Method of harmonic linearization. Harmonic linearization

General linearization method

In most cases, it is possible to linearize non-linear dependencies using the method of small deviations or variations. To consider ᴇᴦο, let's turn to some link in the automatic control system (Fig. 2.2). The input and output quantities are denoted by X1 and X2, and the external perturbation is denoted by F(t).

Let us assume that the link is described by some non-linear differential equation of the form

To compile such an equation, you need to use the appropriate branch of technical sciences (for example, electrical engineering, mechanics, hydraulics, etc.) that studies this particular type of device.

The basis for linearization is the assumption that the deviations of all variables included in the link dynamics equation are sufficiently small, since it is precisely on a sufficiently small section that the curvilinear characteristic can be replaced by a straight line segment. The deviations of the variables are measured in this case from their values in the steady process or in a certain equilibrium state of the system. Let, for example, the steady process is characterized by a constant value of the variable X1, which we denote as X10. In the process of regulation (Fig. 2.3), the variable X1 will have the values where denotes the deviation of the variable X 1 from the steady value X10.

Similar relationships are introduced for other variables. For the case under consideration, we have ˸ and also .

All deviations are assumed to be sufficiently small. This mathematical assumption does not contradict the physical meaning of the problem, since the very idea of automatic control requires that all deviations of the controlled variable during the control process be sufficiently small.

The steady state of the link is determined by the values X10, X20 and F0. Then equation (2.1) should be written for the steady state in the form

Let us expand the left side of equation (2.1) in the Taylor series

where D are higher order terms. Index 0 for partial derivatives means that after taking the derivative, the steady value of all variables must be substituted into its expression.

The higher order terms in formula (2.3) include higher partial derivatives multiplied by squares, cubes and higher degrees of deviations, as well as products of deviations. They will be small of a higher order compared to the deviations themselves, which are small of the first order.

Equation (2.3) is a link dynamics equation, just like (2.1), but written in a different form. Let us discard the higher-order smalls in this equation, after which we subtract the steady-state equations (2.2) from Eq. (2.3). As a result, we obtain the following approximate equation of the link dynamics in small deviations˸

In this equation, all variables and their derivatives enter linearly, that is, to the first degree. All partial derivatives are some constant coefficients in the event that a system with constant parameters is being investigated. If the system has variable parameters, then equation (2.4) will have variable coefficients. Let us consider only the case of constant coefficients.

General linearization method - concept and types. Classification and features of the category "General linearization method" 2015, 2017-2018.

In most cases, it is possible to linearize non-linear dependencies using the method of small deviations or variations. To consider it, let's turn to a certain link in the automatic control system (Fig. 2.2). The input and output quantities are denoted by X1 and X2, and the external perturbation is denoted by F(t).

Let us assume that the link is described by some non-linear differential equation of the form

Similar relationships are introduced for other variables. For the case under consideration, we have: and also .

The steady state of the link is determined by the values X10, X20 and F0. Then equation (2.1) can be written for the steady state in the form

Let us expand the left side of equation (2.1) in the Taylor series

where D are higher order terms. Index 0 for partial derivatives means that after taking the derivative, the steady value of all variables must be substituted into its expression.

Equation (2.3) is a link dynamics equation, just like (2.1), but written in a different form. Let us discard the higher order smalls in this equation, after which we subtract the steady state equations (2.2) from Eq. (2.3). As a result, we obtain the following approximate link dynamics equation in small deviations:

Obtaining equation (2.4) is the goal of the linearization done. In the theory of automatic control, it is customary to write the equations of all links so that the output value is on the left side of the equation, and all other terms are transferred to the right side. In this case, all terms of the equation are divided by the coefficient at the output value. As a result, equation (2.4) takes the form

where the following notation is introduced

In addition, for convenience, it is customary to write all differential equations in operator form with the notation

Etc. (2.7)

Then the differential equation (2.5) can be written in the form

This record will be called the standard form of the link dynamics equation.

The coefficients T1 and T2 have the dimension of time - seconds. This follows from the fact that all terms in equation (2.8) must have the same dimension, and for example, the dimension (or px2) differs from the dimension of x2 by a second to the minus first power (). Therefore, the coefficients T1 and T2 are called time constants .

The coefficient k1 has the dimension of the output value divided by the dimension of the input. It is called transmission ratio link. For links whose output and input values have the same dimension, the following terms are also used: gain - for a link that is an amplifier or has an amplifier in its composition; gear ratio - for gearboxes, voltage dividers, scaling devices, etc.

The transfer coefficient characterizes the static properties of the link, since in the steady state . Therefore, it determines the steepness of the static characteristic at small deviations. If we depict the entire real static characteristic of the link, then linearization gives or . The transmission coefficient k1 will be the tangent of the slope of the tangent at that point C (see Fig. 2.3), from which small deviations x1 and x2 are measured.

It can be seen from the figure that the above linearization of the equation is valid for control processes that capture such a section of the AB characteristic, on which the tangent differs little from the curve itself.

In addition, another, graphical method of linearization follows from this. If the static characteristic and the point C, which determines the steady state around which the regulation process takes place, are known, then the transfer coefficient in the link equation is determined graphically from the drawing according to the dependence k1 = tg, taking into account the scale of the drawing and the dimension x2. In many cases graphical linearization method turns out to be more convenient and leads to the goal faster.

The dimension of the coefficient k2 is equal to the dimension of the gain k1 multiplied by time. Therefore, equation (2.8) is often written in the form

where is the time constant.

The time constants T1, T2 and T3 determine the dynamic properties of the link. This issue will be considered in detail below.

The coefficient k3 is the external disturbance gain.

As an example of linearization, consider an electric motor controlled from the side of the excitation circuit (Fig. 2.4).

To find a differential equation that relates the speed increment to the voltage increment on the excitation winding, we write down the law of equilibrium of electromotive forces (emf) in the excitation circuit, the law of equilibrium of emf in the armature circuit and the law of equilibrium of moments on the motor shaft:

In the second equation, for simplicity, the term corresponding to the self-induction emf in the armature circuit is omitted.

In these formulas, RВ and RЯ are the resistances of the excitation circuit and the armature circuit; ІВ and ІЯ - currents in these circuits; UВ and UЯ are the voltages applied to these circuits; wВ is the number of turns of the excitation winding; Ф – magnetic flux; Ω is the angular speed of rotation of the motor shaft; M is the moment of resistance from external forces; J is the reduced moment of inertia of the engine; CE and
SM - coefficients of proportionality.

Let us assume that before the appearance of an increment in the voltage applied to the excitation winding, there was a steady state, for which equations (2.10) will be written as follows:

If now the excitation voltage will receive an increment UВ = UВ0 + ΔUВ, then all variables that determine the state of the system will also receive increments. As a result, we will have: ІВ = ІВ0 + ΔІВ; Ф = Ф0 + ΔФ; IЯ = IЯ0 + ΔІЯ; Ω = Ω0 + ΔΩ.

We substitute these values into (2.10), discard the higher-order small ones and get:

Subtracting equations (2.11) from equations (2.12), we obtain a system of equations for deviations:

In these equations, a proportionality coefficient is introduced between the flux increment and the excitation current increment, determined from the motor magnetization curve (Fig. 2.5).

The joint solution of system (2.13) gives

where is the transfer coefficient, ,

electromagnetic time constant of the excitation circuit, s,

where LB = a wB is the dynamic coefficient of self-induction of the excitation circuit; electromagnetic time constant of the engine, s,

From expressions (2.15) - (2.17) it can be seen that the system under consideration is essentially non-linear, since the transfer coefficient and time "constant" are, in fact, not constant. They can be considered constant only approximately for a certain mode, provided that the deviations of all variables from the steady-state values are small.

Of interest is a special case when in the steady state UB0 = 0; IB0 = 0; Ф0 = 0 and Ω0 = 0. Then formula (2.14) takes the form

In this case, the static characteristic will link the motor acceleration increment and the voltage increment in the excitation circuit.

test questions

1. Describe linear and non-linear ACS.

2. Give the concept of linearization and explain its necessity.

3. State the general method of linearization.

4. What is the standard form for writing differential equations?

Rice. 2.2. ATS link

In most cases, it is possible to linearize non-linear dependencies using the method of small deviations or variations. To consider it, let's turn to a certain link in the automatic control system (Fig. 2.2). The input and output quantities are denoted by X 1 and X 2 , and the external perturbation is denoted by F(t).

Let us assume that the link is described by some non-linear differential equation of the form

The basis for linearization is the assumption that the deviations of all variables included in the link dynamics equation are sufficiently small, since it is precisely on a sufficiently small section that the curvilinear characteristic can be replaced by a straight line segment. The deviations of the variables are measured in this case from their values in the steady process or in a certain equilibrium state of the system. Let, for example, a steady process is characterized by a constant value of the variable X 1 , which we denote as X 10 . In the process of regulation (Fig. 2.3), the variable X 1 will have the values where
denotes the deviation of the variable X 1 from the steady value of X 10 .

BUT

Rice. 2.3. Link regulation process

tax ratios are introduced for other variables. For the case under consideration, we have: and

Next, you can write:
;
and
, because
and

The steady state of the link is determined by the values of X 10 , X 20 and F 0 . Then equation (2.1) can be written for the steady state in the form

Let us expand the left side of equation (2.1) in the Taylor series

where  are higher order terms. The index 0 for partial derivatives means that after taking the derivative, the steady value of all variables must be substituted into its expression
.

where the following notation is introduced

. (2.6)

In addition, for convenience, it is customary to write all differential equations in operator form with the notation

Then the differential equation (2.5) can be written in the form

This record will be called the standard form of the link dynamics equation.

The coefficients T 1 and T 2 have the dimension of time - seconds. This follows from the fact that all the terms in equation (2.8) must have the same dimension, and for example, the dimension (or px 2) differs from the dimension of x 2 per second to the minus first power (
). Therefore, the coefficients T 1 and T 2 are called time constants .

The coefficient k 1 has the dimension of the output value divided by the dimension of the input. It is called transmission ratio link. For links whose output and input values have the same dimension, the following terms are also used: gain - for a link that is an amplifier or has an amplifier in its composition; gear ratio - for gearboxes, voltage dividers, scaling devices, etc.

The transfer coefficient characterizes the static properties of the link, since in the steady state
. Therefore, it determines the steepness of the static characteristic at small deviations. If we depict the entire real static characteristic of the link
, then the linearization gives
or
. The transmission coefficient k 1 will be the tangent of the slope tangent at that point C (see Fig. 2.3), from which small deviations x 1 and x 2 are measured.

In addition, another, graphical method of linearization follows from this. If the static characteristic and point C are known, which determines the steady state around which the regulation process takes place, then the transfer coefficient in the link equation is determined graphically from the drawing according to the dependence k 1 = tg taking into account the scale of the drawing and dimensions x 2. In many cases graphical linearization method turns out to be more convenient and leads to the goal faster.

The dimension of the coefficient k 2 is equal to the dimension of the gain k 1 times the time. Therefore, equation (2.8) is often written in the form

where
is the time constant.

Rice. 2.4. Independent excitation motor

time constants T 1 , T 2 and T 3 determine the dynamic properties of the link. This issue will be considered in detail below.

The factor k 3 is the gain for external perturbation.

As an example of linearization, consider an electric motor controlled from the side of the excitation circuit (Fig. 2.4).

;

In the second equation, for simplicity, the term corresponding to the self-induction emf in the armature circuit is omitted.

In these formulas, R B and R I are the resistances of the excitation circuit and the armature circuit; І В and І Я - currents in these circuits; U V and U I are the voltages applied to these circuits;  V is the number of turns of the excitation winding; Ф – magnetic flux; Ω is the angular speed of rotation of the motor shaft; M is the moment of resistance from external forces; J is the reduced moment of inertia of the engine; C E and C M - coefficients of proportionality.

Let us assume that before the appearance of an increment in the voltage applied to the excitation winding, there was a steady state, for which equations (2.10) will be written as follows:

(2.11)

If now the excitation voltage will receive an increment U B = U B0 + ΔU B, then all variables that determine the state of the system will also receive increments. As a result, we will have: І В = І В0 + ΔІ В; Ф = Ф 0 + ΔФ; I I \u003d I I0 + ΔІ I; Ω = Ω0 + ΔΩ.

We substitute these values into (2.10), discard the higher-order small ones and get:

(2.12)

Subtracting equations (2.11) from equations (2.12), we obtain a system of equations for deviations:

(2.13)

Rice. 2.5. Magnetization curve

these equations introduced the coefficient of proportionality between the flux increment and the excitation current increment

determined from the magnetization curve of the electric motor (Fig. 2.5).

The joint solution of system (2.13) gives

where is the transfer coefficient, ,

; (2.15)

electromagnetic time constant of the excitation circuit, s,

(2.16)

where L B = a B is the dynamic coefficient of self-induction of the excitation circuit; electromagnetic time constant of the engine, s,

. (2.17)

It can be seen from expressions (2.15) - (2.17) that the system under consideration is essentially non-linear, since the transfer coefficient and time "constant" are, in fact, not constant. They can be considered constant only approximately for a certain mode, provided that the deviations of all variables from the steady-state values are small.

An interesting is the special case when in the steady state U B0 = 0; I B0 = 0; Ф 0 = 0 and Ω 0 = 0. Then formula (2.14) takes the form

. (2.18)

In this case, the static characteristic will relate the increase in engine acceleration
and voltage increment in the excitation circuit.

I was supposed to post this article last night, as promised, but this was prevented by Soviet vinyl technology, which requires a complete disassembly, regardless of the severity of the breakdown.

I will continue to become TAU secrets. This time the question is about linearization. Very often, two parameters are interconnected in a non-linear relationship. Hyperbolic, parabolic, logarithmic, etc. This is very inconvenient when doing calculations. For example, we have an encoder at the output of which is a series of pulses. The encoder speed is inversely proportional to the pulse period. The overall goal is to get speed feedback. The entire scale from 0 to 100% should be relatively linear in order to subsequently stabilize the speed.
According to cut graphics from Calca, a lot of water and a drop of theory:

In openOffice Calc, let's build our curve from the original dependency:

The dependence of the encoder rotation frequency as a percentage of the pulse repetition period in timer ticks.

As you can see, to find the rotational speed, you have to divide. It's resource intensive. Moreover, we have a hyperbole, but somewhere there may be a logarithm. It's even worse. We need to simplify. It needs to be linearized. What is linearization? There may be two approaches here.

Take, for example, the saturation curve of steel:

If you work in the range of 0-a, then we can assume that this element is linear. The meaning of such a task is to limit yourself in the working range. Somewhere it fits. Somewhere not.

In our case, the correct solution will be another way - we will break our curve into intervals. For example, the saturation curve can be divided into sections 0-a, a-b, b- ... Inside this section, the relationship between the magnetic field strength and magnetization is roughly proportional.

Let's break our schedule into two sections. Like this:

Looks rough, I agree. The best option would be to break the curve into three sections. But in our case, this is enough.
Let's use the segment formula:

From the graph, we determine the coordinates:

And let's calculate our functions:
For the low speed section:

For the high speed section:

Let's see what we got:

Yes, it will do just fine. Just at high speeds, a small error. Now let's see how the relationship between absolute and relative speeds looks like:

Well, in the region of low speeds, everything does not look the best, but by eye we really won’t see anything there, but in the region of high speeds it is relatively linear. Personally, I'm quite happy with this result.
Now all that is needed is to use the following code upon arrival of the next pulse from the encoder:
//I have this code called by the timer responsible for the PWM of the drive. tic++; if (Encoder.Impulse)( if (tic>130)//rpm is greater than 22% speed=-0.016*tic+24; else //rpm is less than 22% speed=-0.76*tic+121; tic= 0; ) else(//at zero speed, the pulse repetition period is equal to infinity if (tic>2000)(//therefore, if we exceeded some conceivable value speed=0;//then we consider that the encoder is stationary. tic-=1000;// ticks cannot be equated to zero in this case - if an impulse comes with the next tick, then the drive will calculate a huge speed. ) )

The method described here does not claim to be unique and repeatable. The main point of this article is a recommendation to model and calculate such things.
In the next times, we will consider digital implementations of typical links and gradually create a library of components.

Let's discuss again the choice of scale for representing these data in a graphical form (see Fig. 30). The maximum mark of °C, corresponding to the temperature axis X, fits very well on 40 cells, which corresponds to a very convenient division of 10 cells for every 50°C. How much more risk is needed? In this case, I propose to arrange them through 2 cells, which will make it easier to determine the coordinate, since the interval between such risks will correspond to 10 ° C, which is very convenient.

But on the Y-axis, I placed the risks through 5 cells for every 500 ohms of resistance, which led to incomplete use of the paper area. But, judge for yourself, if the axis is divided into 6 or 7 cells, it would be inconvenient to find the coordinate, and if it is 8 cells, then the maximum risk corresponding to 2000 Ohm would not fit on the axis.

Now we need to discuss the form of the theoretical curve. Let's open the guidelines for performing laboratory work on page 28 and find formula 3 that describes the dependence of semiconductor resistance on temperature,

where is the band gap, is the Boltzmann constant, is some constant with the dimension of resistance, and, finally, is the temperature expressed in Kelvin. Let's start creating a new table. First, let's convert the temperature to Kelvin. Secondly, let's set ourselves the task of not only drawing a new graph, but also finding the band gap using the graph. To do this, we take the logarithm of the exponential dependence and get

Denote , , and . Then we get a linear dependence,

which we will depict on the graph. The data corresponding to the values and will be written in Table 9.

Table 9. Recalculation of data in table 8.

point number
T, K
1/T, 10–3 K–1	3,34	3,19	3,00	2,83	2,68	2,54	2,42	2,31	2,21	2,11
ln R, Ohm	7,62	7,51	7,25	7,06	6,99	6,74	6,61	6,56	6,36	6,34

If, according to Table 9, to build a dependence graph in Fig. 31, then all experimental points will take up very little space on the sheet with a large empty space. Why did it happen? Because the labels on the X and Y axes are placed starting from 0, although the values, for example, start only with the value . Is it necessary to make the initial label equal to 0? The answer to this question depends on the tasks at hand. In the example with the Oberbeck pendulum (see Fig. 28) it was very important to find the intersection of the X-axis of the theoretical line at the point with coordinate Y=0, which corresponded to the value . And in this problem, it is only necessary to find the band gap, which is related to the constant , corresponding to the slope of the straight line in Fig. 31, so it is not at all necessary to place labels on the axes, starting from 0.

Studying the data from Table 9 and choosing a convenient scale, we can say with confidence that the orientation of the graph paper needs to be changed, as shown in Fig. 32. Study the selected scale yourself and make sure that it is very convenient for working with the chart. On the theoretical line (drawn by eye in the best way between the experimental points), we put two points A and B with coordinates and . The slope coefficient is expressed in terms of the coordinates of these points by the formula

And finally, we calculate the band gap

Using the method of paired points, we calculate the same coefficient and its error, for this we consider pairs of points from Table 9:

1-4, 2-5, 3-6, 4-7, 5-8, 6-9 and 7-10.

Calculate for these pairs of points the slope coefficients of the straight lines that pass through them

Mean

Now let's calculate the band gap and its error .

Thus we have arrived at the answer

Independent work.

I suggest you do independent calculations, plotting and processing graphs in the next virtual laboratory work, code-named "Determine the stiffness of the spring." But let's raise the bar of the Experiment to a higher level: it is necessary not only to get a number, but to compare two methods of measuring the stiffness of a spring - static and dynamic.

Let's briefly review these methods.

static method.

If a load of mass is suspended from a fixed vertical spring, then the spring will stretch according to Hooke's law, where is the length of the stretched spring, and is the length of the unstretched spring (initial length).

Note: Hooke's law speaks of the proportionality of the elastic force of the spring to absolute elongation, i.e. , where is the coefficient of elasticity (or stiffness) of the spring.

In a state of equilibrium, the force of gravity of the load will be balanced by the force of elasticity and we can write . Let's open the brackets and see the dependence of the length of the spring on the mass of the load

If you make a change of variables, then you get the equation of a straight line. No need to do linearization!

So, your task is to process the data from table 10, which were entered there by the young Experimenter (he was tired of throwing bricks from the roof of a nine-story building). For experiments, he stocked up with a set of weights, found a dozen or two different springs and, hanging weights of different masses, measured the length of the stretched spring using a millimeter ruler.

Exercise 1.

1. Select a spring number from table 10.

2. Make your table with two columns. Enter the force of gravity in the first column, where is the mass of the load (in kg), m / s 2. In the second column, transfer the lengths of the selected spring (in meters). Provide cells for averages and .

Table 10


m, g	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm
	11,8	15,4	17,6	19,4	13,2	15,4	19,6	21,4	11,2
	12,3	16,5	18,3	21,5	14,3	16,5	21,3	22,4	11,7
	13,6	17,6	19,3	21,6	14,8	16,5	22,1	22,6	12,7
	14,1	18,2	21,5	22,1	15,6	17,3	21,5	23,7	13,1
	16,6	22,3	22,5	24,9	17,6	19,9	23,9	25,5	15,4
	21,6	25,6	27,4	29,5	21,4	23,8	27,7	29,9	18,3
	22,5	26,4	28,8	31,4	22,6	24,2	28,8	32,1	19,6
	23,3	27,9	29,4	31,7	23,8	25,6	29,5	31,7	22,1
	26,2	32,1	32,0	34,3	25,5	27,9	31,9	33,6	22,2
	27,8	31,4	33,7	35,3	27,6	29,1	33,2	35,3	23,1

Table 10 (continued)


m, g	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm	l, cm
	15,1	17,1	19,3	11,4	15,3	19,0	10,8	15,2	19,1
	15,6	17,7	19,7	11,6	15,6	19,6	11,5	15,3	19,3
	16,7	18,5	21,2	12,0	16,1	20,4	12,3	16,3	20,2
	17,3	19,3	21,4	12,5	16,5	20,7	12,4	16,7	20,4
	19,4	21,1	23,5	14,9	18,9	22,4	14,2	18,0	21,8
	22,3	24,6	26,3	17,4	21,4	25,8	16,5	20,7	24,4
	23,5	25,6	27,0	18,2	22,3	26,1	17,2	21,6	25,7
	24,4	26,1	28,5	19,4	23,3	27,0	18,4	22,0	26,4
	26,4	28,5	31,1	20,3	24,5	28,6	19,3	23,5	27,3
	27,0	29,0	31,4	21,9	25,8	29,9	20,7	24,7	28,5

3. Take a sheet of graph paper, mark the coordinate axes on it. Choose according to the data optimal Scale and plot gravity versus spring length , plotting values along the x-axis and values along the y-axis.

4. Make 7 pairs of points: 1-4, 2-5, 3-6, 4-7, 5-8, 6-9, 7-10. Using the paired point method, calculate the 7 slope factors using the formula

Etc.

5. Find the average value , which corresponds to the average value of the coefficient of elasticity of the spring .

6. Find the standard deviation , confidence interval , (because 7 values were obtained). Present the result as

Additional task (optional)

7. Calculate the initial length of the spring. To do this, obtain an expression for the coefficient from the equilibrium equation and substitute the average values into it

8. Calculate the confidence interval for the coefficient

9. Considering that , calculate the initial length of the spring and the confidence interval for it

Dynamic method

Suspend the weight of the mass to the fixed vertical stiffness spring and push it slightly down. Harmonic oscillations will begin, the period of which is (see , page 76). We express the mass of the load through the period of oscillations

Portal for the student. Self-training

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