Control System Engineering

PE-3032Prof. CHARLTON S. INAODefence Engineering College, Debre Zeit , Ethiopia

Modeling

of

Nonlinearities/ Linearization

NonlinearitiesA linear system possesses two properties: superposition and homogeneity. • The property of superposition means that the output response of

a system to the sum of inputs is the sum of the responses to the individual inputs.

Thus, if an input of r1(t) yields an output of c1(t) and an input of r2(t) yields an output of c2(t), then an input of r1(t) + r2(t) yields an output of c1(t) + c2(t).

• The property of homogeneity describes the response of the system to a multiplication of the input by a scalar.

Specifically, in a linear system, the property of homogeneity is demonstrated if for an input of r1(t) that yields an output of c1(t), an input of Ar1(t) yields an output of Ac1(t); that is, multiplication of an input by a scalar yields a response that is multiplied by the same scalar.

We can visualize linearity as shown in Figure 6. Figure 6(a) is a linear system where the output is always ½ of the input, or f(x) = 0.5X:, regardless of the value of X, Thus each of the two properties of linear systems applies.

For example, an input of 1 yields an output of ½ and an input of 2 yields an output of 1. Using superposition, an input that is the sum of the original inputs, or 3, should yield an output that is the sum of the individual outputs, or 1.5. From Figure 6(a), an input of 3 does indeed yield an output of 1.5.

FIGURE 6 a. Linear system; b. nonlinear system

To test the property of homogeneity, assume an input of 2, which yields an output of 1. Multiplying this input by 2 should yield an output of twice as much, or 2. From Figure 6(a), an input of 4 does indeed yield an output of 2. The reader can verify that the properties of linearity certainly do not apply to the relationship shown in Figure 6(b).

Figure 7 shows some examples of physical nonlinearities. An electronic amplifier is linear over a specific range but exhibits the nonlinearity called saturation at high input voltages. A motor that does not respond at very low input voltages due to frictional forces exhibits a nonlinearity called dead zone. Gears that do not fit tightly exhibit a nonlinearity called backlash: The input moves over a small range without the output responding. The reader should verify that the curves shown in Figure 7 do not fit the definitions of linearity over their entire range.

FIGURE 7 Some physical nonlinearities

increasing

decreasing

Non Linear Effects (from Aircraft Engine Controls)

A designer can often make a linear approximation to a nonlinear system. Linear approximations simplify the analysis and design of a system and are used as long as the results yield a good approximation to reality.

For example, a linear relationship can be established at a point on the nonlinear curve if the range of input values about that point is small and the origin is translated to that point. Electronic amplifiers are an example of physical devices that perform linear amplification with small excursions about a point.

Linearization The electrical and mechanical systems covered

thus far were assumed to be linear. However, if any nonlinear components are present, we must linearize the system before we can find the transfer function. After discussing and defining nonlinearities non-linearities, we show how to obtain linear approximations to nonlinear systems in order to obtain transfer functions or to model a physical system.

The first step is to recognize the nonlinear component and write the nonlinear differential equation. When we linearize a nonlinear differential equation, we linearize it for small-signal inputs about the steady-state solution when the small-signal input is equal to zero. This steady-state solution is called equilibrium and is selected as the second step in the linearization process. For example, when a pendulum is at rest, it is at equilibrium. The angular displacement is described by a nonlinear differential equation, but it can be expressed with a linear differential equation for small excursions about this equilibrium point.

Next we linearize the nonlinear differential equation, and then we take the Laplace transform of the linearized differential equation, assuming zero initial conditions. Finally, we separate input and output variables and form the transfer function. Let us first see how to linearize a function; later, we will apply the method to the linearization of a differential equation.

Linearization Rules

If we assume a nonlinear system operating at point A, [x0, f(x0)] in Figure 8, small changes in the input can be related to changes in the output about the point by way of the slope of the curve at the point A. Thus, if the slope of the curve at point A is ma, then small excursions of the input about point A, δx, yield small changes in the output, δf(x), related by the slope at point A. Thus,

FIGURE 8 Linearization about points

Eq. 3.0

Eq. 3.1

Eq. 3.2

This relationship is shown graphically in Figure 8, where a new set of axes, δx and δf(x), is created at the point A, and f(x) is approximately equal to f(x0), the ordinate of the new origin, plus small excursions, maδx, away from point A.

Linearization of NonLinear Systemsby Taylor Series/Expansion

Example: Linearizing a FunctionSample 1: Linearize f(x) = 5cosx about x=π/2.

SOLUTION: • We first find that the derivative of f(x) is df/dx=(-

5sinx). At x=π/2, the derivative is -5. Also f(x0) = f(π /2) = 5cos(π/2) = 0. Thus, from Eq. (3.2), the system can be represented as f(x) = -5δx for small excursions of x about π/2. The process is shown graphically in Figure 9, where the cosine curve does indeed look like a straight line of slope -5 near π/2.

FIGURE 9 Linearization of 5cos x about x = π/2

The previous discussion can be formalized using the Taylor series expansion, which expresses the value of a function in terms of the value of that function at a particular point, the excursion away from that point, and derivatives evaluated at that point. The Taylor series is shown in Eq. (3.3).

For small excursions of x from xo, we can neglect higher-order terms. The resulting approximation yields a straight-line relationship between the change in f(x) and the excursions away from XQ. Neglecting the higher-order terms in Eq. (3.3), we get

Eq.3.3

Eq. 3.4

or

Eq. 35.

which is a linear relationship between δf(x) and δx for small excursions away from x0.It is interesting to note that Eqs. (3.4) and (3.5) are identical to Eqs. (3.0) and (3.1), which we derived intuitively.

Linearization Sample Problem 2

Linearization Sample Problem 2cont……..

a b

Linearization Sample Problem 3

Derivative of nonlinear

Graph of z=x2+4xy+6y2z=x^2+4xy+6y^2

Non linearz x y

131 1 -568 2 -427 3 -38 4 -2

11 5 -136 6 083 7 1

152 8 2243 9 3356 10 4491 11 5648 12 6827 13 7

1028 14 81251 15 91496 16 101763 17 112052 18 122363 19 132696 20 143051 21 153428 22 163827 23 17

X=9, y=3

z=30x+72y+243

z x y-63 6 039 7 1

141 8 2243 9 3345 10 4447 11 5549 12 6651 13 7753 14 8855 15 9957 16 10

1059 17 111161 18 121263 19 131365 20 141467 21 151569 22 161671 23 171773 24 181875 25 191977 26 202079 27 212181 28 222283 29 23

Linearization Sample Problem 4

Derivative of nonlinear

Graph of z=x2+8xy+3y2

Y against z

Y=11, z=636

Graph of z with x value shown(non linear)

3

636 3 11731 3.5 11.5832 4 12939 4.5 12.5

1052 5 131171 5.5 13.51296 6 141427 6.5 14.51564 7 151707 7.5 15.51856 8 162011 8.5 16.52172 9 172339 9.5 17.52512 10 182691 10.5 18.52876 11 193067 11.5 19.53264 12 203467 12.5 20.53676 13 21

z x y

linearz x y

268 1 9360 1.5 9.5452 2 10544 2.5 10.5636 3 11728 3.5 11.5820 4 12912 4.5 12.5

1004 5 131096 5.5 13.51188 6 141280 6.5 14.51372 7 151464 7.5 15.51556 8 161648 8.5 16.51740 9 171832 9.5 17.51924 10 182016 10.5 18.52108 11 192200 11.5 19.52292 12 202384 12.5 20.52476 13 21

Linearization Sample Problem 5

Derivative of nonlinear differential equation

Substituting x=2

Linearized equation