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What are poles & zero in Real.......??
- Thread starter koolguy
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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.
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.
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Last edited:
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:
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.
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))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.
Last edited:
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.
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.
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