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Control System Engineering 
PE-3032 
Prof. CHARLTON S. INAO 
Defence Engineering College, 
Debre Zeit , Ethiopia
Modeling 
of 
Nonlinearities/ 
Linearization
Nonlinearities 
A 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 Systems 
by Taylor Series/Expansion
Example: Linearizing a Function 
Sample 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). 
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. 3.4 
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 2 
cont…….. 
a b
Linearization Sample Problem 3 
Derivative of nonlinear
Graph of z=x2+4xy+6y2 
z=x^2+4xy+6y^2 
Non linear 
z x y 
131 1 -5 
68 2 -4 
27 3 -3 
8 4 -2 
11 5 -1 
36 6 0 
83 7 1 
152 8 2 
243 9 3 
356 10 4 
491 11 5 
648 12 6 
827 13 7 
1028 14 8 
1251 15 9 
1496 16 10 
1763 17 11 
2052 18 12 
2363 19 13 
2696 20 14 
3051 21 15 
3428 22 16 
3827 23 17 
X=9, y=3
z=30x+72y+243 
z x y 
-63 6 0 
39 7 1 
141 8 2 
243 9 3 
345 10 4 
447 11 5 
549 12 6 
651 13 7 
753 14 8 
855 15 9 
957 16 10 
1059 17 11 
1161 18 12 
1263 19 13 
1365 20 14 
1467 21 15 
1569 22 16 
1671 23 17 
1773 24 18 
1875 25 19 
1977 26 20 
2079 27 21 
2181 28 22 
2283 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 
z x y 
636 3 11 
731 3.5 11.5 
832 4 12 
939 4.5 12.5 
1052 5 13 
1171 5.5 13.5 
1296 6 14 
1427 6.5 14.5 
1564 7 15 
1707 7.5 15.5 
1856 8 16 
2011 8.5 16.5 
2172 9 17 
2339 9.5 17.5 
2512 10 18 
2691 10.5 18.5 
2876 11 19 
3067 11.5 19.5 
3264 12 20 
3467 12.5 20.5 
3676 13 21
linear 
z x y 
268 1 9 
360 1.5 9.5 
452 2 10 
544 2.5 10.5 
636 3 11 
728 3.5 11.5 
820 4 12 
912 4.5 12.5 
1004 5 13 
1096 5.5 13.5 
1188 6 14 
1280 6.5 14.5 
1372 7 15 
1464 7.5 15.5 
1556 8 16 
1648 8.5 16.5 
1740 9 17 
1832 9.5 17.5 
1924 10 18 
2016 10.5 18.5 
2108 11 19 
2200 11.5 19.5 
2292 12 20 
2384 12.5 20.5 
2476 13 21
Linearization Sample Problem 5 
Derivative of nonlinear 
differential equation 
Substituting x=2 
Linearized equation
Wk 6 part 2 non linearites and non linearization april 05
Wk 6 part 2 non linearites and non linearization april 05

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Wk 6 part 2 non linearites and non linearization april 05

  • 1. Control System Engineering PE-3032 Prof. CHARLTON S. INAO Defence Engineering College, Debre Zeit , Ethiopia
  • 3. Nonlinearities A 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.
  • 4. 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).
  • 5. 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.
  • 6. FIGURE 7 Some physical nonlinearities increasing decreasing
  • 7. Non Linear Effects (from Aircraft Engine Controls)
  • 8. 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.
  • 9. 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.
  • 10. 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.
  • 12. 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.
  • 13. Linearization of NonLinear Systems by Taylor Series/Expansion
  • 14.
  • 15. Example: Linearizing a Function Sample 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.
  • 16. FIGURE 9 Linearization of 5cos x about x = π/2
  • 17. 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). 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. 3.4 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.
  • 19. Linearization Sample Problem 2 cont…….. a b
  • 20. Linearization Sample Problem 3 Derivative of nonlinear
  • 21. Graph of z=x2+4xy+6y2 z=x^2+4xy+6y^2 Non linear z x y 131 1 -5 68 2 -4 27 3 -3 8 4 -2 11 5 -1 36 6 0 83 7 1 152 8 2 243 9 3 356 10 4 491 11 5 648 12 6 827 13 7 1028 14 8 1251 15 9 1496 16 10 1763 17 11 2052 18 12 2363 19 13 2696 20 14 3051 21 15 3428 22 16 3827 23 17 X=9, y=3
  • 22. z=30x+72y+243 z x y -63 6 0 39 7 1 141 8 2 243 9 3 345 10 4 447 11 5 549 12 6 651 13 7 753 14 8 855 15 9 957 16 10 1059 17 11 1161 18 12 1263 19 13 1365 20 14 1467 21 15 1569 22 16 1671 23 17 1773 24 18 1875 25 19 1977 26 20 2079 27 21 2181 28 22 2283 29 23
  • 26.
  • 27. Y against z Y=11, z=636
  • 28. Graph of z with x value shown(non linear) 3 z x y 636 3 11 731 3.5 11.5 832 4 12 939 4.5 12.5 1052 5 13 1171 5.5 13.5 1296 6 14 1427 6.5 14.5 1564 7 15 1707 7.5 15.5 1856 8 16 2011 8.5 16.5 2172 9 17 2339 9.5 17.5 2512 10 18 2691 10.5 18.5 2876 11 19 3067 11.5 19.5 3264 12 20 3467 12.5 20.5 3676 13 21
  • 29. linear z x y 268 1 9 360 1.5 9.5 452 2 10 544 2.5 10.5 636 3 11 728 3.5 11.5 820 4 12 912 4.5 12.5 1004 5 13 1096 5.5 13.5 1188 6 14 1280 6.5 14.5 1372 7 15 1464 7.5 15.5 1556 8 16 1648 8.5 16.5 1740 9 17 1832 9.5 17.5 1924 10 18 2016 10.5 18.5 2108 11 19 2200 11.5 19.5 2292 12 20 2384 12.5 20.5 2476 13 21
  • 30.
  • 31. Linearization Sample Problem 5 Derivative of nonlinear differential equation Substituting x=2 Linearized equation