《通信原理教程》部分习题解答-前5章(徐家恺)

i
1 3 11 P( y0 ) = × 0.8 + × 0.1 = 4 4 40 1 3 29 P( y1 ) = × 0.2 + × 0.9 = 4 4 40
所以
11 11 29 29 log 2 − log 2 = 0.849 (bit/符号) 40 40 40 40 (2) H ( x / y ) = −∑∑ P( xi , y j ) log 2 P( xi / y j ) H ( y) = −
·5·
2-12
解： I a = − log 2 P(a ) = 3.989bit
I y = − log 2 P( y ) = 6.381bit
2-13
解： H = −∑ P( xi ) log 2 P( xi )
i =1
5
1 1 1 1 1 3 3 5 5⎞ ⎛1 = − ⎜ log 2 + log 2 + log 2 + log 2 + log 2 ⎟ 4 8 8 8 8 16 16 16 16 ⎠ ⎝4 1 3 3 = + + + 0.453 + 0.524 2 8 8
·12·
H L (ω ) =
高通
H H (ω ) =
1 1 + (ω RC )
2
τ (ω ) =
dϕ (ω ) − RC = dω 1 + (ω RC )2
⎡ 1 ⎤
R R+ 1 jω C
=
+ jarctan ⎢ ⎥ jω RC ω RC ⎣ (ω RC ) ⎦ e = 1+jω RC 1 + (ω RC )2
i j
= −∑∑ P( y j ) ⋅ P( xi / y j ) ⋅ log 2 P( xi / y j )
i j
·8·
1 P ( y0 / x0 ) ⋅ P ( x0 ) 4 × 0.8 8 P ( x0 / y0 ) = = = 11 P ( y0 ) 11 40 3 2 27 同理 P( x1 / y0 ) = , P( x0 / y1 ) = , P( x1 / y1 ) = 11 29 29 所以 H ( x / y ) = − P ( y0 ) [ P ( x0 / y0 ) log 2 P ( x0 / y0 ) + P ( x1 / y0 ) log 2 P ( x1 / y0 ) ]
噪声系数 =
10 = 1.43 = 1.55(dB) 7
3-11
解
Si / N i = 1.1, So / N o
So / N o =
−1
[ Po (ω )] =
τ
− n0 1 ⋅ ⋅ e RC 2 2 RC
τ
− n = 0 ⋅ e RC 4 RC
2-11
解： Po (ω ) = P (ω ) ⋅ H (ω ) = ξ
2
1 ⋅ P (ω ) ξ 1 + (ω RC ) 2
P (ω ) = FT[ Rξ (τ )] ξ R0 (τ ) = I FT[ Po (ω )]
2-20
解： I ( X , Y ) = log 2 M + ∑ P( y j / xi ) log 2 P( y j / xi )
j =1 M
·9·
Pe = 0.1, I = log 2 2 + 0.1log 2 0.1 + 0.9 log 2 0.9 = 1 − 0.33 − 0.14 ≈ 0.53 C = IRB = 0.53 × 1000 = 530(bps) Pe = 0.2, I = 1 − 0.2 log 2 0.2 − 0.8log 2 0.8 ≈ 0.28 C = 280bps Pe = 0.3, I = 1 − 0.52 − 0.36 = 0.12 C = 120bps
− P( y1 )[ P( x0 / y1 ) log 2 P( x0 / y1 ) + P( x1 / y1 ) log 2 P( x1 / y1 )] 8 3 3⎤ 2 27 27 ⎤ ⎡8 ⎡2 = −0.275 × ⎢ log 2 + log 2 ⎥ − 0.725 ⎢ log 2 + log 2 ⎥ 11 11 11 ⎦ 29 29 29 ⎦ ⎣11 ⎣ 29 =0.495(bit/符号) 1 3 H ( y / x) = − × [ 0.8 × log 2 0.8 + 0.2 × log 2 0.2] − × [ 0.1 × log 2 0.1 + 0.9 × log 2 0.9] 4 4 =0.532(bit/符号) (3) I ( x, y ) = H ( x) − H ( x / y ) =0.811−0.495=0.316(bit/符号)
( ω 0 附近)
所以 so (t ) = K ⋅ si (t − td ) = A cos Ω (t − td ) ⋅ cos ω 0 (t − td ) 所以结论由此得出
3-10
解：因为 所以
Si = 10, 可得 Ni = 0.1mW Ni
So 1mW = = 5 = 7(dB) N o (0.1+0.1)mW
2-18
解： I = H × 1000 × 3600 = 2.23 × 1000 × 3600
·7·
= 8.028(Mbit)
当仅传 B、C 符号时，可达到最大Biblioteka Baidu息量
I max = 1000 × log 2 8 × 3600 = 10.8(Mbit)
2-19
⎛0 1⎞ ⎛ x , x2 ⎞ ⎜ ⎡0.8 0.2 ⎤ 解： ⎜ 1 = 1 3 ⎟ ⎡ P( y j / xi ) ⎤ = ⎢ ⎟ ⎜ ⎣ ⎦ 0.1 0.9 ⎥ ⎟ ⎟ ⎣ ⎦ ⎝ P P2 ⎠ ⎜ 1 ⎝4 4⎠ (1) P( y j ) = ∑ P( xi ) ⋅ P( y j / xi )
= E[ x(t ) ⋅ cos ω 0 t − x(t + τ ) ⋅ sin ω 0 (t + τ )] = δ 2 cos ω 0τ
2-9
解：利用傅里叶变换对偶性质可直接得出 1 1) R(τ ) = P(ω ) e jω t dω 2π ∫ 1 ⎛ω t ⎞ Sa ⎜ 0 ⎟ cos ωτ = 2π ⎝ 2 ⎠
Pe = 0.4, I = 1 − 0.53 − 0.44 = 0.03 C = 30bps Pe = 0.5, I = 0,
0.55
C = 0bps
0.1
0.5
2-21
解：(1) C = 4 × 103 × log 2 (1 + 1000) ≈ 39.87(kbps)
(2) Rb = C ′ = 4800bps
·10·
B=
4800 = 4kHz S⎞ ⎛ log 2 ⎜1 + ⎟ N⎠ ⎝
所以
S = 1.129(dB) N
2-22
解： C = 3 × 105 × log 2 (16 × 30) = 3.6 × 107 (bps) C 所以 B = = 3.61 × 106 Hz S⎞ ⎛ log 2 ⎜1 + ⎟ N⎠ ⎝
所以 H (ω ) = 2 K cos 当ω =
ωτ
2
(2n + 1)π
τ
= 1000π(2n + 1) 时为零点
·13·
f = 500(2n + 1)
极点 2 nπ ω= = 2000π ⋅ n ,
τ
f = 1000n
3-5
解：接收的信号波形：
3-6
解： τ m = 5ms,
Δf =
1
τm
= 200Hz
1) 信号的平均功率
S = 200 × ∫ 400 1
−
400
A2 ⋅ sin 2 (400πt )dt 4 A2 8
⎛ A⎞ =⎜ ⎟ ⎝2⎠
2
2=
2) 自相关函数
·2·
R(t , t + τ ) = ∫ =
T 2 T − 2
f (t ) f (t + τ )dt
A2 1 ⎡ × cos 400πτ dt − ∫ cos 400π(2t + τ )dt ⎤ ⎥ ⎣ ⎦ 4 2 ⎢∫ 2 2 A A cos 400πτ = cos 400πτ = 8 8 3) 功率谱密度
H H (ω ) =
ω RC
1 + (ω RC )2
, τ (ω ) =
⎡ ⎤ 1 ⎥ ⋅ RC ⋅ ⎢− 2 2 ⎢ ⎥ ⎛ 1 ⎞ ⎣ (ω RC ) ⎦ 1+ ⎜ ⎟ ⎝ ω RC ⎠ RC =− 1 + (ω RC )2 1
3-4
解： H (ω ) =
Fo (ω ) = K (1 + cos ωτ − jsin ωτ ) Fi (ω )
1-5
解： RB = 1.2 × 108 (60 × 2) = 106 (Baud) Pe = 3 /(1.2 × 108 ) = 2.5 × 10−8
·1·
第 二 章
2-4
解： f (t ) = A cos 200πt ⋅ sin 200πt = T= A sin 400πt 2
1 s 200
1
《通信原理教程》 部分习题解答
(仅供参考)
第 一 章
1-4
解： s (t ) = A cos(2π × 105 + ϕ 0 ), T =
1 = 10μs 105
v =30 × 104 km / s, s =80km 80 td = = 267μs, 设 ϕ 0 = 0 3 × 105 2 4 ⎡ 267 ⎤ Δϕ = ⎢ mod1× 2π = × 2π = π 10 ⎥ 3 3 ⎣ ⎦
·11·
第 三 章
3-1
解：已知 H (ω ) = K 0 e − jω td 由傅里叶变换的时延性质
f 0 (t ) = K 0 fi (t − td )
即幅度放大K0倍，信号延时 td 秒。 满足线性不失真条件
3-2
解：低通
1 1 1 jω C H L (ω ) = e− jarctan(ω RC ) = = 2 1 1 + jω RC 1 + (ω RC ) R+ jω C
≈2.23(bit/符号)
2-14
解： H ( x) = −∑ P( xi ) ⋅ log P( xi )
i
1 1 1 1 = log 2 4 + log 2 8 + log 2 8 + log 2 2 4 8 8 2 =1.75(bit/符号)
·6·
2-15
解：点的信息量
3 = 0.415(bit) 4 1 划的信息量 I 2 = log 2 = 2(bit) 4 3 1 所以 H = × 0.415 + × 2 = 0.81 (bit/符号) 4 4
P(ω ) =Y
[ R(τ )] =
A2 π [δ (ω + 400π) + δ (ω − 400π) ] 8
2-5
解：
P=
1 ∞ 2 1 ∞ 2 V (t )dt = Sa (ω t ) dt 100 ∫ −∞ 100 ∫ −∞ 1 1 ∞ 2 = × g (−ω ′) × 2πdω ′ 100 2π ∫ −∞ 1 400 1 1 = ∫ − 400 2ω dω ′ = 100 100
= n0 BSa ( πBτ ) cos ω c t
2) 一维概率密度函数 n σ 2 = N = 0 × 2 × B = Bn0 2
·4·
p( x) =
1 2πσ
e
−
x2 2σ 2
=
1 2πBn0
e
−
x2 2 Bn0
2-10
解： H (ω ) =
n 1 , Pi (ω ) = 0 1 + jω RC 2 n0 n0 1 2 Po (ω ) = H (ω ) = ⋅ 2 2 1 + ω 2 R 2C 2 R(τ ) = Y
200 200 ⎛1 1⎞ B = ⎜ − ⎟ Δf = ~ ≈ (66.7 ~ 40)Baud 3 5 ⎝3 5⎠
·14·
所以
T0 =
1 = (3 ~ 5)τ m = (15 ~ 25)ms B
3-7
解： s (t ) = A cos Ω t ⋅ cos ω 0 t
H (ω ) = Ke− jω td
·3·
2-8
解： E ( Z ) = E ( x) cos ω 0 t − E ( y )sin ω 0 t = 0
D( Z ) = E [ Z − E ( Z )] = E[ x 2 ]cos 2 ω 0t + E[ y 2 ]sin 2 ω 0t
2
=σ2 R(τ ) = E[ Z (t ) ⋅ Z (t + τ )] = E [ x(0) ⋅ x(τ ) ⋅ cos ω 0τ ] t =0
I1 = log 2
2-16
解： H ( x) = 16 ×
1 1 × log 2 32 + 112 × × log 2 224 32 224
=6.4(bit/符号) Rb = 6.4 × 103 bps
2-17
解： Rb 2 = RB = 1200bps
Rb16 = RB × log 2 16 = 2400 × 4 = 9600(bps)