Module 21 - Laplace Transforms

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 21

LAPLACE TRANSFORMS

1. Introduction 2. Transforms of simple functions 3. Properties of Laplace transforms 4. Inverse Laplace transform 5. Solution of differential equations

1. Introduction Laplace transforms prove a very useful tool in analysing engineering systems - particularly in the fields of signal processing and linear systems analysis. Before considering the details of the method let us briefly outline how its use compares with the standard method of solving ordinary differential equations. Standard method (i) Consider a given ODE in the function f (t); (ii) find complementary function fc and particular integral f ∗ ; (iii) general solution = fc + f ∗ ; (iv) apply initial conditions to determine unknown constants; (v) deduce solution. Laplace transform method (i) Consider the given ODE in the function f (t); (ii) take Laplace transform (using tables), incorporating the initial conditions, to give an algebraic equation in the transform F (s); (iii) solve for F (s); (iv) rearrange this expression for F (s) to appropriate form for inversion; (iv) take inverse Laplace transform (using tables) to give solution. The general procedure is outlined above but before we can apply the method it is necessary to return to look in more detail at the various parts of the above process. First we must state the definition of a Laplace transform. Def.

The Laplace transform of a function f (t) is defined by Z L{f (t)} =

∞

e−st f (t) dt = F (s),

0

where L denotes the Laplace transform operator and s is a parameter which is normally complex (i.e. not real). It should be noted that in some texts the notation F (s) is replaced by f (s) and p is used in place of the parameter s. 1

The definition shows that the Laplace operator L{·} transforms the function f (t), in the time or t-domain, to F (s) in the frequency or s-domain. Can we obtain the Laplace transforms of ALL functions f (t)? No, but it can be shown that Laplace transforms exist (for some s) for all “reasonable” functions f (t). (The functions must be piecewise continuous, for instance, and not go to infinity too rapidly as t → ∞. The Laplace transform of the function exp(t2 ), for example, does not exist since the integral will not converge). Precise conditions for the existence of the transform can be stated but these are not necessary in this module. It is sufficient for you to know that transforms can be found for all the standard functions that normally arise. The transform process can be reversed by taking the inverse Laplace transform, denoted by L−1 , so that F (s) = L{f (t)}

f (t) = L−1 {F (s)}.

implies

2. Transforms of simple functions Let us now use the definition stated above to derive the transforms of the simplest functions. (a)

f (t) = 1

On substituting f (t) = 1 into the definition of the Laplace transform Z L{1} =

∞

e−st 1 dt = lim

T →∞

0

and hence

Z

 L{1} = lim

T →∞

e−st −s

T 0

=−

T

e−st 1 dt,

0

  1 lim e−sT − 1 . s T →∞

Put s = σ + jω, where σ and ω are real, then e−sT = e−(σT +jωT ) = e−σT e−jωT = e−σT (cos(ωT ) − j sin(ωT )), and it follows that, provided σ > 0 (i.e. Re(s) > 0), lim e−sT = 0.

T →∞

Using this limit we deduce that 1 1 L{1} = − (0 − 1) = , s s

(b)

provided Re(s) > 0.

f (t) = ekt

In this case use of the definition of the transform leads to kt

Z

L{e } =  = lim

T →∞

∞

e

Z

−st kt

e dt = lim

T →∞

0

e−(s−k)t −(s − k)

T 0

=−

T

e−(s−k)t dt

0

h i 1 lim e−(s−k)T − 1 . s − k T →∞

Introducing s = σ + jω leads to e−(s−k)T = e−(σ+jω−k)T = e−(σ−k)T e−jωT = e−(σ−k)T (cos(ωT ) − j sin(ωT )). 2

It then follows that

lim e−(s−k)T = 0,

T →∞

provided that σ > k, if k is real, and σ > Re(k), if k is complex, therefore L{ekt } = −

1 1 (0 − 1) = , s−k s−k

provided Re(s) > Re(k).

Two simple Laplace transforms that appear in the standard tables have been calculated above. In a similar way the transforms of other simple functions can be found. For example, for f (t) = t the transform is found after integrating by parts. The transforms of a number of functions are listed on the information sheet, and in the Engineering Data book, and in applications it is usual to use these tables and not calculate the transforms from first principles. 3. Properties of Laplace transforms Three properties of Laplace transforms will now be considered in turn. (a) Linearity

If α and β are constants then L{α f (t) + β g(t)} = α L{f (t)} + β L{g(t)}.

The above result, which is easily proved from the integral definition of the transform, is very useful in applications. Ex 1. Find L{2 + 5e3t } Using property (a), and the tables,     2 5 1 1 +5 = + , L{2 + 5e } = 2 L{1} + 5 L{e } = 2 s s−3 s s−3 3t

3t

which can be algebraically combined to produce a single term if required. The first transform above holds when Re(s) > 0 whilst the second holds when Re(s) > 3, and hence the final result only holds when both are satisfied, i.e. when Re(s) > 3. (b) First shift theorem

If L{f (t)} = F (s), for Re(s) > b, then L{eat f (t)} = F (s − a),

provided Re(s) > b + Re(a).

The above property, which is extremely important in applications, can be proved from the original definition and a change of variable. It is useful to note that the theorem can also be expressed L{eat f (t)} = F (s)|s→s−a . Multiplication by eat therefore produces a shift in the variable, but leaves the functional form of the transform unaltered. Ex 2.

Find L{te2t }.

We know from tables that L{t} = L{te2t } =

1 provided Re(s) > 0, and so property (b) implies that s2

1 1 = , s2 s→s−2 (s − 2)2

provided Re(s) > 0 + 2 = 2. 3

If L{f (t)} = F (s), with Re(s) > b, then

(c) Derivative of transform

L{tn f (t)} = (−1)n

dn F dsn

provided Re(s) > b.

This time the result follows from the initial definition and integration by parts (integrating a number of times, or using induction). Ex 3.

Find L{te2t }.

This time we will derive the answer using property (c). It is clear from the tables that L{e2t } = Re(s) > 2, and (c) above then gives L{te2t } = −

d dF =− ds ds



1 s−2



= −{−1(s − 2)−2 } =

1 , (s − 2)2

1 when s−2

provided Re(s) > 2.

4. Inverse transform The inverse Laplace transform was introduced at the end of section 1 and is simplythe inverse operation to  1 1 then L−1 = e−3t . The taking the Laplace transform. That is to say, since L{e−3t } = s+3 s+3 table of results, which is what you normally use to find an inverse, shows for example that L−1



2 s2 + 4



= L−1



2 s2 + 2 2

 = sin(2t) .

The inverse transform, L−1 , is a linear operator and so satisfies the linearity property: if α and β are constants then

L−1 {α F (s) + β G(s)} = α f (t) + β g(t).

 1 3 + 2 . s s +9 It follows from use of the above linearity property for the inverse and tables that

Ex 4.

Find L−1



L−1



1 3 + 2 s s +9



    3 1 + L−1 s s2 + 3 2     1 1 1 3 + L−1 = 3 + sin(3t). = 3 L−1 2 2 s 3 s +3 3 = L−1

Note that in calculating the second term above the expression is rearranged into a form that appears in the given tables. It is this manipulation which often creates the most difficulty in finding solutions, and the method of partial fractions must often be used as part of the rearrangement process, as shown below.  s . Ex 5. Find L (s + 1)(s + 2) In this situation the quantity to be inverted must be rewritten using partial fractions. Using the standard procedure we have A B A(s + 2) + B(s + 1) s = + = (s + 1)(s + 2) s+1 s+2 (s + 1)(s + 2) −1



i.e. s = −2; s = −1;

s = A(s + 2) + B(s + 1) −2 = A(0) + B(−1), −1 = A(1) + B(0), 4

→B=2 → A = −1.

Therefore, it follows that L−1



s (s + 1)(s + 2)



= L−1



2 −1 + s+1 s+2



= −e−t + 2e−2t .

The decomposition into partial fractions is often more difficult than in the above example, and can be the most difficult part of the solution. We considered the first shift theorem in section 3(b), stating it in the form if L{f (t)} = F (s) then

L{eat f (t)} = F (s − a) .

Using the above to obtain Laplace transforms is relatively easy, but it is more difficult using it to find inverse transforms. To facilitate finding the inverse it is useful to rewrite the above result as L−1 {F (s − a)} = eat f (t) = eat L−1 {F (s)}. Ex 6.

Find L−1



 1 . (s + 2)2

1 1 = 2 . Now from the tables we note that (s + 2)2 s s→s+2

The function to be inverted can be written   1 −1 = t , and hence using the above property (with a = −2) it follows that L s2   1 = te−2t . L−1 (s + 2)2

 1 . Ex 7. Find L s2 + 2s + 2 The denominator here does not factorise into linear real factors, so we need to “complete the square”. Since s2 + 2s + 2 = (s + 1)2 + 1 we can obtain the answer as follows:       1 1 1 −1 −1 −t −1 =L =e L = e−t sin t, L s2 + 2s + 2 (s + 1)2 + 1 s2 + 1 2 −1



where this time we have used the inverse form of the shift theorem with a = −1.   s . Ex 8. Find L−1 s2 + 6s + 13 Once again the denominator cannot be factorised into real linear factors, so we begin by completing the square     s s −1 L−1 = L . s2 + 6s + 13 (s + 3)2 + 4 The quotient on the right hand side of the above equation consists of a denominator which is a function of s + 3 whereas the numerator is a function of s. To use the inversion result it is necessary to express both in terms of the same combination of s, so the numerator must be written in terms of s + 3. Hence, we proceed as follows     s (s + 3) − 3 −1 −1 L =L (s + 3)2 + 4 (s + 3)2 + 22     s+3 3 −1 −1 −L =L (s + 3)2 + 22 (s + 3)2 + 22     s 3 −3t −1 −3t −1 −e L L =e s2 + 2 2 s2 + 2 2    3  s −3t −1 −3t −1 2 (2) −e =e L L s2 + 2 2 s2 + 2 2 5

hence the inverse is

e−3t cos(2t) −

3 −3t e sin(2t). 2

5. Solution of differential equations Transform of derivatives The basic properties of transforms have now been briefly discussed but before applying the ideas to solving differential equations it is necessary to consider the transforms of derivatives. The important basic result, as stated in the tables, is  L

df dt

 = sF (s) − f (0).

[Proof: It follows from the definition and integration by parts that  L

df dt



Z

∞

=

e

−st

0

df dt = [e−st f (t)]∞ 0 − dt

Z

∞

f (t) (−se−st ) dt

0

= 0 − f (0) + s F (s) = −f (0) + s F (s) .] The corresponding result for the second derivative is  L

d2 f dt2



= s2 F (s) − s f (0) −

df (0). dt

df [Proof: The easiest way to prove the above is to use the result for the first derivative twice. Put g = dt then    2  dg d f = s G(s) − g(0) . =L L dt2 dt Now using the first derivative result  G(s) = L{g(t)} = L 

and hence L

d2 f dt2

df dt

 = sF (s) − f (0),

 = sG(s) − g(0) = s(sF (s) − f (0)) − = s2 F (s) − sf (0)) −

df (0), dt

df (0) dt

as required.

All the necessary pieces are now in place to apply the method of Laplace transforms to the solutions of ordinary differential equations and we shall consider two examples. dx + x = 1 subject to x = 0 when t = 0. dt Take the Laplace transforms of both sides

Ex 9.

Solve

   dx dx +x =L + L {x} = L{1} dt dt 1 so (sX(s) − x(0)) + X(s) = s 1 (s + 1) X(s) = , s

 L

6

X=

1 s(s + 1)

The function X(s) must now be rewritten using partial fractions A B A(s + 1) + Bs 1 = + = s(s + 1) s s+1 s(s + 1) i.e. s = 0; s = −1;

1 = A(s + 1) + Bs →A=1 → B = −1

1 = A(1) + B(0), 1 = A(0) + B(−1),

Therefore, it follows that x(t) = L−1 {X(s)} = L−1



1 s(s + 1)



= L−1



1 1 − s s+1



= 1 − e−t .

Having worked through the solution of an ODE it is instructive to return to the introduction in section 1 and read through the general description of the transform method, referring to the above example to illustrate each step in the solution process. Finally a more difficult example with a second order equation. dx d2 x dx + 9 x = 50 sin t subject to x = 0 and = 0 when t = 0. +6 2 dt dt dt Taking Laplace transforms and using the linearity property and tables

Ex 10.

Solve

    dx 1 s2 X(s) − sx(0) − (0) + 6{sX(s) − x(0)} + 9X(s) = 50 dt s2 + 1 Recalling the initial conditions and collecting like terms then gives 50 +1 50 2 (s + 3) X(s) = 2 s +1

(s2 + 6s + 9) X(s) = →

therefore X(s) =

s2

50 . (s + 3)2 (s2 + 1)

Partial fractions are again necessary to rewrite the right-hand side into a form which can be inverted. 50 A B Cs + D = + . + 2 2 2 2 (s + 3) (s + 1) s + 3 (s + 3) s +1 Note that the repeated linear factor gives rise above to two factors, in the standard way. The common denominator for the terms on the right-hand side is (s + 3)2 (s2 + 1) so we obtain A(s + 3)(s2 + 1) + B(s2 + 1) + (Cs + D)(s + 3)2 50 = . 2 2 (s + 3) (s + 1) (s + 3)2 (s2 + 1) The latter expression is an identity so equating the two numerators yields 50 = A(s + 3)(s2 + 1) + B(s2 + 1) + (Cs + D)(s + 3)2 . The constants are now found, as usual, through a combination of comparing the coefficients of the powers of s and choosing particular values for s. For the present case with just one linear factor it is appropriate to 7

choose s = −3 which makes the linear terms zero, and then look at coefficients, although other methods of obtaining the answer are possible. s = −3; 3

s ; s; const.

50 = A(0) + B(9 + 1) + (C(−3) + D)(0), 0 = A + C,

→B=5

(1)

0 = A + 6D + 9C, 50 = 3A + B + 9D.

(2) (3)

After substituting the known value for B into equations (1)-(3), the latter form three simultaneous equations for the three unknowns A, C and D. Moreover, using (1) to eliminate C (since it gives C = −A), equations (2) and (3) become −8A + 6D = 0 (or −4A + 3D = 0) and 3A + 9D = 45 (or A + 3D = 15) respectively. The two simplified equations in A and D can now be solved by subtracting to give −5A = −15,

→

A = 3.

Substituting back then leads to D = 4,

C = −3 .

The solution can now be found using linearity of the inverse transform and the tables:  5 −3s + 4 3 + + 2 x(t) = L s + 3 (s + 3)2 s +1         1 1 s 1 −1 −1 −1 −1 +5L +4L −3L = 3L s+3 (s + 3)2 s2 + 1 s2 + 1 −1

Hence



x(t) = 3e−3t + 5te−3t − 3 cos t + 4 sin t.

To conclude it should be emphasised that the Laplace transform method is a black box way of solving a differential equation. Compared with the standard method there is no need, for instance, to consider the type of complementary function that is appropriate for the solution of the auxiliary equation, or the form of the particular integral that follows from the right-hand side of the given differential equation. With the transform method all linear ordinary differential equations are treated with the same approach and the differences in the given ODEs only really appear in the different forms of solution for the transform X(s) and in the details of the partial fraction process. It is the latter rearrangement process which usually forms the major part of the solution. rec/01llt

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