transfer function - Roll-off of a filter

I know that the roll-off of a transfer function gain is given by:

$$20\log_{10}\left(\left|\text{H}\left(\omega_2\right)\right|\right)-20\log_{10}\left(\left|\text{H}\left(\omega_1\right)\right|\right)\tag1$$

And when I have a three pole transfer function, that the roll-off tends to \$-60\space\text{dB}/\text{decade}\$ when \$f\to\infty\$.

Question: In order to calculate the roll-off accurately I must choose \$\omega_2\$ and \$\omega_1\$ much larger than the cut-off frequency. But how much larger?

Answer

Since you are relying on the fact that for \$f \to \infty\$ the slope of the gain function is -60dB/dec, you should be aware that that approximation is valid, at finite frequencies, when you are far away from poles cut-off frequencies.

So you must choose ω1 and ω2 much higher than the highest cut-off frequency of your poles.

How higher? This depends on the precision needed in the calculations. In general, at least one decade higher is the minimum to get reasonable results. With ω2 a decade higher than ω1.

Of course this is a very rough rule of thumb, applicable in general. For a specific gain function H you could calculate exactly the position of those two frequencies so as to achieve the precision you need in your roll-off value.

If you don't want to cope with difficult calculations when dealing with a generic H function, and you don't need more than 2 significant digit accuracy in your roll-off value (rarely needed in most applications), I'd suggest to go for this values:

$$ \omega_2=10\cdot\omega_1=100\cdot\omega_H $$

where \$\omega_H\$ is the highest cut-off frequency among all the poles in H.

BTW, all the preceding discussion assumes you have no zeroes in your gain function, otherwise the roll-off at infinity will depend also on how many zeroes your function has.