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RHHP and complex pole

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ee484

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Hi, all.

When does right half plane pole in analog circuit?


When does a complex pole occur in analog circuit?

For example, passive components cannot give complex poles. In order to have complex pole, what mechanism should be associated with?

Any SIMPLE, but intuitive examples or materials are very welcome!!

Please give me your wisdom!!


B - Thanks in advance!
 

I have the same doubt about the RHP and complex pole...
One more question,why the complex pole are always in pairs and locate in sym. about the real axis?I mean why there are no triple or higher roots for the poles,like the roots of s^3=1,so there are three poles equally distributed on the unit circle,but not sym to the real axis...
 

Hi

Passive circuits consisting of inductors and capacitors can have complex poles.

Try to make a transfer function of a standard parallel circuit consisting of an inductor, capacitor and resistor. Then with different values you can see how the poles and zeroes change.

regards
 

RHPP will always lead to unstable system because its solution will have a positive real that means infinity.
Yes, passive circuit with an inductor and a cap will contribute complex poles. They will degrade phase quickly.
 

so only resistor and capasitor(or inductor) wont contribute complex pole,right?

and how to calculate complex pole's contribution in bode plot?It seems that the complex pole's mag and phase are much more complex than real pole?
 

Hi

Yes it is right, a resistor and a capacitor can not make a complex pole. You need at least two reactive elements.

Actually the complex work in the way as "real" poles. But they always come in conjugate pair. So instead of -20dB/decade, they give -40dB/decade(conjugate pair). The pole frequency is simply the distance from the center of the origin, which is the magnitude of the complex pole.

The phase works in the same manner, giving you +/- 90 degrees(instead of 45 degrees) one decade up and down from the pole frequency.

Hope this helps.
 

I think complex poles will lead to 180 degree phase shift instead of 90 degree.
 

leo_o2 said:
I think complex poles will lead to 180 degree phase shift instead of 90 degree.
Click to expand...

would u plz explain why ?
 

tyassin said:
Hi

The complex poles will start at -90 degree phase shift at the pole frequency. Then the asymtotic behaivior will reach 0 degrees at lower frequencies. It will reach 180 degree at higher frequencies.

This link explains something: https://www.swarthmore.edu/NatSci/e...odeHow.html#A Complex Conjugate Pair of Poles

Regards
Click to expand...

Hi

for a negative real pole A,it is pole frequency at bode plot is A,this is obvious,but for a conjugate pole pair，say A+Bi and A-Bi,what's their pole frequency? I suppose it is sqrt(A^2+B^2),but the mag decay and phase shift are not 6dB and -90degree there,actually it varies with different values of A and B, it really confuse me....
 

Hi, all.

Thank you all for your reply.

Let me clear my question...

How does RHHP occur with capacitor and active device (e.g. MOS) - without inductor!?

How does a complex pole pair occur with capacitor with active devices - again without inductor?

Of course, I can derive all tedios transfer function to find out RHHP and complex pole pair. However, do you have any intuitive explanation or a way to find those?


BTW, I can answer some of question in the reply. Let me know if my explanations are wrong.

1) Why always complex pole are in pair?
--> This is because a real number that has frequency associated with (e.g. sinusoid) in complex plane in math should be "in pair". For example, to represent 3 in real number (for example, this number 3 is coming from sinusoidal function, meaning associated with frequency), we can write 3+jb. To represent this number in complex plane, there should be 3-jb. With these two numbers 1/2(3+jb + 3-jb) = 3. In conclusion, to represent a real number in complex domain requires to have complex conjugate pair!!!
(NOTE) if you are dealing with imaginary number, yes, a complex number can exit alone, not in a pair.

2) Inductor with capacitor = two reactive components will give you complex poles.
Associating with 1), if you have real value of L and C, you will have a complex pole IN pair. The poles cannot be alone.
However, with R and C only....you will have only real poles (I am not 100% sure, but very sure)
For example, if you have two RC lowpass filter cascaded with no buffer between and find the transfer function, you can make sure that two poles must be real and all in the LHP.

3) tyassin answered complex pole gives us -40dB/dec in magnitude and -90degree/dec in phase.
I agree with -40dB/dec in magnitude at high frequency, but phase respone is in doubt.
If you think simple LC lowpass filter (Here, R is omitted), magnitude response is -40dB/dec drop where f>>f0, where f0 = 1/(2*pi*sqrt(LC)), which is resonant frequency. Now, if you solve phase response, you will have 0 on both sides where f<fo and f>0. If you assume +/-delta f, now you can find the phase change occurs 180 degree. However, this phase response occurs instantly and its slope cannot be -90degree/dec. It is more likey infinity at that resonant frequency. Thus, simple LC filter phase respone cannot be -90degree/dec in general.
I understand this is little tricky.


Reply me if you have different viewpoints or you don't agree.

B
 
ee484 said:
Hi, all.

Thank you all for your reply.

Let me clear my question...

How does RHHP occur with capacitor and active device (e.g. MOS) - without inductor!?

How does a complex pole pair occur with capacitor with active devices - again without inductor?

Of course, I can derive all tedios transfer function to find out RHHP and complex pole pair. However, do you have any intuitive explanation or a way to find those?


BTW, I can answer some of question in the reply. Let me know if my explanations are wrong.

1) Why always complex pole are in pair?
--> This is because a real number that has frequency associated with (e.g. sinusoid) in complex plane in math should be "in pair". For example, to represent 3 in real number (for example, this number 3 is coming from sinusoidal function, meaning associated with frequency), we can write 3+jb. To represent this number in complex plane, there should be 3-jb. With these two numbers 1/2(3+jb + 3-jb) = 3. In conclusion, to represent a real number in complex domain requires to have complex conjugate pair!!!
(NOTE) if you are dealing with imaginary number, yes, a complex number can exit alone, not in a pair.

2) Inductor with capacitor = two reactive components will give you complex poles.
Associating with 1), if you have real value of L and C, you will have a complex pole IN pair. The poles cannot be alone.
However, with R and C only....you will have only real poles (I am not 100% sure, but very sure)
For example, if you have two RC lowpass filter cascaded with no buffer between and find the transfer function, you can make sure that two poles must be real and all in the LHP.

3) tyassin answered complex pole gives us -40dB/dec in magnitude and -90degree/dec in phase.
I agree with -40dB/dec in magnitude at high frequency, but phase respone is in doubt.
If you think simple LC lowpass filter (Here, R is omitted), magnitude response is -40dB/dec drop where f>>f0, where f0 = 1/(2*pi*sqrt(LC)), which is resonant frequency. Now, if you solve phase response, you will have 0 on both sides where f<fo and f>0. If you assume +/-delta f, now you can find the phase change occurs 180 degree. However, this phase response occurs instantly and its slope cannot be -90degree/dec. It is more likey infinity at that resonant frequency. Thus, simple LC filter phase respone cannot be -90degree/dec in general.
I understand this is little tricky.


Reply me if you have different viewpoints or you don't agree.

B
Click to expand...

1) I dont quite catch your point,sorry.but I meant what if we have denominator of transfer function like 1+s^3=0,at this situation the roots (poles)is 1, 0.5+j*sqrt(3)/2 ,0.5j*sqrt(3)/2,the poles come in conjugate pair,but what if the 1 in the denominator become j,the poles will not come in conjugate...maybe if we have real value of R,L,C,then the constants in the denominator always be real,so roots(poles) always come in conjugate pair,right?

3)you can check out the link given by tyassin for complex poles pair,at high frequency,the phase shift is 180 degree,but it is 90degree at pole frequency
 

2 cents: One of the problems with complex poles pair is the peaking.
 

Hi,

let me just answer b) first.

I don't mean 2nd order transfer function cannot have -90degree/dec per pole. However, it is all dependent on the damping ratio if you look at any control book.

For example, I ask you about a pure LC lowpass filter.
What are the frequency resonse?
The transfer function is 1/(1+s^2*L*C), right?
You may guess that the magnitude response is similar to RC lowpass filter, but the it attenuate -40dB/dec after fc=1/(2*pi*LC).

Now, try to find the phase response.
If you sustitute s with jw, there is no imaginary part in the transfer function.
What happend?
Does this mean there is no phase change? The phase change occurs at the resonant frequency with infinite degree/dec if I have to say.
The phase change is instant. It is not -90degree/dec per pole even though overall phase change is -180degree/dec.


a) I don't understand your question.
s^3 +1 = 0
will have poles as follow.

s = -1
s = 0.5+j*0.866
s = 0.5 -j*0.866

Here poles come in conjugate pair and the rest is -1, which is real.

How -1 becomes j? You cannot make this kind of thing with real components.

Besides, if we are talking about poles this system is not worth to mention. It has two RHHP poles. The system is dead.

In linear system with L, R, and C...(of course all real values) will have only three cases.
1) real
2) complex conjugate
3) repeated
roots.

We cannot make purely imaginary pole with real L, R, and C

Hope this is clear.

Reply this if I am wrong
 

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