VCO gain Calculation stupid question;

Hello,

I have a doupt regarding the VCO Gain calculation. All we knon that the Gain is computed using the relation W=W0+Gv(Wtun2-Wtun1)/(Vtun2-Vtun1) where W0 is the central freq (in rad/s).

This represents the equation of a line and if this line rises Gv is positive else if this line is falling the Gv is negative.

My question is, if the Gv is negative when I calculate the capacitance for the filter of the PLL I obtain a negative value for the Capacitances and resistances.

Have I to use the absolute value of Gv so that the Capacitance value become
positive ?

Re: VCO gain Calculation Tupid question;

Quote: All we knon that the Gain is computed using the relation W=W0+Gv(Wtun2-Wtun1)/(Vtun2-Vtun1) where W0 is the central freq (in rad/s).
This represents the equation of a line and if this line rises Gv is positive else if this line is falling the Gv is negative.

What you call Gv is the control voltage of the VCO. This can be positive or negative. The other part of the equation - the brackets - is the slope of the VCO function.
But realize that the whole equation is not something like a "gain"; it´s simply the VCO equation and gives the relation between control voltage and output frequency.

In addition, you should realize that the sign of the control voltage has no influence on the filter transfer function and, hence, no influence on the sign of capacitors or resistors. If the output of the filter (control voltage) is negative the input is negative as well !

Added after 47 minutes:

Addendum: I think, in most cases the control voltage will be positive and will only be shiftet around the nominal operational (positive) value.

The linear caracteristic of the VCO is rising or falling depends on the PLL's reference and feedback signals connections to the Phase and Freq detector.
One exemple of falling curve is give nby the figure.

When the Gv is negative the phase will have an addition 180 deg I think ? so all will change.
Please correct me if I'm wrong.

The linear caracteristic of the VCO is rising or falling depends on the PLL's reference and feedback signals connections to the Phase and Freq detector.
One exemple of falling curve is give nby the figure.

1.) The characteristic shown in the figure obviously is not linear
2.) I see no "falling curve" . I see a curve frequnecy vs. control voltage.

When the Gv is negative the phase will have an addition 180 deg I think ? so all will change.

To avoid misunderstandings and confusion you should use in your verbal descriptions the same symbols as in the figure (which shows Vc on the horizontal axis),
Why do you use Gv ???

3.) The curve shows a nominal operating point at Vc=+1Volt. More than that, according to the curve there never will be a negative control voltage.
So, what is the problem ?

I'm sorry members, I uploaded the wrong figure.

I'll try to be more accurate in my questions.
1- The curve is linear.
2- Gvco=(w1-w2)/(v2-v1) which is negative

@LvM

Yes your where rigth when you said that the curve is not falling. Thank you.
But I have two remarks:
1- As far as I know, in order to simplify the study of such VCO,, this kind of curves are linearized. And that is quite easy since the errors introduced are very weak.
2- The f=F(Vtune) and w=F(Vtune) are the same coz we can get the one from the other by a simple multiplication/division by (2*pi).
3- Concerning the 3rd point There is no problem.
So we return to my first question.

It would be a good idea to repeat your "first question". What is the problem now ?

use abs(Gvco) to specify the passive components.