I want to calculate Morlet time and frequency resolution. the Morlet wavelet function is define as : $\psi(t)=\frac{1}{\sqrt{\pi f_b}}e^{j2\pi f_c}e^{-t^2/f_b}$
Note: I know the answers, but I don't know how to achieve it on my own.I want someone to explain the time and frequency resolution relation to me please. here are some hint that I got from paper and might help you.
in some paper I found that Morlet wavelet time and frequency resolution is : $\Delta t=\frac{f_c\sqrt{f_b}}{2f_i}$ $\Delta f=\frac{1}{2\pi f_c \sqrt{f_b}}$
Because generally, the wavelet time and frequency resolution is define as:
$\Delta t=s \Delta t_\psi$ and $\Delta f= \frac {\Delta f_\psi}{s}$ (1)
where 's' is scale. it is also evident that the frequency and scale are related to each other by $f_i=\frac{fc}{s_i}$ . $\Delta t$ and $\Delta f$ are wavelet time and frequency resolution,respectively. $\Delta t_\psi$ and $\Delta f_\psi$ are morlet function time and frequency resolution which are as follow: $\Delta t_\psi=\frac{\sqrt{f_b}}{2}$ and $\Delta f_\psi=\frac{1}{2\pi \sqrt{f_b}}$ (2)
the Heisenberg uncertainty principle also says: $\Delta t \Delta f >= \frac{1}{4\pi}$
Basically i want to prove (1) and (2). other things can be obtained by substituting. Thanks


2 Answers 2


Consider shifted and scaled versions of a mother wavelet $\psi(t)$:

$$\psi_{a,b}(t)=\frac{1}{\sqrt{a}}\psi\left(\frac{t-b}{a}\right),\quad a>0,\;b\in\mathbb{R}\tag{1}$$

By the definition of the Fourier transform

$$\Psi(\omega)=\int_{-\infty}^{\infty}\psi(t)e^{-i\omega t}dt$$

it can be shown that the Fourier transform of $\psi_{a,b}(t)$ is


Note that the scaling factor $a$ appears in the denominator of the argument of $\psi_{a,b}(t)$ and in the "numerator" of the argument of $\Psi_{a,b}(\omega)$. This implies that stretching in the time domain ($a>1$) implies compression in the frequency domain and vice versa. Consequently, if $\Delta t$ and $\Delta f$ denote the "spread" of $\psi(t)$ in time and frequency, it follows from (1) and (2) that the corresponding spreads of $\psi_{a,b}(t)$ - call them $\Delta t_{a}$ and $\Delta f_{a}$ - are given by

$$\Delta t_a=a\Delta t\quad\textrm{and}\quad\Delta f_{a}=\frac{\Delta f}{a}$$

As for the definition of $\Delta t$ and $\Delta f$, there is not way to "prove" the results that you stated. It is just a matter of defining the spread of a function. Since $|\psi(t)|$ is a Gaussian with standard deviation $\sigma=\sqrt{f_b/2}$ it seems natural to define $\Delta t$ as being proportional to the standard deviation, but there is no "correct" way to choose the proportionality constant. Once $\Delta t$ has been defined, the corresponding definition of $\Delta f$ can be obtained from the uncertainty principle.


Consider a wavelet to be 1 bin of a windowed DFT or FT. Scale is roughly proportional to how many cycles of a sine wave is inside the bulk of the wavelet window.

Hold the frequency constant and make the wavelet twice as long, and you have to move a wavelet twice as far before some event goes from centered in the wavelet to outside the bulk of the window. Thus time resolution decreases with window width.

A sinusoid the same frequency as a wavelet correlates well. Change the frequency such that the number of periods of sine wave inside the window bulk differs by one period per window bulk width, and that sinusoid become roughly orthogonal, or correlates poorly. Increase the window width, and the frequency change required to make the number of periods of the same frequency within that wider window differ by one period become smaller. Thus frequency resolution increases with window width.

  • $\begingroup$ Thanks @hotpaw2 I need mathematical proof. I almost find something that proves relationships in (2), they used standard deviation and RMS things to find effective width of Gaussian window which is $\frac{\sqrt{f_b}}{2}$ then they applied HUP to find $\Delta f_\psi$.The mathematics of (2) was roughly hard. but never mind, now I only need a proof for (1) $\endgroup$
    – SAH
    Commented Apr 12, 2014 at 5:04

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