信号与系统课件第九章

9.9 9.319.329.339.35 例如x(t)

这里

y(t)H(s)estH(s)

XX(s)

s X(s)X(

x(t)e(j)t

]ejtx(t)eatu(t) 确定 变换的表达式解 此例也 x(t)

X(s)

eatu(t)est

e(as)t0 a

e(as0

这里即

aX(s)

eatu(t)est e(as)tx(t)x(t)eatu(t)a a

e(as

这里Resa s的取值范围确保 值范围叫做收敛域-----regionofconvergence 变换x(t)3e2tu(t)

X(s)

t (3e2t

3e(2s)tdt2e(1s)t442 442 s s

ss2s x(t)

dt

dt 0x(t)e 0

dt

dt多 变换都用分式表示 X(s)

N

如果x(t)的变换X(s)是有理的，那如果x(t)的变换X(s)是有理的，若变换可能的ROC,并考虑是否 s 逆 ax1(t)bx2(t)aX1(s)bX2(s 例如：x1(t)x2tabax1(tbx2t)0X(s)0

x(tT)esTX(sROC不变s

Re{s

解 s

e2tu(t)

s

e2(t2) 若则

ees0tx(t)(ss0ROC=R+Re{s0u(t)se2tu(t) e2tu(t)

Re{s}0Re{s}2 s3g(t)e2t

若则x(at)1X(s

etu(t)etu(t) e

s se2

e2tu(t)e2tu(t) 1 s s若则

y(t)

y(tX(s)H(s)Yx(t)X(s)dx(t)sX(sdt因为

X(s)

ROC=ROC=故

(t)x(t)estdt若则

tt

s

ROCRI{Re{s若x(t0当t<0t=0处x(t不x(0)limsXslimx(t)limsX

]

dx(t)estdtx(t0tL[ ]

dx(t)est estx(t) x(t)de x(0)

x(t)e 42 x(t)的

(若Re{sx(0)sX(s)dx(t) dtx(0)sX 即：estdx(tx(0sX

estdx(t) sX

x(0

0x(0)limsXx(0)x(0 举例u(t)解L{u(t)}

u(t)estdt

est0s

e

x(t1 xt 变换tu(t)解L{tu(t)}s ••(s2s(s2(s1)2

Re{s}Re{s}Re{s}X(s)

j/6 j/

Re{s}

(sj3)(s s sx1(t)

jej3tu(t)6

jej3tu(t)6

1(sin3X(s)

1/ 1/

(sj3)(s s s Re{s}x2(t)

1ej3tu(t)2

1ej3tu(t)(cos3t)u(t) x2(t)(cos(3t))u(t)(s)2 Re{s}, g(t)cos(3t)u(t),

s2

Re{s}X(s)

s

G(s (s1)2

3 x(t)etg(t)et(cos3t)u(t)3 H H(ej h(t)=0当tY(s)h(t)=0当t 时域为右边函

H(s) 2s解 e解 2

2(t

2)u(t Re{s}h(t)e2(t2)u(th(t)0当t<0,

h(t) 变半平面时，H(s)9.34526页11Hs1HH(s)9.34526页s2s2H(s)2sH(s)2H(s22s2)H(s)

limHlimH(s)次s22sA5s22s Nadky(t)MN

dkk

NNN

dky(t)dtk

MMM

dkx(t)dtkk(askk

)Y(s)(bskkkM

)Xk(bskkNH(s)k Nk(askk

H(s) s H(s)1 s sH(s)

YX

s2

5s3ss2Y(s)3sY(s)2Y(s)s2X(s)5sX(s)5Xd2y(t)

2y(t)

5

5x(t)dt dt 1)y(t9.26506页x(t)=1+h2h2h2

+

H1h2h2H(s)

H1(s)1H1(s)H2H(s)

1H1(s)H2

H(s)

H1(s)H2(s)1H(s) 11s

3反馈连接++

H(s)

H1(s) 1H1(s)H2 H(s)

1H(s)s++

++

2H(s)

1(s2)(sa)并联b)级联c)H(s)

2s24ss23s解 x(t)

at解:因为x(t)=0当t<0,单边和双边的 相等.故可以查表9.2, (sa)n

Re{s}>x(t)ea(t1)u(t a(t 0 u(t (sa)t1Re{s}1Re{s}ea

s

Re{s}

dt

x(t)e dtx(t) 当t 9.9.39.9.3程 变换d3x(t)

)

144444

44444ULT

d2x(ts3X(s)s2x(0)sx(0)x 零状态响 d2yt

2dxy 输入为xt(t)ut 。求 ROC 变换和&LCCDE表示的系统的全响因果系统B由微分方程表征:dy(t)y(t)dw(t) x(te3ty(t2）HA(s)

sH(s)s

(s)

(s)H(s)

(s1)(sx(t)e