Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Whats the zero compensation effect and theory?

Status
Not open for further replies.

huangjw

Member level 2
Joined
May 10, 2005
Messages
53
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Activity points
1,694
about zero

please,the zero compensation's effect,and theory?
 

Re: about zero

Ur doubt is not clear.
 

Re: about zero

in a feedback system,Why we add a zero.what effect of the zero.such as increasing the bandwidth or reponse rate? how does the zero function?
 

Re: about zero

usually adding a zero in order to increase the phase margin of feedback system.
 

Re: about zero

Hi:

If the feedback can creat a zero on the left half plane in the s domain, it can be used to cancel some poles, therefore to reduce the overall phase margin to increase the stability of the circuit.

J
 

Re: about zero

Jenifer_gao said:
Hi:

If the feedback can creat a zero on the left half plane in the s domain, it can be used to cancel some poles, therefore to reduce the overall phase margin to increase the stability of the circuit.

J
hi J,
how can it be "reduce the overall phase margin to increase the stability of the circuit" ?
I think it should be increasing the overall phase margin to increase the stability of the circuit.
 

about zero

hi,
-jw is 90 phase margin,jw is -90 phase margin,so left plane zero is reverse to pole,so it can reduce phase margin.rignt?
like that,can we use a zero cancel the dominate pole to boarden bandwidth
 

Re: about zero

huangjw said:
hi,
-jw is 90 phase margin,jw is -90 phase margin,so left plane zero is reverse to pole,so it can reduce phase margin.rignt?
like that,can we use a zero cancel the dominate pole to boarden bandwidth
Phase Margin (PM)= ΦH(jωu)-(-180°)=180°+ΦH(jωu),
(1)add a LHP zero in H(jw) , u will get PM=180+Φ[H(jwu)*(jwu/ωz+1)]=180+Φ[H(jwu)]+90°, if wu >> wz, where wu is the unity gain bandwidth
(2)add a RHP zero in H(jw) , u will get PM=180+Φ[H(jwu)*(jwu/ωz-1)]=180+Φ[H(jwu)]-90°, if wu >> wz

so (1) will increase PM, (2) will decrease PM
 

    huangjw

    Points: 2
    Helpful Answer Positive Rating
Re: about zero

Btrend said:
huangjw said:
hi,
-jw is 90 phase margin,jw is -90 phase margin,so left plane zero is reverse to pole,so it can reduce phase margin.rignt?
like that,can we use a zero cancel the dominate pole to boarden bandwidth
Phase Margin (PM)= ΦH(jωu)-(-180°)=180°+ΦH(jωu),
(1)add a LHP zero in H(jw) , u will get PM=180+Φ[H(jwu)*(jwu/ωz+1)]=180+Φ[H(jwu)]+90°, if wu >> wz, where wu is the unity gain bandwidth
(2)add a RHP zero in H(jw) , u will get PM=180+Φ[H(jwu)*(jwu/ωz-1)]=180+Φ[H(jwu)]-90°, if wu >> wz

so (1) will increase PM, (2) will decrease PM

I think is's better to use the following way to understand the pole and zero. Considering a transfer function with one zero and pole:

H(s) = Ao(1+s/ωz)/(1+s/ωp), we say this transfer function has a zero at the left half plane, because when the nominator equals to 0, we can get s = -ωz, and we say this transfer function has a pole at the left half plane, because when s = -ωp, the dominator will be 0. But when we calcualte the phase margin, we have to convert this H(s) to the following form:
H(s) = Ao(1+s/ωz)(1-s/ωp)/(1+(s/ωp)^2)
the term of (1+s/ωz) will result in a postive phase, but the term of (1-s/ωp) will lead to a negative phase, so these two can be used to cancel each other. For the right half plane zero, it has the form of (1-s/ωz), which will give a negative phase, therefore will worse the stability. Hope I explain it clearly.

J
 

Re: about zero

zero eliminate pole to get pm
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top