3. Ordinary Linear Differential Equations

3.1.1 Graphical Study of the Solution

According to the previous section, if a > 0, then the natural response goes to zero over time: lim_t→∞ y _n(t) = 0; hence, when time is large enough the solution of the differential equation is equal to the forced response: . This means that the faster the natural response y _n(t) tends toward zero (which occurs as a > 0 is larger), the faster the complete solution y(t) reaches the forced response y _f(t). Hence, the natural response can be seen as a means of transport, allowing the complete solution y(t) to go from the initial value y(0) = y ₀ to the forced response y _f(t). Existence of such a means of transport is justified recalling that the solution of the differential equation in (3.6), i.e., y(t), is a continuous function of time. This means that the solution cannot go from y(t) = y ₀ to y(t) = y _f(t) ≠ y ₀ in a zero time interval.

The graphical representation of the solution in (3.17), when y ₀ = 0 and a > 0, i.e.,

(3.20)

is depicted in Fig. 3.1. An important parameter in first-order differential equations is time constant τ, which is defined as:

(3.21)

This is indicative of how fast the natural response vanishes, i.e., how fast y(t) approaches the forced response y _f(t). Replacing (3.21) into (3.20), it is easy to verify that:

(3.22)

Finally, it is important to stress that y(t) grows without a limit if a < 0. If a = 0, then y(t) = kAt also grows without a limit (see Sect. 3.2).

3.1.2 Transfer Function

Consider a zero initial condition, i.e., y ₀ = 0, then (3.8) can be written as:

or, equivalently:

(3.23)

The function G(s) is known as the transfer function of the differential equation in (3.6). The polynomial in the denominator of G(s), i.e., s + a, is known as the characteristic polynomial and its roots are known as the poles of G(s). In this case, there is only one pole at s = −a. According to the discussion above, the following can be stated with respect to y(t), i.e., the output of the transfer function G(s):

• If pole located at s = −a is negative, i.e., if a > 0, then the natural response y _n(t) vanishes and the complete solution y(t) approaches the forced response y _f(t) as time increases. If u(t) = A is a constant, then .

• The faster y(t) approaches y _f(t), the farther to the left of the origin s = 0 is placed the pole at s = −a. This can be quantified using the time constant . A large time constant implies a slow response whereas a small time constant implies a fast response.

• If k = a > 0, then it is said that the transfer function G(s) has a unitary gain in a steady state because, according to the final value theorem (3.4), if u(t) = A is a constant then:

  

  (3.24)

  In general, the steady-state gain of the transfer function in (3.23) is computed as .
• If the pole located at s = −a is positive , i.e., if a < 0, then the complete solution y(t) grows without limit. Notice that this implies that the pole is located on the right half of the plane s (see Fig. 3.2).

  

• The case when the pole is located at s = a = 0 cannot be studied from the above computations and it is analyzed as a special case in the next section. However, again to summarize all the possible cases, we present here results that are obtained in the next section. When the pole is located at s = a = 0, the natural response neither vanishes nor increases without limit and the forced response is given as the integral of the input .

Example 3.2

Consider the tank containing water depicted in Fig. 3.3. Assume that the tank section C is constant. Water enters the tank at a rate given by the input flow q _i (m³/s). Water leaves the tank at a rate given by the output flow q _o (m³/s) through a valve with hydraulic resistance R. The water level in the tank is represented by h. The mathematical model describing this system is obtained using the mass conservation law . As water is not compressible, this law can be stated in terms of mass or volume.

Let ΔV _i and ΔV _o be the water volumes entering and leaving the tank respectively during a time interval Δt. During this time interval, the water volume ΔV  increases inside the tank at a rate given as:

(3.25)

On the other hand:

Replacing this in (3.25) and considering small increments of time such that Δt → dt and Δh → dh:

(3.26)

Assuming that the water flow is laminar through the output valve [1], pp. 155, then:

(3.27)

To understand the reason for this expression, consider the following situations.

1. a)

   Suppose that the opening of the output valve is kept without change, i.e., R remains constant, and the water level h increases. Everyday experience and the expression in (3.27) corroborate that the output flow q _o increases under this situation.

    
2. b)

   Suppose that the water level h is kept constant, because there is water entering the tank to exactly compensate for water leaving the tank. Then, slowly close the output valve, i.e., slowly increase R. Everyday experience and the expression in (3.27) corroborate that the output flow q _o decreases in this situation.

    

Substituting (3.27) in (3.26), it is found that:

Hence, the mathematical model of the water level system depicted in Fig. 3.3 is given as:

Dividing by C, the following linear, ordinary, first-order differential equation with constant coefficients is finally obtained:

(3.28)

The reason why this differential equation is the mathematical model of the water level system is because, solving it, the evolution over time of the water level h(t) can be known if the input flow q _i(t), the tank section C, the hydraulic resistance R, and the initial water level are known.

Example 3.3

Consider the water level system depicted in Fig. 3.3. It was found in Example 3.2 that the mathematical model is given as:

(3.29)

This differential equation can be written as in (3.6) if it is assumed that y = h, u = q _i and y ₀ = h ₀. Hence, if q _i = A, the solution h(t) is similar to that in (3.16), i.e.,:

(3.30)

Let us analyze (3.30).

1. 1.
   Assume that the tank is initially empty, i.e., h ₀ = 0, and water enters the tank with a constant rate q _i = A > 0. Under these conditions, the water level h(t) evolves as depicted in Fig. 3.1 by assuming that h(t) = y(t). Notice that . From this figure, the solution in (3.30), and using everyday experience, several cases can be studied.

   • If A is chosen to be larger, then the final water level is also larger.

   • If the output valve is slowly closed, i.e., R is slowly increased, then the final water level slowly increases.

   • The tank cross-section has no effect on the final water level, i.e., the final water level is the same in a thin tank and a thick tank.

   • If the product RC is larger, then a is smaller; hence, the system response is slower, i.e., more time is required for the level to reach the value and, as consequence, also to reach the final water level . This is because the function e^−at tends toward zero more slowly when is smaller. Notice that a larger value of RC can be obtained by increasing the tank cross-section C, or decreasing the output valve opening, i.e., increasing R.

    
2. 2.
   Suppose that no water is entering the tank, i.e., q _i(t) = A = 0, and the initial water level is not zero, i.e., h ₀ > 0. From the solution in (3.30), and everyday experience can verify it, the water level decreases (recall that ) until the tank is empty, i.e., lim_t→∞ h(t) = 0 (see Fig. 3.4). This behavior is faster as RC is smaller, i.e., when a is larger, and slower as RC is larger, i.e., when a is smaller.

   

    
3. 3.

   The case when a < 0 is not possible in an open-loop water level system because C and R cannot be negative. However, a < 0 may occur in closed-loop systems as a consequence of the interconnection of diverse components.

    
4. 4.

   The case when a = 0 is studied in the next section.

    

3.3.1 Graphical Study of the Solution

The graphical form of solution presented in (3.64) for the differential equation in (3.53) is studied in the following. Recall that it is assumed that 0 ≤ ζ < 1, ω _n > 0, k > 0 and that the initial conditions are zero y ₀ = 0, . Notice that the natural response tends toward zero as time increases if 0 < ζ < 1:

(3.69)

because ζω _n > 0. Also recall that, in the case when the initial conditions are not zero, the natural response also tends toward zero as time increases. Hence, when time is large, the solution of the differential equation is equal to the forced response:

(3.70)

This means that the faster the natural response vanishes y _n(t), i.e., as ζω _n > 0 is larger, the faster the complete solution y(t) reaches the forced response y _f(t). As was noted for first-order differential equations, the natural response allows the complete solution y(t) to go from the initial conditions (zero in this case) to the forced response y _f(t).

Two important parameters of the response in (3.64) are the rise time t _r and overshoot M _p, which are shown in Fig. 3.13. Overshoot can be measured as:

where y _f = kA stands for the forced response. Recall that it is assumed that . Carefully analyzing (3.64), it can be shown that [1], pp. 232:

(3.71)

Figures 3.14 and 3.15 show how the solution in (3.64) is affected when the parameters ζ and ω _n change. Notice that overshoot M _p is affected only by ζ, whereas rise time t _r is mainly affected by ω _n, although ζ also has a small effect on t _r.

Finally, it is important to point out that, in the case when u = A = 0 but some of the initial conditions or y ₀ are different from zero, y(t) behaves as the time function in (3.65) and is graphically represented as the oscillatory part without the constant component, i.e., it oscillates around y = 0, in any of the Figs. 3.13, 3.14 or 3.15. This is because of function p(t) appearing in (3.62).

3.3.2 Transfer Function

If the initial conditions are zero y ₀ = 0, , then (3.54) can be written as:

or:

(3.72)

where G(s) is the transfer function of the differential equation in (3.53). Notice that G(s), defined in (3.72), has a second-degree characteristic polynomial, i.e., G(s) has two poles. Hence, G(s) is a second-order transfer function. According to (3.55) these poles are located at s = a and , where:

It is important to stress that these poles are complex conjugate only under the condition − 1 < ζ < 1. The reader can verify that these poles are real if ζ does not satisfy this condition. Such a case is studied in the subsequent sections.

Example 3.8

Consider the mass-spring-damper system depicted in Fig. 3.18. According to Example 2.​2, Chap. 2, the mass position x is given by the following second-order differential equation:

(3.73)

where K is the spring stiffness constant, b is the viscous friction coefficient, m is the body mass, and f is an applied external force. The expression in (3.73) can be written as:

which can be rewritten as the expression in (3.53):

(3.74)

where x is the output y, whereas f is the input u. Hence, x(t) is given as y(t) in (3.64) if 0 ≤ ζ < 1 when the initial conditions are zero. Notice that all constants in (3.73) are positive: the mass m, the viscous friction coefficient b, and the spring stiffness coefficient K only can be positive. According to (3.74), the damping coefficient is given as:

(3.75)

If a constant force f = A is applied, the results in Figs. 3.13, 3.14, and 3.15 can be employed to conclude the following.

• If there is no friction, i.e., if b = 0, then ζ = 0 and the mass will oscillate forever around the position . In the case when the external force is zero f = 0 but some of the initial conditions or x ₀ are different from zero, then the mass will oscillate again, though now around the position x = 0, as described by p(t) appearing in (3.62).

  

• If the friction coefficient b > 0 increases, then ζ > 0 also increases and the mass oscillations will disappear because the system will have more damping. It is shown in the subsequent sections that the poles are real and repeated when ζ = 1. In all of these cases, the mass will stop moving at . In the case when the external force is zero f = 0 but some initial conditions, or x ₀, are different from zero, the mass will stop at x = 0. This is again described by the function p(t) appearing in (3.62). Notice that, according to (3.75), the damping ζ also increases if the mass m or the spring stiffness constant K decreases. To understand the reason for this, consider the opposite case: larger values of both m and K produce larger forces ( for the mass and Kx for the spring), which, hence, are less affected by the friction force () and thus, the mass can remain in movement for a longer period of time.

• As the rise time t _r inversely depends on the undamped natural frequency , then a more rigid spring (with a larger K) or a smaller mass produces faster responses (t _r smaller). On the other hand, advantage can be taken of the fact that the oscillation frequency is given by with to adjust the rate of a mechanical clock based on a mass-spring-damper system: if the clock “lags,” it is necessary to increase its oscillation frequency, or rate, which is achieved by applying tension to the spring to increase K.

The case when − 1 < ζ < 0 is not possible in this example because all of the constants b, m, and K are positive. However, the situation when − 1 < ζ < 0 commonly occurs in feedback systems as a result of the interconnection of diverse components.

3.4.1 Real and Different Roots

Suppose that N(s) has k ≤ n real and different roots, which are different from s = 0, i.e.,:

where p _i ≠ 0, i = 1, …, k, with p _j ≠ p _i if i ≠ j, stand for the roots of N(s), whereas N ₀(s) is a polynomial containing the remaining n − k roots of N(s), where none of them is located at s = 0. According to the partial fraction expansion method [2], Ch. 4, [3], Ch. 7, in this case the following must be written:

(3.78)

where c _i, i = 1, …, k, and f are real constants to be computed, whereas Q(s) represents all fractions corresponding to the roots of N ₀(s). Constant c _i can be computed as follows. Multiply both sides of (3.78) by the factor (s − p _i) and evaluate the resulting expression at s = p _i to obtain:

A similar procedure allows us to compute:

Using the tabulated inverse Laplace transform, [4], Ch.32:

the following is finally obtained:

where . Notice that the following can be written:

(3.79)

where the natural response y _n(t) only depends on the roots of the characteristic polynomial N(s), whereas the forced response y _f(t) only depends on the fraction introduced by U(s). Consider the following possibilities.

1. 1.

   If p _i < 0 for all i = 1, …, k, then and lim_t→∞ y(t) = q(t) + y _f(t).

    
2. 2.

   If p _i > 0 for at least one i = 1, …, k, then , as t →∞ and lim_t→∞ y(t) = ∞.

    

Example 3.9

Consider the following second-order differential equation also given in (3.53):

(3.80)

If ζ > 1, then the only two roots, p ₁ and p ₂, of the characteristic polynomial are real and different:

Also, both roots are negative:

Although this is obvious for p ₂, for p ₁, a little observation is required: notice that , and since , then . In this case q(t) = 0, because Q(s) = 0 because N ₀(s) = 1, i.e., N ₀(s) has no roots. In Fig. 3.19 the solution y(t) of the differential equation in (3.80) is shown, when ζ > 1, u = A and the initial conditions are zero. Notice that y(t) does not present oscillations in this case. Recall that, when − 1 < ζ < 1, the oscillation frequency is equal to the imaginary part of both roots. Hence, when ζ > 1, there is no oscillation, because the imaginary part of both roots is zero; thus, the oscillation frequency is also zero.

Example 3.10

Consider again the mass-spring-damper system in Example 3.8. Recall that (3.74) describes the mass movement. Assume that:

Then, according to the previous example, the roots p ₁ and p ₂ of the characteristic polynomial are real, different, and negative. Notice that this case stands when the viscous friction coefficient b > 0 is very large or when the spring stiffness constant K > 0 or mass m is very small. Also notice that one root () approaches the origin as ζ > 1 becomes larger. According to the discussion above in the present section, this means that the movement of the mass is slower as ζ grows, because it takes more time to function to vanish (despite the function vanishing faster because moves farther to the left of the origin). This explains why slower responses are depicted in Fig. 3.19, as ζ > 1 is larger.

Example 3.11

Consider the following differential equation:

with c, d and e some real constants. The roots p ₁ and p ₂ of the characteristic polynomial N(s) = s ² + cs + d are given as:

The following cases have to be considered:

• If d < 0, then:

  

  and both roots are real and different, with one positive and the other negative, no matter what the value of c. It is shown, in Example 7.​5 of Chap. 7, that this case corresponds to a mass-spring-damper system with a negative spring stiffness constant and that this is obtained in practice when a simple pendulum works around its inverted configuration (also called unstable).

• If c > 0 and d > 0, the case studied in Sect. 3.3is retrieved when 1 > ζ > 0, and the case studied in the present section is retrieved when ζ > 1.

• If c < 0 and d > 0, with abs(c) small and d large, the case studied in Sect. 3.3 is retrieved when − 1 < ζ < 0. The situation when ζ < −1 is also obtained in this case when abs(c) is large and d is small, but both roots are positive and different:

  

• The cases when ζ = −1 and ζ = 1 are studied in the next section.

3.4.2 Real and Repeated Roots

Suppose that N(s) has one root that is repeated k times and it is different from zero, i.e.,:

with p ≠ 0 and N ₀(s) a polynomial that contains the remaining n − k roots in N(s), and none of them is located at s = 0. According to the partial fraction expansion method [2], Ch. 4, [3], Ch. 7, the following must be written:

(3.81)

where c _i, i = 1, …, k, and f represent real constants to be computed, whereas Q(s) represents the fractions corresponding to roots of N ₀(s). One method of computing c _k is by multiplying both sides of (3.81) by the factor (s − p)^k and evaluating at s = p to obtain:

On the other hand, the constants c _i, i = 1, …, k − 1, can be computed by multiplying both sides of (3.81) by the factor (s − p)^k, differentiating k − i times with regard to s and evaluating at s = p to obtain:

Finally, the constant f is computed by multiplying both sides of (3.81) by the factor s and evaluating at s = 0 to find:

Using the tabulated inverse Laplace transform [4], Ch. 32:

the following is found:

where . Notice that the following can be written:

(3.82)

It is remarked again that y _n(t) only depends on the roots of the characteristic polynomial N(s) and y _f(t) only depends on the fraction , introduced by U(s). Consider the following possibilities:

1. 1.
   If p < 0, it is useful to compute the following limit, where j is any positive integer number:

   

   which represents an indetermination, because − pt → +∞. Then, L’Hopital’s rule [12], pp. 303, can be employed:

   

   As an indetermination appears again, L’Hopital’s rule can be applied several times to obtain:

   

   Applying this result to the solution obtained above, it is concluded that:

   

   and lim_t→∞ y(t) = q(t) + y _f(t) for any p < 0, no matter how close to zero p is.
    
2. 2.
   If p > 0, it is clear that:

   

   as t →∞ and lim_t→∞ y(t) = ∞.
    
3. 3.
   Finally, consider the case when the characteristic polynomial N(s) has k real and repeated roots at p = 0, i.e., N(s) = N ₀(s)s ^k, where N ₀(s) is a polynomial containing the remaining n − k roots of N(s). Then, according to the partial fraction expansion method [2], Ch. 4, [3], Ch. 7, in this case the following must be written:

   

   where Q(s) contains fractions corresponding to roots of N ₀(s). Using the tabulated Laplace transform [4], Ch. 32:

   

   (3.83)

   yields:

   

   (3.84)

   where . In this case:

   

   since the natural response only depends on the k real and repeated roots at s = 0 that N(s) has, which, according to (3.83), only introduces terms included in y _n(t). It is clear that y _n(t) →∞ and y(t) →∞ when t →∞ for k ≥ 2 and that y _n(t) is a constant if k = 1. On the other hand, the forced response y _f(t) represents the integral of u(t) iterated k times.
    

Example 3.12

Consider the following second-order differential equation, also given in (3.53):

(3.85)

If ζ = 1, then both roots, p ₁ y p ₂, of the characteristic polynomial are real, repeated, and negative:

The solution y(t) of the differential equation in (3.85) is depicted in Fig. 3.19, when ζ = 1, u = A and the initial conditions are zero. Notice that y(t) does not exhibit oscillations in this case because both roots have a zero imaginary part. This case, ζ = 1, is at the edge between an oscillatory response ζ < 1 and a response without oscillations ζ > 1. In fact, the case ζ = 1 represents the fastest response without oscillations because, when ζ > 1, the response is slower as ζ increases.

Example 3.13

Consider again the mass-spring-damper system in Example 3.8. Recall that (3.74) describes the movement of mass. Assume that:

Then, in this case, the characteristic polynomial only has a real and negative root:

which is repeated twice. The mass position x(t) evolves exactly as y(t) in Fig. 3.19. This class of response may be very useful if a fast mass movement is required, but without oscillations.

Example 3.14

Suppose that there is neither a spring nor a damper in Fig. 3.18, i.e., assuming that b = 0 and K = 0, the differential equation in (3.73) becomes:

(3.86)

Notice that in this case the characteristic polynomial is s ², which only has a real root at p = 0 repeated twice. Now suppose that all of the initial conditions are zero and that the mass receives a small disturbance force described by:

(3.87)

where ε ₀ > 0 and t ₁ > 0 are small constant numbers. By direct integration of (3.86), it is found:

Again, integrating yields (assuming that ):

Replacing (3.87), the following is found:

Notice that x(t) →∞ as t →∞ (see Fig. 3.20) despite f = 0 for t > t ₁, i.e., despite the forced response being zero for all t > t ₁. This means that the natural response x _n(t) →∞ when t →∞, which corroborates results obtained in the present section for real roots at zero, which are repeated at least twice. The reader may resort to everyday experience to verify that a slight hit (f in (3.87)) on the mass m is enough for the mass to begin moving to never stop (x(t) →∞) if no friction exists between the mass and the floor.

Example 3.15

Consider the second-order differential equation given in (3.53):

If ζ = −1, then both roots, p ₁ and p ₂, of the characteristic polynomial are real, repeated, and positive:

3.4.3 Complex Conjugate and Nonrepeated Roots

Consider the following second-degree polynomial:

where p ₁ and p ₂ are the complex, but not conjugate roots, i.e., c ₁, c ₂, d ₁, d ₂, are real numbers such that c ₁ ≠ c ₂ if d ₁ = −d ₂ or d ₁ ≠ − d ₂ if c ₁ = c ₂. We have that:

Note that, because of conditions c ₁ ≠ c ₂ if d ₁ = −d ₂ or d ₁ ≠ − d ₂ if c ₁ = c ₂, one of the coefficients a or b is a complex number, i.e., the polynomial s ² + bs + a has at least one complex coefficient. In the previous sections, we have seen that physical systems have characteristic polynomials whose coefficients are given only in terms of properties that can be quantified using real numbers, i.e., mass, inertia, friction coefficients, stiffness constant, electrical resistance, inductance, capacitance, etc. Thus, the characteristic polynomials of physical systems cannot have a complex root without including its corresponding complex conjugate. This means that in automatic control we are only interested in characteristic polynomials that have pairs of complex conjugate roots.

Suppose that N(s) has k pairs of complex conjugate and nonrepeated roots, i.e., according to Sect. 3.3, the following can be written:

where σ _i ≠ σ _j or ω _di ≠ ω _dj if i ≠ j, whereas N ₀(s) is a polynomial containing the remaining n − 2k roots of N(s), none of them located at s = 0. Then, according to the partial fraction expansion method [2], Ch. 4, [3], Ch. 7, in this case the following must be written:

where Q(s) contains fractions corresponding to roots of N ₀(s), f is a constant to be computed as in the previous sections, i.e.,:

whereas F _i and C _i, i = 1, …, k, are constants to be computed as B and C were computed in Sect. 3.3. Hence, using (3.60) the following can be written:

(3.88)

where β _i, ω _ni, ζ _i y ϕ _i, i = 1, …, k, are real constants (see Sect. 3.3), whereas . The evolution of the solution in (3.88) satisfies one of the following cases.

1. 1.

   If 0 < ζ _i < 1 for all i = 1, …, k, i.e., if the real parts of all of the complex conjugate roots are negative, − ζ _i ω _ni < 0 for all i = 1, …, k, then y(t) oscillates such that and lim_t→∞ y(t) = q(t) + y _f(t).

    
2. 2.

   If − 1 < ζ _i < 0 for at least one i = 1, …, k, i.e., if the real part is positive, − ζ _i ω _ni > 0 for at least one of the complex conjugate roots i = 1, …, k, then y(t) oscillates such that y _n(t) and y(t) grow without a limit as time increases.

    
3. 3.

   If ζ _i = 0 for all i = 1, …, k, i.e., if the real parts of all the complex conjugate and nonrepeated roots are zero, − ζ _i ω _ni = 0 for all i = 1, …, k, then does not disappear as time increases, but nor does it grow without a limit. Thus, although y(t) may not grow without a limit (depending on the behavior of q(t)), it does not converge to y _f(t). Notice that for this situation to stand, it is enough that ζ _i = 0 for at least one i and 0 < ζ _i < 1 for all of the remaining i.

    

It is important to point out that, in the cases when ζ ≥ 1 or ζ ≤−1, real roots are obtained and one of the cases in Sects. 3.4.1 or 3.4.2 is retrieved.

Example 3.16

The mathematical model of the mechanical system composed of two bodies and three springs depicted in Fig. 3.21 has been obtained in Example 2.​4 in Chap. 2. From this result, the case when the spring at the left is not present can be studied. This is achieved by assuming that K ₁ = 0, i.e.,:

Applying the Laplace transform (3.2) to each one of these differential equations and assuming that all of the initial conditions are zero, the following is found:

Replacing the second of these equations in the first one, and rearranging, it is found that:

To simplify the required algebra, assume that b = 0, then:

or:

(3.89)

Performing the indicated operation, it is found that:

for some numbers c and e such that:

i.e.,:

Then, the following can be written:

(3.90)

As e and c are positive and different, it is ensured that the characteristic polynomial in (3.90) has two pairs of complex conjugate and different roots (in fact, the roots are imaginary conjugate and different). Hence, mass m ₁, whose position is represented by x ₁(t), moves by oscillating and it never stops. However, the amplitude of this oscillatory movement does not grow without a limit. These oscillations are performed at two angular frequencies defined by and . Recall that the damping of these oscillations is zero.

3.4.4 Complex Conjugated and Repeated Roots

Assume that the characteristic polynomial N(s) has a pair of complex conjugate roots that are repeated k > 1 times, i.e.,:

where N ₀(s) is a polynomial that includes the remaining n − 2k roots of N(s), without any of them located at s = 0. Then, the partial fraction expansion method establishes that in this case [2], Ch. 4:

where Q(s) contains fractions corresponding to the roots of N ₀(s). In the case of real roots, it was found that when roots are not repeated, they introduce functions such as e^pt, where p is the corresponding root. When such a root is repeated j times, it was found that the functions introduced become t ^j−1 e^pt. In the case of complex conjugate and repeated roots, a similar situation occurs. In this case, any formal proof is not presented because of cumbersome notation, and only intuitive ideas are given to understand the reason for the resulting functions.

Using the inverse Laplace transform in (3.60), it was found that a complex conjugate nonrepeated root introduces the time function:

for some constants β and ϕ. Then, in the case of complex conjugated and repeated roots, y(t) is given as:

(3.91)

where . The solution in (3.91) has one of the following behaviors.

1. 1.

   If the complex conjugate root has a positive real part σ = −ζω _n > 0, i.e., if − 1 < ζ < 0, it is easy to see that y _n(t) →∞ and y(t) →∞ as t →∞.

    
2. 2.
   If the complex conjugate root has a negative real part σ = −ζω _n < 0, i.e., if 0 < ζ < 1, it is possible to proceed as in Sect. 3.4.2 to compute the following limit:

   

   for any positive integer j and any real negative σ. Then, for complex conjugate and repeated roots with a negative real part:

   

   (3.92)

   and lim_t→∞ y(t) = q(t) + y _f(t).
    
3. 3.
   If the root has a zero real part, i.e., σ = 0, then:

   

   Notice that, in this case:

   

   (3.93)

   as time increases in the case when the root is repeated at least twice, i.e., if k > 1. But if this root is not repeated, i.e., if k = 1, then:

   

   (3.94)

   is an oscillatory function whose amplitude neither grows nor vanishes, i.e., although y(t) does not grow without a limit (depending on q(t)) it will never reach y _f(t) in a steady state.
    

3.4.5 Conclusions

In the previous sections it has been assumed that all of the initial conditions are zero, i.e., that:

If some of the initial conditions are not zero, then the previous expression is not valid and the solutions y(t) found in the previous sections have to be completed adding the function:

As the denominator of:

is still the characteristic polynomial N(s), then, according to the partial fraction expansion method, w(t) contains the same functions of time contained in the natural response y _n(t) found in every case studied in the previous sections. This means that the complete natural response in every one of the cases studied in the previous sections must also include w(t).

It is said that the differential equation in (3.76) or, equivalently, that transfer function , is:

• Stable if the natural response tends toward zero as time increases.

• Unstable if the natural response grows without a limit as time increases.

• Marginally stable if the natural response neither grows without limit nor tends toward zero as time increases.

Assume that all of the cases studied in the previous sections, i.e., regarding the roots of N(s), appear simultaneously. Recall that the roots of N(s) represent the poles of G(s). The following can be concluded:

Conditions to Ensure lim_t→∞ y(t) = A

The complete response y(t) reaches the desired output u = A as time increases, i.e., y _f(t) = A and y _n(t) → 0, if G(s) is stable and the coefficients of the independent terms of the polynomials B(s) and N(s) are equal, i.e., if B(0) = N(0). This can be verified using the final value theorem (3.4):

(3.95)

Under these conditions, it is said that G(s) is a transfer function with unitary gain in a steady state . When u is not a constant and it is desired that lim_t→∞ y(t) = u(t), G(s) is also required to be stable, i.e., that y _n(t) → 0, but some additional conditions must be satisfied. The determination of these conditions and how to satisfy them is the subject of the study presented in the subsequent chapters of this book (see Sect. 4.​4).

Example 3.17

Consider the situation presented in Example 3.14. From this example, an experiment can be designed that allows us to know whether a system, a differential equation or a transfer function is stable or unstable:

If the system is originally at “rest,” apply a pulse disturbance to it and observe its evolution:

• If the system is stable then it “moves” and after a while it comes back to the configuration where it was originally at “rest.”

• If the system is unstable, then it “moves” more and more such that it goes far away from the configuration where it was originally at “rest.”

Example 3.18

Consider the mass-spring-damper system studied in Example 3.8, but now assume that the spring is not present. Replacing K = 0 in (3.73) yields:

Suppose that a force is applied according to:

(3.96)

where k _p is a positive constant, x _d is a constant standing for the desired position, and x is the mass position. Combining these expressions yields:

and rearranging:

Using the Laplace transform and assuming that all of the initial conditions are zero, the following is found:

As an exercise, the reader is to find the roots of the characteristic polynomial to corroborate that both roots have a negative real part if and . Then, using (3.95), it is found that the position reached in a steady state:

equals the desired position x _d. The reason for this result can be explained using, again, everyday experience: when x = x _d, the force produced according to (3.96) is f = 0; hence, the mass can stop and remain at that position. Moreover, if x < x _d, then f > 0 and the mass increases its position x approaching x _d (see Fig. 3.18 to recall that x increases in the same direction where f is positive). If x > x _d, then f < 0 and the mass position x decreases, approaching x _d again.

As an exercise, the reader is to verify that, in the case where a spring is present, i.e., when K > 0, then lim_t→∞ x(t) ≠ x _d if x _d ≠ 0. Notice, however, that again this can be explained using everyday experience:

If x = x _d then, according to (3.96), the external force applied to the mass is zero f = 0 and the friction force is also zero , if it is assumed that the mass is at rest. However, if x = x _d ≠ 0 then the force of the spring on the mass − Kx is different from zero; hence, the mass will abandon the position x = x _d ≠ 0. The mass will reach a position x such that the force of spring and the force f in (3.96) exactly cancel each other out, i.e., where k _p(x _d − x) = Kx.

3.5.1 Approximate Pole-Zero Cancellation and Reduced Order Models

Consider the following second-order system:

(3.97)

Suppose that U(s) = A∕s, p ₁ ≠ p ₂, p ₁ < 0, p ₂ < 0, d < 0, and p ₁ p ₂ ≈−kd to render approximately the unitary gain in a steady state of the transfer function in (3.97). Using partial fraction expansion:

(3.98)

Multiplying both sides by the factor (s − p ₁) and evaluating at s = p ₁ yields:

Multiplying both sides of (3.98) by the factor (s − p ₂) and evaluating at s = p ₂ yields:

Multiplying both sides of (3.98) by the factor s and evaluating at s = 0 yields:

If p ₁ ≈ d < 0 then, according to p ₁ p ₂ ≈−kd, we have k ≈−p ₂ and, hence:

to finally obtain:

The reader can verify that:

is the solution of:

(3.99)

with U(s) = A∕s and k = −p ₂. Thus, the following is concluded. If one pole and one zero of a transfer function are very close, then they approximately cancel each other out to obtain a reduced order transfer function. This means that (3.99) can be used instead of (3.97) to obtain very similar results. The advantages of using the model in (3.99) are: i) it is a model with a smaller order than (3.97), and ii) it has no zero. The advantage of using a reduced order model, which approximately describes a higher-order system is that the response of a reduced order system can often be handled as that of a first- or a second-order system. On the other hand, a zero in a transfer function modifies the time response in a manner that is not easy to quantify; hence, a transfer function that has no zeros is preferable.

It is important to say that the simplification obtained by the cancellation of one pole and one zero is useful only if the pole and the zero have negative real parts. This is because one pole with a positive real part, which is not exactly cancelled because of parameter uncertainty, has a dangerous effect that becomes larger as time increases.

3.5.2 Dominant Poles and Reduced Order Models

Consider the following transfer function:

(3.100)

where U(s) = A∕s. Using partial fraction expansion, the following is found:

(3.101)

Multiplying both sides by the factor and evaluating at :

Multiplying both sides of (3.101) by the factor s + a and evaluating at s = −a:

Multiplying both sides of (3.101) by the factor s and evaluating at s = 0:

If , then:

Hence, using (3.101) and neglecting C, the following is found:

The reader can verify that:

is the solution of:

(3.102)

with U(s) = A∕s. Thus, (3.102) can be used instead of (3.100). The condition is interpreted as “the pole at s = −a is very fast compared with the pole at ”. The pole at is known as the dominant pole because its effect is more important in the system response. The following is concluded. If one pole is much faster than the others, then a reduced order transfer function can be obtained by neglecting the fastest pole to only maintain the dominant (the slowest) poles. An accepted criterion is that the fastest poles (those to be neglected) have a real part that is five times the real parts of the dominant poles (those to be kept). Notice, however, that it is not just a matter of simply eliminating the factor s + a at the denominator of the transfer function: the constant a still appears at the denominator of (3.102) because a is necessary to keep without change the steady-state gain of the transfer function in (3.100).

A simple way of obtaining (3.102) from (3.100) is the following:

(3.103)

If a is very large it can be assumed that ; hence:

which is the expression in (3.102).

Finally, it is important to say that order reduction, as presented above, is only valid if the neglected pole has a negative real part. If the neglected pole has a positive real part, then the contribution of this pole will grow to infinity as time increases, no matter how small C is. Thus, the contribution of such a pole cannot be neglected.

Example 3.19

According to Exercise 9 and Example 2.​20 in Chap. 2, the mathematical model of a permanent magnet brushed DC motor is given as:

(3.104)

where ω is the motor velocity (also see (10.​9) and (10.​10) in Chap. 10). Suppose that this motor actuates on a hydraulic mechanism, which, by centrifugal effect, produces a water flow, q _i, which is proportional to the motor velocity, i.e., q _i = γω, where γ is a positive constant. Finally, this water flow enters the tank in Example 3.2 in the present chapter, whose mathematical model is given in (3.28), rewritten here for ease of reference:

Using the Laplace transform (3.2) and assuming that all the initial conditions are zero, it is not difficult to verify that the corresponding equations become:

(3.105)

(3.106)

(3.107)

Combining (3.105) and (3.106) yields:

According to (3.103), the following can be written:

In small motors, the inductance L _a and the viscous friction coefficient B are small; hence, . Thus, it can be assumed that and the following can be written:

Rearranging terms, this expression can be rewritten as:

Replacing this and Q _i(s) = γω(s) in (3.107) yields:

Proceeding again as in (3.103), the following can be written:

If the tank cross-section is large enough, the motor will reach its nominal velocity a long time before the water level in tank increases significantly. This can be stated by saying that “the tank time constant is very large compared with the motor time constant,” i.e., . Then, it is possible to say that ; hence, the following approximation is valid:

or:

Then, defining:

and comparing with (3.107), the expression in (3.139) is justified, i.e., that the water flow is proportional to the voltage applied to the motor through a constant k ₁. This is possible if: 1) the motor electrical time constant is smaller than the motor mechanical time constant, i.e., if , and 2) if the tank time constant is very large compared with the complete motor time constant, i.e., if . Finally, it is important to say that this procedure is valid because and , i.e., the neglected poles are negative.

3.5.3 Approximating the Transient Response of Higher-Order Systems

The simplicity of first- and second-order systems allows us to compute their exact solutions. However, in the case of systems represented by differential equations of an order greater than 2, also called higher-order systems, the complexity of the problem prohibits us from obtaining their exact solutions for control purposes. The traditional treatment of higher-order systems in classical control has been to approximate their response as a kind of extrapolation of responses obtained in first- and second-order systems.

According to the studies presented in Sects. 3.1 and 3.3, the poles of first- and second-order systems can be located on the s complex plane, as depicted in Fig. 3.22. Then, the distance of a pole to the origin, s = 0, is representative of the response time: the larger this distance, the shorter the response time. The angle between the imaginary axis and the location vector of the pole is representative of how damped the response is: the closer this angle is to zero, the more oscillatory the response; conversely, the closer this angle is to 90^∘, the less oscillatory the response.

The responses of two first-order systems are compared in Fig. 3.23. The location of the respective real poles is depicted in Fig. 3.22. It is observed that a faster response if obtained, as the pole is located farther to the left of the origin.

The responses of three second-order systems are compared in Fig. 3.24. The location of the respective pair of complex conjugate poles is depicted in Fig. 3.22. It is observed that (i) a shorter response time is achieved as poles are located farther from the origin, and (ii) a less oscillatory response is achieved as the poles are located closer to the real axis.

The above results are applied to higher-order systems by considering the worst case, i.e., by observing the “slowest” pole (the closest pole to the origin) and the “least damped” pole (the closest pole to the imaginary axis).

3.8.1 Proportional Control of Velocity in a DC Motor

According to Chap. 10, the mathematical model of a permanent magnet brushed DC motor is given in (10.​9), (10.​10), which are repeated here for ease of reference:

(3.120)

The reader is encouraged to see Chap. 10 for an explanation of variables in this model. For the purposes of this example, it suffices to say that ω, u, i, T _p represent the load angular velocity, the applied voltage, the electric current through the armature circuit, and an external torque disturbance respectively, and all of the parameters are positive. If the motor is small, then it is widely accepted that L ≈ 0 can be assumed, and the following is obtained:

(3.121)

Replacing the electric current in (3.120) and rearranging:

(3.122)

Consider the following proportional velocity controller :

(3.123)

where k _p is a constant known as the proportional gain, and ω _d represents the desired velocity given as a step signal of magnitude A, i.e., . The closed-loop system is obtained by replacing u, given by the proportional controller, in the previous equation, i.e.,:

(3.124)

Rearranging terms yields:

This is a first-order system with two inputs. Then, we can apply superposition, i.e., define two signals ω ₁ and ω ₂ such that:

First, analyze the equation for ω ₁, applying the Laplace transform with zero initial conditions:

(3.125)

Notice that this transfer function is identical to:

(3.126)

if we define:

Notice that this expression has the same structure as the transfer function in (3.23). Hence, if the parameters and variables b ₁, b ₂, ω ₁ and ω _d are suitably associated with those in (3.23), the solution ω ₁(t) is the same as in (3.16), i.e.,:

Stability is ensured if, and only if, the only pole of G ₁(s), located at s = −b ₁, is negative, i.e., − b ₁ < 0, which implies that:

(3.127)

If this is the case, then the final value of the velocity can be computed using the final value theorem as:

Recall that A represents the magnitude of the desired velocity. As it is desirable for the final value of the velocity to be close to A, it also desirable for to have the same sign as A, i.e., b ₂∕b ₁ > 0. As b ₁ > 0, because of the stability requirement, then b ₂ > 0 must also be true. This is ensured if k _p > 0, which is in complete agreement with (3.127). Hence, we can conclude that:

(3.128)

which means that the final velocity remains below the desired velocity. This result may also be explained using everyday experience as follows. Assume for a moment that lim_t→∞ ω ₁(t) = A. Then, according to (3.123):

the voltage applied at the motor terminals is zero. Hence, the motor tends to stop moving, i.e., ω ₁ decreases to zero and cannot stay at the desired value A. This means that ω ₁ < A will be obtained again. Hence, it is not possible to keep ω ₁ = A forever, which justifies the result in (3.128).

The difference between the final velocity and A is known as the steady-state error e _ss. We remark that, according to (3.128), the steady-state error can be arbitrarily reduced by increasing k _p > 0, although a zero steady-state error will never be accomplished. On the other hand, if a specific steady-state error e _ss is required, the corresponding value for k _p can be computed as:

Notice that to ensure k _p > 0, the steady-state error is required to be a fraction of the desired velocity, i.e., e _ss = ρA with 0 < ρ < 1.

Now, consider the expression for ω ₂. Using the Laplace transform:

Notice that G ₂(s) can be written as , with , i.e., a transfer function such as that in (3.126). Thus, we may proceed as before to conclude that:

The important observation in this expression is that the final value of ω ₂ can be reduced by increasing k _p > 0. This is good news, as ω ₂ is the velocity deviation due to external disturbance T _p; hence, it is desirable to keep ω ₂ close or equal to zero, if possible. In this case, the deviation can be rendered arbitrarily small by increasing k _p > 0, but a zero deviation will never be achieved with a proportional controller.

As the problem at hand is a first-order system, the transient response is completely described by the time constant:

Recall that a faster response is obtained as τ is shorter. This means that a faster response is obtained as k _p > 0 is chosen to be larger. Moreover, if all the system parameters are known and a desired time constant T is specified, then the required proportional gain can be exactly computed as:

This means that, to ensure k _p > 0, the closed-loop system is required to be faster than the open-loop system, i.e.:

The velocity responses of a DC motor, when using two different proportional gains k _p = 12 and k _p = 40, are shown in Fig. 3.32. This was performed using the MATLAB/Simulink diagram shown in Fig. 3.33 to represent the closed-loop system in (3.124). It is assumed that J = 1, nk _m∕R = 1 and . These values were selected merely to obtain representative, clearly appreciable results for the proportional control of velocity. Block ω _d is a step applied at t = 0 with a zero initial value and 2 as a final value. The external disturbance T _p is a step applied at t = 0.5 with a zero initial value and 7 as a final value. Notice that the use of a larger k _p results in a faster response, i.e., T ₂ < T ₁, a smaller steady-state error, i.e., a larger steady-state response β, and a smaller deviation δ because of a constant external torque disturbance.

The following is MATLAB code that is executed in an m-file to draw Fig. 3.32 after simulation in Fig. 3.33 stops:

3.8.2 Proportional Position Control Plus Velocity Feedback for a DC Motor

Consider the motor model in (3.122), but, now, expressed in terms of position θ, i.e., :

(3.129)

Assume that the voltage applied at the motor terminals is computed according to the following expression:

where k _v is a constant known as the velocity gain, and θ _d represents the desired position, which is assumed to be given as a step signal with magnitude A, i.e., . This represents a controller known as proportional position control plus velocity feedback . The closed-loop equation is found by replacing u in (3.129) and rearranging terms, i.e.,:

(3.130)

Using superposition and Laplace transform yields:

First, analyze the equation for θ ₁. By equating:

it is possible to write:

(3.131)

Notice that, under these conditions, the inverse Laplace transform of θ ₁(s) = G ₁(s)θ _d(s), is identical to the function presented in (3.57). Hence, stability is ensured if k _p > 0, to ensure that both ω _n and ζ are real numbers, and:

to ensure that ζ > 0. If this is the case, then we can use the final value theorem to compute the final position as:

Thus, the motor position reaches the constant desired position in a steady state. On the other hand, according to (3.131), we conclude that:

• The closed-loop system is faster as k _p > 0 is chosen to be larger. This is because ω _n increases as k _p > 0 increases. However, as k _p > 0 increases the system becomes more oscillatory. This is because ζ decreases as k _p increases.

• The system response is less oscillatory as k _v > 0 is chosen to be larger. This is because ζ increases as k _v increases.

This means that the response can be rendered as fast and damped as desired merely by suitably selecting both k _p and k _v.

Now, consider the equation for ω ₂. Notice that this variable represents the position deviation produced by an external torque disturbance. Hence, it is desirable for ω ₂ to be close or equal to zero, if possible. Notice that G ₂(s) is stable if is G ₁(s) stable, because both transfer functions have the same characteristic polynomial. Thus, we can compute the final value of ω ₂ using the final value theorem when the disturbance is a step signal, i.e., :

This means that the steady-state deviation is not zero. However, the good news is that this deviation can be arbitrarily reduced by increasing k _p. We conclude that a faster closed-loop system also results in smaller position deviations because of constant external torque disturbances. This is corroborated by simulations shown in Fig. 3.34, which were performed using the MATLAB/Simulink diagram in Fig. 3.35. There, the closed-loop system in (3.130) is simulated using J = 1, , and nk _m∕R = 70. These values were employed merely because they allow representative, clearly appreciable, results to be obtained. The block theta_d is a step applied at t = 0 with a zero initial value and 2 as a final value. The external disturbance T _p is a step applied at t = 0.5[s] with a zero initial value and 7 × 70 as a final value. Once simulation in Fig. 3.35 stops, the following MATLAB code is executed using an m-file to draw Fig. 3.34:

Notice that in the case where some t _r and M _p are specified, the use of expressions in (3.71) allow the required ω _n and ζ to be computed. These data and (3.131) allow the corresponding controller gains to be computed as:

(3.132)

Using this procedure together with t _r = 0.039[s], M _p = 29.9%, for the continuous line in Fig. 3.34, results in k _p = 40 and k _v = 0.45, whereas t _r = 0.071[s], M _p = 29.9%, for the dashed line, results in k _p = 12 and k _v = 0.2.

3.8.3 Proportional–Derivative Position Control of a DC Motor

Consider the motor model in (3.129) in a closed loop with the following proportional–derivative (PD) controller :

i.e.,:

where k _d is known as the derivative gain. Using superposition and the Laplace transform yields:

Notice that except for the term in the numerator of G ₁(s) and the use of k _d instead of k _v in the denominator, we have exactly retrieved the same closed-loop equations as in the previous example. Thus, we can proceed exactly as in the previous example to conclude:

• Expressions in (3.131) stand again with k _d instead of k _v.

• Stability is ensured if k _p > 0 and:

  

• If any external disturbance is not present, then the position reaches the constant desired position in a steady state.

• The closed-loop system is faster as k _p > 0 is chosen to be larger. This is because ω _n increases as k _p > 0 increases. However, as k _p > 0 increases, the system becomes more oscillatory. This is because ζ decreases as k _p increases.

• The system response is less oscillatory as k _d > 0 becomes larger. This is because ζ increases as k _d increases.

• The response can be rendered as fast and damped as desired merely by suitably selecting both k _p and k _d.

• If a step external torque disturbance is present, then a position deviation exists in a steady state and it is given as:

  

On the other hand, to study the transient response notice that the transfer function G ₁(s) can be rewritten as:

where:

(3.133)

have been used. Hence:

(3.134)

As , notice that the inverse Laplace transform of the second term in this expression is the function presented in (3.57), whereas the inverse Laplace transform of the first term is the time derivative of the function in (3.57) multiplied by the factor . This means that, although the shape of θ ₁(t) is similar to that of the function in (3.57), there are some differences. For instance, formulas for the rise time and overshoot given in (3.71) are only approximate in the present control problem.

In the case in which some t _r and M _p are specified, use of the expressions in (3.71) allow only approximate values for the required ω _n and ζ to be computed. These data and (3.133) allow approximate values for the corresponding controller gains to be computed as:

Some simulation results are presented in Fig. 3.36 when J = 1, , and nk _m∕R = 70. The corresponding MATLAB/Simulink diagram is shown in Fig. 3.37. The block theta_d is a step applied at t = 0 with a zero initial value and 2 as a final value. The external disturbance T _p is a step applied at t = 0.5[s] with a zero initial value and 7 × 70 as a final value. When the simulation in Fig. 3.37 stops, the following MATLAB code is executed in an m-file to draw Fig. 3.36:

nn=length(InputOutput(:,1));

n=nn-1;

Ts=1/n;

t=0:Ts:1;

plot(t,InputOutput(:,2),'k-.',t,InputOutput(:,3),

'k-');

hold on

plot(t,InputOutput(:,1),'k--');

axis([-0.02 1 0 3])

xlabel('time [s]')

ylabel('Position [rad]')

hold off

The dashed line represents the desired response, i.e., the inverse Laplace transform of:

(3.135)

which is identical to the response obtained with controller , introduced in the previous section, with k _p = 40, k _v = 0.45. The continuous line represents the inverse Laplace transform of the complete expression for θ ₁(s) in (3.134), i.e., the response when a PD position controller is used with k _p = 40, k _d = 0.45. Notice that the continuous line in Fig. 3.36 has a smaller rise time and a larger overshoot than the dashed line. The reason for this is that the PD control law includes the term , which is not present in control law . Recall that θ _d is not a constant, but it is a step signal whose time derivative is very large at the time when such a step signal is applied. This contributes with a large voltage spike at this point in time, which results in a faster response, including a larger overshoot for the step response when a PD controller is employed.

Notice that a nonzero steady-state deviation results when a step external torque disturbance appears at t = 0.5[s]. We stress that this deviation is the same in both responses, the continuous and the dashed line, because the same k _p is used.

3.8.4 Proportional–Integral Velocity Control of a DC Motor

Consider again the velocity control problem in a permanent magnet brushed DC motor. Now assume that the following proportional–integral (PI) controller is employed:

where k _i is a constant known as the integral gain, whereas ω _d is a step signal standing for the desired velocity, i.e., . The closed-loop system is obtained by replacing u, given above, in the model (3.122), i.e.,:

Differentiating once with respect to time and rearranging yields:

(3.136)

This is a second-order system with two inputs. Then, superposition has to be used, i.e., to define two signals ω ₁ and ω ₂ such that:

First, analyze the equation for ω ₁. Applying the Laplace transform with zero initial conditions:

(3.137)

Since the characteristic polynomial is second-degree, we proceed to equate:

to find:

(3.138)

These expressions require k _i > 0 to ensure that ω _n and ζ are real numbers. On the other hand, ζ > 0 is true if k _p is chosen to satisfy:

Hence, G ₁(s) is a stable transfer function and, thus, the velocity final value can be computed using the final value theorem:

It is concluded that the desired velocity is reached in a steady state, an important property of PI velocity control.

This result can be explained using everyday experience. As ω ₁ = A is achieved in a steady state, the error signal e = A − ω(t) is expected to behave as depicted in Fig. 3.38. Recall that the integral operation represents the area (positive) under the function shown in Fig. 3.38. This means that the above integral operation remains constant at a positive value when ω ₁ = ω _d = A (because the integrand becomes zero in these conditions). Then, according to , a positive voltage given by continues to be applied at the motor terminals. This allows the motor to stay rotating at the desired velocity, i.e., the condition ω ₁ = ω _d = A holds forever.

On the other hand, to study the transient response, notice that the transfer function in (3.137) can be rewritten as:

where (3.138) has been used. Hence, according to (3.137):

As , notice that the inverse Laplace transform of the second term in this expression is the function presented in (3.57), whereas the inverse Laplace transform of the first term is the time derivative of the function in (3.57) multiplied by the factor . This means that, although the shape of ω ₁(t) is similar to that of the function in (3.57), there are some differences. For instance, the rise time and overshoot formulas given in (3.71) are only approximate in the present control problem. This is a similar situation to that found in the study of the transient response of PD position control of a DC motor. Despite this, the expressions in (3.138) are still valid and they can be used to conclude the following:

• A faster response is achieved as k _i > 0 is chosen to be larger. This is because ω _n increases as k _i > 0 is increased. However, a more oscillatory response is obtained as k _i > 0 is chosen to be larger. This is because ζ decreases as k _i increases.

• The oscillation can be reduced by increasing k _p > 0. This is because ζ increases as k _p > 0 increases.

These observations can be used as a tuning procedure to select both k _p and k _i for a PI velocity controller. Note that this is a trial and error procedure.

Finally, consider the expression for ω ₂:

Using the Laplace transform yields:

Then, the final value of ω ₂ can be computed as:

when the external torque disturbance is a step signal, i.e., . Recall that it has already been ensured that the characteristic polynomial in this expression has two poles with a negative real part. Hence, the steady-state velocity deviation due to a constant external torque disturbance is zero, which is very good news.

The closed-loop response of velocity under PI control is shown in Fig. 3.39 where J = 1, , and nk _m∕R = 70 are used. These values are employed merely because they allow representative, clear, responses to be obtained. The corresponding MATLAB/Simulink diagram is shown in Fig. 3.40. The block ω _d is a step applied at t = 0 with a zero initial value and 2 as a final value. The external disturbance T _p is a step applied at t = 0.5[s] with a zero initial value and 0.4 × 70 as a final value. Once the simulation stops, the following MATLAB code is executed in an m-file to draw Fig. 3.39:

nn=length(InputOutput(:,1));

n=nn-1;

Ts=1/n;

t=0:Ts:1;

plot(t,InputOutput(:,1),'k--',t,InputOutput(:,2),

'k-');

axis([-0.02 1 0 2.5])

xlabel('time [s]')

ylabel('Velocity [rad/s]')

Notice that the velocity reaches its desired value before and after a constant external torque disturbance appears at t = 0.5[s].

3.8.5 Proportional, PI, and PID Control of First-Order Systems

Consider the water level system studied in Example 3.2 and depicted in Fig. 3.3. Suppose that a water pump is employed to produce a water flow q _i that is proportional to the voltage u applied at motor terminals actuating on the pump, i.e.,:

(3.139)

where k ₁ is a positive constant. Combining (3.139) and (3.28) it is obtained:

Notice that this differential equation can also be represented by a transfer function with the form:

(3.140)

Hence, identical conclusions are obtained as those for the proportional control of velocity, if a proportional level controller is used, or those for a proportional–integral velocity controller, if a PI level controller is used. Moreover, identical conclusions are valid for any first order plant with transfer function having the same structure as in (3.140).

On the other hand, first order systems represented by (3.140) do not need use of a controller with a derivative action. This can be explained as follows. Suppose that the following proportional–integral–derivative (PID) controller:

is used for plant in (3.140), i.e.,:

and differentiating once with respect to time:

The corresponding closed-loop transfer function is:

(3.141)

Notice that the characteristic polynomial is second degree. Thus, only two free parameters are necessary to assign the corresponding two roots at any location on the s complex plane, i.e., to achieve any desired transient response. However, three free parameters appear in this characteristic polynomial, i.e., k _p, k _d and k _i. Hence, one of them is not necessary, i.e., k _d.

On the other hand, the transfer function in (3.141) has polynomials at the numerator and the denominator having the same degree, i.e., it has the same number of poles and zeros. It has been shown in Examples 3.6 and 3.7 that time response of a system whose transfer function has the same number of poles and zeros is discontinuous when a step input is applied. It is important to remark that such a response is not possible, for instance, in a velocity control system because the velocity cannot be discontinuous. This is due to the fact that such a response would require a very large voltage to be applied at the motor terminals during a very short time interval. Mathematically, this is explained by the time derivative of the desired value y _d (a discontinuous, step signal) which is introduced by the derivative action of a controller. However, such large voltage spikes are not possible in practice. Thus, an unpredictable response is expected. Moreover, in the case of a velocity control system, the derivative action on velocity (equivalent to acceleration measurements, a very noisy signal) results in a strong amplification of noise, as velocity measurements are recognized to be noisy signals.

Thus, it is concluded that any performance improvement is not expected when using a PID controller with respect to performance achieved with a PI controller for first-order plants.