[PDF] Solutions to Exercises An Introduction to Laplace Transforms

View - Download Solutions to Exercises

Previous PDF

Next PDF

A

Solutions to Exercises

Exercises 1.4

1. (a) lnt is singular at t = 0, hence the Laplace Transform does not exist.

(b)

C{e3t} ;:::: e3te-atdt;:::: __ e(3-a)t ;:::: __ .

1

00 [ 1 ] 00 1

o

3-8 0 8-3

(c) > leMtl for any M for large enough t, hence the Laplace Transform does not exist (not of exponential order). (d) the Laplace Transform does not exist (singular at t = 0). (e) the Laplace Transform does not exist (singular at t = 0). (f) does not exist (infinite number of (finite) jumps), also not defined unless t is an integer.

2. Using the definition of Laplace Transform in each case, the integration is

reasonably straightforward: as in part (b) of the previous question. (b) Integrating by parts gives,

Integrating by parts again gives the result

8 185

186 An Introduction to Laplace Transforms and Fourier Series

(c) Using the definition of cosh(t) gives .c{cosh(t)} = {LX! ete-ddt + LX! e- t e- 3t dt}

1{ II} 8

= 2 8 - 1 + 8 + 1 = 82 - 1 .

3. (a) This demands use of the first shift theorem, Theorem 1.3, which with

b = 3 is .c{ e- 3t

F(t)} = f(8 + 3)

and with F(t) = t 2, using part (b) of the last question gives the answer 2 (8+3)3' (b) For this part, we use Theorem 1.1 (linearity) from which the answer 4 6

82 8 -4

follows at once. (c) The first shift theorem with b = 4 and F(t) = sin(5t) gives 5 5 (8 + 4)2 + 25 - 82 + 88 + 41'

4. When functions are defined in a piecewise fashion, the definition integral

for the Laplace Transform is used and evaluated directly. For this problem we get .c{F(t)} = LX! e- st

F(t)dt = 10

1 te-ddt + 12(2 -t)e-Itdt which after integration by parts gives

5. Using Theorem 1.8 we get

12 (1 -e-

s 8

2t} d 1 1

(a) .c te = d8 (8 _ 2) = (8 _ 2)2 d 8 1-8 2 (b) .c{tcos(t)} = d8 1 + 82 = (1 + 82)2

The last part demands differentiating twice,

2 8 28

3 - 68
(c) .c{t cos(t)} = d82 1 + 82 = (1 + 82)3'

A. Solutions to Exercises 187

6. These two examples are not difficuit: the first has application to oscillating

systems and is evaluated directly, the second needs the first shift theorem with b = 5. (a) C{sin(wt + lPH = 1'X) e- Bt sin(wt + lP)dt and this integral is evaluated by integrating by parts twice using the fol lowing trick. Let I = 1 00 e- 8t sin(wt + lP)dt then derive the formula from which

1-ssin(lP)+wcos(lP)

-s2+W2 •

M 8-5 s-5

(b) C{e cosh(6t)} = (s _ 5)2 _ 36 = S2 -lOs -11·

7. This problem illustrates the difficulty in deriving a linear translation plus

scaling property for Laplace Transforms. The zero in the bottom limit is the culprit. Direct integration yields:

C{G(t)} = roo e-otG(t)dt = roo ae{u-b)8Ia F(u)du

io i-bla where we have made the substitution t = au + b so that G(t) = F(u). In terms of f (as) this is ae- 8b f(as) + ae- Bb r O e- aot

F(t)dt.

i-bla

8. The proof proceeds by using the definition as follows:

which gives the result. Evaluation of the two Laplace Transforms follows from using the results of Exercise 5 alongside the change of scale result just derived with, for (a) a = 6 and for (b) a = 7. The answers are

188 An Introduction to Laplace Transforms and Fourier Series

Exercises 2.8

1. If F(t) = cos (at) then F'(t) = -asin(at). The derivative formula thus

gives

L:{ -a sin (at)} = sL:{ cos(at)} -F(O).

Assuming we know that L:{cos(at)} = A then, straightforwardly s +a s a 2

L:{-asin(at)} = s---1 = ---

S2 +a 2

S2 + a

2 i.e L:{sin(at)} = A as expected. s +a

2. Using Theorem 2.2 gives

In the text (after Theorem 2.4) we have derived that

L: {lot dU} = tan-

1 in fact this calculation is that one in reverse. The result is immediate. In order to derive the required result, the following manip ulations need to take place:

L: { } = LXi e-.t dt

and if we substitute ua = t the integral becomes 1

00 -"'1£ sin(au) d

e --u. o u

This is still equal

to tan- 1

Writing p = as then gives the result. (p

is a dummy variable of course that can be re-Iabelled s.)

3. The calculation is as follows:

L: {[ p(v)dv } =

so

L: {lot Io

v

F(U)dUdV} = ;L: {Io

V

F(U)dU} = sI

2/(8} as required.

A. Solutions to Exercises 189

4. Using Theorem 2.4 we get

£. { (t cos(au) -cos(bu) dU} = ! 1

00 __ u ___ u_ du.

10 U 8 3 a

2 + u 2 b 2 + u 2 1 (8 2 +a 2) These integrals are standard "In" and the result -; In 82 + b2 follows at once.

5. This transform is computed directly as follows

£. { 2 Sin(t)tSinh(t) } := £. { e

t } _ £. { e- t s;n(t) } . Using the first shift theorem (Theorem 1.3) and the result of Exercise 2 above yields the result that the required Laplace Transform is equal to tan- 1

C 1) -tan-

1 (8 1) = tan- 1 (The identity tan- 1 (x) -tan-1(y) = tan- 1 (;.; ;y) has been used.)

6. This follows straight from the definition of Laplace Transform:

lim /(8) = lim [00 e- at

F(t)dt = [00 lim e-

at

F(t)dt := o.

3-+00 3-+00 J 0 J 0 ..... 00

It also follows from the final value theorem (Theorem 2.13) in that if lima-+oo 8/(8) is finite then by necessity lim.-+oo /(8) := O.

7. These problems are all reasonably straightforward

2(28 + 7) 3 1

(a) (8 + 4)(8 + 2) = 8 + 2 + 8 + 4 and inverting each Laplace Transform term by term gives the result 3e- 2t e-4t b) S· ·1 1 8 + 9 2 1 Imlar y 82
_9:= 8-3 -8+3 and the result of inverting each term gives 2e 3t -e- 3t and inverting gives the result 1 1

2" + 2" cos(2kt) := cos

2 (kt). 190
which inverts to An Introduction to Laplace Transforms and Fourier Series

1 1 1

9s 9(s + 3) 3(s + 3)2

-+ 1)e- 3t

99·

(d) This last part is longer than the others. The partial fraction decom position is best done by computer algebra, although hand computation is possible. The result is

1 1 3 1 2

(s -2)2(s + 3)3 = -625(s -2) + 25(8 + 3)3 + 3 + 625(s + 3) e 2t e- 3t and the inversion gives 625 (5t -3) + 1250 (25t2 + 20t + 6). 2 8

8. (a) F(t) = 2 + cos(t) -t 3 as t -t 0, and as -+ -2--1 we also have that

s 8 +

81(s) -t 2 + 1 = 3 as s -t 00 hence verifying the initial value theorem.

(b) F(t) = (4+t)2 -t 16 as t -t o. In order to find the Laplace Transform, we expand and evaluate term by term so that sl(s) = 16 + 8/s + 2/s2 which obviously also tends to 16 as s -t 00 hence verifying the theorem once more. 3 1

9. (a) F(t) = 3 + e-

t -t 3 as t -t 00. f(s) = -+ --so that Sf(8} -t 3 as s 8 + 1

8 -t 0 as required by the final value theorem.

(b) With

F(t) = t

3 e- t, we have 1(8) = 6/(s + 1)4 and as F(t) -t 0 as t -t 00 and 81(8) also tends to the limit 0 as 8 -t 0 the final value theorem is verified.

10. For small enough t, we have that

and using the standard form (or the result on p50): with x = 3/2 gives f{3/2}

C{sm(Vt)} = qVt} + ... = --+ ...

83/2
and using that f{3/2} = (1/2)r{1/2} = Vii/2 we deduce that

C{sin(Vt)} = + ....

A. Solutions to Exercises 191

Also, using the formula given,

k 1 k s3/2 exp -4s = s3/2 + .... Comparing these series for large values of s, equating coefficients of s-3/2 gives k== Vi. 2

11. Using the power series expansions for sin and cos gives

and

00 t4n+2

sin(t 2) == (2n + I)!

00 t4n

cos(t 2) = L(-lt- 2 ,. n. n=O

Taking the Laplace Transform term by term gives

'"{ . (2)} )n (4n + 2)!

L, sm t = f='o -1 (2n + 1)!s4n+3

and

2)} )n (4n)!

£ cos(t = f='o -1 (2n)!s4n+l·

12. Given that Q(s) is a polynomial with n distinct zeros, we may write

P(s) Al A2 Ak An

Q(s) s-al s-a2 s-ak s-an

where the AkS are some real constants to be determined. Multiplying both sides by s -ak then letting s -+ ak gives . P(s) . (s -ak)

Ak = hm Q( )(s -ak) == hm P(s) Q() .

• S S

Using I'H6pital's rule now gives

At = P(ak) QI(ak)

for all k == 1,2, ... , n. This is true for all k, thus we have established that

P(s) _ P(ad 1 + ... + P(ak) 1 + ... P(an) 1

Q(s) Q'(ad (s -ad QI(ak) (s -ak) QI(an} (s -an)

Taking the inverse Laplace Transform gives the result

£-1 {P(S)} = t P(ak) ea•t

Q(s) k=1 QI(ak)

sometimes known as Heaviside's expansion formula.

192 An Introduction to Laplace Transforms and Fourier Series

13. All of the problems in this question are solved by evaluating the Laplace

Transform explicitly.

(a) C{H(t -an = e-otdt = _e -. 1

00 -a8

a 8 (b) C{h (tn = 10 2 (t + 1)e- 8t dt + lOO 3e- st dt. Evaluating the right-hand integrals gives the solution 1 1 (28 ) - + -e--1. 8 82 (c) c{h(t)} = 10 2 (t + l)e- st dt + 1 00 6e- ot dt.

Once again, evaluating gives

(d) As the function h(t) is in fact continuous in the interval [0,00) the formula for the derivative of the Laplace Transform (Theorem 2.2) can be used to give the result !(e- 28
-1) at once. Alternative, h can be 8 differentiated (it is 1 -H(t -2)) and evaluated directly.

14. We use the formula for the Laplace Transform of a periodic function The

orem

2.19 to give

_ J;c e- Bt

F(t)dt

C{F(tn -(1 _ e28C) .

The numerator is evaluated directly:

which after routine integration by parts simplifies to

The Laplace Transform is thus

C{F( )} 1 1 (-BC 1)2 1 1 -e-

sc t = 1 _ e2BC 82 e -= 82 1 + eSC which simplifies to 8 12 tanh G8C) .

A. Solutions to Exercises 193

Exercises 3.6

1. (a) IT we substitute u = t -r into the definition of convolution then

9 * f = lot g(r)f(t -r)dr

becomes -1 0 g(u -r)f(u)du = 9 * f· (b) Associativity is proved by effecting the transformation (u, r) -t (x, y) where u = t -x -y, and r = y on the expression l t l t-T f * (g * h) = 0 0 f(r)g(u)h(t -r -u)dudr. The area covered by the double integral does not change under this trans formation, it remains the right-angled triangle with vertices (0, t), (0,0) and (t, 0). The calculation proceeds as follows:

8(u, r) dudr

= 8(x,y) dxdy = -dxdy so that i t rt-z f * (g * h) = 0 10 f(y)g(t -x -y)h(x)dydx = 1t h(x) [1 t-x f(y)g(t -x -Y)dY] dx = l h(x)[f * g)(t -x)dx = h * (f * g) and this is (f * g) * h by part (a) which establishes the result. (c) Taking the Laplace Transform of the expression f * f- 1 = 1 gives

C{f}.C{f-l} = !

8 from which

C{f-l} =

using the usual notation (/(8) is the Laplace Transform of f(t». It mustquotesdbs_dbs42.pdfusesText_42

[PDF] the fourier transform and its applications bracewell pdf

[PDF] fiche pédagogique fourmi

[PDF] nouvelle ? chute cm2

[PDF] documentaire fourmi ce1

[PDF] fourmis maternelle activités

[PDF] texte documentaire sur les fourmis ce2

[PDF] leçon sur les fourmis

[PDF] nouvelles ? chute

[PDF] exposé sur les fourmis cm2

[PDF] fiche documentaire sur les fourmis

[PDF] cycle de vie d'une fourmi

[PDF] tpe fourmis experience

[PDF] algorithme colonie de fourmis pdf

[PDF] organisation sociale des fourmis

[PDF] fourmi reine