Questions about Poles and Zero

[anil555]

Hi all,

I guess i have a very foolish question.

We have all learned that poles and zeros are contributed by capacitor either parasitic or external. Poles tend to increase the gain to infinity and zero to 0.
I have learned that in most cases the output load capacitance , or in the case of Op-amp the capacitance at the output of first stage, is the dominant pole.

The capacitance impedance decreases as the frequency increases. Suppose F0 be the frequency of the dominant pole. when f > F0, the dominant capacitance tend to decrease its impedance, hence the quantum of voltage dropping across it reduces.

Hence according to the above explanation the gain should decrease and that's what happens when u see the Bode plot.

Then why is that poles are defined as those which increase the gain to infinity.


Please correct me.


Thanks

ARJ

 

[naalald]

Re: Poles and Zero

Hi anil555,
In the bode plot we have the magnitude of the transfer function. So for a zero this magnitude increases as the frequency increases, and for a pole this magnitude decreases as the frequency decreases.
For example for a pole we have this:

 

[maninnet]

Poles and Zero

every once in a while i get stuck on this question too. for example, consider 1/(1+s), the pole is s=jw=-1, or w=j,which is on imagniary axis, but in bode plot, the pole frequency is w=1, which we call 3dB frequency, then how do we related the pole w=1 in bode plot to the pole w=j from 1/(1+s)???. And w=1 in bode plot cause gain drop but not infinite gain, but w=j do cause gain to be infinite, when s=-1, i just related it to an opamp configured as an follower and expressed by 1/(1+s), so at for s=-1, i just thought its magnitude is 1, and phase is 180, so just cancel the 1 in 1/(1+s) and the follower oscilliates and goes to infinite if not limit by power supply. hope someone can clarify this

 

[naalald]

Re: Poles and Zero

Hi maninnet,
In the bode plot we plot the magnitude of the transfer function just for the real values of w, and for real w the term w²+p² never equals zero.

 

[anil555]

Re: Poles and Zero

Thanks naalald,

But i am still confused.

In op-amp we use pole splitting to split dominant and non dominant poles. If thats not done then ringing or oscillation can be seen in the output. I presume the reason as gain tending to infinity.

In this case the pole actually makes the output go to infinity. I dont see any feedback here because a Two stage Op-Amp doesnt have any feedback.

Correct me if i am wrong.

ARJ

 

[lurx]

Re: Poles and Zero

s=σ+jω
When we calculate frequency response, let σ=0 (just on imaginary axis on s plane)
H(s)=1/(s+1) => |H(jω)|= |1/(1+jω)|
s=-1, H(-1)=>∞ , Left Half Plane pole
ω=1, |H(jω)|= 1/√2 , 3dB frequency

 

[maninnet]

Poles and Zero

to lurx:
you maths of course makes sense, but the question is: pole is s=jw=-1, or w=j, but why w=1 on bode plot is still called "pole location is w=1", which it just cause gain drops not to infinity?

for example: 1/(s+1)/(s+10)/(s+100), pole are w=j,10j, 100j, but on bode plot, pole are w=1, 10,100, how should we realted them, they are just equal in magnitude, but the so called "pole frequency" on bode plot only cause gain drop, not infinity, then why it is still call "pole"?

 

[sutapanaki]

Re: Poles and Zero

Consider, as I'm sure you all know, that the poles and zeros apply to the Laplace transformed transfer function. That is to say that poles and zeros are in the s-plane in general and only in special cases on the jw axis. When you talk about real circuits and their Bode plot you talk about what happens on the jw axis only. In the first post anil555 says that for f>>Fo the gain drops to 0, which is true, but this doesn't mean that there is a zero at Fo. There is a pole there, the frequency response breaks downwards but it is zero only at infinity, so it is accurate to say that we have a transfer function zero at infinite frequency.
For the case of 1/(1+s) we have a pole at s=-1, which is in the s-plane and on the negative real axis, not on the imaginary jw axis where we evaluate the frequency response of the circuit. Obviously 1/(1+s) defines some surface in the 3-dimensional coordinate system for which the s-plane is just two of the dimensions (real and imaginary part of s are the two variables for the surface). When we intersect that surface with a plane perpendicular to the s plane and passing through the jw axis we get the frequency response. So, Fourier transform is just a special case of the Laplace transform. Anyway, when s=jw 1/(1+s) becomes a complex function 1/(1+jw) which has a phase and magnitude and the magnitude is 1/sqrt(1+w^2) and at w=+-1 we have the magnitude at 0.707 of its original value at DC which by definition is a pole. At infinite frequency we have a zero as I explained above.
However, 1/(1+s^2) has poles on the imaginary axis and and they are +-j1. So at w=+-1 the Laplace transfer function and the frequency transfer function go to infinity, so again we have a pole.
The function (s+2)/(1+s^2) has the same imaginary axis poles but also a zero at s=-2, so this function goes to 0 at this point on the real axis and to infinity on the imaginary axis where we look at the frequency response. On that axis we only see the effect of the zero caused by the 3-dimensional surface being at 0 on the real axis, but the function derived by the intersection of that surface along the imaginary axis (the frequency response) doesn't have to be zero.
Hope all this is clear.

 

[anil555]

Re: Poles and Zero

Thanks all of you for your response

I got a good link which cleared lot of my doubts.

**broken link removed**