Ngày tải lên :
20/01/2014, 20:20
... p + 133 ,33 p + 10 i (t ) = i V1 (t ) = −0,56e− 133 ,33 t + 2[1,28 cos t + 0,96 sin100t ] 100 ≈ −0,56e− 133 ,33 t + 3, 2 cos( t − 37 ) [ A ] 100 I V (p) = 133 ,33 [ − 7,5.10 3 2,1.10 3 9,6.10 3 p + ... p + 0,72 − + ] p p + 133 ,33 p2 + 10 i V ( t ) = i ( t ) = 133 ,33 [ −7,5.10 3 − 2,1.10 3 e− 133 ,33 t + 10 3. 12 cos( t − 37 )] 100 = 1,6 cos( t − 37 ) − (1 + 0,28 e− 133 ,33 t ) 100 [A] i1(t)=i(t)-i2(t)=1+1,6cos(100t -37 0)+0,28e- 133 ,33 t ... + [ −2 cos324t + sin 31 4t] = 31 4 2e−943t + [ −2 cos324t + sin 31 4t ] = 2e−943t + 6 ,32 cos( t + 71,56 ) 31 4 A + B3 = i2(t)=2e-9 43 t +6 ,32 cos (31 4t+71,560).= 2e-9 43 t +6 ,32 sin (31 4t-18, 430 ) I (p)...