Further Laplace Transforms



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20.3

Introduction In this Block we introduce the first and second shift theorems which will ease the determination of Laplace and inverse Laplace transforms of more complicated causal functions. Finally we obtain the Laplace transform of derivatives of causal functions. This will allow us, in a later block, to apply the Laplace transform in the solution of ordinary differential equations.

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Prerequisites Before starting this Block you should . . .

① be able to find Laplace and inverse Laplace transforms of simple causal functions ② be familiar with integration by parts ③ understand what an initial-value problem is

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Learning Outcomes After completing this Block you should be able to . . .

Learning Style To achieve what is expected of you . . .

☞ allocate sufficient study time ✓ use the shift theorems to obtain Laplace and inverse Laplace transforms ✓ take the Laplace transform of the derivative of a causal function

☞ briefly revise the prerequisite material ☞ attempt every guided exercise and most of the other exercises

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1. The First and Second Shift Theorems The shift theorems enable an even wider range of Laplace transforms to be easily obtained from the transforms we have already found and also enable a significantly wider range of inverse transforms to be found.

The first shift theorem If f (t) is a causal function with Laplace transform F (s), i.e. L{f (t)} = F (s), then as we shall see, the Laplace transform of e−at f (t), where a is a given constant, can easily be found in terms of F (s). Using the definition of the Laplace transform:    ∞ −at −st −at L{e f (t)} = e e f (t) dt 0  ∞ e−(s+a)t f (t)dt = 0



But if

∞

F (s) = L{f (t)} =

e−st f (t)dt

0

then by simply replacing ‘s’ by ‘s + a’ on both sides:  ∞ e−(s+a)t f (t)dt F (s + a) = 0

That is, the parameter s is shifted to the value s + a. We have then the statement of the first shift theorem: Key Point If

L{f (t)} = F (s) then L{e−at f (t)} = F (s + a)

For example, we already know (from tables) that L{t3 u(t)} =

6 s4

and so, by the first shift theorem: L{e−2t t3 u(t)} =

6 (s + 2)4

Now do this exercise Use the first shift theorem to determine L{e2t cos 3t.u(t)} Answer We can also employ the first shift theorem to determine some inverse Laplace transforms. Try each part of this exercise Find the inverse Laplace transform of F (s) =

s2

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

3 − 2s − 8 2

Part (a) Begin by completing the square in the denominator Answer Part (b) Realising that L{sinh 3t u(t)} = theorem

3 , complete the inversion using the first shift s2 − 9 Answer

The second shift theorem The second shift theorem is similar to the first except that, in this case, it is the time-variable that is shifted not the s-variable. Consider a causal function f (t)u(t) which is shifted to the right by amount a, that is, the function f (t − a)u(t − a) where a > 0. The following figure illustrates the two causal functions. f (t)u(t)

f (t − a)u(t − a)

t

a

t

The Laplace transform of the shifted function is easily obtained:  ∞ L{f (t − a)u(t − a)} = e−st f (t − a)u(t − a)dt 0 ∞ = e−st f (t − a)dt a

(Note the change in the lower limit from 0 to a resulting from the step function switching on at t = a). We can re-organise this integral by making the substitution x = t − a. Then dt = dx and when t = a, x = 0 and when t = ∞ then x = ∞. Therefore



∞

e a

−st



∞

e−s(x+a) f (x)dx 0  ∞ −sa = e e−sx f (x)dx

f (t − a)dt =

0

The final integral is simply the Laplace transform of f (x), which we know is F (s) and so, finally, we have the statement of the second shift theorem: 3

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

Key Point L{f (t)} = F (s) then L{f (t − a)u(t − a)} = e−sa F (s)

If

Obviously, this theorem has its uses in finding the Laplace transform of time-shifted causal functions but it is also of considerable use in finding inverse Laplace transforms since, using the inverse formulation of the theorem above: Key Point If

L−1 {F (s)} = f (t) then L−1 {e−sa F (s)} = f (t − a)u(t − a)

Try each part of this exercise Find the inverse Laplace transforms of

(i)

e−3s s2

(ii)

s2

s − 2s + 2

Part (i) Use the table of transforms for standard inverses Answer Part (ii) You will need to complete the square in the denominator Answer

2. The Laplace transform of a derivative 2

Here we consider not a causal function f (t) directly but its derivatives df , d f , . . . (which are dt dt2 also causal). The Laplace transform of derivatives will be invaluable when we apply the Laplace transform to the solution of constant coefficient ordinary differential equations. If L{f (t)} is F (s) then we shall seek an expression for L{ df } in terms of the function F (s). dt Now, by the definition of the Laplace transform    ∞ df df L e−st dt = dt dt 0 This integral can be simplified using integration by parts: 

∞

e 0

∞  ∞  −st (−s)e−st f (t)dt dt = e f (t) − dt 0  ∞ 0 = −f (0) + s e−st f (t)dt

−st df

0

(As usual,we assume that contributions arising from the upper limit ∞ are zero). The integral that remains is precisely the Laplace transform of f (t) which we naturally replace by F (s). Thus   df L = −f (0) + sF (s) dt Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

4

As an example, we know that if f (t) = sin t u(t) then L{f (t)} =

s2

1 ≡ F (s) +1

and so, according to the result just obtained,   df = L{cos t u(t)} = −f (0) + sF (s) L dt   1 = 0+s 2 s +1 s = 2 s +1 a result we know to be true. We can find the Laplace transform of the second derivative in a similar way to find:  2  df = −f  (0) − sf (0) + s2 F (s) L dt2 (The reader might wish to derive this result). Here f  (0) is the derivative of f (t) evaluated at t = 0. Key Point If

L{f (t)} = F (s) then   df L = −f (0) + sF (s) dt  2  df L = −f  (0) − sf (0) + s2 F (s) dt2

Try each part of this exercise d2 f df If L{f (t)} = F (s) and − = 3t with initial conditions f (0) = 1, 2 dt dt the explicit expression for F (s).    2  df df and L{3t} , L Part (a) Begin by finding L 2 dt dt

f  (0) = 0, find

Answer Part (b) Now complete the calculation to find F (s) Answer

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Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

More exercises for you to try 1. Find the Laplace transforms of (i) t3 e−2t u(t) (ii) et sinh 3t.u(t) (iii) sin(t − 3).u(t − 3) 2. If F (s) = L{f (t)} find expressions for F (s) if d2 y dy (i) 2 − 3 + 4y = sin t y(0) = 1, y  (0) = 0 dt dt dy (ii) 7 − 6y = 3u(t) y(0) = 0, dt 3. Find the inverse Laplace transforms of 15 3s2 + 11s + 14 6 (ii) (iii) (i) (s + 3)4 s2 − 2s + 10 s3 + 2s2 − 11s − 52

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

(iv)

e−3s s4

(v)

e−2s−2 (s + 1) s2 + 2s + 5 Answer

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End of Block 20.3

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Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

You should obtain (with a = −2)

s s−2 since L{cos 3t.u(t)} = 2 and so by the first shift theorem 2 (s − 2) + 9 s +9 L{e2t cos 3t.u(t)} =

s−2 (s − 2)2 + 9

obtained by simply replacing ‘s’ by ‘s − 2’. Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

8

s2

3 3 = − 2s − 8 (s − 1)2 − 9

Back to the theory

9

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform



You should obtain −1

L

3 (s − 1)2 − 9

 = et sinh 3t u(t)

Here, in the notation of the shift theorem: f (t) = sinh 3t u(t)

F (s) =

s2

3 −9

and a = −1

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

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(i) You should obtain (t − 3)u(t − 3) for the following reasons. We know that the inverse Laplace transform of 1/s2 is tu(t) and so, using the second shift theorem (with a = 3), we have   −1 −3s 1 L e = (t − 3)u(t − 3) s2 This function is graphed in the following figure:

(t-3)u(t-3)

45 3

0

t

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

You should obtain et (cos t + sin t). To obtain this, complete the square in the denominator: s2 − 2s + 2 = (s − 1)2 + 1 and so s s (s − 1) + 1 s−1 1 = = = + s2 − 2s + 2 (s − 1)2 + 1 (s − 1)2 + 1 (s − 1)2 + 1 (s − 1)2 + 1 Now, using the first shift theorem   s−1 −1 = et cos t.u(t) L (s − 1)2 + 1 

and −1

L

 since

−1

L



1 (s − 1)2 + 1

 t

= e sin t.u(t) 

Thus −1

L

s 2 s − 2s + 2

since

−1

L

s 2 s +1 1 s2 + 1

 = cos t.u(t)  = sin t.u(t)

 = et (cos t + sin t)u(t)

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

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You should obtain L{3t} = 3/s2   df L = −f (0) + sF (s) = −1 + sF (s) dt  2  df L = −f  (0) − sf (0) + s2 F (s) = −s + s2 F (s) 2 dt

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

You should find F (s) =

s3 − s 2 + 3 s3 (s − 1)

since, using the transforms we have found: 3 s2 3 s3 − s2 + 3 F (s)[s2 − s] = 2 + s − 1 = s s2

− s + s2 F (s) − (−1 + sF (s)) = so leading to F (s) =

s3 − s 2 + 3 s3 (s − 1)

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform

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1. (i)

6 (s + 2)4

2. (i)

s3 − 3s2 + s − 2 (s2 + 1)(s2 − 3s + 4)

(ii)

3 (s + 1)2 − 9 (ii)

(iii)

e−3s s2 + 1

3 s(7s − 6)

3. (i) e−3t t3 u(t) (ii) 5et sin 3t.u(t) (iii) (2e4t + e−3t cos 2t)u(t) (v) e−t cos 2(t − 2).u(t − 2)

(iv) 16 (t − 3)3 u(t − 3)

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 20.3: The Laplace Transform