Laplaceova transformace. - Katedra matematiky FEL ČVUT
Laplaceova transformace. - Katedra matematiky FEL ČVUT
Laplaceova transformace. - Katedra matematiky FEL ČVUT
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8p<br />
(p2 +4) 2 + p 2(p−2)<br />
p−2 + p−4 − 2.<br />
Použijeme linearitu, vztahy tf(t) −F ′ (p), f ′ (t) pF (p) − f(0+), e 2t f(t) F (p − 2)<br />
a vzorce sin 2t 2<br />
p 2 +4<br />
, cosh 2t p<br />
p 2 −4<br />
19. f(t) = e−3t cos (2t + π<br />
2 ) + sin 3(t − π)<br />
Je cos (2t + π<br />
2<br />
) = cos 2t cos π<br />
2<br />
− sin 2t sin π<br />
2<br />
, 1 1<br />
p .<br />
= − sin 2t<br />
a sin 3(t − π) = sin 3t cos 3π + cos 3t sin 3π = − sin 3t.<br />
Je f(t) = −e −3t sin 2t − sin 3t tedy F (p) =<br />
−2<br />
(p+3) 2 3 − +4 p2 +9 .<br />
Použijeme linearitu, vztah e −3t f(t) F (p + 3) a vzorce<br />
sin ωt ω<br />
p 2 +ω 2 , ω = 3, ω = 2.<br />
20. f(t) = 5.2 −t − 4t3 t<br />
Je f(t) = 5e −t ln 2 − 4te t ln 3 , tudíž F (p) = 5<br />
p+ln 2 −<br />
4<br />
(p−ln 3) 2 .<br />
Použijeme linearitu, vztah eatf(t) F (p−a), a = − ln 2, a = ln 3 a vzorce 1 1 1<br />
p , t p2 .<br />
21. f(t) = t<br />
Je F (p) = 1<br />
p<br />
2<br />
p 2 +<br />
0 (2 + 4eu sinh 3u)du − (3 t − tsinh 5t) ′<br />
2<br />
p +<br />
4.3<br />
(p−1) 2 −9<br />
<br />
− p<br />
<br />
1<br />
p−ln 3 +<br />
12<br />
p(p2 p 10p2<br />
− −2p−8) p−ln 3 − (p2−25) 2 − 1.<br />
Použijeme linearitu, vztahy t<br />
0<br />
f ′ (t) pF (p) − f(0+), tf(t) −F ′ (p) a vzorce 1 1<br />
p<br />
22. f(t) = 6 sin t sinh 3t − 4t 3 − 2 cos t cosh t<br />
Je f(t) = 3 sin t(e 3t − e −3t ) − 4t 3 − cos t(e t + e −t ), tedy<br />
F (p) =<br />
3<br />
p 2 −6p+10 +<br />
3<br />
(p−3) 2 +1 +<br />
<br />
5<br />
p2 <br />
′<br />
− lim<br />
−25 t→0+ (3t − tsinh 5t) =<br />
f(u)du 1<br />
p F (p), eat f(t) F (p − a), a = 1, a = ln 3,<br />
3<br />
(p+3) 2 3! − 4 +1 p4 − p−1<br />
(p−1) 2 p+1<br />
− +1 (p+1) 2 +1 =<br />
3<br />
p2 24 − +6p+10 p4 − p−1<br />
p2 p+1<br />
− −2p+2 p2 +2p+2 .<br />
, sinh ωt ω<br />
p 2 −ω 2 , ω = 3, ω = 5.<br />
Použijeme linearitu, vztah e at f(t) F (p − a), a = 3, a = −3, a − 1, a = −1 a vzorce<br />
t 3 3!<br />
p 4 , sin t 1<br />
p 2 +1<br />
, cos t p<br />
p 2 +1 .<br />
23. f(t) = (e t sinh t + t 3 e 2t − 6) ′ + 3 t<br />
<br />
Je F (p) = p<br />
1<br />
(p−1) 2 −1<br />
p<br />
p2 6p<br />
+ −2p (p−2) 4 − 6 + 6 + 72<br />
0 (u4 − u cos 2u) du<br />
3! + (p−2) 4 − 6<br />
<br />
p<br />
p6 + 3<br />
p<br />
p 2 +4−2p 2<br />
p 2 +4<br />
− lim<br />
t→0+ (et sinh t + t 3 e 2t − 6) + 3<br />
p<br />
1 6p<br />
= p−2 + (p−2) 4 + 72<br />
p6 + 3(4−p2 )<br />
p(p2 +4) .<br />
<br />
4!<br />
p5 <br />
p<br />
+ p2 <br />
′<br />
+4<br />
Použijeme linearitu, vztahy f ′ (t) (pF (p)−f(0+)), t<br />
1<br />
0 f(u)du pF (p), tf(t) −F ′ (p),<br />
e at f(t) F (p − a), a = 1, a = 2 a vzorce t 3 6<br />
p 4 , t 4 4!<br />
p 5 , sinh t 1<br />
p 2 −1<br />
24. f(t) = 3 − 4 t<br />
0 sinh (t − u) cos udu<br />
<br />
1 p<br />
F (p) = 3<br />
p − 4<br />
p2−1 p2 +1<br />
= 3<br />
p −<br />
4p<br />
(p2−1)(p2 +1)<br />
Použijeme linearitu, větu o obrazu konvoluce (f ∗ g)(t) F (p)G(p) a vzorce<br />
1 1<br />
p<br />
, sinh t 1<br />
p 2 −1<br />
, cos t p<br />
p 2 +1 .<br />
=<br />
, cos 2t p<br />
p 2 +4 .