中国大学Principles of Communication_1答案(慕课2023完整答案)

分类：高中发布于：2024-06-02 13:23:51ė39844次浏览650条评论

中国大学Principles of Communication_1答案(慕课2023完整答案)

Chapter 0 Back ground and preview

Chapter 0 Unit Quiz

1、中国整答

2、大学n答

3、案慕案

4、课完

5、中国整答

6、大学n答

Chapter 1 Random processes

Chapter 1 Unit Quiz 1

1、案慕案This 课完random process X (t) =cos(2πfct). the frequencies can be 100, 200, . . . , 600 Hz，then:
A、中国整答The大学n答 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、案慕案The课完 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中国整答 possible values of X (0.001) could be cos(0.2π), cos(0.4π), . . . , cos(l .2π)
D、The大学n答 possible values of X (0.001) could be cos(0.1π), cos(0.2π), . . . , cos(0.6π)

2、
A、案慕案
B、
C、
D、

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

4、

5、

6、

Chapter 1 Random processes

Chapter 1 Unit Quiz 2

1、

2、

3、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?

4、The process X (t) is defned by X(t) = X, where X is a randomvariable unifrmly distrbuted on [- 1, 1] , the autocorelation function and the power spectral density are: Rx(τ) = 1/3，Sx(f) = 1/3

5、Two random processes X(t) and Y(t) arejointly wide-sense stationar, then the cross-corelation RXY(t1, t2) depends ony on r = t1 - t2

6、Assume that Z(t) = X(t) + Y(t), where X(t) and Y(t) are jointly stationary random processes，then Sz(f) = Sx(f) + Sy(f)

Chapter 1 Random processes & Chapter 2 Continuous-wave modulation

Chapter 1 Test

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

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

3、
A、The mean of Y is 0.
B、
C、
D、Y is also Gaussian

4、
A、
B、
C、The power in X(t) is 4W.
D、The power in X(t) is 8W.

5、

6、

7、

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

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

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

chap1 homework

1、

2、

3、

4、

5、

6、

7、

8、

Chapter 2 Continuous-wave modulation

Chapter 2 Unit Quiz 2

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

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

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

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.

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.

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 ____.

Chapter 2 Continuous-wave modulation

Chapter 2 Unit Quiz 3

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

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

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

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.

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.

6、 Then the power of the modulated signal W;

7、 then the frequency deviation of the signal is Hz

8、 Then the deviation ratio is

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

10、An FM signal is modulated by m(t)=sin(2000πt), kf=10000. The bandwidth B of the baseband signal m(t) is __Hz.

Chapter 2 homework

1、Arrange VSB,SSB,DSB, AM and wideband FM in the decreasing order of bandwidth required for transmission .

2、

3、 If A =2， What type of modulation does this correspond to ? If A =0， What type of modulation does this correspond to ?

4、

5、

6、

Chapter 3 Pulse Modulation

Chapter 3 Unit Quiz 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).
A、T=300M, W=30M
B、T=3000M, W=1000M
C、T=3000M, W=300M
D、T=1000M, W=300M

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

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 .
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

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 ().
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

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: ).
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

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: ).
A、T=111.11
B、T=55.56
C、T=27.78
D、T=13.89

Chapter 3 Pulse Modulation

Chapter 3 Unit Quiz 2

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)
A、125
B、15.6
C、1.3
D、10.4

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)
A、2.9, 3.6
B、3.9, 4.6
C、125, 147
D、4.9, 5.6

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).
A、8.16
B、0.1
C、1.3
D、1.5

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).
A、0.125
B、0.785
C、1.57
D、0.3925

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 .
A、6, 100
B、12, 200
C、24, 200
D、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)
A、3
B、4
C、5
D、6

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

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.

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.

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

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.

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

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.

14、Basic operations performed in the transmitter of a pulse code modulation system include , , and .

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}.

Chapter 3 homework

1、Specify the Nyquist rate and the Nyquist interval for each of the following signals: (1) g(t) = sinc(200t) (2) g(t) = (200t) (3) g(t) = sinc(200t)+(200t)

2、Figure shows the idealized spectrum of a message signal m(t).The signal is sampled at a rate equal to 1 kHz using flat-top pulses, with each pulse being of unit amplitude and duration 0.1 ms. Determine and sketch the spectrum of the resulting PAM signal.

3、Twelve different message signals, each with a bandwidth of 10 kHz, are to be multiplexed and transmitted. Determine the minimum bandwidth required for each method if the multiplexing / modulationmethod used is (1) FDM, SSB (2) TDM, PAM

4、Given the data stream 1110010100, sketch the transmitted sequence of pulses for each of the following line codes: (a) Unipolarnonreturn-to-zero (b) Polar nonreturn-to-zero (c) Unipolarreturn-to-zero (d) Bipolar return-to-zero (e) Manchester code

5、A speech signal has a total duration of 10 s. It is sampled at the rate of 8 kHz and then encoded. The signal-to-(quantization) noise ratio is required to be 40 dB. Calculate be minimum storage capacity needed to accommodate this digitized speech signal.

6、A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The specifications of the modulator are as follows: Sampling rate = 10 where is the Nyquistrate of the speech signal. Step size =100 mv The modulator is tested with a 1-kHz sinusoidal signal. Determine the maximum amplitude of this test signal required to avoid slope overload

Chapter 4 Baseband Pulse Transmission

Chapter 4 Unit Quiz 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:
A、
B、
C、
D、

2、he 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:
A、
B、
C、
D、

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:
A、
B、
C、
D、

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:
A、18/N0
B、12/N0
C、8/N0
D、4/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:
A、
B、
C、
D、

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:
A、
B、
C、
D、

Chapter 4 Baseband Pulse Transmission

Chapter 4 Unit Quiz 2

1、An analog signal is sampled, quantized and encoded into a binary PCM wave. The number of representation levels used is 2048. A synchronizing bit is added at the end of each codeword representing a sample of the analog signal. The resulting PCM wave is transmitted over a channel of bandwidth 28kHz, using a 8-ary PAM system with raised-cosine spectrum, where the rolloff factor is 0.75. Then, the information is transmitted at the rate Rb=____ (bit/s) through the channel. The analog signal is sampled at the rate fs=___(Hz). The maximum possible value for the highest frequency component fH of the analog signal is fh=__ (Hz).
A、Rb=96k, fs=8k, fh=4k
B、Rb=48k, fs=8k, fh=4k
C、Rb=96k, fs=16k, fh=8k
D、Rb=48k, fs=16k, fh=8k

2、An analog TV signal with a 6 MHz bandwidth is sampled, quantized, and encoded into a binary PCM wave. The sampling rate is 14 M Sample/Sec., the number of the representation levels is 16, then the minimum transmitting bandwidth is BT= MHz.
A、BT=56
B、BT=28
C、BT=48
D、BT=24

3、There are eight analog signals, each of bandwidth B=4kHz. Samples of these signals are time-division-multiplexed, quantized and binary-coded. The step size Δ of the quantizer cannot be greater than 0.1% of the peak amplitude mmax. Then the minimum number of quantization levels should be L= 2000 . The transmission bandwidth is BT= kHz, if 64-ary PAM system with raised-cosing spectrum (roll-off factor r=1/4) are used, where the sampling rates must be at least 50% above the Nyqusit rate.
A、L=2000, BT=110;
B、L=2000, BT=220;
C、L=1000, BT=110;
D、L=1000, BT=55;

4、The resolution of ultra high definition television (UHDTV) is 3840×2160 and the fresh 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 required to transmit for this video signals is ________, and the minimum bandwidth required to transmit this signal with raised-cosine spectrum (the roll-off factor α=0.5) is ________Hz.
A、16.7G, 25G
B、16.7G, 12.5G
C、278M, 417M
D、278M, 208.5M

5、Inter-symbol interference can be eliminated by a receiver filter which is matched to the catenation of the transmitter filter and the channel.

6、Correlative-level coding uses a partial-response signaling method to avoid inter-symbol interference (ISI).

7、Error propagation is a drawback with correlative-level coding, and is unavoidable.

8、Correlative-level coding (also known as partial-response signaling) can achieve the Nyquist rate of 2W symbols per second over a channel of bandwidth of 2W Hz using realizable and perturbation-tolerant filters.

Chapter 4 homework

1、Consider the signal s(t) shown in figure. a) Determine the impulse response of a filter matched to this signal and sketch it as a function of time. b) Plot the matched filter output as a function of time. c) What is the peak value of the output?

2、Figure shows a pair of pulses that are orthogonal to each other over the interval [0, T]. In this problem we investigate the use of this pulse-pair to study a two-dimensional matched filter. a) Determine the matched filters for the pulse s1(t) and s2(t) considered individually; Form a two-dimensional matched filter by connecting the two matched filters of Part a) in parallel, as shown in figure below. Hence, demonstrate the following: b) When the pulse s1(t) is applied to this two-dimensional filter, the response of the lower matched filter is zero. c) When the pulse s2(t) is applied to this two-dimensional filter, the response of the upper matched filter is zero.

3、The formula for the optimum threshold in the receiver of figure is, in general, given by equation . Discuss, in graphical terms, how this optimum choice affects contributions of the two terms in for the average probability of symbol error P by considering the following two cases: a) b)

4、The binary data stream 001101001 is applied to the input of a duobinary system. a) Construct the duobinary coder output and corresponding receiver output, without a precoder. b) Suppose that owing to error during transmission, the level at the receiver input produced by the second digit is reduced to zero. Construct the new receiver output. c) Repeat problem a) and b), assuming the use of a precoder in the transmitter.

Chapter 5 Signal-space Analysis

Chapter 5 Unit Quiz 1

1、
A、
B、
C、
D、

2、
A、
B、
C、
D、

3、
A、
B、
C、
D、

4、
A、
B、
C、
D、

5、Regarding the ML decision rule for signal transmissions over AWGN channel, the statements _______________ are correct.
A、The ML decision rule is to choose the message point closest to the received signal point.
B、The ML decision rule always equals to the MAP rule.
C、The MAP and ML decision rules can be conditionally equivalent.
D、The ML decision rule can minimize the probability of error.

6、For a given signal constellation, the average probability of symbol error incurred in maximum likelihood signal detection over an AWGN channel is invariant to the _________________ of the signal constellation.
A、compression
B、translation
C、expansion
D、rotation

7、

8、

9、

10、

Chapter 5 homework

1、

2、

3、

4、

5、

6、

7、

8、

Chapter 6 Passband Modulation

Chapter 6 Unit Quiz 1

1、For a digital communication system which can transmit 4800 bits per second with 16-PSK modulation, the symbol rate is ________ symbols per second.
A、1000
B、1200
C、1400
D、1600

2、
A、
B、
C、
D、

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

4、For an M-ary PSK, as the number of states, M, is increased, ________________.
A、the bandwidth efficiency is improved.
B、the error performance becomes worse.
C、the minimum distance of the signal constellation is decreased.
D、the error performance becomes better.

5、Comparing QPSK and BPSK transmitted over AWGN channel, _______________.
A、QPSK can achieve the same bit error rate as BPSK.
B、QPSK can achieve the same symbol error rate as BPSK.
C、the bandwidth efficiency of QPSK is the same as that of BPSK.
D、the bandwidth efficiency of QPSK is twice that of BPSK.

6、In QAM, _______________________ of the carrier are varied.
A、amplitude
B、frequency
C、frequency derivation
D、phase

Chapter 6 Passband Modulation

Chapter 6 Unit Quiz 2

1、Bandwidth efficiency of 8-PSK is __________ bit/s/Hz.
A、1
B、1.5
C、2
D、2.5

2、
A、
B、
C、
D、

3、For M-ary PSK signal, as M increased, the bandwidth efficiency will be ____________.
A、firstly decreased, then increased
B、unchanged
C、increased
D、decreased

4、For 16-QAM, the statement that _____________________ is correct.
A、the noise performance of 16-QAM is better than 16-PSK
B、the noise performance of 16-QAM is worse than 16-PSK
C、the noise performance of 16-QAM is better than QPSK
D、the noise performance of 16-QAM is worse than QPSK

5、_______________ is a type of PSK.
A、4-QAM
B、MSK
C、DPSK
D、BASK

6、The two basic operations at the DPSK transmitter are ____________.
A、differential encoding
B、unipolar nonreturn-to-zero (NRZ) encoding
C、phase-shift keying
D、frequency-shift keying

7、For AWGN channel, QPSK with Gray mapping can achieve the same bit error rate as a coherent BPSK in white Gaussian noise, but the bandwidth efficiency of BPSK is twice that of QPSK.

8、For AWGN channel, 16-PSK requires less signal energy than the 16-FSK for the fixed noise power spectrum density and symbol error rate.

9、A digital communications system with 8-PSK modulation transmits 3600 bits per second, its symbol rate is ________ symbols per second.

10、