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7/18/2019 Ore

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b) obtener con Matlab la transformada de Laplace de las siguientesfunciones:

a) f(t) = 2-3t +5

2 t2

>> syms z t

>> z = 2 - (3t) + ((!"2)t#2)

$ =

(!t#2)"2 % 3t + 2

>> Laplace (z)

&ns =

2"! % 3"s#2 + !"s#3>> pretty(ans)

2 3 !

- -- + - -

! 2 3

! !

b) f(t) = e-'t % (!t % t2)

>> syms * t e

>> *= (e(-'t) % !t +t#2

>> Laplace (*)

&ns =

2"s#3 % !"s#2 + Laplace("e#('t),t,s)

>> pretty(ans)2 !

- - - - -- + transform::Laplace ----,t,s3 2 't

! ! e

c) f(t) =

11

5 e

-3t

-

1

5 e

2t

7/18/2019 Ore


>> syms y e t>> y = (("!)(e#(-3t))) % (("!)!sin(2t))

="(!e#(3t)) % sin(2t)"!>> Laplace(y)

ans =(Laplace("e#(3t),t,s))"! % 2"(!(s#2 + '))>>pretty(ans) " . transform:: laplace / ----,t,s /

/3t / . e "------------------------------------------- - ! 2----------------- 2 ! (s + ')

d)f(t) = 3cos(2t) % "2sen(2t)>> syms 0 t>> 0 = (3cos(2t)) %n("2sin(2t))0 =3cos(2t) % sin(2t)"2>>Laplace(0)&ns

(3s)"(s#2 + ') % "(s#2 + ')>>pretty(ans)3 s ----- - ----- 2 2s + ' + s + '

e) f(t) = e2t (2 % cos(3t) + !"3sen(3t))>> syms 1 t>> 1 = (e#(2t))(2-(cos(3t))) + ((!"3)(sin(3t)))1 =

(!sin(3t))"3 % e#(2t)(cos(3t) % 2)>> Laplace(1)ans =!"(s#2 + ) + 2Laplace(e#(2t), t, s) % laplace(e#(2t), t, s -3i)"2 %Laplace(e#(2t), t, 3i + s)"2>>pretty(ans)

!---------2

+ 4ransform::Laplace(e , t, s % 3 5)

7/18/2019 Ore


---------------------------------------- -2

2t 4ransform::Laplace(e , t, s 35)-------------------------------------- +

2 2t2 transform::Laplace(e , t, s)

f)f(t) = 'e-at cos(0t)>> syms e 0 t s>> s = '(e#(at))cos(0t)s ='e#(at)cos(0t)>>Laplace()ans =2laplace(e#(at), t, s + i0) + 2Laplace(e#(at), t, s % i0)>> pretty(ans) &t2transform::Laplace(e , t, s % 05) + at2transform::Laplace(e , t, s + 05)

6btener con Matlab la transformada in1ersa de Laplace de lassiguientes 784:La transformada mas general de encontrar la transformada in1ersa esdescomponer la funci9n 7(s) en frecciones parciales parciales y luego

aplicar su in1ersa a cada terminoa) 7(s)=

5−3 s+2 s2

s3

>> syms s a>> a = (!-(3s)+(2s#2))"(s#3)a =(2s#2 % 3s + !)"s#3>> ilaplace(a)ans =(!t#2)"2 % 3t + 2

b) 7(s) =2 s+1

s2−4 s+3

>> syms b s>> b = ((2s)+)"((s#2)-('s)+3)

b =(2s + )"(s#2 -'s + 3)>>ilaplace(b)ans =(;e*p(3t))"2 % (3e*p(t))"2

7/18/2019 Ore


c) 7(s) =2

s ( s+0.5 )

>>syms c s

>> c = 2"(s(s + !))c =

2"(s( + <))

>> ilaplace(c)

ans =

' -'"e*p(t"2)

d) (s) =

3 s−5

s2+s−6

>> syms d s>> d = ((3s) % !)"((s#2) + s % )d =(3s % !)"(s#2 + s % )>> ilaplace(d)ans =e*p(2t)"! + '"(!e*p(3t))

e) 7(s) =3 s−1

s2+4

>>syms e s>> e = ((3s) -)"((s#2) + ')e =(3s -)"(s#2 + ')>> ilaplace(e)ans =3cos(2t) % sin(2t)"2

f) ? (s) =5 s

(s+4 )3

>> syms f s>> f = (!s)"((s + ')#3)f =(!s)"(s + ')#3>> ilaplace(e)ans =3cos(2t) % sin(2t)"2@ncuentra las respuestas temporales descritas por lasecuaciones diferenciales usando la transformada de Laplace

a) 2dy

dt + y = 3 * = A y =

>>syms f t y *

7/18/2019 Ore


>>dsol1e(B28y+y-3=C)ans =D!"2e*p(3t)) + <

b) d

2 x

dt 2 +

dy

dt = 3 Adx

dt = , * =

>> syms f t *>> dsol1er(E82* + 8y % 3 = E)ans =D' + t"2 % diF(*(t), t)"

c) d

2

dt 2 % * = 2et , G = ,

dx

dt =

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