KEDF_U3_EA_RACG
3
1 f ( t) =( e 2 t + 3) 4 (4 ) f ( s) = s 2 +8 s ( s 2 + 16 ) 2 f ( t) = [ cos ( π 6 ) −t ] e −t (5 ) f ( s) = e 2 ( s +2) √ π s +2 3 f ( t) =( 1−2 √ t ) 2 e 3 t (3 ) f ( s) = 1 2 ( s 2 −9 ( s 2 −9) 2 ) + 1 2 ( s 2 −49 ( s 2 −49 ) 2 ) 4 f ( t) =cos 2 2 t (1 ) ¿ f ( s) = 1 s−8 + 12 s−6 + 54 s−4 + 108 s−2 + 81 s− 5 f ( t) =e 1−2 t √ t (2 ) f ( s) = 0.86 ( s+ 1 ) ( s+1 ) 2 + 1 + 0.5 ( s+1 ) 2 +1 1.- f ( t) =( e 2 t + 3) 4 =e 8 t + 12 e 6 t +54 e 4t +108 e 2t +81 L ( e 8 t + 12 e 6t +54 e 4t +108 e 2t +81 )=L ( e 8 t ) +L ( 12e 6 t ) +L ( 54 e 4t ) +L ( 108 e 2t ) L ( e 8 t + 12 e 6t +54 e 4t +108 e 2t +81 )= 1 s−8 + 12 s−6 + 54 s−4 + 108 s−2 + 81 s *este tiene un error porque la transformada de laplace de 81 es 81/s y no 81/(s-1) 2.- f ( t) = [ cos ( π 6 −t ) ] e −t = [ cos ( π 6 ) cos ( t ) +sin ( π 6 ) sin ( t) ] e −t f ( t) = [ 0.866 cos ( t ) +.5 sin ( t) ] e −t =0.866 cos ( t ) e −t +.5 sin ( t) e −t L ( 0.866cos ( t) e −t +.5sin ( t ) e −t )=L ( 0.866cos ( t) e −t ) +L ( . 5sin ( t ) e −t ) L ( 0.866cos ( t) e −t )+L ( .5 sin ( t ) e −t )
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Transcript of KEDF_U3_EA_RACG
1 f ( t )=(e2 t+3 )4 (4)f ( s )= s2+8
s ( s2+16 )
2f (t )=[cos( π6 )−t ]e−t (5)
f ( s )= e2 ( s+2 ) √ π
s+2
3 f ( t )=(1−2√t )2 e3 t (3)f ( s )=1
2 ( s2−9
( s2−9 )2 )+ 12 ( s2−49
( s2−49 )2 )4 f (t )=cos22t (1)
¿ f ( s )= 1s−8
+ 12s−6
+ 54s−4
+ 108s−2
+ 81s−1
5 f (t )=e1−2 t√ t (2)f ( s )=0.86 (s+1 )
(s+1 )2+1+ 0.5
(s+1 )2+1
1.-
f ( t )=(e2 t+3 )4=e8 t+12e6 t+54e4 t+108 e2 t+81
L (e8 t+12e6 t+54e4 t+108 e2 t+81 )=L (e8t )+L (12e6 t )+L (54 e4 t )+L (108e2t )
L (e8 t+12e6 t+54e4 t+108 e2 t+81 )= 1s−8
+ 12s−6
+ 54s−4
+ 108s−2
+ 81s
*este tiene un error porque la transformada de laplace de 81 es 81/s y no 81/(s-1)
2.-f (t )=[cos( π6−t)]e−t=[cos ( π6 )cos (t )+sin( π6 )sin ( t )]e−tf ( t )=[0.866 cos (t )+.5 sin ( t ) ] e−t=0.866 cos (t )e−t+ .5sin ( t ) e−t
L (0.866cos ( t ) e−t+.5 sin (t )e−t )=L (0.866cos ( t ) e−t )+L(.5sin ( t ) e−t)
L (0.866cos (t ) e−t )+L (.5 sin ( t )e−t )
Usando
.866 s1+s2
+ .51+s2
∨¿s→s+1→.866 (s+1)1+(s+1)2
+ .51+(s+1)2
¿
3.-f (t )=(1−2√t )2 e3 t=(1−2√ t+4 t )e3 t=e3 t−2√ t e3 t+4 t e3 t
L (f ( t ) )=L (e3 t )−2L (√ t e3 t )+4 L ( t e3 t )
Usando
L (f (t ) )= 1s−3
−2 √ π2 s3/2
∨¿s→s−3+41s2
∨¿s→s−3⋱¿¿
1s−3
− 2√π2(s−3)3 /2
+ 4¿¿
4.-
f ( t )=cos2 (2t )
f ( t )=12+ 12cos (4 t )
L (f ( t ) )=L( 12 )+L( 12 cos (4 t ))L (f ( t ) )= 1
2 s+ s2(s2+16)
= s2+8s ( s2+16 )
5.-
f ( t )=e1−2 t√ t= e
e2 t√t=e e−2t √t
usando
L (f ( t ) )=e √π2 s3 /2
∨¿s→s+2→e√π
2(s+2)3/2= e2 ( s+2 ) √ π
s+2¿
