PS:


This is the curve of a series oscillation circuit:


This is the curve of a parrael resonat circuit:

 

So the peak of the magnitue curve and the falling of frequency curve must be the resonance frequency of my LC circuit right?

Yes, for this peak in the magnitude of a transfer function. The L1C1 parallel resonance circuit will take responsibility.

As far as I could see this circiut is a low pass filter. Because low frequncy can pass, and high frequceny is going be filtred. But please correct me if I am wrong.

This is not a low pass filter for sure.

In other word. What kind of a problem is resulting with that frequency for my circuit?

Well, the resonance is the problem. Simply plug the s = i/√(L1 C1) into the transfer function and should be able to see the problem.

It seems to me that Your main problem is a lack of understanding of basic circuit theory and basic op-amp circuits.

 

Good morning,

I searched in the internet. And now I have more inforamtions about this.

First of all the this is a lossless or a perfect resonat circiut which is parallel. (L1 || C1)

f < f0 predominantly inductive resistance
f > f0 predominantly capacitive resistance

f=f0 resonat frequency, facts:

|z|= ∞, (resistant is going to be ∞ )

Therefore sum of the current is also 0 so the current flow is I=0. That means,with that frequency we have no more current flow from voltage source in our circiut. But we our saved current in L and C.

So the final answer to this questions above is:

The problem with that frequency is, that we have no more current flow in our circiut.

I'm not sure about this. Maybe some can confirm that.

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When I plug the frequency s=i/sqrt(L1C1) in my transfer function:
\[ G(s)=-\frac{sL_1}{\frac{1+s^2L_1C_1}{sC_2}+R_2+s^2L_1C_1R_2} \ \]


As reslut:

\[ G(\frac{j}{\sqrt(L_1C_1})=-\frac{jL_1}{\sqrt(L_1C_1} \]

Or in other words, what is now the consultion ?

And after I inserted this frequency in G(s) I get a result. And I cant see there any problem.

Or straight talking:

What kind for an information does I have with the expression, after I inserte the frequency f0 in to G(s) ?

 

Good morning. For me, without doing any math I see a problem at s = i/√(L1 C1) due to the parallel resonance LC tank circuit and that the z of a tank circuit goes to infinity. Thus we have a brake in a negative feedback loop. So, if plug this L1 = 1H, C1 = 1F, C2 = 1F, R2 = 1Ω

We have s = j/√(L1 C1) = j/ √(1) = j

And now back to the transfer function:

G(s) = (1 s^2)/(1 + 1 s + 1 s^2 + 1 s^3) = (1 j^2)/(1 + 1 j + 1 j^2 + 1 j^3) = -1/(1 + j + (-1) + (-j) ) = ?

Do you see the problem?

 

Yes, if we put the frequency f0 into G(s) -> G(s)=-1

But I cant see a problem. But I know that z is going to infinity. And G(s) = -1 tells me the same ?

So we have no current flow ?

Or does we have a negativ current in our negative feedback actions ?

 

Ohh shit Jonny, I saw it. The problem is, that the reinforcement is going to infinity. That is not possible, in the real life our circurt will burn like fire. Mathematicly is also now allowed to divide vale to zero.

\[ G(s)=\frac{-1}{0} \]

And this expression is not deffined. Please correct me Jony.

 

And the next questions are:

1.) What what is the behavoir of G(S) our transfer function in S-> 0 and S- infinity
2.) The time constants T1 from L1, C1 and T2 from C2, R2 are the same size and the capacitances C1 = C2 = C are also the same. What does this mean for the inductance L1?

 

PUSH- Does no one got an idea ?

 

Does no one got an idea ?

About what?

 

@ Jony130

This questions:

1.) What is the behavoir of G(S) the transfer function for S-> 0 and S- infinity ?
2.) The time constants T1 from L1, C1 and T2 from C2, R2 are the same size and the capacitances C1 = C2 = C are also the same. What does this mean for the inductance L1?

 

1.) What is the behavoir of G(S) the transfer function for S-> 0 and S- infinity ?

Why don't you just "plug" this into the transfer function and see by yourself what you get?

2.) The time constants T1 from L1, C1 and T2 from C2, R2 are the same size and the capacitances C1 = C2 = C are also the same. What does this mean for the inductance L1?

So you want to calculate the L1 value based on the circuit time constant?

 

Why don't you just "plug" this into the transfer function and see by yourself what you get?

Yes I know but I was confused because of an other post of you. (I'm not lazy, I'm just unsure about.)

But we can answer this just by looking at your circuit. We don't need to know the transfer function.

For s -> 0 we have Xc = ∞, XL = 0Ω

and for s ->∞ we have Xc = 0Ω, XL = ∞

And when I insert s- > 0 and s - infinity I get something which doesn't compare, with that you said here.

So you want to calculate the L1 value based on the circuit time constant?

Yes you got it. This is what I want to calculate.

 

And when I insert s- > 0 and s - infinity I get something which doesn't compare, with that you said here.

Show me what doesn't compare with what I said

Yes you got it. This is what I want to calculate.

So, what is the problem?

 

Show me what doesn't compare with what I said

\[ G(s)=-\frac{sL_1}{\frac{1+s^2L_1C_1}{sC_2}+R_2+s^2L_1C_1R_2} \ \]

You told me that the transfer function is ok. And so I belived you. And If we put now 0 in it for example.
\[ G(0)=\frac{0}{ \frac{1}{0} +R_2+0} \]

So, what is the problem?

I don't know where to start. Because I don't have real values. That makes it complicated for me.


PS: Sry for the headache

 

If you don't know how to take "limits" in math. Then try to analyze these two equivalent circuits:






And this one

I don't know where to start. Because I don't have real values. That makes it complicated for me.

Do the math

\[ \tau_1 = \sqrt{L_1 C_1} \]

\[\tau_2 = R_2C_2 \]

 

If you don't know how to take "limits" in math. Then try to analyze these two equivalent circuits

\[ \tau_1 = \sqrt{L_1 C_1} \]

\[\tau_2 = R_2C_2 \]

Ohh I guess, I lost my head yesterday. Thanks a lot for your help Jony130. You helped me a lot. And please excuse me for amount of headaches because of me. Now I have no more question about this.