# What are poles & zero in Real.......??

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#### koolguy

##### Active Member
Hi,

I have seen poles and zero in course book in many subject, it is said they define stability of function. in actual what is this??

#### crutschow

##### Well-Known Member
The location of the poles and zeros in the transfer function of a close-loop feedback control system can be used to determine its stability. Read this for starters.

#### koolguy

##### Active Member
I have listen when we make any mechanical or signal we check it stability by using z transform or other method but how in practical??

#### MrAl

##### Well-Known Member
Hi there,

The picture attached shows a pole zero constellation diagram to the left and to the right it shows a 'rubber sheet' model of that diagram, where the model is of some impedance for example. The pole zero constellation is a 2 dimensional illustration of a 3 dimensional model. You can see where the model is zero the left diagram shows a circle (blue) and that represents a zero, while the places where the model tends to infinity the left diagram shows an 'X" (red). The rubber sheet model only shows the upper half of the pole zero constellation for simplicity.

From your second post however I have a feeling though that you are looking for a way to test a real physical system. In that case you would probably use a network analyzer.
You can input a step and watch the output of the system on a scope too, but probably better would be to input a sweep of sine waves and observe the amplitude and phase shift for each frequency. You can look for the phase and gain margin and try to determine if there is some place where either margin is not large enough, which would indicate that if something physical changes a little in the system it might start to oscillate.

#### Attachments

• PoleZero.jpg
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#### koolguy

##### Active Member
Hi,
as In z transform we see ROC, what the real use of it. as i am not getting right answer to these topics all bookish Lang. i want to see it in practical how to use Z transform??

#### koolguy

##### Active Member
Is there any one to reply my question??

#### MrAl

##### Well-Known Member
Hi again,

You might try writing your question a little more clear as to what you actually want to know. Use a lot more detail in your question so we know what you are really asking. More than a few lines would be nice

#### koolguy

##### Active Member
Hi again,

I am studying Z transform in this there ROC to be found in diff. question but i am not clear how they are used in practical life ??

#### MrAl

##### Well-Known Member
Hi again,

The unit circle in the z plane corresponds to the left half of the s plane. In other words, the sampled system is stable if all of the poles of the closed loop transfer function T(z) are found to be within the unit circle.

#### koolguy

##### Active Member
Can you please, tell me in more detail as there are numerical question given how the question is know. is it is any signal??

#### MrAl

##### Well-Known Member
Hello again,

Here is a complete numerical example.
We want to look at a system as the gain changes so we can tell if the system will become unstable.

The system block diagram:

Code:
       +
Vin >---O--->--S----ZOH----Gs----+--->Vout
|-                       |
|                        |
+--------<---------------+

(Note the system has negative feedback as shown by the minus sign)

O = summing junction with one plus and one minus input
S = sampler
ZOH =  zero order hold
Gs = G(s) = K/(s*(s+1))
and we want to look at stability as the gain K is varied.

The z transform of G(s) with unit impulse and sampling period of 1 second is:
G(z)=K*(0.36787944025834*z+0.26424111948332)/(z^2-1.367879441171442*z+0.36787944117144)

If the system is stable the roots of 1+G(z)=0 all lie within the unit circle, so
we compute 1+G(z)=0 for various gains.

With K=1 we have:
1+G(z)=0
where G(z) evaluates to:
(0.36787944025834*z+0.26424111948332)/(z^2-1.367879441171442*z+0.36787944117144)+1=0

which simplifies to:
z^2-z+0.63212056394153=0

which has two roots:
z=0.61815901185822*j+0.5, z=0.5-0.61815901185822*j

which have two amplitudes:
0.79506010083611, and 0.79506010083611

which are both less than 1, so the system with gain K=1 is stable.

Now we do the same thing with K=10.
With K=10 we have:
1+G(z)=0
where G(z) now evaluates to:
(3.6787944025834*z+2.6424111948332)/(z^2-1.367879441171442*z+0.36787944117144)+1=0

which simplifies to:
z^2+2.310914925501759*z+3.01029067709018=0

which has two roots:
z=1.294298547037537*j-1.155457462750879, and z=-1.294298547037537*j-1.155457462750879

which have two amplitudes:
1.735018927012089, and 1.735018927012089

which are both greater than 1, so this system has now become unstable.

Since we moved from a system that was stable to a system that was unstable when we varied the gain from K=1 to K=10, we can assume that the system becomes unstable somewhere between K=1 and K=10. We can compute more roots with other values of K such as K=5 and try to determine when we are close to the value of K that causes the system to transition from stable to unstable, or we can plot the roots.
If we were to plot more roots we would find that the system becomes unstable near
the point where K becomes equal to approximately 2.4.

Computing the roots with K=2.39 we get the roots:
z=0.96939239056609*j+0.24432379944186, and z=0.24432379944186-0.96939239056609*j

which have amplitudes:
0.99970782024607, and 0.99970782024607

where we can see they are both pretty close to 1 now so the system is very close to being unstable. If we compute the roots with K=2.40 we would find that the system again becomes unstable. Thus, the roots move outside the unit circle near the point where K=2.4 approximately.
This means there would have to be some way to ensure that the gain of G(s) could not get too close to 2.4 or else this system would go unstable.

We could also go back and try a shorter sampling time period.

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#### koolguy

##### Active Member
Hi,
firstly thanks for the reply, please come to low level example it will be easy for me...

#### MrAl

##### Well-Known Member
Hello,

Well, that's a fairly simple system so im not sure we can get any simpler than that. Maybe a simpler Gs perhaps.

You might go through my previous reply and see what you dont understand yet and tell me which line isnt clear for you.

#### koolguy

##### Active Member
It is very tough to understand it i am not getting it..

#### birdman0_o

##### Active Member
The ROC determines the stability of a signal. If the signal is out of the ROC, the Laplace integral will no longer converge, meaning, as time goes to infinity so will the output of the system (not stable).

Also you are probably one of the rudest posters I have seen on this forum, we are not a service you pay for, you cannot demand us to reply.

#### koolguy

##### Active Member
Also you are probably one of the rudest posters I have seen on this forum, we are not a service you pay for, you cannot demand us to reply.

Sorry for misunderstanding but i am not!!

#### MrAl

##### Well-Known Member
Hi again,

Maybe you just want to know how to calculate the poles and zeros or something?
Did you ever factor a polynomial that had one or more complex roots?

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