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Laboratory 5. Dynamics of Simple Systems5.1 Integrating systems5.2 First-order systems

5.2.1 Transient response5.2.2 Identification from step response

5.3 Second-order systems5.3.1 Transient response5.3.2 Identification of overdamped system5.3.3 Identification of underdamped system

5.4 Time-delay systems5.5 Inverse-response systems

KEH Process Dynamics and Control 5–1

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Laboratory 5. Dynamics of Simple SystemsIn Chapter 3, we derived models for some simple technical systems. In all cases, the systems could be described by ordinary differential equations (ODEs) of relatively low order.

In many cases, the ODEs were nonlinear, but they can be linearized around an operating point.

In this chapter we shall study the properties of some simple, linear, dynamical systems.

In particular, we shall derive the time response to well-defined inputs such as impulses and steps; analysis

based on such responses is called transient analysis.

We shall also study simple, graphical techniques for determining linear low-order models from

step-response data; these are simple examples of system identification.


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5. Dynamics of Simple Systems

5.1 Integrating systemsAn integrating system is the simplest dynamical system that can be described by a differential equation. The most typical process example of such a system is probably a liquid tank.

Example 5.1. A liquid tank.Consider the liquid tank in Fig. 5.1. 𝑉𝑉 = volume of liquid in the tank 𝐹𝐹1 = volumetric flow rate of liquid

into the tank 𝐹𝐹2 = volumetric flow rate of liquid

out of the tank Fig. 5.1. A liquid tank.

Note that 𝑉𝑉 is the system output (i.e., a dependent variable), 𝐹𝐹1 and 𝐹𝐹2 are system inputs (“independent” variables).


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5.1 Integrating systems

A mass balance around the tank gives, under the assumption of constant density (which cancels out), the model

d𝑉𝑉(𝑡𝑡)d𝑡𝑡

= 𝐹𝐹1(𝑡𝑡) − 𝐹𝐹2(𝑡𝑡) (1)

This equation is linear and we can directly replace the variables with Δ-variablesto get

d∆𝑉𝑉(𝑡𝑡)d𝑡𝑡

= ∆𝐹𝐹1(𝑡𝑡) − ∆𝐹𝐹2(𝑡𝑡) (2)

The Laplace transform, considering that the initial state is zero, yields

𝑠𝑠∆𝑉𝑉 𝑠𝑠 = ∆𝐹𝐹1 𝑠𝑠 − ∆𝐹𝐹2(𝑠𝑠) or ∆𝑉𝑉 𝑠𝑠 = 1𝑠𝑠∆𝐹𝐹1 𝑠𝑠 − 1

𝑠𝑠∆𝐹𝐹2(𝑠𝑠) (3)

The two transfer functions of the system are∆𝑉𝑉(𝑠𝑠)∆𝐹𝐹1(𝑠𝑠)

= 1𝑠𝑠

and ∆𝑉𝑉(𝑠𝑠)∆𝐹𝐹2(𝑠𝑠)

= −1𝑠𝑠

(4)

which, according to the Laplace transform, corresponds to integrals in the time domain.


Example 5.1. A liquid tank

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Generally, a linear integrating system with input signal 𝑢𝑢 and output signal 𝑦𝑦can be described by the differential equation

d𝑦𝑦d𝑡𝑡

= 𝐾𝐾𝑢𝑢 or 𝑇𝑇 d𝑦𝑦d𝑡𝑡

= 𝑢𝑢 (5.1)

The transfer function of the system is

𝐺𝐺 𝑠𝑠 ≡ 𝑌𝑌 𝑠𝑠𝑈𝑈 𝑠𝑠

= 𝐾𝐾𝑠𝑠

= 1𝑇𝑇𝑠𝑠

(5.2)

Exercise 5.1Derive and sketch(a) the impulse response(b) the step response(c) the ramp response

for ∆𝑉𝑉(𝑡𝑡) when the input flow 𝐹𝐹1 is changed in Example 5.1.


5.1 Integrating systems

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5.2 First-order systemsA non-integrating linear first-order system can be described by the differential equation

𝑇𝑇 d𝑦𝑦d𝑢𝑢

+ 𝑦𝑦 = 𝐾𝐾𝑢𝑢 (5.3)

where 𝐾𝐾 is the static system gain and 𝑇𝑇 is its time constant. The transfer function of the system is


= 𝐾𝐾𝑇𝑇𝑠𝑠+1

(5.4)

5.2.1 Transient responseThe time response 𝑦𝑦(𝑡𝑡) of the system output to a given input signal 𝑢𝑢(𝑡𝑡) is called a transient response. It can be determined by finding the inverse Laplace transform of 𝑌𝑌(𝑠𝑠) for the given input 𝑈𝑈 𝑠𝑠 via a table of Laplace transforms, possibly preceded by a partial fraction expansion.

Two common inputs for transient analysis are the (unit) impulse function step function


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5.2 First-order systems

Impulse responseAn input impulse with the time integral (area, volume, mass, etc.) 𝐼𝐼, i.e., 𝑢𝑢 𝑡𝑡 = 𝐼𝐼𝛿𝛿(𝑡𝑡), where 𝛿𝛿(𝑡𝑡) is the unit impulse (Dirac delta function), has the Laplace transform (LT Table, pt 6) 𝑈𝑈 𝑠𝑠 = 𝐼𝐼. Substitution into (5.4) gives

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑈𝑈 𝑠𝑠 = 𝐾𝐾𝐾𝐾𝑇𝑇𝑠𝑠+1

(5.5)

The inverse Laplace transform (LTT, pt 25) gives the impulse response

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾𝑇𝑇

e−𝑡𝑡/𝑇𝑇 (5.6)

Step responseAn input step change of size 𝑢𝑢step, i.e., 𝑢𝑢 𝑡𝑡 = 𝑢𝑢step𝜎𝜎(𝑡𝑡), where 𝜎𝜎(𝑡𝑡) is the unit step, has the Laplace transform (LT Table, pt 1) 𝑈𝑈 𝑠𝑠 = ⁄𝑢𝑢step 𝑠𝑠. Substitution into (5.4) gives

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑈𝑈 𝑠𝑠 = 𝐾𝐾𝑢𝑢step𝑠𝑠(𝑇𝑇𝑠𝑠+1)

(5.7)

The inverse Laplace transform (LTT, pt 26) gives the step response

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝑢𝑢step 1 − e−𝑡𝑡/𝑇𝑇 (5.8)


5.2.1 Transient response

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Transient response properties

Fig. 5.2. Normalized impulse Fig. 5.3. Normalized stepresponse of first-order system. response of first-order system.

The initial slope (i.e., the derivative) of the responses is the slope of a line drawn from the start of the response to the line denoting the new steady state at time 𝑡𝑡 = 𝑇𝑇.

The vertical distance between the response and the point of inter-section at 𝑡𝑡 = 𝑇𝑇 is ⁄1 𝑒𝑒 = 0.368 of the total change of the output.

In practice, the new steady-state value (within 2%) is reached at 𝑡𝑡 ≈ 4𝑇𝑇 (but in theory, it takes an ∞ long time).



stepKuKI T

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5.2.2 Identification from step responseFrom Fig. 5.2 and 5.3 it is obvious that the static gain and time constant can be identified (i.e., determined) from the transient response, which can be generated experimentally by a suitable change in the input signal.

Because a well-defined impulse tends to be difficult to create, identification using a step change in the input signal is often preferred.

The static gain is determined by

𝐾𝐾 = ⁄𝑦𝑦∞ 𝑢𝑢step (5.9)where 𝑢𝑢step is the size of the step change of the input 𝑦𝑦∞ is the total change of the output signal when 𝑡𝑡 → ∞

There are different methods for determining the time constant and a possible time delay. Some simple “graphical” methods are presented next.



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The method of 63 % change The step response of a first-order system reaches 63.2 % of the total change at

the time 𝑡𝑡 = 𝑇𝑇 (see Fig. 5.3). Thus, the time constant 𝑇𝑇 can be estimated as the time it takes for a step

response to reach 63.2 % of the total change. This estimate is often used even when the true system is not a first-order

system. It is then called the equivalent time constant 𝑇𝑇eq. In practice, a system often contains a time delay. A first-order system with a

time delay 𝐿𝐿 has the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾e−𝐿𝐿𝐿𝐿

𝑇𝑇𝑠𝑠+1(5.10)

and step response𝑦𝑦 𝑡𝑡 + 𝐿𝐿 = 𝐾𝐾𝑢𝑢step 1 − e−𝑡𝑡/𝑇𝑇 (5.11)

– The time delay 𝐿𝐿 is estimated as the time until the outputs starts to respond (input change at 𝑡𝑡 = 0).

– The time to reach 63.2 % of the total change gives 𝐿𝐿 + 𝑇𝑇.


5.2.2 Identification from step response

Fig. 5.4. Step-response identificationof first-order system with time delay.

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The tangent methodIn practice, a system is hardly ever a perfect first-order system. Often the step response does not have its steepest slope immediately at the

beginning of the response. This means that the system is of higher order than first order.

In spite of this, we might be interested in approximating the system as a first-order system with a time delay described by (5.10). The static gain is calculated by Eq. (5.9). To determine the time delay and

time constant, a tangent throughthe inflection point (i.e., the pointof the steepest slope) of the stepresponse 𝑡𝑡i, 𝑦𝑦i is drawn. Thepoint of intersection with– the time axis gives the

time delay 𝐿𝐿– the final asymptotic value

gives the time coordinate𝑡𝑡 = 𝐿𝐿 + 𝑇𝑇.



Fig. 5.5. Approximate identification as first-order system by the tangent method.

true stepresponse—— step response

of model - - -

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Comments on the tangent methodFor model identification As shown by Fig. 5.5, the step response of the identified model can differ

considerably from the true step response; the model response will always be slower than the true response.

This means that the time constant determined by the tangent method will always be too large.

Thus, the tangent method is not a good method for estimating an approximate first-order time constant.

For controller tuning There are controller tuning methods, e.g., Ziegler-Nichols’s step-response base

recommendations, that assume that the tangent method has been used to characterize the dynamics.

In such cases, it is OK to use the tangent method, because the time constant error is taken into account in the controller tuning rule.


The tangent method

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The modified tangent methodIn practice, a better first-order system identification is obtained by a combination of the 63% method and the tangent method. The time delay 𝐿𝐿 is determined

by to the tangent method. The time constant is determined

according to the 63% methodsuch that 𝐿𝐿 + 𝑇𝑇 is the time ittakes to reach 63.2 % of thetotal output change.

This procedure is better than thepure tangent method, because it gives a model whose step res-

ponse matches the true stepresponse much better;

is more robust against disturbances and measurement errors becausetwo response points are used instead of only one.



Fig. 5.6. Identification as first-order sys-tem by the modified tangent method.

true stepresponse—— step response

of model - - -

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The Sundaresan-Krishnaswamy methodThe inflection point and the pointwhere the step response reaches63.2 % of the total change are oftenclose to each other. It would be better to have a

method where the two pointsare farther apart.

In a method developed by Sunda-resan and Krishnaswamy, thepoints at 35 % and 85 % of thetotal output change are used.

According to the K-S method, thetime constant and time delay arecalculated by

𝑇𝑇 = 0.682(𝑡𝑡85 − 𝑡𝑡35) , 𝐿𝐿 = 𝑡𝑡35 − 0.431𝑇𝑇 (5.12)

where 𝑡𝑡35 and 𝑡𝑡85 are the times it takes to reach 35 % and 85 % of the total output change.

As always, the static gain is calculated by (5.9).



Fig. 5.7. Identification as first-order system by the method of Sundaresan and Krishnaswamy.

step responseof model - - -

true stepresponse——

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5.3 Second-order systemsA linear, strictly proper, second-order system can be describe by the DE

d2𝑦𝑦d𝑡𝑡2

+ 𝑎𝑎1d𝑦𝑦d𝑡𝑡

+ 𝑎𝑎2𝑦𝑦 = 𝑏𝑏1d𝑢𝑢d𝑡𝑡

+ 𝑏𝑏2𝑢𝑢 (5.13)

and the transfer function


= 𝑏𝑏1𝑠𝑠+𝑏𝑏2𝑠𝑠2+𝑎𝑎1𝑠𝑠+𝑎𝑎2

(5.14)

We will only consider systems with 𝑏𝑏2 ≠ 0. In this section only the case 𝑏𝑏1 = 0, i.e., transfer functions without zeros. In

Section 5.5, systems with zeros are considered.


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The transfer function is often expressed in terms of more meaningful parametersthan those in (5.14) as

𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝜔𝜔n2

𝑠𝑠2+2𝜁𝜁𝜔𝜔n𝑠𝑠+𝜔𝜔n2 (5.15)

where 𝐾𝐾 is the static system gain, 𝜁𝜁 is the relative damping, and 𝜔𝜔n is the undamped natural frequency. Sometimes, the form

𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑇𝑇n2𝑠𝑠2+2𝜁𝜁𝑇𝑇n𝑠𝑠+1

(5.16)

where 𝑇𝑇n = ⁄1 𝜔𝜔n. There is no universally accepted term for 𝑇𝑇n, both natural period and second-order time constant are used.

The relative damping is an important parameter for characterizing the main properties of the system. The system is overdamped if 𝜁𝜁 > 1 critically damped if 𝜁𝜁 = 1 underdamped if 0 ≤ 𝜁𝜁 < 1 unstable if 𝜁𝜁 < 0


5.3 Second-order systems

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5.3.1 Transient responseThe transient response to a given input can be determined in the normal way using the inverse Laplace transform. However, the form of the solution, and the applicable inverse transform, is different depending on whether the system is overdamped critically damped underdamped

The reason for this is that an overdamped system (𝜁𝜁 > 1) has real poles a critically damped system (𝜁𝜁 = 1) has a real double pole an underdamped system (0 ≤ 𝜁𝜁 < 1) has complex poles

The partial fraction expansion (Section 4.4.2) showed why the solutions are different in the three cases.



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Overdamped systemThe transfer function of an overdamped system, without a time delay, is usually written in the form

𝐺𝐺 𝑠𝑠 = 𝐾𝐾(𝑇𝑇1𝑠𝑠+1)(𝑇𝑇2𝑠𝑠+1)

, 𝑇𝑇1 > 𝑇𝑇2 > 0 (5.17)

where the time constants 𝑇𝑇1 and 𝑇𝑇2 are related to 𝜔𝜔n and 𝜁𝜁 by

𝑇𝑇1 = 1𝜔𝜔n

𝜁𝜁 + 𝜁𝜁2 − 1 , 𝑇𝑇2 = 1𝜔𝜔n

𝜁𝜁 − 𝜁𝜁2 − 1 (5.18)

𝜔𝜔n = 1𝑇𝑇1𝑇𝑇2

, 𝜁𝜁 = 𝑇𝑇1+𝑇𝑇22 𝑇𝑇1𝑇𝑇2

(5.19)

The system has the impulse response (input 𝑢𝑢 𝑡𝑡 = 𝐼𝐼𝛿𝛿(𝑡𝑡))

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾𝑇𝑇1−𝑇𝑇2

e−𝑡𝑡/𝑇𝑇1 − e−𝑡𝑡/𝑇𝑇2 (5.20)

and the step response (input 𝑢𝑢 𝑡𝑡 = 𝑢𝑢step𝜎𝜎(𝑡𝑡))

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝑢𝑢step 1 − 1𝑇𝑇1−𝑇𝑇2

𝑇𝑇1e−𝑡𝑡/𝑇𝑇1 − 𝑇𝑇2e−𝑡𝑡/𝑇𝑇2 (5.21)

The responses are shown in Fig. 5.8 and 5.9 (𝜁𝜁 > 1).



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Critically damped systemThe transfer function of an critically damped system, without a time delay, is often written in the form

𝐺𝐺 𝑠𝑠 = 𝐾𝐾(𝑇𝑇𝑠𝑠+1)2

, 𝑇𝑇 > 0 (5.22)

where 𝑇𝑇 = ⁄1 𝜔𝜔n.

The system has the impulse response (input 𝑢𝑢 𝑡𝑡 = 𝐼𝐼𝛿𝛿(𝑡𝑡))

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾𝑡𝑡𝑇𝑇2𝑒𝑒−𝑡𝑡/𝑇𝑇 (5.23)

and the step response (input 𝑢𝑢 𝑡𝑡 = 𝑢𝑢step𝜎𝜎(𝑡𝑡))

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝑢𝑢step 1 − 1 − 𝑡𝑡𝑇𝑇

e−𝑡𝑡/𝑇𝑇 (5.24)

The responses are shown in Fig. 5.8 and 5.9 (𝜁𝜁 = 1).



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Underdamped systemThe transfer function of an underdamped system, without a time delay, is usually represented in the form (5.15).

Because the roots of the characteristic equation are complex, the analytical expressions for transient responses contain trigonometric functions.

The impulse response (input 𝑢𝑢 𝑡𝑡 = 𝐼𝐼𝛿𝛿(𝑡𝑡)) is

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐼𝐼𝜔𝜔n𝛽𝛽−1𝑒𝑒−𝜁𝜁𝜔𝜔n𝑡𝑡 sin 𝛽𝛽𝜔𝜔n𝑡𝑡 (5.25)where

𝛽𝛽 = 1 − 𝜁𝜁2 , 0 ≤ 𝜁𝜁 < 1 (5.26)

The step response (input 𝑢𝑢 𝑡𝑡 = 𝑢𝑢step𝜎𝜎(𝑡𝑡)) is

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝑢𝑢step 1 − 𝛽𝛽−1e−𝜁𝜁𝜔𝜔n𝑡𝑡 sin 𝛽𝛽𝜔𝜔n𝑡𝑡 + 𝜑𝜑 (5.27)where

𝜑𝜑 = cos−1 𝜁𝜁 (5.28)

The responses are shown in Fig. 5.8 and 5.9 (0 ≤ 𝜁𝜁 < 1).



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Normalized transient responseFigure 5.8 shows the impulse response and Figure 5.9 the step response of allsecond-order systems without zeros. A time delay would not affect the form of the responses, only delay them.When the output and the time are normalized as shown in the figures, the responses are parametrized by the relative damping 𝜁𝜁.The figures show that overdamped and critically damped systems have monotonic responses underdamped systems have oscillatory responses

Fig. 5.8. Normalized impulse response Fig. 5.9. Normalized step response ofof second-order system without zeros. second-order system without zeros.



step

yKun

yKIω

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5.3.2 Identification of overdamped systemA simple method for the identification of an overdamped second-order system without zeros, but possibly including a time delay, is described. Thus, the transfer function of interest is

𝐺𝐺 𝑠𝑠 = 𝐾𝐾e−𝐿𝐿𝐿𝐿

(𝑇𝑇1𝑠𝑠+1)(𝑇𝑇2𝑠𝑠+1), 𝑇𝑇1 > 0, 𝑇𝑇1 ≥ 𝑇𝑇2 ≥ 0 (5.29)

where 𝐾𝐾 is the static gain 𝐿𝐿 is a time delay (which may be 0) 𝑇𝑇1 and 𝑇𝑇2 are time constants

The assumption 𝑇𝑇1 ≥ 𝑇𝑇2 ≥ 0 is only for convenience. As special cases it includes a critically damped system (𝑇𝑇1 = 𝑇𝑇2 > 0) a first-order system (𝑇𝑇1 > 0, 𝑇𝑇2 = 0)



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A modification of Harriott’s methodThe identification method is a modification of Harriott’s method for identification of the time constants 𝑇𝑇1 and 𝑇𝑇2 from a step response. The static gain 𝐾𝐾 and the time delay 𝐿𝐿 are determined separately. The gain 𝐾𝐾 is calculated by Eq. (5.9). The time delay 𝐿𝐿 is determined “visually” from the step response (a time

delay determined by the tangent method tends to be too long).The identification method isbased on two observations.The step response of thestudied class of systems reaches 72 % of the total

output change at a time𝑡𝑡72 ≈ 1.25(𝑇𝑇1 + 𝑇𝑇2)

is well separated for diffe-rent ratios z ≡ ⁄𝑇𝑇2 𝑇𝑇1 atthe time 𝑡𝑡𝑧𝑧 = 0.5(𝑇𝑇1 + 𝑇𝑇2)

Fig. 5.10. Normalized step response of overdamped second-order system parameterized by 𝑧𝑧.


5.3.2 Identification of overdamped system

0z =

1z =

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Comments In Fig. 5.10, the time axis is normalized by 𝑇𝑇Σ ≡ 𝑇𝑇1 + 𝑇𝑇2 (and 𝐿𝐿). When the

system is to be identified, 𝑇𝑇Σ is unknown, but the identification procedure can be formulated in terms of the “real” time 𝑡𝑡.

The factor 1.25 in 𝑡𝑡72 ≈ 1.25𝑇𝑇Σ is not exact, but it varies only by about 1 % as a function of z = ⁄𝑇𝑇2 𝑇𝑇1. Although 𝑧𝑧 is unknown initially, this variation can also be taken into account.

Identification procedure Calculate the static gain 𝐾𝐾 by (5.9). Determine the time delay 𝐿𝐿 visually from the step response. Measure the time it takes for the step response to reach 72 % of the total

output change 𝑦𝑦∞. Denote this time 𝑡𝑡72. Calculate the time 𝑡𝑡𝑧𝑧 for good curve separation by

𝑡𝑡𝑧𝑧 = 0.4𝑡𝑡72 + 0.6𝐿𝐿 (5.30)

Read the value of the output, 𝑦𝑦 = 𝑦𝑦𝑧𝑧, at time 𝑡𝑡𝑧𝑧.


Modified Harriott’s method

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Calculate the ratio ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞, and if the value is in the range0.27 ≤ ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ < 0.40 , (5.31)

read the corresponding 𝑧𝑧 value from Fig. 5.11. Alternatively, 𝑧𝑧 can be calculated by

𝑧𝑧 = −0.167 ln ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ − 0.2687 − 0.85 ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ (5.32) If (5.31) is not satisfied,

the step response doesnot belong to the classof studied systems with the chosen time delay.To fit a model of thisclass, the time delay– 𝐿𝐿 has to be increased

if ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ < 0.27,– 𝐿𝐿 has to be decreased

if ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ ≥ 0.40.After this, a new 𝑡𝑡𝑧𝑧 iscalculated by (5.30). Fig. 5.11. ⁄𝑦𝑦𝑧𝑧 𝑦𝑦∞ as a function of 𝑧𝑧.


Identification procedure

zyy∞

z

72is step response at 0.4 0.6zy t t L= +

2 1/z T T≡

ProcessControl

Laboratory


Determine 𝜏𝜏72 from Fig. 5.12, or calculate it by

𝜏𝜏72 = 1.273 − 0.27𝑧𝑧 1−0.8𝑧𝑧1+18𝑧𝑧2

(5.33)

Calculate the sum of the time constants by𝑇𝑇Σ = ⁄(𝑡𝑡72 − 𝐿𝐿) 𝜏𝜏72 (5.34)

Calculate the time constants by𝑇𝑇1 = ⁄𝑇𝑇Σ (1 + 𝑧𝑧) (5.35a)𝑇𝑇2 = 𝑧𝑧𝑇𝑇1 (5.35b)

Now all parameters of thetransfer function (5.29) havebeen determined!

Fig. 5.12. 𝜏𝜏72 as a function of 𝑧𝑧.


Identification procedure

z

72τ 72 72 1 2( ) / ( )t L T Tτ = − +

2 1/z T T≡

ProcessControl

Laboratory


Example 5.2. Approximate identification of first- and second-order systems.We shall approximate the system

𝐺𝐺 𝑠𝑠 = 5(6𝑠𝑠+1)(4𝑠𝑠+1)(2𝑠𝑠+1)

(1)

with first- and second-order systems by fitting their step responses. These approximation procedures can be used if 𝐺𝐺(𝑠𝑠) is unknown, but its step response is known 𝐺𝐺(𝑠𝑠) is known, and its step response is calculated

Calculation of step responseIt is first shown how the step response is calculated. Assume a step change of size 𝑢𝑢step = 2 is used. Its Laplace transform is 𝑈𝑈 𝑠𝑠 = 2/𝑠𝑠, yielding the Laplace-domain output

𝑌𝑌 𝑠𝑠 = 10𝑠𝑠(6𝑠𝑠+1)(4𝑠𝑠+1)(2𝑠𝑠+1)

(2)

This function has the PFE

𝑌𝑌 𝑠𝑠 = 𝐶𝐶1𝑠𝑠

+ 𝐶𝐶26𝑠𝑠+1



(3)



ProcessControl

Laboratory

Example 5.2

Because all poles of (2) are real and distinct, the constants 𝐶𝐶𝑘𝑘 can conveniently be calculated by (4.40). This yields

𝐶𝐶1 = lim𝑠𝑠→0

𝑠𝑠𝑌𝑌(𝑠𝑠) = lim𝑠𝑠→0

10(6𝑠𝑠+1)(4𝑠𝑠+1)(2𝑠𝑠+1)

= 10

𝐶𝐶2 = lim𝑠𝑠→−16

(𝑠𝑠 + 16)𝑌𝑌(𝑠𝑠) = lim

𝑠𝑠→−16

106𝑠𝑠(4𝑠𝑠+1)(2𝑠𝑠+1)

= 10−1⋅26⋅

46

= −45



𝑠𝑠→−14

54𝑠𝑠(6𝑠𝑠+1)(2𝑠𝑠+1)

= 5−1⋅ −24 ⋅24

= 40



𝑠𝑠→−12

52𝑠𝑠(6𝑠𝑠+1)(4𝑠𝑠+1)

= 5−1⋅ −42 ⋅(−1)

= −5

Thus, we obtain𝑌𝑌 𝑠𝑠 = 10

𝑠𝑠− 45

6𝑠𝑠+1+ 40

4𝑠𝑠+1− 5

2𝑠𝑠+1(4)

which has the inverse Laplace transform

𝑦𝑦 𝑡𝑡 = 10 1 − 4.5e−𝑡𝑡/6 + 4e−𝑡𝑡/4 − 0.5e−𝑡𝑡/2 (5)

This step response is plotted in Fig. 1 (Ex. 5.2).


Calculation of step response

ProcessControl

Laboratory


Identification of static gainThe step response is shown in Fig. 1 (Ex. 3.2). The total output change is 𝑦𝑦∞ = 10. Since the size of the input step change was 𝑢𝑢step = 2, the static gain is

𝐾𝐾 = ⁄𝑦𝑦∞ 𝑢𝑢step = 5

Fig. 1 (Ex. 5.2). Stepresponse of 𝐺𝐺(𝑠𝑠) with𝑢𝑢step = 2.


Example 5.2

y

t

ProcessControl

Laboratory


Identification of first-order systemThe time delay is determined by the intersection of the time axis and the tangent through the inflection point of the step response. inflection point 𝑡𝑡i, 𝑦𝑦i ≈ 7.3, 3 time delay 𝐿𝐿 ≈ 3

The time constant 𝑇𝑇 isdetermined as the time𝑡𝑡63 to reach 63.2 % of thetotal output change re-duced by the time delay. 𝑡𝑡63 ≈ 13 𝑇𝑇 = 𝑡𝑡63 − 𝐿𝐿 ≈ 10

The transfer function is

𝐺𝐺1 𝑠𝑠 = 5e−3𝐿𝐿

10𝑠𝑠+1(6)

Fig. 2 (Ex. 5.2). Identification of first-order dynamics.


Example 5.2

t

y

63tL

63y

ProcessControl

Laboratory

Example 5.2

For the input step 𝑈𝑈 𝑠𝑠 = 2/𝑠𝑠, the identified system has the step response𝑦𝑦1 𝑡𝑡 + 3 = 10 1 − e−𝑡𝑡/10 (7)

Fig. 3 shows this step response together with the original step response. As can be seen, the “fit” is not so good.If the model parameters are determined by numerical optimization over the whole visible step response, the result is 𝐾𝐾 = 5.1, 𝐿𝐿 = 3.8, 𝑇𝑇 = 9.1. The step response is shown in Fig. 4. The overall fit is better, but clearly the time delay seems too large (and 𝐾𝐾 > 5).

Fig. 3 (Ex. 5.2). Standard first-order fit. Fig. 4 (Ex. 5.2). Optimized first-order fit.


Identification of first-order system

tt

yy

ProcessControl

Laboratory


Identification of second-order systemThe time delay should be shorter than in the identification of a 1st order system since the “smooth” start of the step response can be handled by the 2nd order dynamics. However, it must be long enough to satisfy (5.31). This suggests the choice 𝐿𝐿 = 1.5

Fig. 5 plus calculations yield Fig. 5: 𝑡𝑡72 ≈ 15 (5.30): 𝑡𝑡𝑧𝑧 = 6.9 Fig. 5: 𝑦𝑦𝑧𝑧 ≈ 2.7 (5.32): 𝑧𝑧 = 0.88 (5.33): 𝜏𝜏72 = 1.268 (5.34): 𝑇𝑇Σ ≈ 10.7 (5.35a): 𝑇𝑇1 ≈ 5.7 (5.35b): 𝑇𝑇2 ≈ 5

The transfer function is

𝐺𝐺2 𝑠𝑠 = 5e−1.5𝐿𝐿

(5.7𝑠𝑠+1)(5𝑠𝑠+1)(8) Fig. 5 (Ex. 5.2). Identification of second-

order dynamics.


Example 5.2

t

y

72y

72tzt

zy

ProcessControl

Laboratory

Example 5.2

For the input step 𝑈𝑈 𝑠𝑠 = 2/𝑠𝑠, the identified system has the step response

𝑦𝑦2 𝑡𝑡 + 1.5 = 10 1 − 5.70.7

e−𝑡𝑡/5.7 + 50.7

e−𝑡𝑡/5 (9)

Fig. 6 shows this step response together with the original step response. As can be seen, the fit is almost perfect.

If the model parameters aredetermined by numericaloptimization over the wholevisible step response, theresult is 𝐾𝐾 = 5.01, 𝐿𝐿 = 1.38,𝑇𝑇1 = 𝑇𝑇2 = 5.35. The stepresponse of this systemcannot be distinguishedfrom that in Fig. 6. In fact,choosing 𝑧𝑧 = 1 from Fig.5.11 would result in thesame time constant forthe standard procedure.

Fig. 6 (Ex. 5.2). Standard second-order fit.


Identification of 2nd order system

t

y

ProcessControl

Laboratory


5.3.3 Identification of underdamped systemAs shown in Fig. 5.9, a step response of an underdamped system (0 ≤ 𝜁𝜁 < 1) is characterized by oscillations.

Obviously, the amplitude and frequency of the oscillations can be used for identification of a second order underdamped system.

Systems with oscillatory step response can be characterized by different parameters that can be determined from the step response. A number of such parameters are indicated in Fig. 5.13.

To facilitate the definition of the parameters, it is assumed that

the initial value of the output signal is zero (i.e., deviation variables are used); the final value of the step response is positive (i.e., the input step is in the

direction that increases the output).



ProcessControl

Laboratory


Step response parameters𝑦𝑦∞ Final value (>0) of output signal𝑦𝑦max Maximal value of output signal (peak of first overshoot)𝑀𝑀 Maximum relative overshoot, 𝑀𝑀 = (𝑦𝑦max − 𝑦𝑦∞)/𝑦𝑦∞𝑃𝑃 Period of oscillation (first period is easiest to measure)𝑡𝑡r Rise time — time to reach 𝑦𝑦∞ for the first time𝑡𝑡p Time to first peak —

time to reach 𝑦𝑦max𝑡𝑡s Settling time —

time requiredfor the output toreach and remainbetween (1 − 𝛿𝛿)𝑦𝑦∞and (1 + 𝛿𝛿)𝑦𝑦∞Usually, 𝛿𝛿 = 0.05 (5%)or 𝛿𝛿 = 0.02 (2%).

Fig. 5.13. Step response parameters of underdamped system.


5.3.3 Identification of underdamped system

ptrt st

maxy

y∞

(1 )yδ ∞+

(1 )yδ ∞−

P

ProcessControl

Laboratory


The analytical expression for the step response of an underdamped system, Eq. (5.27), can be used to derive relationships between the step response parameters and the parameters of the transfer function (5.15). The parameter 𝛽𝛽 ≡ 1 − 𝜁𝜁2 (5.26) is used in the relationships.

Maximum relative overshoot: 𝑀𝑀 ≡ 𝑦𝑦max−𝑦𝑦∞𝑦𝑦∞

= e−𝜋𝜋𝜁𝜁/𝛽𝛽 (5.35)

Period of oscillation: 𝑃𝑃 = 2𝜋𝜋𝛽𝛽𝜔𝜔n

(5.36)

Rise time: 𝑡𝑡r = 𝜋𝜋−arctan( ⁄𝛽𝛽 𝜁𝜁)𝛽𝛽𝜔𝜔n

(5.37)

Time to first peak: 𝑡𝑡p = 𝑃𝑃2

= 𝜋𝜋𝛽𝛽𝜔𝜔n

(5.38)

Settling time: 𝑡𝑡s ≈ − ln 𝛿𝛿𝜁𝜁𝜔𝜔n

, 𝛿𝛿 < 𝑀𝑀 (5.39)

Note that Eqs (5.35) – (5.38) are exact relationships whereas (5.39) is only approximate. Note also that 𝛿𝛿 is expressed as a fraction, not %.

Note that the expressions for 𝑡𝑡r , 𝑡𝑡p and 𝑡𝑡s assume no time delay. If there is a time delay 𝐿𝐿 in the step response, it has to be excluded (initially).


Step response parameters

ProcessControl

Laboratory


Identification It is easiest, and sufficient for a complete identification of the dynamics, to

measure 𝑀𝑀 and 𝑃𝑃. Then,– the relative damping 𝜁𝜁 can be calculated by (5.26) and (5.35);– the undamped natural frequency 𝜔𝜔n is calculated by (5.36).

The time to the first peak, 𝑡𝑡p , can be measured instead of 𝑃𝑃 and (5.38) used instead of (5.36).

The rise time 𝑡𝑡r can be measured instead of 𝑀𝑀 or 𝑃𝑃 and (5.37) used instead of (5.35) or (5.36).

The static gain 𝐾𝐾 is calculated by (5.9).The step response of a strongly underdamped system tends to be sensitive to disturbances, variations in parameters, and nonlinear behaviour. In such cases it might be useful to base the identification on several periods of oscillations. Equation (5.35) is then replaced by

𝑀𝑀R𝑘𝑘 = 𝑦𝑦max,𝑛𝑛+𝑘𝑘−𝑦𝑦∞

𝑦𝑦max,𝑛𝑛−𝑦𝑦∞= 𝑦𝑦∞−𝑦𝑦min,𝑛𝑛+𝑘𝑘

𝑦𝑦∞−𝑦𝑦min,𝑛𝑛= e−2𝜋𝜋𝑘𝑘𝜁𝜁/𝛽𝛽 (5.40)

where 𝑦𝑦max,𝑖𝑖 (𝑦𝑦min,𝑖𝑖) is the peak value of the 𝑖𝑖th overshoot (undershoot). 𝑀𝑀R𝑘𝑘 denotes the ratio between the (𝑛𝑛+𝑘𝑘)th and 𝑛𝑛th over/undershoot value.



ProcessControl

Laboratory


Example 5.3. Identification of an underdamped second-order system.The identification of an underdamped second-order system is illustrated by use of the step response in Fig. 5.13. Because of missing data, the following assumptions are used: the time axis goes from 0 to 20 seconds the output signal axis goes from 0 to 2.5.

The figure yields𝑀𝑀 = 𝑦𝑦max−𝑦𝑦∞

𝑦𝑦∞≈ 2.17−1.5

1.5= 0.447 (1)

𝑃𝑃 ≈ 9.75 − 3.25 = 6.5 (2)Equation (5.26) and (5.35) can be solved with respect to 𝜁𝜁 to yield

𝜁𝜁 = − ln 𝑀𝑀𝜋𝜋2+ ln 𝑀𝑀 2 (3)

Numerically, (3) yields 𝜁𝜁 = 0.2485. Eq. (5.36) yields 𝜔𝜔n = 0.998. The gain 𝐾𝐾cannot be determined since the size of the input step is not given.

The true values used to calculate the step response are 𝜁𝜁 = 0.25 and 𝜔𝜔n = 1.



ProcessControl

Laboratory


5.4 Time-delay systemsThe output 𝑦𝑦(𝑡𝑡) from a system consisting of a pure time delay 𝐿𝐿 is the input signal 𝑢𝑢(𝑡𝑡) delayed by 𝐿𝐿 time units, i.e.,

𝑦𝑦 𝑡𝑡 + 𝐿𝐿 = 𝑢𝑢(𝑡𝑡) (5.41)

In practice, a time delay is often caused by a transport delay. A conveyor belt is a typical example of a transport delay. For liquid and gas flows in a pipeline, the properties of the flowing medium

(e.g., temperature and concentration) are delayed when the output is located some distance from the input. However, the mass flow rate of an incompressible fluid is not delayed.

Measurement instruments can sometimes cause a time delay, for example when samples are analysed.

The transfer function of a time delay 𝐿𝐿 is

𝐺𝐺 𝑠𝑠 = e−𝐿𝐿𝑠𝑠 (5.42)

In principle, this function is simple, but it causes a lot of problems in analysis, computation, and control.


ProcessControl

Laboratory


Problems caused by time delays A time-delay system belongs to the class of non-minimum phase systems.

Such systems are more difficult to control than corres-ponding minimum-phase systems.– Another example of a non-minimum phase system is an inverse-response

system (Section 5.5). This is a system with positive zeros. The exponential function is an irrational function. This causes analysis

problems when it is combined with a rational function (polynomial). For this reason, it is often desirable to approximate the exponential function by a rational expression.

ApproximationsSimple rational approximations of the exponential function can be derived via Taylor series expansions of e−𝐿𝐿𝑠𝑠. A straightforward expansion of e−𝐿𝐿𝑠𝑠 gives

e−𝐿𝐿𝑠𝑠 = 1 − 𝐿𝐿𝑠𝑠 + 𝐿𝐿𝑠𝑠 2

2!− 𝐿𝐿𝑠𝑠 3

3!+ ⋯ (5.43)

This expression is exact when infinitely many terms are included, but useful approximations are obtained by a finite number of terms.


5.4 Time-delay systems

ProcessControl

Laboratory


The first two terms of the (5.43) give the simple, but relatively inexact, first-order Taylor series approximation

e−𝐿𝐿𝑠𝑠 ≈ 1 − 𝐿𝐿𝑠𝑠 (5.44)

Using more terms will improve the approximation, but it has the drawback of increasing the order of the resulting polynomial when the approximation is combined with another polynomial (e.g., the numerator polynomial of a transfer function).

Another class of approximations is obtained by a Taylor series expansion of e𝐿𝐿𝑠𝑠in e−𝐿𝐿𝑠𝑠 = ⁄1 e𝐿𝐿𝑠𝑠, i.e.,

e−𝐿𝐿𝑠𝑠 = 1e𝐿𝐿𝐿𝐿

= 1

1+𝐿𝐿𝑠𝑠+ 𝐿𝐿𝐿𝐿 22! + 𝐿𝐿𝐿𝐿 3

3! +⋯(5.45)

A first-order approximation of e𝐿𝐿𝑠𝑠 yields

e−𝐿𝐿𝑠𝑠 ≈ 11+𝐿𝐿𝑠𝑠

(5.46)

i.e., a first-order system with the time constant 𝐿𝐿. It is obvious that this is not a good approximation of a time delay, but it is still used a lot because a better approximation would increase the order of the system more.


Approximations

ProcessControl

Laboratory


The Taylor series expansions (5.43) and (5.45) can be combined to yield

e−𝐿𝐿𝑠𝑠 = e−𝛼𝛼𝐿𝐿𝐿𝐿

e(1−𝛼𝛼)𝐿𝐿𝐿𝐿 =1−𝛼𝛼𝐿𝐿𝑠𝑠+ 𝛼𝛼𝐿𝐿𝐿𝐿 2

2! − 𝛼𝛼𝐿𝐿𝐿𝐿 3

3! +⋯

1+(1−𝛼𝛼)𝐿𝐿𝑠𝑠+ (1−𝛼𝛼)𝐿𝐿𝐿𝐿 22! + (1−𝛼𝛼)𝐿𝐿𝐿𝐿 3

3! +⋯(5.47)

where 𝛼𝛼 is any parameter in the range 0 ≤ 𝛼𝛼 ≤ 1. Using 𝛼𝛼 = 0.5 and a first-order approximation of the numerator and the denominator gives

e−𝐿𝐿𝑠𝑠 ≈ 1−0.5𝐿𝐿𝑠𝑠1+0.5𝐿𝐿𝑠𝑠

= −1 + 21+0.5𝐿𝐿𝑠𝑠

(5.48)

This approximation gives a proper, but not strictly proper, first-order system with somewhat “peculiar” properties (consider, e.g., the step response of (5.48)).

Padé approximations are another kind of polynomial approximations that are derived under some optimizing frequency-domain conditions. A first-order Padéapproximation is identical to (5.48), but the second-order Padé approximationcannot be obtained from (5.47); the approximation is

e−𝐿𝐿𝑠𝑠 ≈1−12𝐿𝐿𝑠𝑠+

112 𝐿𝐿𝑠𝑠

2

1+12𝐿𝐿𝑠𝑠+112 𝐿𝐿𝑠𝑠

2 (5.49)


Approximations

ProcessControl

Laboratory


The exponential function can be defined through

e𝑥𝑥 = lim𝑛𝑛→∞

1 + 𝑥𝑥𝑛𝑛

𝑛𝑛(5.50)

Writing e−𝐿𝐿𝑠𝑠 as ⁄1 e𝐿𝐿𝑠𝑠 and application of (5.50) for some value of 𝑛𝑛 yields the approximation

e−𝐿𝐿𝑠𝑠 ≈ 1

1+1𝑛𝑛𝐿𝐿𝑠𝑠𝑛𝑛 (5.51)

i.e., an nth order system. The order 𝑛𝑛 = 1 gives the same approximation as (5.46). High orders give very good approximations of a time delay (but high order can

have drawbacks).


Approximations

ProcessControl

Laboratory


5.5 Inverse-response systemsSystems with inverse response have step responses whose direction changes one or more times at the beginning of the step response.

This is not to be confused with the oscillations of an underdamped system, whose step response oscillates around the final steady-state value.

Systems with inverse response belong to the class of systems that are known as non-minimum phase systems. Such systems tend to be difficult to control.

Systems with inverse response are not uncommon. A simple example is the mercury thermometer. If the ambient temperature

increases, the temperature of the glass tube starts to increase before the mercury temperature. The glass tube will then expand, causing the mercury column to fall. When the mercury temperature starts to increase, the mercury expands and the column will rise.

Another example with the same type of behaviour is the liquid level in a boilerwhen the feed water supply increases.

This examples suggest that inverse response may occur when the system contains subsystems that affect the output in opposite directions.


ProcessControl

Laboratory


Inverse-response systems are characterized by a transfer function with one ore more positive zeros, which are equivalent to negative numerator time constants. This is illustrated by the step responses in Fig. 5.14.

𝐺𝐺0 =𝑇𝑇1.5𝑠𝑠+1

𝑇𝑇2.5𝑠𝑠+1

𝑇𝑇1𝑠𝑠+1

𝑇𝑇2𝑠𝑠+1

𝑇𝑇3𝑠𝑠+1

𝐺𝐺1 =− 𝑇𝑇1.5𝑠𝑠+1

𝑇𝑇2.5𝑠𝑠+1

𝑇𝑇1𝑠𝑠+1

𝑇𝑇2𝑠𝑠+1

𝑇𝑇3𝑠𝑠+1

𝐺𝐺2 =− 𝑇𝑇1.5𝑠𝑠+1 − 𝑇𝑇

2.5𝑠𝑠+1𝑇𝑇1𝑠𝑠+1

𝑇𝑇2𝑠𝑠+1

𝑇𝑇3𝑠𝑠+1

𝑇𝑇 > 0

Fig. 5.14. Step responses of systems with differentnumber of negative numerator time constants.

Exercise 5.2Two systems 𝐺𝐺1 = 𝐾𝐾1

𝑇𝑇1𝑠𝑠+1and 𝐺𝐺2 = 𝐾𝐾2

𝑇𝑇2𝑠𝑠+1are connected in parallel so that a

system 𝐺𝐺 = 𝐺𝐺1 + 𝐺𝐺2 is obtained. Show that 𝐺𝐺 is non-minimum phase if and only if 𝑇𝑇1

𝑇𝑇2> −𝐾𝐾1

𝐾𝐾2> 1 when 𝑇𝑇1 > 𝑇𝑇2 > 0.


5.5 Inverse-response systems

( )y t

/t T

process 5. dynamics of simple systems - Åbo akademi | … · · 2017-03-30laboratory 5. dynamics...

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