Open Loop gain in Feedback control system

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KillaKem

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Consider a standard feedback control system with controller gain K(s) and plant transfer function G(s) with unity negative feedback( ie H(s) = 1 ).

The characteristic equation of the system is 1 + KGH(s) = 1 + K(s)G(s).

Now when finding the root locus you find the poles and zeros at K=0 ( Open loop poles and zeros ) by finding the poles and zeros of G(s).This is what I really don't understand, if K= 0 then the characteristic equation should be 1 + 0 * G(s) = 0 meaning that they don't exist, the poles should also be undefined.Why does it seem like we are actually taking K to be 1?
 

Yes, of course, Barry is right.
And it is also true that solving 1+KG=0 you get the poles (no zero´s) of the CLOSED-LOOP by variation of K.
 

barry said:
If I remember correctly, poles are found where 1+KG(s)=0 NOT K=0
Click to expand...

The way I understand it is that in order to solve 1 + KGH(s) = 0, you first solve GH(s) = 0 which will give you the open loop poles, those are the poles were the root locus plot will begin ( ie the poles for K=0 )

Or you can consider the fact that in root locus plots the value of K can be found by

\[ k = \frac{\prod |s - p_i| }{ \prod | s - z_i|}\]

so if we try to compute the value of K at the point were the root locus begins, then the numerator will be zero => K = 0.

- - - Updated - - -

LvW said:
Yes, of course, Barry is right.
And it is also true that solving 1+KG=0 you get the poles (no zero´s) of the CLOSED-LOOP by variation of K.
Click to expand...

I thought that at K = 0, open loop poles are at the same position as closed loop poles, after all closed loop poles travel from OL pole positions to OL zero positions as the open loop gain (K) is increased.
 
Last edited:

KillaKem said:
The way I understand it is that in order to solve 1 + KGH(s) = 0, you first solve GH(s) = 0 which will give you the open loop poles, those are the poles were the root locus plot will begin ( ie the poles for K=0 )
Click to expand...

I am afraid, your understanding is not correct.
The term (1+GK) is the denominator of the closed-loop transfer function (H=1 for unity feedback) and, thus, the equation 1+GK=0 gives the system roots, which are the poles of the closed-loop transfer function.
If you set G=0 you get the zero`s of G - nothing else.

KillaKem said:
I thought that at K = 0, open loop poles are at the same position as closed loop poles, after all closed loop poles travel from OL pole positions to OL zero positions as the open loop gain (K) is increased.
Click to expand...
I don`t understand. For K=0 also the product GK=0
 


I meant to write characteristic Equation ( GH(s) ) = 0 instead of just GH(s) = 0.

Lets take a simple example of G = (s+2)/(s+9) , H = 1.

Open Loop Pole = -9
Open loop Zero = -2

The root locus of the closed loop equation ( G/(1 + GH) ) will thus be a straight line travelling from -9 to -2.Thats OK right?

Now the question I'm asking myself is what point does the start point of the root locus correspond to, because if you use the formula it gives K = 0.
 


So your "root locus" consists of a line connecting a pole with a zero. Where comes this instruction from?
 

LvW said:
So your "root locus" consists of a line connecting a pole with a zero. Where comes this instruction from?
Click to expand...

Isn't that how closed loop poles travel, how does the root locus of G(s) = (s+2)(s+9) look like?

Consider this, lets assume my plot is correct, if i wanted to find the value of K at the point s = -3 in the root locus plot, I would use the formula

\[ k = \frac{\prod |s - p_i| }{ \prod | s - z_i|} = \frac{6}{1} = 6\]

Now, if i checked the validity of this value of K by finding the poles of the CL system i would get

\[\frac{ KG(s) }{1 + KGH(s) } = \frac{ \frac{6 ( s+2 )}{s+9} }{1 + \frac{6 ( s+2) }{s+9}} = \frac{ 6(s+2) }{ (s + 9) + 6(s+2)} \]

==> Pole is @ 7s + 21 = 0
==> Pole = -3

Which is consistent because -3 is the pole we were examining.

If we were to pick s = -8, then K =1/6.
If we were to pick s = -2.1, then K = 69

==> the pole seems to be travelling towards -2 with increasing K.

Which is consistent.


Just went on Matlab and simulated it (
Code: 
rlocus( tf([ 1, 2], [1 , 9]))
) and matlab drew a line from -9 to -2.
 
Last edited:

Killakem,

in your second equation above the denominator of the left side must be (1+KG).
But you are right, for H=1 (unity feedback) the closed-loop pole approaches -2 for very large K value.
 

LvW said:
Killakem,

In your second equation above the denominator of the left side must be (1+KG).
But you are right, for H=1 (unity feedback) the closed-loop pole approaches -2 for very large K value.
Click to expand...

You make it sound like it's true for only H(s) = 1 when it works for all H(s). If G(s) = (s+2)(s+9) and H(s) = 1/(s+6), then

OL poles = -9,-6
OL zeros = -2

Root locus of CL system will travel from [ -9 and -6 ] to [ -2 and \[\pm\]Infinity ], H(s) just changes the path the locus will take, but anyway, the question is what is the value of K at the start position of root locus plot? K = 0 ?
 

KillaKem said:
the question is what is the value of K at the start position of root locus plot? K = 0 ?
Click to expand...

KillaKem, regarding the relation between open and closed loop properties, you are right. There was a misunderstanding between us.
Regarding the K range, I think that it depends on your specific requirements.
If it makes sense in your system to use K values smaller than 1 you can start for K approaching zero (in practice perhaps K=0.01, for calculation of the root locus K=0).
The root locus shows how the pole moves for parameter variation - and you can select the part of the locus which is relevant for your design.
 

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