signal analysis - Morlet wavelet time and frequency resolution

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

Answer

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

$$\Psi_{a,b}(\omega)=\sqrt{a}e^{-ib\omega}\Psi(a\omega)\tag{2}$$

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.