信号与系统2014chap2第2章连续时间时域分析

连续时间系统的时域分时域分析：对系统的分析与计算均以时间t为变量LTI2.1LTIe(t

y(t 描述输入、输出关系的连续系统数学模型为n e(t

+

y(t0 0y(t)a1y(t)a0y(t)b2e(t)b0e(tay(n)(t) y(n1)(t)Lay(t)ay(t be(m)(t) e(m1)(t)Lbe(t)be(t m ay(i)(t) be(j)(t i j ay(i)(t) be(j)(t (用时域法求解连续系统的流程图 求特征根i求特征根i

全解

y(t)yh(t)yp(t ay(i)(t) be(j)(t

aiy(i)(t)

j

La 阶微分方程有n个特征根i(i1y(t)3y(t)2y(t)2e(ty(t)2y(t)y(t)e(ty(t)2y(t)5y(t)2e(t

ay(i)(t) be(j)(t j特解yp(t)的形式由激励e(t)的形式决定查P34表2说明：激励e(t)是在tt00时刻加入系统，因此特解yp(t)存在的时间为tt00例：已知某⼀系统的微分方程为y(t)3y(t2y(t)2e(te(t求当激励为(1)e(t)(t2e(t)e3t(t) ay(i)(t) be(j)(t 全解：y(tyh(typ(t齐次 特

j确定,n阶微分方程需要n个初始条件。例：已知系统的微分方程为y(t)4y(t4y(t)e(t3e(ty(0) y(0 e(t)et(t 例：已知某⼀系统的微分方程为y(t4y(t3y(t)e(ty(0) y(0 e(t10cos 关于系统在t=0-与t=0+状态的讨论（难点 ay(i)(t) be(j)(t 0t t<0 et0时系统的响应y ay(i)(t) be(j)(t j初始条件：yj)(0（由yj0)和e(t)共同决定从0: y(j)(t)可能发生跳即yj0yj0 令yj0yj0y = y(j)(0)y(j)(0= ay(i)(t) be(j)(t jy(t)4y(t)4y(t)e(t)3e(t(1)e(t)(t (2)e(t)et(ty(t)4y(t)4y(t)(t)3(ty(t)4y(t)4y(t)et(t)et(t)3et(t(t)2et(t阶导数项，使之方程两端平衡。而右端冲激函数的产生意味着左端y(i)(t)中的某些项在t=0处有跳变。方程左端(t)方程右端(t420242021

t0处跳跃量y(0y(0 例：y(t3y(t3e(t)y(0求(1)e1(t)(t (2)e2(t)(t)时y(0例：y(t4y(t3y(te(t2e(te(t已知e(t)(t)，y(0) y(0) 求y(0),y(0y(i)(t)同阶次y(i)(t) 当平衡完成后,y(i)(t)(i=0,1…n-1)项中所含有的(t)项的系数即例：y(t3y(t)3(t),求ye(0注意注意例：已知y(t3y(t)3e(t)，y(0)0，e(t)(t)求y(t)ye(0) y(t)3e3ty(t)y(t)

3e3t(t)9e3t

3(t)9e3t y(0+),y(0+)y(t)4y(t)5y(t)2y(t)(t)3(t)求),n ay(i)(t) nCy(t t C yi(0)yi(0)yi(0 ay(i)(t) be(j)(t 解由yh(t)yp(t) y(t)y(t)y(t)Ceit y(j)(0)0(j0,1,L.ny(j)(0)y(j)(0 由yh(t)yp(t)

aiy(i)(t)bje(j)(t y(t) yh(t)yp(t) yzi(t) y(t) Ci

it

(t CeitCeity(t)p p y(t pyh(t

初始条件=初始状态+跳变量初始条件=初始状 初始条件=跳变y(t)=yzi(t)+y(t)yh(t(t)yh(t)yp(t完全 齐次 特解(t)yh(t)yp(t完全 齐次 特解y(i)(0)y(i)(0)y(i)(0)y(i)(0)y(i)(0 y(i)(0)y(i)e(iy(i)(0)y(i)(0) 例:某LTI系统数学模型 y(t)3y(t)2y(t)2(t)6(t已知y(0) y(0) 求yzi yzs y(t)说明说明2-12-22-4，（1）（3）2-6 e(t) yzs(t)h(t

T0,(t 系统方程的⼀般形式为 ay(i)(t) be(j)(t ah(i)(t)

b(j)(t

j？ (j0,1,2Ln？ 讨论：t>0时系统属于什么响应t>0时系统的冲激响应为零输入响 例：某⼀系统数学模型为y(t5y(t6y(te(t2e(t求该系统的冲激响应h(t) h(t)5h(t)6h(t)(t)2(t h(t)满足h(0h(0 可求得h(01h(0)当t>0时，h(t5h(t6h(t h(t)(c ce3t)(t

h(0)c1c2 h(0)2c

解 h(t)

h(t)5h(t)6h(t)(t)2(t 方法设h1(t)满足h(0 (ch(t) ce3t)((c 可求得h1(00h1(0 h1(0)c1c2 c1ch(0)c

13c2

1h(t)(e2te3t)(t1h(t)h(t)2h(t)e3t(t 例：某⼀系统数学模型为y(t5y(t6y(th(t)5h(t)6h(t)(t)2(th(0) h(0) h(t)e3t(t

e(t)2e(t例：某⼀系统数学模型为y(t5y(t6y(te(t2e(th(t)5h(t)6h(t)(t)2(th(0) h(0) 例：某⼀系统数学模型为y(t5y(t6y(te(t2e(th(t)5h(t)6h(t)(t)2(t h(0)21,h(0)h(t)(t)5(t)(12e2t33e3t)(th(t)的形式与(i) ah(i)(t) (j)(t j当nm

Cieit(tn当nm h(t)

Cieit(t)C0(tn当nm h(t)n

Ceit(t)C(t)C(t)L

(te(t) ng(t)T0,(tn

g(t)

ay(i)(t)bmj)

ag(i)(t) b(j)(t ？ (j0,1,2Ln？初始条件：gj)(0

h(t)与g(t)的关 g(t)h( Q(t)d(t) h(t)dg(tdt例：某⼀系统的数学模型为y(t5y(t6y(te(t2e(t求该系统的阶跃响应g(t)g(t)5g(t)6g(t)(t)2(tg(0)1,g(0)g(t)e3t(t例：某⼀系统的数学模型为y(t5y(t)6y(t)e(t)2e(t求该系统的冲激响应h(t) h(t)5h(t)6h(t)(t)2(th(0) h(0)h(t)(t)3e3t h(t)g(t)(t)3e3t例：某⼀系统的数学模型为y(t)求该系统的冲激响应h(t

3y(t)4y(t)e(t)2e(th(t)(3et4e2t)(t求该系统的阶跃响应g(tg(t)

h(x)dx

(3ex4e2x)(x)dx3ett tt t(3e0

4e2x)dx(3et2e2t1)(t设f1(t)与f2t)是定义在（）f1(t)f2(t)的卷积积分，积分结果仍是以时间t为变量的函数f(t)。f(t) f1()ff1(t)f2(t)的卷积积分，积分结果仍是以时间t为变量的函数f(t)。f(t)f1(t)f2(tff1(t)f2(t)当f1(t)和f2(t)的时间没有限制时，卷积积分的积分限从–+f1(t和f2(t)f(t)f1(t)f2(t)当f1(t)和f2(t)的时间没有限制时，卷积积分的积分限从–+f1(t和f2(t) 1)若t0则f(t

f1(t f1()f2(t (20若t0t

f2(t f(t) f1()f2(t (2若t0时 f1(t)f2(t) f(t)

f1()f2(t (2f(t)f1(t)f2(t) f1()f2(t (267)例 f1(t)=t(t f2(t)(t 求f(t)f1(t)f2(t f(t)f(t)f(t)1t2(t 例 f1(t)=e2t(t f2(t)(t 求f(t)f1(t)f2(t f(t)f(t)f(t)1(1e2t)(t f(t)f1(t)f2(t)f1()f2(t从卷积的(数学)变量置换t反折f2(f2(-将

(-)在轴上平移t得

(t–f1()和f2(t–f1(t)f2(t)

f2( f2(t f1(t)f2(t2

f2(t

f1(t)2[(t)(t3

f2(t) 4

变量置换t 3)将f2(-)在轴上平移t得反折f2()f2(- t0 f( t0 1

f2(t–的波形在 有参变量t4)将f1()和f2(t–)当t从－∞逐渐增大时，f2(t–)沿f2(t

2

t

f(t) tf2(t

t0t 3t

f(t)024(t)dt3t 2

t

2t f(t)f(t)

t2242t

434

)d t-2

4t

4

422f2(t2f2(t0t4t

t0tf(t) 2t2

34

4)2 4ttf2(t 在t的可积函数(即t<t1时f1(t)=0,t<t2时f2(t)=0)例：已知信 f1(t)(t)(t3) f2(t)ett试求y(tf1(tf2t)

当0

3 f(t)f(t)

1e( -1 f2(t

当

133

e f(t)f(t)

1e(t 0

(e3

e10123y(t

f(t)f(t)

0,t 1

,0t

(e3 et,t1

y(t) t)0h(tt00 (tt0 h(tta(t)(tt0 ah(t)h(tt00

gV(tf(t) f(n)g

0n)

f(t

V当)宽)于无穷小量用d表示，nV趋于连续变量用

n)

n

f(t) f(tf(t)limf(n)(tn)V0

f()(t (2 f(t)

f(n)(tn) (2对于LTI

e(t) x(0) y(t e(t) e(n)(tn) e()(tV0 e(t)(t) yzs(t) e(t)(t yzs(t)h(te(t)e(nV)(tnV) yzs(t)e(nV)h(tnV) e(t)lime(n)(tn)yzs(t)lime(n)h(tn)V0

V0yzs(t)e( )d(2yzs(t)e()h(t (2yzs(t)e(t) 结论:系统在激励信号e(t)作用下的零状态响应yzs(t)e(t)与系统冲激响应h(t)的卷积积分(卷积积分的物理意义)对因果系统，若激励e(t)在t=0 e(t) h(t)tyzs(t)e(t)h(t)0e()h(tt例：已知某LTI连续系统的h(t)=(t)，激励信号e(t)=(t-解：yzs(te(th(te()h(tt 例 h(t)(t)，e(t)e2t(t)求yzs(t解：yzs(te(th(te()h(t

0

1(1e2t)(t) f(t)f1(t)f2(t)f2(t)f1(t 2f2f1(t02 选反折函数时要考虑：1)f1(t)f2(t)f3(t)f1(t)f2(t)f1(t)f3(t (2 e(tf2(tf3(te1(te2(th(t)f1(t则1y1zs(te(t)[e1(t)e2tLen(t

2y2zs(t1y1zs(t y2zs(t ynzs(tb)设e(tf1(th(tf2(tf3(th1(th2(t系统的冲激响应则yzs(te(th(te(t[h1(th2(t)]若h(t)[1(h2(t)Lhn(t)]e(t则yzs(th(te(t

e(t)1()e(t)h2(t)Le(t)hn(te(t

(t) yzs(t)

e(t …h(t)h(t)h(t)Lh(t)…(t)(t)f1(t)f2(t)f3(t)f1(t)f2(t)f3(t (2物理意义：设e(tf1(th1(t)h2(t)

f2(tf3(t

yzs(te(th1(t)h2(te(th1(th2(t)e(th(te(t e(t)1

yzs(t e(th(t)

1(th2

(t)h(t)1()h2(tLhn(t)e(e(t(t)(t)(t)

yzs(tf(t)(t)f(t (2任意函数f(t)与(t)卷积的结果为该函数f(t)f(t)(tt0)f(tt0 (2任意函数f(t)与延时t0f(t102ff(t102 f(t10f(t102 若f(t)(t则(t(t(t(t)(tt1)(tt1f(tt1)(tt2)f(tt1t2f(t)(tt1)(tt2)f(tt1t2例：f1(t)和f2(t)的波形如图所示，求f1(t)*f2(t)f2(t02f1(tf2(t02f1(t101 32 01 3 T(t)(tnT)梳状函fT(t)

f0(t)Tf0(t)(tnT

fT(t)1

f0(t)T(t

(tnT (2 TfT(t)f0(t)T(t fT(t)

f0(t)T(t

T Tttf

df(t)

f(

(t

f(f(2)(t)设f(tf(tf(tf(f(2)(t) fff(1)(t)f(1)(t) (1)(t) f f(t) fff(1)(t)f(1)(t) f(2)(t) 设f(tf(tf(f(2)(t) 则f(1)(t) (t) ( f(t)f(1)(t)f(1)(t)f(t) f(t)(t)f(t (2t函数f(t)f(t)tf(t)[(t)](1)f(t)(t) f( (2函数f(t)与(t)卷积，相当于对f(t)从∞到tf(t)f1(t)f2(tf(1)(t)f(1)(t)f(1)(t)f(1)(t (2 f1(t)与f2(t卷积等于先对其中任⼀函数求导数对另⼀函数f(t)f(i)(t)f(i)(t)f(i)(t)f(i)(t f1fi(t)f(j)(t) (ijf1

(t)f(ij

(t)

f(j)(t2例f1(t)和f2(t)的波形如图所示， f1(t)f2(t21

f(1)(t111

f2(t

f(1)(t

f1(t)21

f2(t)2 2

2 例求e2t(t(t(t例:f1(t)cost(t f2(t)(t)(t4π),求f1(t)f2(t解：f(t)f(t) f(1)(t)f(t sint(t)[(t)(t4sint(t)sin(t4π)(tsint(t)[(t) (t4例：已知某⼀系统的数学模型为y(t)2y(t)e(t3e(t)，其中e(t)t(t),求yzs(t)。系统对任意激励系统对任意激励e(t)的零状态响应yzs(t)yzs(t)e(t)h(th(t)(t)e2t(tyzs(t)

e(t)h(t

f(