That is not a simple question, it is a mind reading exercise. You have not said where that code comes from or what any of the variables are or how it relates to the “reson” filter.
Surprisingly, my crystal ball is working quite well today and I see that you are referring to this code:

;control f "medium frequency" int "[Hz]" 1000 100 4000
;control fl "lfo frequency" real "[Hz]" 1 0.1 5
(reson s (sum f (mult f 0.5 (lfo fl)))(/ f 2.0) 2)

In future, please do not expect my crystal ball to work. Please give the full context of your question (as above) so that other people on the forum can understand what you are asking.

I thank steve for the answer.
f corresponds to the “center”. But, here is a code:

(reson (noise) 100 20 2)

In this code the “center” correspond to 100, the “bandwith” to 20 and “n” to 2. It is right?
Thus, in the complex code, it is the complicated operation [100 + (100 X 0.5) x 2] ( if, for example “f” = 100 and if “fl” = 2) which gives the value of the “center”.
Thus, my question is the following one: why to note in a so complicated way the value of the “center”?
I hope that my question is understandable.

;control f "medium frequency" int "[Hz]" 1000 100 4000
;control fl "lfo frequency" real "[Hz]" 1 0.1 5
(reson s (sum f (mult f 0.5 (lfo fl)))(/ f 2.0) 2)

The centre of this reson filter is defined by (sum f (mult f 0.5 (lfo fl)))
Note the expression (lfo fl) http://www.cs.cmu.edu/~rbd/doc/nyquist/part8.html#index337
(lfo fl) is a low frequency oscillator. It produces a low frequency sine wave, the frequency of which is set by “fl”.
The amplitude of lfo is between 1.0 and -1.0
Try running this:

(force-srate *sound-srate*
(lfo 3))

LFO generates a sine wave at 3 Hz. FORCE-SRATE is used to force the sample rate of the sine wave to match the audio track sample rate (LFO would normally generate the sine wave at the “control rate”, which is 1/20th of the sound sample rate.

So, let’s say that “f” = 100 and “fl” = 1.0
(lfo fl) will produce a sine wave, amplitude +/- 1.0 with a frequency of 1.0 Hz.
(mult 100 0.5 (lfo fl)) will produce a sine wave with an amplitude +/- 50.0
(sum 100 (mult 100 0.5 (lfo fl))) will produce a sine wave with an amplitude +50 to +150.
Thus we have a control signal that rises and falls at 1 cycle per second (1 Hz) between +50 and +150 (the average value is thus 100, which is the value “f”).
We then use this control signal to set the centre frequency of the “reson” filter. So the filter does not have a fixed frequency, but has a frequency that rises and falls cyclically at a rate set by “fl” and with an average centre frequency of “f”.

Hello.
I have a question to ask. it concerns the value of the “center” of the filter. The value (amplitude) between + 50 and + 150.
These values (+ 50, + 150) correspond to radians?

.It produces a sine wave with an amplitude which oscillates between +50 and +150.I wanted to know if this amplitude which oscillates between 50 and 150 corresponds has a value in radians?
I hoped that my question is understandable.

That code on its own will only produce an error.
If you add a “)” to the end, it will generate a sine wave with a frequency of 1 Hz, a peak-to-peak amplitude of 50 and a DC off-set of 100 at the control sample rate.

Thank you.
I try to understand. Generally, DC offset is a problem.
My question is the following one: in this precise code, as it was able to choose a DC offset of “100” as the following reason: the amplitude is very exactly equal to “50”.
Beyond this value (“50”), there would be a problem.