A Very Brief History of Feedback: Before and After Maxwell Jean-Michel Coron Laboratoire J.-L

A very brief history of feedback: before and after Maxwell Jean-Michel Coron Laboratoire J.-L. Lions, Université Pierre et Marie Curie http://www.ann.jussieu.fr/coron/ BCAM OPTPDE summer school, Bilbao, July 4-8 2011 The cart-inverted pendulum

C F The cart-inverted pendulum: An equilibrium

C Instability of this equilibrium

C Stabilization of the unstable equilibrium

F : depending on the state C F is a feedback Doubleinverted pendulum (CAS, ENSMP/La Villette)

The Clepsydra (water clock) The Clepsydra (water clock) Ctésibius (3th century BC) towards tank 1

tank 2

tank 3

James Watt

1736. Birth, Greenock, Scotland. 1769. Patent of the separate condensation chamber. 1782. Double action. 1788. Adaptation to the steam engine of a regulator used in windmills. 1819. Death, Heathﬁeld, England. Thomas Newcomen 1663-1729 *** Thomas Savery 1650-1715

History of the ﬂy ball (centrifugal governor)

In fact, this regulator is directly inspired from regulators used in windmills. There exists a letter, dated May 28 1788, where Boulton, Watt’s partner in the ﬁrm Boulton & Watt, describes this regulator: [the regulation of the top mill stone] is produced by the centrifugal force of 2 lead weights which rise up horizontal when in motion and fall down when the motion is decreased, by which means they act on a lever that is divided as 30 to 1, but to explain it requires a drawing. In the windmills, these regulators where used to regulate • the distance between the two mill stones, • the speed of rotation of the top mill stone.

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39 42 Maxwell’s article “On governors”

On governors, Proceedings of the Royal Society, vol. 16, 1868, pages 270–283, James Clerk Maxwell (1831-1879) James Clerk Maxwell

1831. Birth, Edinburgh, Scotland. 1857. Adams Prize for his article “On the Stability of the Motion of Saturn’s Rings”.

Alexis Bouvard

1821 : Uranus does not follow the expected trajectory (that one can compute using the Laplace perturbations theory and the informations on the known planets)! John Couch Adams and Urbain Le Verrier

Le Verrier (1811-1877) Adams (1819-1892) 1845 : beginning of the 1843: Beginning of the computations. computations. 1846 (June) : First results. 1845 (October): Results given to G. B. Airy and J. Challis. 1846 (August) : New results. 1846 (August) : J. Challis is 1846 (September) : J. G. Galle looking for the planet. sees the planet at the place indicated by Le Verrier.

John Couch Adams and Urbain Le Verrier

Le Verrier (1811-1877) Adams (1819-1892) 1845 : beginning of the 1843: Beginning of the computations. computations. 1846 (June) : First results. 1845 (October):Results given to G. B. Airy and J. Challis. 1846 (August) : New results. 1846 (August): J. Challis is 1846 (September) : J. G. Galle looking for the planet. sees the planet at the place indicated by Le Verrier.

A picture of “la belle inconnue”

Sonde Voyager 2; 1989

Distance to sun : 4,5 milliards de kms. Radius : 25 000 kms. Temperature : -226 degrés C. Period of revolution 164 ans 323 jours. James Clerk Maxwell

1831. Birth, Edinburgh, Scotland. 1857. Adams Prize for his article “On the Stability of the Motion of Saturn’s Rings”.

James Clerk Maxwell

1831. Birth, Edinburgh, Scotland. 1857. Adams Prize for his article “On the Stability of the Motion of Saturn’s Rings”. 1861. First color photograph. Works on the perception of color 1855 to 1872. 1866. Kinetic theory of gases. 1868. Article: “On governors”. 1873. Book: “A treatise on electricity and magnetism” (équations de Maxwell). 1879. Death, Cambridge, England. Aleksandr Lyapunov (1857-1918)

Thesis: The general problem of the stability of motion (1892 in Russian, 1907 in French -see Numdam, English 1992) • Deﬁnition of the stability and of the asymptotic stability.

Remark In the case of Mechanical systems, there were important previous works, in particular by J.-L. Lagrange, P.-S. Laplace, G. Dirichlet. Stability Stability Stability Stability Stability Attractor Attractor Attractor Attractor Attractor does not imply stable Attractor does not imply stable Attractor does not imply stable Attractor does not imply stable Attractor does not imply stable Attractor does not imply stable Attractor does not imply stable Asymptotically stable

Déﬁnition (Asymptotically stable) n n n Let xe be an equilibrium point of X : R → R , i.e. a point in R such that X(xe) = 0. One says that xe is asymptotically stable for x˙ = X(x) if xe is stable and attractor. One says that xe is unstable for x˙ = X(x) if it is not stable. First Lyapunov’s theorem

Theorem 1 Let us assume that X is of class C and that xe is an equilibrium point of ′ x˙ = X(x). If all the eigenvalues of X (xe) have a strictly negative real ′ part, then xe is locally asymptotically stable for x˙ = X(x). If X (xe) has an eigenvalue with a strictly positive real part, then the equilibrium xe is unstable for x˙ = X(x).

Remark This theorem was assumed to be true before Lyapunov proved it. See, for example, the fundamental paper “On governors” by J.C. Maxwell (1868). Second Lyapunov’s theorem

Theorem

The point xe is asymptotically stable for x˙ = X(x) if and only if there n ∞ exists η> 0 and V : B(xe, η) : {x ∈ R ; |x − xe| < η}→ R of class C such that

(1) V (x) > V (xe), ∀x ∈ B(xe, η) \ {xe} n ∂V (2) ∇V (x) X(x)= X (x) < 0, ∀x ∈ B(x , η) \ {x }. ∂x i e e i=1 i Remark • In the cases of mechanical systems; there are important previous results by J.-L Lagrange (1788) and G. Dirichlet (1847). • For the only if part: J. Massera and J. Kurzweil. • This theorem explains why “asymptotic stability” is the good notion. Robustness and Lyapunov function Robustness and Lyapunov function Key innovations of Maxwell’s paper

1. Regulation is a dynamical problem. Before Maxwell’s paper, the engineers and scientists were looking only to equilibria. 2. Linearization of the dynamical system. 3. The stability holds if all the roots of the characteristic polynomial have a strictly negative real part.

Remark In fact, part of the above innovations already appeared in two papers by G.B. Airy (1840 and 1850). G.B. Airy was “Astronomer Royal”. For a telescope, having a very good regulation of speed is an essential point: One needs to compensate exactly the earth rotation in order to make astronomical observations. The two Airy papers are on the dynamical properties of regulators used for telescopes. Watt’s regulator: Notations S

ϕ Dynamic equations

Let x1 = θ, x2 = θ˙ and x3 =ϕ ˙. The dynamic equations are

(1) x˙ 1 = x2, 2 g C (2) x˙ 2 = sin(x1) cos(x1)x3 − sin(x1) − x2, l 2ml2 Γr Γ0 k (3) x˙ 3 = − + − (1 − cos(x1)), J J J

where g is the gravitational constant, Cx2 = Cθ˙, with C > 0 is the friction term (it includes the friction at the pivot point S), Γr > 0 is the load torque applied to the steam engine, Γ0 − k(1 − cos(x1)) is the engine torque produced by the steam J is the moment of inertia. The feedback law is Γ0 −k(1 − cos(x1)). The constants Γ0 and k are part of the design of the feedback law. They can be essentially be chosen freely by modifying the throttle valve and beam linkage. Watt’s regulator: The equilibrium point

x1 = x1, x2 = x2 and x3 = x3 is an equilibrium of the regulator if and only if

2 g (1) x2 = 0, sin(x1) cos(x1)x3 − sin(x1) = 0, l Γr Γ0 k (2) − + − (1 − cos(x1)) = 0. J J J

For physical reasons sin(x1) = 0. Hence, using (1), g (3) cos(x1)= 2 . lx3

′ The desired angular velocity is x3 = ϕ = ω. Then, using (2), one imposes on Γ0 and k to satisfy g (4) Γ =Γ0 − k 1 − . r lω2 Maxwell’s key idea: The linearization

One writes x1 = x1 + y1, x2 = x2 + y2 = y2, x3 = x3 + y3 = ω + y3. We assume that the yi’s are small. Up to the order 1 in the yi’s, we have y˙ = Ay, with

0 1 0 2 2 C (1) A = −ω sin (x1) − ω sin(2x1) .  2ml2  k − sin(x1) 0 0  J    The eigenvalues of A are the root of the following polynomial (called characteristic polynomial):

3 C 2 2 2 2kg 2 (2) z + z + ω sin (x1)z + sin (x1) = 0. 2ml2 Jlω In his 1868 paper, J. C. Maxwell presents other regulators for which the characteristic polynomial is of larger degree (n ∈ {3, 4, 5}). This leads naturally to the following problem. Let n be a positive integer. Let a1, a2, ..., an be n real numbers. Under which condition on the ai’s, all the roots of the polynomial

n n−1 n−2 (1) P (z)= z + a1z + a2z + . . . + an−1z + an have a strictly negative real part. Remark J.C. Maxwell says “the possible roots and the possible parts of the impossible roots” for the real roots and the real parts of the nonreal roots. A necessary condition on the ai’s

One easily checks that the following theorem holds. Theorem (n ∈{1, 2}) 2 If P = z + a1 or P (z)= z + a1z + a2, all the roots have a strictly negative real part if and only if the coeﬃcients of P are all strictly positive.

Corollary (For every positive integer n> 0) n n−1 n−2 1 If all the roots of P (z)= z + a1z + a2z + . . . + an−1z + an have a strictly negative real part, then all the coeﬃcients a1, a2, ..., an−1, an are strictly positive.

Unfortunately, as soon as n > 3, the converse of the above corollary is false. 3 2 The degree 3 case: P (z)= z + a1z + a2z + a3 Theorem (Maxwell (1868)) 3 2 All the roots of P = z + a1z + a2z + a3 have a strictly negative real part if and only if

(1) a1 > 0, a1a2 > a3 and a3 > 0. Proof. Clearly P vanishes on R. Hence there are three real numbers a, b and c such that P (z) = (z + a)(z2 + bz + c). Using the previous theorem for n = 1 and n = 2, all the roots of P have a strictly negative real part if and only if a> 0, b> et c> 0. Expanding the above expression of P , one gets

(2) a1 = a + b, a2 = ab + c, a3 = ac. The key point is the following identity: (3) a1a2 − a3 = b(a(a + b)+ c). Hence, if a> 0, b> 0 and c> 0, then one has (1). Conversely, let us assume that (1) holds. If a 6 0... Application to the Watt regulator

Let us recall that, for the Watt regulator, the characteristic polynomial is

3 C 2 2 2 2kg 2 (4) z + z + ω sin (x1)z + sin (x1) = 0. 2ml2 Jlω The Maxwell conditions in this case are

C 2 2kg (5) C > 0, ω > . 2ml Jω

Remark It is important to have C > 0 (use if necessary a dashpot). It is better to have m small. What to do if n> 3 ? In his 1868-paper, J.C. Maxwell gives, for n = 5, a n n n−i necessary condition on the ai’s for P (z)= z + i=1 aiz to have only roots with strictly negative real part. But this condition is not suﬃcient. In 1875, following Maxwell’s suggestion, the following subject is proposed for the 1877 Adams prize: The Criterion of Dynamical Stability.

E. Routh won this Adams prize. He found a necessary and suﬃcient n n n−i condition on the ai’s for P (z)= z + i=1 aiz to have only roots with strictly negative real part. Edward John Routh

1831. Birth in Quebec, Canada. 1842. Arrival in England. 1854. Senior Wrangler (J.C. Maxwell was Second Wrangler the same year). 1864. Marriage with with Hilda Airy, George Biddell Airy’s eldest daughter. They will have 6 children together. 1877. Adams Prize for his book “A treatise on the stability of motion”. 1907. Death, Cambridge, England. E. Routh was an outstanding teacher (27 of his students were Senior Wranglers). He developed the mathematical theory of mechanics. His main work is his stability criterion (Routh’s stability criterion). Routh’s stability criterion (1877)

Let P le real polynomial of degree n n n−1 n−2 (1) P (z)= z + a1z + a2z + . . . + an−1z + an.

One builds the following table of numbers, with a0 := 1,

a0 a2 a4 a6 a8 a10 . . . a1 a3 a5 a7 a9 a11 . . . b1 b2 b3 b4 b5 b6 . . . c1 c2 c3 c4 c5 c6 ...... where a1a2 − a0a3 a1a4 − a0a5 a1a6 − a0a7 (2) b1 := , b2 := , b3 = ,... a1 a1 a1

a3b1 − a1b2 a5b1 − a1b3 a7b1 − a1b4 (3) c1 := , c2 := , c3 = ,... b1 b1 b1 Routh’s stability criterion (1877)

yz − xt (1) u = z Routh’s stability criterion: a necessary and sufficient condition for all roots of P to have a strictly negative real part is that all the n + 1 first elements of the first column are strictly positive. In fact this criterion was already in a paper by Charles Hermite (1856)! Independently of Hermite and Routh, Adolf Hurwitz found in 1895 another criterion. The two criteria were shown to be identical by Enrico Bompiani in 1911. Routh’s stability criterion (1877)

x y z t u

yz − xt (1) u = z Routh’s stability criterion: a necessary and sufficient condition for all roots of P to have strictly negative real part plane is that all the n + 1 first elements of the first column are strictly positive. In fact this criterion was already in a paper by Charles Hermite (1856)! Independently of Hermite and Routh, Adolf Hurwitz found in 1895 another criterion. The two criteria were shown to be identical by Enrico Bompiani in 1911. Routh’s stability criterion (1877)

x y z t u

The Routh table is

1 a2 a4 0 0 0 a1 a3 0 000 a1a2 − a3 a4 0 000 a1 2 2 a1a2a3 − a3 − a1a4 0 0 000 a1a2 − a3 a4 0 0 000 0 00000

Therefore, all the roots of P have a strictly negative real part if and only if

2 2 (1) a1 > 0, a1a2 > a3, a1a2a3 − a3 − a1a4 > 0, a4 > 0. 5 4 3 2 Application to P (z) := z + a1z + a2z + a3z + a4z + a5

The Routh criterion is now

(1) a1 > 0, a1a2 − a3 > 0, (a1a2 − a3)a3 − a1(a1a4 − a5) > 0, 2 (2) (a1a2 − a3)a3 − a1(a1a4 − a5) (a1a4 − a5) − (a1a2 − a3) a5 > 0. Remark Even if P cannot, in general, be solvable by radicals (Evariste Galois, 1831), one gets algebraic conditions. Hermite’s proof

For simplicity we deal only with n = 4. Hence

4 3 2 (1) P (z)= z + a1z + a2z + a3z + a4.

One has

1 P (x)P (y) − P (−x)P (−y) 2 3 y (2) = (1,x,x ,x )H  2 , 2(x + y) y y3     with

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3 (3) H :=   . a1a4 0 a1a2 − a3 0  0 a3 0 a1     Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1      Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1     

∆1 = a1 =: γ1. Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1     

∆1 = a1 =: γ1. ∆2 = γ1γ2, with γ2 := a1a2 − a3. Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1     

∆1 = a1 =: γ1. ∆2 = γ1γ2, with γ2 := a1a2 − a3. 2 ∆3 = γ2γ3, with γ3 := (a2a3 − a1a4)a1 − a3. Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1     

∆1 = a1 =: γ1. ∆2 = γ1γ2, with γ2 := a1a2 − a3. 2 ∆3 = γ2γ3, with γ3 := (a2a3 − a1a4)a1 − a3. 2 ∆4 = γ3γ4, with γ4 := a4 a3(a1a2 − a3) − a1a4 = a4γ3. Conditions for H > 0

a3a4 0 a1a4 0 0 a2a3 − a1a4 0 a3   a1a4 0 a1a2 − a3 0  0 a3 0 a1     

∆1 = a1 =: γ1. ∆2 = γ1γ2, with γ2 := a1a2 − a3. 2 ∆3 = γ2γ3, with γ3 := (a2a3 − a1a4)a1 − a3. 2 ∆4 = γ3γ4, with γ4 := a4 a3(a1a2 − a3) − a1a4 = a4γ3. One recovers the Routh condition for n = 4:

2 2 (1) a1 > 0, a1a2 > a3, a1a2a3 − a3 − a1a4 > 0, a4 > 0. P.C. Parks’s method (1992)

tr 4 Let us consider the dynamical system x˙ = Ax, x = (x1,x2,x3,x4) ∈ R with

0 1 0 0 0 0 1 0 (1) A :=   0 0 0 1 −a4 −a3 −a2 −a1     It is classical and easy to check that 4 3 2 det (zId − A)= z + a1z + a2z + a3z + a4 =: P (z). Let tr V (x) := x Hx. Let us compute the time derivative V˙ of V along the trajectories of x˙ = Ax. One gets 2 (2) V˙ = −2(a3x2 + a1x4) . We ﬁrst assume that H > 0 and want to prove that 0 is asymptotically stable for x˙ = Ax. This will show that the Routh criterion is a suﬃcient condition for asymptotic stability. We apply the LaSalle invariance principle. Let x : R → R4 be a trajectory of x˙ = Ax such that

(1) a3x2 + a1x4 = 0.

Diﬀerentiating (1) with respect to time and using (1), one gets

(2) − a1a4x1 + (a3 − a1a2)x3 = 0.

Diﬀerentiating (2) with respect to time and using once more (1) one gets

(3) − a1a4x2 + (a3 − a1a2)x4 = 0.

Note that the determinant of (1) and (3) considered as linear equations in tr 2 (x2,x4) is a3(a3 − a1a2)+ a1a4. By the Routh conditions a3 > 0, (a3 − a1a2) > 0, a4 > 0. Hence this determinant is strictly positive and therefore not 0. Hence x2 = 0, from which one easily gets that x = 0. Conversely... Oﬀset

Unfortunately the Watt regulator leads to a static error if the load torque is not the one which is expected. (This is the oﬀset phenomena.) Indeed, let us recall that the equilibrium of the Watt regulator is given by g (1) Γ0 − Γr = k 1 − . lx2 3

We had assumed that Γr was known and we have chosen the coeﬃcients Γ0 et k of the feedback laws so that x3 = ω is a solution of (1). If the load torque is not Γr but Γr + δΓr, the angular velocity at the equilibrium is x3 given by 1 (2) x3 = ω 2 = ω. lω δΓr 1+ kg Oﬀset

The quest for the Holy Grail: Find a regulator with no oﬀset.

Isochronous regulators

f(r)

r Isochronous governors: The equilibrium

The new dynamical equations are now

′ 2 ′ ′′ 2 2 ′ (1) 1+ f (r) r¨ + f (r)f (r)r ˙ − rϕ˙ + gf (r)= −C(r)r, ˙ (2) Mϕ¨ = −Γr + u(r), where, as above C(r)r ˙ is a damping term with C(r) > 0. We want that, at the equilibrium, ϕ˙ = ω what ever is the load torque Γr. Looking at (1), one sees that this possible if and only if−rω2 + gf ′(r) = 0, i.e. if and only if there exists a ∈ R such that ω2r2 (3) f(r)= + a. 2g

For u, one can then simply take, for example, u(r):=Γ0 − k(r − r0). With this parabolic proﬁle there is no more oﬀset: the equilibrium is ϕ˙ = ω, r˙ = 0, r = r0 + (Γr − Γ0)/k.

Villarceau’s regulator (1872)

ϕ Villarceau’s regulator (1872)

Instability of the isochronous pendulum (Wischnegradski, 1876)

Let us introduce y1 = x1 − x1 = r − r¯, y2 =r ˙, y3 =ϕ ˙ − ω. Assuming that y1, y2 and y3 and developing to the order 1 the dynamic equation variables,

2 2 g C(x1) 2ωg x1 k (1) y˙1 = y2, y˙2 = − 2 4 2 y2 + 2 4 2 y3, y˙3 = − y1. g + ω x1 g + ω x1 M

The characteristic polynomial of (1) is

2 2 3 g C(x1) 2 k 2ωg x1 (2) P (z)= z + 2 4 2 z + 2 4 2 . g + ω x1 M g + ω x1

The z is term 0: there exists therefore at least one root of P whose real part is nonnegative. In fact there exists at least one root of P whose real part is strictly positive. Hence the equilibrium is unstable for the linearized system (1) and for the nonlinear system. How to avoid the oﬀset: a simple example

Let us consider the simple control system

(1) x˙ = u,

where the state is x ∈ R and the control is u ∈ R. In order to stabilize the origin for the control system (1), it suﬃces to take the following feedback u(x)= −kx where k is a positive constant. Then the closed loop system is

(2) x˙ = −kx,

and the origin is indeed asymptotically stable for this system. We now assume that there exists a constant perturbation: the control system is not (1) but

(3) x˙ = u + p,

where p is the constant perturbation. This perturbation is unknown: one cannot use it to deﬁne the feedback. If one uses on the true system x˙ = u + p the feedback u(x)= −kx, the closed loop system is

(1) x˙ = −kx + p.

For the initial data x(0) = a, the solution of (1) is

p p − (2) x(t)= + a − e kt. k k Hence, as t → +∞, x(t) does not converge to 0 but to p/k, which is not the desired value. To deal with this problem (the oﬀset), one adds an integration: one considers the control system

(3) x˙ = u + p, y˙ = v, where the state is now (x,y)tr ∈ R2 and the control is (u, v)tr ∈ R2. One tries a feedback of the form u = k1x + k2y, v = k3x + k4y. The closed loop system is now

(4) x˙ = p + k1x + k2y, y˙ = k3x + k4y. Study of the closed loop system x˙ = p + k1x + k2y, y˙ = k3x + k4y

The point (x, y)tr is an equilibrium of the closed loop system if an only if

(1) p + k1x + k2y = 0, k3x + k4y = 0.

Ones assume that k1k4 = k2k3. Then, (1) has one and only one solution:

k4 k3 (2) x = − p, y = p. k1k4 − k2k3 k1k4 − k2k3 In order to have a x = 0 which independent of p it is (necessary and) tr suﬃcient to take k4 = 0. Let Y = y − y. The dynamic equation of (x,Y ) is simply

(3) x˙ = k1x + k2Y, Y˙ = k3x. It follows from what we have seen above that the origin of R2 is asymptotically stable for (3) if and only if

(4) k1 < 0 and k2k3 < 0. For example, one can takek1 = −1, k2 = −1 and k3 = 1. For these values of the ki’s, the closed loop system is

(1) x˙ = −x − y + p, y˙ = x.

Remark Under the above assumption (k1 < 0 et k2k3 < 0), as t → +∞, y(t) does not converge to 0 but to y = −p/k2. Roughly speaking the oﬀset has been transferred to y. But we are interested in x and not in y.

Conclusion. For the application to the regulation of the steam engine one needs to ﬁnd a way to integrate the diﬀerence between the speed engine and the desired speed. Various solutions to this problem were proposed. Let us mention, in particular, solutions proposed by 1790 The Périer brothers, 1863 Fleeming Jenkin. (See J.C. Maxwell’s 1868 paper.)

Another drawback of the Watt regulator

The Watt regulator was to slow to react to a change of the load torque. The solution: The inertial regulators (the D term of the PID-controller). Usually, one says that the ﬁrst inertial controller is due to the Siemens brothers (1845). However there is already a inertial term in the Mead controller.

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2a is the length of the blue rod, l is the distance between the centers of the blue balls and the blue rod, m is the mass of the each blue ball. Dynamic of the Mead regulator

a a 2 (1) θ¨ = cos(θ)ϕ ¨ − sin(θ)θ˙ϕ˙ + sin(θ) cos(θ)ϕ ˙ l l a g C + sin(θ)ϕ ˙θ˙ − sin(θ) − θ,˙ l l 2ml2 Dynamic of the Watt regulator

a 2 g C (2) θ¨ = cos(θ) + sin(θ) ϕ˙ − sin(θ) − θ.˙ l l 2ml2 The ϕ¨ is a key term: It allows the Mead regulator to react to the disturbances much quicker than the Watt regulator. This is the D term of “PID” (Proportional-Integral-Derivative) controller. N. Minorsky (1922) introduced a controller with the the three terms (Proportional-Integral-Derivative) for the steering of ships. Dynamic of the Mead regulator

a a 2 (1) θ¨ = cos(θ)ϕ¨ − sin(θ)θ˙ϕ˙ + sin(θ) cos(θ)ϕ ˙ l l a g C + sin(θ)ϕ ˙θ˙ − sin(θ) − θ,˙ l l 2ml2 Dynamic of the Watt regulator

a 2 g C (2) θ¨ = cos(θ) + sin(θ) ϕ˙ − sin(θ) − θ,˙ l l 2ml2 The ϕ¨ is a key term: It allows the Mead regulator to react to the disturbances much quicker than the Watt regulator. This is the D term of “PID” (Proportional-Integral-Derivative) controller. N. Minorsky [1922] introduced a controller with the the three terms ((Proportional-Integral-Derivative)) for the steering of ships.