测试技术课后题答案6信号分析.docx
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测试技术课后题答案6信号分析
测试技术课后题答案6信号分析
习题6
⎧Ae-at
6.1求x(t)=⎨
⎩0
(t≥0,a>0)(t
∞
的自相关积分。
解
R(τ)=
=
⎰
02
x(t)x(t+τ)dt=A
e
-aτ
2
⎰
∞
e
-at
e
-a(t+τ)
dt
A
2a
6.2求初始相角ϕ为随机变量的正弦函数x(t)=Acos(ωt+ϕ)的自相关函数,如果
x(t)=Asin(ωt+ϕ),Rx(τ)有何变化?
解
Rx(τ)=
===A1T
⎰
T
T
x(t)x(t+τ)dt=
AT
2
⎰
T
cos(ωt+ϕ)cos[ω(t+τ)+ϕ]dt
AA
2
2
⎰
2
T
cosωtcosω(t+τ)dt12
[cosωτ+cos(2ωt+ωτ)]d(ωt)
2π
⎰
2π
2
cosωτ
当x(t)=Asin(ωt+ϕ)时
Rx(τ)=
1T
⎰
2
T
x(t)x(t+τ)dt=
T
1T
⎰
T
sin(ωt+ϕ)sin[ω(t+τ)+ϕ]dt
==
ATA
⎰
2
cosωtcosω(t+τ)dt
2
cosωτ
6.3一线性系统,其传递函数为H(s)=
11+Ts
,当输入信号为x(t)=x0sin2πf0t时,
求:
(1)Sy(f);
(2)Ry(τ);(3)Sxy(f);(4)Rxy(f)。
解
H(f)=
11+j2πfT
(1)
1
y(t)=H(f)x(t+t0)=
式中ϕ=-arctan(2πfT)
Ry(τ)=
1T
x0
+(2πfT)
2
sin(2πf0t+ϕ)
⎰
T
x0
+(2πfT)
2
sin(2πf0t+ϕ)⋅
x0
+(2πfT)
2
sin[2πf0(t+τ)+ϕ]dt
=
x0
2
2
2[1+(2πfT)]
cos(2πf0τ)
(2)
Rx(τ)=
=1T
⎰
2
T
x0sin(2πf0t)⋅x0sin[2πf0(t+τ)]dt
x02
cos(2πf0τ)
x04
2
Sx(f)=F[Rx(τ)]=[δ(f+f0)+δ(f-f0)]
2
2
Sy(f)=F[Ry(τ)]=
2
x0
4[1+(2πfT)]
x0
2
[δ(f+f0)+δ(f-f0)]
[δ(f+f0)+δ(f-f0)]
或Sy(f)=H(f)Sx(f)=(3)
Rxy(τ)=
1T
4[1+(2πfT)]
x0
+(2πfT)
2
2
⎰
T
x0sin(2πf0t)⋅sin[2πf0(t+τ)+ϕ]dt
=
x0
2
2
2+(2πfT)
cos(2πf0τ+ϕ)
式中ϕ=-arctan(2πfT)
(4)
Sxy(f)=H(f)Sx(f)=
11+j2πfT
⋅x04
2
[δ(f+f0)+δ(f-f0)]
=
x0
2
4(1+j2πfT)
[δ(f+f0)+δ(f-f0)]
6.5已知限带白噪声的功率谱密度为Sx(f)=⎨
⎧S0⎩0
f≤Bf>B
2
求其自相关函数Rx(τ)。
解
∞
B
Rj2πfτ
x(τ)=
⎰
-∞
Sx(f)e
df=⎰
j2πfτ
-B
S0e
df
=2SB
S00⎰cos(2πfτ)df=
πτ
sin(2πB0
τ)
=2BS
sinc(2πBτ)
2-1已知信号的自相关函数R60
2
x=(τ
)sin(50τ),求该信号的均方值ψx。
解:
ψ
2=R)=lim3000
sin(50τ)x
x(0)=lim(
60
=3000
τ→0
τ
)sin(50ττ→0
50τ
2-4求指数衰减函数x(t)=e
-at
cosω0t的频谱函数X(f),(a>0,
t≥0)。
求单边指数函数y(t)=⎧e-at
解(t≥0,a>0)⎨
的傅里叶变换及频谱
⎩0
(t
Y(ω)=
⎰
∞
-jωt
-jωt
-∞
y(t)e
dt=⎰∞
-at
-∞
e
edt=
1a+jω
因为cosωe
-jω0t
+ejω0t
0t=
2
由位移性质,有
X(ω)=
1
2[
1
1a+j(ω+ω]
0)
+
a+j(ω-ω0)
于是,有
X(f)=
11
1
2[
+
]a2
+4π2
(f+f0)
2
a2+4π2(f-f2
0)
3