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Laboratory4. Laplace Transform Methods4.1 Linear systems

4.1.1 Linear differential equations4.1.2 Principle of superposition

4.2 The Laplace transform4.2.1 Definition4.2.2 Laplace transforms of common functions4.2.3 Properties of the Laplace transform

4.3 Modelling in the Laplace domain4.3.1 The transfer function4.3.2 Combination of systems4.3.3 Notational conventions

4.4 Applying the inverse Laplace transform4.4.1 Solving linear differential equations4.4.2 Partial fraction expansion

4.5 Table of Laplace transforms

KEH Process Dynamics and Control 4–1

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Laboratory

4. Laplace Transform Methods

4.1 Linear systemsIn practice, all real processes (systems) are nonlinear to some degree. However, there are several reasons why we want to study linear systems. It is often difficult to include the right nonlinearity in a model; a linear model

might be the best available approximation. A system operating close to a stationary operating point — as controlled

systems tend to do — often behaves as a linear system. There are powerful methods based on linear algebra and operator theory for

analysis, synthesis and design of linear systems.

As we have seen, differential equations give a mathematical description of continuous-time dynamical systems describe how a given variable, the output, depends on one or several other

variables, inputs

Linear DEs are therefore suitable for describing linear continuous-time systems mathematically.


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4.1.1 Linear differential equationsA linear ODE has the general form

𝑎𝑎0d𝑛𝑛𝑦𝑦d𝑡𝑡𝑛𝑛

+ 𝑎𝑎1d𝑛𝑛−1𝑦𝑦d𝑡𝑡𝑛𝑛−1

+ ⋯+ 𝑎𝑎𝑛𝑛−1d𝑦𝑦d𝑡𝑡

+ 𝑎𝑎𝑛𝑛𝑦𝑦

= 𝑏𝑏0d𝑚𝑚𝑢𝑢d𝑡𝑡𝑚𝑚

+ 𝑏𝑏1d𝑚𝑚−1𝑢𝑢d𝑡𝑡𝑚𝑚−1 + ⋯+ 𝑏𝑏𝑚𝑚−1

d𝑢𝑢d𝑡𝑡

+ 𝑏𝑏𝑚𝑚𝑢𝑢 (4.1)

𝑦𝑦 is the output from the system, 𝑢𝑢 is the input to the system. 𝑛𝑛, the order of the highest output derivative, is the system order. The system is proper if 𝑛𝑛 ≥ 𝑚𝑚, it is strictly proper if 𝑛𝑛 > 𝑚𝑚; physical systems

are practically always proper (but an ideal controller might be nonproper). The coefficients 𝑎𝑎0,𝑎𝑎1, … , 𝑎𝑎𝑛𝑛−1,𝑎𝑎𝑛𝑛, 𝑏𝑏0, 𝑏𝑏1, … , 𝑏𝑏𝑚𝑚−1, 𝑏𝑏𝑚𝑚 are system

parameters that completely characterize the properties of the system.

The system parameters can be rescaled (if desired) by multiplying (or dividing) them all by the same factor. Rescaling to get 𝑎𝑎0 = 1 is always possible (if 𝑎𝑎0 = 0, the system is not 𝑛𝑛th order) 𝑎𝑎𝑛𝑛 = 1 is possible if the static gain is nonzero (which usually applies)


4.1 Linear systems

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Laboratory

4.1 Linear systems

If the system is proper, we can always use 𝑛𝑛 as the order of the highest derivative of the input by letting the coefficients 𝑏𝑏0, … , 𝑏𝑏𝑛𝑛−𝑚𝑚−1 be zero in





= 𝑏𝑏0d𝑛𝑛𝑢𝑢d𝑡𝑡𝑛𝑛

+ ⋯+ 𝑏𝑏𝑛𝑛−𝑚𝑚−1d𝑚𝑚+1𝑢𝑢d𝑡𝑡𝑚𝑚+1 + 𝑏𝑏𝑛𝑛−𝑚𝑚

d𝑚𝑚𝑢𝑢d𝑡𝑡𝑚𝑚

+ ⋯+ 𝑏𝑏𝑛𝑛−1d𝑢𝑢d𝑡𝑡

+ 𝑏𝑏𝑛𝑛𝑢𝑢

Thus, we can without loss of generality write





= 𝑏𝑏0d𝑛𝑛𝑢𝑢d𝑡𝑡𝑛𝑛

+ 𝑏𝑏1d𝑛𝑛−1𝑢𝑢d𝑡𝑡𝑛𝑛−1

+ ⋯+ 𝑏𝑏𝑛𝑛−1d𝑢𝑢d𝑡𝑡

+ 𝑏𝑏𝑛𝑛𝑢𝑢 (4.2)

Note how the subscripts of the coefficients are related to the order of the corresponding time derivative. (This will be useful later.)


4.1.1 Linear differential equations

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4.1.2 Principle of superpositionFor linear systems, the principle of superposition applies. Assume that

𝑦𝑦1 𝑡𝑡 = 𝐹𝐹 𝑢𝑢1 𝑡𝑡 (4.3a)

𝑦𝑦2(𝑡𝑡) = 𝐹𝐹(𝑢𝑢2(𝑡𝑡)) (4.3b)

are two solutions to (4.2). According to the principle of superposition, the input

𝑢𝑢 𝑡𝑡 = 𝛼𝛼𝑢𝑢1 𝑡𝑡 + 𝛽𝛽𝑢𝑢2 𝑡𝑡 (4.4)

where 𝛼𝛼 and 𝛽𝛽 are arbitrary constants, gives the solution

𝑦𝑦 𝑡𝑡 = 𝛼𝛼𝑦𝑦1 𝑡𝑡 + 𝛽𝛽𝑦𝑦2 𝑡𝑡 (4.5)

For a linear system with more than one input, this implies that we can consider one input at a time. The combined effect of several inputs is then obtained by combining the respective outputs in the same way.

This is a reason why it is (usually) sufficient to include only one input in the DE (4.1) or (4.2).


4.1 Linear systems

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4.2 The Laplace transformIt is possible to solve the linear ODE (4.1) or (4.2) “analytically” using basic mathematics if the system parameters are constant the input 𝑢𝑢(𝑡𝑡) has a reasonable simple form

The full solution, i.e. the function 𝑦𝑦(𝑡𝑡), is obtained as the sum of a particular solution (any solution satisfying the DE), and the general solution to the corresponding autonomous DE (𝑢𝑢(𝑡𝑡) ≡ 0)

However, this way of solving DEs is cumbersome: the mathematics tend to be complicated for systems of high order there are no convenient short-cuts do deal with systems composed of simple

subsystems

The Laplace transform offers a practical way of solving linear DEs. Furthermore, it plays a fundamental role in analysis, synthesis, and design of linear systems. It is especially useful for systems with one output (and one input).


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4.2.1 DefinitionThe signals in dynamical systems are functions of time. Consider a fairly arbitrary function 𝑓𝑓(𝑡𝑡). For “technical reasons” concerning the Laplace transform we need to assume that 𝑓𝑓 𝑡𝑡 = 0 for 𝑡𝑡 < 0, 𝑓𝑓(𝑡𝑡) can be integrated for 𝑡𝑡 ≥ 0.

The Laplace transform 𝐹𝐹 𝑠𝑠 = ℒ 𝑓𝑓(𝑡𝑡) of such a time function 𝑓𝑓(𝑡𝑡) is defined by the integral

𝐹𝐹 𝑠𝑠 = ℒ 𝑓𝑓(𝑡𝑡) ≡ ∫0∞ e−𝑠𝑠𝑡𝑡𝑓𝑓 𝑡𝑡 d𝑡𝑡 (4.6)

where 𝑠𝑠 is a complex variable, whose real part Re(𝑠𝑠) has to be large enough for the integral to have a finite value. 𝑓𝑓(𝑡𝑡) is a function in the time domain. 𝐹𝐹(𝑠𝑠) is a function in the Laplace domain or 𝑠𝑠 domain. It is recommended to use small (lower-case) letters for functions in the time

domain and the corresponding large (upper-case) letter for a function in the Laplace domain.– This recommendation is not always obeyed; we might use the same letter

followed by the domain variable 𝑡𝑡 or 𝑠𝑠 as argument (e.g. 𝐹𝐹(𝑡𝑡) and 𝐹𝐹(𝑠𝑠)).


4.2 The Laplace transform

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The inverse Laplace transformIn order for the Laplace transform to be useful, it is necessary that we can also transform the other way, i.e. to calculate the time function 𝑓𝑓(𝑡𝑡) that corresponds to a given Laplace function 𝐹𝐹(𝑠𝑠).

Formally, this can be done by means of the integral formula

𝑓𝑓 𝑡𝑡 = ℒ−1 𝐹𝐹(𝑠𝑠) ≡ 12𝜋𝜋j ∫𝜎𝜎−j∞

𝜎𝜎+j∞ e𝑠𝑠𝑡𝑡𝐹𝐹 𝑠𝑠 d𝑠𝑠 , 𝑡𝑡 ≥ 0 (4.7)

where j = −1 is the imaginary unit and 𝜎𝜎 is a real number large enough so that 𝐹𝐹(𝑠𝑠) is finite for all Re 𝑠𝑠 > 𝜎𝜎.

In practical calculations (4.7) is not needed (4.6) is seldom needed

because tabulated Laplace transform/time domain function pairs can be used(see Laplace Transform Table, LTT, Section 4.5). However, there is a need to apply the superposition principle partial fraction expansion (section 4.4.2)

to be able to use tabulated function relationships.


4.2.1 Definition

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4.2.2 Laplace transforms of common functionsA rectangular pulse

𝑓𝑓(𝑡𝑡)𝑎𝑎

0 𝑇𝑇 𝑡𝑡Fig. 4.1. A rectangular pulse.

𝐹𝐹 𝑠𝑠 = ∫0𝑇𝑇 e−𝑠𝑠𝑡𝑡𝑎𝑎 d𝑡𝑡 = 𝑎𝑎 − 1

𝑠𝑠e−𝑠𝑠𝑡𝑡

0

𝑇𝑇= 𝑎𝑎 1−e−𝑠𝑠𝑠𝑠

𝑠𝑠(4.8)

A unit pulse has the area 1, i.e. 𝑎𝑎𝑇𝑇 = 1. The Laplace transform of the unit pulse is thus

𝐹𝐹 𝑠𝑠 = 1−e−𝑠𝑠𝑠𝑠

𝑠𝑠𝑇𝑇(4.9)



A rectangular pulse has a constant amplitude (height) 𝑎𝑎 given pulse length 𝑇𝑇

It is assumed to start at 𝑡𝑡 = 0.

By means of the Laplace transformdefinition (4.6) we can derive

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Laboratory


The unit impulse — Dirac’s delta functionAn impulse is defined as a (rectangular) pulse, whose pulse length 𝑇𝑇 → 0 𝛿𝛿(𝑡𝑡) amplitude 𝑎𝑎 → ∞ 𝑎𝑎𝑇𝑇 is finite

For the unit impulse, denoted 𝛿𝛿(𝑡𝑡), 0 𝑡𝑡 𝑎𝑎𝑇𝑇 = 1 (with some physical unit according to the application)

The Laplace transform of the unit impulse can be obtained by letting 𝑇𝑇 → 0 in the Laplace transform of the unit pulse. A Taylor series expansion then gives

ℒ 𝛿𝛿(𝑡𝑡) = lim𝑇𝑇→0

1−e−𝑠𝑠𝑠𝑠

𝑠𝑠𝑇𝑇= lim

𝑇𝑇→0

𝑠𝑠𝑇𝑇−12 𝑠𝑠𝑇𝑇2+⋯

𝑠𝑠𝑇𝑇= lim

𝑇𝑇→01 − 1

2(𝑠𝑠𝑇𝑇) = 1

ℒ 𝛿𝛿(𝑡𝑡) = 1 (4.10)

Impulses are important in many practical applications. Input signals of very short duration can usually be approximated by impulses. E.g.– voltage and current pulses in electrical systems– sudden forces in mechanical systems, e.g. wind gusts– injection of tracers in medical and process technical applications


4.2.2 Laplace transforms of common functions

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A unit stepA step function can be considered a rectangular pulse with infinite pulse length 𝑇𝑇

For a unit step, denoted 𝜎𝜎(𝑡𝑡), 𝜎𝜎 𝑡𝑡 = 1 𝑎𝑎 = 1 (with some physical unit) 1

The Laplace transform of a unit step can be derivedby letting 𝑇𝑇 → ∞ in the Laplace transform of a pulse 0 𝑡𝑡with 𝑎𝑎 = 1. A Taylor series expansion gives

ℒ 𝜎𝜎(𝑡𝑡) = lim𝑇𝑇→∞

1−e−𝑠𝑠𝑠𝑠

𝑠𝑠= 1

𝑠𝑠(4.11)

A unit rampA ramp function is a function whose value changes 𝜌𝜌 𝑡𝑡 = 𝑡𝑡linearly with time. For a unit ramp, denoted 𝜌𝜌(𝑡𝑡), the slope coefficient is 1, i.e. 𝜌𝜌 𝑡𝑡 = 𝑡𝑡, 𝑡𝑡 ≥ 0.

The Laplace transform of the unit ramp can be derivedfrom the definition (4.6). Integration by parts gives 0 𝑡𝑡

ℒ 𝜌𝜌(𝑡𝑡) = 1𝑠𝑠2

(4.12)



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Relationships between the simple unit functionsConsider the unit impulse, unit step, and unit ramp in Fig. 4.2. the impulse is the time derivative of the step function the step function is the time derivate of the ramp function

Mathematically:𝛿𝛿 𝑡𝑡 = d

d𝑡𝑡𝜎𝜎(𝑡𝑡) , 𝜎𝜎 𝑡𝑡 = d

d𝑡𝑡𝜌𝜌(𝑡𝑡)

In the Laplace domain we have derived the relationships:

ℒ 𝛿𝛿 𝑡𝑡 = 1 = 𝑠𝑠 1𝑠𝑠

= 𝑠𝑠ℒ 𝜎𝜎(𝑡𝑡) , ℒ 𝜎𝜎 𝑡𝑡 = 1𝑠𝑠

= 𝑠𝑠 1𝑠𝑠2

= 𝑠𝑠ℒ 𝜌𝜌(𝑡𝑡)This means that a time derivative corresponds to multiplication by 𝑠𝑠 in the ℒ-domain thus, a time integral corresponds to division by 𝑠𝑠 in the ℒ-domain

𝜌𝜌 𝑡𝑡 = 𝑡𝑡𝜎𝜎 𝑡𝑡 = 1

𝛿𝛿 𝑡𝑡 = 1

0 𝑡𝑡 0 𝑡𝑡 0 𝑡𝑡Fig. 4.2. A unit impulse, unit step, and unit ramp.



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Exponential functionAn exponential function is defined 𝑓𝑓 𝑡𝑡 = e−𝑎𝑎𝑡𝑡, 𝑡𝑡 ≥ 0. If 𝑎𝑎 > 0, the function is exponentially decaying 𝑎𝑎 < 0, the function is exponentially increasing

The Laplace transform can be derived from the definition (4.6)

𝐹𝐹 𝑠𝑠 = ℒ e−𝑎𝑎𝑡𝑡 = ∫0∞ e−𝑠𝑠𝑡𝑡e−𝑎𝑎𝑡𝑡 d𝑡𝑡

= ∫0∞ e−(𝑠𝑠+𝑎𝑎)𝑡𝑡 d𝑡𝑡 = − 1

𝑠𝑠+𝑎𝑎e−(𝑠𝑠+𝑎𝑎)𝑡𝑡

0

∞

⇒ 𝐹𝐹 𝑠𝑠 = ℒ e−𝑎𝑎𝑡𝑡 = 1𝑠𝑠+𝑎𝑎

(4.13)

It is not obvious from the final result, but the integral exists (is finite) if and only if Re 𝑠𝑠 + 𝑎𝑎 > 0

This is a limiting restriction mainly if 𝑎𝑎 < 0 , i.e., if the function is exponentially increasing. Such a function is unstable, because it approaches infinity as time goes towards infinity.



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Sine and cosine functionsThe sine function (with amplitude one) is defined 𝑓𝑓 𝑡𝑡 = sin(ω𝑡𝑡), 𝑡𝑡 ≥ 0, where 𝜔𝜔 is the oscillation frequency in radians per time unit.

The Laplace transform can be derived using integration by parts, defined

∫𝑡𝑡1𝑡𝑡2 𝑔𝑔′ 𝑡𝑡 𝑓𝑓 𝑡𝑡 d𝑡𝑡 = 𝑔𝑔 𝑡𝑡 𝑓𝑓(𝑡𝑡) 𝑡𝑡1

𝑡𝑡2 − ∫𝑡𝑡1𝑡𝑡2 𝑔𝑔 𝑡𝑡 𝑓𝑓′ 𝑡𝑡 d𝑡𝑡

By applying partial integration twice to the definition (4.6), we can derive

ℒ sin(𝜔𝜔𝑡𝑡) = 𝜔𝜔𝑠𝑠2+𝜔𝜔2 (4.14)

The cosine function (amplitude one) is defined 𝑓𝑓 𝑡𝑡 = cos(ω𝑡𝑡), 𝑡𝑡 ≥ 0. Similarly as above, we can derive

ℒ cos(𝜔𝜔𝑡𝑡) = 𝑠𝑠𝑠𝑠2+𝜔𝜔2 (4.15)

We could have derived (4.15) more easily by observing that

cos 𝜔𝜔𝑡𝑡 = 𝜔𝜔−1 dd𝑡𝑡

sin(𝜔𝜔𝑡𝑡)and using

ℒ cos 𝜔𝜔𝑡𝑡 = 𝜔𝜔−1𝑠𝑠ℒ sin 𝜔𝜔𝑡𝑡



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4.2.3 Properties of the Laplace transformSuperpositionIf ℒ 𝑓𝑓1(𝑡𝑡) = 𝐹𝐹1(𝑠𝑠) and ℒ 𝑓𝑓2(𝑡𝑡) = 𝐹𝐹2(𝑠𝑠)

Then ℒ 𝛼𝛼𝑓𝑓1 𝑡𝑡 + 𝛽𝛽𝑓𝑓2(𝑡𝑡) = 𝛼𝛼𝐹𝐹1 𝑠𝑠 + β𝐹𝐹2(𝑠𝑠) (4.16)

𝛼𝛼𝑓𝑓1 𝑡𝑡 + 𝛽𝛽𝑓𝑓2 𝑡𝑡 = ℒ−1 𝛼𝛼𝐹𝐹1 𝑠𝑠 + β𝐹𝐹2(𝑠𝑠) (4.17)

where 𝛼𝛼 and 𝛽𝛽 are arbitrary constants.

This means that we can easily calculate the Laplace transform of the linear combination 𝑓𝑓 𝑡𝑡 = 𝛼𝛼𝑓𝑓1 𝑡𝑡 + 𝛽𝛽𝑓𝑓2(𝑡𝑡)

according to (4.16) inverse Laplace transform of 𝐹𝐹 𝑠𝑠 = 𝛼𝛼𝐹𝐹1 𝑠𝑠 + 𝛽𝛽𝐹𝐹2(𝑠𝑠) according to (4.17)

if we know the Laplace transforms 𝐹𝐹1(𝑠𝑠) and 𝐹𝐹2(𝑠𝑠) of (the simpler functions) 𝑓𝑓1(𝑡𝑡) and 𝑓𝑓2(𝑡𝑡), respectively.



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DerivativesIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), then the Laplace transform of the first-order time derivative𝑓𝑓 ≡ d𝑓𝑓

d𝑡𝑡is given by

ℒ 𝑓𝑓(𝑡𝑡) = 𝑠𝑠𝐹𝐹 𝑠𝑠 − 𝑓𝑓(0−) (4.18)

where 𝑓𝑓 0− = lim𝑡𝑡→0−

𝑓𝑓(𝑡𝑡)

i.e. 𝑓𝑓 0− is the initial value of the time domain function 𝑓𝑓(𝑡𝑡) when 𝑡𝑡approaches 0 from the negative side. Negative or positive side makes adifference if 𝑓𝑓(𝑡𝑡) has a discontinuity at 𝑡𝑡 = 0 (e.g. a step function).

The Laplace transform of an 𝑛𝑛:th order derivative 𝑓𝑓(𝑛𝑛) ≡ d𝑛𝑛𝑓𝑓d𝑡𝑡𝑛𝑛

is

ℒ 𝑓𝑓 𝑛𝑛 (𝑡𝑡) = 𝑠𝑠𝑛𝑛𝐹𝐹 𝑠𝑠 − 𝑠𝑠𝑛𝑛−1𝑓𝑓 0− − 𝑠𝑠𝑛𝑛−2 𝑓𝑓 0− − ⋯−𝑠𝑠𝑓𝑓(𝑛𝑛−2) 0− − 𝑓𝑓(𝑛𝑛−1) 0− (4.19)

where 𝑓𝑓 0− , 𝑓𝑓 0− , … are the values of 𝑓𝑓(𝑡𝑡), 𝑓𝑓 𝑡𝑡 , … when 𝑡𝑡 → 0−.

Note that all initial values are 0 for systems, where the variables are deviations from a steady-state operating point.


4.2.3 Properties of the Laplace transform

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Laboratory


IntegralsIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), the Laplace transform of the time integral of 𝑓𝑓(𝑡𝑡) is given by

ℒ ∫0−𝑡𝑡 𝑓𝑓 𝜏𝜏 d𝜏𝜏 = 1

𝑠𝑠𝐹𝐹 𝑠𝑠 (4.20)

For a multiple integral, consisting of 𝑛𝑛 integrals, the Laplace transform is

ℒ ∫0−𝑡𝑡 ∫0−

𝑡𝑡 ⋯∫0−𝑡𝑡 𝑓𝑓 𝜏𝜏 d𝜏𝜏𝑛𝑛 = 1

𝑠𝑠𝑛𝑛𝐹𝐹 𝑠𝑠 (4.21)

Note that initial values are not an issue in this case.

DampingIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), the Laplace transform of an exponentially damped function e−𝑎𝑎𝑡𝑡𝑓𝑓(𝑡𝑡) is given by

ℒ e−𝑎𝑎𝑡𝑡𝑓𝑓(𝑡𝑡) = 𝐹𝐹 𝑠𝑠 + 𝑎𝑎 (4.22)

Note that it is required that Re 𝑠𝑠 + 𝑎𝑎 > 0 , which is a limiting restriction if 𝑎𝑎 <0. In practice, 𝑎𝑎 > 0 (otherwise there is no damping).



ProcessControl

Laboratory


Time delayIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), the Laplace transform of the function 𝑓𝑓(𝑡𝑡 − 𝐿𝐿) , i.e. the function 𝑓𝑓(𝑡𝑡) delayed by 𝐿𝐿 time units, is given by

ℒ 𝑓𝑓(𝑡𝑡 − 𝐿𝐿) = e−𝐿𝐿𝑠𝑠𝐹𝐹 𝑠𝑠 (4.23)

Fig. 4.3. Undelayed and delayed time function.



( )f t ( )f t L− L

t t

ProcessControl

Laboratory


Limit theoremsIn general small values of the time variable 𝑡𝑡 correspond to large values of the Laplace

variable 𝑠𝑠 and vice versa

However, there is no exact relationship between time and the Laplace variable, except for two limit conditions. They are useful for calculating the initial and final steady-state values of a time function whose Laplace transform is known without doing an inverse Laplace transformation.

The initial value theoremIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), where 𝐹𝐹(𝑠𝑠) is strictly proper,

lim𝑡𝑡→0+

𝑓𝑓 𝑡𝑡 = lim𝑠𝑠→∞

𝑠𝑠𝐹𝐹(𝑠𝑠) (4.24)

Here 𝑡𝑡 → 0+ means that 𝑡𝑡 approaches zero from the positive side of 𝑡𝑡.

The final value theoremIf ℒ 𝑓𝑓(𝑡𝑡) = 𝐹𝐹(𝑠𝑠), where s𝐹𝐹(𝑠𝑠) is bounded for all Re(𝑠𝑠) ≥ 0,

lim𝑡𝑡→∞

𝑓𝑓 𝑡𝑡 = lim𝑠𝑠→0

𝑠𝑠𝐹𝐹(𝑠𝑠) (4.25)



ProcessControl

Laboratory


Exercise 4.1Determine the Laplace transform of the function

𝑓𝑓 𝑡𝑡 = 6 + 8e−𝑡𝑡 − 5e−2𝑡𝑡

Validate the result by means of the initial and final value theorems.

Exercise 4.2Determine the time function whose Laplace transform is

𝐹𝐹 𝑠𝑠 = 2.4e−0.8𝑠𝑠

2𝑠𝑠+3.6

Validate the result by means of the initial and final value theorems.

Exercise 4.3Determine the Laplace transform of the 𝑓𝑓(𝑡𝑡)delayed saw-tooth pulse in the figure. 3

0 3 5 𝑡𝑡



ProcessControl

Laboratory


4.3 Modelling in the Laplace domain

Assume that the DE (4.1) is satisfied by the variable values

�𝑦𝑦(𝑛𝑛), �𝑦𝑦(𝑛𝑛−1), … , �𝑦𝑦, �𝑦𝑦, �𝑢𝑢 𝑚𝑚 , �𝑢𝑢 𝑚𝑚−1 , … , �𝑢𝑢, �𝑢𝑢

i.e., this is a solution to the DE. If this solution is of particular interest, we can call it a reference state or an

operating point. This point is often a steady state or an equilibrium point, where all time

derivatives are zero.

Any solution 𝑦𝑦(𝑛𝑛),𝑦𝑦(𝑛𝑛−1), … , ��𝑦,𝑦𝑦,𝑢𝑢 𝑚𝑚 ,𝑢𝑢 𝑚𝑚−1 , … , ��𝑢,𝑢𝑢 to the DE can be related to the reference state by

𝑦𝑦(𝑛𝑛) = �𝑦𝑦(𝑛𝑛) + ∆𝑦𝑦(𝑛𝑛), … , ��𝑦 = �𝑦𝑦 + ∆��𝑦, 𝑦𝑦 = �𝑦𝑦 + ∆𝑦𝑦 (4.26a)

𝑢𝑢(𝑚𝑚) = �𝑢𝑢(𝑚𝑚) + ∆𝑢𝑢(𝑚𝑚), … , ��𝑢 = �𝑢𝑢 + ∆��𝑢, 𝑢𝑢 = �𝑢𝑢 + ∆𝑢𝑢 (4.26b)

where the ∆-variables indicate deviations from the reference state.


4.3.1 The transfer function

ProcessControl

Laboratory


Substitution of (4.26) into the DE (4.1), cancelling out of the reference state (which satisfies the DE separately), and the choice 𝑎𝑎0 = 1, yield

d𝑛𝑛∆𝑦𝑦d𝑡𝑡𝑛𝑛

+ 𝑎𝑎1d𝑛𝑛−1∆𝑦𝑦d𝑡𝑡𝑛𝑛−1

+ ⋯+ 𝑎𝑎𝑛𝑛−1d∆𝑦𝑦d𝑡𝑡

+ 𝑎𝑎𝑛𝑛∆𝑦𝑦

= 𝑏𝑏0d𝑚𝑚∆𝑢𝑢d𝑡𝑡𝑚𝑚

+ 𝑏𝑏1d𝑚𝑚−1∆𝑢𝑢d𝑡𝑡𝑚𝑚−1 + ⋯+ 𝑏𝑏𝑚𝑚−1

d∆𝑢𝑢d𝑡𝑡

+ 𝑏𝑏𝑚𝑚∆𝑢𝑢 (4.27)

Considering that all initial values of the ∆-variables are zero, the Laplace transform of (4.27) gives,

𝑠𝑠𝑛𝑛∆𝑌𝑌 𝑠𝑠 + 𝑎𝑎1𝑠𝑠𝑛𝑛−1∆𝑌𝑌 𝑠𝑠 + ⋯+ 𝑎𝑎𝑛𝑛−1𝑠𝑠∆𝑌𝑌 𝑠𝑠 + 𝑎𝑎𝑛𝑛∆𝑌𝑌(𝑠𝑠) (4.28)

= 𝑏𝑏0𝑠𝑠𝑚𝑚∆𝑈𝑈 𝑠𝑠 + 𝑏𝑏1𝑠𝑠𝑚𝑚−1∆𝑈𝑈 𝑠𝑠 + ⋯+ 𝑏𝑏𝑚𝑚−1𝑠𝑠∆𝑈𝑈 𝑠𝑠 + 𝑏𝑏𝑚𝑚∆𝑈𝑈(𝑠𝑠)

where ∆𝑌𝑌(𝑠𝑠) and ∆𝑈𝑈(𝑠𝑠) are the Laplace transforms of ∆𝑦𝑦(𝑡𝑡) and ∆𝑢𝑢(𝑡𝑡), respectively.

Thus, the Laplace transform of an 𝑛𝑛th order derivative yields a factor 𝑠𝑠𝑛𝑛 when the initial state is zero.



ProcessControl

Laboratory


Equation (4.28) can also be written

𝑠𝑠𝑛𝑛 + 𝑎𝑎1𝑠𝑠𝑛𝑛−1 + ⋯+ 𝑎𝑎𝑛𝑛−1𝑠𝑠 + 𝑎𝑎𝑛𝑛 ∆𝑌𝑌(𝑠𝑠)= 𝑏𝑏0𝑠𝑠𝑚𝑚 + 𝑏𝑏1𝑠𝑠𝑚𝑚−1 + ⋯+ 𝑏𝑏𝑚𝑚−1𝑠𝑠 + 𝑏𝑏𝑚𝑚 ∆𝑈𝑈(𝑠𝑠)

or more compactly ∆𝑌𝑌 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)∆𝑈𝑈(𝑠𝑠) (4.29)

where

𝐺𝐺 𝑠𝑠 = 𝑏𝑏0𝑠𝑠𝑚𝑚+𝑏𝑏1𝑠𝑠𝑚𝑚−1+⋯+𝑏𝑏𝑚𝑚−1𝑠𝑠+𝑏𝑏𝑚𝑚𝑠𝑠𝑛𝑛+𝑎𝑎1𝑠𝑠𝑛𝑛−1+⋯+𝑎𝑎𝑛𝑛−1𝑠𝑠+𝑎𝑎𝑛𝑛

≡ 𝐵𝐵(𝑠𝑠)𝐴𝐴(𝑠𝑠)

(4.30)

is the transfer function of the system described by the DE (4.27).

As (4.29) shows, the output defined in the Laplace domain is obtained when the input, also defined in the Laplace domain, is multiplied by the transfer function of the system in question.

Thus, when working in the Laplace domain, the solution of a DE (i.e., finding the output for a given input) can be obtained by algebraic operations only.



ProcessControl

Laboratory


In (4.30) , 𝐴𝐴(𝑠𝑠) is the denominator polynomial and 𝐵𝐵(𝑠𝑠) is the numerator polynomial of the transfer function. 𝐴𝐴 𝑠𝑠 = 0 is the characteristic equation of the system. The roots (i.e., solutions) of 𝐴𝐴 𝑠𝑠 = 0 are called system poles. The roots of 𝐵𝐵 𝑠𝑠 = 0 are called system zeros.

The importance of poles and zeros is treated in Ch. 5 and 6.

If the system contains a pure time delay, also called a dead time, it takes some time before an input begins to affect an output. If the input signal is 𝑣𝑣, and there is a time delay of 𝐿𝐿 time units, 𝑣𝑣(𝑡𝑡 − 𝐿𝐿) takes the place of 𝑢𝑢(𝑡𝑡) in (4.27). By using the substitution 𝑢𝑢 𝑡𝑡 = 𝑣𝑣(𝑡𝑡 − 𝐿𝐿) in (4.27) , (4.29) and (4.30) are obtained.

The use of ∆-variables and the Laplace transform gives

∆𝑈𝑈 𝑠𝑠 = e−𝐿𝐿𝑠𝑠∆𝑉𝑉(𝑠𝑠) (4.31)

where ∆𝑉𝑉(𝑠𝑠) is the Laplace transform of ∆𝑣𝑣(𝑡𝑡). The transfer function (TF) from ∆𝑉𝑉(𝑠𝑠) to ∆𝑌𝑌(𝑠𝑠) is then 𝐺𝐺(𝑠𝑠)e−𝐿𝐿𝑠𝑠.

To calculate the TF of a system with a time delay, we can first calculate 𝐺𝐺(𝑠𝑠)assuming no time delay, then include the time delay as 𝐺𝐺(𝑠𝑠)e−𝐿𝐿𝑠𝑠.



ProcessControl

Laboratory


Example 4.1. Transfer function of a first-order system.A mercury thermometer can approximately be described by the DE

𝑇𝑇 d𝜗𝜗2d𝑡𝑡

+ 𝜗𝜗2 = 𝜗𝜗1 (1)

where 𝜗𝜗1 is the temperature outside the thermometer and𝜗𝜗2 is the temperature of mercury inside the thermometer.The time constant 𝑇𝑇 of the thermometer is ≈ 2 min.We introduce ∆-variables to denote deviations from anequilibrium point 𝜗𝜗1 = ��𝜗1 and 𝜗𝜗2 = ��𝜗2, i.e.,

𝜗𝜗1 = ��𝜗1 + ∆𝜗𝜗1 , 𝜗𝜗2 = ��𝜗2 + ∆𝜗𝜗2 (2)

where ∆𝜗𝜗1 and ∆𝜗𝜗2 denote the size of the deviations. Substitution into (1) yields

𝑇𝑇 d �𝜗𝜗2+∆𝜗𝜗2d𝑡𝑡

+ ��𝜗2 + ∆𝜗𝜗2 = ��𝜗1 + ∆𝜗𝜗1 (3)

In the equilibrium point it must apply that 𝑇𝑇 𝑑𝑑�𝜗𝜗2𝑑𝑑𝑡𝑡

+ ��𝜗2 = ��𝜗1 (even more strongly, ��𝜗2 = ��𝜗1 and ⁄d��𝜗2 d𝑡𝑡 = 0). Combination with (3) gives

𝑇𝑇 d∆𝜗𝜗2(𝑡𝑡)d𝑡𝑡

+ ∆𝜗𝜗2(𝑡𝑡) = ∆𝜗𝜗1(𝑡𝑡) (4)

where the time argument 𝑡𝑡 is included for clarity.



ProcessControl

Laboratory


The Laplace transform of (4) yields

𝑇𝑇 𝑠𝑠∆Θ2 𝑠𝑠 − ∆𝜗𝜗2(0−) + ∆Θ2 𝑠𝑠 = ∆Θ1(𝑠𝑠) (5)

where ∆Θ1(𝑠𝑠) and ∆Θ2(𝑠𝑠) are the Laplace transforms of ∆𝜗𝜗1(𝑡𝑡) and ∆𝜗𝜗2(𝑡𝑡), respectively.

The initial value ∆𝜗𝜗2(0−) is the value of ∆𝜗𝜗2 when the 𝑡𝑡 approaches zero from the negative side. Since we use ∆-variables that indicate deviations from the initial state, ∆ϑ2(0ˉ) = 0. Equation (5) the reduces to

𝑇𝑇𝑠𝑠∆Θ2 𝑠𝑠 + ∆Θ2 𝑠𝑠 = ∆Θ1(𝑠𝑠) (6)or

∆Θ2 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)∆Θ1(𝑠𝑠) (7)where

𝐺𝐺 𝑠𝑠 = 1𝑇𝑇𝑠𝑠+1

(8)

is the transfer function of the mercury thermometer.


Ex. 4.1. First-order transfer function

ProcessControl

Laboratory


Exercise 4.4A system is described by the differential equation

d2𝑦𝑦d𝑡𝑡2

+ 5 d𝑦𝑦d𝑡𝑡

+ 6𝑦𝑦 = 𝑢𝑢

where 𝑢𝑢 and 𝑦𝑦 indicate deviations from an equilibrium point. Determine the transfer function of the system.



ProcessControl

Laboratory


4.3.2 Combination of systemsBlock diagrams, and especially combinations of signals in block diagrams, were introduced in Section 2.2. Here some typical block connections are considered.

Series connectionSeries (or cascade) connection is a common type of block interconnection. According to (4.29), the connection yields

𝑌𝑌 𝑠𝑠 = 𝐺𝐺2 𝑠𝑠 𝑋𝑋 𝑠𝑠 = 𝐺𝐺2 𝑠𝑠 𝐺𝐺1 𝑠𝑠 𝑈𝑈(𝑠𝑠)

from which it follows that𝐺𝐺 𝑠𝑠 = 𝐺𝐺2(𝑠𝑠)𝐺𝐺1(𝑠𝑠) (4.32)

is the transfer function for a series connection of two systems. Obviously, this can be extended to more than two systems in series.

Note that this requires that 𝐺𝐺1(𝑠𝑠)and 𝐺𝐺2(𝑠𝑠) are not changed bythe interconnection, i.e., that thesystems are “non-interacting”.

Fig. 4.4. Series connection.



ProcessControl

Laboratory

4.3.2 Combination of systems

Example 4.2. Non-interacting tanks.The liquid tank in the figure can be described bythe model (see Example 3.5)

𝐴𝐴 dℎ(𝑡𝑡)d𝑡𝑡

= 𝐹𝐹0(𝑡𝑡) − 𝐹𝐹1(𝑡𝑡), 𝐹𝐹1(𝑡𝑡) = 𝛽𝛽 ℎ(𝑡𝑡)

where 𝐹𝐹1 is a free outflow governed by gravityand the liquid level ℎ . 𝛽𝛽 and 𝐴𝐴 are constants. Linearization at ℎ = �ℎ and replacement of ℎ by 𝐹𝐹1 as dependent variable yields the transfer function∆𝐹𝐹1(𝑠𝑠)∆𝐹𝐹0(𝑠𝑠)

= 𝐺𝐺 𝑠𝑠 = 1𝑇𝑇𝑠𝑠+1

, 𝑇𝑇 = 2𝐴𝐴 �ℎ𝛽𝛽

.

The two tanks connected in serieshave the transfer functions𝐺𝐺1(𝑠𝑠) = 1

𝑇𝑇1𝑠𝑠+1and 𝐺𝐺2 𝑠𝑠 = 1

𝑇𝑇2𝑠𝑠+1

with 𝑇𝑇1 = 2/𝛽𝛽 𝐴𝐴1 �ℎ1 and 𝑇𝑇2 = 2/𝛽𝛽 𝐴𝐴2 �ℎ2 .The transfer function from 𝐹𝐹0 to 𝐹𝐹2 is given by (4.32) as

∆𝐹𝐹2(𝑠𝑠)∆𝐹𝐹0(𝑠𝑠)

= 𝐺𝐺 𝑠𝑠 = 𝐺𝐺2 𝑠𝑠 𝐺𝐺1 𝑠𝑠 = 1(𝑇𝑇1𝑠𝑠+1)(𝑇𝑇2𝑠𝑠+1)

.


Series connection

0F

1Fh

1F

2F2h

0F

1h

ProcessControl

Laboratory


Interacting systemsIn the modelling of series-connected systems, it is important to know if they are interacting or non-interacting. A series connection is non-interacting, if each subsystem is only affected by previous subsystems in

the series interacting, if a subsystem is affected by a subsequent subsystem in the series

The transfer function from the input to the first subsystem to the output from the last subsystem in a series connection is the product of the transfer functions of the individual subsystems if the series

connection is non-interacting, Eq. (4.32); not directly obtained from the transfer functions of the individual subsystems

if the series connection is interacting; the complete series connection has to be treated as a whole.


Series connection

ProcessControl

Laboratory

Series connection

Example 4.3. Interacting tanks.The liquid-tank series in the figure isinteracting because 𝐹𝐹1 (the outflowfrom tank 1) is affected not only byℎ1 but also by ℎ2 (a variable in tank 2).Tank 2 is described by the model

𝐴𝐴2dℎ2(𝑡𝑡)d𝑡𝑡

= 𝐹𝐹1 𝑡𝑡 − 𝐹𝐹2 𝑡𝑡 , 𝐹𝐹2(𝑡𝑡) = 𝛽𝛽 ℎ2(𝑡𝑡)yielding the transfer function

∆𝐹𝐹2(𝑠𝑠)∆𝐹𝐹1(𝑠𝑠)

= 𝐺𝐺2 𝑠𝑠 = 1𝑇𝑇2𝑠𝑠+1

, 𝑇𝑇2 = 2𝐴𝐴2 �ℎ2𝛽𝛽

as for non-interacting tanks. However, because 𝐹𝐹1(𝑡𝑡) = 𝛽𝛽 ℎ1 𝑡𝑡 − ℎ2(𝑡𝑡)(assuming ℎ1 ≥ ℎ2), 𝐺𝐺1(𝑠𝑠) is different from the non-interacting case. Linearization and elimination ℎ1 and ℎ2 from the model for tank 1 yields

𝑇𝑇1d∆𝐹𝐹1(𝑡𝑡)d𝑡𝑡

+ ∆𝐹𝐹1 𝑡𝑡 = 𝐾𝐾1∆𝐹𝐹0 𝑡𝑡 ⟹ ∆𝐹𝐹1(𝑠𝑠)∆𝐹𝐹0(𝑠𝑠)

= 𝐺𝐺1 𝑠𝑠 = 𝐾𝐾1𝑇𝑇1𝑠𝑠+1

𝐾𝐾1 = 2𝐴𝐴2 �ℎ2+𝛽𝛽2(𝐴𝐴1+𝐴𝐴2) �ℎ2+𝛽𝛽

, 𝑇𝑇1 = 2𝐴𝐴1 �ℎ1−�ℎ2𝛽𝛽

𝐾𝐾1.

Now the transfer function from 𝐹𝐹0 to 𝐹𝐹1 is given by 𝐺𝐺 𝑠𝑠 = 𝐺𝐺2(𝑠𝑠)𝐺𝐺1(𝑠𝑠).


Interacting systems

2F2h

0F

1F1h

ProcessControl

Laboratory


Parallel connectionA parallel connection is illustrated in Fig. 4.5. It contains a signal branch and a summation. Standard block algebra yields

𝑌𝑌 𝑠𝑠 = 𝑌𝑌1 𝑠𝑠 + 𝑌𝑌2 𝑠𝑠 = 𝐺𝐺1 𝑠𝑠 𝑈𝑈 𝑠𝑠 + 𝐺𝐺2 𝑠𝑠 𝑈𝑈 𝑠𝑠

= 𝐺𝐺1 𝑠𝑠 + 𝐺𝐺2 𝑠𝑠 𝑈𝑈(𝑠𝑠)from which it follows that

𝐺𝐺 𝑠𝑠 = 𝐺𝐺1 𝑠𝑠 + 𝐺𝐺2(𝑠𝑠) (4.33)

is the transfer function for a parallel connection of two subsystems.

An inverse-response system(Section 5.5) is often obtainedby a parallel connection of twosystems.

Fig 4.5. Parallel connection.



ProcessControl

Laboratory


Feedback connectionThe most fundamental system interconnection in control theory is (negative) feedback, which is illustrated in Fig. 4.6. Standard block algebra yields

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐸𝐸 𝑠𝑠 = 𝐺𝐺(𝑠𝑠) 𝑅𝑅 𝑠𝑠 − 𝐻𝐻 𝑠𝑠 𝑌𝑌(𝑠𝑠)from which

𝑌𝑌 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)1+𝐺𝐺 𝑠𝑠 𝐻𝐻(𝑠𝑠)

𝑅𝑅(𝑠𝑠)Thus,

𝐺𝐺c 𝑠𝑠 = 𝐺𝐺(𝑠𝑠)1+𝐺𝐺 𝑠𝑠 𝐻𝐻(𝑠𝑠)

(4.34)

is the transfer function of a feedbackconnection, where 𝐺𝐺(𝑠𝑠) is the transferfunction in the “forward” directionand 𝐻𝐻(𝑠𝑠) in the feedback part. The product 𝐺𝐺ℓ 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝐻𝐻(𝑠𝑠)

is called the loop-transfer function. The equation 1 + 𝐺𝐺ℓ 𝑠𝑠 = 0 is the

characteristic equation of the system. Fig 4.6. Feedback connection.



ProcessControl

Laboratory


Exercise 4.5Determine the transfer function from the input 𝑢𝑢 to the output 𝑦𝑦 in the block diagram below.



x

ProcessControl

Laboratory


4.3.3 Notational conventionsAs previously noted for calculations in the Laplace domain, the output of a system is obtained by multiplying the input by the its transfer function — no other terms can be included in the expression (if there is one input). By applying the Laplace transform to a DE, such a linear expression is obtained

only if the initial values of the signal and derivatives are zero. The condition is satisfied automatically when ∆-variables are used because

they denote deviations from a reference state that apply at 𝑡𝑡 = 0−, i.e., when time zero is approached from the negative side. (This is relevant if the function is discontinuous at t = 0.)

Since it is an essential requirement in calculations with transfer functions that the signals have the above characteristics, it is consid-ered to be the case even if it is not mentioned. Thus, as shown in Exercise 4.4, the symbol ∆ can be omitted to simplify the notation.

If ∆-variables with the symbol ∆ are used, it is often to emphasize the physical meaning of the signals. In such cases the symbol without ∆ is often used to denote a “true” physical value of a variable, for example, a measurement in the process.



ProcessControl

Laboratory


It is recommended to denote time functions with “small” letters (lower-case letters) and their Laplace transforms with the corres-ponding “large” letter(capital letter).

However, it is not unusual to use the same symbol for both the time function and its Laplace transform; sometimes this is due to lack of available letters.– This can be done because it is usually clear from the context which the

function type is. For example, in calculations with transfer func-tions, it is clear that the Laplace transform of the signals are used.

– If there is a danger of misunderstanding, the argument “𝑡𝑡” or “𝑠𝑠” can be included to indicate the type of function. This may be needed if the same symbol is used for the time function and its Laplace transform.

It should be noted that time-domain signals and their Laplace trans-form generally have a physical unit.– Mathematical operations should therefore satisfy unit dimensions both in

the time domain and the Laplace domain.– In particular it should be noted that the gain of a system is not

dimensionless if the input and the output have different units.


4.3.3 Notational conventions

ProcessControl

Laboratory


4.4 Applying the inverse Laplace transformThe inverse Laplace transform is used when we want to find the time-domain function that corresponds to a given Laplace-domain function.

The general procedure for solving time-domain problems in the Laplace domain is illustrated in the figure.

Fig. 4.7. Solving time-domain problems in the Laplace domain.


ProcessControl

Laboratory


4.4.1 Solving linear differential functionsA convenient way to solve linear ordinary differential equations is to use Laplace transform methods. When the Laplace transform is applied to a differential equation term by term,

the Laplace transform of the dependent variable, i.e. the output signal, can be solved by purely algebraic methods.

If the differential equation describes a dynamic system, it has an input that is also transformed.

The time function of the dependent variable can then be obtained by using the inverse transform of its Laplace transform.

Tables of Laplace transforms and their corresponding time functions can be used to find Laplace transforms as well as inverse Laplace transforms. If the table does not include a needed Laplace function, it can usually, by using

a partial fraction expansion, be written as a sum of simpler functions whose time functions are found in the table.

According to the superposition theorem (see Section 4.2.3), the full time function is then obtained as the sum of the time functions of the simpler Laplace functions.


4.4 Applying the inverse Laplace transform

ProcessControl

Laboratory


Example 4.4. An initial-value problem.Solve the differential equation ��𝑦 + 5��𝑦 + 6𝑦𝑦 = 1 with the initial values 𝑦𝑦 0− =0 and ��𝑦 0− = 1.

The Laplace transform of the DE with the given initial values is obtained by (4.18) and (4.19) as

𝑠𝑠2𝑌𝑌 𝑠𝑠 − 𝑠𝑠𝑦𝑦 0− − ��𝑦(0−) + 5 𝑠𝑠𝑌𝑌 𝑠𝑠 − 𝑦𝑦 0− + 6𝑌𝑌 𝑠𝑠 = 1𝑠𝑠

(1)

Substitution of the initial values yields after some rearrangement

𝑌𝑌 𝑠𝑠 = 𝑠𝑠+1𝑠𝑠(𝑠𝑠2+5𝑠𝑠+6)

= 𝑠𝑠+1𝑠𝑠(𝑠𝑠+2)(𝑠𝑠+3)

= 1(𝑠𝑠+2)(𝑠𝑠+3)

+ 1𝑠𝑠(𝑠𝑠+2)(𝑠𝑠+3)

(2)

The Laplace transforms having 𝑠𝑠 + 1 in the numerator are not listed in the Table in Section 4.5. However, the Laplace transforms of the last two terms are listed in pt 17 and 18. Because of the superposition principle, the inverse transforms of the two terms can be added to obtain

𝑦𝑦 𝑡𝑡 = 13−2

e−2𝑡𝑡 − e−3𝑡𝑡 + 12⋅3

+ 12(2−3)

e−2𝑡𝑡 − 13(2−3)

e−3𝑡𝑡

= 12

e−2𝑡𝑡 − 23

e−3𝑡𝑡 + 16

(3)

The solution should be verified by checking if it satisfies the DE + initial conditions.


4.4.1 Solving linear DEs

ProcessControl

Laboratory


Example 4.5. Step response of a first-order system.A linear first-order system is described by the differential equation

𝑇𝑇 d𝑦𝑦d𝑡𝑡

+ 𝑦𝑦 = 𝐾𝐾𝑢𝑢 (1)

where 𝑢𝑢 is the input, 𝑦𝑦 is the output, 𝐾𝐾 is the static gain and 𝑇𝑇 is the time constant of the system.

If 𝑢𝑢 = 0, 𝑦𝑦 = 0 is a solution to the DE. This equilibrium point can be assumed to apply at 𝑡𝑡 = 0−. Laplace transformation of (1) then gives

𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 𝑈𝑈(𝑠𝑠) (2)where

𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑇𝑇𝑠𝑠+1

(3)

is the transfer function of the system.

Assume that the input 𝑢𝑢 suddenly changes from 0 to 𝑢𝑢step at 𝑡𝑡 = 0, i.e.,

𝑢𝑢 𝑡𝑡 = 0 , 𝑡𝑡 < 0 ; 𝑢𝑢 𝑡𝑡 = 𝑢𝑢step , 𝑡𝑡 ≥ 0 (4)

The size of this step change is 𝑢𝑢step times the size of a unit step. According to pt 1 in the Laplace table (and the superposition principle),

𝑈𝑈 𝑠𝑠 = ⁄𝑢𝑢step 𝑠𝑠 (5)



ProcessControl

Laboratory


Substitution of (5) and (3) into (2) gives

𝑌𝑌 𝑠𝑠 = 𝐾𝐾𝑢𝑢step𝑠𝑠(𝑇𝑇𝑠𝑠+1)

(6)

Pt 26 in the Laplace transform table now yields the time function

𝑦𝑦 𝑡𝑡 = 𝐾𝐾𝑢𝑢step 1 − e−𝑡𝑡/𝑇𝑇 (7)

which is the step response of a first-order system.

Exercise 4.6Determine the unit step response for the system in Exercise 4.4, i.e., the differential equation

d2𝑦𝑦d𝑡𝑡2

+ 5 d𝑦𝑦d𝑡𝑡

+ 6𝑦𝑦 = 𝑢𝑢

Note the similarities and differences with Example 4.4.


Example 4.3. Step response of first-order system

ProcessControl

Laboratory


4.4.2 Partial fraction expansionThe Laplace transform of a time function f(t), such as the solution to a differential equation with a given input and given initial conditions, can typically be written in the form

𝐹𝐹 𝑠𝑠 = 𝑏𝑏0𝑠𝑠𝑚𝑚+𝑏𝑏1𝑠𝑠𝑚𝑚−1+⋯+𝑏𝑏𝑚𝑚−1𝑠𝑠+𝑏𝑏𝑚𝑚𝑠𝑠𝑛𝑛+𝑎𝑎1𝑠𝑠𝑛𝑛−1+⋯+𝑎𝑎𝑛𝑛−1𝑠𝑠+𝑎𝑎𝑛𝑛

≡ 𝐵𝐵(𝑠𝑠)𝐴𝐴(𝑠𝑠)

(4.35)

For a Laplace function 𝑌𝑌(𝑠𝑠) containing a time delay 𝐿𝐿, so that

𝑌𝑌 𝑠𝑠 = 𝐹𝐹(𝑠𝑠)e−𝐿𝐿𝑠𝑠 (4.36)

one can first determine 𝑓𝑓(𝑡𝑡) from 𝐹𝐹(𝑠𝑠), then obtain 𝑦𝑦 𝑡𝑡 = 𝑓𝑓(𝑡𝑡 − 𝐿𝐿)according to the time-delay theorem (4.23).

The time function 𝑓𝑓(𝑡𝑡) corresponding to the Laplace transform 𝐹𝐹(𝑠𝑠) can often be found directly in a table (Section 4.5) or after a simple separation of the numerator terms according to the superposition theorem as in Example 4.4.

If this does not solve the problem, 𝐹𝐹(𝑠𝑠) can be rewritten as a sum of simpler terms by a partial fraction expansion (PFE).



ProcessControl

Laboratory


Extracting a non-strictly proper part from 𝐹𝐹(𝑠𝑠)Most physical systems are proper, but in the rare case when 𝐹𝐹(𝑠𝑠) is not strictly proper, i.e., 𝑚𝑚 ≥ 𝑛𝑛 in (4.35), the non-strictly proper part has to be extracted before proceeding with the PFE.

This is done by performing a polynomial long division of 𝐹𝐹(𝑠𝑠) to obtain

𝐹𝐹 𝑠𝑠 = 𝐹𝐹0 𝑠𝑠 + 𝐵𝐵0(𝑠𝑠)𝐴𝐴(𝑠𝑠)

(4.37)

where 𝐹𝐹0(𝑠𝑠) is a polynomial of order 𝑚𝑚 − 𝑛𝑛 ≥ 0, 𝐴𝐴(𝑠𝑠) is the 𝑛𝑛th order denominator polynomial of 𝐹𝐹(𝑠𝑠), and 𝐵𝐵0(𝑠𝑠) is a polynomial of order 𝑚𝑚0 < 𝑛𝑛. Thus, ⁄𝐵𝐵0(𝑠𝑠) 𝐴𝐴(𝑠𝑠) is strictly proper.

According to the superposition principle, the inverse Laplace transform of 𝐹𝐹(𝑠𝑠)is the sum of the inverse transforms of 𝐹𝐹0(𝑠𝑠) and ⁄𝐵𝐵0(𝑠𝑠) 𝐴𝐴 𝑠𝑠 . The inverse transform 𝑓𝑓0(𝑡𝑡) of 𝐹𝐹0(𝑠𝑠) will consist of one or more terms such

as an impulse and time derivatives of impulses (a factor 𝑠𝑠𝑖𝑖 results in an 𝑖𝑖th

order derivative). The inverse transform of ⁄𝐵𝐵0(𝑠𝑠) 𝐴𝐴 𝑠𝑠 can be obtained via a PFE.


4.4.2 Partial fraction expansion

ProcessControl

Laboratory


Example 4.6. Long division of polynomials.

Consider the rational function 𝐹𝐹 𝑠𝑠 = 𝑠𝑠3−2𝑠𝑠2−4𝑠𝑠−3

. Write it in the form of (4.37) by doing a polynomial long division. Find the inverse Laplace of 𝐹𝐹(𝑠𝑠).

The long division procedure yields

𝐵𝐵 𝑠𝑠 = 𝑠𝑠3 − 2𝑠𝑠2 −4 �𝑠𝑠 − 3 = 𝐴𝐴(𝑠𝑠)−𝑠𝑠3 + 3𝑠𝑠2 𝑠𝑠2 + 𝑠𝑠 + 3 = 𝐹𝐹0(𝑠𝑠)

𝑠𝑠2 −4−𝑠𝑠2 + 3𝑠𝑠

3𝑠𝑠 − 4−3𝑠𝑠 + 9

5 = 𝐵𝐵0(𝑠𝑠)

Thus, 𝐹𝐹 𝑠𝑠 = 𝑠𝑠2 + 𝑠𝑠 + 3 + 5𝑠𝑠−3

.

Equations (4.18), (4.19) and pts 6 and 8 in the Laplace Transform Table yield the inverse Laplace transform 𝑓𝑓 𝑡𝑡 = d2𝛿𝛿(𝑡𝑡)

d𝑡𝑡2+ d𝛿𝛿(𝑡𝑡)

d𝑡𝑡+ 3𝛿𝛿 𝑡𝑡 + 5e3𝑡𝑡, where 𝛿𝛿(𝑡𝑡) is

the unit impulse.


Extracting a non-strictly proper part

ProcessControl

Laboratory


Factorization of 𝐴𝐴(𝑠𝑠)The next step is to factorize 𝐴𝐴(𝑠𝑠) into 𝑛𝑛 first-order factors

𝐴𝐴 𝑠𝑠 = 𝑠𝑠 − 𝑝𝑝1 𝑠𝑠 − 𝑝𝑝2 … (𝑠𝑠 − 𝑝𝑝𝑛𝑛) (4.38)

where 𝑝𝑝𝑘𝑘, 𝑘𝑘 = 1,2, … ,𝑛𝑛, are the roots of the characteristic equation 𝐴𝐴 𝑠𝑠 = 0. Complex roots always appear as pairs of complex-conjugated roots, but otherwise any combination of real and complex roots is possible. The form of the PFE depends on the properties of the roots.

𝑑𝑑 distinct (unequal) real roots yield the PFE

∑𝑘𝑘=1𝑑𝑑 𝐶𝐶𝑘𝑘(𝑠𝑠−𝑝𝑝𝑘𝑘)

(4.39)

where 𝐶𝐶𝑘𝑘, 𝑘𝑘 = 1,2, … ,𝑑𝑑, are constants that need to be determined.

𝑟𝑟 repeated (equal) real roots yield the PFE

∑𝑘𝑘=1𝑟𝑟 𝐶𝐶𝑘𝑘(𝑠𝑠−𝑝𝑝𝑟𝑟)𝑘𝑘

(4.40)

where 𝐶𝐶𝑘𝑘, 𝑘𝑘 = 1,2, … , 𝑟𝑟, are constants that need to be determined and 𝑝𝑝𝑟𝑟 = 𝑝𝑝𝑘𝑘 are the 𝑟𝑟 repeated roots. In practice, repeated roots are not very common.



ProcessControl

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A pair of complex-conjugated roots 𝑝𝑝 = 𝜎𝜎 ± j𝜔𝜔, where j = −1 is the imaginary unit, yields the PFE

𝐶𝐶1𝑠𝑠+𝐶𝐶2(𝑠𝑠−𝜎𝜎)2+𝜔𝜔2 (4.41)

where 𝐶𝐶1 and 𝐶𝐶2 are constants that need to be determined. Repeated complex-conjugated roots result in rising powers of the numerator similarly as for repeated real roots. In practice, such roots are very rare.

The full PFE for 𝑑𝑑 distinct real roots (𝑝𝑝1, … ,𝑝𝑝𝑑𝑑), 𝑟𝑟 repeated real roots (𝑝𝑝𝑟𝑟 = 𝑝𝑝𝑑𝑑+1 = ⋯ = 𝑝𝑝𝑑𝑑+𝑟𝑟), and one pair of complex-conjugated roots (𝑝𝑝𝑛𝑛−1 and 𝑝𝑝𝑛𝑛) is

𝐹𝐹 𝑠𝑠 = 𝐹𝐹0 𝑠𝑠 + ∑𝑘𝑘=1𝑑𝑑 𝐶𝐶𝑘𝑘(𝑠𝑠−𝑝𝑝𝑘𝑘)

+ ∑𝑘𝑘=𝑑𝑑+1𝑑𝑑+𝑟𝑟 𝐶𝐶𝑘𝑘(𝑠𝑠−𝑝𝑝𝑟𝑟)𝑘𝑘

+ 𝐶𝐶𝑛𝑛−1𝑠𝑠+𝐶𝐶𝑛𝑛(𝑠𝑠−𝜎𝜎)2+𝜔𝜔2 (4.42)

All terms in the PFE (4.42) are such that their inverse transforms are easily found in the Laplace Transform Table in Section 4.5. According to the superposition theorem, the final function 𝑓𝑓(𝑡𝑡) is the sum of the individual inverse transforms.

Note that the numerator of 𝐹𝐹(𝑠𝑠) does not affect the form of the PFE.


Factorization of 𝐴𝐴(𝑠𝑠)

ProcessControl

Laboratory


Calculation of the constants 𝐶𝐶𝑘𝑘The constants 𝐶𝐶𝑘𝑘, 𝑘𝑘 = 1,2, … ,𝑛𝑛, can be determined in several ways. Because the PFE must apply for any value of the variable 𝑠𝑠, 𝑛𝑛 different,

“appropriately” chosen, values of 𝑠𝑠 can be substituted into the PFE to yield 𝑛𝑛equations, from which the constants 𝐶𝐶𝑘𝑘 can be determined.

Another, more systematic method is to multiply both sides of the PFE by 𝐴𝐴(𝑠𝑠)and to cancel out the denominators against the same factor in 𝐴𝐴(𝑠𝑠). The expression thus obtain must be equal to 𝐵𝐵0(𝑠𝑠). The constants 𝐶𝐶𝑘𝑘 can then be determined from the 𝑛𝑛 number of equations that arise when the PFE is required to apply separately for each power of 𝑠𝑠.

If the roots are distinct and real, 𝐶𝐶𝑘𝑘 can be determined according to

𝐶𝐶𝑘𝑘 = lim𝑠𝑠→𝑝𝑝𝑘𝑘

(𝑠𝑠 − 𝑝𝑝𝑘𝑘) 𝐵𝐵0(𝑠𝑠)𝐴𝐴(𝑠𝑠)

(4.43)

Note that the factor (𝑠𝑠 − 𝑝𝑝𝑘𝑘) is cancelled out against the same factor in 𝐴𝐴(𝑠𝑠).



ProcessControl

Laboratory


Example 4.7. Ramp response of a first-order system.The ramp response of a system is the output as a function of time when the input changes as a ramp, i.e., linearly with time. The input is thus

𝑢𝑢 𝑡𝑡 = 𝑏𝑏𝑡𝑡, 𝑡𝑡 ≥ 0 (1)

where 𝑏𝑏 is the slope of the input ramp. The function has the Laplace transform (pt 2 in the Laplace transform table)

𝑈𝑈 𝑠𝑠 = ⁄𝑏𝑏 𝑠𝑠2 (2)

The Laplace-domain output of a first-order system is given by

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

(3)

Substitution of (2) into (3) yields

𝑌𝑌 𝑠𝑠 = 𝐾𝐾𝑏𝑏𝑠𝑠2(𝑇𝑇𝑠𝑠+1)

(4)

The time-domain output 𝑦𝑦(𝑡𝑡) is given by the inverse Laplace transform of (4). This can be found directly in the Laplace Transform Table (pt 27), but here a partial fraction expansion of (4) will be used to find the solution.



ProcessControl

Laboratory


The function (4) is strictly proper (i.e., no need to do a long division) and the denominator is already factored. There is a simple root 𝑠𝑠 = −1/𝑇𝑇, and a double root 𝑠𝑠 = 0.

In accordance with (4.42), the PFE is

𝑌𝑌 𝑠𝑠 = 𝐾𝐾𝑏𝑏𝑠𝑠2(𝑇𝑇𝑠𝑠+1)

≡ 𝐶𝐶1𝑠𝑠

+ 𝐶𝐶2𝑠𝑠2

+ 𝐶𝐶3(𝑇𝑇𝑠𝑠+1)

(5)

Multiplication of both sides by 𝑠𝑠2(𝑇𝑇𝑠𝑠 + 1), and cancellation of common factors, yields

𝐾𝐾𝑏𝑏 ≡ 𝐶𝐶1𝑠𝑠 𝑇𝑇𝑠𝑠 + 1 + 𝐶𝐶2 𝑇𝑇𝑠𝑠 + 1 + 𝐶𝐶3𝑠𝑠2 (6)

This expression must apply separately for each power of 𝑠𝑠:

𝑠𝑠0: 𝐾𝐾𝑏𝑏 = 𝐶𝐶2 ⟹ 𝐶𝐶2 = 𝐾𝐾𝑏𝑏

𝑠𝑠1: 0 = 𝐶𝐶1 + 𝐶𝐶2𝑇𝑇 ⟹ 𝐶𝐶1 = −𝐾𝐾𝑏𝑏𝑇𝑇 (7)

𝑠𝑠2: 0 = 𝐶𝐶1𝑇𝑇 + 𝐶𝐶3 ⟹ 𝐶𝐶3 = 𝐾𝐾𝑏𝑏𝑇𝑇2


Ex. 4.5. Ramp response of first-order system

ProcessControl

Laboratory


Substitution of (7) into (5) gives

𝑌𝑌 𝑠𝑠 = −𝐾𝐾𝑏𝑏𝑇𝑇𝑠𝑠

+ 𝐾𝐾𝑏𝑏𝑠𝑠2

+ 𝐾𝐾𝑏𝑏𝑇𝑇2

(𝑇𝑇𝑠𝑠+1)(8)

Points 1, 2 and 25 in the Laplace Transform Table now yield

𝑦𝑦 𝑡𝑡 = −𝐾𝐾𝑏𝑏𝑇𝑇 + 𝐾𝐾𝑏𝑏𝑡𝑡 + 𝐾𝐾𝑏𝑏𝑇𝑇e−𝑡𝑡/𝑇𝑇

= 𝐾𝐾𝑏𝑏(𝑡𝑡 − 𝑇𝑇 + 𝑇𝑇e−𝑡𝑡/𝑇𝑇), 𝑡𝑡 ≥ 0 (7)

Since e−𝑡𝑡/𝑇𝑇 → 0 as 𝑡𝑡 → ∞, the output approaches a ramp with the ramp coefficient 𝐾𝐾𝑏𝑏.

Exercise 4.7Determine, by the use of partial fraction expansion, the inverse Laplace transform of the following functions:

a) 𝐹𝐹a 𝑠𝑠 = 𝑠𝑠+32𝑠𝑠(𝑠𝑠−4)2

b) 𝐹𝐹b 𝑠𝑠 = 3𝑠𝑠+5𝑠𝑠(𝑠𝑠2+6𝑠𝑠+25)


Ex. 4.5. Ramp response of 1st order system

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4.5 Table of Laplace transformsFrom Tore Gustafsson: Ingenjörsmatematisk formelsamling, 1999–2013.


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(also ( ))tσ

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