A interesting problems about analysis of the amplifier with negtive feedback

[lily1981216]

Hello, everybody! I wanna ask a question about the stability about the negative feedback system, shown as the figure.


As the book of "Signals and Systems" said, if all the poles of a system transfer function are in the left part of the S-plane, the system is steady. So can I judge the stability of the system in the figure just by analyzing the poles of the closed-loop transfer function? a(s)/(1+a(s)*f)

 

[LvW]

As the book of "Signals and Systems" said, if all the poles of a system transfer function are in the left part of the S-plane, the system is steady. So can I judge the stability of the system in the figure just by analyzing the poles of the closed-loop transfer function? a(s)/(1+a(s)*f)

Yes, correct.
However, as you probably know, in most cases it is much easier to evaluate the loop gain (transfer function -a*f) for a stability check using the Nyquist criterion.

 

[lily1981216]

Yes, correct.
However, as you probably know, in most cases it is much easier to evaluate the loop gain (transfer function -a*f) for a stability check using the Nyquist criterion.

Thanks for replying! However, is it easier to evaluate the loop gain for a stability check? I think the loop gain f need caculating. If the f isn't a constant but a variable of frequency, that's not easier than caculating the transfer function of the closed loop. And another problem is that when i calculated the transfer function of the a(s)/(1+a(s)*f), supposing the transfer function is a single pole transfer function, I found the criterion to make the poles in the left part dosen't equal to the loop gain criterion (mag(a(w0)f)=1,arg(a(w0)f)<180degree).Why?

 

[LvW]

However, is it easier to evaluate the loop gain for a stability check? I think the loop gain f need caculating.

The loop gain is not simply "f" but LG=-a(s)*f(s).

If the f isn't a constant but a variable of frequency, that's not easier than caculating the transfer function of the closed loop.

For my opinion, it is easier to calculate resp. assess a function LG rather than H(s)=a(s)/(1-LG(s)).
More than that, in addition you have to find the roots if you want to check stability. Therefore, Nyquist/Strecker have invented the versatile method of loop gain evaluation which means: Determination of closed-loop properties by means of open-loop evaluation.

And another problem is that when i calculated the transfer function of the a(s)/(1+a(s)*f), supposing the transfer function is a single pole transfer function, I found the criterion to make the poles in the left part dosen't equal to the loop gain criterion (mag(a(w0)f)=1,arg(a(w0)f)<180degree).Why?

I don't understand your question. From mathematics, a single pole function has a maximum phase shift of -90 deg.
What do you mean with "why"?
LvW

 

[lily1981216]

For example, the pole of a(s)/(1+a(s)*f) is p1(1+a0*f) and p1 is negative. If the pole is in the left part of the s-plane, a0*f should be bigger than -1. So the criterion of system stability is a0*f>1. Do you think that equals to the loop gain criterion? Or there is something wrong with my derivation. Where is it ? I'm confused!

---------- Post added at 10:42 ---------- Previous post was at 10:35 ----------

And another problem is that when i calculated the transfer function of the a(s)/(1+a(s)*f), supposing the transfer function is a single pole transfer function, I found the criterion to make the poles in the left part dosen't equal to the loop gain criterion (mag(a(w0)f)=1,arg(a(w0)f)<180degree).Why?

I don't understand your question. From mathematics, a single pole function has a maximum phase shift of -90 deg.
What do you mean with "why"?
LvW[/QUOTE]

I'm sorry! The a(s ) is a single pole transfer function!

 

[LvW]

For example, the pole of a(s)/(1+a(s)*f) is p1(1+a0*f) and p1 is negative. If the pole is in the left part of the s-plane, a0*f should be bigger than -1. So the criterion of system stability is a0*f>1. Do you think that equals to the loop gain criterion? Or there is something wrong with my derivation. Where is it ? I'm confused!
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Me too.
Sorry, but I don't understand your "explanation". How can you compute poles without knowing the function a(s)?
What is p1 and what is a0 ?

 

[lily1981216]

Me too.
Sorry, but I don't understand your "explanation". How can you compute poles without knowing the function a(s)?
What is p1 and what is a0 ?

I'm so sorry for not giving all the conditions. And a(s) is the transfer function of the amplifier in the figure. It is supposed to be a single pole transfer function which is a(s)=a0/(1-s/p1) . p1 is the pole of the amplifier and a0 is the DC gain of the amplifier. And thank you very much for taking time to discuss that with me.

 

[LvW]

And a(s) is the transfer function of the amplifier in the figure.
It is supposed to be a single pole transfer function which is a(s)=a0/(1-s/p1) .


At first, I suppose you mean (1+s/p1) in the denumerator, don't you? Otherwise, the basic amplifier is unstable.
Secondly, what is now the problem? Stability if feedback is applied?
In this case, you need to know the frequency dependence of the feedback factor f(s).
If f(s) is a constant the resulting circuit is again a lowpass of first order - of course with another gain and another corner frequency.