现代通信原理 - 仿真作业3 - 加性高斯白噪声(AWGN)通过DSB-SC解调器
0. 作业要求
1 n i ( t ) \color{Brown}n_i(t) ni(t) 加性高斯白噪声(AWGN)
1.1 表达式
由于
n
i
(
t
)
n_i(t)
ni(t) 的单边功率谱密度为
n
0
=
1
0
−
6
W
/
H
z
n_0 = 10^{-6} W/Hz
n0=10−6W/Hz
所以
n
i
(
t
)
n_i(t)
ni(t) 的双边功率谱密度为
P
N
i
(
f
)
=
n
0
2
=
5
×
1
0
−
7
(
W
/
H
z
)
P_{Ni}(f) = \frac{n_0}{2} = 5 \times 10^{-7} \ (W/Hz)
PNi(f)=2n0=5×10−7 (W/Hz)
1.2 平均功率
P N i = ∫ − ∞ ∞ P N i ( f ) d f = ∞ P_{Ni} = \int_{- \infty}^{\infty} P_{Ni}(f) df = \infty PNi=∫−∞∞PNi(f)df=∞
2 n ( t ) \color{Brown}n(t) n(t)
2.1 表达式
由
H
1
(
f
)
=
R
e
c
t
(
f
−
f
0
B
)
+
R
e
c
t
(
f
+
f
0
B
)
H_1(f) = Rect(\frac{f - f_0}{B}) + Rect(\frac{f + f_0}{B})
H1(f)=Rect(Bf−f0)+Rect(Bf+f0)
得到带通滤波器的传递函数为
∣
H
1
(
f
)
∣
2
=
{
1
,
90
<
∣
f
∣
<
110
0
,
o
t
h
e
r
\left| H_1(f) \right|^2 = \begin{cases} 1,& 90 < |f| < 110 \\ 0,& other \\ \end{cases}
∣H1(f)∣2={1,0,90<∣f∣<110other
所以
n
(
t
)
n(t)
n(t) 的双边功率谱密度为
P
N
(
f
)
=
P
N
i
(
f
)
⋅
∣
H
1
(
f
)
∣
2
=
{
5
×
1
0
−
7
,
90
<
∣
f
∣
<
110
0
,
o
t
h
e
r
(
W
/
H
z
)
P_N(f) = P_{Ni}(f) \cdot |H_1(f)|^2 = \begin{cases} 5 \times 10^{-7},& 90 < |f| < 110 \\ 0,& other \\ \end{cases} \ (W/Hz)
PN(f)=PNi(f)⋅∣H1(f)∣2={5×10−7,0,90<∣f∣<110other (W/Hz)
2.2 平均功率
P N = ∫ − ∞ ∞ P N ( f ) d f = 2 × 1 0 − 5 ( W / H z ) P_N = \int_{- \infty}^{\infty} P_{N}(f) df = 2 \times 10^{-5} \ (W/Hz) PN=∫−∞∞PN(f)df=2×10−5 (W/Hz)
3 n d ( t ) \color{Brown}n_d(t) nd(t)
3.1 表达式
由于
n
d
(
t
)
=
n
(
t
)
cos
(
2
π
f
c
t
)
n_d(t) = n(t) \cos(2 \pi f_c t)
nd(t)=n(t)cos(2πfct)
故
P
N
d
(
f
)
=
1
4
[
P
N
(
f
−
f
c
)
+
P
N
(
f
+
f
c
)
]
=
{
2.5
×
1
0
−
7
,
∣
f
∣
<
10
1.25
×
1
0
−
7
,
190
<
∣
f
∣
<
210
0
,
o
t
h
e
r
(
W
/
H
z
)
P_{Nd}(f) = \cfrac{1}{4} \big [ P_N(f - f_c) + P_N(f + f_c) \big ] = \begin{cases} 2.5 \times 10^{-7},& |f| < 10 \\ 1.25 \times 10^{-7},& 190 < |f| < 210 \\ 0,& other \\ \end{cases} \ (W/Hz)
PNd(f)=41[PN(f−fc)+PN(f+fc)]=⎩⎪⎨⎪⎧2.5×10−7,1.25×10−7,0,∣f∣<10190<∣f∣<210other (W/Hz)
3.2 平均功率
P N d = ∫ − ∞ ∞ P N d ( f ) d f = 1 0 − 5 ( W / H z ) P_{Nd} = \int_{- \infty}^{\infty} P_{Nd}(f) df = 10^{-5} \ (W/Hz) PNd=∫−∞∞PNd(f)df=10−5 (W/Hz)
4 n o ( t ) \color{Brown}n_o(t) no(t)
4.1 表达式
由
H
2
(
f
)
=
R
e
c
t
(
f
B
)
H_2(f) = Rect(\frac{f}{B})
H2(f)=Rect(Bf)
得到带通滤波器的传递函数为
∣
H
2
(
f
)
∣
2
=
{
1
,
∣
f
∣
<
10
0
,
o
t
h
e
r
\left| H_2(f) \right|^2 = \begin{cases} 1,& |f| < 10 \\ 0,& other \\ \end{cases}
∣H2(f)∣2={1,0,∣f∣<10other
所以
n
o
(
t
)
n_o(t)
no(t) 的双边功率谱密度为
P
N
o
(
f
)
=
P
N
d
(
f
)
⋅
∣
H
2
(
f
)
∣
2
=
{
2.5
×
1
0
−
7
,
∣
f
∣
<
10
0
,
o
t
h
e
r
(
W
/
H
z
)
P_{No}(f) = P_{Nd}(f) \cdot |H_2(f)|^2 = \begin{cases} 2.5 \times 10^{-7},& |f| < 10 \\ 0,& other \\ \end{cases} \ (W/Hz)
PNo(f)=PNd(f)⋅∣H2(f)∣2={2.5×10−7,0,∣f∣<10other (W/Hz)