22.08.2015   Frequency Response Testing. in Nuclear Reactors

Laplace Transforms*

Laplace transforms have a key role in dynamic system analysis and in dynamic testing. The Laplace transform is defined as follows:

/* CO

F(s) = L{f(t)} = f(t)e~s, dt (2.2.1)

Jo

where/(f) is some function of t, F(s) the Laplace transform of/(f), s a param­eter [it is not necessary to specify a value for s, but there must be some value of s that makes the integral in Eq. (2.2.1) converge], and L the Laplace trans­form operator. This simple definition permits the development of a table of Laplace transforms of functions and operators. The process of determining an f(t) whose Laplace transform is L{/(t)} is called inversion of the Laplace transform.

See Aseltine (1).

Example 2.2.1. Determine the Laplace transform of eal.

image1

1 00 1

Example 2.2.2. Determine the Laplace transform of df/dt.

image2 image003

Integrate by parts to obtain

A table of Laplace transforms appears in Table 2.1.

Laplace transforms are useful for solving differential equations. The procedure for linear, ordinary differential equations is:

1. Laplace transform all terms in the differential equation. This gives an algebraic equation.

2. Solve the algebraic equation for the Laplace transform of the desired solution.

3. Obtain the solution by inverting the expression for the Laplace trans­form of the solution. This is done using the table or by using a general inversion theorem (described below).

Example 2.2.3. Solve the following differential equation using Laplace transforms:

dx/dt = R — ax

where R and a are constants.

image3

Step 1.

image4

Step 2.

image5

Step 3.

Подпись: TABLE 2.1 LAPLACE TRANSFORM PAIRS m F(s) df dt sF(s) - /(0) d2f dt2 S2F(S) - sf(0) - f(0) dt dj df s"F(s)-s*-'m----^1( 0) $’of(x) dx F(s)/s f(t - T)u(t - T) e~'sF(s) for T > 0 (u is the unit step function) F(s + a) <L(f) 1 (<5+ is the impulse or delta function) U(t) t 1/s (u is the unit step function) 1Д2 t2 2/s3 f n!/s"+1 e~“ l/(s + a) te~at l/(s + a)2 fe~°' n!/(s + a)"+1 sin fit Ms2 + P1) cos fit s/(s2 + fj2) sinh fit P/(s2 - P2) cosh fit s/(s2 - P2) sin fit P (s + a)2 + P2 e~“ cos fit (s + a)/(s + a)2 + P2 1 (<?-“' е~ы) b — a l/(s + a)(s + b)

In general, the Laplace transform of the solution for a lumped-parameter system will be a ratio of polynomials in s.

Подпись: (2.2.2)sm + bm_ і sm 1 + • • • + bls 4- b(, s" + a„_ 1s"~1 + ■ ■ ■ + ats + a0

image009 Подпись: (2.2.3)

These polynomials may be written in factored form to give

The values zl, z2, ■ ■ ■, zm are called the zeros of X(s). The values Pi, p2, ■ ■ •, p„ are called the poles of X(s). The Laplace transform of the solution of a system of ordinary differential equations with equations of various orders will have Np poles, where

Np = I 0, (2.2.4)

all

equations

where 0,- is the order of the ith equation in the set. For example, if all the equations are first order, then the number of poles is equal to the number of equations. It is possible for several poles to have the same value. Poles that appear once are called simple poles. Poles that appear more than once are called multiple poles.

The inversion of Laplace transforms may be handled by a general inver­sion method based on the residue theorem. The residue theorem gives the following:

L-^{F(s)}= Y Ri (2-2.5)

;= 1

image011 Подпись: £^ї~ PiTFW Подпись: (2.2.6)

where I is the number of distinct poles, and Rt the residue of the ith pole. The residue is given by

where n is the number of times the pole p; is repeated. For a simple pole, this simplifies to

Ri = [(S — рте3‘1=рі (2.2.7)

Example 2.2.4. Invert the following:

Подпись: F(s) =(s+ 1)

(s + 2)(s + 3)

Подпись: -2 + 1 -2 + 3 image016 Подпись: R2 Подпись: -3 + 1' -3 + 2 Подпись: e

The residues are

Thus

L~l{F(s)} = — e~2′ + 2e“3′