How to remove/reduce noise by using the Morlet wavelet?

Question

I'm trying to use Morlet wavelet to reduce noise in my signal. I found formula for Morlet wavelet

$$\Psi(x) = \frac{1}{\sqrt{\pi\cdot \textrm{bandwidth}}} \cdot \exp\left(i \cdot 2\pi \cdot \textrm{centerFreq} \cdot x\right)\cdot\exp\left(\frac{x^2}{\textrm{bandwidth}}\right)$$

(sb told me I shoud use formula for complex Morlet wavelet)

I also got CWT formula

$$C = \int f(t) \cdot \frac{1}{\sqrt{\textrm{scale}}} \cdot \Psi\left(\frac{t-\textrm{shift}}{\textrm{scale}}\right)dt$$

I convert both formulas to java code

public double morletRe(double x)
{
return (1/Math.sqrt(Math.PI*fb))*Math.cos(2*Math.PI*fc*x)* Math.exp(x*x/fb); 
}

public double[] cwt(double[] data,double scale, double position)
{
double[] newData = new double[data.length];
    for (int t = 0; t<data.length; t++ )
    {
    newData[t] = data[t]*(morletRe(t-position)/scale)[0])/Math.sqrt(scale);
    }
return newData;
}

But in the results I get a strange chart which is completely not related to my data, is anyone can point me what I'm doing wrong? regards.

Answer 1

I'm not sure of the reliability of your source... I worked with wavelets on my master's thesis and the basic wavelet transform is a convolution of the wavelet an your signal. In most practical cases, it's best to go through Fast Fourier Transform (FFT), as it's mathematically equivalent.

With WT being the wavelet transform, and FFT() being the Fast Fourier Transform of what's inside the brackets, iFFT() is the inverse-FFT. You would compute the WT as such:

FFT(WT) = FFT(signal) * FFT(wavelet)
WT = iFFT(WT)

Now, for the wavelet formula itself, I believe there has to be a specific scaling factor to get a reliable result. Here's what we used:

$$\Psi = exp\Big(-\frac{ik_0x}{\lambda} - \frac{x^2}{2\lambda^2}\Big)$$

Where $\lambda$ is the wavelength (or scale) and $k_0 = 5.336$ (source: J.F. Kirby, Computers & Geosciences 31 (2005) 846–864) I did not include the normalization factor because you just have to divide your wavelet by it's highest value, and subtract its average, and you get the same result with less headhache. Off course, you can interchange $1/\lambda$ by $2\pi*frequency$ in the formula.

In order to qualify for the "Morlet" name, I think you have to use the same wavelength in both places in the formula. Else, you over-sample or under-sample the frequency part of your signal. $k_0$ is the best ratio between the two, as explained by Kirby in his article.