servo system and transfer function etc.

[PG1995]

Hi

Could you please help me with these queries? Thank you very much.

Reference #1: https://www.electro-tech-online.com/attachments/lcs_ogata_example_3-3-jpg.72409/. (The attachment is from this post).

Regards
PG

 

[misterT]

Q1: Everything you say is correct. It depends how the system is designed. There can be a switch, but in this particular example there is an amplifier that simply amplifies the error voltage ev. When error is zero, the output from the amplifier is zero and the motor stops. Slow systems like thermostats use a simple switch to control the power.

Q2: In this context La is negligible compared to the mechanical properties of the system. The current through armature responses very fast compared to the Inertia of the system. Because of this we can practically ignore the inductance La. It is like ignoring the weight of a gas pedal in a car model.. it does not affect the system at all in practice.

Q3: I don't quite understand the question. Both b and b0 are viscous friction. The other example is for linear motion, the other one is for rotational motion. So the speed units are a bit different. Well yeah, that would make the viscous friction units different also. Also the inertia units (mass) are different for rotational and linear motion. http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

Q4: Yes, Inertia is just a common term for: moment of inertia, mass moment of inertia, rotational inertia, polar moment of inertia, or the angular mass. It is easy to just say "inertia". Sometimes inertia is used to mean linear mass also.

Q5: Gs also determines the closed loop transfer function which in this case is Gs/(Gs + 1). It is easiest just to calculate Gs.. that is what is interesting to us in this example. Not the closed loop equation.

Q6: The friction coefficient b0 is constant.. and yes, so is the term K2K3/R. The text only says that this increases the viscous friction. It does not say that it dynamically increases the viscous friction. It is only a mathematical observation. It is hard to imagine how this happens in the actual system. Well.. larger the back emf constant the faster the motor will spin. Though efectively resulting larger frictions.

Q7: Speed translates linearily through gears. Inertia is more like a "two dimensional" unit (rotational energy).

Rotational energy: 0.5 * J1 * w^2
w is the angular speed. You can see where the 1/n^2 comes. After the gear the rotational energy is: 0.5 * J2 *(w/n)^2
Rotational energy on both sides need to be the same. So: J1 = J2 / n^2

 

[MrAl]

Hi,

Just to add a little here to a good post already here by MisterT...

The units of Torque would be kilogram meters not Newton meters if working in SI units. The energy is in Newton meters.

The angular velocity for angle A is dA/dt and has units of rads/sec.

The friction would be in units of torque per unit velocity, or kg*m/(rads/sec) in SI units.

 

[PG1995]

Thank you, misterT, MrAl.

@misterT: Your reply was very helpful and really straightforward that I didn't even have to read it twice to understand it. Though I have one follow-on query related to Q2. But I think it's better to hold it back for some time until I complete reading of the chapter which that text was taken from.

Regards
PG

 

[misterT]

There is an excellent video blog about automation and control systems from Brian Douglas on youtube. https://www.youtube.com/user/ControlLectures/videos

 

[PG1995]

Q1:
This servo system, let's call it Servo1, is from my first post above (in the book this system was presented in chapter #3). In later chapter on transient response, this new servo system, let's call it Servo2, is introduced which I have failed to understand. I have good understanding of Servo1 but how do I relate Servo2 with Servo1? Could you please help me? Thank you.


Q2:

Q2: In this context La is negligible compared to the mechanical properties of the system. The current through armature responses very fast compared to the Inertia of the system. Because of this we can practically ignore the inductance La. It is like ignoring the weight of a gas pedal in a car model.. it does not affect the system at all in practice.

I'm still confused about La, the armature winding. I don't see why it can be ignored. It's the main inductance which makes the motor turn. You could ignore La if, say, it was 5 μH but I don't think it has such a low value in this case. Perhaps, this text about transformer could help you understand my confusion better. Why is La ignored? Please help me with it. Thanks.

Regards
PG

 

[steveB]

PG,

When considering the circuit effects, we don't need to worry about how inductance relates flux and current, despite the fact that this is important in its own right. For circuit effects, consider inductive reactance as X=ωL. Here, both the inductance and the frequency are important to determining if the inductive effects are significant in the circuit. For example if X is much less than the resistance R in the circuit, it can be neglected.

In a motor, the electrical frequency is related to the mechanical frequency by the number of pole pairs per revolution N. Hence, ω=N dθ/dt, and it is this frequency you can consider when determining how significant inductance is in a particular case.

Note that you said that you can ignore 5 μH, but consider that 5 μH is not usually negligible at radio frequencies.

 

[PG1995]

Thank you, Steve.

I'm sorry to ask this but I still don't see any particular reason why La can be ignored. My apologies if the answer is there in your post. Perhaps, then I'm acting more dumb than usual!

Please also note that I have updated my previous post with a new question. Kindly help me.

Regards
PG

 

[steveB]

PG,

I didn't follow this thread initially, so I won't try to answer any detailed questions about this. - At least not until I've caught up on all that has been said previously.

However, in the particular latest question that I did try to address, can you explain why my explanation didn't hit the mark. Without some feedback, I'm not sure how to clarify my answer.

 

[MrAl]

Thank you, Steve.

I'm sorry to ask this but I still don't see any particular reason why La can be ignored. My apologies if the answer is there in your post. Perhaps, then I'm acting more dumb than usual!

Please also note that I have updated my previous post with a new question. Kindly help me.

Regards
PG

Hi there PG,


If you are talking about what i think you are talking about, about the inductance of the armature being somewhat insignificant, then yes, that is true for many (but not all) motors, and the short answer is because the armature inductance acts in a manner similar to how the rotational inertia acts toward the overall response of the motor. So that when you have inductance La plus rotational inertia J in the 'circuit' (electrical analog) you end up with almost the same response as when you take out the La and just leave the rotational inertia J.

Another pseudo electrical analogy is like this: say you have a two pole low pass RC filter, where you have R1,C1 for the first section and it is followed immediately by R2,C2 which is the second section. Now say we make R1=1 and C1=0.1, and make R2=10 and C2=10. That means the time constant of R1 and C1 is 0.1 seconds, and the time constant of R2 and C2 is 100 seconds. That means that the effect of R1 and C1 is completely over in approximately 0.5 seconds, while the effect of R2 and C2 is not over until about 500 seconds later. Obviously the R1 and C1 does almost nothing to help the circuit, so if we removed it we would still have almost the same response. It would not be exactly the same, but it would be very close to the same response.

So the inductor La and associated resistance acts very fast relative to the rotational inertia and associated friction, which does almost the same thing as the inductance, so we can sometimes simplify the overall analysis by removing the inductance. It also helps to note the position of La and the position of J in the motor equation to see how similar they are in the way they affect the response. J in a system often has a profound effect on the start up time for a motor, while La often has little effect in a regular type motor. Electrically and mechanically, this means that the inductance allows the MAX current to flow almost right away without too much delay, while the rotational inertia makes the spin up long and drawn out, so when we remove the inductance we still see the effect of the rotational inertia and dont notice the removed start up current delay because it happens so fast anyway. So if J causes a spin up of 10 seconds and the inductance causes a peak current delay of 0.1 seconds, the total spin up time (very very roughly) would be 10.1 seconds, but without the inductance it would be 10.0 seconds...obviously not much different

You may not be able to take this liberty however with a motor like a stepper where the fastest response is desired. That's because the inductance is used as part of the current limiting mechanism (and hence power dissipation) of the device while still getting the fastest speed possible. That's where we want very fast current rise time which is very unlike most motor applications which dont require this kind of very fast action.

 

[misterT]

You could ignore La if, say, it was 5 μH but I don't think it has such a low value in this case.

This was allready answered correctly, but I want to (try) highlight the point. The inductance is very small compared to the inertia J. I am talking about response times. The current through the inductor shoots up very fast when voltage is applied across it, but the load inertia barely has begun to react to that change.

 

[steveB]

PG,

I don't disagree with misterT and MrAl about the comparison of electrical and mechanical response times, but perhaps one thing that is confusing you is that L and J have different units and saying one is negligible compared to the other is confusing when the units don't match. It would be better to compare time constants that use L and J to make this comparison, because then the units match and comparisons are meaningful.

However, my thought was to compare ωL and R which have the same units. We can see in the text that they arrive at eqn. 3.120 by neglecting sL compared to R. Then, they lump parameters together to get a simple expression. I think this is the easiest way to understand the neglecting of inductance in this case. This is what I mentioned above, and I'm not sure why you didn't understand it.

Whenever you see a Laplace frequency "s", you can think of this as "jω", and if jωL is small compared to R, then the complex value is very close to the real value R, and any voltage driving the motor does not see a significant inductance. This also means the electrical response of the motor is fast (time constant is L/R which is small, meaning fast dynamics), compared to the mechanical system. One problem for you might be to find ω, but typically motors (at least in steady state) have electrical frequency matched to the mechanical frequency if there is one pole pair per revolution. If there are more pole pairs per revolution of the motor, the electrical frequency is higher by a multiple of the number of pole pairs.

 

[MrAl]

Hello Steve, MisterT, and PG,


Steve:
I think i follow your argument which shows a concern for comparing J to La when the units are different. However, inertia is the mechanical equivalent of electrical inductance, so if they are in the correct units to begin with they should be easily comparable. In other words, we should be able to take the approximate ratio r=J/La and note that if r is less than 1 then the inductance has more influence over the response and if r is greater than 1 then the inertia has more influence. We can see this must be so by considering the motor speed:
Speed=K1*K2/((La*s+Ra)*(J0*s+b0)+K2*K3)

[LATEX]\[Speed=\frac{K1\,K2}{\left( s\,La+Ra\right) \,\left( s\,J0+b0\right)+K2\,K3 }\][/LATEX]

and noting that if we had low resistance Ra and low friction b0 that we would have something similar to:
Speed=K1*K2/((La*s)*(J0*s)+K2*K3)

[LATEX]\[Speed=\frac{K1\,K2}{\left( s\,La\right) \,\left( s\,J0\right)+K2\,K3 }\][/LATEX]


and now note that La and J0 play the same role exactly.

Comparing w*La to Ra does not help because w*J0 compared to b0 may also be small, and then the inductance still plays a significant role in the total response. Comparing w*La to Ra just says for example that "The time for the current to rise is very fast", but if the time for the inertia for example, "The time for the inertia to be overcome is very fast" then we still do not know if the inductance is insignificant or not when it comes to determining the spin up time for example.

However, it does make sense to compare the time constants themselves such as La/Ra and J0/b0. Is that what you meant to show?
This would make a lot of sense because in a two inductor two resistor RL circuit we see the same thing and it's almost the same equation too.
So although we say "compare inductance to rotational inertia" the more accurate statement would be:

r=J*Ra/(La*b0)

[LATEX]\[r=\frac{J\,Ra}{La\,b0}\][/LATEX]

and when r>>1 we see the inductance play a more insignificant role in the response of the motor.

In my previous post i quoted a 0.1 second current rise time and a 10 second inertial response time, for a total spin up time of 10.1 seconds. And that's if we considered the inductive resistive part to be a complete 0.1 second delay. So with this delay we get a spin up time of 10.1 seconds but without the inductive part we get a spin up time of 10 seconds, which is almost exactly the same as the response when we do include the inductance.

 

[steveB]

However, it does make sense to compare the time constants themselves such as La*Ra and J0*b0.

Hi MrAl

I basically agree. It's important to have units that match. I think you meant to divide to find the time constants? For example, τ=L/R, not τ=L*R. In making the ratio r, a unitless definition is better than r=J/L because J/L has units, and you can't compare something with units to 1 meaningfully in general.

However, I disagree that "Comparing La to Ra does not help", because this seems to be the exact argument that the textbook is making, but obviously it is sL compared to R. Hence, this is the best answer to give to PG, unless you disagree with the treatment given by the book. Also, comparing Jo and bo suffers from the same problem of units that don't match. We must compare sJ with b, to compare, and this depends on frequency. However, the book treatment assumes bo and Jo are both important and they are retained in the analysis.

 

[MrAl]

Hi MrAl

I basically agree. It's important to have units that match. I think you meant to divide to find the time constants? For example, τ=L/R, not τ=L*R. In making the ratio r, a unitless definition is better than r=J/L because J/L has units, and you can't compare something with units to 1 meaningfully in general.

However, I disagree that "Comparing La to Ra does not help", because this seems to be the exact argument that the textbook is making, but obviously it is sL compared to R. Hence, this is the best answer to give to PG, unless you disagree with the treatment given by the book. Also, comparing Jo and bo suffers from the same problem of units that don't match. We must compare sJ with b, to compare, and this depends on frequency. However, the book treatment assumes bo and Jo are both important and they are retained in the analysis.

Hello again Steve,

Yes i meant to state this:
[LATEX]r=\frac{J\,Ra}{La\,b0}[/LATEX]

which is easier to understand as:
[LATEX]r=\frac{J/b0}{La/Ra}[/LATEX]

From the above you can see i am comparing the two times constants which is more exact than simply comparing the inertia to the inductance.
We sometimes shorten this in discussion to r=J/La even though they do not have the same units, but of course that is not as correct.

But i can not find a way to state what we are talking about in terms of only La and Ra because La and Ra are ONLY responsible for the current rise time, and if the current rise time is short compared to the time it takes to overcome the rotational inertia then we still dont have to consider La in the approximation to the response.

You are right in that it does not matter to me whether or not the 'book' wants to admit to this or not, because it can be easily proved by plugging some numbers in and calculating the response time with a few different values for each component.
For a really quick example, if L=1 and R=1 and J=100 and f=1, we have the motor current rise to very near max in 5 seconds, which means the input power is already at a max and the torque is at a max and is now constant, while it takes the rotational inertia about 500 seconds to be nearly completely overcome, so we have as a very rough estimate 500+5=505 seconds to spin up to full speed, and making L=10 and R=10 we see the same thing except the max current is now less, so takes even longer for the rotational inertia to be overcome.
But also, if we keep the same L and R and make J=1 and f=1 as before we see that the current comes up in the same time as before, but now it takes much less time for the inertia to be overcome, so the motor spins up faster.
So with L=1, R=1, J=10 and f=10, we see a long time to spin up, but with L=1, R=1, J=1, and f=1, we see a short time to spin up yet we see the same comparison of L and R.

Also, what did you mean when you wanted to use wL to compare to R, what do you intend to make 'w' here?


Better numerical example follows...

The true test is:

[LATEX]\frac{\partial f(t)}{\partial \,var}[/LATEX]

where 'var' is the variable in question and f(t) is the speed with time, and we should find something like this:
[LATEX]\frac{\partial f(t)}{\partial \,var1}>\frac{\partial f(t)}{\partial \,var2}[/LATEX]

Start with L=0.001, R=1, J=1, f=1, speed after 1 second is: 0.432332 rads/second.
Now change L=0.002, speed after 1 second is: 0.432332 rads/second.
Now put L back to 0.001 and change J=2, speed after 1 second is 0.315968 rads/second.
Now leave J=2 and make L=0.002 again, speed after 1 second is 0.315876 rads/second.

So doubling the inductance did not change the speed after 1 second much (it did change out to many decimal places) but
doubling the rotational inertia made a much bigger difference in the speed after 1 second.

It also is worth noting that the response exponential multiplier time exponent turns out to be the average of the reciprocals of the two time constants, where one is L/R and the other is J/f:
e^-at, with a=0.5*(R/L+f/J)

So although R/L (or L/R) does in fact affect the spin up time, it does not act alone.

 

[steveB]

Also, what did you mean when you wanted to use wL to compare to R, what do you intend to make 'w' here?

I intended the omega (ω), not w, to mean the frequency. Basically sL+R shows up in the transfer function. The book replaces sL+R with R to make the approximation that L is negligible. This one simplification allows a very simple transfer function to be established in eqn. 3.120. However, s is really jω for typical sinusoidal driving. Hence, the complex value sL+R is being replaced by the real value of R at low frequency. In other words, there is a high frequency pole that can be ignored for typical frequencies in motor control.

As far as what values to use for ω to see if the approximation is valid, I tried to answer this twice already. The mechanical frequency of the motor can be related to the electrical frequency as ω=N ωr, where ωr is the mechanical frequency, ω is the electrical frequency and N is the number of pole pairs per revolution on the motor construction. This is strictly correct for synchronous motors in steady state, and approximately correct for all motors in typical transient cases. In other words, the motor operating speed gives some hint at the electrical bandwidth required for control, and this sets the frequency range to consider for the inductance, to see if it can be neglected.

 

[misterT]

All this math etc. Sorry I did not read all of it. I just want to say that, in general, poles near origin dominate the system and you can pretty much ignore all poles that are far away from the origin (in laplace plane). In this case the inertia of the motor/load creates a pole that is much closer to origin than the pole created by the inductance.

https://www.electro-tech-online.com/custompdfs/2013/06/ME360DominantPoles.pdf
https://www.electro-tech-online.com/custompdfs/2013/06/PoleZero.pdf

 

[steveB]

misterT,

I agree, that is the essence of the question and answer. If the value of L does not control the low frequency poles we care about, then it can be neglected. Still, we don't always have clear formulas for the pole locations in higher order systems with many poles. Often, it is easier to identify the transfer function with polynomials in numerator and denominator, and then cross off any negligible terms in both polynomials. What the book does is take a shortcut and uses neither of these approaches. It provides a simple observation to eliminate a high frequency pole from the analysis by comparing sL with R. But, whatever method is used, it all amounts to the same end result.

The thing missing here, is PG's feedback to tell us if any of these perspectives and responses are helping him.

 

[MrAl]

I intended the omega (ω), not w, to mean the frequency. Basically sL+R shows up in the transfer function. The book replaces sL+R with R to make the approximation that L is negligible. This one simplification allows a very simple transfer function to be established in eqn. 3.120. However, s is really jω for typical sinusoidal driving. Hence, the complex value sL+R is being replaced by the real value of R at low frequency. In other words, there is a high frequency pole that can be ignored for typical frequencies in motor control.

As far as what values to use for ω to see if the approximation is valid, I tried to answer this twice already. The mechanical frequency of the motor can be related to the electrical frequency as ω=N ωr, where ωr is the mechanical frequency, ω is the electrical frequency and N is the number of pole pairs per revolution on the motor construction. This is strictly correct for synchronous motors in steady state, and approximately correct for all motors in typical transient cases. In other words, the motor operating speed gives some hint at the electrical bandwidth required for control, and this sets the frequency range to consider for the inductance, to see if it can be neglected.

Hi Steve,

I dont see any reference in the text that suggests that La is being eliminated due to a "high frequency component" of any kind. I checked it again, and i dont see it. Did i miss it anyway? That's possible, so if you can point it out i can understand what you are getting at a little better perhaps.

Until then, i still see the fact that La and R are not enough to look at in order to determine what is significant and what is not. You mentioned that "5uH is not significant at radio frequencies" but i believe that is the wrong view to take, even though it might seem correct, because the basic issue is the response due to R and La and the response due to J and f, and both need to be considered in order to determine what is negligible and what is not. Suppose J is negligible instead of L, how would we know looking at only R and L (or R and wL)?

The book however does not seem to go into detail about THEIR reason for doing this (if i didnt miss it), and they unfortunately dont bother to include the full block diagram of the motor so i will do that here. It is from the block diagram of the motor itself that we can see this much more clearly.

Note also in the block diagram that we could simulate the two forward path blocks (one with R,L and the other with J,f) using instead two cascaded passive RL low pass filter circuits, with different values of R and L for each. It should be easy to show that if one filter section has a large time constant and one has a very small time constant that the one with the small time constant can be eliminated, but what can not be told is which one should be eliminated without examining them both.

I also took the liberty to show the armature current and how it comes from R and L, and this can be used to show that the torque comes from the current times the motor constant Km to drive the load, and the load is the block with the inertia J and friction f. But also notice that we can swap the positions of the two blocks and get the same results.

The better motor block diagram now...

 

[steveB]

I dont see any reference in the text that suggests that La is being eliminated due to a "high frequency component" of any kind. I checked it again, and i dont see it. Did i miss it anyway? That's possible, so if you can point it out i can understand what you are getting at a little better perhaps.

The problem with the book is that they do not explain what they are doing very well. I simply am drawing a logical conclusion from the step they make deriving eqn. 3.120. They say "when La is small it can be neglected", which doesn't give much information. But, look at what they do in that step. They replace (s La + Ra) with (Ra). Well, that's fine in some cases, but we are not really comparing La to Ra, but need to compare (s La) to Ra, which includes frequency. The question is "what frequency", obviously it is the frequencies relevant for driving the motor, which tends to be low, especially when the mechanical system is slow to react due to inertia and when the motor speed is slow. So, s La does not become comparable to Ra until high frequency. But, we don't care about high frequency. Or, as MisterT pointed out, we dont care about the high frequency pole caused by keeping (s La + R) in the transfer function.

Until then, i still see the fact that La and R are not enough to look at in order to determine what is significant and what is not.

It may very well be that La and R are not enough to look at to see if La can be neglected. However, this is what the book is doing, and my quick look at it didn't reveal a case where La can be neglected if R is small compared to sLa. However, if you have identified another case where R is small and La can still be neglected, then that's fine. You have looked at this in more detail than I have, and I trust your analysis over my lack of analysis.

So, take my answer for PG to be an explanation of what the book is doing, and we can take your additional information as a comment on the incompleteness of the book to consider other cases. I'll have to look at what you provided in more detail.

You mentioned that "5uH is not significant at radio frequencies" but i believe that is the wrong view to take, even though it might seem correct, because the basic issue is the response due to R and La and the response due to J and f, and both need to be considered in order to determine what is negligible and what is not. Suppose J is negligible instead of L, how would we know looking at only R and L (or R and wL)?

I was talking about the general case in response to PG's comment that 5 uH is small. Forgetting about motors and this specific case, a 5 uH inductor is not usually negligible at radio frequencies. I was just trying to point out that an inductor is a frequency dependent device. Of course, for a motor at low frequencies, typical for mechanical systems, the 5 uH would usually be negligible.

Sorry, i should have been clearer in what I was trying to say.