-第二章连续系统的时域分析习题解答

1、第二章 连续系统的时域分析习题解答 f(t)i (t)和u (t)00+-+4kf(t)f(t)u (t)02 Fu (t)0解：(a)(b)图题 2-1ft)41416(a)( 0.002p u t) pu t) 375ft);00i 2106 pu (3pi 7.5104 pf ;000fp1 p1up11(b)u t) f , i f ,010p20p2 p1p1 p(p p1)u (p f ; (p pi f .2200 i (t)和u (t)f(t)H()。00u (t)f(t) 3p625i (t)07.510 p3754(a)H (p) u f; H (p) i f;：0f(t)

2、3p625u (t)01i (t)f(t)pp p2(b)H (p) u f; H (p) i f.0f(t)p2 1pp p 12 H()pp3p3p3 H(p) ;(2)H(p) ;p32p3p(p(p(p2)H(p) dy;(4)H(p) dy.d fdtd f3y ;(2)3y 3f ;dt解：dtdtdydtd fdtd ydydtd fdfdt2223y 3f ;(4)3 2y 3.dt2dt2 0 2p4(pp yt) (2) yt) yt) ft), y(0 ) y(0 ) 1;-(2pft), y(0 ) (0 ) (0 ) 0;p(p4p2-3p1p(pft), y(0 )

3、 (0 ) (0 ) 4.2-x p p y(t) Ae A e ,t12122 A A yt) , t121t12(2) p p 2 , yt) A A e cos(2t A 11230 A A cosA12311 2A (cosA sinA ) A 0.5 yt) sint, t 0.23320 4A 2cosA233 p p 2, yt) A (At A )e ,12,31230 A A1310 A 2A A 2 yt) 1(2t , t 0.2t2324 4A 4A233、(a)u (t) 6e t 4e tV, t 0;x(b)u (t) 2e t cost 6e tsintV, t

4、 0.x$-x161 3x(b)u (0 ) u (0 ) 20 2V;xxx )=)=u ()；-x )=)=i (；-x )=)=u ().-x2!i，(a)# p 2, p 3u (t) Ae t A e t23x12125 A Ai(0 )u (0 ) 5V, u (0 ) 012xxC0 2A 3A12 A A u (t) t tV, 0 .23x12111(b) Y(p) 0 p0 p 2p2 021 p p 1 1, p 1 1i (t) Ae t t A )x12120 A Ai (0 ) 0, i (0 ) 104 412xx4 A A A A1212 A A / i (t)

5、 t t 0 .x124p(c): p5 0 p 5p4 p p 212u Ae t A e t, u (0 ) u (0 ) 530 4xxx12A A u t e t t 0 .4x12； H()2p4(pp(2p3p1;(2);.p(p 4pp(p2)22(。11：H(p) h(t) (e e ) (t);3t tp1 p3181812(A )p (B )p12Ap B1(2)H(p) A ,B 1.5pp 4p82p(p 4p821 1(p2)0.8752(p2) 28 8 H(p) p221 18 8 h(t) ( e cos2t sin2t)(t);2t2t14p14p22.5(p

6、2)21 54 214H(p) h(t) ( e e (t).2t2t ()()。+-u+1 -(a)111418ut1222p1 p23 p p2 ht) (2e t)V.3 1p10.5 p2p【1;0.4 2.4tp1 p6p+-2p p2$ tt2g(t) h)d (t) cos(t) cos(t) .2400_1p1 h(t) (t) e (t),2p）t12th( )d (t) e(t) e (t) .200_1p21h(t)(2e e (t),tt2pttg(t)( ) e e ( ) (et t00_1h ()22#( ) ( t th(t) (t)(t f (t)* f (t

7、);(2) f (t)* f (t);(3) f (t)* f (t).121314【tttt0 24(-1)-2 02-1 0 1%(a)(b)(d)？(c)图题 2-11：f (t) (t 2) t1212( 2) ( ) ( ( 2)f ()*f ()t ttt121t f (t)* f (t) f (t 2) f (t 2)1211-402 412 (t (t 4)(t (t 2)t(t)？f ()*f ()1312(t (t 2) (t (t 4);t(2) f (t)* f (t) f (t 2) f (t f (t 4)0 1 3 4 5 6*1311111212 t(t) t

8、(t (t t 2)2121212(t t (t (t 4) (t (t (t (t 6);f ()*f ) f (t)* f (t) f (t f (t 141411121212 (t (t (t (t t (t t|-3 -10-112 (t (t (t (t (t (t 12323212 (t (t (t (t (t (t (t t 3). f (t)(t), f (t) (t;(2) f (t) (t), f (t) e (t);t1212 f (t) e (t), f (t) e (t);(4) f (t) e (t), f (t) (t);t2tt1212 f (t) t(t)(t

9、, f (t) (t (t2);12T(6) f (t) (t), f (t) t(t).12n0 yt) tt1);(2) yt) ( te t) e t);-0121 yt) (e e t) (e e t);-2t-2t()11(4) f t) (e e )jtjt2j|2t 01211 yt)(e e t)(e e t)-tjt-tjt2j2j(j2je e )(e e )-tjtjtjtjt12t) (e cost sintt);-t4j yt) f t f t2)11 sint t t 2) sintt 2) t ;()(6) yt)f tnT) sin tnTtnT)10Tt2n0

10、n03T 4Tsin t(sin tt).【TT ()=f (f () 。121010tt1424tt】02-1 0(b)图题 2-13： t(a) y(t) e )d * (t 2)(t 4)0 (e ) t (t) * (t 2) (t 4) e t (t 2) e t (t 4) ( 2)( 4)0(b)注意: f () , 可借助于图解分段计算:2t12t 1, y(t) t 1, y(t) d d et 2(e et ) 2e et ;12121f (t112 e ;2t或: y(t) e (e et t)%221f (-2f (t)*t(t) (t e (t) tf() 。+10f

11、 (t)*(t) e (t)(0e (t) e (t)t0t；f (t)*(t) e (t)e ) e (t) f (t)t0t.t02某LTI系统的激励f()( 所y (f10 t 1ty (t) ) d * (t)(t2)：22f0图题 2-15 1 t t)tt * t)t24|1414 t t)tt t ttt22t024 -y(t)f(t)4020123234t4(a) 输入(b) 输出图题 2-16y(t) 2f(t)2f(t2)h(t) (t)(t2)：y(t)3y(t)2 f (t)3f(t) ，0 -y(0 ) 1, y(0 ) 2 y (；xf(t)y ()(。tf (pp

12、2)y(t) (p f(t)y (t) Ae A e 121A 32x1212xp321t22et e ( )t2tfxf y (t) h(t)*e (t) (e e (t)tt3fy(t) y (t) y (t) t 4e t(t)2xf*(t)t(.(b)1221213(t) (2t ( ) ( ) ( )t ttp1y (t) 2 t ( ) ( )*e ( ) (2 3 )e ( )t tt t t t t f 令f()= ()=()，212333233p11f( ) e ( )t tt2p2p2ay (t) e2t tte ( )t t t 2tfS12+-FbuC+-+ u -11

13、- t0 u ( u ()xf()。i) u )：f11p22p22H(p) 12p12 pp p21ph(t) (t) ( t 2t tu (t)t) ( ( ) e t dt e ( )t 2t t2f0(t) Ae A e) u (：u0.5t2txx12u (0 ) 1112V, u (0 ) i (0 ) xLxC x12i (0 ) , u (0 ) 2u (0 ) 1i (0 )u (0 ) 2.5V/sLxC xxLxC x2 A AA 1( ) ee t u(0 ) ttu tx2121 A 2AA 1 u(t) u (t)u (t) 1e t e t(t).2122xf已知某系统的微分方程为 y(t)3y(t)2yt) f(t)3ft)，当激励2-20(t)时，系统的全响应 ( ) ( e e e ) ( ); 试求y t t t721ft)=e4ttt2436自由响应与强迫响应、暂态响应与稳态响应。y (y ()、xfp321H(p) , h