Automation Systems

Lecture 2 - Mathematical Models of Dynamical Systems

Jakub Mozaryn

Institute of Automatic Control and Robotics, Department of Mechatronics, WUT

Warszawa, 2019

Jakub Mozaryn Automation Systems Mathematical modelling for control systems

Real processes, and thus control systems, have nonlinear properties: turbulences, multiple stable states, hysteresis, energy losses due to friction. In practice, to simplify the mathematical description, there is carried lin- earization, enabling the formulation of the approximate description of a linear phenomenon, in vicinity of the operating point (this point corre- sponds to the nominal or average operating conditions of the system).

Linearization steps

1 description of the phenomenon in the form of differential equations, 2 linearization, 3 operational calculus: differential equations → algebraic equations.

Jakub Mozaryn Automation Systems 6 step approach to modelling

STEP 1: Define the system and its components. STEP 2: Formulate the mathematical model and fundamental necessary assumptions based on basic physical principles. STEP 3: Obtain differential equations representing the mathematical model. STEP 4: Solve equations for the desired output variables. STEP 5: Examine the assumptions and solutions. STEP 6: If necessary, reconsider and redesign the system.

Jakub Mozaryn Automation Systems Description of linear models

The basic forms of mathematical description of the system dynamical properties are: Equations of Motion: equations of system dynamics in form of differential equations. Transfer function. State Space Equations (not covered in the course). In the case of dynamical system (process) with one input signal x(t) and one output signal y(t) equation of motion describes the relationship between the output signal y(t) and the input signal x(t) in a following form: y(t) = f (x(t)) (1)

Jakub Mozaryn Automation Systems Description of linear models / systems

Principle of superposition:

f (x1 + x2) = f (x1) + f (x2), and f (0) = 0. (2)

Space of solutions of the equation that satisfies (2) is a linear space. Homogeneity (implies scale invariance): Function f (x, y) is said to be homogeneous of degree k if

f (βx, βy) = βk f (x, y), and f (0) = 0, (3)

where: β - constatnt coefficient. Linear system Homogenous system, which preserve the principle of superposition.

Nonlinear system The system, which does not preserve the principle of superposition and/or is not homogenous.

Jakub Mozaryn Automation Systems Linearity - example

Using the appropriate theorems and assummptions, check whether the system described by the equation

y(t) = ax(t) + b (4)

where: y - output signal, x - input signal, a = const, b = const - constant parameters, is linear.

Jakub Mozaryn Automation Systems Description of linear models / systems

General form of the differential equation describing linear system:

d ny d n−1y d mx d m−1x a +a +···+a y = b +b +···+b x (5) n dtn n−1 dtn−1 0 m dtm m−1 dtm−1 0

where: y - output signal, x - input signal, ai , bi - constant coefficients.

Jakub Mozaryn Automation Systems Proportional elements

Dynamics equation - relationship between input and output signal:

R2 U2(t) = U1(t) (6) R1 + R2 Figure 1: Proportional element - voltage divider General equation of proportional Input signal x(t) - voltage element U1(t). Output signal y(t) - voltage y(t) = kx(t) (7) U2(t).

Jakub Mozaryn Automation Systems First order lag elements

Dynamics equation - relationship between input and output signal:

L dU (t) 2 + U (t) = U (t) (8) R dt 2 1

General equation of first order Figure 2: First order lag element - lag element RL filter Input signal x(t) - voltage dy(t) T + y(t) = kx(t) (9) U1(t). dt Output signal y(t) - voltage U2(t).

Jakub Mozaryn Automation Systems First order lag elements - equivalents

Figure 3: First order lag elements

V dp2(t) a) αRΘ dt + p2(t) = p1(t) J dω(t) 1 b) R dt + ω(t) = R M(t) L dU2(t) c) R dt + U2(t) = U1(t)

General equation of first order lag element dy(t) T + y(t) = kx(t) (10) dt

Jakub Mozaryn Automation Systems Static characteristic

Static characteristics

Static characteristic fs describes the dependence of the output signal y of the system from the input signal x in steady state.

Steady state Steady state is a state in which all derivatives of the input signal and output signal are equal to zero. In such a situation the output Figure 4: Static characteristics of linear system. signal has a steady value.

Jakub Mozaryn Automation Systems Linearization

Creation of linear description of the system, based on nonlinear description of this system is called linearization.

Linearization of nonlinear description in the form of nonlinear algebraic equations is called static linearization. (There are no derivatives)

Linearization of nonlinear description in the form of nonlinear differential equations is called the dynamic linearization.

Methods of static linearization secant method: obtain the best relation between the linear and nonlinear description of a system in the specified range of changes of the independent variable (input). tangent method: obtain the best relation between the linear and nonlinear description of a system for a given value of the independent variable (input), and hence a particular value of the dependent variable (output).

Jakub Mozaryn Automation Systems Static linearization

Figure 5: Static linearization; a) secant method, b) tabgent method.

In control system design, there is considered the behavior of plant/system in a vicinity of a specified operating point. Therefore in practical appli- cations tangent linearization method is much useful.

Jakub Mozaryn Automation Systems Tangent method

Process of linearization using tangent method involves: replacement of the curve representing nonlinear relationship y = f (x) with its tangent at operating point, transfer the origin to operating point, replacement in mathematical model absolute variables x and y with deviations of these variables from operating point - incremental variables ∆x and ∆y. Static characteristics obtained using linearized equation, in terms of the specified operating point, is a linear function. It can be also obtained by linearization of real characteristics in terms of the same operating point.

Jakub Mozaryn Automation Systems Static linearization

Example [homework] Determine the linearized function that describes the dependence of the mass flow Q of the fluid flowing through the valve, from the pressures p1 and p2 and the distance x of the plug from the valve seat. p Q(t) = απd · x(t) 2ρ(p1(t) − p2(t)) (11) We look for the linear function in following form (find coefficients)

QL(t) = b1∆x(t) + b2∆p1(t) + b3∆p2(t) (12)

Figure 6: Example of the system (valve). Jakub Mozaryn Automation Systems Dynamic linearization - example

Example: non-homogenous function

y = mx + b (13)

The nominal operating point - {x0, y0}, y0 = f (x0)

Taylor series expansion about the operating point

df (x − x ) d 2f (x − x )2 y = f (x) = f (x ) + | 0 + | 0 + ... (14) 0 dx x=x0 1! dx 2 x=x0 2! The slope (first derivative) over the operating point isa good approxima- tion of the curve over small range.

Therefore df y = f (x ) + | (x − x ) = y + m(x − x ) (15) 0 dx x=x0 0 0 0 and finally y − y0 = m(x − x0) → ∆y = m∆x (16)

Jakub Mozaryn Automation Systems Dynamic linearization

An example of differential equation, which describes linear relationship between functions x(t), y(t) and their derivatives.

F [y(t), y˙(t), y¨(t),..., y (n)(t), x, x˙(t), x¨(t),..., x (m)(t)] = 0 (17)

During dynamic linearization, functions x(t), y(t) and their derivatives are treated analogously to variables of implicit function.

n ( ) m ( ) X  ∂F  X  ∂F  ∆y (i) + ∆x (j) = 0 (18) (i) (j) ∂y (i) ∂x (j) i=0 y0 j=0 x0

where: d∆y d n∆y ∆y = y(t) − y , ∆y (1) = ,..., ∆y (n) = 0 dt dtn

d∆x d m∆x ∆x = x(t) − x , ∆x (1) = ,..., ∆x (m) = 0 dt dtm

Jakub Mozaryn Automation Systems Dynamic linearization - example

Linearize the following differential function y(t) = 2x(t)2 + x(t)x ˙(t) + 2¨x(t)2 (19)

The nominal operating point - {x0, y0}, x0 = 1, x˙0 = 0, x¨0 = 0.

Using Taylor series expansion around the operating point

∆y(t)+[−4x(t)−x˙(t)]0∆x(t)−[x(t)]0∆¨x(t)−[4¨x(t)]0∆¨x(t) = 0. (20) In the nominal operating point, the linearized model of the system, has the following form ∆y(t) − 4∆x(t) − ∆x ˙(t) = 0. (21) Nonlinear static characteristic of the system (14) y = 2x 2. (22) Linear static characteristic (the tangent of nonlinear characteristic in op- erating point, all time derivatives are equal to zero) ∆y = 4∆x. (23)

Jakub Mozaryn Automation Systems Laplace transform

Replacing differential equation with transfer function (algebraic equation) needs transition from the time domain (t) to the complex plane (s).

f (t) ⇔ f (s), where s = c + jω (24)

where: c - real part coefficient, ω - conjugate part coefficient.

Laplace transform

∞ Z f (s) = L[f (t)] = f (t)e−st dt (25)

0 Inverse Laplace transform - Riemann-Mellin integral

c+jω 1 Z f (t) = L−1[f (s)] = F (s)est ds (26) 2πj c−jω

Jakub Mozaryn Automation Systems Laplace transform of the linear systems

Laplace transform is used for an analysis of control systems. As a tool for graphical analysis, complex plane S is used, where multiplication by s has the effect of differentiation and division by s has the effect of integration. Analysis of complex roots of a linear equation, may disclose information about the frequency characteristics and the stability of the system. To determine the function’s Laplace transform the following conditions must be met: f (t) has a finite value in any finite interval,

df (t) f (t) has a derivative dt in any finite interval, ∞ there exists a set of real numbers X for which the integral R e−ct is 0 absolutely convergent.

Jakub Mozaryn Automation Systems Laplace transform of the linear systems

Linear system is described by following differential equation

d ny d n−1y d mx d m−1x a +a +···+a y = b +b +···+b x (27) n dtn n−1 dtn−1 0 m dtm m−1 dtm−1 0 Using the n-th derivative property of the Laplace transform

d ny  L = sny(s) − sn−1y(0+) − · · · − y n−1(0+) (28) dtn

and assuming that initial conditions are zero, one obtains

d ny  L = sny(s) (29) dtn

Laplace transform of the linear dynamic system (22) with zero initial conditions take the following form

n n−1 m m−1 y(s)(ans +an−1s +···+a0) = x(s)(bms +bm−1s +···+b0) (30)

Jakub Mozaryn Automation Systems Transfer function

Transfer function For continuous-time input signal x(t) and output y(t), the transfer func- tion G(s) is the linear mapping of the Laplace transform of the input, X (s) = L[x(t)], to the Laplace transform of the output Y (s) = L[y(t)] at zero initial conditions:

n n−1 m m−1 y(s)(ans +an−1s +···+a0) = x(s)(bms +bm−1s +···+b0) (31)

m m−1 y(s) bms + bm−1s + ··· + b0 G(s) = = n n−1 (32) x(s) ans + an−1s + ··· + a0 Numerator m m−1 M(s) = bms + bm−1s + ··· + b0 (33) Denominator - characteristic equation

n n−1 N(s) = ans + an−1s + ··· + a0 (34)

Jakub Mozaryn Automation Systems Determination of static characteristics from transfer function

x0 = lim x(t), y0 = lim y(t), (35) t→∞ t→∞ using the final value theorem

y0 = lim y(t) = lim sy(s) = lim sG(s)x(s) (36) t→∞ s→0 s→0 For the input signal in the form of the unit step 1 x = const ⇒ x(s) = x (37) 0 s 0 y 0 = lim G(s). (38) x0 s→0 Finally, the static characteristcs has a form

b0 y0 = x0 (39) a0

Jakub Mozaryn Automation Systems Methods for determining the transient response of the system

d ny d n−1y d mx d m−1x a +a +···+a y = b +b +···+b x (40) n dtn n−1 dtn−1 0 m dtm m−1 dtm−1 0 Classic: Assumption of the initial conditions x(0), y(0). Solution of differential equations. Using transfer function:

f (t) = L−1[y(s)] = L−1[G(s)x(s)] (41)

To perform Laplace transform and its reverse, which are the basic oper- ations of a transfer function calculus, it is often sufficient to know basic properties of transfer fuctions and tables of transfer fuctions.

Jakub Mozaryn Automation Systems Typical input signals

Unit step (Heaveside function)

 1(t) for t ≥ 0 1 x(t) = x(s) = 0 for t < 0 s

Step with constant value

 x 1(t) for t ≥ 0 1 x(t) = st x(s) = x 0 for t < 0 st s

Impulse - Dirac delta function

 0 for t 6= 0 x(t) = δ(t) = ∞ for t = 0 x(s) = 1

Ramp

a x(t) = at x(s) = s2

Jakub Mozaryn Automation Systems System properties

Changes of output signal y(t) as a response to a specific change of an input signal x(t)

Figure 7: Example of a transient response of the dynamical system

Jakub Mozaryn Automation Systems Table of transfer functions

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Figure 8: Table of transfer functions

Jakub Mozaryn Automation Systems Automation Systems

Lecture 2 - Mathematical Models of Dynamical Systems

Jakub Mozaryn

Institute of Automatic Control and Robotics, Department of Mechatronics, WUT

Warszawa, 2019

Jakub Mozaryn Automation Systems