自动控制原理及其应用课后习题答案第二章.pdf

第二章习题课第二章习题课 (2-1a)(2-1a)(2-1a)(2-1a) u u u uo o o o i i i i2 2 2 2= = = = R R R R2 2 2 2 C C C C u u u ui i i iu u u uo o o o R R R R1 1 1 1 R R R R2 2 2 2 2-1(a) 2-1(a) 2-1(a) 2-1(a) 试建立图所示电路的动态微分方试建立图所示电路的动态微分方 程。程。 解：解： 输入量为输入量为u u u ui i i i，输出量为，输出量为u u u uo o o o。u u u ui i i i=u=u=u=u1 1 1 1+u+u+u+uo o o o u u u u1 1 1 1=i=i=i=i1 1 1 1R R R R1 1 1 1i i i ic c c c=C=C=C=C dudududuc c c c dt dt dt dt = = = = dt dt dt dt d d d d( ( ( (u u u ui i i i-u-u-u-uo o o o) ) ) ) i i i i1 1 1 1=i=i=i=i2 2 2 2-i -i -i -ic c c c u u u u1 1 1 1= = = = R R R R1 1 1 1+ + + +u u u uo o o o u u u uo o o o - - - -C C C C d d d d( ( ( (u u u ui i i i-u-u-u-uo o o o) ) ) ) dt dt dt dt R R R R2 2 2 2 R R R R2 2 2 2u u u ui i i i=u=u=u=uo o o oR R R R1 1 1 1-C R-C R-C R-C R1 1 1 1R R R R2 2 2 2+C R+C R+C R+C R1 1 1 1R R R R2 2 2 2+u+u+u+uo o o oR R R R2 2 2 2 dudududui i i i dt dt dt dtdt dt dt dt dudududuo o o o u u u uo o o oR R R R1 1 1 1+C R+C R+C R+C R1 1 1 1R R R R2 2 2 2+u+u+u+uo o o oR R R R2 2 2 2=R=R=R=R2 2 2 2u u u ui i i i+C R+C R+C R+C R1 1 1 1R R R R2 2 2 2 dudududuo o o odudududui i i i dt dt dt dtdt dt dt dt C C C C u u u uc c c c R R R R1 1 1 1 R R R R2 2 2 2 u u u ui i i i i i i i1 1 1 1 i i i i2 2 2 2 u u u uo o o o i i i ic c c cC C C C 第二章习题课第二章习题课 (2-1b)(2-1b)(2-1b)(2-1b) 2-1(b) 2-1(b) 2-1(b) 2-1(b) 试建立图所示电路的动态微分方试建立图所示电路的动态微分方 程。程。 u u u uo o o o u u u ui i i i R R R R1 1 1 1L L L L R R R R2 2 2 2 C C C C i i i i1 1 1 1= = = =i i i iL L L L+i+i+i+ic c c c u u u uL L L L=L=L=L=Ldi di di diL L L L dt dt dt dt u u u uo o o o i i i iL L L L=i=i=i=i2 2 2 2= = = = R R R R2 2 2 2 u u u uL L L L= = = = L L L L R R R R2 2 2 2 dudududuo o o o dt dt dt dt i i i ic c c c= = +C= = +C= = +C= = +C dudududuc c c c dt dt dt dt CLCLCLCL R R R R2 2 2 2 d d d d2 2 2 2u u u uo o o o dt dt dt dt2 2 2 2 dudududuo o o o dt dt dt dt + + + + u u u uo o o o R R R R2 2 2 2 CLCLCLCL R R R R2 2 2 2 d d d d2 2 2 2u u u uo o o o dt dt dt dt2 2 2 2 dudududuo o o o dt dt dt dt i i i i1 1 1 1= = = = +C +C +C +C u u u uo o o o i i i i2 2 2 2= = = = R R R R2 2 2 2 输入量为输入量为u u u ui i i i，输出量为，输出量为u u u uo o o o。 u u u ui i i i=u=u=u=u1 1 1 1+u+u+u+uo o o o u u u u1 1 1 1=i=i=i=i1 1 1 1R R R R1 1 1 1 i i i ic c c c=C=C=C=C dudududuc c c c dt dt dt dt = = = = dt dt dt dt d d d d( ( ( (u u u ui i i i-u-u-u-uo o o o) ) ) ) 习题课一习题课一 (2-2)(2-2)(2-2)(2-2) 求下列函数的拉氏变换。求下列函数的拉氏变换。 (1) (1) (1) (1) f(tf(tf(tf(t)=sin4t+cos4t)=sin4t+cos4t)=sin4t+cos4t)=sin4t+cos4t 解解: : : :L L L L sinwtsinwtsinwtsinwt = = = = w w w w w w w w2 2 2 2+s+s+s+s2 2 2 2 s s s s w w w w2 2 2 2+s+s+s+s2 2 2 2 L L L L sin4t+cos4tsin4t+cos4tsin4t+cos4tsin4t+cos4t = = = = 4 4 4 4 s s s s2 2 2 2+16+16+16+16 s s s s s s s s2 2 2 2+16+16+16+16 = = = = s+4s+4s+4s+4 s s s s2 2 2 2+16+16+16+16 + + + + L L L L coswtcoswtcoswtcoswt = = = = (2) (2) (2) (2) f(tf(tf(tf(t)=t)=t)=t)=t3 3 3 3+e+e+e+e4t 4t 4t 4t 3!3!3!3! 解解: : : :L L L L t t t t3 3 3 3+e+e+e+e4t 4t 4t 4t = + = += + = += + = += + = + 3!3!3!3! s s s s3+1 3+13+13+1 1 1 1 1 s-4s-4s-4s-4 s s s s4 4 4 4 1 1 1 1 s-4s-4s-4s-4 (3) (3) (3) (3) f(tf(tf(tf(t)=)=)=)=t t t tn n n ne e e eat at at at 解解: : : :L L L L t t t tn n n ne e e eat at at at= n!n!n!n! (s-a)(s-a)(s-a)(s-a)n+1 n+1n+1n+1 (4) (4) (4) (4) f(tf(tf(tf(t)=(t-1)=(t-1)=(t-1)=(t-1)2 2 2 2e e e e2t 2t 2t 2t 解解: : : :L L L L (t-1)(t-1)(t-1)(t-1)2 2 2 2e e e e2t 2t 2t 2t=e e e e-(s-2) -(s-2)-(s-2)-(s-2) 2 2 2 2 (s-2)(s-2)(s-2)(s-2)3 3 3 3 2-3-12-3-12-3-12-3-1 函数的拉氏变换。函数的拉氏变换。 F(sF(sF(sF(s)=)=)=)= s+1s+1s+1s+1 (s+1)(s+3)(s+1)(s+3)(s+1)(s+3)(s+1)(s+3) 解解：A A A A1 1 1 1=(s+2)=(s+2)=(s+2)=(s+2) s+1s+1s+1s+1 (s+1)(s+3)(s+1)(s+3)(s+1)(s+3)(s+1)(s+3) s=-2s=-2s=-2s=-2 = -1= -1= -1= -1 (s+1)(s+3)(s+1)(s+3)(s+1)(s+3)(s+1)(s+3) A A A A2 2 2 2=(s+3)=(s+3)=(s+3)=(s+3) s+1s+1s+1s+1 s=-3s=-3s=-3s=-3 = 2= 2= 2= 2 F(sF(sF(sF(s)= - )= - )= - )= - 2 2 2 2 s+3s+3s+3s+3 1 1 1 1 s+2s+2s+2s+2f(tf(tf(tf(t)=2e)=2e)=2e)=2e-3t -3t-3t-3t-e -e -e -e-2t-2t-2t-2t F(sF(sF(sF(s)=)=)=)= s s s s (s+1)(s+1)(s+1)(s+1)2 2 2 2(s+2)(s+2)(s+2)(s+2) 2-3-22-3-22-3-22-3-2 函数的拉氏变换。函数的拉氏变换。 解解: : : :f(tf(tf(tf(t)= )= )= )= e e e est st st st + + + +lim limlimlim e e e est st st st s s s s (s+1)(s+1)(s+1)(s+1)2 2 2 2 s=-2s=-2s=-2s=-2 d d d d dsdsdsds s s s s s+2s+2s+2s+2 s -1s -1s -1s -1 =-2e=-2e=-2e=-2e-2t -2t-2t-2t+lim( +lim( +lim( +lim( e e e est st st st+ + + + e e e est st st st) ) ) ) s -1s -1s -1s -1 st st st st s+2s+2s+2s+2 2 2 2 2 (s+2)(s+2)(s+2)(s+2)2 2 2 2 =-2e=-2e=-2e=-2e-2t -2t-2t-2t-te -te-te-te-t -t -t -t+2e+2e+2e+2e-t -t -t -t =(2-t)e=(2-t)e=(2-t)e=(2-t)e-t -t -t -t-2e-2e-2e-2e-2t -2t-2t-2t F(sF(sF(sF(s)=)=)=)= 2s2s2s2s2 2 2 2-5s+1-5s+1-5s+1-5s+1 s(ss(ss(ss(s2 2 2 2+1)+1)+1)+1) 2-3-32-3-32-3-32-3-3 函数的拉氏变换。函数的拉氏变换。 解解:F(s)(s:F(s)(s:F(s)(s:F(s)(s2 2 2 2+1)+1)+1)+1)s=+j s=+js=+js=+j=A =A=A=A1 1 1 1s+As+As+As+A2 2 2 2 s=+js=+js=+js=+j A A A A1 1 1 1=1, A=1, A=1, A=1, A2 2 2 2=-5=-5=-5=-5 A A A A3 3 3 3= = = =F(s)sF(s)sF(s)sF(s)s = = = =1 1 1 1 s=0s=0s=0s=0 f(tf(tf(tf(t)=1+cost-5sint)=1+cost-5sint)=1+cost-5sint)=1+cost-5sint F(sF(sF(sF(s)= + +)= + +)= + +)= + + 1 1 1 1s s s s s s s s2 2 2 2+1+1+1+1 s s s s -5-5-5-5 s s s s2 2 2 2+1+1+1+1 2-3-42-3-42-3-42-3-4 函数的拉氏变换。函数的拉氏变换。 (4) (4) (4) (4) F(sF(sF(sF(s)=)=)=)= s+2s+2s+2s+2 s(s+1)s(s+1)s(s+1)s(s+1)2 2 2 2(s+3)(s+3)(s+3)(s+3) 解解: : : :f(tf(tf(tf(t)= )= )= )= e e e est st st st + + + + e e e est st st st s+2s+2s+2s+2 (s+1)(s+1)(s+1)(s+1)2 2 2 2(s+3)(s+3)(s+3)(s+3) s=0s=0s=0s=0 s+2s+2s+2s+2 s(s+1)s(s+1)s(s+1)s(s+1)2 2 2 2 s=-3s=-3s=-3s=-3 + + + + limlimlimlim s -1s -1s -1s -1 d d d d e e e est st st st s+2s+2s+2s+2 s(s+3)s(s+3)s(s+3)s(s+3) dsdsdsds = + e= + e= + e= + e-3t -3t-3t-3t+lim +lim+lim+lim + + + + 2 2 2 2 3 3 3 3 1 1 1 1 12121212 s -1s -1s -1s -1 (-s(-s(-s(-s2 2 2 2-4s-6)e-4s-6)e-4s-6)e-4s-6)est st st st (s (s (s (s2 2 2 2+3)+3)+3)+3)2 2 2 2 (s+2)te(s+2)te(s+2)te(s+2)test st st st s s s s2 2 2 2+3s+3s+3s+3s = + e= + e= + e= + e-3t -3t-3t-3t- - - - e e e e-t -t -t -t- - - - e e e e-t -t -t -t 2 2 2 2 3 3 3 3 1 1 1 1 12121212 3 3 3 3 4 4 4 4 t t t t 2 2 2 2 (2-4-1)(2-4-1)(2-4-1)(2-4-1) 求下列微分方程。求下列微分方程。 d d d d2 2 2 2y(t)y(t)y(t)y(t) dt dt dt dt2 2 2 2 +5 +6y(t)=6 +5 +6y(t)=6 +5 +6y(t)=6 +5 +6y(t)=6 ，初始条件：初始条件： dy(tdy(tdy(tdy(t) ) ) ) dt dt dt dt y(0)=y(0)=2 y(0)=y(0)=2 y(0)=y(0)=2 y(0)=y(0)=2 。 解解: : : :s s s s2 2 2 2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)= 1 1 1 1 s s s s A A A A1 1 1 1= = = =sY(ssY(ssY(ssY(s) ) ) ) s=0s=0s=0s=0 y(ty(ty(ty(t)=1+5e)=1+5e)=1+5e)=1+5e-2t -2t-2t-2t-4e -4e-4e-4e-3t -3t-3t-3t A A A A2 2 2 2=(s+2)Y(s) =(s+2)Y(s) =(s+2)Y(s) =(s+2)Y(s) s=-2s=-2s=-2s=-2 A A A A3 3 3 3=(s+3)Y(s)=(s+3)Y(s)=(s+3)Y(s)=(s+3)Y(s) s=-3s=-3s=-3s=-3 A A A A1 1 1 1=1 , A=1 , A=1 , A=1 , A2 2 2 2=5 , A=5 , A=5 , A=5 , A3 3 3 3=-4=-4=-4=-4 Y(sY(sY(sY(s)=)=)=)= 6+2s6+2s6+2s6+2s2 2 2 2+12s+12s+12s+12s s(ss(ss(ss(s2 2 2 2+5s+6)+5s+6)+5s+6)+5s+6) (2-4-2)(2-4-2)(2-4-2)(2-4-2)求下列微分方程。求下列微分方程。 d d d d3 3 3 3y(t)y(t)y(t)y(t) dt dt dt dt3 3 3 3 +4 +29 =29,+4 +29 =29,+4 +29 =29,+4 +29 =29, d d d d2 2 2 2y(t)y(t)y(t)y(t) dt dt dt dt2 2 2 2 dy(tdy(tdy(tdy(t) ) ) ) dt dt dt dt 初始条件初始条件: : : : y(0)=0 , y(0)=17 , y(0)=-122y(0)=0 , y(0)=17 , y(0)=-122y(0)=0 , y(0)=17 , y(0)=-122y(0)=0 , y(0)=17 , y(0)=-122 解解: : : : 2-5-a 2-5-a 2-5-a 2-5-a 试画题试画题2-12-12-12-1图所示电路的动态结构图图所示电路的动态结构图, , , ,并并 求传递函数。求传递函数。 C C C C u u u uc c c c R R R R1 1 1 1 R R R R2 2 2 2 u u u ui i i i i i i i1 1 1 1 i i i i2 2 2 2 u u u uo o o o i i i ic c c cC C C C 解解: : : :u u u ui i i i=R=R=R=R1 1 1 1i i i i1 1 1 1+u+u+u+uo o o o ， ，i i i i2 2 2 2=i=i=i=ic c c c+i+i+i+i1 1 1 1 U U U UI I I I(s (s (s (s)=R)=R)=R)=R1 1 1 1I I I I1 1 1 1(s)+U(s)+U(s)+U(s)+UO O O O(s)(s)(s)(s) dudududuc c c c i i i ic c c c=C=C=C=C dt dt dt dt I I I I2 2 2 2(s)=I(s)=I(s)=I(s)=IC C C C(s)+I(s)+I(s)+I(s)+I1 1 1 1(s)(s)(s)(s) I I I IC C C C(s (s (s (s)=)=)=)=CsUCsUCsUCsUC C C C(s (s (s (s) ) ) ) 即即: =: =: =: =I I I I1 1 1 1(s)(s)(s)(s) U U U UI I I I(s)-U(s)-U(s)-U(s)-UO O O O(s (s (s (s) ) ) ) R R R R1 1 1 1 U U U UI I I I(s)-U(s)-U(s)-U(s)-UO O O O(s)(s)(s)(s) CsCsCsCs= = = =I I I IC C C C(s (s (s (s) ) ) ) U U U UO O O O(s (s (s (s) ) ) ) U U U UI I I I(s (s (s (s) ) ) ) = = = = 1 1 1 1 R R R R1 1 1 1 ( sC)R( sC)R( sC)R( sC)R2 2 2 2 1+1+1+1+ 1 1 1 1 R R R R1 1 1 1 ( sC)R( sC)R( sC)R( sC)R2 2 2 2 = = = = R R R R2 2 2 2+R+R+R+R1 1 1 1R R R R2 2 2 2sCsCsCsC R R R R1 1 1 1+R+R+R+R2 2 2 2+R+R+R+R1 1 1 1R R R R2 2 2 2sCsCsCsC 1 1 1 1 R R R R1 1 1 1 sCsCsCsC R R R R2 2 2 2 U U U UI I I I(s (s (s (s) ) ) ) - U U U UO O O O(s (s (s (s) ) ) ) I I I IC C C C(s (s (s (s) ) ) ) I I I I1 1 1 1(s)(s)(s)(s) I I I I2 2 2 2(s)(s)(s)(s) 1 1 1 1 R R R R1 1 1 1 sCsCsC sC R R R R 2 2 2 2 ( )( )( )( ) U U U UI I I I(s (s (s (s) ) ) ) - U U U UO O O O(s (s (s (s) ) ) ) 2-5-b 2-5-b 2-5-b 2-5-b 试画出题试画出题 2-12-12-12-1图所示的电路图所示的电路 的动态结构图的动态结构图, , , ,并并 求传递函数。求传递函数。 u u u uo o o o u u u ui i i i R R R R1 1 1 1L L L L R R R R2 2 2 2 C C C C 解：解：u u u ui i i i=R=R=R=R1 1 1 1I I I I1 1 1 1+u+u+u+uc c c cu u u uc c c c= = = =u u u uo o o o+u+u+u+uL L L Lu u u uL L L L=L=L=L=Ldi di di diL L L L dt dt dt dt i i i iL L L L= = = = u u u uo o o o R R R R2 2 2 2 i i i i1 1 1 1= = = =i i i iL L L L+i+i+i+ic c c ci i i ic c c c=C=C=C=C dudududuc c c c dt dt dt dt U U U Ui i i i(s (s (s (s)=R)=R)=R)=R1 1 1 1I I I I1 1 1 1(s)+U(s)+U(s)+U(s)+UC C C C(s)(s)(s)(s)U U U UC C C C(s (s (s (s)=)=)=)=U U U UO O O O(s)+U(s)+U(s)+U(s)+UL L L L(s (s (s (s) ) ) ) U U U UL L L L(s (s (s (s)=)=)=)=sLIsLIsLIsLIL L L L(s (s (s (s) ) ) ) I I I I1 1 1 1(s)=(s)=(s)=(s)=I I I IL L L L(s)+I(s)+I(s)+I(s)+IC C C C(s (s (s (s) ) ) ) 1 1 1 1 R R R R1 1 1 1 CsCsCsCs sLsLsLsL R R R R2 2 2 2 I I I I1 1 1 1 U U U UO O O O U U U Ui i i i I I I IC C C C - - - - - - - - U U U UC C C C=U=U=U=UO O O O+U+U+U+UL L L L I I I IL L L L U U U UL L L L I I I I2 2 2 2(s)=(s)=(s)=(s)= U U U UO O O O(s (s (s (s) ) ) ) R R R R2 2 2 2 I I I IC C C C(s (s (s (s)=)=)=)=CsUCsUCsUCsUC C C C(s (s (s (s) ) ) ) I I I I1 1 1 1(s)=(s)=(s)=(s)= U U U UO O O O(s (s (s (s) ) ) ) R R R R2 2 2 2 I I I I1 1 1 1(s)=(s)=(s)=(s)= U U U UI I I I(s)+U(s)+U(s)+U(s)+UC C C C(S(S(S(S) ) ) ) R R R R1 1 1 1 即：即： I I I IL L L L(s (s (s (s)=I)=I)=I)=I1 1 1 1(s)-I(s)-I(s)-I(s)-IC C C C(s)(s)(s)(s) I I I IC C C C(s (s (s (s)=)=)=)= U U U UC C C C(s (s (s (s) ) ) ) CsCsCsCs 解解: : : :电路等效为电路等效为: : : : 2-6-a 2-6-a 2-6-a 2-6-a 用运算放大器组成的有源电网络如图用运算放大器组成的有源电网络如图 所示所示, , , ,试采用复数阻抗法写出它们的传递函数。试采用复数阻抗法写出它们的传递函数。 U U U UO O O O = = = = R R R R3 3 3 3 SCSCSCSCR R R R2 2 2 2 R R R R2 2 2 2 1 1 1 1 U U U UI I I I R R R R1 1 1 1 U U U UO O O O R R R R3 3 3 3 SCSCSCSC R R R R2 2 2 2 R R R R2 2 2 2 1 1 1 1 SCSCSCSC 1 1 1 1 = = = = R R R R1 1 1 1+R+R+R+R3 3 3 3+R+R+R+R2 2 2 2R R R R3 3 3 3CSCSCSCS = = = = R R R R1 1 1 1(R(R(R(R2 2 2 2SC+1)SC+1)SC+1)SC+1) R R R R2 2 2 2R R R R3 3 3 3 = = = =( + )( + )( + )( + ) R R R R1 1 1 1(R(R(R(R2 2 2 2SC+1)SC+1)SC+1)SC+1)R R R R1 1 1 1 R R R R1 1 1 1 R R R R2 2 2 2 = = = = ( +R( +R( +R( +R3 3 3 3 ) ) ) ) (R(R(R(R2 2 2 2SC+1)SC+1)SC+1)SC+1) 1 1 1 1 = = = = R R R R2 2 2 2 1 1 1 1 R R R R3 3 3 3 R R R R2 2 2 2SCSCSCSC R R R R1 1 1 1 C(S)=C(S)=C(S)=C(S)= U U U UO O O O(S)(S)(S)(S) U U U UI I I I(S)(S)(S)(S) C C C C R R R R1 1 1 1 R2R2R2R2R3R3R3R3 u u u ui i i i u u u uo o o o C C C C R R R R1 1 1 1 R2R2R2R2R3R3R3R3 u u u ui i i i u u u uo o o o C C C C R R R R1 1 1 1 R2R2R2R2R3R3R3R3 u u u ui i i i u u u uo o o o R R R R4 4 4 4 R R R R5 5 5 5 2-6-b 2-6-b 2-6-b 2-6-b 用运算放大器组成的有源电网络如用运算放大器组成的有源电网络如 力所示力所示, , , ,试采用复数阻抗法写出它们的传试采用复数阻抗法写出它们的传 递函数。递函数。 = = = = R R R R5 5 5 5 R R R R4 4 4 4+ R+ R+ R+ R5 5 5 5 U U U UO O O O(R(R(R(R3 3 3 3SC+1)SC+1)SC+1)SC+1) R R R R2 2 2 2R R R R3 3 3 3SC+RSC+RSC+RSC+R2 2 2 2+R+R+R+R3 3 3 3 U U U UO O O O U U U UI I I I = = = = (R(R(R(R2 2 2 2R R R R3 3 3 3SC+RSC+RSC+RSC+R2 2 2 2+R+R+R+R3 3 3 3)(R)(R)(R)(R4 4 4 4+R+R+R+R5 5 5 5) ) ) ) R R R R1 1 1 1(R(R(R(R3 3 3 3SC+1)RSC+1)RSC+1)RSC+1)R5 5 5 5 = = = = (R(R(R(R4 4 4 4+R+R+R+R5 5 5 5)(R)(R)(R)(R2 2 2 2+R+R+R+R3 3 3 3)( SC+1) )( SC+1) )( SC+1) )( SC+1) R R R R2 2 2 2R R R R3 3 3 3 R R R R2 2 2 2+R+R+R+R3 3 3 3 R R R R1 1 1 1R R R R5 5 5 5(R(R(R(R3 3 3 3SC+1)SC+1)SC+1)SC+1) U U U UI I I I R R R R1 1 1 1= = = = R R R R5 5 5 5 R R R R4 4 4 4+ R+ R+ R+ R5 5 5 5 U U U UO O O O R R R R2 2 2 2 R R R R3 3 3 3 SCSCSCSCSCSCSCSC R R R R3 3 3 3 SCSCSCSC 1 1 1 1 R R R R5 5 5 5 R R R R4 4 4 4+ R+ R+ R+ R5 5 5 5 U U U UO O O O R R R R2 2 2 2 R R R R3 3 3 3 R R R R3 3 3 3SCSCSCSC1 1 1 1 c(tc(tc(tc(t) ) ) ) t t t t0 0 0 0T T T T K K K K (t)(t) 2-8 2-8 2-8 2-8 设有一个初始条件为零的系统，系设有一个初始条件为零的系统，系 统的输入、输出曲线如图，求统的输入、输出曲线如图，求G(sG(sG(sG(s) ) ) )。 c(tc(tc(tc(t) ) ) ) t t t t0 0 0 0T T T T K K K K (t)(t) c(tc(tc(tc(t)=)=)=)= K K K K T T T T t- t- t- t-(t-T)(t-T)(t-T)(t-T) K K K K T T T T C(sC(sC(sC(s)=)=)=)= K K K K (1-e )(1-e )(1-e )(1-e ) TsTsTsTs2 2 2 2 -TS-TS-TS-TS C(sC(sC(sC(s)=G(S)=G(S)=G(S)=G(S) 第二章习题课第二章习题课 (2-8)(2-8)(2-8)(2-8) 解解: : : : 2-9 2-9 2-9 2-9 若系统在单位阶跃输入作用时，已若系统在单位阶跃输入作用时，已 知初始条件为零的条件下系统的输出响知初始条件为零的条件下系统的输出响 应，求系统的传递函数和脉冲响应。应，求系统的传递函数和脉冲响应。 r(tr(tr(tr(t)=)=)=)=I(tI(tI(tI(t) ) ) )c(t c(tc(tc(t)=1-e +e)=1-e +e)=1-e +e)=1-e +e -2t-2t-2t-2t -t -t -t -t 解解: : : : R R R R( ( ( (s s s s)=)=)=)= 1 1 1 1 s s s s G(S)=G(S)=G(S)=G(S)=C(s)/R(sC(s)/R(sC(s)/R(sC(s)/R(s) ) ) ) 1 1 1 1 s+2s+2s+2s+2 1 1 1 1 s s s s - - - - C(sC(sC(sC(s)=)=)=)= 1 1 1 1 s s s s+ + + +1 1 1 1 + + + += = = = s(s+1)(s+2)s(s+1)(s+2)s(s+1)(s+2)s(s+1)(s+2) (s (s (s (s2 2 2 2+4s+2)+4s+2)+4s+2)+4s+2) = = = = (s+1)(s+2)(s+1)(s+2)(s+1)(s+2)(s+1)(s+2) ( ( ( (s s s s2 2 2 2+4s+2+4s+2+4s+2+4s+2) ) ) ) C(sC(sC(sC(s)=)=)=)= (s+1)(s+2)(s+1)(s+2)(s+1)(s+2)(s+1)(s+2) ( ( ( (s s s s2 2 2 2+4s+2+4s+2+4s+2+4s+2) ) ) ) 脉冲响应脉冲响应: : : : 2 2 2 2 s s s s+ + + +2 2 2 2 =1+=1+=1+=1+ 1 1 1 1 s s s s+ + + +1 1 1 1 - - - - c(tc(tc(tc(t)=)=)=)=( (t t) )+2e +e+2e +e+2e +e+2e +e -2t-2t-2t-2t -t -t -t -t 第二章习题课第二章习题课 (2-(2-(2-(2-) ) ) ) 2-10 2-10 2-10 2-10 已知系统的微分方程组的拉氏变换已知系统的微分方程组的拉氏变换 式，试画出系统的动态结构图并求传递式，试画出系统的动态结构图并求传递 函数。函数。 解解: : : : X X X X1 1 1 1(s)=(s)=(s)=(s)=R R R R(s)(s)(s)(s)G G G G1 1 1 1(s)-(s)-(s)-(s)-G G G G1 1 1 1(s)(s)(s)(s)G G G G7 7 7 7(s)-(s)-(s)-(s)-G G G G8 8 8 8(s)(s)(s)(s)C(s)C(s)C(s)C(s) X X X X2 2 2 2(s)=(s)=(s)=(s)=G G G G2 2 2 2(s)(s)(s)(s)X X X X1 1 1 1(s)-(s)-(s)-(s)-G G G G6 6 6 6(s)(s)(s)(s)X X X X3 3 3 3(s)(s)(s)(s) X X X X3 3 3 3(s)=(s)=(s)=(s)=G G G G3 3 3 3(s)(s)(s)(s)X X X X2 2 2 2(s)-(s)-(s)-(s)-C C C C(s)(s)(s)(s)G G G G5 5 5 5(s)(s)(s)(s) C C C C(s (s (s (s)=)=)=)=G G G G4 4 4 4(s)(s)(s)(s)X X X X3 3 3 3(s)(s)(s)(s) G G G G1 1 1 1G G G G2 2 2 2G G G G3 3 3 3 G G G G5 5 5 5 - - - - - - - - - - - - C(s)C(s)C(s)C(s) - - - - R(s)R(s)R(s)R(s) G G G G4 4 4 4 G G G G6 6 6 6 G G G G8 8 8 8 G G G G7 7 7 7 X X X X1 1 1 1(s)=(s)=(s)=(s)=R R R R(s)-C(s)(s)-C(s)(s)-C(s)(s)-C(s)G G G G7 7 7 7(s)-(s)-(s)-(s)-G G G G8 8 8 8(s)(s)(s)(s)GGGG1 1 1 1(s)(s)(s)(s) C(s)C(s)C(s)C(s)G G G G7 7 7 7(s)-(s)-(s)-(s)-G G G G8 8 8 8(s)(s)(s)(s) G G G G6 6 6 6(s)(s)(s)(s)X X X X3 3 3 3(s)(s)(s)(s) X X X X1 1 1 1(s)(s)(s)(s) X X X X2 2 2 2(s)(s)(s)(s) C C C C(s)(s)(s)(s)G G G G5 5 5 5(s)(s)(s)(s) X X X X3 3 3 3(s)(s)(s)(s) G G G G1 1 1 1G G G G2 2 2 2G G G G3 3 3 3 G G G G5 5 5 5 - - - - - - - - - - - - C(s)C(s)C(s)C(s) - - - - R(s)R(s)R(s)R(s) G G G G4 4 4 4 G G G G2 2 2 2G G G G6 6 6 6 G G G G8 8 8 8 G G G G7 7 7 7 G G G G1 1 1 1G G G G2 2 2 2 G G G G5 5 5 5 - - - - C(s)C(s)C(s)C(s) - - - - R(s)R(s)R(s)R(s) G G G G7 7 7 7- - - -G G G G8 8 8 8 1+1+1+1+G G G G3 3 3 3G G G G2 2 2 2G G G G6 6 6 6 G G G G3 3 3 3G G G G4 4 4 4 - - - - C(s)C(s)C(s)C(s)R(s)R(s)R(s)R(s) G G G G7 7 7 7-G-G-G-G8 8 8 8 1+1+1+1+G G G G3 3 3 3G G G G2 2 2 2G G G G6 6 6 6 +G+G+G+G3 3 3 3G G G G4 4 4 4G G G G5 5 5 5 G G G G1 1 1 1G G G G2 2 2 2G G G G3 3 3 3G G G G4 4 4 4 1+1+1+1+G G G G3 3 3 3G G G G2 2 2 2G G G G6 6 6 6 +G+G+G+G3 3 3 3G G G G4 4 4 4G G G G5 5 5 5+G+G+G+G1 1 1 1G G G G2 2 2 2G G G G3 3 3 3G G G G4 4 4 4(G(G(G(G7 7 7 7 -G-G-G-G8 8 8 8) ) ) ) G G G G1 1 1 1G G G G2 2 2 2G G G G3 3 3 3G G G G4 4 4 4 R R R R( ( ( (s s s s) ) ) ) C C C C( ( ( (s s s s) ) ) ) = = = = 第二章习题课第二章习题课 (2-10)(2-10)(2-10)(2-10) 解解: : : : 2-11(a)2-11(a)2-11(a)2-11(a) G G G G1 1 1 1(s)(s)(s)(s)G G G G2 2 2 2(s) (s)(s)(s) G G G G3 3 3 3(s)(s)(s)(s) H H H H1 1 1 1(s)(s)(s)(s) _ _ _ _ _ _ _ + + + +R(s) R(s)R(s)R(s)C(s)C(s)C(s)C(s) H H H H2 2 2 2(s)(s)(s)(s) G G G G1 1 1 1(s)(s)(s)(s)G G G G2 2 2 2(s) (s)(s)(s) H H H H1 1 1 1(s)(s)(s)(s) _ _ _ _ _ _ _ + + + + R(s)R(s)R(s)R(s)C(s)C(s)C(s)C(s) H H H H2 2 2 2(s)(s)(s)(s) G G G G3 3 3 3(s)(s)(s)(s) 求系统的求系统的 传递函数传递函数 1+1+1+1+G G G G2 2 2 2H H H H1 1 1 1 G G G G2 2 2 2 G G G G1 1 1 1+ + + +G G G G3 3 3 3 1+1+1+1+G G G G1 1 1 1H H H H2 2 2 21+ 1+1+1+G G G G2 2 2 2H H H H1 1 1 1 G G G G2 2 2 2 1+1+1+1+G G G G2 2 2 2H H H H1 1 1 1 G G G G2 2 2 2 = = = =