Harald, perhaps the following applies only to me - however: To me, this sounds as if a so called "rail-to-rail opamp" would be an alternative solution to single-supply operation. I am sure, that´s not what you wanted to express, correct?
In short:
Each opamp needs a voltage between the power pins in order to work properly - that means: To be able to amplify in its linear transfer characteristics.
1.) Then, the best quiescent operating point (DC output voltage) is in the middle of this region. Then, we have equal room (above and...
You have mentioned to use split supply. In this case, you do NOT need a virtual ground node.
Instead, use the pos. supply voltage between the corresponding opamp pin and ground - and do the same with the negative supply voltage.
In this case, the ouput voltages of the opamps will be centered at...
Fish - I fully agree with you. My answer concerns only the point of "changing the phase of an oscillatory signal" - nothing else.
If such an approach can be a starting point for any noise cancellation system is outside of my region of experience.
All you need to do is slowly tune up and down the frequency of your signal generator. This will change the phase - if compared with any external signal. However - this is, more or less, a trial-and-error procedure because the amount of phase changed is unknown.
Hi Ratch, I didn`t check your calculation - and I was too lazy for calculating by myself.
However, apparently there is a conflict beween your result (C*Rc²>1) and my simulation.
As I have mentioned above, according to simulation (symbolic analyzer as well as PSpice) the imaginary part of the...
Perhaps it is interesting to numerically compare the various alternatives (definitions):
Example circuit:
Rc||C=1kohm||1nF
Rs=250 Ohms; L=10uH.
Results (Simulation):
|Imax| at 1.76 MHz
IMG(I)=0 at 1.583 MHz
|Vout,max| at 2.78 MHz.
Yes, that is another possible - perhaps the best(?) - criterion for defining resonance. It is intersting to note (prooved by simulation) that the peak of the total current through the network is NOT identical with the frequency where the current phase is zero (no reactive component).
As I have mentioned - it depends on the definition (max. current or max. output voltage).
The current (magnitude) through the whole circuit has a maximum at a certain the frequency f1, but this does NOT mean that the magnitude of the voltage drop across Rs+L has a maximum at the same frequency...
So you have tried different concepts which didn`t work satisfactorily (for "some reason") - and you ask us for alternatives.
Obviously, it is not easy to answer without knowing what you have done already.
I don`t think that this is true for the shown circuit. The (different) peaks of the current and output voltage depend on the R values. Therefore my question how the resonance point is defined for this circuit.
Yes - each R-C lowpass charges UP to 63.2% of the final value within a time t=T=RC.
However, you were asking for a discharging process!
It is even possible (without calculating) to make a good estimate of the result.
Ask yourself what the output - according to the following equation - will be for a discharging time t approaching infinite. Is the result OK?
Vc / V = 1 - e^(-t/T)