectrum X

Spectrogram X

Morlet Wavelet X

Instantaneous frequency X

Instantaneous frequency X : Details

Marginal Spectra X

Data: Sin (πt/360) + 0.8* Sin (πt/320) : t=0:23040

Hilbert Spectrum X08

Marginal Spectra X08

Two ways to view modulated wave

a b a bsina sin b = 2 sin cos .

2 2

New developments

• G. RILLING, P. FLANDRIN, 2008 :  "One or Two Frequencies? The Empirical Mode Decomposition Answers,“ IEEE Trans. on Signal Proc., Vol. 56, No. 1, pp. 85-95.

“….close tones are no longer perceived as such by the human ear but are rather considered as a whole, one can wonder whether a decomposition into tones is a good answer if the aim is to get a representation matched to physics (and/or perception) rather than to mathematics.”

Example

General case

1 1 1 2 2 2

Consider

Though we have 6 parameters, but there is no loss of

generality to consider the cas

x(t) = a cos (2 f t + ) + a cos (2 f t + )

x(t) = cos 2 t + a cos (2 ft + ), for 0 <

f < 1 .

e

2 22 1

1 1

a fwhere a = ; f = ; and =

a f

He

x'(t) sin 2 t + af sin (2 ft + ), for 0 < f < 1 .

cos 2 t re we have as the high frequency component and

a cos (2 ft + as the l)

ow frequency component.

Derivatives of HF and LF components

Af < 1

Af2 > 1

Numerical experiments

2

2

L

L

We define a criterion as

d( t ;a , f ) cos 2 tC( a , f , )

a cos( 2 ft )

where is the first IMF component derived from EMD from x(t).

C 0, when the extraction of high frequency is complete.

C 1, wh

d

en the ex

traction misses totally.

Numerical Experiments : C

Numerical Experiments : C

One or two-frequency?

• Mathematically, if we select strict Fourier basis, it is two-frequency signal.

• Physically, it is a modulated one frequency signal.• Using EMD, we could separate the signal, if the

amplitude-frequency combination satisfies certain condition*, the condition coincides with physical perception.

*The condition: if frequency separation more than a factor of 2; and the amplitude of the low frequency is relatively small.

Example 2

Duffing’s Pendulum

Duffing Pendulum

2

22( co .) s1

d xx tx

dt

x

Duffing Type Wave : Data: x = cos(wt+0.3 sin2wt)

Duffing Type Wave : Perturbation Expansion

For 1 , we can have

x( t ) cos t sin 2 t

cos t cos sin 2 t sin t sin sin 2 t

cos t sin t sin 2 t ....

1 cos t cos 3 t ....2 2

This is very similar to the solutionof Duffing equation .

Duffing Type Wave :Wavelet Spectrum

Duffing Type Wave : Hilbert Spectrum

Duffing Type Wave : Marginal Spectra

Duffing Equation

23

2.

Solved with for t 0 to 200 with

1

0.1

od

0.04 Hz

Initial condition :

[ x( o ) ,

d xx x c

x'( 0 ) ] [1

os t

, 1]

3

t

e2

d

tb

Duffing Equation : Data

Duffing Equation : IMFs

Duffing Equation : IMFs

Duffing Equation : Hilbert Spectrum

Duffing Equation : Detailed Hilbert Spectrum

Duffing Equation : Wavelet Spectrum

Duffing Equation : Hilbert & Wavelet Spectra

Duffing Equation : Marginal Hilbert Spectrum

Rössler Equation

x ( y z ),

1y x y ,

51

z z

Rossler Equation solved with ode23 :

Initital conditions :

3.5

[ x , y , z ]

( x ) .

[1, 1 , 0 ]

For

t 0 : 200 .

5

Rössler Equation : Data

Rössler Equation : 3D Phase

Rössler Equation : 2D Phase

Rössler Equation : IMF Strips

Rössler Equation : IMF

Rössler Equation : Hilbert Spectrum

Rössler Equation : Hilbert Spectrum & Data Details

Rössler Equation : Wavelet Spectrum

Rössler Equation : Hilbert & Wavelet Spectra

Rössler Equation : Marginal Spectra

Rössler Equation : Marginal Spectra

Resolution for Paradox 3

Negative Frequency

Examples of Negative Frequency 1

Different references

Hilbert Transform a cos + b : Data

Hilbert Transform a cos + b : Phase Diagram

Hilbert Transform a cos + b : Phase Angle Details

Hilbert Transform a cos + b : Frequency

The Empirical Mode Decomposition Method and

Hilbert Spectral Analysis

Sifting

Examples of Negative Frequency 2

FM and AM Frequencies

a sin ω t + b sin φ t

sin ω t + 0.4 sin 4 ω t

Hilbert : sin ω t + 0.4 sin 4 ω t

sin ω t + sin 4 ω t

Hilbert : sin ω t + sin 4 ω t

a sin ωt + b sin φt

• The data need to be sifted first.

• Whenever Hilbert Transform has a loop away from the original (negative maximum or positive minimum), there will be negative frequency.

• Whenever the Hilbert pass through the original (both real and imaginary parts are zero), there will be a frequency singularity.

• Hilbert Transform is local to a degree of 1/t.

IMF : sin ω t + 0.2 sin 4 ω

IMF : sin ω t + 0.4 sin 4 ω

IMF : sin ω t + sin 4 ω

Negative Frequency

• Negative instantaneous frequency values are mostly due to riding waves.

• IMF is a necessary (but not a sufficient) condition for having non-negative frequency.

• There are occasion when abrupt amplitude change in an IMF (but no riding waves) can also generate negative frequency. The amplitude induced problem is covered by Bedrosian theorem; normalized HHT will take care of it.

• Physically, the abrupt amplitude change also shows the non-local characteristics of the Hilbert Transform.

Resolution for Paradox 5

Non-local influence does exist, they may come from Gibbs Phenomenon, end effects, and the limitation of the 1/t window in the Hilbert Transform. But most of the problems can be rectified through the Normalized HHT.

In fact, the non-local property of Hilbert transform is fully resolved by Quadrature method, though the solution is no longer a ‘Hilbert Spectrum’.

Data with magnitude jump : Signal

Data with magnitude jump : Signal

Hilbert Spectrum

Spectrogram

Morlet Wavelet

Data with magnitude jump

Data with magnitude jump : Details

Normalized Hilbert Spectrum

Amplitude Effects on Marginal Hilbert & Fourier Spectra

Instantaneous frequency

Instantaneous frequency : Details

Resolution for Paradox 5

Hilbert Transform is Non-local; therefore, the instantaneous

frequency is not local.

Instantaneous Frequency

• Hilbert transform might not be local, but it is very close to be so, for the window is 1/t. Therefore, the instantaneous frequency through Hilbert Transform is only nearly local.

• We can use the Empirical AM/FM decomposition, normalization and quadrature to compute the instantaneous frequency. Then, it is perfectly local.

Summary: The so called paradoxes are really not

problems, once some misconceptions are clarified

• Instantaneous Frequency (IF) has very different meaning than the Fourier frequency.

• IF for special mono-component functions only: IMFs; a necessary but not a sufficient condition.

• Even for IMFs, there are still problems associated with IF through Hilbert Transform. We can rectify most of them with the Normalized HHT.

• The better solution is through quadrature.