Design with root locusmaapc/static/files/SYSTHEORY/Slid… · Design with root locus 1. Systems and...

Systems and Control Theory

STADIUS - Center for Dynamical Systems, Signal Processing and Data Analytics

Lead and lag compensators

Design with root locus

1



Definition Both a lead compensator and a lag compensator have the same

shape:Lead compensators:

𝐶𝐶 𝑠𝑠 = 𝐾𝐾𝑠𝑠+1𝜏𝜏𝑠𝑠+ 1

𝛼𝛼𝜏𝜏with 0 < 𝛼𝛼 < 1

So they have a zero at 𝑠𝑠 = −1𝜏𝜏

and a pole at 𝑠𝑠 = − 1𝛼𝛼𝜏𝜏

or − 1𝛽𝛽𝜏𝜏

For lead compensators the pole lies more to the left in the complex plane than the zero and vice versa for lag compensators

Lag compensators:

𝐶𝐶 𝑠𝑠 = 𝐾𝐾𝑠𝑠+1𝜏𝜏𝑠𝑠+ 1

𝛽𝛽𝜏𝜏with 𝛽𝛽 > 1

2



Lead compensators

3



Lead compensators: design with root locus Design based on time-domain quantities that can be

expressed in terms of the dominant pole locations (overshoot, rise time, settling time, damping ratio, …)

This is done based on the root locus method Previously we’ve seen how to use the root locus method to

tune the position of the closed loop poles based on one parameter, but now we have a problem of an entirely different order: we have to determine the position of a pole and a zero and tune a parameter (𝐾𝐾)

4



Lead compensators: design with root locus Our strategy will be the following: We will look at where we want our dominant poles to lie

(based on the time domain quantities) We will place our pole and zero (1/𝛼𝛼𝛼𝛼 and 1/𝛼𝛼) such that the

root locus of 𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 moves through these desired positions

Then we’ll find the corresponding 𝐾𝐾 from the root locus of 𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠

5



Lead compensators: design with root locus A contribution of a lead compensator is that it moves the

closed loop poles to the leftYou just have to look at the effect on the centroid (with 𝑝𝑝𝑖𝑖 and 𝑧𝑧𝑖𝑖 the poles and zero’s of 𝑃𝑃 𝑠𝑠 and 𝑝𝑝 and 𝑧𝑧 of 𝐶𝐶 𝑠𝑠 ):

∑𝑖𝑖 𝑝𝑝𝑖𝑖+𝑝𝑝−∑𝑖𝑖 𝑧𝑧𝑖𝑖−𝑧𝑧#𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑠𝑠+1−#𝑧𝑧𝑝𝑝𝑧𝑧𝑝𝑝′𝑠𝑠−1

= ∑𝑖𝑖 𝑝𝑝𝑖𝑖−∑𝑖𝑖 𝑧𝑧𝑖𝑖+𝑝𝑝−𝑧𝑧#𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑠𝑠−#𝑧𝑧𝑝𝑝𝑧𝑧𝑝𝑝′𝑠𝑠

, with 𝑝𝑝 < 𝑧𝑧

So the centroid is moved to the left, which means the asymptotic values of the branches will be more negative

This is how you can stabilize a system which the root locus has no value 𝐾𝐾 for which it resides entirely in the left halve plane

6



Lead compensators: design with root locus How can we now find a way to let the root locus go exactly

through a certain point? We can actually do that quite easily by noticing the following

property of every point on the root locus:As you know the root locus display the roots of 1 + 𝐾𝐾𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 = 0 for all 𝐾𝐾 (with 𝐶𝐶 𝑠𝑠 = 1+𝑧𝑧

1+𝑝𝑝from

now)So for a point to be on the root locus there are two requirements: ∠ 𝐾𝐾𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 has to be equal to ±180° 2𝑘𝑘 + 1 and 𝐾𝐾𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 has to equal 1

Let’s call our desired closed loop position 𝑞𝑞 (and its complex conjugate, �𝑞𝑞), then you have to find 𝑧𝑧, 𝑝𝑝 and 𝐾𝐾 such that ∠ 𝐾𝐾𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞 = ±180° 2𝑘𝑘 + 1 and 𝐾𝐾𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞 = 1

7



Lead compensators: design with root locus The first requirement becomes:

∠ 𝑞𝑞 + 𝑧𝑧 − ∠ 𝑞𝑞 + 𝑝𝑝 = ±180° 2𝑘𝑘 + 1 − ∠ 𝑃𝑃 𝑞𝑞 And the second requirement becomes:

𝐾𝐾 𝑞𝑞+𝑧𝑧𝑞𝑞+𝑝𝑝

= 1𝑃𝑃 𝑞𝑞

These two requirements determine two of the degrees of freedom: For small enough 𝑧𝑧/𝑝𝑝 (= 𝛼𝛼) and reasonable ±180°(2𝑘𝑘 +

8



Lead compensators: design with root locus So in theory we could again put restrictions on the steady-

state error (which leads to a requirement on 𝐾𝐾𝑧𝑧/𝑝𝑝) But we are not going to do that, since that gives us a

complicated system of equations ∠ 𝑞𝑞 + 𝑧𝑧 − ∠ 𝑞𝑞 + 𝑝𝑝 = 𝜙𝜙 (= the angle condition)

𝐾𝐾 𝑞𝑞+𝑧𝑧𝑞𝑞+𝑝𝑝

= 1𝑃𝑃 𝑞𝑞

𝐾𝐾𝛼𝛼 = . . . This system is difficult to solve, a solution is not guaranteed

and we lose control over the position of the new pole (which might overtake the role of the dominant poles)

So we’ll use our extra degree of freedom to place the zero 𝑧𝑧 at a location such that the angle condition can most likely be met and without 𝑝𝑝 becoming a dominant pole

9



𝜃𝜃1𝜃𝜃2𝜃𝜃𝑧𝑧𝜃𝜃𝑝𝑝

Lead compensators: design with root locus Let’s look at the following: we have a system with two poles

( ), which we want to replace to another location ( ) Where should you position the zero of the lead compensator? It should not be far to the left, since then the risk exists that

there is no valid position of the pole:∠ 𝑞𝑞 + 𝑧𝑧 − ∠ 𝑞𝑞 + 𝑝𝑝 = 𝜙𝜙, so the zero adds phase and the pole subtracts phase; so the added phase thanks to 𝑧𝑧 should be large enough to make a sum of 𝜙𝜙 possibleAs you can see ∠ 𝑃𝑃 𝑞𝑞 = −𝜃𝜃1 − 𝜃𝜃2 ≅ −230° and 𝜃𝜃𝑧𝑧 < 𝜙𝜙 ≅ 50°, so there exists no 𝑝𝑝 such that ∠ 𝐶𝐶 𝑞𝑞 = 𝜙𝜙

Here you can also see graphically that 𝜃𝜃𝑧𝑧 > 𝜃𝜃𝑝𝑝for a lead compensator

10



Lead compensators: design with root locus You should also not place it too far to the right, otherwise the

risk exists that the pole will also be placed too far to the right and interfere too strongly with the dominant poles, strongly influencing the closed loop system behavior

𝜃𝜃𝑝𝑝 𝜃𝜃1𝜃𝜃2𝜃𝜃𝑧𝑧

11



Lead compensators: design with root locus The last thing we have to know, is how to determine 𝑝𝑝 such

that ∠ 𝐾𝐾𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞 = ±180° 2𝑘𝑘 + 1 ? We’ll deduce that here:

∠ 𝐾𝐾𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞 = ∠ 𝑃𝑃 𝑞𝑞 + ∠ 𝑞𝑞 + 𝑧𝑧 − ∠ 𝑞𝑞 + 𝑝𝑝

∠ 𝑞𝑞 + 𝑝𝑝 = ∓180° 2𝑘𝑘 + 1 + ∠ 𝑃𝑃 𝑞𝑞 + ∠ 𝑞𝑞 + 𝑧𝑧Now you have to pick 𝑘𝑘 and pick which sign ∓ has to be for the right hand side to be the smallest positive number; we’ll call the resulting right hand side 𝜃𝜃Now we’ll rewrite ∠ 𝑞𝑞 + 𝑝𝑝arctan ℑ 𝑞𝑞 /ℜ 𝑞𝑞 + 𝑝𝑝 = 𝜃𝜃ℑ 𝑞𝑞 / ℜ 𝑞𝑞 + 𝑝𝑝 = tan 𝜃𝜃𝑝𝑝 = cot 𝜃𝜃 ℑ 𝑞𝑞 − ℜ 𝑞𝑞

12



Lead compensators: design with root locus We’ve now seen enough to draw up this action plan:1. Determine the desired locations of the closed loop poles (𝑞𝑞

and �𝑞𝑞) from the time domain qualifications; you should take a safety margin, as the compensator’s pole and zero will have an impact (and even more so if it was not a 2nd order system to start with), if 𝜙𝜙 > 60° continue with 2 lead compensators

2. Check whether a proportional controller won’t do the trick3. If not, continue with the design of a lead

compensator by placing its zero somewhere in the region indicated by the green circle:

4. Find 𝑝𝑝 (see previous slide)5. Find the corresponding 𝐾𝐾-value: 𝐾𝐾 = 𝑞𝑞 + 𝑝𝑝 / 𝑞𝑞 + 𝑧𝑧 𝑃𝑃 𝑞𝑞6. Verify the result and if it does not suffice begin anew

13



Lead compensators: design with root locus We repeat a previous slide here, which links the time-domain

design criteria with the position of the (dominant) poles: 𝛼𝛼 + 𝑗𝑗𝛽𝛽, with 𝛼𝛼 < 0 and 𝛽𝛽 > 0 The damping ratio: 𝜁𝜁 = 𝛽𝛽/ 𝛼𝛼 + 𝑗𝑗𝛽𝛽 (0 ≤ 𝜁𝜁 ≤ 1)

The natural frequency: 𝜔𝜔𝑛𝑛 = −𝛼𝛼/𝜁𝜁 = 𝛽𝛽/ 1 − 𝜁𝜁2

The rise time: 𝑡𝑡𝑧𝑧 ≅ 1.8/𝜔𝜔𝑛𝑛 The settling time: 𝑡𝑡𝑠𝑠 = 4.6/𝜁𝜁𝜔𝜔𝑛𝑛 The peak time: 𝑡𝑡𝑝𝑝 = 𝜋𝜋/𝜔𝜔𝑑𝑑, with 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁2

The overshoot: 𝑀𝑀𝑝𝑝 = 𝑒𝑒−𝜋𝜋𝜋𝜋/ 1−𝜋𝜋2

14



Lead compensators: design with root locus We’ll now illustrate our action plan with a simple example: 𝑃𝑃 𝑠𝑠 = 1/𝑠𝑠 𝑠𝑠 + 2 The overshoot has to be less than 20 % The rise time has to be less than 0.65 s

1. We first extract desired values for 𝜁𝜁 and 𝜔𝜔𝑛𝑛 out of the requirements:

𝑀𝑀𝑝𝑝 = 𝑒𝑒−𝜋𝜋𝜋𝜋/ 1−𝜋𝜋2 ≤ 0.2This will lead to 𝜁𝜁 ≥ 0.4558𝑡𝑡𝑧𝑧 ≅ 1.8/𝜔𝜔𝑛𝑛 ≤ 0.65 s𝜔𝜔𝑛𝑛 ≥ 2.77

1. So with some safety margin we get: 𝜁𝜁 = 0.5 and 𝜔𝜔𝑛𝑛 = 3.5, or 𝑞𝑞 = −𝜔𝜔𝑛𝑛𝜁𝜁 + 𝑗𝑗 1 − 𝜁𝜁2𝜔𝜔𝑛𝑛 = −1.75 + 3.03𝑗𝑗

X: 0.4558Y: 0.2000

𝜁𝜁

𝑀𝑀𝑝𝑝

15



Lead compensators: design with root locus2. To check whether a proportional controller isn’t a possibility,

we’ll look at the root locus of 𝑃𝑃 𝑠𝑠 (with the desired closed loop pole locations indicated with orange x’es)

The root loci do not move through the desired locations, but to the right of it; hence a lead compensator is a good option

16



Lead compensators: design with root locus3. Now let’s place the zero at −z = −2.5, which is, hopefully,

far enough to the left in order to have no interference of the new pole with the dominant poles

4. The value of 𝑝𝑝 can now be determined:

𝜃𝜃 = ∓180° 2𝑘𝑘 + 1 + ∠ 𝑃𝑃 𝑞𝑞 + ∠ 𝑞𝑞 + 𝑧𝑧𝜃𝜃 = ∓180° 2𝑘𝑘 + 1 − ∠ 𝑞𝑞 − ∠ 𝑞𝑞 + 2 + ∠ 𝑞𝑞 + 2.5

𝜃𝜃 = ∓180° 2𝑘𝑘 + 1 − tan−1 3.03−1.75

− tan−1 3.030.25

+

tan−1 3.030.75

𝜃𝜃 = ∓180° 2𝑘𝑘 + 1 − 120° − 85° + 76 = 51°This is very dangerous, tan−1(3.03/−1.75) gives −60°, but if you look at the geometry, you’ll see it has to be 180° + −60° = 120°

1. So now we can find 𝑝𝑝 = cot 𝜃𝜃 ℑ 𝑞𝑞 − ℜ 𝑞𝑞 = 4.2217



Lead compensators: design with root locus

5. Now we can find 𝐾𝐾 = 𝑞𝑞+𝑝𝑝𝑞𝑞+𝑧𝑧 𝑃𝑃 𝑞𝑞

= 13.33

6. Let’s see how the root locus of 𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 looks:

1. The root locus moves exactly through 𝑞𝑞 and �𝑞𝑞, as expected

18



Lead compensators: design with root locus1. But what we really need to verify is if our time-domain

requirements are met, so let’s see how the system reacts to a step input:

2. step(K*P*C/(1+K*P*C))

3. stepinfo(K*P*C/(1+K*P*C))RiseTime: 0.4441

SettlingTime: 2.3230

SettlingMin: 0.9017

SettlingMax: 1.1878

Overshoot: 18.7759

Undershoot: 0

Peak: 1.1878

PeakTime: 1.0000

1. Each of our requirements is nicely met19



Lead compensators: design with root locus An interesting example of how you can use a Matlab tool to

design lead compensators with root locus (check it out!): http://ctms.engin.umich.edu/CTMS/index.php?example=AircraftPitch&se

ction=ControlRootLocus

It is a more realistic design example: The design does not aim to get the dominant closed loop

poles at a certain position, but the goal is to get them in a certain region

They note the current root locus is too far to the right, hence they resort to a lead compensator to bend it to the left

Both 𝑧𝑧 and 𝑝𝑝 are placed in a more intuitive manner We stick to an action plan, but it is meant for you to build up

your intuition; but remember that design is normally carried out with tools like sisotool

20

http://ctms.engin.umich.edu/CTMS/index.php?example=AircraftPitch&section=ControlRootLocus



Lag compensators

21



Lag compensators: design with root locus The goal of lag compensators, when you’re designing them

with root locus: Reduce the steady state error significantly With a marginal impact on (the relevant part of) the root locus While maintaining the freedom to adjust your position on

that root locus! Let’s show this graphically: Let’s say we have a system 𝑃𝑃 𝑠𝑠 with the

following root locus:The orange poles indicate the desired dominant pole locations (which we’ll call 𝑞𝑞and �𝑞𝑞), and as you can see they are on the root locus, for a gain value we’ll call 𝐾𝐾𝑞𝑞

22



Lag compensators: design with root locus Unfortunately a gain 𝐾𝐾𝑞𝑞 does not suffice, since it corresponds

to a steady state error that is too large A lag compensator is capable of decreasing the steady state

error (you can’t see that on the root locus plot), but while leaving the root locus practically unaltered:

As the positions of the pole and zero onlydetermine 𝛽𝛽 and 𝛼𝛼; we still have one degreeof freedom to chose our position on the rootlocus: 𝐾𝐾

Now we get to the question of how a lag compensator succeeds in doing that; as you probably expected, it will be thanks to the fact that we placed the compensator’s pole ( ) and zero ( ) close to the imaginary axis

23



Lag compensators: design with root locus As you know, a point 𝑐𝑐 in the s-plane is on the root locus of 𝐺𝐺 𝑠𝑠 if ∠ 𝐾𝐾𝐺𝐺 𝑠𝑠 = ±180° 2𝑘𝑘 + 1 , and if 𝐾𝐾𝐺𝐺 𝑐𝑐 = 1

So if 𝑞𝑞 has to (approximately) lie on the root locus of both 𝑃𝑃 𝑠𝑠 and 𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 (with 𝐶𝐶 𝑠𝑠 = 1+1/𝜏𝜏

1+1/𝛽𝛽𝜏𝜏in this section; so with the 𝐾𝐾

outside of 𝐶𝐶 𝑠𝑠 ) and if this has to be for (approximately) the same 𝐾𝐾 then: ∠ 𝑃𝑃 𝑞𝑞 ≅ ∠ 𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞 𝑃𝑃 𝑞𝑞 ≅ 𝑃𝑃 𝑞𝑞 𝐶𝐶 𝑞𝑞(every root locus that contains 𝑞𝑞 automatically also contains �𝑞𝑞, its complex conjugate) This is the case when the lag compensator’s pole ( ) lies much

closer to its zero ( ) than either lies to 𝑞𝑞 ( )

24



Lag compensators: design with root locus It is quite readily visible from the following root locus plot that

the condition for a small impact on (a region of) the root locus by a lag compensator is that its pole and zero lie much closer to each other than to that region: The effect of this pole and

zero on the angle at 𝑞𝑞 is only the angle between the dotted and the dashed red line

The effect of this pole and zero on the magnitude at 𝑞𝑞is only the ratio between the length of the dashed and the dotted red lines

25



Lag compensators: design with root locus We now know how a lag compensator can leave the root

locus almost unaltered The second thing we need to know is how a lag compensator

with its pole and zero very close to each other can still result in a significant rise in the DC gain

That’s quite simple in fact, you just have to look at the formula of the lag compensator and its DC gain:

lim𝑠𝑠→0

𝐾𝐾𝐶𝐶 𝑠𝑠 = lim𝑠𝑠→0

𝐾𝐾 1+1/𝜏𝜏1+1/𝛽𝛽𝜏𝜏

= 𝐾𝐾𝛽𝛽

This shows we can move freely on the root locus by adjusting 𝐾𝐾 and then we can tune the DC gain by adjusting 𝛽𝛽

26



Lag compensators: design with root locus Let’s summarize this by looking at our three degrees of

freedom in the equation of the lag compensator (𝐾𝐾𝐶𝐶 𝑠𝑠 ):

𝐾𝐾𝐶𝐶 𝑠𝑠 = 𝐾𝐾 𝑠𝑠+1/𝜏𝜏𝑠𝑠+1/𝛽𝛽𝜏𝜏

with 1 < 𝛽𝛽

We use 𝐾𝐾 to position ourselves on the root locus We use 𝛽𝛽 to tune the DC gain We use 𝛼𝛼 to push the poles close enough to the imaginary

axis so the pole (at 1/𝛽𝛽𝛼𝛼) is close enough to the zero (at 1/𝛼𝛼) so they hardly change the root locus in the neighborhood of 𝑞𝑞 and �𝑞𝑞 (the desired dominant poles)

The only question that is left open is how to determine the size of 𝛼𝛼: we recommend to take 𝛼𝛼 = 50/ ℜ 𝑞𝑞

27



Lag compensators: design with root locus Side note: now −1/𝛽𝛽𝛼𝛼 lies much closer to the imaginary axis

than 𝑞𝑞, doesn’t that make it dominant (which means it dominates the transient behavior)?

The answer is: yes and no, depending what you mean with ‘transient behavior’

We will try to leave this ambiguity behind us with an example:

Take 𝑃𝑃 𝑠𝑠 = 1𝑠𝑠+2+3𝑗𝑗 𝑠𝑠+2−3𝑗𝑗

= 1𝑠𝑠2+4𝑠𝑠+13

And 𝐶𝐶 𝑠𝑠 = 𝑠𝑠+0.04𝑠𝑠+0.004

, hence, we used 𝛼𝛼 = 50ℜ 𝑞𝑞

= 25 and

𝛽𝛽 = 10 (a typical value) These poles and zero are located as

follows in the s-plane:

28



Lag compensators: design with root locus Let’s first look at how the steady-state is reached in several

situations:

29



Lag compensators: design with root locus It is clear that the pole at −0.004 completely determines

when and how the eventual steady-state is reached, even if the zero at −0.04 is present

So does this mean that 𝑞𝑞 and �𝑞𝑞 are no longer the dominant poles in the system, but −1/𝛽𝛽𝛼𝛼 is?

It is, if you find the reaching of the eventual steady-state the most important characteristic of the transient behavior

It is clear that we do not do that;to us there is a more importantcharacteristic; which is when and howthe first significant step towards equilibrium is made

This changes markedly thanks to the presence of the zero at −0.04

30



Lag compensators: design with root locus If you zoom in, you can see that the effect of the zero is

indeed that this interpretation of the transient behavior is unaltered:

31



Lag compensators: design with root locus So due to its location much closer to the imaginary axis, the

pole at 1/𝛽𝛽𝛼𝛼 will eventually be the determining factor in approaching the steady state behavior

But due to the fact that there is a zero that is very close to that pole, its effect will not be visible as long as there are large frequencies that dominate the input, which is at the beginning of the behavior, when the poles at 𝑞𝑞 and �𝑞𝑞dominate the behavior

So 𝑠𝑠+0.04𝑠𝑠2+4𝑠𝑠+13 𝑠𝑠+0.004

has two different time-scales:

A fast one, which comes from the original system and which is to us the most important in the transient behavior

A slow one, which is added by the lag compensator and which only comes into effect after the transient period

32



Lag compensators: design with root locus This allows us to explain once more how a lag compensator

can succeed in increasing the DC gain without impacting the time-domain behavior:

You add a very slow component ( 1+1/𝜏𝜏1+1/𝛽𝛽𝜏𝜏

with 𝛼𝛼 and 𝛽𝛽fairly large), which only takes effect after the (fast part of the) transient behavior is finishedThe effect of that component is to increase the gain with a factor 𝛽𝛽 after longer time

A factor 𝛽𝛽

33



Lag compensators: design with root locus Now we’re ready to provide a recipe to design a lag

compensator with root locus:1. Determine the desired dominant pole locations, 𝑞𝑞 and �𝑞𝑞2. Determine what gain 𝑃𝑃 𝑠𝑠 should be provided with in order

for its closed loop poles to be equal to 𝑞𝑞 and �𝑞𝑞 this is 𝐾𝐾 You can do this graphically, from the root locus plot Or analytically from the demand 𝐾𝐾𝑃𝑃 𝑞𝑞 = 1

3. Translate the steady-state requirement into a requirement on the DC gain of the lag compensator (lim

𝑠𝑠→0𝐾𝐾𝐶𝐶 𝑠𝑠 = 𝐾𝐾𝛽𝛽) and

determine 𝛽𝛽4. Determine 𝛼𝛼 as 50/ ℜ 𝑞𝑞5. Verify the behavior

34



Lag compensators: design with root locus, example

Take the system 𝑃𝑃 𝑠𝑠 = 0.3𝑠𝑠 𝑠𝑠+1 𝑠𝑠+2

We want to increase the DC gain with a factor 30 with respect to the DC gain of 𝑃𝑃 𝑠𝑠 , and we want the dominant poles to be such that 𝜁𝜁 = 0.5 and 𝜔𝜔𝑛𝑛 = 0.67 rad/s

1. This corresponds to dominant poles at −0.33 ± 𝑗𝑗0.582. You can find the required value of 𝐾𝐾 for this:

𝐾𝐾 = 1𝑃𝑃 𝑠𝑠

= 3.5

35



Lag compensators: design with root locus, example3. Our steady-state requirement is already translated into a

requirement on the DC gain; it needs to be 30 times the DC gain of 𝑃𝑃 𝑠𝑠

So lim𝑠𝑠→0

𝐾𝐾𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 = 30 lim𝑠𝑠→0

𝑃𝑃 𝑠𝑠

Or lim𝑠𝑠→0

𝐾𝐾𝐶𝐶 𝑠𝑠 = 𝐾𝐾𝛽𝛽 = 30

we find 𝛽𝛽 = 30/𝐾𝐾 = 8.64. Finding 𝛼𝛼 is easy: 𝛼𝛼 = 50/ ℜ 𝑞𝑞 = 50/0.33 = 1505. Let’s verify our result now, we will do this by looking at the

old and new positions of the closed loop poles (we will not do it here, but you should of course also always verify the steady-state behavior)

36



Lag compensators: design with root locus, example3. The closed loop pole and zero locations of 𝐾𝐾𝑃𝑃 𝑠𝑠 and

𝐾𝐾𝑃𝑃 𝑠𝑠 𝐶𝐶 𝑠𝑠 are shown here below:

4. As you can see there is hardly any distinction (that’s why they weren’t plotted on the same graph, because they almost entirely overlap)

5. The reason is that the factor 50 we use in determining 𝛼𝛼 is very large; this gives very small pole-zero values; this also means the steady-state value is very slowly approached, so sometimes you’ll want a smaller factor than 50 and as you can see in this example, there is quite some margin left

37

Design with root locusmaapc/static/files/SYSTHEORY/Slid… · Design with root locus 1. Systems and...

Documents

Transcript of Design with root locusmaapc/static/files/SYSTHEORY/Slid… · Design with root locus 1. Systems and...