8. Frequency Response Analysis - Åbo Akademi | Startsida · · 2017-03-308. Frequency Response Analysis ... Frequency Response Analysis 8.4 Controller design in the frequency domain

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Laboratory8. Frequency Response Analysis8.0 Overview8.1 The frequency response for a stable system

8.1.1 Simple system elements8.1.2 First-order system8.1.3 Second-order systems8.1.4 Systems of arbitrary order8.1.5 Systems in series8.1.6 Summary

8.2 Graphical frequency-response representation8.2.1 Overview8.2.2 Bode diagram

8.3 Stability analysis of feedback systems8.3.1 Bode’s stability criterion8.3.2 Stability margins8.3.3 Numerical solution of frequency relationships

KEH Process Dynamics and Control 8–1

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8. Frequency Response Analysis

8.4 Controller design in the frequency domain 8.4.1 Design of PI controllers8.4.2 Design of PD controllers8.4.3 Design of PID controllers


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8.0 OverviewSo far in this course system properties have been studied in the time domain (e.g., step response) Laplace domain (e.g., stability)

In this chapter, system properties are studied in the frequency domain

by studying the stationary behaviour when the system is excited by a sinusoidal input of given frequency. Stationary behaviour refers to the situation when initial effects have vanished, i.e., when the time 𝑡𝑡 → ∞ .

The output from a linear system will then also change sinusoidally with certain characteristic properties that depend on the system as well as the amplitude and frequency of the input sinusoidal. These properties expressed as function of the input frequency are referred to as the frequency response of the system. System analysis based on the frequency response is called frequency analysis.

The study of sinusoidal inputs is useful because measurement noise and time-varying disturbances can often be approximated by sinusoidal signals.


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8.1 The frequency response for a stable systemIn this section, the frequency response for an arbitrary linear, stable, system is derived.

In order to introduce new concepts step-wise, some simple system elements that essentially lack dynamics are treated first. These elements are often part of more complex systems.

After this, systems of first and second order are considered. In these systems, dynamics play a prominent role.

For the derivation of the frequency response of high-order systems, the frequency response of simpler systems is utilized. The decomposition of a high-order system into low-order components connected in series is fundamental in frequency analysis.


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8.1.1 Simple system elementsStatic systemA linear static system is described by

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾(𝑡𝑡) (8.1)

where 𝐾𝐾(𝑡𝑡) is an input, 𝑦𝑦(𝑡𝑡) is an output, and 𝐾𝐾 is the system gain. Let the input change sinusoidally as

𝐾𝐾 𝑡𝑡 = 𝐴𝐴 sin𝜔𝜔𝑡𝑡 (8.2)

where 𝜔𝜔 ≥ 0 is the angular frequency (expressed in radians per time unit) and 𝐴𝐴 > 0 is the amplitude of the sinusoidal. The output is then

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴 sin𝜔𝜔𝑡𝑡 (8.3)

There are now two situations regarding the phase of the output: If 𝐾𝐾 > 0, it is the same as the phase of the input. If 𝐾𝐾 < 0, it is opposite to the phase of the input. In this case, 𝑦𝑦(𝑡𝑡) has a

maximum when 𝐾𝐾(𝑡𝑡) has a minimum, and vice versa.Thus, the negative system gain causes a phase shift of ±𝜋𝜋 radians, or ±180°, in the output.


8.1 Frequency response for a stable system

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8.1.1 Simple system elements

It is of interest to rewrite (8.3) so that the phase shift is explicitly seen in the equation the output amplitude is always a positive quantity

To accomplish this, (8.3) is written as

𝑦𝑦 𝑡𝑡 = 𝐾𝐾 𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) (8.4)where

𝜑𝜑 = � 0 if 𝐾𝐾 > 0−𝜋𝜋 if 𝐾𝐾 < 0 (8.5)

is the phase shift of the system.The ratio between the amplitudes of the output and input signals is also of interest. In this case, the amplitude ratio is

𝐴𝐴R = 𝐾𝐾 (8.6)

Generally, the output can be written as

𝑦𝑦 𝑡𝑡 = 𝐴𝐴R𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) (8.7)

The stationary output of any stable linear system with an input (8.2) can be written as (8.7), where 𝐴𝐴R is the amplitude ratio and 𝜑𝜑 is the phase shift of the system in question.KEH Process Dynamics and Control 8–6

Static system

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Derivative systemA system with the transfer function 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑠𝑠 has an output that is the time derivative of the input amplified by the factor 𝐾𝐾, i.e.,

𝑦𝑦 𝑡𝑡 = 𝐾𝐾 d𝑢𝑢(𝑡𝑡)d𝑡𝑡

(8.8)

When the input is sinusoidal as in (8.2), the output becomes

𝑦𝑦 𝑡𝑡 = 𝐾𝐾 d(𝐴𝐴 sin 𝜔𝜔𝑡𝑡)d𝑡𝑡

= 𝐾𝐾𝐴𝐴𝜔𝜔 cos𝜔𝜔𝑡𝑡 (8.9)

where the relationship d(sin𝜔𝜔𝑡𝑡)/d𝑡𝑡 = 𝜔𝜔 cos𝜔𝜔𝑡𝑡 has been used.

Here the output is a cosine function, but by using the trigonometric identity cos𝜔𝜔𝑡𝑡 = sin(𝜔𝜔𝑡𝑡 + 𝜋𝜋/2) , (8.9) can be written as

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝜔𝜔𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜋𝜋/2) (8.10)

Here the phase shift +𝜋𝜋/2 is used, not −3𝜋𝜋/2, although they both give the same output. The reason is not that +𝜋𝜋/2 is closer to 0, but that the derivative yields a prediction of the input. Hence, the output can be considered to correspond to a future value of the input meaning that the phase of the output is before the phase of the input. Thus, the derivative is a phase-advancing element.



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If the gain 𝐾𝐾 < 0, this has to be taken into account as for a static system. Thus, the output is given by (8.7) with

𝐴𝐴R = 𝐾𝐾 (8.11)

𝜑𝜑 = � 𝜋𝜋/2 if 𝐾𝐾 > 0−𝜋𝜋/2 if 𝐾𝐾 < 0 (8.12)

Parallel connection of static and derivative systemA system composed of a parallel connection of a static and a derivative system has the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾(1 + 𝑇𝑇𝑠𝑠) (8.13)

This can be a PD controller, but other systems with zeros also have such factors in the numerator of the transfer function.

In the time domain, the output is given by

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾 𝑡𝑡 + 𝐾𝐾𝑇𝑇 d𝑢𝑢(𝑡𝑡)d𝑡𝑡

(8.14)

When the input is a sinusoidal as in (8.2), this gives

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴(sin𝜔𝜔𝑡𝑡 + 𝑇𝑇𝜔𝜔 cos𝜔𝜔𝑡𝑡) (8.15)


Derivative system

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Derivative system

The right-hand side of (8.15) can be written as a single sine function by means of the trigonometric identity

sin𝜔𝜔𝑡𝑡 + 𝑥𝑥 cos𝜔𝜔𝑡𝑡 = 1 + 𝑥𝑥2 sin(𝜔𝜔𝑡𝑡 + arctan 𝑥𝑥) (8.16)This gives

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴 1 + (𝑇𝑇𝜔𝜔)2sin(𝜔𝜔𝑡𝑡 + arctan𝑇𝑇𝜔𝜔) (8.17)

The output is thus given by (8.7) with

𝐴𝐴R = 𝐾𝐾 1 + (𝑇𝑇𝜔𝜔)2 (8.18)

𝜑𝜑 = �arctan𝑇𝑇𝜔𝜔 if 𝐾𝐾 > 0arctan𝑇𝑇𝜔𝜔 − 𝜋𝜋 if 𝐾𝐾 < 0 (8.19)


Parallel connection of static and derivative system

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Integrating systemA system with the transfer function 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑠𝑠−1 has an output that is the time integral of the input amplified by the factor 𝐾𝐾. This is equivalent to the time derivative of the output being equal to the input amplified by 𝐾𝐾, i.e.,

d𝑦𝑦(𝑡𝑡)d𝑡𝑡

= 𝐾𝐾𝐾𝐾(𝑡𝑡) (8.20)

The sinusoidal input (8.2) givesd𝑦𝑦(𝑡𝑡)d𝑡𝑡

= 𝐾𝐾𝐴𝐴 sin𝜔𝜔𝑡𝑡 (8.21)

Because d cos 𝜔𝜔𝑡𝑡d𝑡𝑡

= −𝜔𝜔 sin𝜔𝜔𝑡𝑡, it is easy to verify by differentiation that

𝑦𝑦 𝑡𝑡 = −𝐾𝐾𝐴𝐴𝜔𝜔−1 cos𝜔𝜔𝑡𝑡 (8.22)

satisfies (8.20). Because cos𝜔𝜔𝑡𝑡 = sin(𝜔𝜔𝑡𝑡 + 𝜋𝜋/2) = − sin(𝜔𝜔𝑡𝑡 − 𝜋𝜋/2), substitution into (8.22) yields (8.7) with

𝐴𝐴R = 𝐾𝐾 𝜔𝜔−1 (8.23)

𝜑𝜑 = � −𝜋𝜋/2 if 𝐾𝐾 > 0−3𝜋𝜋/2 if 𝐾𝐾 < 0 (8.24)



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Parallel connection of static and integrating systemA system composed of a parallel connection of a static and an integrating system has the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾 1 + 1𝑇𝑇𝑇𝑇

(8.25)

This can, e.g., be the transfer function of a PI controller. From (8.3) and (8.22) it is clear that the sinusoidal input (8.2) gives

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴 sin𝜔𝜔𝑡𝑡 − (𝑇𝑇𝜔𝜔)−1cos𝜔𝜔𝑡𝑡 (8.26)

Taking the trigonometric identity (8.16) and the sign of 𝐾𝐾 into account yields (8.7) with

𝐴𝐴R = 𝐾𝐾 1 + (𝑇𝑇𝜔𝜔)−2 (8.27)

𝜑𝜑 = �arctan (𝑇𝑇𝜔𝜔)−1 if 𝐾𝐾 > 0arctan (𝑇𝑇𝜔𝜔)−1 − 𝜋𝜋 if 𝐾𝐾 < 0

(8.28)


Integrating system

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Time delayA time delay of length 𝐿𝐿 has the transfer function 𝐺𝐺 𝑠𝑠 = e−𝐿𝐿𝑇𝑇. In the time domain, the relationship between the output and the input is

𝑦𝑦 𝑡𝑡 = 𝐾𝐾(𝑡𝑡 − 𝐿𝐿) (8.29)

The sinusoidal input (8.2) then gives

𝑦𝑦 𝑡𝑡 = 𝐴𝐴 sin 𝜔𝜔 𝑡𝑡 − 𝐿𝐿 = 𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) (8.30)where

𝜑𝜑 = −𝐿𝐿𝜔𝜔 (8.31)

As seen from (8.30), a pure time delay has the amplitude ratio

𝐴𝐴R = 1 (8.32)

Note that the phase shift of a time delay is unlimited — the higher the frequency, the more negative the phase shift.



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8.1.2 First-order systemA first-order system is described by the differential equation

𝑇𝑇 d𝑦𝑦(𝑡𝑡)d𝑡𝑡

+ 𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐾𝐾(𝑡𝑡) (8.33)

where 𝐾𝐾(𝑡𝑡) is an input, 𝑦𝑦(𝑡𝑡) is an output, 𝐾𝐾 is the gain, and 𝑇𝑇 is the time constant of the system.

Frequency response via the Laplace domainAs for the previous cases, the frequency response could be derived directly in the time domain. However, it is instructive to derive the frequency response via the Laplace domain; this will help derive more general relationships for the frequency response, which are useful when systems of arbitrary order are considered.



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8.1.2 First-order system

Laplace transformation of (8.33) yields

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑈𝑈(𝑠𝑠) , 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑇𝑇𝑇𝑇+1

(8.34)

where 𝑈𝑈(𝑠𝑠) is the Laplace transform of 𝐾𝐾(𝑡𝑡), 𝑌𝑌(𝑠𝑠) is the Laplace transform of 𝑦𝑦(𝑡𝑡), and 𝐺𝐺(𝑠𝑠) is the transfer function of the system. The Laplace transform of the sinusoidal input (8.2) is

𝑈𝑈 𝑠𝑠 = 𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2 (8.35)

which substituted into (8.34) gives

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2 = 𝐾𝐾

𝑇𝑇𝑇𝑇+1⋅ 𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2 (8.36)

Since the second-order factor 𝑠𝑠2 + 𝜔𝜔2 has complex-conjugated zeros, (8.36) has the partial fraction expansion

𝑌𝑌 𝑠𝑠 = 𝐵𝐵𝑇𝑇𝑇𝑇+1

+ 𝐶𝐶𝑇𝑇+𝐷𝐷𝑇𝑇2+𝜔𝜔2 (8.37)

where the coefficients 𝐵𝐵, 𝐶𝐶 and 𝐷𝐷 have to be determined so that (8.36) and (8.37) are equivalent.


Frequency response via the Laplace domain

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It is useful to consider the inverse Laplace transform of (8.37) before 𝐵𝐵, 𝐶𝐶 and 𝐷𝐷are determined. The transforms 25, 38 and 39 in the Laplace transform table in Section 4.5 yield

𝑦𝑦 𝑡𝑡 = 𝐵𝐵𝑇𝑇

e−𝑡𝑡/𝑇𝑇 + 𝐶𝐶 cos𝜔𝜔𝑡𝑡 + 𝐷𝐷𝜔𝜔−1sin𝜔𝜔𝑡𝑡 (8.38)

We are interested in the stationary solution when 𝑡𝑡 → ∞ . Because 𝑇𝑇 > 0 for a stable system, and 𝐵𝐵 is finite, the first term on the right-hand side will vanish as 𝑡𝑡 → ∞ . Thus, the stationary solution is

�𝑦𝑦 𝑡𝑡 ≡ lim𝑡𝑡→∞

𝑦𝑦 𝑡𝑡 = 𝐶𝐶 cos𝜔𝜔𝑡𝑡 + 𝐷𝐷𝜔𝜔−1 sin𝜔𝜔𝑡𝑡 (8.39)

This means that the coefficient 𝐵𝐵 does not affect the stationary solution and it is sufficient to determine only 𝐶𝐶 and 𝐷𝐷 if it can be done independently of 𝐵𝐵.

Combination of the first part of (8.36) with (8.37) gives𝐵𝐵

𝑇𝑇𝑇𝑇+1+ 𝐶𝐶𝑇𝑇+𝐷𝐷

𝑇𝑇2+𝜔𝜔2 = 𝐺𝐺 𝑠𝑠 𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2

from which

𝐶𝐶𝑠𝑠 + 𝐷𝐷 ≡ 𝐺𝐺 𝑠𝑠 𝐴𝐴𝜔𝜔 − 𝐵𝐵(𝑇𝑇2+𝜔𝜔2)𝑇𝑇𝑇𝑇+1

(8.40)



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The identity (8.40) has to apply for arbitrary values of 𝑠𝑠. Choosing 𝑠𝑠 = j𝜔𝜔means that 𝑠𝑠2 + 𝜔𝜔2 = 0. Then (8.40) yields

𝐶𝐶𝜔𝜔j + 𝐷𝐷 ≡ 𝐺𝐺 j𝜔𝜔 𝐴𝐴𝜔𝜔 (8.41)

The identity (8.41) requires that the real part and the imaginary part are satisfied independently. Because 𝐺𝐺(j𝜔𝜔) is a complex number with the real part Re𝐺𝐺 j𝜔𝜔 and the imaginary part Im𝐺𝐺 j𝜔𝜔 , i.e.,

𝐺𝐺 j𝜔𝜔 = Re𝐺𝐺 j𝜔𝜔 + Im𝐺𝐺 j𝜔𝜔 j(8.41) yields

𝐶𝐶 = 𝐴𝐴 Im𝐺𝐺(j𝜔𝜔) , 𝐷𝐷 = 𝐴𝐴𝜔𝜔 Re𝐺𝐺(j𝜔𝜔) (8.42)

The first-order system (8.34) yields

𝐺𝐺 j𝜔𝜔 = 𝐾𝐾𝑇𝑇𝜔𝜔j+1

= 𝐾𝐾𝑇𝑇𝜔𝜔j+1

⋅ 1−𝑇𝑇𝜔𝜔j1−𝑇𝑇𝜔𝜔j

= 𝐾𝐾−𝐾𝐾𝑇𝑇𝜔𝜔j1+ 𝑇𝑇𝜔𝜔 2 (8.43)

from whichRe𝐺𝐺 j𝜔𝜔 = 𝐾𝐾

1+ 𝑇𝑇𝜔𝜔 2 , Im𝐺𝐺 j𝜔𝜔 = −𝐾𝐾𝑇𝑇𝜔𝜔1+ 𝑇𝑇𝜔𝜔 2 (8.44)



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Substitution of (8.44) into (8.42) and further into (8.39) yields the stationary solution

�𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴1+(𝑇𝑇𝜔𝜔)2

sin𝜔𝜔𝑡𝑡 − 𝑇𝑇𝜔𝜔 cos𝜔𝜔𝑡𝑡 (8.45)

The trigonometrical identity (8.16) applied to (8.45) yields

�𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝐴𝐴1+(𝑇𝑇𝜔𝜔)2

sin(𝜔𝜔𝑡𝑡 − arctan𝑇𝑇𝜔𝜔) (8.46)

Thus, the stationary solution has the same form as (8.7), i.e.,

�𝑦𝑦 𝑡𝑡 = 𝐴𝐴R𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) (8.47)

For a first-order system (8.47) applies with

𝐴𝐴R = 𝐾𝐾1+(𝑇𝑇𝜔𝜔)2

(8.48)

𝜑𝜑 = �− arctan(𝑇𝑇𝜔𝜔) if 𝐾𝐾 > 0− arctan(𝑇𝑇𝜔𝜔) − 𝜋𝜋 if 𝐾𝐾 < 0 (8.49)



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8.1.3 Second-order systemsA second-order system without zeros and with no time delay has the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝜔𝜔n2

𝑇𝑇2+2𝜁𝜁𝜔𝜔n𝑇𝑇+𝜔𝜔n2 (8.50a)

where 𝐾𝐾 is the static gain of the system, 𝜁𝜁 is its relative damping and 𝜔𝜔n > 0is its undamped natural frequency. Since only stable systems are considered, 𝜁𝜁 > 0. If the poles of the system are real, the transfer function is often written in the form

𝐺𝐺 𝑠𝑠 = 𝐾𝐾(𝑇𝑇1𝑇𝑇+1)(𝑇𝑇2𝑇𝑇+1)

(8.50b)

The parameters of (8.50a) are then given by

𝜁𝜁 = 𝑇𝑇1+𝑇𝑇22 𝑇𝑇1𝑇𝑇2

, 𝜔𝜔n = 1𝑇𝑇1𝑇𝑇2

(8.51)

It is here sufficient to consider second-order systems without zeros because systems with zeros can be decomposed into a series connection of two (or more) subsystems and handled by the methods in Section 5.1.5.



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Derivation of frequency responseAnalogously to (8.36), the Laplace transform of the sinusoidal input yields

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2 = 𝐾𝐾𝜔𝜔n

2

𝑇𝑇2+2𝜁𝜁𝜔𝜔n𝑇𝑇+𝜔𝜔n2 ⋅

𝐴𝐴𝜔𝜔𝑇𝑇2+𝜔𝜔2 (8.52)

When 𝜁𝜁 > 0, there always exists a partial fraction expansion

𝑌𝑌 𝑠𝑠 = 𝐵𝐵1𝑇𝑇+𝐵𝐵2𝑇𝑇2+2𝜁𝜁𝜔𝜔n𝑇𝑇+𝜔𝜔n

2 + 𝐶𝐶𝑇𝑇+𝐷𝐷𝑇𝑇2+𝜔𝜔2 (8.53)

According to our Laplace transform table, the first term on the right-hand side has an inverse transform containing the factor e−𝜁𝜁𝜔𝜔n𝑡𝑡, where 𝑡𝑡 denotes time. Because 𝜁𝜁𝜔𝜔n > 0, e−𝜁𝜁𝜔𝜔n𝑡𝑡 → 0 as 𝑡𝑡 → ∞ and the term containing this factor will vanish.

This means that also in this case the stationary solution to (8.53) is given by (8.39) the coefficients 𝐶𝐶 and 𝐷𝐷 are given by (8.42)


8.1.3 Second-order systems

ProcessControl

Laboratory


In this case

𝐺𝐺 j𝜔𝜔 = 𝐾𝐾𝜔𝜔n2

(j𝜔𝜔)2+2𝜁𝜁𝜔𝜔nj𝜔𝜔+𝜔𝜔n2 = 𝐾𝐾𝜔𝜔n

2

𝜔𝜔n2−𝜔𝜔2+2𝜁𝜁𝜔𝜔n𝜔𝜔 j

= 𝐾𝐾𝜔𝜔n2

𝜔𝜔n2−𝜔𝜔2+2𝜁𝜁𝜔𝜔n𝜔𝜔 j

⋅ 𝜔𝜔n2−𝜔𝜔2−2𝜁𝜁𝜔𝜔n𝜔𝜔 j

𝜔𝜔n2−𝜔𝜔2−2𝜁𝜁𝜔𝜔n𝜔𝜔 j

= 𝐾𝐾𝜔𝜔n2(𝜔𝜔n

2−𝜔𝜔2−2𝜁𝜁𝜔𝜔n𝜔𝜔 j)(𝜔𝜔n

2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2(8.54)

from which

Re𝐺𝐺 j𝜔𝜔 = 𝐾𝐾𝜔𝜔n2(𝜔𝜔n

2−𝜔𝜔2)(𝜔𝜔n

2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2, Im𝐺𝐺 j𝜔𝜔 = −2𝐾𝐾𝜔𝜔n

3𝜔𝜔(𝜔𝜔n

2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2(8.55)

Substitution into (8.42) and further into (8.39) gives

�𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝜔𝜔n2(𝜔𝜔n

2−𝜔𝜔2)(𝜔𝜔n

2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2𝐴𝐴 sin𝜔𝜔𝑡𝑡 − 2𝜁𝜁𝜔𝜔n𝜔𝜔

𝜔𝜔n2−𝜔𝜔2 cos𝜔𝜔𝑡𝑡 (8.56)

where we (for now) assume that 𝜔𝜔 ≠ 𝜔𝜔n. Application of the trigonometric identity (8.16) yields the stationary response as

�𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝜔𝜔n2 sgn(𝜔𝜔n

2−𝜔𝜔2)

(𝜔𝜔n2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2

𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) , 𝜑𝜑 = −arctan 2𝜁𝜁𝜔𝜔n𝜔𝜔𝜔𝜔n2−𝜔𝜔2 (8.57)


Derivation of frequency response

ProcessControl

Laboratory


Eq. (8.57) can be expressed in the form of (8.47) with

𝐴𝐴R = 𝐾𝐾 𝜔𝜔n2

(𝜔𝜔n2−𝜔𝜔2)2+(2𝜁𝜁𝜔𝜔n𝜔𝜔)2

= 𝐾𝐾1−( ⁄𝜔𝜔 𝜔𝜔n)2 2+(2𝜁𝜁 ⁄𝜔𝜔 𝜔𝜔n)2

(8.58)

𝜑𝜑 = �− arctan 2𝜁𝜁𝜔𝜔n𝜔𝜔

𝜔𝜔n2−𝜔𝜔2 if 𝐾𝐾𝜔𝜔n > 𝐾𝐾𝜔𝜔

− arctan 2𝜁𝜁𝜔𝜔n𝜔𝜔𝜔𝜔n2−𝜔𝜔2 − 𝜋𝜋 if 𝐾𝐾𝜔𝜔n < 𝐾𝐾𝜔𝜔

(8.59a)

= �− arctan 2𝜁𝜁 ⁄𝜔𝜔 𝜔𝜔n

1− ⁄𝜔𝜔 𝜔𝜔n 2 if 𝐾𝐾𝜔𝜔n > 𝐾𝐾𝜔𝜔

− arctan 2𝜁𝜁 ⁄𝜔𝜔 𝜔𝜔n1− ⁄𝜔𝜔 𝜔𝜔n 2 − 𝜋𝜋 if 𝐾𝐾𝜔𝜔n < 𝐾𝐾𝜔𝜔

(8.59b)

For 𝜔𝜔 = 𝜔𝜔n, these reduce to

𝐴𝐴R = 𝐾𝐾2𝜁𝜁

, 𝜑𝜑 = � − ⁄𝜋𝜋 2 if 𝐾𝐾 > 0−3𝜋𝜋/2 if 𝐾𝐾 < 0 , 𝜔𝜔 = 𝜔𝜔n (8.60–61)

If 𝜁𝜁 < 1/ 2 ≈ 0.7, the amplitude ratio has a peak (i.e., maximum) at the peak frequency 𝜔𝜔p, given by

𝜔𝜔p = 𝜔𝜔n 1 − 2𝜁𝜁2 ⇒ 𝐴𝐴R,peak = 𝐾𝐾2𝜁𝜁 1−2𝜁𝜁2

(8.62–63)


Derivation of frequency response

ProcessControl

Laboratory


8.1.4 Systems of arbitrary orderFor a stable system of arbitrary order, but no time delay, the response to a sinusoidal input can be expressed by a partial fraction expansion as in (8.37) and (8.53). As before, the stationary solution in the time-domain is given by (8.39) with 𝐶𝐶 and 𝐷𝐷 given by (8.42).

To simplify the notation, we define

𝑅𝑅(𝜔𝜔) ≡ Re𝐺𝐺(j𝜔𝜔) , 𝐼𝐼 𝜔𝜔 = Im𝐺𝐺(j𝜔𝜔) (8.64)

Substitution of (8.42) into (8.39) then gives

�𝑦𝑦 𝑡𝑡 = 𝐴𝐴 𝑅𝑅 𝜔𝜔 sin𝜔𝜔𝑡𝑡 + 𝐼𝐼(𝜔𝜔) cos𝜔𝜔𝑡𝑡

= 𝐴𝐴𝑅𝑅(𝜔𝜔) sin𝜔𝜔𝑡𝑡 + 𝐼𝐼(𝜔𝜔)𝑅𝑅(𝜔𝜔)

cos𝜔𝜔𝑡𝑡 (8.65)

Application of the trigonometrical identity (8.16) yields

�𝑦𝑦(𝑡𝑡) = 𝑅𝑅(𝜔𝜔) 1 + ⁄𝐼𝐼(𝜔𝜔) 𝑅𝑅(𝜔𝜔) 2𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑)

= sgn 𝑅𝑅(𝜔𝜔) 𝑅𝑅(𝜔𝜔)2 + 𝐼𝐼(𝜔𝜔)2𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜑𝜑) (8.66)where

𝜑𝜑 = arctan ⁄𝐼𝐼(𝜔𝜔) 𝑅𝑅(𝜔𝜔) (8.67)



ProcessControl

Laboratory


Because 𝐺𝐺(jω) is a complex number, it can be characterized by its magnitude𝐺𝐺(j𝜔𝜔) and argument ∠𝐺𝐺(j𝜔𝜔), also denoted arg𝐺𝐺(j𝜔𝜔). According to the

theory of complex number,

𝐺𝐺(j𝜔𝜔) = 𝑅𝑅(𝜔𝜔)2 + 𝐼𝐼(𝜔𝜔)2 , tan arg𝐺𝐺(j𝜔𝜔) = 𝐼𝐼(𝜔𝜔)𝑅𝑅(𝜔𝜔)

(8.68a,b)

The sign of sgn 𝑅𝑅(𝜔𝜔) in (8.66) has the same effect on the phase shift as the sign of the gain 𝐾𝐾 in previous sections. Substitution of (8.68) into (8.66) and (8.67) then yields

�𝑦𝑦 𝑡𝑡 = 𝐺𝐺(j𝜔𝜔) 𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + arg𝐺𝐺 j𝜔𝜔 ) (8.69)

arg𝐺𝐺 j𝜔𝜔 = �arctan ⁄𝐼𝐼(𝜔𝜔) 𝑅𝑅(𝜔𝜔) if 𝑅𝑅(𝜔𝜔) ≥ 0arctan ⁄𝐼𝐼(𝜔𝜔) 𝑅𝑅(𝜔𝜔) − 𝜋𝜋 if 𝑅𝑅 𝜔𝜔 < 0 (8.70)

From (8.69) it is obvious that the magnitude 𝐺𝐺(j𝜔𝜔) is identical to the amplitude ratio the argument arg𝐺𝐺(j𝜔𝜔) is identical to the phase shift

Thus𝐴𝐴R = 𝐺𝐺(j𝜔𝜔) , 𝜑𝜑 = arg𝐺𝐺(j𝜔𝜔) (8.71)


8.1.4 Systems of arbitrary order

ProcessControl

Laboratory


Because the function arctan only takes values between −𝜋𝜋/2 and +𝜋𝜋/2, direct calculation from (8.70) gives an argument in the range

−32𝜋𝜋 < arg𝐺𝐺(j𝜔𝜔) < 1

2𝜋𝜋 (8.72)

However, because of the periodicity of the trigonometric functions, (8.68b) would be satisfied for any integer multiple of 2𝜋𝜋 added to arg𝐺𝐺 j𝜔𝜔 . This means that the solution calculated from (8.70) is ambiguous by an integer multiple of 2𝜋𝜋.The figure shows a sinusoidal input 𝐾𝐾(𝑡𝑡) and the stationary output 𝑦𝑦(𝑡𝑡), which is also sinusoidal. When the axes are normalized as indicated, the values of 𝐺𝐺and arg𝐺𝐺 can be directly readfrom the plot for the appliedinput frequency. Because thesinusoidal signals can beshifted by an integer multipleof 2𝜋𝜋 without any detectabledifference, arg𝐺𝐺 is similarlyambiguous also here.

This ambiguity can be solved by the method in Section 5.1.5.KEH Process Dynamics and Control 8–24

8.1.4 Systems of arbitrary order

Normalizedtime

Nor

mal

ized

sign

al

ProcessControl

Laboratory


8.1.5 Systems in seriesConsider a system composed of 𝑁𝑁 subsystems connected in series. If the subsystems have the transfer functions 𝐺𝐺𝑖𝑖(𝑠𝑠), 𝑖𝑖 = 1, … ,𝑁𝑁, the transfer function 𝐺𝐺(𝑠𝑠) of the full system is given by

𝐺𝐺 𝑠𝑠 = 𝐺𝐺1 𝑠𝑠 𝐺𝐺2 𝑠𝑠 ⋯𝐺𝐺𝑁𝑁 𝑠𝑠 = ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(𝑠𝑠) (8.73)

Conversely, a system with the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾 (𝑇𝑇𝑛𝑛+1𝑇𝑇+1)⋯(𝑇𝑇𝑛𝑛+𝑚𝑚𝑇𝑇+1)(𝑇𝑇1𝑇𝑇+1)(𝑇𝑇2𝑇𝑇+1)⋯(𝑇𝑇𝑛𝑛𝑇𝑇+1)

e−𝐿𝐿𝑇𝑇 (8.74)

can be decomposed into the series-connected subsystems

𝐾𝐾, 1𝑇𝑇1𝑇𝑇+1

, 1𝑇𝑇2𝑇𝑇+1

,…, 1𝑇𝑇𝑛𝑛𝑇𝑇+1

, (𝑇𝑇𝑛𝑛+1𝑠𝑠 + 1) ,…, (𝑇𝑇𝑛𝑛+𝑚𝑚𝑠𝑠 + 1) , e−𝐿𝐿𝑇𝑇.

If the system has complex-conjugated poles or zeros, they can be included as second-order factors in the decomposition.

Thus, a high-order transfer function can be decomposed into factors of at most second order and a possible time delay.



ProcessControl

Laboratory


From (8.73) it follows that the frequency response of the full system is

𝐺𝐺 j𝜔𝜔 = ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(j𝜔𝜔) (8.75)

where 𝐺𝐺𝑖𝑖(j𝜔𝜔), 𝑖𝑖 = 1, … ,𝑁𝑁, are the frequency responses of the individual subsystems.

According to the theory of complex numbers, 𝐺𝐺𝑖𝑖(j𝜔𝜔) can be expressed in terms of its magnitude 𝐺𝐺𝑖𝑖(j𝜔𝜔) and its argument arg𝐺𝐺𝑖𝑖(j𝜔𝜔) as

𝐺𝐺𝑖𝑖 j𝜔𝜔 = 𝐺𝐺𝑖𝑖(j𝜔𝜔) ej arg𝐺𝐺𝑖𝑖(j𝜔𝜔) (8.76)

Substitution into (8.75) yields

𝐺𝐺 j𝜔𝜔 = ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(j𝜔𝜔) ej arg𝐺𝐺𝑖𝑖(j𝜔𝜔) = ∏𝑖𝑖=1

𝑁𝑁 𝐺𝐺𝑖𝑖(j𝜔𝜔) ∏𝑖𝑖=1𝑁𝑁 ej arg𝐺𝐺𝑖𝑖(j𝜔𝜔)

= ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(j𝜔𝜔) ej ∑𝑖𝑖=1

𝑁𝑁 arg𝐺𝐺𝑖𝑖(j𝜔𝜔) (8.77)

Naturally, 𝐺𝐺(j𝜔𝜔) can also be expressed as (8.76). From that and (8.77),

𝐺𝐺 j𝜔𝜔 = ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(j𝜔𝜔) (8.78)

arg𝐺𝐺 j𝜔𝜔 = ∑𝑖𝑖=1𝑁𝑁 arg𝐺𝐺𝑖𝑖(j𝜔𝜔) (8.79)


8.1.5 Systems in series

ProcessControl

Laboratory


Equations (8.78) and (8.79) mean that for the full system the amplitude ratio (or magnitude) is obtained as the product of the amplitude

ratios (or magnitudes) of the subsystems phase shift (or argument) is obtained as the sum of the phase shifts (or

arguments) of the subsystems

The user is allowed to decompose the full system into subsystems as desired. However, it is advantageous to decompose into subsystems of first or second order and a possible time delay because we know the frequency responses of such systems. In particular, we know that a first-order system has a phase shift in the range − ⁄𝜋𝜋 2 < 𝜑𝜑 < 0 a second-order system has a phase shift in the range −𝜋𝜋 < 𝜑𝜑 < 0

when the gain is positive.

By adding up the phase shifts (arguments) of the individual subsystems according to (8.79), the correct phase shift (argument) for the full system is obtained. This means that a phase shift outside the range of (8.72) can be obtained, i.e., the correct integer multiple of 2𝜋𝜋 is obtained.


8.1.5 Systems in series

ProcessControl

Laboratory


8.1.6 SummaryTable 8.1. Frequency response of low-order systems.



( )G s R | ( j ) |A G ω= arg ( j )Gϕ ω=

1− 1 π−

0K > K 0

s ω / 2π

1 Ts+ 21 ( )Tω+ arctanTω

1 / s 1 / ω / 2π−

1 1 / Ts+ 21 1 / ( )Tω+ arctan(1 / )Tω−

e Ls− 1 Lω−

11Ts +

2

1

1 ( )Tω+ arctanTω−

2n

2 2n n2s s

ωζω ω+ +

2 2 2n n

1

(1 ( / ) ) (2 / )ω ω ζω ω− +

n2

n

2 /arctan1 ( / )

ζω ωω ω

−−

, nω ω≤

n2

n

2 /arctan1 ( / )

ζω ωπω ω

− −−

, nω ω≥

ProcessControl

Laboratory


8.2 Graphical frequency-response representation

The complex-valued function 𝐺𝐺(j𝜔𝜔) of a system with the transfer function 𝐺𝐺(𝑠𝑠)contains all information about the frequency response of the system—except for a multiple of 2𝜋𝜋 in the phase shift. 𝐺𝐺(j𝜔𝜔) can be considered a system property 𝐺𝐺(j𝜔𝜔) is called the frequency function

Since 𝐺𝐺(j𝜔𝜔) is a complex number, it can be represented in two ways:

𝐺𝐺(j𝜔𝜔) = Re𝐺𝐺(j𝜔𝜔) + jIm𝐺𝐺(j𝜔𝜔) = | )𝐺𝐺(j𝜔𝜔 |e )jarg𝐺𝐺(j𝜔𝜔

𝑅𝑅 𝜔𝜔 ≡ Re𝐺𝐺(j𝜔𝜔) is the real part of 𝐺𝐺(j𝜔𝜔) 𝐼𝐼 𝜔𝜔 ≡ Im𝐺𝐺(j𝜔𝜔) is the imaginary part of 𝐺𝐺(j𝜔𝜔) | )𝐺𝐺(j𝜔𝜔 | ≡ 𝑅𝑅(𝜔𝜔)2 + 𝐼𝐼(𝜔𝜔)2 is the magnitude of 𝐺𝐺(j𝜔𝜔) )arg𝐺𝐺(j𝜔𝜔 ≡ arctan[ ⁄𝐼𝐼(𝜔𝜔) 𝑅𝑅(𝜔𝜔)] ± 𝑛𝑛𝜋𝜋 is the argument of 𝐺𝐺(j𝜔𝜔)

This gives several possibilities of representing 𝐺𝐺 j𝜔𝜔 graphically, e.g., Nyquist diagram: 𝐼𝐼 𝜔𝜔 vs 𝑅𝑅 𝜔𝜔 as 𝜔𝜔 varies Bode diagram: | )𝐺𝐺(j𝜔𝜔 | vs 𝜔𝜔 and )arg𝐺𝐺(j𝜔𝜔 vs 𝜔𝜔 in two diagrams


8.2.1 Overview

ProcessControl

Laboratory


8.2.2 Bode diagramIn a Bode diagram, | )𝐺𝐺(j𝜔𝜔 | and )arg𝐺𝐺(j𝜔𝜔 are plotted against the frequency 𝜔𝜔in two diagrams.

The absolute value | )𝐺𝐺(j𝜔𝜔 | is plotted on a logarithmic scale, either expressed as a pure amplitude ratio or with the logarithmic “unit” decibel (dB), defined

| )𝐺𝐺(j𝜔𝜔 |dB = 20 log10 | )𝐺𝐺(j𝜔𝜔 | (8.80)

In this course, the pure amplitude ratio is used.

The phase shift )arg𝐺𝐺(j𝜔𝜔 is plotted on a linear scale, either expressed in degrees or radians, defined

1 rad = ⁄180° 𝜋𝜋 (8.81)In this course, degrees are used in the Bode diagram, but in calculations radians are used.

The frequency 𝜔𝜔 is expressed on a logarithmic scale in both diagrams.


8.2 Graphical representation

ProcessControl

Laboratory


First-order systemThe transfer function of a first-order system is 𝐺𝐺 𝑠𝑠 = 𝐾𝐾

𝑇𝑇𝑇𝑇+1. The following

expressions have been derived for the amplitude ratio and the phase shift

amplitude ratio: 𝐴𝐴R 𝜔𝜔 = )𝐺𝐺(j𝜔𝜔 = 𝐾𝐾1+(𝜔𝜔𝑇𝑇)2

phase shift: 𝜑𝜑(𝜔𝜔) = )arg𝐺𝐺(j𝜔𝜔 = −arctan𝜔𝜔𝑇𝑇

The Bode diagram appliesto all first-order systemsbecause the normalizedamplitude ratio (obtainedthrough division by 𝐾𝐾) anda normalized frequency(multiplication by 𝑇𝑇) areplotted.

At high frequency slope of ⁄𝐴𝐴R 𝐾𝐾 is −1 𝜑𝜑 → −90°


8.2.2 Bode diagram

Tω

Tω

RAK

ϕ

ProcessControl

Laboratory


Second-order systemsThe transfer function of a second-order system is 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝜔𝜔n

2

𝑇𝑇2+2𝜁𝜁𝜔𝜔n𝑇𝑇+𝜔𝜔n2.

The Bode diagram displays the amplitude ratio:𝐴𝐴R = 𝐾𝐾

1−( ⁄𝜔𝜔 𝜔𝜔n)2 2+ ⁄2𝜁𝜁𝜔𝜔 𝜔𝜔n 2

phase shift:

𝜑𝜑 = �− arctan ⁄2𝜁𝜁𝜔𝜔 𝜔𝜔n

1− ⁄𝜔𝜔 𝜔𝜔n 2 if 𝐾𝐾𝜔𝜔n ≥ 𝐾𝐾𝜔𝜔

− arctan ⁄2𝜁𝜁𝜔𝜔 𝜔𝜔n1− ⁄𝜔𝜔 𝜔𝜔n 2 − 𝜋𝜋 if 𝐾𝐾𝜔𝜔n ≤ 𝐾𝐾𝜔𝜔

resonance peak and frequency:𝐴𝐴R 𝜔𝜔p = 𝐾𝐾

2𝜁𝜁 1−2𝜁𝜁2at 𝜔𝜔p = 𝜔𝜔n 1 − 2𝜁𝜁2

At high frequency slope of ⁄𝐴𝐴R 𝐾𝐾 is −2 𝜑𝜑 → −180°

The diagram applies to all second-order systems due to normalized axes.KEH Process Dynamics and Control 8–32

8.2.2 Bode diagram

RAK

ϕ

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Time delayThe transfer function of a time delayis 𝐺𝐺 𝑠𝑠 = e−𝐿𝐿𝑇𝑇. The Bode diagramdisplays the amplitude ratio: 𝐴𝐴R 𝜔𝜔 = 1 phase shift: 𝜑𝜑 𝜔𝜔 = −𝜔𝜔𝐿𝐿

At high frequency 𝜑𝜑 → −∞° as 𝜔𝜔 → ∞

If the frequency axis 𝜔𝜔𝐿𝐿 werelinear, the slope of the phase shiftplot would be −1.


8.2.2 Bode diagram

RA

ϕ

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Numerator time constantA time constant in the numerator of a transfer function corresponds to a subsystem with the transfer function 𝐺𝐺 𝑠𝑠 = 1 + 𝑇𝑇𝑠𝑠. It is also the transfer function of a PD controller (with the gain 1 and derivative time 𝑇𝑇). The transfer function has the following characteristics

amplitude ratio: 𝐴𝐴R 𝜔𝜔 = 1 + (𝜔𝜔𝑇𝑇)2

phase shift: 𝜑𝜑 𝜔𝜔 = arctan𝜔𝜔𝑇𝑇At high frequency slope of 𝐴𝐴R is +1 𝜑𝜑 → +90° if 𝑇𝑇 > 0, 𝜑𝜑 → −90° if 𝑇𝑇 < 0

Note the similarity (and the difference) to a first-order system. For the numerator time constant the amplitude plot is symmetrical to the normalized amplitude plot of a first-

order system with respect to the line 𝐴𝐴R = 1 (i.e., its “mirror”) if 𝑇𝑇 > 0, the phase plot is symmetrical to the phase plot of a first-order

system with respect to the frequency line 𝜑𝜑 = 0 (i.e., its mirror); if 𝑇𝑇 < 0, the plots are identical


8.2.2 Bode diagram

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Integrating systemAn integrator has the transfer function 𝐺𝐺 𝑠𝑠 = 𝑠𝑠−1 with the following characteristics amplitude ratio: 𝐴𝐴R 𝜔𝜔 = 1/𝜔𝜔 phase shift: 𝜑𝜑 𝜔𝜔 = −𝜋𝜋/2

At all frequencies slope of 𝐴𝐴R is −1 𝜑𝜑 = −90°

Derivative systemA derivative system has the transfer function 𝐺𝐺 𝑠𝑠 = 𝑠𝑠 with the following characteristics amplitude ratio: 𝐴𝐴R 𝜔𝜔 = 𝜔𝜔 phase shift: 𝜑𝜑 𝜔𝜔 = +𝜋𝜋/2

At all frequencies slope of 𝐴𝐴R is +1 𝜑𝜑 = +90°


8.2.2 Bode diagram

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Systems in seriesA system composed of 𝑁𝑁 subsystems in series with the transfer functions 𝐺𝐺1(𝑠𝑠), 𝐺𝐺2(𝑠𝑠), …, 𝐺𝐺𝑁𝑁(𝑠𝑠), has the transfer function

𝐺𝐺 𝑠𝑠 = ∏𝑖𝑖=1𝑁𝑁 𝐺𝐺𝑖𝑖(𝑠𝑠)

with the following characteristics

amplitude ratio: 𝐴𝐴R 𝜔𝜔 = ∏𝑖𝑖=1𝑁𝑁 𝐴𝐴R,𝑖𝑖(𝜔𝜔)

log𝐴𝐴R 𝜔𝜔 = ∑𝑖𝑖=1𝑁𝑁 log𝐴𝐴R,𝑖𝑖(𝜔𝜔)

phase shift: 𝜑𝜑 𝜔𝜔 = ∑𝑖𝑖=1𝑁𝑁 𝜑𝜑𝑖𝑖 (𝜔𝜔)

Because the amplitude plot in a Bode diagram is logarithmic, the logarithmic amplitude plot of the overall system is obtained by adding the logarithmic amplitude plots of the subsystems. This is a useful property if the Bode plot is drawn “by hand”.

Because the phase plot is a Bode diagram is linear, the phase plot of the overall system is obtained by adding the phase plots of the subsystem.


8.2.2 Bode diagram

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8.3 Stability analysis of feedback systems

The system in the blockdiagram has the followingtransfer functions: 𝐺𝐺p is a process to be controlled 𝐺𝐺m is a measurement device 𝐺𝐺v is an actuator (e.g., a valve) An open loop. 𝐺𝐺c is a controller

The loop transfer function of the system is

𝐺𝐺ℓ = 𝐺𝐺m𝐺𝐺p𝐺𝐺v𝐺𝐺c (8.82)

Consider a system with 𝐺𝐺c = 𝐾𝐾c , 𝐺𝐺v = 1 , 𝐺𝐺p = e−0.1𝑠𝑠

0.5𝑇𝑇+1, 𝐺𝐺m = 1 , where the

time unit in 𝐺𝐺p is “minutes”. The loop transfer of the system is

𝐺𝐺ℓ = 𝐾𝐾ce−0.1𝑠𝑠

0.5𝑇𝑇+1(8.83)


8.3.1 Bode’s stability criterion

pGvGcG

mG

r e u y

my−+

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A thought experiment Assume that the setpoint 𝑟𝑟 changes sinusoidally with the angular frequency 𝜔𝜔 = 17 rad/min. At stationary conditions (i.e., when the initial effects from starting the sinusoidal have vanished) the phase shift through the loop transfer is𝜑𝜑 = −0.1 ⋅ 17 − arctan 0.5 ⋅ 17≈ −𝜋𝜋 = −180°

This means that 𝑦𝑦m will oscillate exactlyin opposite phase to 𝑟𝑟 (and 𝑒𝑒), asillustrated in the figure to the right.

Assume now that the feedback loop Before loop is closed.is closed. Because of the minus signin 𝑒𝑒 = 𝑟𝑟 − 𝑦𝑦m, −𝑦𝑦m will be exactly inphase with 𝑟𝑟, as shown in the figureto the right. Even if 𝑟𝑟 is set to zeroat the same moment the loop is closed,𝑒𝑒 will continue to oscillate with thesame frequency because 𝑒𝑒 = −𝑦𝑦m.

After loop is closed.



r

my

r

my−

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The amplitude of 𝑦𝑦m can be determined from the amplitude ratio of the system at the frequency 17 rad/min. In this case

𝐴𝐴R 𝜔𝜔 = 𝐾𝐾c1+(𝜔𝜔𝑇𝑇)2

⇒ 𝐴𝐴R 17 = 𝐾𝐾c1+(17⋅0.5)2

≈ ⁄𝐾𝐾c 8.56 (8.84)

There are now three possibilities. If the amplitude of 𝑦𝑦m is smaller than the amplitude of 𝑟𝑟 and 𝑒𝑒 before the

loop is closed (i.e., if 𝐾𝐾c < 8.56 ⇒ 𝐴𝐴R < 1), the amplitude of 𝑒𝑒 will decrease when the loop is closed. This, in turn, will reduce the amplitude of 𝑦𝑦m, making the amplitude of 𝑒𝑒 even smaller, and so on. Eventually, the oscillation will die out.

If the amplitude of 𝑦𝑦m is equal to the amplitude of 𝑟𝑟 and 𝑒𝑒 before the loop is closed (i.e., if 𝐾𝐾c = 8.56 ⇒ 𝐴𝐴R = 1), the amplitude of 𝑒𝑒 will not change when the loop is closed. This means that the oscillation will continue with the same amplitude “for ever”.

If the amplitude of 𝑦𝑦m is larger than the amplitude of 𝑟𝑟 and 𝑒𝑒 (i.e., if 𝐾𝐾c >8.56 ⇒ 𝐴𝐴R > 1), the amplitude of 𝑒𝑒 will increase when the loop is closed. This, in turn, will increase the amplitude of 𝑦𝑦m, making the amplitude of 𝑒𝑒even larger, and so on. The amplitude of the oscillation will become larger and larger and the system is unstable.


A though experiment

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The stability criterionThe smallest frequency such that the loop transfer function 𝐺𝐺ℓ has a phase shift equal to −180° is determined. This frequency, 𝜔𝜔c, is the critical frequency of the system.

The amplitude ratio of the loop transfer function at this frequency, i.e., 𝐴𝐴R,ℓ 𝜔𝜔c , is determined.

The closed-loop system is stable, if 𝐴𝐴R,ℓ 𝜔𝜔c < 1 unstable, if 𝐴𝐴R,ℓ 𝜔𝜔c > 1

for the loop transfer function. Note that the stability of the closed-loop system is determined by considering the loop transfer function in open loop.



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How to determine 𝜔𝜔c and 𝐴𝐴R(𝜔𝜔c)The critical frequency 𝜔𝜔c and the amplitude ratio 𝐴𝐴R(𝜔𝜔c) can be determined in different ways.

Graphically, by drawing the Bode diagram for the loop transfer function 𝐺𝐺ℓ. From such a diagram, it is easy to read off 𝜔𝜔c at −180° for the phase curve and 𝐴𝐴R 𝜔𝜔c from the amplitude curve.

Numerically, by solving the equation

𝜑𝜑ℓ(𝜔𝜔c) ≡ arg𝐺𝐺ℓ 𝜔𝜔c = −𝜋𝜋 (8.85)for 𝜔𝜔c and calculating

𝐴𝐴R,ℓ 𝜔𝜔c = 𝐺𝐺ℓ(𝜔𝜔c) (8.86)

It will be shown how 𝜑𝜑ℓ(𝜔𝜔c) = −𝜋𝜋 can be solved iteratively to find 𝜔𝜔c.

By simulation of feedback control using a P controller. The critical frequency 𝜔𝜔c and the maximum controller gain 𝐾𝐾c,ma𝑥𝑥 can be determined as described in Section 7.4.1. The amplitude ratio of the loop transfer without a controlleris then 𝐴𝐴R 𝜔𝜔c = ⁄1 𝐾𝐾c,max.



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Example 8.3.1Determine the critical frequency and the amplitude ratio at this frequency for a system with the loop transfer function

𝐺𝐺ℓ 𝑠𝑠 = 𝐺𝐺1(𝑠𝑠)𝐺𝐺2(𝑠𝑠)𝐺𝐺3(𝑠𝑠)𝐺𝐺4(𝑠𝑠)where

𝐺𝐺1 𝑠𝑠 = e−4𝑇𝑇 , 𝐺𝐺2 𝑠𝑠 = 1.52𝑇𝑇+1

,

𝐺𝐺3 𝑠𝑠 = 210𝑇𝑇+1

, 𝐺𝐺4 𝑠𝑠 = 0.85𝑇𝑇+1

The solution can be found graphicallyby means of a Bode diagram. Thediagram is constructed by plotting

𝐴𝐴R,ℓ 𝜔𝜔 = 1.51+(2𝜔𝜔)2

⋅ 21+(10𝜔𝜔)2

⋅ 0.81+(5𝜔𝜔)2

and𝜑𝜑ℓ 𝜔𝜔 = −4𝜔𝜔 − arctan2𝜔𝜔

− arctan10𝜔𝜔 − arctan5𝜔𝜔against the frequency 𝜔𝜔.From this diagram it is easy to read off𝜔𝜔c ≈ 0.21 rad/min at 𝜑𝜑ℓ = −𝜋𝜋 = −180°and 𝐴𝐴R,ℓ 𝜔𝜔c ≈ 0.7.



rad/min

rad/min0.21

0.7

R,A

ϕ

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Exercise 8.3.1A process that can be modelled as a pure time delay is controlled by a P controller. The control valve and the measurement device have negligible dynamics, but their gains are 𝐾𝐾v = 0.5 and 𝐾𝐾m = 0.8, respectively. When a small change in the setpoint is made, the controlled process starts to oscillate with constant amplitude and the period 10 min.a) Which is the controller gain?b) How large is the time delay?



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8.3.2 Stability marginsGain marginThe gain margin 𝐴𝐴m is the factor by which the gain of the loop transfer function has to be changed in order to reach the stability limit. Mathematically,

𝐴𝐴m = 1𝐴𝐴R,ℓ(𝜔𝜔c)

(8.87)

Stability of the closed-loop system requires 𝐴𝐴m > 1.

The stability margin yields robustness not only against variations in the loop gain, but also against variations in other process parameters.

Example 8.3.2Consider a system with the loop transfer function 𝐺𝐺ℓ = 𝐾𝐾ce−0.1𝑠𝑠

0.5𝑇𝑇+1, which was used

in the introduction of the Bode stability criterion.a) Determine a P controller to give the closed-loop system the (designed) gain

margin 𝐴𝐴m = 1.7.b) Is the closed-loop system stable with the designed P controller if the time

delay changes to 0.15 min?



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8.3.2 Stability margins

a) It was previously stated that an input sinusoidal with the frequency𝜔𝜔 = 17 rad/min resulted in a phase shift of −180°. Thus, this is the critical frequency 𝜔𝜔c of the loop transfer function. We want a P controller such that 𝐴𝐴m = 1.7, i.e., 𝐴𝐴R,ℓ 𝜔𝜔c = ⁄1 1.7 . This is obtained if

𝐴𝐴R,ℓ 𝜔𝜔c = 𝐾𝐾c1+(0.5⋅17)2

= 11.7

⇒ 𝐾𝐾c = 5.0

b) To check if the system is stable with 𝐾𝐾c = 5 and 𝐿𝐿 = 0.15 min, a Bode diagram could be drawn using these parameters. The diagram gives the new critical frequency and the amplitude ratio at that frequency.Here, the solution is obtained numerically. The phase shift equation is

−𝜋𝜋 = −0.15𝜔𝜔c − arctan 0.5𝜔𝜔cfrom which 𝜔𝜔c can be calculated iteratively by direct substitution in

𝜔𝜔c = (𝜋𝜋 − arctan 0.5𝜔𝜔c)/0.15This converges to 𝜔𝜔c = 11.6 rad/min using, e.g., 𝜔𝜔c = 17 rad/min as starting value. The new gain margin is found by

𝐴𝐴m = 1𝐴𝐴R,ℓ(𝜔𝜔c)

= 1+(0.5⋅11.6)2

5≈ 1.18 > 1

Because 𝐴𝐴m > 1, the closed-loop system is stable with 𝐿𝐿 = 0.15 min.


Example 8.3.2

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Phase marginThe phase margin 𝜑𝜑m denotes how much the phase shift at the cross-over frequency of the loop transfer function has to change in order to reach the stability limit. The cross-over frequency 𝜔𝜔g is the frequency, where the amplitude ratio of the loop transfer function is 1. Mathematically, the phase margin is defined

𝜑𝜑m = 𝜑𝜑 𝜔𝜔g + 𝜋𝜋 (8.88)where 𝜔𝜔g is the frequency that satisfies

𝐴𝐴R,ℓ 𝜔𝜔g = 1 (8.89)Stability of the closed-loop system requires 𝜑𝜑m > 0. The phase margin yields robustness not only against variations in the phase

shift of the loop transfer, but also against variations in other process parameters.

Example 8.3.3Consider the same system as in Example 8.3.2. a) Determine a P controller to yield a designed phase margin 𝜑𝜑m = 30°.b) Is the closed-loop system stable if the time delay changes to 0.15 min?



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a) We need to find 𝜔𝜔g such that 𝜑𝜑(𝜔𝜔g) = 𝜑𝜑m − 𝜋𝜋 = 30° − 180° = −150° = −5𝜋𝜋/6

𝜔𝜔g can be read off the phase curve of the loop transfer function at −150°. We can also find it numerically by solving

−5𝜋𝜋/6 = −0.1𝜔𝜔g − arctan 0.5𝜔𝜔gThis can be done iteratively by direct substitution in the expression

𝜔𝜔g = (5𝜋𝜋/6 − arctan 0.5𝜔𝜔g)/0.1The result is 𝜔𝜔g = 12.11 rad/min using 11.6 rad/min as starting value. Next, 𝐾𝐾c has to be chosen such that 𝐴𝐴R,ℓ 𝜔𝜔g = 1, i.e.,

𝐾𝐾c1+(0.5⋅12.11)2

= 1 ⇒ 𝐾𝐾c = 6.14

b) The stability of the system with 𝐾𝐾c = 6.14 and 𝐿𝐿 = 0.15 min can be checked in many ways. Here, it is simplest to use the result from Ex. 8.3.2, where 𝐿𝐿 = 0.15 min gave the critical frequency 𝜔𝜔c = 11.6 rad/min. (Note that the phase curve is independent of gains.) Thus,

𝐴𝐴R,ℓ ωc = 6.141+(0.5⋅11.6)2

= 1.04 > 1

Since 𝐴𝐴R,ℓ > 1, the closed-loop system is unstable with 𝐿𝐿 = 0.15 min.


Example 8.3.3

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Example 8.3.3The gain and phase margins can easily be found from a Bode diagram when the controller gain is given. The figure shows the Bode diagram for the loop transfer function 𝐺𝐺ℓ = 𝐾𝐾ce−0.1𝑠𝑠

0.5𝑇𝑇+1with 𝐾𝐾c = 5.

The gain margin isfound as follows.𝜔𝜔c is obtained fromthe phase curve at𝜑𝜑ℓ = −180°. Thegain curve yieldsthe amplitude ratio𝐴𝐴R,ℓ(𝜔𝜔c) at 𝜔𝜔c.𝐴𝐴m = ⁄1 𝐴𝐴R,ℓ(𝜔𝜔c).

The phase marginis found as follows.𝜔𝜔g is obtained fromthe gain curve at𝐴𝐴R,ℓ = 1. 𝜑𝜑m is thedifference betweenthe phase shift at𝜔𝜔g and −180°.



mphase margin ϕ

rad/min

rad/min

ϕ

R,A

1mA−

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8.3.3 Numerical solution of frequency relationshipsIn Example 8.3.2 and 8.3.3 the phase equation was solved by a simple iterative method, where the frequency appearing together with a time delay was solved out. However, there is no guarantee that the solution will converge, and if there is no time delay, the method cannot even be used.

In this section, better methods for solving both the phase equation and the gain equation are developed.

The system can be a general system with the transfer function

𝐺𝐺 𝑠𝑠 = 𝐾𝐾 𝑇𝑇𝑛𝑛+1𝑇𝑇+1 ⋅…⋅(𝑇𝑇𝑁𝑁𝑇𝑇+1)𝑇𝑇1𝑇𝑇+1 ⋅…⋅(𝑇𝑇𝑛𝑛𝑇𝑇+1)

𝑠𝑠𝑚𝑚e−𝐿𝐿𝑇𝑇 (8.90)

Note that 𝑚𝑚 > 0 if a factor 𝑠𝑠 appears in the numerator, and 𝑚𝑚 < 0 if it is in the denominator. Complex-conjugated poles and zeros will also be considered.

Eq. (8.90) may be any transfer function of interest, but in practice it will be the loop transfer function of the system.



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The phase equationThe system (8.90) has the phase shift

𝜑𝜑r = 𝑚𝑚𝜋𝜋2− 𝐿𝐿𝜔𝜔 − ∑𝑖𝑖=1𝑛𝑛 arctan𝑇𝑇𝑖𝑖𝜔𝜔 + ∑𝑖𝑖=𝑛𝑛+1𝑁𝑁 arctan𝑇𝑇𝑖𝑖𝜔𝜔 (8.91)

The phase shift 𝜑𝜑r, expressed in radians, is assumed to be known, and it is desired to find the frequency 𝜔𝜔 satisfying (8.91). The phase shift can have any value relevant for the calculation, but two typical choices are 𝜑𝜑r = −𝜋𝜋, if the critical frequency 𝜔𝜔 = 𝜔𝜔c is to be found 𝜑𝜑r = 𝜑𝜑m − 𝜋𝜋, if the cross-over frequency 𝜔𝜔 = 𝜔𝜔g is to be found

The following function is defined:

𝑓𝑓 𝜔𝜔 ≡ 𝜑𝜑r − 𝑚𝑚 𝜋𝜋2

+ 𝐿𝐿𝜔𝜔 + ∑𝑖𝑖=1𝑛𝑛 arctan𝑇𝑇𝑖𝑖𝜔𝜔 − ∑𝑖𝑖=𝑛𝑛+1𝑁𝑁 arctan𝑇𝑇𝑖𝑖𝜔𝜔 (8.92)

Since 𝑓𝑓 𝜔𝜔 = 0 at the solution, a possibility is to calculate 𝜔𝜔 iteratively by

𝜔𝜔𝑘𝑘+1 = 𝜔𝜔𝑘𝑘 − 𝜌𝜌𝑘𝑘𝑓𝑓(𝜔𝜔𝑘𝑘) (8.93)

where 𝜔𝜔𝑘𝑘 is the solution at iteration step 𝑘𝑘 and 𝜌𝜌𝑘𝑘 is a parameter that depends on the iteration method.


8.3.3 Numerical solution

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8.3.3 Numerical solution of frequency relationships

According to the Newton-Raphson method,

𝜌𝜌𝑘𝑘 = ⁄1 𝑓𝑓𝑓(𝜔𝜔𝑘𝑘) (8.94)

where 𝑓𝑓′ 𝜔𝜔𝑘𝑘 ≡ ⁄d𝑓𝑓(𝜔𝜔𝑘𝑘) d𝜔𝜔𝑘𝑘 . Differentiation of (8.92) yields

𝑓𝑓′ 𝜔𝜔𝑘𝑘 = 𝐿𝐿 + ∑𝑖𝑖=1𝑛𝑛 𝑇𝑇𝑖𝑖1+(𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2

− ∑𝑖𝑖=𝑛𝑛+1𝑁𝑁 𝑇𝑇𝑖𝑖1+(𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2

(8.95)

An initial guess of the frequency is required to start the iteration. This can be any frequency which might be close to the solution. If no such frequency is known, 𝜔𝜔0 = 0 can be used as starting value. However, this will always result in

𝜌𝜌0 = 𝐿𝐿 + ∑𝑖𝑖=1𝑛𝑛 𝑇𝑇𝑖𝑖 − ∑𝑖𝑖=𝑛𝑛+1𝑁𝑁 𝑇𝑇𝑖𝑖−1

(8.96a)

𝜔𝜔1 = −𝜌𝜌0(𝜑𝜑r − ⁄𝑚𝑚𝜋𝜋 2) (8.96b)

which thus can be used as a better starting value. If desired, the iteration can be continued with 𝜌𝜌𝑘𝑘 = 𝜌𝜌0, but better convergence is usually achieved if 𝜌𝜌𝑘𝑘 is recalculated at every step, or at least occasionally — the improvement can, in fact, be quite dramatic. If divergence occurs, 𝜌𝜌𝑘𝑘 can be reduced “manually”, e.g., by halving 𝜌𝜌𝑘𝑘.


The phase equation

ProcessControl

Laboratory


Complex poles and zerosIf there are complex poles or zeros, they occur as complex-conjugated pairs. Such a pair has the corresponding complex-conjugated time constants 𝑇𝑇𝑗𝑗 and 𝑇𝑇𝑗𝑗+1 , which satisfy

(𝑇𝑇𝑗𝑗𝑠𝑠 + 1)(𝑇𝑇𝑗𝑗+1𝑠𝑠 + 1) = ⁄𝑠𝑠2 + 2𝜁𝜁𝜔𝜔n𝑠𝑠 + 𝜔𝜔n2 𝜔𝜔n2 (8.97)

From this is obtained𝑇𝑇𝑗𝑗 + 𝑇𝑇𝑗𝑗+1 = ⁄2𝜁𝜁 𝜔𝜔n (8.98)

arctan𝑇𝑇𝑗𝑗𝜔𝜔 + arctan𝑇𝑇𝑗𝑗+1𝜔𝜔 = arctan 2𝜁𝜁𝜔𝜔n𝜔𝜔𝜔𝜔n2−𝜔𝜔2 (8.99)

which can substituted into the appropriate positions in (8.92) and (8.96a). In (8.95), the substitution

𝑇𝑇𝑗𝑗1+(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2

+ 𝑇𝑇𝑗𝑗+11+(𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2

= 2𝜁𝜁𝜔𝜔n(𝜔𝜔n2+𝜔𝜔𝑘𝑘

2)𝜔𝜔n4+2 2𝜁𝜁2−1 𝜔𝜔n

2𝜔𝜔𝑘𝑘2+𝜔𝜔𝑘𝑘

4 (8.100)

is used. Since the exact value of 𝜌𝜌𝑘𝑘 is not very important, an approximation such as

𝑇𝑇𝑗𝑗1+(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2

+ 𝑇𝑇𝑗𝑗+11+(𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2

≈ 2𝜁𝜁 1+( ⁄𝜔𝜔𝑘𝑘 𝜔𝜔n)2

𝜔𝜔n 1+( ⁄𝜔𝜔𝑘𝑘 𝜔𝜔n)4(8.101)

might also be used.


The phase equation

ProcessControl

Laboratory


The gain equationThe system (8.90) has the amplitude ratio

𝐴𝐴r = 𝐾𝐾 𝜔𝜔𝑚𝑚 1+(𝑇𝑇𝑛𝑛+1𝜔𝜔)2 ⋅…⋅ 1+(𝑇𝑇𝑁𝑁𝜔𝜔)2

1+(𝑇𝑇1𝜔𝜔)2 ⋅…⋅ 1+(𝑇𝑇𝑛𝑛𝜔𝜔)2(8.102)

The amplitude ratio 𝐴𝐴r, expressed as a “pure ratio”, is assumed to be known, and it is desired to find the frequency 𝜔𝜔 satisfying (8.102). The amplitude ratio can have any value relevant for the calculation, but a typical choices is 𝐴𝐴r = 1, if the cross-over frequency 𝜔𝜔 = 𝜔𝜔g is to be found

The following function is defined:

𝑔𝑔 𝜔𝜔 ≡ 𝐴𝐴r − 𝐾𝐾 𝜔𝜔𝑚𝑚 1+(𝑇𝑇𝑛𝑛+1𝜔𝜔)2 ⋅…⋅ 1+(𝑇𝑇𝑁𝑁𝜔𝜔)2

1+(𝑇𝑇1𝜔𝜔)2 ⋅…⋅ 1+(𝑇𝑇𝑛𝑛𝜔𝜔)2(8.103)

Since 𝑔𝑔 𝜔𝜔 = 0 at the solution, 𝜔𝜔 can be calculated iteratively by

𝜔𝜔𝑘𝑘+1 = 𝜔𝜔𝑘𝑘 − 𝜎𝜎𝑘𝑘𝑔𝑔(𝜔𝜔𝑘𝑘) (8.104)

where 𝜔𝜔𝑘𝑘 is the solution at iteration step 𝑘𝑘 and 𝜎𝜎𝑘𝑘 is a parameter that depends on the iteration method.



ProcessControl

Laboratory


According to the Newton-Raphson method,

𝜎𝜎𝑘𝑘 = ⁄1 𝑔𝑔𝑓(𝜔𝜔𝑘𝑘) (8.105)

where 𝑔𝑔′ 𝜔𝜔𝑘𝑘 ≡ ⁄d𝑔𝑔(𝜔𝜔𝑘𝑘) d𝜔𝜔𝑘𝑘 . Differentiation of (8.103) yields

𝑔𝑔′ 𝜔𝜔𝑘𝑘 = −𝑚𝑚 + ∑𝑖𝑖=1𝑛𝑛 (𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2− ∑𝑖𝑖=𝑛𝑛+1𝑁𝑁 (𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑖𝑖𝜔𝜔𝑘𝑘)2𝐴𝐴r−𝑔𝑔(𝜔𝜔𝑘𝑘)

𝜔𝜔𝑘𝑘 (8.106)

An initial guess of the frequency is required to start the iteration. This can be any frequency which might be close to the solution. However, 𝜔𝜔0 = 0 can not be used as a starting value. The critical frequency 𝜔𝜔c, which is often known, would usually be a good starting value.

If desired, the iteration can be continued with 𝜎𝜎𝑘𝑘 = 𝜎𝜎0, but better convergence is probably achieved if 𝜎𝜎𝑘𝑘 is recalculated at every step, or at least occasionally. If divergence occurs, 𝜎𝜎𝑘𝑘 can be reduced “manually”, e.g., by halving 𝜎𝜎𝑘𝑘.


The gain equation

ProcessControl

Laboratory


Complex poles and zerosIf there are complex poles or zeros, they occur as complex-conjugated pairs. If the corresponding complex-conjugated time constants are 𝑇𝑇𝑗𝑗 and 𝑇𝑇𝑗𝑗+1,

1 + (𝑇𝑇𝑗𝑗𝜔𝜔)2 1 + (𝑇𝑇𝑗𝑗+1𝜔𝜔)2 = 1 + 2 2𝜁𝜁2 − 1 𝜔𝜔𝜔𝜔n

2+ 𝜔𝜔

𝜔𝜔n

4(8.107)

is substituted into (8.103). In (8.106), the substitution(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2+ (𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2= 2𝜔𝜔𝑘𝑘

2[ 2𝜁𝜁2−1 𝜔𝜔n2+𝜔𝜔𝑘𝑘

2]𝜔𝜔n4+2 2𝜁𝜁2−1 𝜔𝜔n

2𝜔𝜔𝑘𝑘2+𝜔𝜔𝑘𝑘

4 (8.108)

is used. Since the exact value of 𝜎𝜎𝑘𝑘 is not very important, an approximation such as

(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑗𝑗𝜔𝜔𝑘𝑘)2+ (𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2

1+(𝑇𝑇𝑗𝑗+1𝜔𝜔𝑘𝑘)2≈ 2𝜔𝜔𝑘𝑘

4

𝜔𝜔n4+𝜔𝜔𝑘𝑘

4 (8.109)

might also be used.


The gain equation

ProcessControl

Laboratory


Exercise 8.3.2Calculate 𝐾𝐾c,max for the system below using frequency analysis.

𝐺𝐺m = 1𝑇𝑇+1

, 𝐺𝐺p = 15𝑇𝑇+1

, 𝐺𝐺v = 12𝑇𝑇+1

, 𝐺𝐺c = 𝐾𝐾c



ProcessControl

Laboratory


8.4 Controller design in the frequency domainIn this section it is shown how PI, PD and PID controllers can be designed to satisfy stability and performance criteria in the frequency domain.

The used stability criteria are the gain margin 𝐴𝐴m and the phase margin 𝜑𝜑m. Usually, 𝐴𝐴m ≈ 2 and 𝜑𝜑m ≈ 45° ( ⁄𝜋𝜋 4) are good values.

The cross-over frequency 𝜔𝜔g is related to performance — the higher the cross-over frequency, the better the performance. Usually, a good value is 𝜔𝜔g ≈ 0.3𝜔𝜔c , where 𝜔𝜔c is the critical frequency of the uncontrolled (or P-controlled) system.

8.4.1 Design of PI controllersA PI controller has the transfer function

𝐺𝐺PI 𝑠𝑠 = 𝐾𝐾c 1 + 1𝑇𝑇i𝑇𝑇

= 𝐾𝐾c𝑇𝑇i𝑇𝑇+1𝑇𝑇i𝑇𝑇

(8.110)

If the system to be controlled has the transfer function 𝐺𝐺(𝑠𝑠), the loop transfer function is

𝐺𝐺ℓ 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐺𝐺PI 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)𝐾𝐾c𝑇𝑇i𝑇𝑇+1𝑇𝑇i𝑇𝑇

(8.111)


ProcessControl

Laboratory

8.4 Controller design in the frequency domain

The amplitude ratio and the phase shift of the loop transfer function are

𝐴𝐴R,ℓ 𝜔𝜔 = 𝐺𝐺(j𝜔𝜔) 𝐾𝐾c𝑇𝑇i𝜔𝜔

1 + (𝑇𝑇i𝜔𝜔)2 (8.112)

𝜑𝜑ℓ 𝜔𝜔 = arg𝐺𝐺(j𝜔𝜔) + arctan𝑇𝑇i𝜔𝜔 − 𝜋𝜋2

(8.113)

Design for desired phase marginThe integral time 𝑇𝑇i ≈ ⁄5 𝜔𝜔g , where 𝜔𝜔g is the cross-over frequency, is usually a good choice for a PI controller. Based on this choice, a PI controller for a desired phase margin 𝜑𝜑m can be designed as follows.

1. Solve (8.113) for 𝜔𝜔 = 𝜔𝜔g with 𝜑𝜑ℓ = −𝜋𝜋 + 𝜑𝜑m and 𝑇𝑇i𝜔𝜔g = 5, i.e.,

𝜑𝜑m − 𝜋𝜋2− arctan 5 − arg𝐺𝐺 j𝜔𝜔g = 0 (8.114)

2. Solve (8.112) for 𝐾𝐾c with 𝐴𝐴R,ℓ = 1, 𝜔𝜔 = 𝜔𝜔g, and 𝑇𝑇i𝜔𝜔g = 5, i.e.,

𝐾𝐾c = 526

𝐺𝐺(j𝜔𝜔g) −1(8.115)

3. The integral time is 𝑇𝑇i = ⁄5 𝜔𝜔g.


8.4.1 PI controller

ProcessControl

Laboratory


Example 8.4.1Design a PI controller for a system with the transfer function

𝐺𝐺 𝑠𝑠 = e−𝑠𝑠

10𝑇𝑇+1to achieve the phase margin a) 𝜑𝜑m = 30°, b) 𝜑𝜑m = 60°. Also calculate controller tunings by the classical methods in Section 7.4 and 7.5.

a) Eq. (8.114) with the pertinent expression for arg𝐺𝐺 j𝜔𝜔g (see Section 8.1.6) has to be solved with 𝜑𝜑m = 30° = ⁄𝜋𝜋 6. This can be done iteratively according to (see Section 8.3.3)

𝜔𝜔𝑘𝑘+1 = 𝜔𝜔𝑘𝑘 − 𝜌𝜌𝑘𝑘𝑓𝑓(𝜔𝜔𝑘𝑘)

where 𝑓𝑓 𝜔𝜔𝑘𝑘 = ⁄−𝜋𝜋 3 − arctan 5 + 𝜔𝜔𝑘𝑘 + arctan 10𝜔𝜔𝑘𝑘

𝜌𝜌𝑘𝑘 = 1 + 101+100𝜔𝜔𝑘𝑘

2

−1

Using 𝜌𝜌0 = ⁄1 11 and 𝜔𝜔1 = ⁄7𝜋𝜋 (11 ⋅ 9) ≈ 0.22 as starting values yields 𝜌𝜌1 = 0.37, 𝜔𝜔2 = 0.61, 𝜌𝜌2 = 0.79, 𝜔𝜔3 = 0.93, 𝜌𝜌3 = 0.90, 𝜔𝜔4 = 0.9542 = 𝜔𝜔g.

The controller gain is 𝐾𝐾c = 5 1 + 10 ⋅ 0.9542 2/ 26 ≈ 9.41 and the integral time is 𝑇𝑇i = ⁄5 𝜔𝜔g = ⁄5 0.9542 = 5.24.


8.4.1 PI controller

ProcessControl

Laboratory

8.4.1 Design of PI controllers

b) 𝜑𝜑m = 60° = ⁄𝜋𝜋 3 means that𝑓𝑓 𝜔𝜔𝑘𝑘 = ⁄−𝜋𝜋 6 − arctan 5 + 𝜔𝜔𝑘𝑘 + arctan 10𝜔𝜔𝑘𝑘

A larger phase margin means a lower cross-over frequency. Based on a), we might therefore choose 𝜔𝜔0 = 0.9 as starting value for the iteration. This gives 𝜌𝜌0 = 0.9, which will be used throughout the calculation to get 𝜔𝜔1 = 0.48, 𝜔𝜔2 = 0.53, 𝜔𝜔3 = 0.514, 𝜔𝜔4 = 0.5179, 𝜔𝜔4 = 0.5170, 𝜔𝜔6 = 0.5172 = 𝜔𝜔g.

The controller gain is 𝐾𝐾c = 5 1 + 10 ⋅ 0.5172 2/ 26 ≈ 5.17 and the integral time is 𝑇𝑇i = ⁄5 𝜔𝜔g = ⁄5 0.5172 ≈ 9.67.

The result of all considered methods is summarized below.


Example 8.4.1

PIParam.

𝜑𝜑m30° Z-N ITAE

Regul.CHR 20%

Regul.CHR 0%Regul.

𝜑𝜑m60°

ITAETrack.

CHR 20%Track.

CHR 0%Track.

𝐾𝐾c 9.41 9.00 8.15 7.00 6.00 5.17 4.83 3.50 6.00

𝑇𝑇i 5.24 3.33 3.10 2.30 4.00 9.67 9.87 12.00 10.00

ProcessControl

Laboratory

8.4.1 Design of PI controllers

The figure shows simulated setpoint responses fora) 𝜑𝜑m = 30° (full line),b) 𝜑𝜑m = 60° (dashed line).


Example 8.4.1

t

y

ProcessControl

Laboratory


8.4.2 Design of PD controllersA PI controller can be designed to yield a desired phase margin, but it might require a low cross-over frequency, which means that the performance might not be good enough. A PD controller can be designed for a desired phase margin and a desired cross-over frequency.

A PD controller with a derivative filter has the transfer function

𝐺𝐺PDf 𝑠𝑠 = 𝐾𝐾c 1 + 𝑇𝑇d𝑇𝑇𝑇𝑇f𝑇𝑇+1

= 𝐾𝐾c(𝑇𝑇d+𝑇𝑇f)𝑇𝑇+1

𝑇𝑇f𝑇𝑇+1(8.116)

where 𝑇𝑇f is the filter time constant. The PD controller causes a phase shift

𝜑𝜑PDf = arctan(𝑇𝑇d+𝑇𝑇f)𝜔𝜔 − arctan𝑇𝑇f𝜔𝜔 = arctan 𝑇𝑇d𝜔𝜔1+(𝑇𝑇d+𝑇𝑇f)𝑇𝑇f𝜔𝜔2 (8.117)

For positive 𝑇𝑇d and 𝑇𝑇f, this phase shift is positive. The maximum phase lift is obtained at the frequency 𝜔𝜔 = 𝜔𝜔max, where

𝜔𝜔max = (𝑇𝑇d + 𝑇𝑇f)𝑇𝑇f −1/2 (8.118)This gives the phase lift

𝜑𝜑max = arctan(0.5𝑇𝑇d𝜔𝜔max) (8.119)


8.4 Design in the frequency domain

ProcessControl

Laboratory


Design for desired phase margin and cross-over frequencyIt is the derivative of the PD controller that produces the phase lift. The phase lift should be enough to satisfy the phase margin requirement, but excessive derivative action is not desired. This means that the maximum phase lift should be at the cross-over frequency, i.e., 𝜔𝜔max = 𝜔𝜔g.

If the system to be controlled has the transfer function 𝐺𝐺(𝑠𝑠), the loop transfer function is

𝐺𝐺ℓ 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐺𝐺PDf 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)𝐾𝐾c(𝑇𝑇d+𝑇𝑇f)𝑇𝑇+1

𝑇𝑇f𝑇𝑇+1(8.120)

With the maximum phase lift at the cross-over frequency, (8.118) and (8.119) with 𝜔𝜔max = 𝜔𝜔g apply. The amplitude ratio and phase shift equations at the cross-over frequency then become

𝐺𝐺(j𝜔𝜔g) 𝐾𝐾c𝑇𝑇f𝜔𝜔g

= 1 (8.121)

𝜑𝜑m − 𝜋𝜋 − arg𝐺𝐺(j𝜔𝜔g) − arctan(0.5𝑇𝑇d𝜔𝜔g) = 0 (8.122)


8.4.2 PD controller

ProcessControl

Laboratory

8.4.2 PD controller

A PD controller for desired phase margin and cross-over frequency can now be designed as follows.

1. Calculate the derivative time from (8.122), i.e.,

𝑇𝑇d = 2𝜔𝜔g

tan 𝜑𝜑m − 𝜋𝜋 − arg𝐺𝐺(j𝜔𝜔g) (8.123)

2. Calculate the derivative filter time constant from (8.118), i.e.,

𝑇𝑇f = −0.5𝑇𝑇d + 0.25𝑇𝑇d2 + 𝜔𝜔g−2 (8.124)

3. Calculate the controller gain from (8.121), i.e.,

𝐾𝐾c = 𝑇𝑇f𝜔𝜔g 𝐺𝐺(j𝜔𝜔g) −1(8.125)


Desired phase margin and cross-over frequency

ProcessControl

Laboratory


Example 8.4.2Design a PD controller with a derivative filter for a system with the transfer function


10𝑇𝑇+1

to achieve the phase margin 𝜑𝜑m = 60° and the cross-over frequency 1 rad/time unit.

Eq. (8.123) gives

𝑇𝑇d = 2𝜔𝜔g

tan 13𝜋𝜋 − 𝜋𝜋 + 𝜔𝜔g + arctan 10𝜔𝜔g

= 2tan −23𝜋𝜋 + 1 + arctan 10 = 0.79

Eq. (8.124) gives

𝑇𝑇f = −0.5𝑇𝑇d + 0.25𝑇𝑇d2 + 𝜔𝜔g−2

= −0.5 ⋅ 0.79 + 0.25 ⋅ 0.792 + 1 = 0.68Eq. (8.125) gives

𝐾𝐾c = 𝑇𝑇f𝜔𝜔g 1 + (10𝜔𝜔g)2= 0.68 101 = 6.83


8.4.2 PD controller

ProcessControl

Laboratory


8.4.3 Design of PID controllersA PD controller can be designed for stability (phase margin) and performance (cross-over frequency), but a drawback is that the steady-state control error will not be zero due to the lack of integral action. This can be remedied by connecting a PI and a PD controller in series. However, the previous PI and PD controller designs are not necessarily optimal for that purpose.

A PID controller on series form with a derivative filter has the transfer function

𝐺𝐺PIPDf 𝑠𝑠 = 𝐾𝐾c′ 1 + 1𝑇𝑇i′𝑇𝑇

1 + 𝑇𝑇d′𝑇𝑇

𝑇𝑇f′𝑇𝑇+1

= 𝐾𝐾c′(𝑇𝑇i

′𝑇𝑇+1)((𝑇𝑇d′+𝑇𝑇f

′)𝑇𝑇+1)𝑇𝑇i′𝑇𝑇(𝑇𝑇f

′𝑇𝑇+1) (8.126)

where “primed” parameter symbols are used to distinguish them from the corresponding parameters in the parallel form of the PID controller. The amplitude ratio and the phase shift equations of (8.126) are

𝐴𝐴R,PIPDf 𝜔𝜔 = 𝐾𝐾c′

𝑇𝑇i′𝜔𝜔

[1+(𝑇𝑇i′𝜔𝜔)2][1+ 𝑇𝑇d

′+𝑇𝑇f′ 2

𝜔𝜔2][1+(𝑇𝑇f

′𝜔𝜔)2]

1/2

(8.127)

𝜑𝜑PIPDf 𝜔𝜔 = −𝜋𝜋2

+ arctan𝑇𝑇i′𝜔𝜔 + arctan 𝑇𝑇d′ + 𝑇𝑇f′ 𝜔𝜔 − arctan𝑇𝑇f

′𝜔𝜔 (8.128)


8.4 Design in the frequency domain

ProcessControl

Laboratory


Design for desired phase margin and cross-over frequencyAs for the PD controller design, it is reasonable to choose the maximum phase lift of the PD part of the PID controller to occur at the cross-over frequency. From (8.118) it follows that this is achieved if

𝑇𝑇d′ + 𝑇𝑇f′ 𝑇𝑇f

′𝜔𝜔g2 = 1 (8.129)

Similarly to (8.121) and (8.122), when the integral part is added, the amplitude ratio and phase shift equations for the loop transfer function at the cross-over frequency become

𝐺𝐺(j𝜔𝜔g) 𝐾𝐾c′[1+(𝑇𝑇i′𝜔𝜔g)2]1/2

𝑇𝑇i′𝑇𝑇f

′𝜔𝜔g2 = 1 (8.130)

𝜑𝜑m − 0.5𝜋𝜋 − arg𝐺𝐺(j𝜔𝜔g) − arctan (𝑇𝑇i′+0.5𝑇𝑇d

′)𝜔𝜔g

1−0.5𝑇𝑇i′𝑇𝑇d

′𝜔𝜔g2 = 0 (8.131)

In (8.131), arctan𝑇𝑇i′𝜔𝜔g and arctan(0.5𝑇𝑇d𝜔𝜔g) have been combined. Solution of (8.131) for the integral and derivative times gives

(𝑇𝑇i′+0.5𝑇𝑇d

′)𝜔𝜔g

1−0.5𝑇𝑇i′𝑇𝑇d

′𝜔𝜔g2 = tan 𝜑𝜑m − 0.5𝜋𝜋 − arg𝐺𝐺(j𝜔𝜔g) (8.132)

The solution of 𝑇𝑇d′ from (8.132) depends on how 𝑇𝑇i′ is specified.


8.4.3 PID controller

ProcessControl

Laboratory


𝑇𝑇i′ or 𝑇𝑇i′𝜔𝜔g is knownIf 𝑇𝑇i′𝜔𝜔g is known, 𝑇𝑇i′𝜔𝜔g ≈ 5 is a typical value.A PID controller for desired phase margin and cross-over frequency can now be designed as follows.1. Calculate the derivative time from (8.132), which can be rewritten as

𝑇𝑇d′ = 2𝜔𝜔g

tan 𝜑𝜑m − 0.5𝜋𝜋 − arctan𝑇𝑇i′𝜔𝜔g − arg𝐺𝐺(j𝜔𝜔g) (8.133)

2. Calculate the derivative filter time constant from (8.129), i.e.,

𝑇𝑇f′ = −0.5𝑇𝑇d′ + 0.25𝑇𝑇d′2 + 𝜔𝜔g−2 (8.134)

3. Calculate the controller gain from (8.130), i.e.,

𝐾𝐾c′ = 𝑇𝑇i′𝑇𝑇f

′𝜔𝜔g2

[1+(𝑇𝑇i′𝜔𝜔g)2]1/2 𝐺𝐺(j𝜔𝜔g) −1

(8.135)

The ratio 𝑇𝑇i′/𝑇𝑇d′ is knownIf 𝑇𝑇i′/𝑇𝑇d′ is known, 𝑇𝑇i′/𝑇𝑇d′ ≈ 4 is a typical value. Eq. (8.133) is replaced by

𝑇𝑇d′ = 𝜔𝜔g−1 −𝑝𝑝 + 𝑝𝑝2 + 2𝑇𝑇d′

𝑇𝑇i′ , 𝑝𝑝 = 1+0.5𝑇𝑇d

′/𝑇𝑇i′

tan 𝜑𝜑m−0.5𝜋𝜋−arg 𝐺𝐺(j𝜔𝜔g)(8.136)


Desired phase margin and cross-over frequency

ProcessControl

Laboratory


Example 8.4.3Design a PID controller with a derivative filter for a system with the transfer function


10𝑇𝑇+1

to achieve the phase margin 𝜑𝜑m = 60° and the cross-over frequency 1 rad/time unit. Use 𝑇𝑇i′𝜔𝜔g = 5.

Eq. (8.133) gives

𝑇𝑇d′ = 2𝜔𝜔g

tan 13𝜋𝜋 − 1

2𝜋𝜋 − arctan 5 + 𝜔𝜔g + arctan 10𝜔𝜔g

= 2tan −16𝜋𝜋 − arctan 5 + 1 + arctan 10 = 1.29

Eq. (8.134) gives

𝑇𝑇f′ = −0.5𝑇𝑇d′ + 0.25𝑇𝑇d′2 + 𝜔𝜔g−2

= −0.5 ⋅ 1.29 + 0.25 ⋅ 1.792 + 1 = 0.54Eq. (8.135) gives

𝐾𝐾c′ = 𝑇𝑇i′𝑇𝑇f′𝜔𝜔g2

1+(10𝜔𝜔g)2

1+(𝑇𝑇i′𝜔𝜔g)2

= 5 ⋅ 0.54 10126

= 5.36



ProcessControl

Laboratory


Exercise 8.4.1Design a PID controller with a derivative filter for a system with the transfer function

𝐺𝐺 𝑠𝑠 = 4(4𝑇𝑇+1)3

to achieve the phase margin 𝜑𝜑m = 35° and the cross-over frequency 2 rad/time unit. Use 𝑇𝑇i′𝜔𝜔g = 5.



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8. Frequency Response Analysis - Åbo Akademi | Startsida · · 2017-03-308. Frequency Response Analysis ... Frequency Response Analysis 8.4 Controller design in the frequency domain