通信原理(常州工学院) 中国大学mooc慕课答案2024版 m86879

Chapter 0 Back ground and preview 第零章单元测验

1、

答案: 错误

2、

答案: 正确

3、

答案: 正确

4、

答案: 错误

5、

答案: 错误

6、

答案: 错误

7、

答案: 错误

8、

答案: 错误

9、

答案: 正确

10、

答案: 错误

11、
答案: transmitter, channel, receiver

12、
答案: broadcasting, point-to-point communication

13、
答案: transmitted power, channel bandwidth

14、
答案: a sinusoidal wave

15、
答案: a periodic sequence of rectangular pulses

16、
答案: the average signal power, the average noise power

17、
答案: band-limited

18、
答案: FDM, TDM, CDM

19、
答案: telephone channel, coaxial cable, optical fiber

20、
答案: wireless broadcast channel, mobile radio channel, satellite channel

Chapter 1 Random processes 第一章单元测试

1、

答案: 错误

2、

答案: 错误

3、

答案: 正确

4、

答案: 正确

5、

答案: 正确

6、

答案: 错误

7、

答案: 错误

8、

答案: 错误

9、

答案: 正确

10、

答案: 正确

Chapter 2 Continuous-wave modulation 第二章单元测试-1

1、 带宽为ＷＨz的基带调制信号经过ＡＭ调制后，已调信号的带宽为

答案: 2WHz

2、 DSB信号采用相干解调，相干的意义是

答案: 同频同相

3、 关于单边带调制说法正确的是

答案: 单边带调制节约传输带宽

4、 下列属于线性调制方案的是

答案: ＡＭ调制;
ＳＳＢ调制

5、 AM信号可以进行包络检测的条件是

答案: 调制百分比总是小于1;
载波的中心频率大于基带信号的带宽

6、 下列说法正确的是

答案: 包络检测器是非相干检测器;
AM可以采用相干检测

Chapter 2 Continuous-wave modulation 第二章单元测试-2

1、 5个不同的消息信号，带宽均为10kHz，先进行SSB调制，再经FDM复用后，所需要的最小带宽为

答案: 50kHz

2、 窄带FM和ＡＭ调制相比

答案: 窄带FM包络恒定而ＡM包络随信号变化而变化

3、 ＦＭ信号的频偏和调制信号的哪个分量有个关？

答案: 幅度

4、 VSB可采用的解调方式有

答案: 相干解调;
插入载波的包络检测;
相关器

5、 关于FM调制描述正确的是

答案: 载波的瞬时频偏随时间的变化而变化;
FM是非线性调制方案

6、 关于角度调制描述正确的是

答案: 角度调制是非线性调制;
角度调制的抗干扰能力强于幅度调制;
角度调制的包络是恒定的

Chapter 2 Continuous-wave modulation 第二章单元测试-3

1、 对于商用广播，频偏最大值为75kHz，若调制频率为15kHz，则相应偏移率为

答案: 5

2、 对信号进行FM调制，当考虑到频率的稳定性时，一般采用

答案: 间接调频，非相干解调

3、 关于FM信号解调描述正确的是

答案: 宽带FM采用非相干解调;
窄带FM可采用相干解调，也可采用非相干解调

4、 接收机的解调增益取决于

答案: 输出信噪比;
信道信噪比

5、 在FM调制中，非正弦调制中偏移率的作用和正弦信号调制中的调制指数相同。

答案: 正确

6、 从实际相同看，由卡逊准则得到的带宽小于FM系统需要的带宽

答案: 正确

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,

Chapter 2 Continuous-wave modulation Chapter 2 Continuous-wave modulation -4

1、 关于DSB与SSB解调增益描述正确的是

答案: DSB与SSB解调增益都是1

2、 对各线性调制方案的抗噪声性能进行比较，正确的是

答案: DSB=SSB>AM

3、 关于FM包络检测器描述正确的是

答案: 包络检测器是非相干解调;
包络检测器抑制弱信号，支持强信号;
宽带FM接收机有门限效应

4、 对基带信号m(t)进行连续波调制，对各线性调制方案的已调信号带宽进行比较，正确的是：

答案: 宽带FM>DSB=AM>VSB>SSB;
宽带FM>窄带FM=AM>VSB>SSB

5、 线性调制系统的解调增益

答案: 与调制方式有关，与解调方式有关;
AM的相干解调与非相干解调增益是一致的。

6、 相干解调会出现门限效应。

答案: 错误

7、 DSB可以通过提高信噪比的方式改善解调增益

答案: 错误

8、 解调增益越高，说明接收机的抗噪声性能越好。

答案: 正确

Chapter 2 Continuous-wave modulation Chapter 2　章节测试

1、

答案: 正确

2、

答案: 错误

3、

答案: 错误

4、

答案: 正确

5、

答案: 正确

6、

答案: 正确

7、

答案: 错误

8、

答案: 正确

9、

答案: 正确

10、
答案: 90000

11、
答案: (以下答案任选其一都对)200k;
200000

12、
答案: 20

13、
答案: 7.2

14、
答案: 204

Chapter 3 Pulse Modulation 第三章单元测试-1

1、 The resolution of full high definition television (Full HD) is 1920×1080 and the frame rate is 60 Hz. For each pixel, the quantization levels of different color samplings (R, G, B) are all set to 4096. Then, the number of bits per second and the minimum bandwidth required to transmit this video signals are about T and W (Hz), respectively, where the channel SNR=30(dB). Hint: C=W*log2(1+SNR).

答案: T=3000M, W=300M

2、 In a digital communication system, the Nyquist interval of signal  should be millisecond (ms).

答案: 1.25

3、 An analog signal m(t) with zero mean is a stationary process. Its frequency range is 0(Hz) ~ 8000(Hz) and its amplitude is uniformly distributed between -5(V) ~ +5(V). Then, the minimum Nyquist sampling frequency for this signal m(t) is f (Hz) , the average power is P (W). Assuming a uniform quantizer for this signal and the step size is D=0.04(V), the minimum number of quantization bits is R .

答案: f=16000, P=25/3, R=8

4、 A signal s(t)=3cos(20pt) (V) is quantized by a uniform quantizer, the step size of the quantizer is 0.1(V), then the minimal number of quantization level is L , the number of bits per sample is at least R , and the variance of the quantization error is T ().

答案: L=60, R=6, T=0.000833

5、 A compact disk (CD) is used to store music. Suppose that both the two independent channels of the true stereo music with the highest frequency 22.05 kHz are sampled at the Nyquist sampling rate, then the sampling rate for each channel is f (kHz). The encoded PCM is to have an average SNR of at least 96 dB. Then, the minimum number of the uniform quantization of the sampled data should be R bits. If the Beethoven’s Symphony No. 9 with 74 minutes in PCM data can be stored in CD, the minimum storage capacity of the CD should be T (MB, Megabyte). (Hint: ).

答案: f=44.1, R=16, T=783

6、 A 1G bytes flash memory is used to store PCM data. Suppose that a VF (voice-frequency) signal is sampled at 8kHz and the PCM data with R bits quantization is to have an average SNR of at least 30dB. Then, about T hours of VF signal in PCM data can be stored in this flash memory. (Hint: ).

答案: T=55.56

Chapter 3 Pulse Modulation 第三章章节测试

1、 In a voice transmission system, the sampling frequency is 8000Hz, and it is used to multiplex 12 independent voice inputs based on an 8-bit PCM word. The bit duration is about (ms, microsecond)

答案: 1.3

2、 32 voice signals are sampled uniformly and then time-division multiplexed. The sampling operation uses flat-top with 1(μs) duration. The highest frequency component of each voice signal is 3.4 kHz. Assuming a sampling rate of 8 kHz, the spacing between successive pulses of the multiplexed signal is (μs). If the Nyquist rate sampling rate is used, repeat the calculation is (μs)

答案: 2.9, 3.6

3、 A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 12 fs, where fs is the Nyquist rate of speech signal. The step size Δ=100(mV). The modulator is tested with a 1 (kHz) sinusoidal signal. The maximum amplitude of this test signal required to avoid slope overload is ______(V).

答案: 1.3

4、 In a Delta Modulation (DM) system, A sinusoidal test signal of amplitude Am=1(V) and frequency fm = 1 (kHz) is applied to the system. The sampling rate is fs = 8 (kHz) ，to avoid slope overload, the step size Δ is at least (V).

答案: 0.785

5、 There are eight analog signals, each of bandwidth W=2(kHz). Samples of these signals are time-division-multiplexed, quantized and binary-coded. The step size Δ of the quantizer cannot be greater than 0.5% of the peak amplitude mp. The sampling rates (kHz) is 50% above the Nyquist rate and the minimum number of quantization levels should be .

答案: 48, 400

6、 In a voice transmission system (the highest frequency fh = 4kHz), the Nyquist sampling frequency is used, and it is used to multiplex _ independent voice inputs. Each input voice signal is quantized by an R-bits PCM word to achieve average SNR of at least 48(dB), where the maximum bit duration is set to Tb = 3.125(ms)

答案: 5

7、 DM requires a sampling rate much higher than the Nyquist Rate.

答案: 正确

8、 Aliasing refers to the phenomenon of a high-frequency component in the spectrum of the signal seemingly taking on the identity of a lower frequency in the spectrum of its sampled version.

答案: 正确

9、 For both m-law and A-law, the signal-to-noise (SNR) of low-level and high-level signals improve with the increasing m and A.

答案: 错误

10、 Performance of a PCM system is only affected by quantization noise.

答案: 错误

11、 In the sampling process, if the sampling period is larger than the Nyquist interval, then the sampled signal can be completely described by the sample values.

答案: 错误

12、 One of the important advantages of pulse-code modulation (PCM) is the robustness to channel noise and interference.

答案: 正确

13、 Noise in a pulse-code modulation (PCM) system includes noise and noise. While, in a linear delta modulator, there is a special quantization error caused by distortion.
答案: Quantization, Channel, Slope-overload

14、 Basic operations performed in the transmitter of a pulse code modulation system include , , and .
答案: Sampling, Quantizing, Encoding

15、 In a differential encoding system, a transition denotes symbol 0 and no transition denotes symbol 1. Symbol 1 is used as reference bit. If a binary data sequence with length eight bits { 1011 } is the input to this system, the output sequence is { 1 , 10000110}.
答案: 1010

Chapter 4 Baseband Pulse Transmission 第四章单元测试-1

1、 The signal s(t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a matched filter with impulse response h(t) is used at the receiver, the matched filter h(t) should be: 

答案:  

2、 The
signal s(t) shown in figure is transmitted through an AWGN channel with noise
PSD N0/2. Suppose a matched filter with impulse response h(t) is
used at the receiver, the matched filter output should be:

答案: ‍

3、 The signal s(t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a filter with impulse response h(t) is used at the receiver, the filter output should be:

答案:

4、 The
signal s(t) shown in figure is transmitted through an AWGN channel with noise
PSD N0/2. Suppose a matched filter with impulse response h(t) is
used at the receiver (coefficient K=1), the peak pulse signal-to-noise ratio
is: 

答案: 18/N0

5、 A binary PCM wave using unipolar NRZ signaling to transmit symbols “1” and “0”, symbol “1” is represented by a rectangular pulse of amplitude A and duration Tb and symbol “0” is represented by transmitting no pulse. The channel noise is modeled as additive, white and Gaussian, with zero mean and power spectral density N0/2. Assuming that symbols “1” and “0” occur with equal probability and the average transmitted signal energy per bit is defined as Eb. Using a matcher filter, the average probability of error at the receiver output is:

答案:

6、 A binary PCM wave using unipolar NRZ signaling
to transmit symbols “1” and “0”, symbol “1” is represented by a rectangular
pulse of amplitude A and duration Tb/2 and symbol “0” is represented
by transmitting no pulse. The channel noise is modeled as additive, white and
Gaussian, with zero mean and power spectral density N0/2. Assuming
that symbols “1” and “0” occur with equal probability and the average
transmitted signal energy per bit is defined as Eb. Using a matcher filter,
the average probability of error at the receiver output is:

答案:

Chapter 6 Passband Modulation Ch6单元测验

1、 For a digital communication system which can transmit 4800 bits per second with 16-PSK modulation, the symbol rate is __ symbols per second.

答案: 1200

2、

答案:

3、 The approximate average probability of symbol error for coherent 32-PSK is _____.

答案:

4、 For an M-ary PSK, as the number of states, M, is increased, ____.

答案: the bandwidth efficiency is improved.;
the error performance becomes worse.;
the minimum distance of the signal constellation is decreased.

5、 Comparing QPSK and BPSK transmitted over AWGN channel, ___.

答案: QPSK can achieve the same bit error rate as BPSK.;
the bandwidth efficiency of QPSK is twice that of BPSK.

6、 In QAM, _____ of the carrier are varied.

答案: amplitude;
phase

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