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
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>> 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)
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---------------------------------------- -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
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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
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>>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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