Ex 1: Determiner transfo laplace des fonctions proposer L[f(t-T)U(t-T)] = F(p) * e**-Tp h(t)= t*U(t) - 2(t-2)*U(t-2) + (t-4)*U(t-4) H(p)= 1/p**2 - 2*(1/p**2 * e**-2p) + 1/p**2 * e**-4p L[f(t+7)U(t-9)] = [(t-9+16)U(t-9)] = 1/p**2 * e**-9p + 16/p * e**-9p Ex 2:Retrouver l'original de chaque transfo laplace G(p)= 1/p - 1/p+1 = g(t)= (1-e**-t)U(t) J(p)=G(p) * e**-3p = (1-e**-3p)* U(t-3) Ex 3:Détermine les reels a et b: 1) 1/p(p+1/4) = a/p + b/p+1/4 => a(p+0,25) + bp / p(p+1/4) = ap + 0,25a + bp / p(p+1/4) = (a+b)p + 0,25a / p(p+1/4) a+b = 0 ; 1/4a = 1 => a = 4 b=-4 2) Completer le tableau F(p) f(t) 1/p 1U(t) 1/p * e**-2p U(t) * 1/p+2 1/p+1/4 e**-1/4t U(t) 1/p+1/4 * e**-2p e**-1/4p**(t-2) U(t-2) 3) On donne S(p)= 1/p(p+1/4) * (1-e**-2p) = 4(1/p - 1/p+1/4) * (1-e**-2p) (formule avec a et b) s(t)= 4( U(t)-e**-1/4t U(t) - U(t-2)-e**-1/4(t-2) U(t-2)] 4[(1-e**-1/4t)U(t) - (1-e**-0,25(t-2) U(t-2)] Ex4: determiner la transfo F1(p)= 10/ p**2 +10p+34 = 10/ (p+5)**2 +9 = 10/3 * 3/(p+5)**2 +3**2 f1(t)= 10/3 sin(3t)e**-5t**2 12p=12(p+8-8) = 12(p+8)+12*(-8) -> 2*6 F2(p)=12p/p**2 +16p+100 = 12p/(p+8)**2 +36 = 12* p+8/(p+8)**2 + 6**2 -8*2* 6/(p+8)**2 +6 f2(t)= 12cos(6t)e**-8t U(t) - 16sin(6t)e**-8t U(t) = 4[3cos(6t)-4sin(6t)]e**-8t U(t) On cherche : p**2 + 23 p + 120 = 0 du type : a p**2 + bp + c = 0 avec, a = 1 ; b = 23 ; c = 120. Calcul du discriminant : ∆ = b**2 – 4 a c = 23**2 – 4 × 1 × 120 = 529 – 480 = 49 > 0 Il y a donc deux racines : p1= -b-racdelta /2a = -23-rac49/ 2*1 = -30/2 = -15 p2= -b+racdelta /2a = -23+rac49/ 2*1 = -16/2 = -8 Ainsi 1p**2 + 23 p + 120 = 1 (p–(–15)) (p–(–8)) = (p+15) (p+8) F3(p)= p+10/p(p**2 +23p+120) = p+10/p(p+15)(p+8) = a/p + b/p+8 + c/p+15 a= (p*F3(p)) p=0 = (p+10/(p+8)(p+15)) = 1/12 b= ((p+8)*F3(p)) p=-8 = (p+10/p(p+15)) p=-8 = -1/28 c= ((p+15)*F3(p)) p=-15 = (p+10/p(p+8)) p=-15 = 1/21 f3(t)= (1/12 - 1/28 e**-8t +1/21 e**-15t) U(t) On cherche : p**2 + 23 p + 60 = 0 du type : a p**2 + bp + c = 0 avec, a = 1 ; b = 23 ; c = 60. Calcul du discriminant : ∆ = b**2 – 4 a c = 23**2 – 4 × 1 × 60 = 529 – 240 = 289 > 0 Il y a donc deux racines : p1= -b-racdelta /2a = -23-rac289/ 2*1 = -40/2 = -20 p2= -b+racdelta /2a = -23+rac289/ 2*1 = -6/2 = -3 Ainsi 1p**2 + 23 p + 60 = 1 (p–(–3)) (p–(-20)) = (p+3) (p+20) F4(p)= 16p/(p+16)(p**2 +23p+60) = 16p/(p+16)(p+3)(p+20) = a/p+16 + b/p+3 + c/p+20 a= ((p+16)*F4(p)) p=-16 = (16p/(p+3)(p+20)) = 64/13 b= ((p+3)*F4(p)) p=-3 = (16p/(p+16)(p+20)) p=-3 = -48/221 c= ((p+20)*F4(p)) p=-20 = (16p/(p+16)(p+3)) p=-20 = -80/17 f4(t)= (64/13 e**-16t - 48/221 e**-3t -80/17 e**-20t) U(t)