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


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?


2 Answers 2


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.


Consider a simple 2nd order low pass filter made from a resistor in series with an inductor that is in series with a capacitor. The voltage output taken across the capacitor. If you vary the resistor value you could make the roll-off slope converge on the perfect (at f = infinity) value much much earlier: -

enter image description here

So, the answer to your question is that it depends entirely on the nature of the filter and how much accuracy you are prepared to tolerate. The resistor values for the above simulation varied from 1 ohm to 128 ohm in powers of 2. I would say that when R = 64 ohms (2nd blue curve from the bottom) you have a nearly perfect roll-off slope starting at quite close to \$\omega_n\$.

With lower resistor values you have a steeper initial roll-off and with higher value resistors you have a shallower roll-off. It all depends on your filter and the more complex it is, the more complexity there is in fomulating a numerical answer.

  • \$\begingroup\$ Andy aka - I know it`s perhaps only semantics or a matter of definition - however, for my opinion, the term "roll-off" is restricted to the magnitude response for frequencies very far from the pole frequenvy only. Therefore, all 2nd-order low pass responses - independent on the Q value - have a roll-off of -40db/Dec. \$\endgroup\$
    – LvW
    Apr 2, 2018 at 16:45
  • \$\begingroup\$ @LvW I'm unsure about the point you are making. \$\endgroup\$
    – Andy aka
    Apr 2, 2018 at 17:24
  • \$\begingroup\$ OK - my fault. You spoke about a "steeper roll-off" for lower resistor values. This touches the quality factor Q of the pole (the peaking characteristic around the pole frequency). For my opinion, in this context (peaking) we must not use the term "roll-off" because this term depends on the filter order only (far above the pole frequency) - as can be seen in your graph for f>700kHz. \$\endgroup\$
    – LvW
    Apr 2, 2018 at 17:59
  • \$\begingroup\$ I might be wrong, but think what LvW refers to is the selectivity, the slope of the attenuation near fc. For example, a Chebyshev (type I) filter will have a higher selectivity the larger the ripples. \$\endgroup\$ Apr 14, 2018 at 8:30

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