# Steady-State error – Control Systems

BEFORE: Control System Stability

### Introduction

Errors in a control system can be attributed to many factors. Changes in the reference input will cause unavoidable errors during transient periods and may also cause steady-state errors. Imperfections in the system components, such as static friction, backlash, and amplifier drift, as well as aging or deterioration, will cause errors at steady state. In this section, however, we shall not discuss errors due to imperfections in the system components. Rather, we shall investigate a type of steady-state error that is caused by the incapability of a system to follow particular types of inputs.

Steady-state error is the difference between the input and the output for a prescribed test input as time tends to infinity. Test inputs used for steady-state error analysis and design are summarized in Table 7.1. In order to explain how these test signals are used, let us assume a position control system, where the output position follows the input commanded position.

Step inputs represent constant position and thus are useful in determining the ability of the control system to position itself with respect to a stationary target. An antenna position control is an example of a system that can be tested for accuracy using step inputs.

Ramp inputs represent constant-velocity inputs to a position control system by their linearly increasing amplitude. These waveforms can be used to test a system’s ability to follow a linearly increasing input or, equivalently, to track a constant velocity target. For example, a position control system that tracks a satellite that moves across the sky at a constant angular velocity.

Parabolas inputs, whose second derivatives are constant, represent constant acceleration inputs to position control systems and can be used to represent accelerating targets, such as a missile.

Any physical control system inherently suffers steady-state error in response to certain types of inputs. A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input. (The only way we may be able to eliminate this error is to modify the system structure.) Whether a given system will exhibit steady-state error for a given type of input depends on the type of open-loop transfer function of the system.

### Definition of the error in steady state depending on the configuration of the system.

The steady-state errors of linear control systems depend on the type of the reference signal and the type of system. Before undertaking the error in steady state, it must be clarified what is the meaning of the system error.

The error can be seen as a signal that should quickly be reduced to zero, if this is possible. Consider the system of Figure 7-5:

Where r (t) is the input signal, u (t) is the acting signal, b (t) is the feedback signal and y (t) is the output signal. The error e (t) of the system can be defined as:

We must remember that r (t) and y (t) do not necessarily have the same dimensions. On the other hand, when the system has unit feedback, H (s) = 1, the input r (t) is the reference signal and the error is simply:

That is, the error is the acting signal, u (t). When H (s) is not equal to 1, u (t) may or may not be the error, depending on the form and purpose of H (s). Therefore, the reference signal must be defined when H (s) is not equal to 1.

The error in steady state is defined as:

To establish a systematic study of the error in steady state for linear systems, we will classify the control systems as follows:

1. Unit feedback systems,
2. Non-unit feedback systems.

### Steady-State Error in Unity-feedback control systems

Consider the system shown in Figure 5-49:

The closed-loop transfer function for this can be obtained as:

The transfer function between the error signal e(t) and the input signal r(t) is:

Where the error e(t) is the difference between the input signal and the output signal. The final-value theorem provides a convenient way to find the steady-state performance of a stable system. Since:

This last equation allows us to calculate the steady-state error Ess, given the input R(s) and the transfer function G(s). We then substitute several inputs for R(s) and then draw conclusions about the relationship that exists between the open-loop system G(s)  and the nature of the steady-state error Ess.

• Step Input: Using R(s) =1/S, we obtain:

Where:

Is the gain of the forward transfer function. In order to have zero steady-state error,

To satisfy the last condition, G(s) must have the following form:

And for the limit to be infinite the denominator must be equal to zero a S goes to zero. So n>=1, that is, at least one pole must be at the origin, equal to say that at least one pure integration must be present in the forward path. The steady-state respond for this case of zero steady-state  error is similar to that shown in Figure 7-2a, ouput 1.

If there are no integrations, the n=0, and it yields a finite error. This is the case shown in Figure 7-2a, output 2.

In summary, for a step input to a unity feedback system, the steady-state error will be zero if there is at least one pure integration in the forward path.

• Ramp Input:  Using R(s) =1/Sˆ2, we obtain:

To have zero steady-state error to a ramp input we must have:

To satisfy this G(s) must take the form where n>=2. In other words, there must be at least two integrations in the forward path. An example of a steady-state error for a ramp input is shown in Figure 7.2b, output 1:

If only one integrator exist in the forward path then lim sG(s) is finite rather tan infinite and this lead to a constant error, as shown in Figure 7.2b, output 2. If there is only one integrator in the forward path then lim sG(s) =0, and the steady-state error will be infinite and lead to diverging ramp, as shown in Figure 7.2b, output 3.

• Parabolic Input: Using R(s) =1/Sˆ3, we obtain:

In order to have zero steady-state error for a parabolic input, we must have:

To satisfy this G(s) n must be n>=3. In other words, there must be at least three integrations in the forward path. If there only two integrators in the forward path then lim sˆ2G(s) is finite rather tan infinite and this lead to a constant error. If there one or zero integrators in the forward path then e() is infinite.

### Classification of Control Systems (System Types) and Static Errors Constant.

System Type.  Control system may be classified according to their ability to follow step inputs, ramp inputs or parabolic inputs and so on. This is a reasonable classification scheme because most of the actual inputs can be considered a combination of such inputs. Consider the unity-feedback control system with the following open-loop transfer function G(s):

It involves the term SˆN in the denominator, representing a pole of multiplicity N at the origin. A system is called type 0, type 1, type 2,…if N=0, 1, 2…respectively. As the type increases, accuracy is improved. However, this agraves the stability problem. If  G(s) is written so that each term in the numerator and denominator, except the term SˆN, approaches unity as s approaches zero, then the open-loop gain K is directly related to the steady-state error.

Static Error Constant. The Static Error Constants defined in the following are figures of merit of control systems. The higher the constants, the smaller the steady-state error.

• Static Position Error Constant Kp. The steady-state error of a system for a unit-step input is:

The Static Position Error Constant Kp is defined by:

Thus the steady-state error in terms of the Static Position Error Constant Kp is given by:

For a type 0 system:

For a type 1 or higher system:

• Static Velocity Error Constant Kv. The steady-state error of a system for a unit-ramp input is given by:

The Static Velocity Error Constant Kv is defined by:

Thus the steady-state error in terms of the Static Velocity Error Constant Kv is given by:

For a type 0 system:

For a type 1 system:

For For a type 2 system or higher:

• Static Acceleration Error Constant Ka. The steady-state error of a system for a unit-parabolic input is given by:

The Static Acceleration Error Constant Kv is defined by:

Thus the steady-state error in terms of the Static Acceleration Error Constant Ka is given by:

For a type 0 system:

For a type 1 system:

For a type 2 system:

For a type 3 system or higher:

Table 7.2 ties together the concepts of steady-state error, static error constants and system type. The table shows the static error constants and the steady-state error as a functions of the input waveform and the system type.

### Steady-State Error for Non-unity Feedback Systems.

Control systems often do not have unity feedback because of the compensation used to improve performance or because of the physical model of the system. In these cases the most practical way to analyze the steady-state error is to take the system and form a unity feedback system by adding and subtracting unity feedback paths as shown in Figure 7.15:

Donde G(s)=G1(s)G2(s) y H(s)=H1(s)/G1(s). Notice that these steps require that input and output signals have the same units.

BEFORE: Control System Stability

Sources:

1. Control Systems Engineering, Nise pp 340, 353
2. Sistemas de Control Automatico Benjamin C Kuo pp 390, 395
3. Modern_Control_Engineering, Ogata 4t pp 301,305

Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contact: Caracas, Quito, Guayaquil, Cuenca – Telf. 00593998524011

WhatsApp: +593984950376

 Attention:If what you need is to solve urgently a problem of a “Mass-Spring-Damper System ” (find the output X (t), graphs in Matlab of the 2nd Order system and relevant parameters, etc.), or a “System of Electromechanical Control “… to deliver to your teacher in two or three days, or with greater urgency … or simply need an advisor to solve the problem and study for the next exam … send me the problem…I Will Write The Solution To any Control System Problem… , …I will give you the answer in digital and I give you a video-conference to explain the solution … it also includes simulation in Matlab. In the link above, you will find the description of the service and its cost.

Related:

The Block Diagram – Control Engineering

Block Diagram of Electromechanical Systems – DC Motor

Transient-response Specifications

Control System Stability

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# Control System Stability

### Introduction

The most important problem in linear control systems concerns stability. It is the most important system specification among the three requirements enter into the design of a control system: transient response, stability and steady-state error. That is, under what conditions will a system become unstable? If it is unstable, how should we stabilize the system?

The total response of a system is the sum of the forced and natural responses:

Using this concept, we present the following definition of stability, instability and marginal stability:

• A linear time-invariant system is stable if the natural response approaches to zero as time approaches infinity.
• A linear time-invariant system is unstable if the natural responses grows without bound as time approaches infinity.
• A linear time-invariant system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates as time approaches infinity.

Thus the definition of stability implies that only the forced response remains as the natural response approaches zero. An alternate definition of stability is regards the total response and implies the first definition based upon the natural response:

• A system is stable if every bounded input yields a bounded output. We call this statement a bounded-input, bounded-output (BIBO) definition of stability.

We now realize that if the input is bounded but the output is unbounded, the system is unstable. If the input is unbounded we will see an unbounded total response and we cannot draw any about conclusion about stability.

Physically, an unstable system whose natural response grows without bound can cause damage to the system, to adjacent property, or to human life. Many times systems are designed with limited stops to prevent total runaway.

Negative feedback tends to improve stability. From the study of the system poles we must recall that poles in the left half-plane (lhp) yield either pure exponential decay or damped sinusoidal natural responses. These natural responses decay to zero when time approaches infinity. Thus, if the closed-loop system poles are in the left half of the plane and hence have a negative real part, the system is stable. That is:

• Stable systems have closed-loop transfer functions with poles only in the left half-plane.

Poles in the right half-plane (rhp) yield either pure exponentially increasing or exponentially increasing sinusoidal natural responses, which approche infinity when time approaches infinity. Thus:

• Unstable systems have closed-loop transfer functions with at least one pole in the right half-plane and /or poles of multiplicity greater than 1 on the imaginary axis.

Finally, the system that has imaginary axis poles of multiplicity 1 yields pure sinusoidal oscillations as a natural response. These responses neither increase nor decrease in amplitude. Thus,

• Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity one and poles in the left half-plane.

Figure 6.1a shows a unit step response of a stable system, while Figure 6.1b shows an unstable system.

[1]

### Routh’s Stability Criterion

Routh’s Stability criterion tell us whether or not there are unstable roots in a polynomial equation without actually solving for them. This stability criterion applies to polynomials with only a finite number of terms. When the criterion is applied to a control system, information about absolute stability can be obtained directly from the coefficients of the characteristic equation.

The method requires two steps: 1) Generate a data table called a Routh Table and 2) Interpret the to tell how many closed-loop system poles are in the left half-plane, the right half-plane, and on the jw-axis. The power of the method lies in design rather than analysis. For example, if you have an unknown parameter in the denominator of a transfer function, it is difficult to determine via a calculator the range of this parameter to yield stability. We shall see that The Routh-Hurwitz criterion can yield a closed-form expression for the range of the unknown parameter.

• Generating a basic Routh Table. Considering the equivalent closed-loop transfer function in Figure 6.3. Since we are interested in the system poles, we focus our attention on the denominator. We first create the Routh Table shown in Table 6.1:

Begin by labeling the rows with powers of s from the highest power of the denominator of the closed-loop transfer function to s^0. Next start with the coefficient of the highest power of s in the denominator and list, horizontally in the first row every other coefficient.

In the second row, list horizontally, starting with the next higher power of s, every coefficient that was skipped in the first row. The remain entries are filled as follows:

Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row. The left-hand column of the determinant is always the first column of the previous two rows, and the right-hand column is the elements of the column above and to the right.

Figures 6.4 shows an example of building the Routh Table :

The complete array of coefficients is triangular. Note that in developing the array an entire row maybe divided or multiplied by a positive number in order to simplify the subsequent numerical calculation without altering the stability conclusion.

Consider the following characteristic equation, example 5-13:

The first two rows can be obtained directly from the given polynomial. The second is divided by two but we arrive to the same conclusion:

• Interpreting the basic Routh Table. Simply stated, the Routh-Hurwitz criterion declares that the number of roots of the polynomial that are in the right-half plane is equal to the number of sign changes in the first column.

If the closed-loop transfer function has all poles in the left-hand plane the system is stable. Thus, the system is stable if there are no sign changes in the first column of the Routh Table.

The last case, Example 5-13, is one of an unstable system. In that example the number of sign changes in the first column is equal to two. This means that there are two roots with positive real parts. Table 6.3 is also an unstable system. There, the first change occurs from 1 in the s^2 row to -72 in the s^1 row. The second occurs from -72 in the  s^1 row to 103 in the  s^0 row. Thus, the system has two poles in the right-half plane.

Routh’s stability criterion is of limited usefulness when applying to control system analysis because it does not suggest how to improve relative stability or how to stabilize a unstable system. It is possible, however, to determine the effects of changing one or two parameters of the system by examining the values that causes instability. In the following we consider the problem of determining the stability range of a parameter value. Consider the system of the Figure 5-38. Let us determine the range of K for stability.

The characteristic equation is:

And the Routh Table:

For stability, K must be positive and all coefficient in the first column must be positive. Therefore:

When K=14/9 the system becomes oscillatory and, mathematically, the oscillation is sustained at constant amplitude.

### Routh-Hurwitz Criterion Special Cases

Two special cases can occur: (1) The Routh table sometimes will have a zero only in the first column of a row, (2) The Routh table sometimes will have an entire row that consists of zeros

• Zero only in the first column. If the first elemento of a row is zero, division by zero will be required to form the next row. To avoid this phenomenon, an epsilon ε is assigned to replace the zero in the first column. The value is then allowed to approach zero from either the positive or the negative side, after which the signs of the entries in the first column can be determined. To see the application of this, let us look the follow example: determine the stability of the closed-loop transfer function T(s):

The solution is shown in table 6.4:

We must begin by assembling the Routh table down to the row where a zero appears only in the first column (the s^3 row). Next, replace the zero by a small number ε complete the table. To begin the interpretation we must first assume a sign, positive or negative for the quantity  ε. Table 6.5  shows the first column of table 6.4 along with the resulting signs for choices of  ε positive and ε negative.

If is chosen ε positive Table 6.5 shows a sign change from the s^3 row to the s^2 row, and there will be another sign change from the s^2 row to the s^1 row. Hence the system is unstable and has two poles in the right half-plane. Alternatively, we could chose ε negative.  Table 6.5 then shows a sign change from the s^4 row to the s^3 row. Another sign change would occur from the s^3 row to the s^2 row. Our result would be exactly the same as that for a positive choice for ε. Thus, the system is unstable.

• Entire Row is zero. We now look at the second special case. Sometimes while making a Routh table, we can find that an entire row consists of zeros because there is a even polynomial that is a factor of the original polynomial. This case must be handled differently from the previous case. Next example shows how to construct and interpret the Routh table wne an entire row of zeros is present.

Determine the number of right half-plane poles in the closed-loop transfer function T(s):

Start by forming the Routh table for the denominator. We get Table 6.7:

At the second we multiply by 1/7 for convenience. We stop at the third row since the entire row consist of zeros and use the following procedure. First we return to the row immediately above the row of zeros and form an auxiliary polynomial using the entries in that row as coefficients. The polynomial will start with the power of s in the label column and continue by skipping every other power of s. Thus, the polynomial formed for this example is:

Next we differentiate the polynomial with respect to s and we obtain:

Finally we use the coefficients of this last equation to replace the row of zeros. Again, for convenience the third row is multiplied by ¼ after replacing the zeros. The remainder of the table is formed in a straightforward manner by following the standard form shown in Table 6.2

We get Table 6.7. It shows that all entries in the first column are positive. Hence, there are no right half-plane poles and the system is stable.

Source:

Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contact: Caracas, Quito, Guayaquil, Cuenca – Telf. 00593998524011

WhatsApp: +593984950376

 Attention: If what you need is to solve urgently a problem of a “Mass-Spring-Damper System ” (find the output X (t), graphs in Matlab of the 2nd Order system and relevant parameters, etc.), or a “System of Electromechanical Control “… to deliver to your teacher in two or three days, or with greater urgency … or simply need an advisor to solve the problem and study for the next exam … send me the problem…I Will Write The Solution To any Control System Problem… , …I will give you the answer in digital and I give you a video-conference to explain the solution … it also includes simulation in Matlab. In the link above, you will find the description of the service and its cost.

Related:

The Block Diagram – Control Engineering

Block Diagram of Electromechanical Systems – DC Motor

Transient-response Specifications

# Transient-response Specifications

### Introduction

The response in time of a control system is usually divided into two parts: the transient response and the steady-state response. Let y (t) be the response of a system in continuous time, then:

where yt (t) is the transient response, while yss (t) is the steady state response.

The transient response of a control system is important since both its amplitude and its duration must be kept within tolerable or prescribed limits. It is defined as the part of the response in time that tends to zero when the time becomes very large. Thus,

All real stable control systems present a transient phenomenon before reaching the steady state response. For analysis and design purposes it is necessary to assume some basic types of test inputs to evaluate the performance of a system. The proper selection of these test signals allows the prediction of system performance with other more complex inputs. The following signals are used: Step function, which represents an instantaneous change in the reference input; Ramp function, which represents a linear change over time; Parabolic function, which represents a faster order than the ramp. These signals have the common characteristic that they are simple to write in mathematical form, it is rarely necessary or feasible to use faster functions. In Figure 7-1 you can see these functions:

[2]

For a linear control system, the analysis and characterization of the transient response is performed frequently using the unit step function Us (t), shown in Figure 7-1a with R = 1. A typical response of a control system to a unit step input is shown in Figure 7-11:

[2]

The transient response of a practical control system often exhibits damped oscillations before reaching the steady state. That’s happens because systems have energy storage and cannot responds immediately. The transient-response to a unit step input depends on the initial conditions. That’s why it is a common practice to use the standard initial conditions that the system is at rest initially with the output an all time derivatives thereof zero.

## Second-order systems and Transient-response specifications.

Figure 5-5a shows a Servo System as an example of a second-order system. It consists of a proportional controller and load elements (inertia and viscous friction elements):

[3]

The closed-loop Transfer Function of the system shown in Figure 5-5c is:

In the transient-response analysis it is convenient to write:

Where σ is called the attenuation; ωn is the undamped natural frequency; and ζ  the damping ratio of the system.  ζ  is the ratio of the actual damping B to the critical damping Bc equal to two times the square-root of JK:

In terms of ωn y σ, the system shown in Figure 5-5c can be expressed as Figure 5-6:

[3]

Now, the Transfer Function C(s)/R(s) can be written as:

This form is called The Standard Form. The dynamic behavior of a second-order system can be now described in terms of the two parameters ωn and σ. In short, the cases of second-order response as a function of σ are summarized in Figure 4.11 (for a better review see FIRST and SECOND ORDER SYSTEMS):

[1]

In specifying the transient-response characteristics of a control system to a unit-step input, it is common to specify the following parameters associated with the underdamped response:

1. Delay time, Td
2. Rise time, Tr
3. Peak time, Tp
4. Percent overshoot (%OS) or Maximum overshoot (Mp)
5. Settling time, Ts

These specifications are defined as follows:

Delay time (Td): it is the time required for the response to reach half the final value the very first time.

Rise time (Tr): it is the time required for the response to rise from 10% to 90%. In other words, to go from 0.1 of the final value to 0.9 of the final value.

Peak time (Tp): it is the time required for the response to reach the first peak of the overshoot.

Maximum overshoot (Mp): it is the maximum peak value of the response curve measured from unity. It is also the amount that the waveform overshoots the final value, expressed as a percentage of the steady-state value.

Settling time (Ts):  it is the time required for the transient damping oscillations to reach and stay within ±2% or ±5% of the final or steady-state value.

These specifications are graphically shown in Figure 5-8:

[3]

It is important to remark that these specifications don’t necessarily apply to any given case. For example, the terms peak time and maximum overshoot do not apply to overdamped systems.

Except for certain applications where oscillations can’t be tolerated, it is desirable that the transient-response be sufficiently fast and sufficiently damped. Thus, for a desirable transient response of a second-order system, the damping ratio must be between 0.4 and 0.8. Small values of σ (σ<0.4) yields excessive overshoot in the transient response, and systems with a large value of σ (σ>0.8) responds sluggishly. We will also see that the maximum overshoot and the rise time conflict with each other. In other words, they cannot be made smaller simultaneously.

Analytically:
Rise time (Tr):

where ωd is the damped natural frequency:

and ß is defined by the Figure 5-9:

[3]

Peak time (Tp):

Settling time (Ts):

## Transient-response of Higher-order systems.

It could be seen that the transient response of a system higher than a second-order is the sum of the responses of first-order and second order systems.

## Transient-response of a First-order system.

We briefly discuss the transient response of a first-order system. A first-order system without zeros can be described by the transfer function shown in Figure 4.4(a).

[1]

If the input is a unit step, where R(s)=1, the Laplace transform of the step response is C(s), where:

Taking the inverse transform:

Figure 4-5 shows a typical response of this system to a unit step input:

[1]

We call 1/a the time constant of the response. The parameter a is the only one needed to describe the transient response for a first-order system. Thus, the time constant can be considered a transient response specification for a first order system, since it is related to the speed at which the system responds to a step input. Since the pole of the transfer function is at a, we can say the pole is located at the reciprocal of the time constant, and the farther the pole from the imaginary axis, the faster the transient response.

The other specifications for a first-order system are:

Rise time (Tr):

Settling time (Ts):

Source:

1. Control Systems Engineering, Nise pp 177-181
2. Sistemas de Control Automatico Benjamin C Kuo p 385,
3. Modern_Control_Engineering, Ogata 4t pp 224, 232

Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contact: Caracas, Quito, Guayaquil, Cuenca – Telf. 00593998524011

WhatsApp: +593984950376

 Attention: If what you need is to solve urgently a problem of a “Mass-Spring-Damper System ” (find the output X (t), graphs in Matlab of the 2nd Order system and relevant parameters, etc.), or a “System of Electromechanical Control “… to deliver to your teacher in two or three days, or with greater urgency … or simply need an advisor to solve the problem and study for the next exam … send me the problem…I Will Write The Solution To any Control System Problem… , …I will give you the answer in digital and I give you a video-conference to explain the solution … it also includes simulation in Matlab. In the link above, you will find the description of the service and its cost.

Related:

The Block Diagram – Control Engineering

Block Diagram of Electromechanical Systems – DC Motor

Control System Stability

# UNDERDAMPED SECOND-ORDER SYSTEM

Fuentes:

Control Systems Engineering, Norman Nise

1. Introduction Chapter 4 pp 162 (162)
2. Poles and Zeros 4.1 pp 162 –
3. First Order System 4.3 pp 165-168
4. Second Order System 4.4 pp 168-177
5. Underdamped Second-Order System 4.6 pp 177-186
1. Modern_Control_Engineering__4t
1. Introduction Chapter 5 pp 219 (232)
2. First Order Systems 221 (234)-224
3. Second Order System pp 224 (237)-234

Literature Review, Martes 14 noviembre 2017, 05:07 am – Caracas, Quito, Guayaquil.

Introduction

Now that we have become familiar with second-order systems and their responses, we generalize the discussion and establish quantitative specifications defined in such a way that the response of a second-order system can be described to a designer without the need for sketching the response. We define two physically meaningful specifications for second-order systems. These quantities can be used to describe the characteristics of the second-order transient response just as time constants describe the first-order system response.

Natural Frequency, Wn

The natural frequency of a second-order system is the frequency of oscillation of the system without damping. For example, the frequency of oscillation of a series RLC circuit with the resistance shorted would be the natural frequency.

Damping Ratio,

We have already seen that a second-order system’s underdamped step response is characterized by damped oscillations. Our definition is derived from the need to quantitatively describe this damped oscillations regardless of the time scale.Thus, a system whose transient response goes through three cycles in a millisecond before reaching the steady state would have the same measure as a system that went through three cycles in a millennium before reaching the steady state. For example, the underdamped curve in Figure 4.10 has an associated measure that defines its shape. This measure remains the same even if we change the time base from seconds to microseconds or to millennia.

A viable definition for this quantity is one that compares the exponential decay frequency of the envelope to the natural frequency. This ratio is constant regardless of the time scale of the response. Also, the reciprocal, which is proportional to the ratio of the natural period to the exponential time constant, remains the same regardless of the time base.

We define the damping ratio, , to be:

Consider the general system:

Without damping, the poles would be on the jw-axis, and the response would be an undamped sinusoid. For the poles to be purely imaginary, a = 0. Hence:

Assuming an underdamped system, the complex poles have a real part, , equal to -a/2. The magnitude of this value is then the exponential decay frequency described in Section 4.4. Hence,

from which

Our general second-order transfer function finally looks like this:

Now that we have defined and Wn, let us relate these quantities to the pole location. Solving for the poles of the transfer function in Eq. (4.22) yields:

From Eq. (4.24) we see that the various cases of second-order response:

Underdamped Second-Order System

Now that we have generalized the second-order transfer function in terms of and Wn, let us analyze the step response of an underdamped second-order system.

Not only will this response be found in terms of and Wn, but more specifications
indigenous to the underdamped case will be defined. The underdamped second order system, a common model for physical problems, displays unique behavior that
must be itemized; a detailed description of the underdamped response is necessary
for both analysis and design. Our first objective is to define transient specifications
associated with underdamped responses. Next we relate these specifications to the
pole location, drawing an association between pole location and the form of the
underdamped second-order response. Finally, we tie the pole location to system
parameters, thus closing the loop: Desired response generates required system
components.

Let us begin by finding the step response for the general second-order system of Eq. (4.22). The transform of the response, C(s), is the transform of the input times the transfer function, or:

where it is assumed that < 1 (the underdamped case). Expanding by partial fractions, using the methods described, yields:

Taking the inverse Laplace transform, which is left as an exercise for the student, produces:

where:

A plot of this response appears in Figure 4.13 for various values of , plotted along a time axis normalized to the natural frequency.

We now see the relationship between the value of and the type of response obtained: The lower the value of , the more oscillatory the response.

The natural frequency is a time-axis scale factor and does not affect the nature of the response other than to scale it in time.

Other parameters associated with the underdamped response are rise time, peak time, percent overshoot, and settling time. These specifications are defined as follows (see also Figure 4.14):

1. Rise time, Tr. The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.
2. Peak time, TP. The time required to reach the first, or maximum, peak.
3. Percent overshoot, %OS. The amount that the waveform overshoots the steady-state, or final value at the peak time, expressed as a percentage of the steady-state value.
4. Settling time, Ts. The time required for the transient’s damped oscillations to reach and stay within 2% of the steady-state value.

All definitions are also valid for systems of order higher than 2, although analytical expressions for these parameters cannot be found unless the response of the higher-order system can be approximated as a second-order system.

Rise time, peak time, and settling time yield information about the speed of the transient response. This information can help a designer determine if the speed and the nature of the response do or do not degrade the performance of the system.

For example, the speed of an entire computer system depends on the time it takes for a hard drive head to reach steady state and read data; passenger comfort depends in part on the suspension system of a car and the number of oscillations it goes through after hitting a bump.

Evaluation of Tp

Tp is found by differentiating c(t) in Eq. (4.28) and finding the first zero crossing after t = 0.

Evaluation of %OS.

From Figure 4.14 the percent overshoot, %OS, is given by:

Evaluation of Ts

In order to find the settling time, we must find the time for which c(t) in Eq. (4.28) reaches and stays within ₎±2% of the steady-state value, C final.

Evaluation of Tr

A precise analytical relationship between rise time and damping ratio cannot be found. However, using a computer and Eq. (4.28), the rise time can be found. Let us look at an example.

We now have expressions that relate peak time, percent overshoot, and settling time to the natural frequency and the damping ratio. Now let us relate these quantities to the location of the poles that generate these characteristics. The pole plot for a general, underdamped second-order system is reproduced in Figure 4.17.

Now, comparing Eqs. (4.34) and (4.42) with the pole location, we evaluate peak time and settling time in terms of the pole location. Thus:

where is the imaginary part of the pole and is called the damped frequency of oscillation, and is the magnitude of the real part of the pole and is the exponential damping frequency part.

At this point, we can understand the significance of Figure 4.18 by examining the actual step response of comparative systems. Depicted in Figure 4.19(a) are the step responses as the poles are moved in a vertical direction, keeping the real part the same. As the poles move in a vertical direction, the frequency increases, but the envelope remains the same since the real part of the pole is not changing.

Let us move the poles to the right or left. Since the imaginary part is now constant, movement of the poles yields the responses of Figure 4.19(b). Here the frequency is constant over the range of variation of the real part. As the poles move to the left, the response damps out more rapidly.

Moving the poles along a constant radial line yields the responses shown in Figure 4.19(c). Here the percent overshoot remains the same. Notice also that the responses look exactly alike, except for their speed. The farther the poles are from the origin, the more rapid the response.

Literature Review by: Larry Francis Obando – Technical Specialist

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

WhatsApp: 00593984950376

# Sistemas Lineales e Invariantes en el tiempo.

### Introducción

Un sistema lineal, en tiempo continuo o discreto, es aquel que posee la importante propiedad de la superposición: si una entrada consiste en la suma ponderada de varias señales, entonces la salida es simplemente la superposición (es decir, la suma ponderada) de las respuestas del sistema a cada una de estas señales.

Sea y1(t) la respuesta del sistema continuo a una entrada x1(t), y sea y2(t) la salida correspondiente a la entrada x2(t). Entonces, el sistema es lineal si:

• La respuesta a x1(t) + x2(t) es y1(t) + y2(t)
• La respuesta a k*x1(t) es k*y1(t), donde k es una constante compleja cualquiera.

Consideremos un sistema S cuya entrada x(t) y salida y(t) están relacionadas mediante:

Ahora consideramos dos entradas arbitrarias x1(t) y x2(t). Ellas generan las siguientes respuestas:

Consideremos una tercera entrada x3(t)=a*x1(t)+b*x2(t), la cual genera una salida y3(t) igual a:

Concluimos entonces que el sistema es lineal.

### El modelo matemático y sus términos

Los sistemas lineales, dinámicos, causales, invariantes en el dominio y deterministas (LDCID) definidos en el dominio del tiempo continuo constituyen parte importante en el estudio de los sistemas eléctricos, debido al hecho de sus innumerables aplicaciones dentro de la ingeniería eléctrica.

En general podría decirse que los sistemas lineales son el resultado de aproximaciones en el modelaje de sistemas. No obstante, aun cuando los sistemas eléctricos forman parte de los llamados sistemas no lineales, su tratamiento como sistemas lineales permiten dar respuestas acertadas a las preguntas que pudiera requerir los profesionales del área.

Los modelos matemáticos de sistemas dinámicos definidos en el dominio continuo presentan términos asociados a operaciones de derivadas de las cantidades externas con respecto a la variable independiente, que por lo general será el tiempo. Estos modelos matemáticos se denominan ecuaciones diferenciales, y sus respectivas respuestas son totalmente definidas por las condiciones de cada sistema representado por el modelo matemático.

Un aspecto importante a estudiar la representación de un sistema a través de su modelo matemático es la identificación de los términos que son expresados en el modelo matemático de un sistema LDCID, el cual es representado por una ecuación diferencial ordinaria ( (ODE) involves derivatives of a function of only one variable) de orden m-ésimo en relación a la señal de excitación x(t), y de orden n-ésimo con respecto a su señal de respuesta y(t), es decir, en general un modelo matemático asociado a un sistema LDCID viene dado por:

donde y(t) representa la señal de respuesta también denominada señal de salida, x(t) representa la señal de excitación o de entrada, y los coeficientes an;a1; …;a0 y bm;b1; … ;b0 representan los parámetros del sistema, que alteran respectivamente la señal de excitación y la señal de respuesta, así como sus derivadas ordinarias, y la variable independiente t, en este caso puede significar el tiempo, con el propósito de contextualizar el dominio en el cual está definido el modelo matemático.

### Modelo matemático de primer orden

Un sistema LDCID en el dominio continuo de primer orden es representado mediante una ecuación diferencial dada por:

Note que el modelo debe ser de primer orden en lo que respecta a los operadores ,

es decir, en mayor orden de derivadas de la señal de respuesta y(t) debe ser n = 1.

Sin embargo, podría ser de cualquier orden con relación a los operadores de la excitación para , debido al hecho de que las operaciones de derivadas sobre la señal de excitación no son consideradas parte del sistema.

Note que las operaciones definidas sobre la señal de excitación no forman parte del sistema, por cuanto las operaciones matemáticas definidas sobre la excitación constituyen el modelo matemático de la señal de excitación.

Por razones de simplificación en la nomenclatura y mediante la propiedad de superposición, se estudiará la solución de la ecuación diferencial:

Note que para obtener la solución del sistema debe conocerse al menos una condición de la respuesta del sistema, la cual usualmente es especificada a través de su condición inicial, y(0).

Luego de manipulaciones algebraicas convenientes (demostración en Fuente 1), se concluye que la solución a la ecuación diferencial representada por la Ecuación (2.12) viene dada por:

donde:

1. Respuesta transitoria:

1. Respuesta permanente

Ejemplo 2.2

La Figura 2.3 muestra un sistema compuesto por una resistencia y un capacitor, y cuyos valores son representados respectivamente por R y C. Además, la figura muestra que el sistema eléctrico es excitado por una señal x(t) = u(t) y su respuesta es medida a través de la tensión sobre el capacitor, donde u(t) representa la función escalón unitario:

El modelo matemático asociado al sistema representado por la Figura 2.3 puede obtenerse empleando elementales ecuación de redes eléctricas:

Entonces, al comparar el modelo matemático definido por la Ecuación (2.12) con el modelo obtenido, se tiene que el coeficiente a0 y la señal de excitación son:

,

Al aplicar la solución expresada por medio de la Ecuación (2.21), se puede afirmar que:

Al operar la Ecuación (2.26) se tiene que la respuesta del sistema es dada por:

Note que:

por cuanto el elemento de memoria representado por el capacitor no permite cambios bruscos y por tal motivo y(0-) = y(0) = y(0+). Además, para buscar una respuesta a la pregunta debe tomarse en cuenta que la excitación tiene un valor de cero y ella ha permanecido en cero desde mucho tiempo atrás, es decir, desde menos infinito, obviamente y(0) = 0.

### Modelo matemático de orden superior.

En este apartado se introducirá el operador p, el cual será empleado para representar el orden de la derivada que está operando en cada término de la ecuación diferencial ordinaria bajo estudio.

la derivada n-ésima con respecto a la variable del dominio continuo. Es decir,

Por otra parte, se debe introducir dos definiciones que conforman la solución completa de una ecuación diferencial ordinaria.

Definición 2.2 (Respuesta transitoria) La respuesta transitoria o, también denominada natural o solución homogénea, es la solución de toda ecuación diferencial ordinaria cuando su señal de excitación viene definida por la función nula.

Definición 2.3 (Respuesta permanente) La respuesta permanente o, también denominada forzada o solución particular, es la solución de la ecuación diferencial ordinaria ante una señal de excitación que actúa sobre el sistema.

Observación 2.1 La respuesta transitoria, natural u homogénea es intrínseca del sistema y no de la excitación, a diferencia de que la respuesta permanente, forzada o particular, que además de depender del sistema, depende de la excitación.

Se conocen condiciones del sistema, bien sean condiciones iniciales a través del valor de la respuesta y(t) para t = 0 y sus primeras n-1 derivadas para t = 0, ó n valores conocidos de la respuesta completa y(t) en n distintos instantes de t, o combinación de lo anterior.

Se tiene que:

donde el coeficiente o también denominado parámetro an = 1 (ODE with leading coefficient equal to 1 is called standard ODE form)

Aplicando las Ecuaciones (2.46), se puede escribir el modelo matemático definido por la Ecuación (2.45) como:

donde D(p) es el ampliamente conocido polinomio característico del sistema.

### Respuesta Transitoria

Existen diversos métodos para determinar la respuesta transitoria de un modelo matemático asociado a un sistema LDCID en el dominio continuo, el cual es representado por una ecuación diferencial ordinaria.

Método 2.1 (Determinación de la Respuesta Transitoria) Dada la ecuación diferencial ordinaria definida por la Ecuación (2.44),

Ejecute:

Paso 1. Asegúrese de que el término an de la ecuación diferencial sea igual a uno. Si no es así, divida toda la ecuación diferencial entre an.

Paso 2. Aplique el operador “p” a la ecuación diferencial.

Paso 3. Determine las n raíces que anulen el polinomio D(p) y denote las raíces reales como ri para cada i = 1; … ;nr, y las raíces complejas conjugadas como

para cada i = nr +1; ..;n, donde

EJEMPLO 2.5 Respuesta transitoria de un sistema de quinto orden

Suponga el modelo matemático de un sistema LDCID en tiempo continuo definido

por:

donde y(t) es la señal de respuesta del sistema, y x(t) representa la señal de excitación. Para el modelo matemático definido mediante la Ecuación (2.48), determine la solución homogénea del sistema aplicando el Método 2.1.

Solución. Debido a que el término a5 no es igual a 1, se debe dividir toda la ecuación diferencial entre a5, para luego aplicar el operador p, obteniéndose:

Al calcular las cinco raíces que anulan D(p), se tiene que sus raíces son: r1 = -2, r22= -3 y

z3 = -1 +- j. Entonces, se puede afirmar que las soluciones asociadas a cada raíz viene dada por:

### Respuesta Permanente

Al despejar y(t) de la Ecuación (2.47):

se tiene que

donde la fracción N(p)/D(p) representa el operador del sistema L(p).

A fin de estudiar el caso más general de las señales de excitaciones más comúnmente presentes en los sistemas eléctricos, se analizará cuando la señal de excitación es considerada una exponencial definida por:

donde en general s es un parámetro o coeficiente complejo, y cuyo valor es

y B es un parámetro constante de la señal de excitación, que pertenece al conjunto de los números reales

Por otra parte, los casos en los cuales pueden ser aplicado el método que será descrito en este punto, corresponden a aquellos en donde D(s) no es igual a 0.

La Ecuación (2.50) permite representar diversas situaciones para la señal de excitación x(t), y cuyos casos son mostrados a continuación mediante la Tabla 2.1

Es importante hacer notar que la operación

ejecuta mediante la operación límite, es decir,

EJEMPLO 2.6 Considere un sistema LDCID con modelo matemático definido por:

Para el sistema representado por la Ecuación (2.52), determine la respuesta permanente del sistema si la señal de excitación es:

Solución. Dado que el coeficiente a3 es igual a uno, se puede aplicar el operador p a la Ecuación (2.52) obteniéndose:

### Respuesta Completa

La respuesta completa del sistema se consigue sumando la respuesta transitoria u homogénea con la respuesta permanente o solución particular, es decir:

donde los coeficientes ci para todo i = 1; …. ;n se obtiene de n condiciones conocidas, en concordancia con el grado de la ecuación característica N(p), es decir, los coeficientes ci

para todo i = 1; …. ;n son determinados por el conocimiento de:

Por ejemplo, el problema ahora es hallar la respuesta completa del sistema, bajo las condiciones:

Solución. Claramente se tiene que el término a3 = 1, hecho que permite aplicar el operador p directamente a la Ecuación (2.52), arrojando el polinomio característico

D(p) = p3 +8p2 +19p+12, y cuyas raíces que lo anulan son r1 = -1, r2 = -3 y r3 = -4.

Como consecuencia del análisis hecho, se tiene que la solución homogénea está dada por:

De las Ecuaciones (2.53) y (2.57) se puede afirmar que la solución completa es:

Al resolver el sistema de ecuaciones lineales definido por la Ecuación (2.60)

se obtiene que c1 = 1/3, c2 = -3/2 y c3 = 5/3, los cuales al ser sustituido en la Ecuación (2.58) se llega a:

Fuentes:

1. Análisis de Sistemas Lineales – Prof. Ebert Brea
1. Análisis de Sistemas en el Dominio Continuo p 29
2. Control Systems Engineering, Norman Nise
1. First Order System 4.3 p 165-168
3. Oppenheim – Señales y Sistemas

Literature Review by: Larry Francis Obando – Technical Specialist, Education Content Writer

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

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# FIRST and SECOND ORDER SYSTEMS

FIRST and SECOND ORDER SYSTEMS

Fuentes:

1. Control Systems Engineering, Norman Nise
1. Introduction Chapter 4 pp 162 (162)
2. Poles and Zeros 4.1 pp 162 –
3. First Order System 4.3 pp 165-168
4. Second Order System 4.4 pp 168-177
2. Modern_Control_Engineering__4t
1. Introduction Chapter 5 pp 219 (232)
2. First Order Systems 221 (234)-224
3. Second Order System pp 225(238)-229

TIME DOMAIN CONTROL SYSTEMS ANALYSIS

Analisis de sistemas de control en el dominio del tiempo

FIRST ORDER SYSTEMS

We now discuss first-order systems without zeros to define a performance specification for such a system…

We now use Eqs. (4.6), (4.7), and (4.8) to define three transient response performance specifications:

• Time Constant: We call 1/a the time constant of the response. From Eq. (4.7), the time constant can be described as the time for to decay to 37% of its initial value. Alternately, from Eq. (4.8) the time constant is the time it takes for the step response to rise to 63% of its final value.

The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we can call the parameter a the exponential frequency. Thus, the time constant can be considered a transient response specification for a first order system, since it is related to the speed at which the system responds to a step input.Since the pole of the transfer function is at a, we can say the pole is located at the reciprocal of the time constant, and the farther the pole from the imaginary axis, the faster the transient response.

• Rise Time (Tr): Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.

• Settling Time (Ts): Settling time is defined as the time for the response to reach, and stay within, 2% of its final value.2

Fuente [1]

Fuente [3]

Fuente [3]

SECOND-ORDER SYSTEMS

Literature Review by: Larry Francis Obando – Technical Specialist

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.