3.3 Introduction to Laplace Transforms

Most control system analysis and design techniques are based on linear systems theory. Although we could develop these procedures using the state space models, it is generally easier to work with transfer functions. Basically, transfer functions allow us to make algebraic manipulations rather than working directly with linear differential equations (state space models). To create transfer functions, we need the notion of the Laplace transform.

The Laplace transform of a time-domain function, f(t), is represented by L[f(t)] and is defined as

Equation 3.6

The Laplace transform is a linear operation, so the Laplace transform of a constant (C) multiplying a time-domain function is just that constant times the Laplace transform of the function,

Equation 3.7

The Laplace transforms of a few common time-domain functions are shown next.

Exponential Function

Exponential functions appear often in the solution of linear differential equations. Here

The transform is defined for t > 0 (we also use the identity that e^x⁺^y = e^xe^y)

graphics/03equ07b.gif

So we now have the following relationship:

Equation 3.8

Derivatives

This will be important in transforming the derivative term in a dynamic equation to the Laplace domain (using integration by parts),

graphics/03equ08a.gif

so we can write

Equation 3.9

For an nth derivative, we can derive

Equation 3.10

n initial conditions are needed: f(0),..., f^{(n – 1)} (0)

One reason for using deviation variables is that all of the initial condition terms in Equation (3.10) are 0, if the system is initially at steady-state.

Time Delays (Dead Time)

Time delays often occur owing to fluid transport through pipes, or measurement sample delays. Here we use q to represent the time delay. If f(t) represents a particular function of time, then f(t – q) represents the value of the function q time units in the past.

graphics/03equ10a.gif

We can use a change of variables, t* = t – q, to integrate the function. Notice that the lower limit of integration does not change, because the function is defined as f(t) = 0 for t < 0.

Equation 3.11

So the Laplace transform of a function with a time delay (q) is simply e^–^q^s times the Laplace transform of the nondelayed function.

Step Functions

Step functions are used to simulate the sudden change in an input variable (say a flow rate being rapidly changed from one value to another). A step function is discontinuous at t = 0. A "unit" step function is defined as

and using the definition of the Laplace transform,

Equation 3.12

Similarly, the Laplace transform of a constant, C, is

Equation 3.13

Pulse

Consider a pulse function, where a total integrated input of magnitude P is applied over t_p time units, as shown in Figure 3-3.

Figure 3-3. Pulse function.

graphics/03fig03.gif

The function is f(t) = P/t_p for 0 < t < t_p and f(t) = 0 for t > t_p. The Laplace transform is

Equation 3.14

graphics/03equ14.gif

Impulse

An impulse function can viewed as a pulse function, where the pulse period is decreased while maintaining the pulse area, as shown in Figure 3-4. In the limit, as t_p approaches 0, the pulse function becomes (using L'Hopital's rule)

Equation 3.15

Figure 3-4. Concept of an impulse function.

graphics/03fig04.gif

If we denote a unit impulse as f(t) = d, then the Laplace transform is

Equation 3.16

Examples of common impulse inputs include a "bolus" (shot or injection) of a drug into a physiological system, or dumping a bucket of fluid or bag of solids into a chemical reactor.

Other Functions

It is rare for one to derive the Laplace transform for a function; rather a table of known transforms (and inverse transforms) can be used to solve most dynamic systems problems.

Table 3-1 presents solutions for most common functions. If you desire to transform a function from the time domain to the Laplace domain, then look for the time-domain function in the first column and write the corresponding Laplace domain function from the second column. Similarly, if your goal is to "invert" a Laplace domain function to the time domain, then look for the Laplace domain function in the second column and write the corresponding time-domain function from the first column. This notion of the inverse Laplace transform can be written

Equation 3.17

For example

Initial- and Final-Value Theorems

The following theorems are very useful for determining limiting values in dynamics and control studies. The long-term behavior of a time-domain function can be found by analyzing the Laplace domain behavior in the limit as the s variable approaches zero. The initial value of a time-domain function can be found by analyzing the Laplace domain behavior in the limit as s approaches infinity.

Table 3-1. Laplace Transforms for Common Time-Domain Functions
Time-domain function
Laplace domain function
f(t)
F(s)
d(t) (unit impulse)
1
C (constant)
f(t - q)
e-^qsF(s)
t
tⁿ
sF(S) - f(0)
Sⁿf(S) - S^n-1f(0) - L - sf^(n-2)(0) - f^(n-1)(0)
e^-at
te^-at
1-e^-t/^t
sin wt
cos wt
e^–^at sin wt
e^–^atcos wt

where , graphics/03tainequ12b.gif

The final-value theorem is

Equation 3.18a

The initial-value theorem is

Equation 3.18b

It should be noted that these theorems only hold for stable systems.

Example 3.2: Application of Initial- and Final-Value Theorems

Find the long-term and short-term behavior of the time-domain function, y(t), using the final- and initial-value theorems on the Laplace domain function Y(s) (we see later that this arises from a step input applied to a second-order process):

The long-term behavior, y(t ), is found using the final-value theorem,

The short-term behavior, y(t 0), is found using the initial-value theorem,

The reader should verify that the time-domain function, y(t), can be found by applying Table 3-1 to find

and that the values of y(t ), and y(t 0), are consistent with the final- and initial-value theorems.

General Solution Procedure

To obtain analytical solutions for differential equation-based models, the general procedure is composed of several steps.

Start with an nth order linear differential equation

and known initial conditions
Transform each element of the differential equation to the Laplace domain,
Use algebraic manipulations to solve for the transformed variable.

where N(s) and D(s) are polynomials in s.
Perform a partial fraction expansion to isolate the individual elements.

where a, b, and so forth are the roots of D(s). If roots are repeated the partial fraction expansion has the form (for an example where three roots all have a value of a).
Invert each element back to the time domain to find the final solution for y(t).

For the case of repeated roots (say 3 that have a value of a), the solution has the form

It should be noted that chemical process systems are rarely described by an nth order differential equation. Usually, a set of n first-order differential equations is transformed to a single nth order equation, as shown in Example 3.3.

Example 3.3: Second-Order Differential Equation

Consider the following state space model of an isothermal CSTR (Module 5).

Equation 3.19

graphics/03equ19.gif

where x₁ and x₂ represent the concentrations of two components (in deviation variable form) in an isothermal reactor; the initial conditions are x₁(0) = x₂(0) = 0. Solve for the output (concentration of component 2) response to a unit step input.

The reader should show (see exercise 6) that this can be arranged to a second-order differential equation,

with the initial conditions y(0) = dy(0)/dt = 0, and the input is initially u(0)= 0.
Taking the Laplace transform of each element,

and substituting back into the differential equation, we find^[1]
^[1] Alternatively, one could take the Laplace transform of each equation in (3.19) and use algebraic substitution to find this second-order equation in s.
Solving for Y(s), we find

Here we consider a unit step input, U(s) = 1/s.
Performing a partial fraction expansion using the roots of the denominator polynomial,

Equation 3.20

Solve for C₁ by multiplying by s,

Setting s = 0 to find

Similarly, solve for C₂ by multiplying Equation (3.20) by s + 2.4053 and setting s = –2.4053

Also, solve for C₃ by multiplying Equation (3.20) by s + 2.2376 and setting s = –2.2376

which yields
Inverting each element back to the time domain,

Equation 3.21

Although it is nice to have an analytical solution, it is generally more pleasing to engineers to view a plot of a variable as it changes with time, as shown in Figure 3-5. In this plot we notice that the concentration initially decreases, before increasing to a new steady-state value.

Figure 3-5. Response of the concentration as a function of time. See Equation (3.21).

In Example 3.3 we solved for the coefficients (C₁, C₂, C₃) by selecting values of s to minimize the multiplications performed during each step. An alternative is to solve for three equations in three unknowns by using various values of s (say s = 1, 2, and 3, for example) in Equation (3.20).

Differential equations textbooks, such as Boyce and DiPrima (1992) present many examples of applications of Laplace transforms to solve differential equations. In practice, it is rare for process engineers to seek analytical solutions to differential equations; it is far easier to solve these numerically. The primary goal of this section was to provide background material to understand the transfer functions and dynamic responses shown in the next section.

[ Team LiB ]