8 15-5 Several basis theorems for the Laplace transform 1.The linearity theorem Lf(t)+f(0=F(s)+F2(s) Example: Iv(s) Find(t)=L-V(s)) (s+a)(s+B) a-B a-B 十 (s+a)(s+B) S+a u(t=L v(s)=La-B a-B S+a s十 B cptu(t e eu()+ e B-a B-a B-a ko(t)<kv(s)
§15-5 Several basis theorems for the Laplace transform 1.The linearity theorem ( ) ( ) ( ) ( ) 1 2 1 2 L f t + f t = F s + F s Example: ( ) ( ) ( )( ) 1 ( ) 1 Find t L V s s s If V s − = + + = + − + + − − = + + = s s s s V s 1 1 ( )( ) 1 ( ) ( ) 1 ( ) 1 e u t e u t t t − − − − + − = u(t) e e t t − − = − − + − + + − − = = − − s s t L V s L 1 1 ( ) ( ) 1 1 k(t) kV(s)
2. The time differentiation theorem du(t) s(s)-b(0-) d u(t) dt dt 2 冷→s(s)-sb(0-)-U(0-) 3. Time integration theorem (s) D(x)d分> 4. The sinusoidal function Jat-e-jot Rsin@ tu(t)=L 2j j's-jo S+j@ S+O LcoS otu(o)=L3 lejot +e-jot 2 2 S-jo S+jo S+O d sinat Lcos o tu(t)9=L o dt S"+ s+O
2. The time differentiation theorem sV(s) ( ) dt d (t ) − 0− s V(s) s ( ) ( ) dt d (t ) ' − 0− − 0− 2 2 2 3. Time integration theorem − t s V s x dx 0 ( ) ( ) 4. The sinusoidal function − = − j e e L t u t L j t j t 2 sin ( ) + = − 2 cos ( ) j t j t e e L t u t L = dt d t L t u t L sin cos ( ) 2 2 1 1 2 1 + = + − − = s ) s j s j ( j 2 2 1 1 2 1 + = + + − = s s ) s j s j ( 2 2 2 2 1 + = + = s s s s
5. Time shift theorem If a time function is delayed by a time a in the time domain, the result in the frequency domain is a multiplication bye Iff(t)+ F(s)we have Lf(t-a)u(t-a=e-as F(s)(a20) Example 1: Iff(0)=(a)5() (b)5u(t-5) (c)5u(t+5). Find F(s Solution (a)F(s)=5/;(b)F(s=(5S)es e 5 (c)F(s)=。5n(t+5)edt 5 e-s=5 s|0
5. Time shift theorem If a time function is delayed by a time in the time domain, the result in the frequency domain is a multiplication by . If we have s e − f (t) F(s) ( − ) ( − )= ( ) ( 0) − L f t u t e F s s Example 1: If f (t) = (a) 5u(t) (b) 5u(t-5) (c) 5u(t+5). Find F(s) Solution: (a) F(s)=5/s; (b) F(s)=(5/s)e-5s ; s s e e dt st st 5 0 5 5 0 = − = = − − dt c F s u t e st − = + 0 ( ) ( ) 5 ( 5)
Example 2: Iff(=2e-3i/u(t+1)-u(t-2)/, find F(s) Solution: f(t=2e -3tu(t+1)-2e-3tu(t-2) =2e-3u(t+1)-2e6e3x2)u(t2) 2 2 F(s)= 2e (1 s+3 s+3 s+3 Example 3: Find F(s) f() Solution f( =Au(-Au(t-dy .F(S) ds e
Example 2: If f (t)=2e -3 t[ u(t+1)- u(t-2)], find F(s). Solution: f (t) =2e -3 tu(t+1)- 2e -3 tu(t-2) =2e –3 tu(t+1)- 2e -6e -3(t-2)u(t-2) (1 ) 3 2 3 1 2 3 2 ( ) 6 2s 6 2s e s e s e s F s − − − − − + = + − + = Example 3: Find F(s) t f (t) 0 A d f(t)=Au (t)-Au (t-d) Solution: ( ) (1 ) ds ds e s A e s A s A F s − − = − = −
Example 4: Find F(s) f(t) 3 3 3-23-23 Solution: f(t)==tu(t) ()-tu(t-2)+3u(t-2) 2 =t()-(t-2)u(t-2)-3u(t-2)+3(t-2) t() (t-2)u(t-2) 31 S)=-× 3-232 3
Example 4: Find F(s) ( 2) 3 ( 2) 2 3 ( ) 2 3 f (t) = tu t − tu t − + u t − ( 2) ( 2) 3 ( 2) 3 ( 2) 2 3 ( ) 2 3 = t u t − t − u t − − u t − + u t − ( 2) ( 2) 2 3 ( ) 2 3 = tu t − t − u t − ( e ) s e s s F(s) s 2s 2 2 2 2 1 2 1 3 2 1 3 2 3 − − = − = − Solution: 2 3 f (t) t
6. Convolution f(0)*f2()分F(s)F2(s)分f2()*f( (s)= V1(s)2(S) (s+a(s+B) s+a S+B H1()= U1(t)=e(t) S+al V2(s)= U2()=e-Bt( S+B D()=C1(sH2(s分2=01()+(O)=D(2(t-2 So e -adu(aye B(-au(t-a)di=le BeBida e-ae(p-a)ddn =e Bt 1 (B-a) e-pt (B-a)t eat B t u(t) B B-a
6. Convolution ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 2 1 f t f t F s F s f t f t ( ) ( ) 1 1 ( )( ) 1 ( ) 1 2 V s V s s s s s V s = + • + = + + = ( ) ( ) 1 ( ) 1 1 t e u t s V s t − = + = ( ) ( ) 1 ( ) 2 2 t e u t s V s t − = + = ()2 (t )d 0 = 1 − − − = t t e e e d 0 0 1 ( ) t e e t − − − = u(t) e e t t − − = − − ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 t = L V S V s = t t − e u e u t d t ( ) ( ) ( ) 0 = − − − − − − = t t e e d 0 ( ) 1 ( ) ( ) e u t e t t − − = − −
