# How to narrow the frequency band of a wavelet

I have a ricker wavelet with a dominant frequency of 15 Hz. The fourier transform shows its frequency band is almost to 50 Hz. How can I narrow the frequency band of this 15 Hz ricker wavelet? I have added a picture of the wavelet and its fftshifted frequency band (The x axis is not the frequency samples).

• Why don't you increase $\sigma$? Or am I missing something? – Matt L. Jan 11 '15 at 13:30
• @MattL.I don't understand, what do you mean by $\sigma$ ? – user3482383 Jan 11 '15 at 13:34
• I'm referring to the definition of the Ricker wavelet as expressed here. – Matt L. Jan 11 '15 at 13:35
• @MattL. does it change the frequency band or dominant frequency? – user3482383 Jan 11 '15 at 13:44
• A larger $\sigma$ makes the wavelet wider in the time domain and narrower in the frequency domain. – Matt L. Jan 11 '15 at 13:55

## 1 Answer

Center frequency and bandwidth are linked in a single "scale" parameter for wavelets. That means if you change the bandwidth you also change the center frequency and vice versa.

For the discrete wavelet transform on a dyadic lattice the relative bandwidth is also fixed at approximately one octave.

If you want a continuous wavelet transform frame with a smaller relative bandwidth you need to construct a different wavelet. Intuitively, it will come with more oscillatory cycles, i.e. more zero crossings.

The Ricker wavelet you have coincides with the 2nd so called Hermite function. Higher order Hermite functions also have an increasing number of zero crossings. Unfortunately they're one of the examples of where the intuition of adding more cycles will not work. The Hermite functions grow in both frequency as well as time support with increasing order. So that's not the way to go. I'm only making this clear because you might find a reference to them in relation to the Ricker Wavelet.

So instead I would try to adapt the construction of the wavelet to your specific needs, which you will have to outline in your question to receive a more detailed answer.