通信原理(东南大学) 中国大学mooc慕课答案2024版 m44568

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

1、

答案: a periodic sequence of rectangular pulses

2、 The reverse of modulation is          .

答案: demodulation

3、 A common problem with electromagnetic wave propagation via sky wave in thehigh fequency (HF) range is          .

答案: signal multipath

4、 Two primary resources employed in communication systems are              and          .       

答案:  power;
bandwidth  

5、 The three commonly used method of  multiplexing are         ，           and          .

答案: FDM;
TDM;
CDM

6、

答案: transmitter;
channel;
receiver

7、

答案: coaxial cable;
telephone channel;
optical fiber

8、

答案: wireless broadcast channel;
 satellite channel;
 mobile radio channel

9、

答案: 错误

10、

答案: 正确

11、

答案: 正确

12、

答案: 错误

13、

答案: 错误

14、

答案: 错误

15、

答案: 错误

16、

答案: 正确

17、

答案: 错误

18、
答案: transmitter, channel, receiver

19、
答案: broadcasting, point-to-point communication

20、
答案: transmitted power, channel bandwidth

21、
答案: a sinusoidal wave

22、
答案: a periodic sequence of rectangular pulses

23、
答案: the average signal power, the average noise power

24、
答案: band-limited

Chapter 1 Random processes 第一章单元测验-1

1、 This random process X (t) =cos(2πfct). the frequencies can be 100, 200, . . . , 600 Hz，then: 

答案: The possible values of X (0.001) could be cos(0.2π), cos(0.4π), . . . , cos(l .2π) 

2、

答案:

3、 Which one of the following functions can be the autocorrelation function of a random process?

答案: ;
;

4、

答案: 错误

5、

答案: 正确

6、

答案: 错误

Chapter 1 Random processes 第一章单元测验-2

1、 The process X (t) is defned by X(t) = X, where X is a randomvariable unifrmly distrbutedon [- 1, 1] , the autocorelation function and the power spectral density are  :

答案: Rx(τ) = 1/3，Sx(f) = δ(f)/3

2、 Two random processes X(t) and Y(t) arejointly wide-sense stationar, then 

答案: X(t) is  stationary. ;
Y(t) is stationary.;
Cross-corelation RXY(t1, t2) depends ony on r = t1 – t2 

3、 Assume that Z(t) = X(t) + Y(t), where X(t) and Y(t) are jointly stationaryrandom processes.

答案: Z(t) is a stationary process ;
Sz(f) = Sx(f) + Sy(f) + Sxy(f) + Syx(f) ;
If the two processes X(t) and Y(t) are uncorelated and at least one of the processes is zero mean，then  Sz(f) = Sx(f) + Sy(f) 

4、

答案: 错误

5、

答案: 正确

6、 We have proved that when the input to an LTI system is stationary, the output is also stationary. If we know that the output process is stationary, can we conclude that the input process is necessarily stationary? 

答案: 错误

作业Chapter 1 Random processes 第一章作业

1、
评分规则:  每空5分

2、
评分规则:  每空5分

3、
评分规则:  每空2.5分

4、
评分规则:  每空5分

5、
评分规则:  

6、
评分规则:  

7、
评分规则:  

8、
评分规则:  

Chapter 1 Random processes 第一章测试

1、 Considering the properties of the autocorrelation function Rx(τ) of a random process X(t). 

答案: ;

2、 Considering the properties of the sine wave plus narrowband noise.

答案: If the narrowband noise is Gaussian, the in-phase and quadrature components are jointly Gaussian.;
Both the in-phase and quadrature component have the same variance.  

3、

答案: The mean of Y is 0.;
;
;
Y is also Gaussian

4、

答案: ;
The power in X(t) is 4W.

5、

答案: 正确

6、

答案: 错误

7、

答案: 错误

8、 Thermal noise is a stationary, zero-mean Gaussian process.

答案: 正确

9、 If two processes are statistically independent, they are orthogonal.

答案: 错误
分析：If two processes are statistically independent and at least one of them has zero mean, they are orthogonal.

10、 For stationary processes, means, variances and covariance are independent of time.

答案: 错误

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

1、 In Amplitude Modulation (AM), if the carrier wave is over-modulated, envelop detection cannot be used to recover the message signal.

答案: 正确

2、 For the same modulating wave, the transmission bandwidth of double sideband-suppressed carrier (DSB-SC) modulation is the same as that of amplitude modulation (AM).  

答案: 正确

3、 The effect of a small, constant phase error between the incoming carrier and the local oscillator in the coherent demodulation of SSB is overmodulation.  

答案: 错误

4、 AM, DSB-SC, SSB, VSB can use coherent demodulation.    

答案: 正确

5、 If the highest frequency component of the message signal is at 10 kHz, then the lowest value of carrier frequency that keeps the Double Sideband-Suppressed Carrier (DSB-SC) modulation from sideband overlap is            kHz
答案: 10

6、
答案: (以下答案任选其一都对)12kHz 14kHz;
14kHz 12kHz

7、
答案: (以下答案任选其一都对)6kHz 8kHz;
8kHz 6kHz

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

1、 Both AM and NBFM have same transmission bandwidth, they are linear modulation schemes. 

答案: 错误

2、  Angle modulation can provide better discrimination against noise and interference than AM modulation.

答案: 正确

3、 Among these modulation schemes including AM，DSB，SSB and VSB ,         is easiest to demodulation.
答案: AM

4、 12 different message signals, each with a bandwidth of 20 kHz, are to be multiplexed and transmitted.  If the multiplexing and modulation methods are frequency-division multiplexing (FDM) and sideband-suppressed carrier (SSB-SC) modulation, respectively, then the minimum bandwidth required is          kHz.
答案: 240

5、 In a single tone Frequency Modulation (FM), the modulating wave is m(t) = Amcos(2pfmt) = 4cos(200pt) V,  the frequency sensitivity of the modulator kf = 0.5, then  the frequency deviation of FM is            Hz.
答案: 2

6、 A single tone frequency modulation (FM) signal with carrier frequency fc = 1 MHz is described by the equation s(t)=50cos(2πfct+5sin(2000πt)). Then the modulation index β is ____.
答案: 5

Chapter 2 Continuous-wave modulation 第二章章节测试

1、  FM can improve noise performance via increasing transmission bandwidth. 

答案: 正确

2、 Coherent detection of Double Sideband-Suppressed Carrier (DSB-SC) signal and envelope detection of Frequency Modulation (FM) signal both have the threshold effect.    

答案: 错误

3、 AM can improve noise performance via increasing transmission bandwidth.  

答案: 错误

4、 10 different message signals, each with a bandwidth of 20 kHz, are to be multiplexed and transmitted.  If the multiplexing and modulation methods are frequency-division multiplexing (FDM) and double sideband-suppressed carrier (DSB-SC) modulation, respectively, then the minimum bandwidth required is 200 kHz.               

答案: 错误

5、 The envelop of a PM or wideband FM signal is constant,whereas the envelop of an AM or narrowband FM signal is depent on the message signal.

答案: 错误

6、   Then the power of the modulated signal         W;
答案: 50

7、 then the frequency deviation  of the signal is         Hz           
答案: 12500

8、 Then the deviation ratio is        
答案: 15

9、   by Carson’rule, the transmission bandwidth of s(t) is       kHz 
答案: 27

10、 An FM signal is modulated by m(t)=sin(2000πt), kf=100.  The bandwidth B of the baseband signal m(t) is __Hz.
答案: 1000

作业Chapter 2 Continuous-wave modulation 第二章作业

1、 Arrange VSB,SSB,DSB, AM and wideband FM in the decreasing order of bandwidth required for transmission                             . 
评分规则:   FM > AM=DSM>VSB>SSB        

2、
评分规则:  the power P of the modulated signal is 1250 w
the frequency deviation Δf is  __5000 HZ
the modulation index β is 5
transmission bandwidth BT is 12000 Hz

3、 If A =2， What type of modulation does this correspond to ?       If A =0， What type of modulation does this correspond to ?       
评分规则:  If A =2，correspond to  AM
If A =0，correspond to  DSB

4、
评分规则:  

5、
评分规则:  

6、
评分规则:  

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