文档介绍：第二章****题课(2-1) 2-1 试建立图所示电路的动态微分方程解: 输入 u i输出 u ou 1=u i -u oi 2 =C du 1dt) -R 2 (u i -u o )=R 1u 0 -CR 1R 2( du idtdt du o C+- +- u iu oR 1R 2i 1i i 2u 1i 1 =i-i 2u o i=R 2u 1i 1=R 1= u i -u oR 1dt d(u i -u o) =CC d(u i -u o)dt u o-R 2= u i -u oR 1 CR 1R 2du odt du idt +R 1u o +R 2u 0 =CR 1R 2 +R 2u i (a) 第二章****题课(2-1) i=i 1 +i 2i 2 =C du 1dt u oi 1=R 2u 1 -u o= LR 2du odt R 1 i= (u i -u 1) (b) C+- +- u iu oR 1R 2i 1ii 2Lu 1=R 1u i -u 1u o +C R 2du 1dt u 1=u o+ LR 2du odtdu odt R 1R 2L du odt + CL R 2d 2u odt 2= - - u iR 1u oR 1u oR 2 +C)u oR 1R 2L du odt ) CL R 2d 2u odt 2= + +( u iR 11R 11R 2 +(C+ 解: 2-2 求下列函数的拉氏变换。(1) f(t )=sin4t+cos4t L[ sin ωt] = ωω 2 +s 2= s+4 s 2 +16 L[ sin4t+cos4t ] = 4s 2 +16 ss 2 +16 + sω 2 +s 2L[ cos ωt]= 第二章****题课(2-2) 解: (2) f(t )=t 3 +e 4t解: L[t 3 +e 4t] = 3!s 41 s-4 + 6s+24+s 4s 4 (s+4) = (3) f(t )=t ne atL[t ne at ]= n! (s-a) n+1 (4) f(t )=1(t- τ)e 2tL[ 1(t- τ)e 2t ]=e -τs1 s-2 第二章****题课(2-2) 解: 解: 2-3 求下列函数的拉氏反变换。 A 1 =(s+2) s+1 (s+2)(s+3) s=-2 =-1 =2 f(t )=2e -3t -e -2t (1) F(s )= s+1 (s+2)(s+3) 第二章****题课(2-3) 解: A 2 =(s+3) s+1 (s+2)(s+3) s=-3 F(s )= 2 s+3 1 s+2 - = A 1 s+2 s+3 + A 2 (2) F(s )= s (s+1) 2 (s+2) f(t )=-2e -2t -te -t +2e -t 第二章****题课(2-3) 解: = A 2 s+1 s+2 + A 3 + A 1 (s+1) 2A 1 =(s+1) 2s (s+1) 2 (s+2) s=-1 A 3 =(s+2) s (s+1) 2 (s+2) s=-2 dds s s+2 ][A 2 = s=-1 =-1 =2 =-2 (3) F(s )= 2s 2 -5s+1 s(s 2 +1) F(s)(s 2 +1) s=+j =A 1 s+A 2 s=+j A 2 =-5 A 3= F(s)s s=0 f(t )=1+cost-5sint 第二章****题课(2-3) 解: = s + A 3s 2 +1 A 1 s+A 2 =1 s 2s 2 -5s+1 =A 1 s+A 2 s= j s= j j -2-5j+1 =jA 1 +A 2 -5j-1=-A 1 +jA 2 A 1 =1 F(s )= 1ss 2 +1 s -5s 2 +1 + + (4) F(s )= s+2 s(s+1) 2 (s+3) 第二章****题课(2-3) 解: = + s+1 A 1 s+3 A 2 (s+1) 2+s A 3+ A 4 -12 A 1 = 23 A 3 = 112 A 4 = A 2 = d[ s=-1 ds ] (s+2) s(s+3) -34 = -34 A 2 = + -4 3+ f(t )= e -t3 2e -3t 2 -te -t12 1 =