附录K 部分习题答案

附录 K 部分习题答案

第 1 章

1-1 3.25 b;u8.97 b

1-2 1.75b / 符号

1-3 (1) 200 b/s; (2) 198.5 b/s

1-4 (1) 3.02 b / 键；(2) 6.04 b/s

1-5 \(6.405 \times 10^{3}\) b/s

1-6 (1) 2500 b/s ; (2) 10000 b/s

1-7 (1) 2.23 b / 符号；

(2) \(8.028 \times 10^{6}\) b ;

(3) \(8.352 \times 10^{6}\) b

1-8 \(10^{-4}\)

第 2 章

2-1 (1) 非周期信号，能量信号；(2) 周期信号，功率信号；(3) 非周期信号，既不是能量信号也不是功率信号。

2-2 略

2-3 (1) \(|C_{n}| = 1\) , \(n = \pm 1\) ;

(2) \(P(f) = \delta (f - f_0) + \delta (f + f_0)\)

2-4 能量信号，能量谱密度

\[ \mid X (f) \mid^ {2} = 4 / \left(1 + 4 \pi^ {2} f ^ {2}\right) \]

2-5 \(S(\omega) = AT\mathrm{Sa}\left(\frac{\omega T}{2}\right)\)

\[ E (\omega) = | S (\omega) | ^ {2} \]

\[ R (\tau) = \]

\[ \left\{ \begin{array}{c c} A ^ {2} T (1 - | \tau | / T) & | \tau | \leqslant T \\ 0 & \text {其他} \end{array} \right. \]

\[ E = A ^ {2} T \]

2-6 \(R_{s}(\tau) = 1 - |\tau |(-1\leqslant \tau \leqslant 1)\)

2-7 (1) \(P_{s}(f) = \frac{k^{2}}{k^{2} + 4\pi^{2}f^{2}}\)

\[ P = k / 2 \]

2 - 8 \(P(f) = \frac{1}{T}\sum_{n = -\infty}^{\infty}\mathrm{Sa}^{2}(\pi \frac{n}{T})\delta (f - \frac{n}{T}) =\)

\[ \frac {1}{2} \sum_ {n = - \infty} ^ {\infty} \mathrm{Sa} ^ {2} \left(\frac {1}{2} \pi n\right) \delta \left(f - \frac {n}{2}\right) \]

2-9 (1) \(\mathrm{j}\pi [\delta (\omega +\omega_0) - \delta (\omega -\omega_0)]\)

(2) \(\frac{1}{2\mathrm{j}} [S(\omega - \omega_0) - S(\omega + \omega_0)]\)

第 3 章

3-1 \(f(y) = \frac{1}{\sqrt{2\pi c^2}}\exp \left(-\frac{(y - d)^2}{2c^2}\right)\)

3-2 \(E_{\xi}(1) = 1;\cup R_{\xi}(0,1) = 2\)

3-3 (1) \(E[Y(t)] = 0, \cup E[Y^2(t)] = \sigma^2\)

(2) \(f(y) = \frac{1}{\sqrt{2\pi}\sigma}\exp \left(-\frac{y^2}{2\sigma^2}\right)\)

(3) \(B(t_{1},t_{2}) = R(t_{1},t_{2}) = \sigma^{2}\cos \omega_{0}\tau\)

3-4 因为 \(E[Z(t)] = a_{X} + a_{Y}, R_{Z}(t_{1}, t_{2})\)

\(= R_{X}(\tau) + R_{Y}(\tau) + 2a_{X}a_{Y}\) ，所以 \(z(t)\) 平稳

3-5 \(P_{e}(f) = \frac{1}{4}\left[P_{s}(f + f_{c}) + P_{s}(f - f_{c})\right]\)

3-6 (1) 因为 \(E[z(t)] = 0\)

\(R_{z}(t_{1},t_{2}) = R_{z}(\tau)\) ，所以 \(z(t)\) 平稳；

(2) \(R_{z}(\tau) = \frac{1}{2} R_{m}(\tau)\cos \omega_{0}\tau\)

(3) \(P_{z}(\omega)=\)

\[ \frac {1}{4} \left[ \mathrm{Sa} ^ {2} \left(\frac {\omega + \omega_ {0}}{2}\right) + \mathrm{Sa} ^ {2} \left(\frac {\omega - \omega_ {0}}{2}\right) \right], S = \]

\[ R _ {z} (0) = \frac {1}{2} \]

3-7 (1) 略；

(2) \(R_{Y}(\tau) = 2R_{X}(\tau) + R_{X}(\tau -\)

\[ T) + R _ {1} (\tau + T), P _ {Y} (\omega) = 2 (1 + \]

\[ \cos \omega T) P _ {x} (\omega); \]

(3) \(R_{Y}(\tau) = 2R_{X}(0) + R_{X}(-T) +\)

\[ R _ {x} (+ T) = 2 R _ {x} (0) + 2 R _ {x} (+ T) \]

3-8 (1) \(R_{\mathrm{o}}(\tau)\)

\[ = n _ {0} B \mathrm{Sa} (\pi B \tau) \cos 2 \pi f _ {c} \tau ; \]

(2) \(N_{0}=n_{0}B;\)

(3) \(f(x) = \frac{1}{\sqrt{2\pi n_0B}}\exp \left[-\frac{x^2}{2n_0B}\right]\)

3-9 (1) \(P_{\mathrm{o}}(\omega) = \frac{n_{0}}{2}\cdot \frac{1}{1 + (\omega RC)^{2}},R_{\mathrm{o}}(\tau)\)

\[ = \frac {n _ {0}}{4 R C} e ^ {- \frac {1}{R C} | \tau |}; \]

(2) \(f(x) = \sqrt{\frac{2RC}{\pi n_0}}\exp \left(-\frac{2RC}{n_0} x^2\right)\)

3-10 略

3-11 \(P_{12}(\omega) = P_{\eta}(\omega)\cdot H_1^* (\omega)\cdot H_2(\omega)\)

3-12 (1) \(Y(t)\) 平稳；

(2) \(P_{\gamma}(\omega) = 2(1 + \cos \omega T)\)

\[ \omega^ {2} P _ {\lambda} (\omega) \]

3-13 \(P_{x}(\omega) = \pi \sum_{n}\mathrm{Sa}^{2}(\frac{n\pi}{2})\delta (\omega -n\pi)\)

第 4 章

4-1 44.7 km

4-2 5274 km

4-3 583 km

4-4 幅频失真；群迟延失真

4-5 7.72 μV

4-6 1.967b / 符号

4-7 1967 b/s

4-8 801.6 s =13.36 min

4-9 56.6W

第 5 章

5-1 (1) \(s_{\mathrm{AM}}(t) = [100 + 60\cos (2\pi \times\)

\[ \left. \left. 1 0 ^ {3} t\right) \right] \cos \left(1 0 ^ {5} \pi t\right); \]

(2) \(P_{\mathrm{e}} = 100\mathrm{W}, P_{\mathrm{USB}} = P_{\mathrm{LSB}} = \frac{m^2}{4} P_{\mathrm{e}}\)

\[ = 9 \mathrm{W}, P _ {\mathrm{s}} = P _ {\mathrm{USB}} + P _ {\mathrm{LSB}} = \frac {m ^ {2}}{2} P _ {\mathrm{c}} \]

\[ = 1 8 \mathrm{W}, P _ {\mathrm{AM}} = P _ {\mathrm{c}} + P _ {\mathrm{s}} = (1 + \]

\[ \frac {m ^ {2}}{2}) P _ {\mathrm{c}} = 1 1 8 \mathrm{W}; \]

(3) \(\eta_{\mathrm{AM}} = \frac{P_{\mathrm{s}}}{P_{\mathrm{AM}}} = \frac{9}{59}\) ;

(4) \(P_{\mathrm{AM}} = (1 + \frac{m^2}{2})P_{\mathrm{c}}\)

\[ = P _ {r} = 1 0 0 \mathrm{W} \]

5-3 略

5-4 \(s_{USB}(t)=\frac{1}{2}\cos(12000\pi t)+\)

\[ \frac {1}{2} \cos (1 4 0 0 0 \pi t), s _ {\mathrm{LSB}} (t) = \]

\[ \frac {1}{2} \cos (8 0 0 0 \pi t) + \frac {1}{2} \cos (6 0 0 0 \pi t) \]

5 - 5 \(s_{\mathrm{VSB}}(t) = \frac{1}{2} m_0\cos 20000\pi t +\)

\[ \frac {A}{2} [ 0. 5 5 \sin 2 0 1 0 0 \pi t - \]

\[ 0. 4 5 \sin 1 9 9 0 0 \pi t + \sin 2 6 0 0 0 \pi t ] \]

5-6 \(s(t) = \frac{1}{2} m(t)\cos (\omega_2 - \omega_1)t -\)

\(\frac{1}{2}\dot{m}(t)\sin(\omega_{2}-\omega_{1})t\) ，它是载频为 \((\omega_{2}-\omega_{1})\) 的上边带信号

5-7 \(c_{1}(t) = \cos \omega_{0}t, c_{2}(t) = \sin \omega_{0}t\)

5-8 (1) 100kHz,10kHz;

(2) \(S_{i} / N_{i} = 1000\)

(3) \(S_{\mathrm{o}} / N_{\mathrm{o}} = 2000\)

(4) \(P_{\mathrm{no}}(f) = 2.5 \times 10^{-9} \mathrm{~W/Hz}\) ,

\[ \mid f \mid \leqslant 5 \mathrm{kHz} \]

5-9 (1)102.5kHz,5kHz;

(2) \(S_{i} / N_{i} = 2000\)

(3) \(S_{\mathrm{o}} / N_{\mathrm{o}} = 2000\)

(4) \(P_{\mathrm{no}}(f) = 1.25 \times 10^{-9} \mathrm{~W/Hz}, |f| \leqslant 5 \mathrm{kHz}\)

5-10 (1) \(S_{i} = \frac{\alpha}{2} f_{m}\) ; (2) \(S_{o} = \frac{\alpha}{4} f_{m}\) ;

(3) \(\frac{S_{o}}{N_{o}} = \frac{\alpha}{2n_{0}}\)

5-11 (1)2000W;(2)4000W

附录 K 部分习题答案

5-12 略

5-13 (1) \(S_{i} / N_{i} = 5000\) ，即 \(37\mathrm{dB}\)

(2) \(S_{o}/N_{o}=2000\) ，即 33dB；

(3) G = 2/5

5-14 略

5-15 (1) \(S_{\mathrm{FM}}(t) = 10\cos (2\pi \times 10^{6}t +\)

\[ 1 0 \sin 2 \pi \times 1 0 ^ {3} t) \]

(2) \(\Delta f = 10kHz, m_{f} = 10, B \approx 22kHz;\)

(3) \(\Delta f = 10kHz, m_{f} = 5,\) \(B \approx 24kHz;\)

(4) 为原来的两倍

5-16 (1) 16MHz, 1200W;

(2) 96MHz,10.67W

5-17 (1) 60kHz\~108kHz,48 kHz;

(2) 420kHz, 468kHz, 516kHz, 564kHz, 612kHz

第 6 章

6-1\~6-3 略

6-4 (1) \(P_{\mathrm{s}}(f) = 4f_{\mathrm{B}}P(1 - P)\mid G(f)\mid^{2} + f_{\mathrm{B}}^{2}\)

\[ (1 - 2 P) ^ {2} \sum_ {- x} ^ {+ \infty} | G (m f _ {\mathrm{B}}) | ^ {2} \delta (f - m f _ {\mathrm{B}}), \]

\[ S = 4 f _ {\mathrm{B}} P (1 - P) \int_ {- \infty} ^ {\infty} | G (f) | ^ {2} \mathrm{d} f + f _ {\mathrm{B}} ^ {2} \]

\[ (1 - 2 P) ^ {2} \sum_ {- \infty} ^ {\infty} | G (m f _ {\mathrm{B}}) | ^ {2} \]

(2) 不存在；(3) 存在

6-5 (1) \(P_{\mathrm{c}}(\omega) = \frac{A^{2}T_{\mathrm{B}}}{16}\mathrm{Sa}^{4}\left(\frac{\pi fT_{\mathrm{B}}}{2}\right)+\)

\[ \frac {A ^ {2}}{1 6} \sum_ {- x} ^ {x} \mathrm{Sa} ^ {4} \left(\frac {m \pi}{2}\right) \delta (f - m f _ {\mathrm{B}}); \]

(2) 可以；功率为 \(\frac{2A^{2}}{\pi^{4}}\)

6-6 (1) \(P_{s}(f) = f_{\mathrm{B}}|G(f)|^{2} =\)

\[ \left\{ \begin{array}{l l} \frac {T _ {\mathrm{B}}}{1 6} (1 + \cos \pi / T _ {\mathrm{B}}) ^ {2} & | f | \leqslant \frac {1}{T _ {\mathrm{B}}} \\ 0 & \text {其他} \end{array} \right. \]

(2) 双极性等概时离散谱消失，故不存在定时分量；

(3) \(R_{B}=1000\) 波特，\(B=1000Hz\)

6-7 AMI 码：+10-1 +100000000

\[ 0 - 1 0 + 1 \]

\(\mathrm{HDB}_{3}\) 码： \(+10 - 1 + 1000\mathrm{V}_{+}\mathrm{B}_{-}0\)

\[ 0 \mathrm{V} - 0 + 1 0 - 1 \]

6-8 双相码：10 01 10 10 01 01 10 01 10

CIM 码：11 01 00 11 01 01 00 01 11

6-9 (1) \(H(\omega) = \frac{T_{\mathrm{B}}}{2}\mathrm{Sa}^{2}\left(\frac{\omega T_{\mathrm{B}}}{4}\right)\mathrm{e}^{-\frac{\omega T_{\mathrm{B}}}{2}};\)

\[ (2) G _ {\mathrm{T}} (\omega) = G _ {\mathrm{R}} (\omega) = \sqrt {H (\omega)} \]

\[ = \sqrt {\frac {T _ {\mathrm{B}}}{2}} \mathrm{Sa} \left(\frac {\omega T _ {\mathrm{B}}}{4}\right) \mathrm{e} ^ {- \mathrm{j} \frac {\omega T _ {\mathrm{B}}}{4}} \]

6-10 (1) \(h(t) = \frac{\omega_0}{2\pi}\mathrm{Sa}^2\left(\frac{\omega_0t}{2}\right)\) ;

(2) 不能实现

6-11 只有图 (c) 满足无码间串扰条件

6-12 应从三个方面考虑：(1) 无码间串扰；(2) 频带利用率高；(3) 单位冲激响应的尾部收敛快。选择传输函数 (c) 较好

6-13 (1) 可以实现；

(2) \(R_{\mathrm{B}} = \frac{\omega_0}{\pi},\eta = \frac{R_{\mathrm{B}}}{B} = \frac{2}{1 + \alpha}\)

6-14 (1) \(B = \frac{1}{2} (1 + \alpha)R_{\mathrm{B}} = 840\mathrm{Hz}\)

\[ \eta = \frac {R _ {\mathrm{B}}}{B} = \frac {2}{1 + \alpha} \approx 1. 4 3 \mathrm{Baud/Hz}, \]

\[ \eta_ {\mathrm{b}} = \frac {R _ {\mathrm{b}}}{B} = 2. 8 6 \mathrm{b/(s·Hz)} \]

(2) \(B = 1200\mathrm{Hz}\) ， \(\eta = 1\) Baud/Hz，

\[ \eta_ {\mathrm{b}} = 2 \mathrm{h} / (\mathrm{s} \cdot \mathrm{Hz}) \]

(3) 无 ISI

6-15 略

6-16 (1) \(R_{\mathrm{B}} = \frac{1}{2\tau_0}\) ;

(2) \(N_{\mathrm{o}} = \frac{n_{0}}{2}\mathrm{W}\)

(3) \(P_{e} = \frac{1}{2}\exp \left(-\frac{A}{2\lambda}\right)\)

6-17 (1) \(P_{\mathrm{e}} = 6.21\times 10^{-3}\)

(2) \(A \geqslant 8.6\sigma_{R}\)

6-18 (1) \(P_{e} = 2.87 \times 10^{-7}\) ;

(2) \(A \geqslant 4.3\sigma_{\mathrm{R}}\)

6-19 略

6-20 \(h(t) = \mathrm{Sa}\left(\frac{\pi}{T_{\mathrm{B}}} t\right) - \mathrm{Sa}\left(\frac{\pi}{T_{\mathrm{B}}}(t - 2T_{\mathrm{B}})\right);\)

\[ \mid H (\omega) \mid = \left\{ \begin{array}{c c} 2 T _ {\mathrm{B}} \sin \omega T _ {\mathrm{B}} & \mid \omega \mid \leqslant \frac {\pi}{T _ {\mathrm{B}}} \\ 0 & \text {其他} \end{array} \right. \]

附录 K 部分习题答案

6-21 3 个，7 个

6-22 略

6-23 \(\frac{37}{48}\) , \(\frac{71}{480}\)

6-24 (1) \(C_{-1} = -0.1779\) ， \(C_0 = 0.8897\) \(C_1 = 0.2847\)

(2) 均衡后的峰值失真 (0.06766) 比均衡前的峰值失真 (0.6) 减小 8.87 倍

第 7 章

7-1 (1) 2 个；(3) \(B_{2PSK}=B_{2DPSK}=B_{2ASK}=\) \(2R_{B}=4000Hz\)

7-2 (2) \(6000\mathrm{Hz}\) ; (3) 相干解调或相关接收

7-3 略

7-4 \(P_{2\mathrm{PSK}}(f) = 288[\mid G(f + f_c)\mid^2 +\) \(\mid G(f - f_c)\mid^2 ] + 0.01[\delta (f + f_c) + \delta (f - f_c)]\)

7-5 略

7-6 (1) \(b^{*} = a / 2\) ， \(P_{e} = 0.0146\) （2） \(b^{*} > a / 2\)

7-7 (1)110.8dB;(2)111.8dB

7-8 (1)113.9dB;(2)114.8dB

7-9 (1) \(3.37 \times 10^{-3}\) , \(2.27 \times 10^{-5}\) , \(10^{-9}\) ; (2) \(8.5 \times 10^{-5}\) , \(4.05 \times 10^{-6}\) , \(1.26 \times 10^{-9}\) , \(2.52 \times 10^{-9}\)

7-10 (1) \(2.24 \times 10^{-8} \mathrm{~W}, 1.12 \times 10^{-8} \mathrm{~W}\) , \(0.56 \times 10^{-8} \mathrm{~W}, 0.61 \times 10^{-8} \mathrm{~W}\) (2) \(2.72 \times 10^{-8} \mathrm{~W}, 1.36 \times 10^{-8} \mathrm{~W}\) , \(0.68 \times 10^{-8} \mathrm{~W}\)

7-11 \(4\times 10^{-6},8\times 10^{-6},2.27\times 10^{-5}\) 7-12 \(4\times 10^{-2},3.93\times 10^{-6}\)

7-13 略、7-14 略

7-15 \(8.1\times 10^{-6}\) ， \(6.66\times 10^{-4}\)

7-16 (1) \(4800\mathrm{Hz}\) , \(1/2(\mathrm{b/s})/\mathrm{Hz}\) ; (2) \(3360\mathrm{Hz}\) , \(0.71(\mathrm{b/s})/\mathrm{Hz}\) (3) 进制数 M 应提高，可以采用 8PSK 或 16PSK

7-17 (1) \(4800\mathrm{Hz}, 1(\mathrm{b/s}) / \mathrm{Hz}\) ; (2) \(3200\mathrm{Hz}, 1.5(\mathrm{b/s}) / \mathrm{Hz}\) 第 8 章

8-1 (1) 2400Hz, 1(b/s)/Hz;

(2) 1680Hz, 1.43(b/s)/Hz;

(3) 应使 M = 16，调制方式可采用 16QAM

8-2 略

8-3 \(f_{0} = 750\mathrm{Hz}\)

8-4、8-5 略

第 9 章

9-1 略

9-2 \(P_{\mathrm{e}} = \frac{1}{2}\mathrm{erfc}\sqrt{\frac{E_{\mathrm{b}}}{4n_0}} = \frac{1}{2}\mathrm{erfc}\sqrt{\frac{A^2T}{8n_0}}\)

9-3 \((3)P_{e} = \frac{1}{2}\mathrm{erfc}\sqrt{\frac{E_{b}}{2n_{0}}} = \frac{1}{2}\mathrm{erfc}\sqrt{\frac{A^{2}T_{s}}{4n_{0}}}\)

9-4 最佳接收机: \(P_{e1} = 4 \times 10^{-6}\) , 普通接收机: \(P_{e2} = 3.4 \times 10^{-2}, P_{e2} / P_{e1} = 8500\)

9-5 \(R_{\mathrm{B}} = 1 / T_{\mathrm{B}} = 5.55\times 10^{6}\) 波特

9-6 \(1.008 \times 10^{-9} \mathrm{~W}\)

9-7 略

9-8 (1) \(t_0 \geqslant T\) ;

\[ \begin{array}{l} (2) h (t) = f (T - t) \\ = \left\{ \begin{array}{l l} - A & 0 \leqslant t \leqslant T / 2 \\ A & T / 2 < t \leqslant T \\ 0 & \text {其他} \end{array} \right. \\ \end{array} \]

\[ y (t) = h (t) * f (t) = \]

\[ \left\{ \begin{array}{l l} - A ^ {2} t & 0 \leqslant t \leqslant T / 2 \\ A ^ {2} (3 t - 2 T) & T / 2 < t \leqslant T \\ A ^ {2} (4 T - 3 t) & T < t \leqslant 3 T / 2 \\ A ^ {2} (t - 2 T) & 3 T / 2 < t \leqslant 2 T \\ 0 & \text {其他} \end{array} \right. \]

(3) \(r_{\mathrm{omax}} = \frac{2E}{n_0} = \frac{2A^2T}{n_0}\)

9-9 都可以

9 - 10 (2) \(h_1(t) = s_1(T - t), h_2(t) = s_2(T - t)\) ; 上支路输出 \(\left\{ \begin{array}{l} s_1(t) * h_1(t) \\ s_2(t) * h_1(t) \end{array} \right.\) , 下支路输出 \(\left\{ \begin{array}{l} s_1(t) * h_2(t) \\ s_2(t) * h_2(t) \end{array} \right.\) ;

附录 K 部分习题答案

(3) \(P_{r} = \frac{1}{2}\mathrm{erfc}\left[\sqrt{\frac{A_{0}^{2}T}{4n_{0}}}\right]\)

9-11 (2) 广义瑞利分布，瑞利分布；

(3) \(P_{\mathrm{e}} = \frac{1}{2}\exp \left(-\frac{E_{\mathrm{b}}}{2n_{0}}\right)\)

\[ = \frac {1}{2} \exp \left(- \frac {A ^ {2} T}{4 n _ {0}}\right) \]

第 10 章

10-1 (1) \(M_{\mathrm{s}}(f) = 300\sum_{-x}^{x}M(f - 300n)\)

(2) \(M_{*}(f) = 400\sum_{n = 1}^{\infty}M(f - 400n)\)

10-2 (1) 不超过 0.25 ms;

(2) \(M_{1}(f) = 5\sum_{n = -\infty}^{\infty}M(f - 5n)\)

10-3 1000 Hz

10-4 (1) 抽样速率应大于 \(2f_{1}\) ;

(3) \(H_{2}(\omega) = \left\{ \begin{array}{ll}\frac{1}{H_{1}(\omega)} & |\omega |\leqslant \omega_{1}\\ 0 & |\omega | > \omega_{1} \end{array} \right.\)

10-5 \(M_{\mathrm{s}}(\omega) = \frac{1}{2\pi} M(\omega)*Q(\omega)\)

\[ = \frac {\tau}{T} \sum_ {n = - x} ^ {x} \mathrm{Sa} ^ {2} (2 \pi n \tau f _ {\mathrm{m}}) M \]

\[ \left(\omega - 4 \pi n f _ {\mathrm{m}}\right) \]

10-6 (1) \(m_{\mathrm{z}}(t) = m(t)p(t)\) ， \(M_{\mathrm{z}}(f) =\)

\[ 4 0 0 \tau \sum_ {n = - x} ^ {x} S a (4 0 0 \pi n \tau) M (f - 4 0 0 n) \]

(2) \(m_{\mathrm{H}}(t)=m_{\mathrm{s}}(t)*h(t)\) ， \(h(t)\) 是宽度为 \(\tau\) ，幅度为 1 的矩形脉冲，

\[ \begin{array}{l} M _ {\mathrm{H}} (f) = 4 0 0 \tau \sum_ {n = - \infty} ^ {\infty} \mathrm{Sa} (\pi \tau f) M (f - \\ 4 0 0 n) \end{array} \]

10-7 N = 6, Δv = 0.5

10-8 8

10 - 9 输出为 0、1.5、5.2、7.06、8.52、9.38、10

10-10 (1) 编码码组为 11100011，编码电平为 \(I_{c}=512+3\times32=608\Delta\) ，量化误差为 \(I_{a}-I_{c}=27\Delta\) ;

(2) 01001100000;

(3) 译码电平为 \(I_{D} = I_{C} + \frac{\Delta v_{7}}{2}\) = 608 +16 = 624 Δ, 量化误差

为 \(\left|I_{s}-I_{D}\right|=635-624=\) 11 Δ

10-11 (1) \(I_{\mathrm{D}} = -\left(256 + 3 \times 16 + \frac{16}{2}\right)\)

\[ = - 3 1 2 \Delta ; \]

(2) 001 0011 1000 0 (不带极性)

10-12 (1) 00110111, 误差 3Δ;

(2) 00001011100

10-13 略

10-14 (1) \(N = 7\) \(\mathrm{M} = 2^{7} = 128\)

(2) \(R_{\mathrm{b}} = 2f_{\mathrm{H}}N = 58.8\mathrm{Mb / s}\) ;

(3) 2169(33.4dB)

10-15 (1) 240 kb/s;

(2) \(B = 240\mathrm{kHz},B_{\min}\) \(= 120\mathrm{kHz};\)

(3) \(B = 480\mathrm{kHz},B_{\min}\) \(= 240\mathrm{kHz}\)

10-16 (1)560 kb/s;

(2) \(B = 560\mathrm{kHz},B_{\min}\) \(= 280\mathrm{kHz};\)

(3) \(B = 1120\mathrm{kHz},B_{\min}\) \(= 560\mathrm{kHz}\)

10-17 \(B_{\mathrm{min}} = 17\mathrm{kHz}\)

10-18 (1)1011 1100,0;

(2) \(R_{\mathrm{b}} = 640\mathrm{kb / s}\) \(B = 640\mathrm{kHz}\)

第 11 章

11-1 \(d_0 = 3\)

11-2 检 2, 纠 1, 不能同时检错和纠错

11-3 检 3, 纠 1, 纠 1 同时检 2

11-4 不能，因为其行与列的错码均为偶数。

11-5 r = 4; 码率 = 11/15

11-8 \(g(x) = x^3 + x + 1\)

11-13 码多项式 \(A(x) = x^{14} + x^{11} + x^{10} + x^8 + x^7 + x^6 + x\)

11-14 有错误

11-15 (1) 最小码距 \(d_{0} = 7\) ，可检测 6 位错码；可纠正 3 位错码；纠 1 同时检 5 或纠 2 同时检 4；

(2) \(d_{0}=4\) , 可检 3, 可纠 1, 纠 1 同时检 2

11-18 输出码流为 11100101010

附录 K

部分习题答案

0 ...

11-20 111 111 100 111 001

11-21 解码输出为 00000

11-22 110011

第 12 章

12-1 略

12-2 11110101…;11101001…

12-3\~12-5 略

12-6 - + + - + - - …; - + - + + + - - - + - …

12-7 略

12-8 20 μs

第 13 章

13-1 \(\mathrm{P}_0.6\times 10^{-5}\)

13-2 \(P_{r} = \frac{1}{2}\left[1 - (\operatorname {erf}\sqrt{r}\cos (\varphi -\theta)^{2}\right]\)

13-3 0.023(1/T)

13-4 略

13-5 (1) \(7\times 10^{-4}\) (2) \(2.1\times 10^{-7}\)

13-6 (1) 1/128; (2) 1/16

13-7 (1) 32; (2) \(6.4 \times 10^{-4}\)

13-8 应选用高质量的晶体振荡器

附录 K

部分习题答案

内容简介

《通信原理 (第 7 版)》是在第 6 版的基础上，参考百余所用书院校教师的反馈意见对全书内容进行了全面修订，其主要目的是使论述更为准确、严谨、简明易读。

本书的修订着眼于基本理论、核心内容和应用背景的论述；章节之间相关内容的融合贯通；理论与实际的紧密联系；尽多采用对比或物理概念诠释的写法替代繁琐的公式推导；改进图表曲线绘制；更好地统一名词和符号；增添了信源编码内容；调整了个别章节顺序；增添和更改习题、例题；为方便演算推导，在附录中增加了常用数学公式。

本书共 13 章，分为 3 部分。第一部分 (第 1 章 \~ 第 5 章) 阐述通信基础知识和模拟调制原理。其中第 2 章和第 3 章在教学中视需要可作为复习性讲述。第二部分 (第 6 章 \~ 第 10 章) 主要论述数字通信、数字信号最佳接收和信源编码的原理。第三部分 (第 11 章 \~ 第 13 章) 讨论纠错编码、正交编码和同步等技术。

本书可作为信息与通信工程、电子技术、计算机技术等专业本科生和研究生的教科书，也可作为从事通信及相关专业的工程技术人员的参考书。