

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The closed-loop control system is stable, i.e., limt→∞ c n(t) = 0, which is ensured if all of poles of the closed-loop transfer function M(s) have a negative real part. This is important because it ensures that c(t) → c f(t) as time increases.
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The convergence c n(t) → 0 is fast.
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The forced response is equal to the desired output
.

Closed-loop system
The design of a controller is performed by suitably assigning the poles (the roots of 1 + G(s)H(s)) and the zeros (the roots of G(s)) of the closed-loop transfer function M(s). In classical control, the plant poles and zeros are always assumed to be known, whereas the structure of the controller is proposed to satisfy the steady-state response specifications, for instance. However, the exact location of the poles and zeros of the controller must be selected to ensure that the closed-loop poles, i.e., the roots of 1 + G(s)H(s), are placed at the desired values and that the closed-loop zeros, i.e., G(s) = 0, are also suitably located.1 In this chapter, one of the main classical control techniques for control design is studied: the root locus method (see [1] for a historical perspective). This method assigns poles of the closed-loop transfer function M(s) through a suitable modification of the open-loop transfer function G(s)H(s). Moreover, it is also possible to place the zeros of the closed-loop transfer function M(s) such that their effects on the closed-loop system response are small.
5.1 Drawing the Root Locus Diagram
Assuming that the poles and zeros of the plant are known, the root locus method is a graphical tool that is useful to locate the closed-loop poles by suitably selecting poles and zeros of a controller. Moreover, this can also be performed by cancelling some undesired (slow) closed-loop poles with some controller zeros. According to Sect. 3.5.1, this achieves system order reduction and the elimination of some zeros such that the closed-loop system transient response can be approximated by some of the simple cases studied in Sects. 3.1.1 and 3.3.1. This simplifies the task of designing a controller such that the closed-loop transfer function M(s) has poles and zeros that ensure the desired transient response specifications.
The root locus diagram is represented by several curves whose points constitute the closed-loop system poles parameterized by the open loop-gain, k, which takes values from k = 0 to k = +∞. According to (5.1), every closed-loop pole s, i.e., a point in the s plane belonging to the root locus, must satisfy 1 + G(s)H(s) = 0. The root locus diagram is drawn by proposing points s on the complex plane s, which are checked to satisfy 1 + G(s)H(s) = 0. To verify this condition in a simple manner, a set of rules are proposed, which are listed next. Following these rules, the root locus diagram is drawn using the poles and zeros of the open-loop transfer function G(s)H(s) as data. This means that the closed-loop poles are determined by the open-loop poles and zeros.








Graphical representation of factors s − z j and s − p i
5.1.1 Rules for Drawing the Root Locus Diagram
- 1.
The root locus starts ( k = 0) at the open-loop poles.
- 2.
Root locus ends ( k →∞ ) at the open-loop zeros.
From (5.8):if k →∞. This means that the closed-loop poles (those satisfying 1 + G(s)H(s) = 0) are identical to the open-loop zeros (s = z j, j = 1, …, m) when k →∞.
- 3.
When k →∞ there are n − m branches of the root locus that tend toward some point at the infinity of the plane s. This means that the open-loop transfer function G(s)H(s) has n − m zeros at infinity.
These branches can be identified by the angle of the asymptote where the root locus diagram approaches as k →∞. The angle that each asymptote forms with the positive real axis can be computed as:(5.9)Fig. 5.3Asymptote angle in rule 3
- 4.
Point σ a where asymptotes intersect the real axis is computed as [ 5 ], Chap. 6:
- 5.
Consider a point on the real axis of the plane s. Suppose that the number of real open-loop poles plus the number of real open-loop zeros on the right of such a point is equal to an odd number. Then such a point on the real axis belongs to the root locus. If this is not the case, then such a point does not belong to the root locus.
Observe Fig. 5.4 and recall the angle condition (5.7). Notice that two open-loop complex conjugate poles or zeros produce angles that added to result in 0∘ or 360∘ and, hence, have no contribution to the angle condition (5.7). This means that, in this rule, only real open-loop poles and zeros must be considered in (5.7). On the other hand, notice that any real pole or zero located at the left of the test point s contributes with a zero angle. This means that, in this rule, in the expression (5.7), only real open-loop poles and zeros located on the right of the test point s must be considered. Notice that each real open-loop zero on the right of the point s contributes with a + 180∘ angle, whereas each real open-loop pole at the right of the same point contributes with a − 180∘ angle. Hence, the total angle contributed by all open-loop poles and zeros appearing on the right of the test point s is:Fig. 5.4Open-loop poles and zeros used to study rule 5
- 6.
The root locus is symmetrical with respect to the real axis.
This is easily understood by recalling that all complex poles appear simultaneously with their complex conjugate pair (see first paragraph in Sect. 3.4.3).
- 7.
The departure angle from an open-loop complex pole is computed as:(5.10)Consider Fig. 5.5. Assume that the point s belongs to the root locus and it is very close to open-loop pole at p v. The angle θ pv is the departure angle from the open-loop complex pole at p v. According to the angle condition (5.7):Fig. 5.5
Open-loop poles and zeros in the study of rule 7
- 8.
The arrival angle to an open-loop complex zero is computed as:(5.11)Consider Fig. 5.6. Assume that the point s belongs to the root locus and is very close to zero at z v. The angle θ zv is the arrival angle to the open-loop zero at z v. According to the angle condition (5.7):Fig. 5.6
Open-loop poles and zeros in the study of rule 8
- 9.
The open-loop gain, k, required for a point s, belonging to root locus, to actually be selected as a desired closed-loop pole is computed according to the magnitude condition ( 5.6 ):
- 10.
The points where the root locus passes through the imaginary axis can be determined using Routh’s criterion (see Sect. 4.3).
- 11.
If an additional open-loop pole located on the left half-plane is considered, “instability tends to increase” in the closed-loop system. This effect is stronger as the additional pole is placed closer to the origin.
Consider Fig. 5.7, where s represents a point that belongs to the root locus and the open-loop transfer function is assumed to have only the shown two poles without zeros. According to the angle condition (5.7):Fig. 5.7A point belonging to the root locus for an open-loop system with two poles
Fig. 5.8The root locus is pushed toward the right when an additional open-loop pole is considered
- 12.
If an additional open-loop zero located on the left half-plane is considered, “stability tends to increase” in the closed-loop system. This effect is stronger as the additional zero is placed closer to the origin.
Consider again Fig. 5.7. According to the angle condition (5.7):Fig. 5.9The root locus is pulled toward the left when an additional open-loop zero is considered
5.2 Root Locus-Based Analysis and Design
5.2.1 Proportional Control of Position








Proportional position control system

Root locus for
On the other hand, according to the angle condition, − (θ 1 + θ 2) = −180∘ = −(α + θ 1) in Fig. 5.11, it is concluded that α = θ 2, i.e., both triangles t 1 and t 2 must be identical for any closed-loop pole s. Hence, both branches referred above, which are shown in Fig. 5.11, must be parallel to the imaginary axis. This means that the breakaway point is located at the middle point between s = 0 and s = −a, i.e., at (0 − a)∕2 = −a∕2. This can also be verified using rules 3 and 4. Finally, according to rule 6, both branches are symmetrical with respect to the real axis.






The above discussion shows that it is not possible to achieve a closed-loop system response that is simultaneously fast and well damped. This is a direct consequence of the fact that closed-loop poles cannot be assigned at any arbitrary location of the s plane, as they can only be located on the thick straight line shown in Fig. 5.11. In the next example, it is shown that the introduction of an additional zero in the open-loop transfer function of the present example allows the closed loop poles to be assigned at any point on the s plane.
Example 5.1

Root locus for closed-loop system in Fig. 5.10

where d is a vector containing all of the specific values of the gain k p, which the user wants to be considered to plot the root locus.

Root locus in Fig. 5.12 when some closed-loop pole pairs are selected
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Step response of the closed-loop system in Fig. 5.10, when the closed-loop poles are located as indicated in Fig. 5.13. The + line corresponds to poles at “o.” The continuous line corresponds to poles at the triangle up. The dashed line corresponds to poles at the triangle down. The dash–dot line corresponds to poles at “*.” The dotted line corresponds to poles at the square
5.2.2 Proportional–Derivative Control of Position









Proportional–derivative (PD) control of position

Root locus for . (a) c > a. (b) c < a
On the other hand, if the zero at s = −c is placed between the open-loop poles at s = 0 and s = −a, then, according to rule 5, the root locus exists on two segments on the negative real axis located between points s = −c and s = 0 and on the left of the pole at s = −a. Notice that, now, the branch beginning (k d = 0) at s = 0 approaches the zero at s = −c, whereas the branch beginning (k d = 0) at s = −a tends toward infinity along the negative real axis. Also notice that this is possible without the necessity for any branch to exist outside the real axis, as is shown in Fig. 5.16b.
From the study of the resulting root locus, it is possible to realize that the closed-loop system is stable for any k p > 0 and k d > 0 because both possibilities for the root locus that have been presented in Fig. 5.16 show that the closed-loop poles are always located on the left half-plane s, i.e., the closed-loop poles have a negative real part. Finally, according to Sect. 3.8.3, there always exist gains k p and k d allowing any values to be chosen for both ω n and ζ. This means that it is always possible to assign both closed-loop poles at any desired point on the left half-plane s.
Notice that the open-loop zero at s = −c is also a zero of the closed-loop transfer function; hence, it also affects the closed-loop system transient response. This means that the transient response will not have the exact specifications computed using (3.71).
Example 5.2


Step response of the closed-loop system in Fig. 5.15, when a closed-loop first-order response with 0.05[s] time constant is desired. Dashed: desired response and closed-loop response when c = a. Dotted: closed-loop response when c = 4. Continuous: closed-loop response when c = 10

Root locus corresponding to Fig. 5.15 when c = a
When c = 4, the closed-loop poles are at s = −20, as desired, and s = −3.25. The latter pole is slow and does not cancel with zero at s = −c = −4; hence, its effects are important in the closed-loop transient response. This explains why, in this case, the response in Fig. 5.17 is slow. When c = 10, the closed-loop poles are located at s = −20 and s = −13, i.e., the pole at s = −13 cannot be cancelled by the zero at s = −c. This results in an overshoot, i.e., the desired response is not accomplished again. The reason for this overshoot is explained in Sect. 8.1.2, where it is stated that overshoot is unavoidable whenever all of the closed-loop poles are on the left of an open-loop zero despite all the closed-loop poles being real, i.e., as in the present case.
Example 5.3


Root locus diagram corresponding to Fig. 5.15 when c > a and two closed-loop complex conjugate poles are specified (triangle up)

Step response of the closed-loop system in Fig. 5.15, when c > a and two closed-loop complex conjugate poles are specified. Dashed: desired response. Continuous: closed-loop response
5.2.3 Position Control Using a Lead Compensator






Position control using a lead compensator
5.2.4 Proportional–Integral Control of Velocity
![$$\displaystyle \begin{aligned} \begin{array}{rcl} \omega(s)&\displaystyle =&\displaystyle \frac{1}{s+a}[kI^*(s)-\frac{1}{J}T_p(s)],\\ a&\displaystyle =&\displaystyle \frac{b}{J}>0, k=\frac{nk_m}{J}>0,\vspace{-4pt} \end{array} \end{aligned} $$](/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_5_Chapter/454499_1_En_5_Chapter_TeX_Equac.png)




Proportional–integral control of velocity. (a) The complete control system. (b) T p(s) = 0. (c) ω d(s) = 0

- i)
There always exist some gains k p and k i allowing both closed-loop poles to be placed at any point on the left half-plane; hence, it is possible to tune the PI controller using a trial and error procedure (see Sect. 3.8.4).
- ii)
The open-loop zero located at s = −c is also a zero of the closed-loop transfer function G 1(s), i.e., it also affects the closed-loop transient response. This means that the transient response does not have the specifications designed using (3.71) to choose the closed-loop poles. This poses the following two possibilities.
- 1.
The problem pointed out at ii) can be eliminated if it is chosen:(5.21)
. Then, if a time constant τ is specified:
(5.22)) is shown in Fig. 5.23. This can be easily verified using rules 3 and 5. As there is only one open-loop pole, there is only one closed-loop pole, which must be real and tends toward one zero at infinity (on an asymptote forming − 180∘ with the positive real axis) because there is no open-loop zero. Furthermore, the magnitude condition:
is reached when:
(5.23)Fig. 5.23Root locus for c = a,
- 2.
Trying to solve the above problem, we might abandon the tuning rule in (5.22) and (5.21). As we now have c ≠ a, the closed-loop transfer function corresponding to the block diagram in Fig. 5.22c is:(5.24)
, the steady-state deviation is zero again: limt→∞ ω(t) = 0, because the transfer function in (5.24) has a zero at s = 0. On the other hand, poles of the transfer function in (5.24), which determine how fast the velocity deviation vanishes, are identical to the poles of the transfer function
; hence, they can be assigned using the root locus method from (5.20). As has been explained, the corresponding root locus is identical to both cases shown in Fig. 5.16 replacing k d by k p. It is important to stress that the transfer function in (5.24) has no zero at s = −c shown in the root locus diagram in Fig. 5.16. This means that none of the closed-loop poles obtained using the root locus method can cancelled with the zero at s = −c and the slowest pole will have the most significant effect on the time required for the velocity deviation due to the disturbance to vanish. With these ideas in mind, the following is concluded.
-
To render small the effect of the zero at s = −c on the transient response to a given velocity reference, one closed-loop pole must be placed close to the zero at s = −c. The other closed-loop pole (the fastest one) moves away to the left and tends toward infinity as the slow pole approaches to s = −c.
-
This means that the fast pole determines the transient response, i.e., the time constant, to a given velocity reference. Then, if it is desired to fix some finite value for the time constant (such that the fast pole is placed at a finite point on the negative real axis), the slow pole is always relatively far from zero at s = −c. This means that the transient response is always affected by both poles and the zero at s = −c. Hence, if c ≠ a, a tuning rule cannot be determined such that transient response to a given velocity reference satisfies the desired specifications. Notice that this is also true if the response to a reference is specified by two complex conjugate poles, as the zero at s = −c cannot be cancelled; hence, it has the effect of modifying the transient response with respect to that specified by the two complex conjugate poles.
-
It is more convenient to select c > a because the slowest closed-loop pole (that approaching s = −c) is shifted further to the left (it is faster) compared with the case when c < a is chosen.
-
This also implies that, if it is desired that the velocity deviation due to a disturbance vanishes faster, we must choose c ≠ a with c > 0 larger. According to c = k i∕k p, this means that a larger integral gain is obtained.
Thus, although it is possible to render shorter the time it takes to the velocity deviation due to disturbance to vanish, it is concluded that the controller gains k p y k i cannot be exactly computed to ensure that, simultaneously, the desired transient response specifications to a given velocity reference are accomplished and that the disturbance effects vanish as fast as desired. This statement is verified experimentally in Chap. 10 and, because of this, in that chapter a modified PI velocity controller is introduced, solving this situation.
-





Example 5.4

MATLAB/Simulink diagram corresponding to the closed-loop system in Fig. 5.22a



Response to a step command in desired velocity ω d(s)


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Response to a step disturbance T p(s)
In Figs. 5.25 and 5.26 the following data are presented. (i) The line marked with “+” corresponds to c = 1, (ii) Continuous line: c = 5, (iii) Dashed line: c = 7, (iv) Dash–dot line: c = 10, and (v) Dotted line: c = 15. Notice that the disturbance effect is rejected very slowly when c = 1. In this case, the closed-loop poles are at s = −20 and s = −0.6842, which can be verified using the MATLAB command “pole(M)”. Recall that the transfer function in (5.24) has no zero to cancel the pole at s = −0.6842.2 This explains the slow rejection of the disturbance effects in this case. Also notice that the response to the velocity reference is also slow. This is because the effect of the pole at s = −0.6842 is not suitably cancelled by the zero at s = −c = −1.
Notice that the disturbance rejection and the response to the reference improve as c increases. Moreover, the system response exactly matches the desired response when c = 7 = a. However, the disturbance effects are still important in this case. When c > 7 = a, the disturbance rejection further improves, but the response to the reference exhibits overshoot, i.e., it is no longer a first-order response. According to Sect. 8.1.2, this behavior appears, and it is unavoidable, when the closed-loop poles are on the left of an open-loop zero. In this respect, notice that the closed-loop poles are at s = −20 and s = −13 when c = 10 and at s = −20 and s = −39 when c = 15.
The simulation results in this example corroborate the system behavior predicted in the above discussion, i.e., the classical PI control has limitations when it is required to satisfy simultaneously a specified response to a reference and a specified performance for disturbance rejection.
5.2.5 Proportional–Integral–Derivative Control of Position
![$$\displaystyle \begin{aligned} \begin{array}{rcl} \theta(s)&\displaystyle =&\displaystyle \frac{1}{s(s+a)}[kI^*(s)-\frac{1}{J}T_p(s)], \end{array} \end{aligned} $$](/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_5_Chapter/454499_1_En_5_Chapter_TeX_Equal.png)








Proportional–integral–derivative (PID) position control. (a) The complete control system. (b) T p(s) = 0. (c) θ d(s) = 0




It is stressed that the three controller gains are positive. As a rule of thumb, if − p 3 > 6|σ 1| then the real and imaginary parts of the poles at p 1 and p 2 can be computed using (3.71) such that the desired rise time t r and overshoot M p(%) are obtained. However, the system response presents some important differences with respect to these values because of the pair of zeros contained in the transfer function in (5.26). Although this is an important drawback of the tuning rule in (5.28), these values can be used as rough approximations of the required gains for the controller to perform fine adjustments afterward, by trial and error, until the desired transient response specifications are accomplished. To this aim, as the closed-loop system is third order, it is important to recall the stability rule obtained in Example 4.12 of Sect. 4.3. The possibility of adjusting the gains of a PID controller by trial and error (see Sect. 5.3) is one of the main reasons for the success of PID control in industry.

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Different possibilities for root locus diagrams of the PID control of position
-
Design a PD controller with transfer function:
-
Given the open-loop transfer function designed in the previous step, introduce the following factor:
-
Compute the PID controller gains using (5.29), i.e.,(5.30)









Proportional–derivative control of the position

Root locus diagram for the system in Fig. 5.29

Root locus for the open-loop transfer function shown in (5.34)


However, it is important to point out a drawback of this design approach: according to (5.30) a small α results in a small integral gain k i. As a consequence, the deviation of the position produced by the external disturbance vanishes very slowly, which may be unacceptable in practice. This can also be explained from the root locus diagram in Fig. 5.31. Notice that a closed-loop pole (located, say, at s = −ε, ε > 0) approaches the zero at s = −α, which means that both cancel each other out in the transfer function shown in (5.26), i.e., the effect of none of them is observed in the transient response to a reference of the position. However, the zero at s = −α does not appear in the transfer function shown in (5.27). But the pole at s = −ε is still present in this transfer function; hence, it significantly affects the transient response: this slow pole (close to the origin) is responsible for a slow transient response when an external disturbance appears.
Despite this drawback, the traditional criteria for root locus-based design suggest proceeding as above when designing PID controllers. Moreover, it is interesting to say that these problems in traditional design methods remain without a solution despite the fact that they have been previously pointed out in some research works. See [10] for instance.
On the other hand, according to (5.30), a larger integral gain can be obtained (to achieve a faster disturbance rejection) by choosing a larger value for α. However, according to the previous discussion, this will result in a transient response to a reference of position that does not satisfy the desired specifications. Hence, it is concluded that there is no tuning rule allowing the gains of a PID controller for position to be computed exactly, simultaneously satisfying the desired transient response specifications to a given reference of the position and producing a satisfactory rejection of the effects of an external torque disturbance. These observations are experimentally verified in Chap. 11 where, given the described drawback, some new controllers solving these problems are designed and experimentally tested.
Finally, despite the above drawback, it is important to recall what was indicated just after (5.28): PID control is one of the most frequently employed controllers in industry because it can be tuned by trial and error (see Sect. 5.3). This has to be performed by taking into account the stability rule obtained in the Example 4.12, Sect. 4.3.
Example 5.5

MATLAB/Simulink diagram for the closed-loop system in Fig. 5.27a

Response to a step command in the desired position θ d(s)

Response to a step disturbance T p(s)
Running the above code and the simulation in Fig. 5.32 several times, the responses are obtained when using different values for α. In Figs. 5.33 and 5.34: (i) Continuous lines stand for α = 0.1, (ii) Dash–dot: α = 1, (iii) Dotted: α = 10, and the fastest dashed line: α = 20.
It is observed that, when α = 0.1, the response is different from
the desired one, i.e., the slowest dashed line in Fig. 5.33. This is because of
the closed-loop zero at − β,
located at s = −54.5225 in this
case, which deviates the system response from the desired one (See
also Fig. 5.20. It is remarked that the desired
closed-loop poles are located at , as desired, in this case. Also
notice that the disturbance rejection is so slow that it seems that
a steady-state deviation might exist.
As α is chosen to be larger, the performance observed in Fig. 5.34 improves with respect to the disturbance rejection, but the price to be paid is that the response to the position reference is increasingly different from the desired one, as observed in Fig. 5.33.
Thus, the simulation results in this example corroborate the predictions stated in the above discussion in the present section.
5.2.6 Assigning the Desired Closed-Loop Poles






Root locus for G(s)H(s) in (5.36)








Root locus diagram for G(s)H(s) in (5.37)




An additional pole moves the breakaway point to the right
This description of the root locus diagram in Fig. 5.36 allows us to conclude that there is always at least one closed-loop pole with a positive real part, i.e., the closed-loop system is unstable for any positive value of k. As k can be interpreted as the gain of a proportional controller, it is concluded that it is not possible to render the closed-loop system stable using any proportional controller and, thus, another controller must be proposed. According to rule 12 it is required to use a controller introducing an open-loop zero because this bends the root locus to the left accomplishing closed-loop stability.

![$$\displaystyle \begin{aligned} \begin{array}{rcl} G(s)H(s)=\frac{116137\;k\;l_4}{l_1l_2l_3}\angle [\theta_4-(\theta_1+\theta_2+\theta_3)],{} \end{array} \end{aligned} $$](/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_5_Chapter/454499_1_En_5_Chapter_TeX_Equ39.png)









Root loci for G(s)H(s) defined in (5.38) obtained by changing b. (a) b = 31.24 < 36.50. (b) b > 36.50. (c) b = 36.50
It is observed, in Fig. 5.38c, that only two closed-loop poles exist. It is observed, in Figs. 5.38a and b, that three closed-loop poles exist and one of them is on the right half-plane (closed-loop instability) if the gain k is too small. On the other hand, if k is too large, two complex conjugate poles exist with too large an imaginary part, i.e., producing a fast oscillation. Although in theory this can work because the three poles have negative real parts, a large k produces a number of practical problems such as noise amplification and power amplifier saturation, i.e., the control system may not work correctly in practice. Hence, a good design is that enabling the designer to locate as desired the closed-loop poles avoiding the use of values that are too large and values that are too small for the gain k.







Open-loop poles and zeros when trying to locate the desired closed-loop poles at s = −25 ± j40
It is important to say that this control system is employed to regulate the output at a zero value. This implies that no matter what the system type, the desired output (i.e., zero) is reached. Hence, consideration with respect to the steady-state response is not necessary.


Closed-loop response of the system (5.38) from an initial output that is different from zero. Continuous line: designed response. Dashed line: desired response. Vertical axis: y[m]. Horizontal axis: time in seconds
5.2.7 Proportional–Integral–Derivative Control of an Unstable Plant







Different possibilities for the root locus diagram when using (5.44) as the open-loop transfer function






Applying Routh’s criterion to the polynomial in (5.47)
s 4 |
1 |
a 2 |
a 0 |
s 3 |
a 3 |
a 1 |
0 |
s 2 |
|
a 0 |
0 |
s 1 |
|
0 |
|
s 0 |
a 0 |



Example 5.6


5.2.8 Control of a Ball and Beam System


Closed-loop system. Proportional control with gain γ
Applying Routh’s criterion to the polynomial s 4 + as 3 + γA x kρ
s 4 |
1 |
0 |
γA x kρ |
s 3 |
a |
0 |
0 |
s 2 |
0 ≈ ε |
γA x kρ |
|
s 1 |
|
0 |
|
s 0 |
γA x kρ |




Closed-loop system. Use of a lead compensator

Closed-loop system. Use of two internal loops and a lead compensator
Applying Routh’s criterion to polynomial s 5 + (a + c)s 4 + acs 3 + γA x kρs + γA x kρδ
s 5 |
1 |
ac |
γA x kρ |
s 4 |
a + c |
0 |
γA x kρδ |
s 3 |
ac |
|
|
s 2 |
|
γA x kρδ |
|
s 1 |
|
0 |
|
s 0 |
γA x kρδ |





-
A greater σ > 0 is chosen, i.e., if the system:(5.52)
-
b > 0 Approaches to zero and c > 0 are large, i.e., if c > b. Note that this is in agreement with rules 11 and 12.

Root locus possibilities for a ball and beam system

-
b and c are proposed such that c > b > 0.
-
MATLAB is employed to draw the root locus diagram.
-
γ > 0 is chosen as that value placing all the closed-loop system poles on the left half-plane. This means that γ is the parameter used by the method to plot the root locus diagram.
-
If such a γ > 0 does not exist, then we go back to the first step.


Root locus diagram for the ball and beam system that has been designed

The ball and beam closed-loop response
-
Once the values for γ, c and b are chosen, some experimental tests must be performed to verify that a good performance is obtained. If this is not the case, then we go back again to the first step in the procedure detailed above.
5.2.9 Assigning the Desired Closed-Loop Poles for a Ball and Beam System

![$$\displaystyle \begin{aligned} \begin{array}{rcl} \zeta&\displaystyle =&\displaystyle \sqrt{\frac{\ln^2{\left(\frac{Mp(\%)}{100}\right)}}{ \ln^2{\left(\frac{Mp(\%)}{100}\right)}+\pi^2} },\\ \omega_d&\displaystyle =&\displaystyle \frac{1}{t_r}\left[ \pi-\arctan\left(\frac{\sqrt{1-\zeta^2}}{\zeta} \right)\right],\\ \omega_n&\displaystyle =&\displaystyle \frac{\omega_d}{\sqrt{1-\zeta^2}},\vspace{-4pt} \end{array} \end{aligned} $$](/epubstore/G/V-M-Guzman/Automatic-Control-With-Experiments/OEBPS/images/454499_1_En_5_Chapter/454499_1_En_5_Chapter_TeX_Equbw.png)














Open-loop pole–zero contributions to the root locus diagram for a ball and beam system

Root locus diagram for the ball and beam system when using the block diagram in Fig. 5.46 and the controller gains b = 0.7, c = 2.8658, γ = 1.3531

Closed-loop time response of the ball and beam system when using the block diagram in Fig. 5.46 and the controller gains b = 0.7, c = 2.8658, γ = 1.3531 (continuous). Dashed: desired response


Use of the lead compensator in the feedback
path. The factor on the left is included to ensure
that limt→∞ x(t) = x d

Closed-loop time response of the ball and beam system when using the block diagram in Fig. 5.53 and the controller gains b = 0.7, c = 2.8658, γ = 1.3531 (continuous). Dashed: desired response

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Root locus for the ball and beam system according to the block diagram in Fig. 5.53 and controller gains b = 1.4, c = 5.1138, γ = 2.1868
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Closed-loop time response of the ball and beam system according to the block diagram in Fig. 5.53 and controller gains b = 1.4, c = 5.1138, γ = 2.1868 (continuous). Dashed: desired response
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Root locus diagram for the ball and beam system when using the block diagram in Fig. 5.53 and the controller gains b = 2, c = 11.7121 and γ = 4.6336
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Closed-loop time response of the ball and beam system when using the block diagram in Fig. 5.53 and the controller gains b = 2, c = 11.7121 and γ = 4.6336 (continuous). Dashed: desired response
5.3 Case Study: Additional Notes on the PID Control of Position for a Permanent Magnet Brushed DC Motor
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Closed-loop pole location for PID position control in a DC motor
-
k p > 0 and k d > 0 are small, whereas k i > 0 is large.
-
Suppose that p 1 and p 2 are complex conjugate. Then,
in Fig. 5.59. According to (5.68), the product
is large, which implies a faster response because a real pole produces faster responses as p 3 moves to the left and a second-order system is faster as ω n is larger. Note that k i can be kept constant and, according to (5.67), k p > 0 can be rendered small if both σ 1 and σ 2 approach zero, i.e., if θ > 0 in Fig. 5.59 approaches zero and, hence, damping approaches zero (without affecting ω n nor − p 3). According to (5.66), this also renders k d > 0 small. Also notice that this is in agreement with the second inequality in (5.69), i.e.,
. If k i > 0 is large and k p > 0, k d > 0 are small, then this condition tends not to be valid, i.e., the closed-loop system approaches instability and becomes more oscillatory.
-
Suppose that p 1 and p 2 are real and different, i.e., ω 1 = 0. Thus, a more damped and, hence, slower response is obtained. According to (5.68), this results in a smaller k i > 0. However, according to (5.66), (5.67), k p > 0, and k d > 0 can be rendered larger because |σ 1| may become larger despite σ 2 approaching zero. Thus, from the second inequality in (5.69), it is concluded too that a more damped response is obtained.
Notice that an increment in k d > 0 has a greater effect on the increment of |σ 1| than the effect that an increment in k p > 0 has on the increment of |σ 1| because this real part is affected by p 3 in (5.67), which is assumed to be large. Thus, it is concluded that k d has a greater effect on the damping than k p, i.e., the slower response predicted above is more affected by k d.
-
-
k i > 0 is small and k p > 0 is large.
This may be possible if p 3 is close to zero and p 1 and p 2, are complex conjugate. Thus, according to (5.67) and (5.66), k p > 0 can be large and k d > 0 small if
is large and σ 1 and σ 2, are close to zero. This means that the damping is small because θ > 0 in Fig. 5.59 is close to zero. Thus, a more oscillatory but faster closed-loop system response is obtained.
However, |σ 1| and |σ 2| may be rendered larger, i.e., to increase k d > 0, without affecting ω n if θ > 0 in Fig. 5.59 is increased. Thus, a more damped but slower system response is obtained. Notice that k p > 0 is less affected by the changes in |σ 1| and |σ 2| because they are affected by p 3 in (5.67), which is assumed to be close to zero.
-
Larger positive values for the integral gain, k i, result in a faster response that becomes more oscillatory.
-
Large positive values of the proportional gain, k p, result in either a) a slightly more damped response if k i > 0 is large, or b) a more oscillatory response if k i > 0 is small. In both cases, a fast response is expected.
-
Larger positive values of the derivative gain, k d, produce a more damped and slower response.
It is convenient to say that the transfer function between the position and its desired value (G 1(s) in (5.26)) has two zeros whose effect on the transient response is not completely clear. Hence, it is possible that sometimes, some variations appear regarding the effects that have just been described for the gains of the PID controller. Furthermore, the fact that the external disturbance is present because the desired position is commanded in some mechanisms may favor these variations.
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Simple pendulum used as a load for a DC motor
- 1.
Upper figure:
-
Continuous: k p = 0.5, k i = 0, k d = 0
-
Dash–dot: k p = 1, k i = 0, k d = 0
-
Dotted: k p = 1, k i = 0, k d = 0.05
-
- 2.
Bottom figure:
-
Continuous: k p = 1, k i = 2, k d = 0.05
-
Dash–dot: k p = 1, k i = 5, k d = 0.05
-
Dotted: k p = 2, k i = 5, k d = 0.05
-
Dashed: k p = 2, k i = 5, k d = 0.1
-
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The PID control of position for a DC motor with a pendulum as a load. Experimental results
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The PID control of position for a DC motor with a pendulum as a load. Experimental results
- 1.
Upper figure:
-
Continuous: k p = 2, k i = 5, k d = 0.1
-
Dash–dot: k p = 2, k i = 5, k d = 0.2
-
Dotted: k p = 2, k i = 8, k d = 0.2
-
- 2.
Bottom figure:
-
Continuous: k p = 2, k i = 8, k d = 0.2
-
Dash–dot: k p = 2.5, k i = 8, k d = 0.2
-
Dotted: k p = 3, k i = 8, k d = 0.2
-
Dashed: k p = 3.5, k i = 8, k d = 0.2
-
It is important to stress the
following. When connecting the pendulum to the motor shaft, the
control system becomes nonlinear. This means that the control
system behavior is not correctly predicted by the analysis
presented above at certain operation regions. For instance, if the
desired position is close to θ
d = ±π, a simple PD controller may be unstable
if the proportional gain (positive) is not greater than a certain
lower threshold. This result is obtained using a PD controller for
the case when ,
in Example 7.5 studied in Chap. 7 by finding the poles for such a
linear approximation. The reader may also see the works reported in
[2, 4], ch. 8, [3], ch.
7. This situation is worse for the case of a PID controller, but
the problem does not appear if the desired position value is kept
far from θ d = ±π, as in the results presented in this
section.
Finally, notice that the selection of the PID controller gains has been performed without requiring knowledge of the numerical value of any motor or pendulum parameter. However, the minimal knowledge that must be available is to verify that the motor can produce the required torque to perform the task. This must include the required torque to compensate for the gravity effect plus some additional torque to achieve the desired rise time.
5.4 Summary
The most general method of controller design using the time response approach is the root locus method. This means that the plants that can be controlled are of arbitrary order with any number of zeros as long as they are less than the number of poles. This method provides the necessary tools to determine, in a graphical way, the location of the closed-loop poles from the location of the open-loop poles and zeros. Some of the open-loop poles and zeros are due to the controller and the idea of the method is to select the location of the controller poles and zeros such that the desired closed-loop poles are assigned. The desired closed-loop poles are chosen from the knowledge of how they affect the corresponding transient response. The study presented in Chap. 3 is important for this. On the other hand, the controller structure is chosen from the knowledge of how the open-loop poles and zeros affect the closed-loop system steady- state response. The study presented in Chap. 4 is very important for this.
Although the root locus has been presented as a controller design method, it is also a powerful tool for control systems analysis. This means that it can be used to determine: (i) The relative stability of a control system, (ii) How the control system response changes as one of its parameters changes, (iii) What has to be done to modify the control system properties, etc. The use of the root locus method in these applications depends, to a large extent, on a good understanding of the method and the material presented in Chaps. 3 and 4.
5.5 Review Questions
- 1.
When would you use each one of the following controllers?
-
Proportional.
-
Proportional–derivative.
-
Proportional–integral.
-
Proportional–integral–derivative.
-
- 2.
Why do you think it is not advised to use the following controllers: (i) Derivative (alone), (ii) Integral (alone), and (iii) Derivative–integral (alone), i.e., without including a proportional part? Explain.
- 3.
How do the requirements of the steady-state error determine the poles and/or zeros of a controller, i.e., the controller structure?
- 4.
What is the main component that a controller must possess to improve the closed-loop system stability? What is the effect of this on the shape of the root locus diagram?
- 5.
What is the main component that a controller must possess to improve the steady-state error? What is the effect of this on the shape of the root locus diagram?
- 6.
Why does the root locus begin at the open-loop poles and end at the open-loop zeros? What do the words “begin” and “end” mean?
- 7.
If the open-loop transfer function has no zeros, where does the root locus end?
- 8.
It is often said that closed-loop system instability appears as the loop gain increases. However, this is not always true, because this depends on the properties of the plant to be controlled. Review the examples presented in this chapter and give an example of a plant requiring the loop gain to be large enough to render the closed-loop system stable.
- 9.
What is a lead compensator and what is its main advantage?
- 10.
Read Appendix F and Sect. 9.2 in Chap. 9. Use this information to explain how to practically implement a PID controller and a lead compensator using both software and analog electronics.
5.6 Exercises
- 1.
Consider the control system in Fig. 5.63 where:Fig. 5.63
A cascade connection of the PI control and lead compensator
- 2.
Consider the following plant:
- a)
- b)
- a)
- 3.
Consider a closed-loop system such as that shown in Fig. 5.1 with H(s) = 1 and
.
-
Using the rules presented in Sect. 5.1.1, draw the root locus diagram when using a proportional controller. Use Routh’s criterion to determine the values of the proportional gain ensuring closed-loop stability.
-
To stabilize the closed-loop system and to achieve a zero steady-state error when the reference is a step, design a PID controller proceeding as follows. (i) Propose k d s 2 + k p s + k i = k d(s − z 1)(s − z 2), with z 1 = −0.5 + 1.3j, z 2 = −0.5 − 1.3j. Using the rules presented in Sect. 5.1.1, draw the root locus diagram to verify that no positive value exists for the derivative gain rendering the closed-loop system stable. (ii) Propose k d s 2 + k p s + k i = k d(s − z 1)(s − z 2), with z 1 = −0.25 + 1.3j, z 2 = −0.25 − 1.3j. Using the rules presented in Sect. 5.1.1, draw the root locus diagram to verify that a range of positive values exists for the derivative gain k d, rendering the closed-loop system stable. Use Routh’s criterion to find the range of values for k d, rendering the closed-loop system stable.
-
How should the zeros of a PID controller be selected to improve the stability of the closed-loop system?
-
- 4.
Verify that the transfer function of the circuit shown in Fig. 5.64 is:Fig. 5.64
A lead compensator
- 5.
Consider the following plant:
- 6.
Consider the PD control of the position in Fig. 5.2.2. Prove that the value of the derivative gain k d does not have any effect on the steady-state error when the reference is a step.
- 7.
Consider the PID control of the position studied in Sect. 5.2.5. Prove that the values of the proportional k p and the derivative k d gains do not have any effect on the steady-state error, despite a constant disturbance being present if the reference is also constant.
Assume that the integral part of the controller is not present, i.e., that k i = 0. Prove that the value of the derivative gain k d does not have any effect on the steady-state error.
- 8.
Consider the mass-spring-damper system shown in Fig. 5.65, which has been modeled in Example 2.1, Chap. 2. In Example 3.18, Chap. 3, it is explained why a steady-state error that is different from zero exists when a proportional controller is employed and the desired position x d is a constant that is different from zero. Recall that the force F(t) applied to the mass is the control signal.Fig. 5.65
A mass-spring-damper system
Now suppose that a PID controller is used to regulate the position. Find the steady-state error when the desired position x d is a constant that is different from zero. Use your everyday experience to explain your response, i.e., try to explain what happens with the force applied to the mass and the effect of the spring.
- 9.
Consider the simple pendulum studied in Example 2.6, Chap. 2, which is shown in Fig. 5.66. Notice that the gravity exerts a torque that is different from zero on the pendulum if the angular position θ is different from zero. Suppose that it is desired to take the pendulum position θ to the constant value θ d = 90∘. Note that torque T(t) is the input for the pendulum. Using the experience in the previous exercise, state which controller you would employ to compute T(t) such that θ reaches θ d. Explain why. This problem is contrived to be solved without using a mathematical model; you merely need to understand the problem. In fact, the mathematical model is in this case nonlinear; hence, it cannot be analyzed using the control techniques studied in this book so far.Fig. 5.66
A simple pendulum
- 10.
Consider an arbitrary plant G(s) in cascade with the following controllers:
-
What is the effect of each one of these compensators on the steady-state error when the reference is a step?
-
What is the effect of each one of these compensators on the closed-loop stability?
-
Which compensator is related to PI control and which to PD control? Explain why.
-
How would you employ these compensators to construct a controller with similar properties to those of PID control? Explain.
-
- 11.
In a ship, it is common to have a unique compass to indicate the course. However, it is important to know this information at several places in the ship. Hence, it is common to transmit this information to exhibit it using a needle instrument. The angular position of the needle is actuated using a DC motor by using the information provided by the compass as the desired position. Design a closed-loop control system such that the needle position tracks the orientation provided by the compass according to the following specifications:
-
When a step change of 8∘ appears in the orientation provided by the compass, the system error decreases and remains at less than 1∘ in 0.3 s or less and overshoot is less than or equal to 25%.
-
When the orientation provided by the compass changes as a ramp of 5∘ per second, the steady-state error must be 0.3∘ or less.
-
There are no external disturbances.
(i) On the basis of the specifications indicated, choose a controller to design. (ii) Using the first specification, determine the zone in the time domain where the closed-loop system response must lay. (iii) Recall that, according to Fig. 3.16, Chap. 3, the response of a second-order system remains between two exponential functions depending on ζ and ω n. With this information and using the desired damping, determine the zone on the complex plane s where the dominant (complex conjugate) poles of the closed-loop system must be located. (iv) Use the root locus rules presented in this chapter to find the controller gains, ensuring that all the closed-loop poles are inside the desired zone in the plane s. (v) Verify that the second specification is satisfied and, if this is not the case, redesign the controller. (vi) Perform some simulations to corroborate that all the requirements are satisfied.
-