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

1、 问题:
选项:
A:a continuous rectangular pulses
B:a periodic sequence of rectangular pulses
C:a continuous sinusoidal signal
D:a periodic sinusoidal signal
答案: 【a periodic sequence of rectangular pulses

2、 问题:The reverse of modulation is .
选项:
A:encoding
B:demodulation
C:decoding
D:modulator
答案: 【demodulation

3、 问题:A common problem with electromagnetic wave propagation via sky wave in thehigh fequency (HF) range is .
选项:
A:Wide transmission bandwidth
B:signal multipath
C:Linear time-varying channel
D:Broad-area coverage
答案: 【signal multipath

4、 问题:Two primary resources employed in communication systems are and .
选项:
A: power
B:bandwidth
C:frequency
D:source of Information
答案: 【 power;
bandwidth

5、 问题:The three commonly used method of multiplexing are , and .
选项:
A:FDM
B:TDM
C:CDM
D:FDMA
答案: 【FDM;
TDM;
CDM

6、 问题:
选项:
A:transmitter
B:channel
C:receiver
D:source of imformation
答案: 【transmitter;
channel;
receiver

7、 问题:
选项:
A:coaxial cable
B:telephone channel
C:universal Serial Bus
D:optical fiber
答案: 【coaxial cable;
telephone channel;
optical fiber

8、 问题:
选项:
A:wireless broadcast channel
B:underwater acoustic channel
C: satellite channel
D: mobile radio channel
答案: 【wireless broadcast channel;
satellite channel;
mobile radio channel

9、 问题:
选项:
A:正确
B:错误
答案: 【错误

10、 问题:
选项:
A:正确
B:错误
答案: 【正确

11、 问题:
选项:
A:正确
B:错误
答案: 【正确

12、 问题:
选项:
A:正确
B:错误
答案: 【错误

13、 问题:
选项:
A:正确
B:错误
答案: 【错误

14、 问题:
选项:
A:正确
B:错误
答案: 【错误

15、 问题:
选项:
A:正确
B:错误
答案: 【错误

16、 问题:
选项:
A:正确
B:错误
答案: 【正确

17、 问题:
选项:
A:正确
B:错误
答案: 【错误

18、 问题:
答案: 【transmitter, channel, receiver

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

20、 问题:
答案: 【transmitted power, channel bandwidth

21、 问题:
答案: 【a sinusoidal wave

22、 问题:
答案: 【a periodic sequence of rectangular pulses

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

24、 问题:
答案: 【band-limited

Chapter 1 Random processes 第一章单元测验-1

1、 问题:This random process X (t) =cos(2πfct). the frequencies can be 100, 200, . . . , 600 Hz,then:
选项:
A:The possible values of X (0.001) could be cos(0.2π), cos(0.4π), . . . , cos(l .2π),and each has a probability 1/6.
B:The possible values of X (0.001) could be cos(0.1π), cos(0.2π), . . . , cos(0.6π), and each has a probability 1/6.
C:The possible values of X (0.001) could be cos(0.2π), cos(0.4π), . . . , cos(l .2π)
D:The possible values of X (0.001) could be cos(0.1π), cos(0.2π), . . . , cos(0.6π)
答案: 【The possible values of X (0.001) could be cos(0.2π), cos(0.4π), . . . , cos(l .2π)

2、 问题:
选项:
A:
B:
C:
D:
答案: 【

3、 问题:Which one of the following functions can be the autocorrelation function of a random process?
选项:
A:
B:
C:
D:
答案: 【;
;

4、 问题:
选项:
A:正确
B:错误
答案: 【错误

5、 问题:
选项:
A:正确
B:错误
答案: 【正确

6、 问题:
选项:
A:正确
B:错误
答案: 【错误

Chapter 1 Random processes 第一章单元测验-2

1、 问题: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 :
选项:
A:Rx(τ) = 1/3,Sx(f) = 1/3
B:Rx(τ) = 1/3,Sx(f) = δ(f)/3
C:Rx(τ) =δ(τ)/3, Sx(f) = 1/3
D:Rx(τ) =δ(τ)/3, Sx(f) = δ(f)/3
答案: 【Rx(τ) = 1/3,Sx(f) = δ(f)/3

2、 问题:Two random processes X(t) and Y(t) arejointly wide-sense stationar, then
选项:
A:X(t) is stationary.
B:Y(t) is stationary.
C:Cross-corelation RXY(t1, t2) depends ony on r = t1 – t2
D:as long as both X(t) and Y(t) are individually stationary
答案: 【X(t) is stationary. ;
Y(t) is stationary.;
Cross-corelation RXY(t1, t2) depends ony on r = t1 – t2

3、 问题:Assume that Z(t) = X(t) + Y(t), where X(t) and Y(t) are jointly stationaryrandom processes.
选项:
A:Z(t) is a stationary process
B:Sz(f) = Sx(f) + Sy(f)
C:Sz(f) = Sx(f) + Sy(f) + Sxy(f) + Syx(f)
D: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)
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、 问题:
选项:
A:正确
B:错误
答案: 【错误

5、 问题:
选项:
A:正确
B:错误
答案: 【正确

6、 问题: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?
选项:
A:正确
B:错误
答案: 【错误

Chapter 1 Random processes 第一章测试

1、 问题:Considering the properties of the autocorrelation function Rx(τ) of a random process X(t).
选项:
A:
B:
C:
D:
答案: 【;

2、 问题:Considering the properties of the sine wave plus narrowband noise.
选项:
A:If the narrowband noise is Gaussian, the in-phase and quadrature components are jointly Gaussian.
B:Both the in-phase and quadrature component have the same variance.
C:Both the in-phase and quadrature component have the same mean.
D:The envelope and phase components are statistically independent.
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、 问题:
选项:
A:The mean of Y is 0.
B:
C:
D:Y is also Gaussian
答案: 【The mean of Y is 0.;
;
;
Y is also Gaussian

4、 问题:
选项:
A:
B:
C:The power in X(t) is 4W.
D:The power in X(t) is 8W.
答案: 【;
The power in X(t) is 4W.

5、 问题:
选项:
A:正确
B:错误
答案: 【正确

6、 问题:
选项:
A:正确
B:错误
答案: 【错误

7、 问题:
选项:
A:正确
B:错误
答案: 【错误

8、 问题:Thermal noise is a stationary, zero-mean Gaussian process.
选项:
A:正确
B:错误
答案: 【正确

9、 问题:If two processes are statistically independent, they are orthogonal.
选项:
A:正确
B:错误
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、 问题:For stationary processes, means, variances and covariance are independent of time.
选项:
A:正确
B:错误
答案: 【错误

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

1、 问题:In Amplitude Modulation (AM), if the carrier wave is over-modulated, envelop detection cannot be used to recover the message signal.
选项:
A:正确
B:错误
答案: 【正确

2、 问题: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).
选项:
A:正确
B:错误
答案: 【正确

3、 问题:The effect of a small, constant phase error between the incoming carrier and the local oscillator in the coherent demodulation of SSB is overmodulation.
选项:
A:正确
B:错误
答案: 【错误

4、 问题:AM, DSB-SC, SSB, VSB can use coherent demodulation.
选项:
A:正确
B:错误
答案: 【正确

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

7、 问题:
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、 问题:Both AM and NBFM have same transmission bandwidth, they are linear modulation schemes.
选项:
A:正确
B:错误
答案: 【错误

2、 问题: Angle modulation can provide better discrimination against noise and interference than AM modulation.
选项:
A:正确
B:错误
答案: 【正确

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

4、 问题: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、 问题: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、 问题: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、 问题: FM can improve noise performance via increasing transmission bandwidth.
选项:
A:正确
B:错误
答案: 【正确

2、 问题:Coherent detection of Double Sideband-Suppressed Carrier (DSB-SC) signal and envelope detection of Frequency Modulation (FM) signal both have the threshold effect.
选项:
A:正确
B:错误
答案: 【错误

3、 问题:AM can improve noise performance via increasing transmission bandwidth.
选项:
A:正确
B:错误
答案: 【错误

4、 问题: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.
选项:
A:正确
B:错误
答案: 【错误

5、 问题: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.
选项:
A:正确
B:错误
答案: 【错误

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

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

8、 问题:Then the deviation ratio is
答案: 【15

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

10、 问题: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

Chapter 3 Pulse Modulation 第三章单元测试-1

1、 问题: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 256. 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).
选项:
A:T=300M, W=30M
B:T=3000M, W=1000M
C:T=3000M, W=300M
D:T=1000M, W=300M
答案: 【T=3000M, W=300M

2、 问题:In a digital communication system, the Nyquist interval of signal should be millisecond (ms).
选项:
A:1.25
B:2.5
C:5
D:10
答案: 【1.25

3、 问题: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 .
选项:
A:f=8000, P=25, R=8
B:f=16000, P=25/2, R=7
C:f=8000, P=25/3, R=9
D:f=16000, P=25/3, R=8
答案: 【f=16000, P=25/3, R=8

4、 问题: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 ().
选项:
A:L=60, R=5, T=0.000833
B:L=30, R=5, T=0.000833
C:L=60, R=5, T=0.001667
D:L=60, R=6, T=0.000833
答案: 【L=60, R=6, T=0.000833

5、 问题: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: ).
选项:
A:f=44.1, R=16, T=783
B:f=22.05, R=16, T=392
C:f=44.1, R=15, T=700
D:f=44.1, R=16, T=392
答案: 【f=44.1, R=16, T=783

6、 问题: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: ).
选项:
A:T=111.11
B:T=55.56
C:T=27.78
D:T=13.89
答案: 【T=55.56

Chapter 3 Pulse Modulation 第三章章节测试

1、 问题: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)
选项:
A:125
B:15.6
C:1.3
D:10.4
答案: 【1.3

2、 问题: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)
选项:
A:2.9, 3.6
B:3.9, 4.6
C:125, 147
D:4.9, 5.6
答案: 【2.9, 3.6

3、 问题: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).
选项:
A:8.16
B:0.1
C:1.3
D:1.5
答案: 【1.3

4、 问题: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).
选项:
A:0.125
B:0.785
C:1.57
D:0.3925
答案: 【0.785

5、 问题: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 .
选项:
A:6, 100
B:12, 200
C:24, 200
D:48, 400
答案: 【48, 400

6、 问题: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)
选项:
A:3
B:4
C:5
D:6
答案: 【5

7、 问题:DM requires a sampling rate much higher than the Nyquist Rate.
选项:
A:正确
B:错误
答案: 【正确

8、 问题:Aliasing refers to the phenomenon of a high-frequency component in the spectrum of the signal seemingly taking on the identity

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