文档介绍:第一章
14(b)xn+4
(d)x-n+2
+4<-2
-n+2<2
Or n+4>4
Or-n+2>4
(b)x(1-t)+x(2-1)
1-1<3
2-t<3
19(,非周期信号(0x=,周期N=2x2,
(d)x[n]=3e
3丌n/537/0
10
周期N==k=-3=10
12=1∑0n21-]=1∑n-k]=∑m小]
二00
=ln+3]
所以M=1,n1=-3
(t)=6(t+2)-d(t-2)
)=x(zdz=(t+2)-(t-2)
E
d=「=4
{]=2x[n-2]+5x{n-3]+2xn-4
(e)y(t)
sin(rtu(t)+sin(-4rt) u(t)
sin (4tt[u(t-u(t)
()x()=Ee(2垂周期信号
(b)x(m)=cos(-n)非周期信号
(c)周期N=8
:(a线性、稳定(b)无记忆、线性、因果、稳定
(c)线性(d)线性、因果、稳定
(e)时不变、线性、因果、稳定
(f)线性、稳定(g)时不变、线性、因果
O else
输入x(t)=ax(n时,y(t)=
x(1)+x(t-2),x1(1)20
O else
a()+ax(t-2,ax(t)≥0
(5)稳定系统。
:(a)线性、稳定(b)时不变、线性、因果、稳定
(c)无记忆、线性、因果(d)线性、稳定
(e)线性、稳定
(f)无记忆、线性、因果、稳定(g)线性、稳定
:(a)(d)(e)(g)(h)(i)(l)(n)是可逆系统
. (a) Invertible. Lavers system ytt=x(# +4,
(b)Non invertible. The signais r()and (c=z/e)+2r give the same output
(e)Non insertible. #fn] and 20sl give the same autput,
(d) Inventible, Inverse sytem: w(e)=(i)/r
(e)=z;1on20-d洲每an<0
(O)Noo invertible. ai and-fml give the same resull
g) avalible Inverse system则-项-可
(h)Invertible. Irmvemx srstem y(n)=z(t)+dz(o/dr.
(I)Invertible, Iuverse system: yim=an]-(/2)=m-I
U)Noa invertible. I =4)is any comstant, then ye)0
(k)Non invertible. eIn] and 2n]result ia yin]=o
Il)Invertible levers systen: y 4)=z(t/2)
m)Non +酢- a and rr=列到=
(n) Invertible, Inverse system r)==(2n).
第二章
23:解:令x(n)=(yu,h1()=Mnl
x(m)=x1(n-2),h(n)=h(n+2),
y(m)=x(m)*h(n)=x(m-2)*h(n+2)
x1(n)*(n-2)*h(m)*(n+2)
x()*(n)=∑(
u(m)uln-m
∑
:解:
xkl
yIn]=x[n]*h[n]
xk]h[n-k1
∑h[n-k
hIn-k
1
h[4-k]
N+40
N+4≤0→N≥4