东南大学通信原理试卷及参考答案

东南大学考试卷（ A 卷）

课程名称通信原理考试学期 04-05-3

得分

适用专业

考试形式

闭卷

考试时间长度

150

分钟

Section A(30%): 2% for each question)

1. A typical mobile radio channel is a free propagation, linear, and time invariant channel.

( )

2. The power spectral density of a stationary process is always nonnegative. ( ) 3. In a communication system, noise is unwanted and over which we have incomplete

control. ( ) 4. If a random process is stationary, it is ergodic; if a Gaussian random process is stationary,

then it is also strictly stationary. ( ) 5. Double Sideband-Suppressed Carrier (DSB-SC), Single Sideband (SSB), and Frequency

Modulation (FM) are all linear modulation schemes. ( ) 6. Figure of merit (defined as (SNR)O/(SNR)C) of AM of DSB-SC is 1/3, and figure of merit

of Amplitude Modulation (AM) is less than or equal to 1/3. ( ) 7. -law is a nonlinear compression law and A-law is a linear compression law. ( ) 8. The matched filter at the receiver maximizes the peak pulse signal-to-noise ratio, thus is

optimal in a baseband data transmission system with Inter-Symbol Interference (ISI).

( )

9. Correlative-level coding (also known as partial-response signaling) schemes are used to

avoid ISI. ( ) 10. Time-Division Multiplexing (TDM) is used in Asymmetric Digital Subscriber Lines

(ADSL) to separate voice signals and data transmission. ( ) 11. If coefficients of an equalizer is adjusted using the Least-Mean-Square (LMS) algorithm

adaptively, then the matched filter in front of the equalizer is not necessary. ( ) 12. In an M-ary Phase-Shift Keying (M-PSK) system, if the average probability of symbol

error is Pe, then the average Bit Error Rate (BER) of the system is Pe/log2M. ( ) 13. With the same Signal-to-Noise Ratio (SNR), 16-ary Quadrature Amplitude Modulation

(16-QAM) has better performance than 16-ary Phase-Shift Keying (16-PSK). The reason is that 16-QAM has constant envelop. ( ) 14. With the same SNR, Minimum Shift Keying (MSK) has better performance than Sunde’s

Frequency-Shift Keying (FSK). They are both Continuous-Phase Frequency-Shift Keying (CPFSK). ( ) 15. If the largest frequency component of an band-limited signal X(t) is at 100 Hz, then the

corresponding Nyquist rate is 200 Hz. ( )

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Section B(30%): Fill in the Blanks (3% for each question)

1. The power spectral density of a stationary process X(t) is SX(f), then the autocorrelation

function of X(t) is RX(t) = 2. A random process Y(t) is defined as Y(t) = X(t) cos(2fct + ), where X(t) is a stationary

process, fc is a constant frequency, and the phase  is randomly distributed over the interval [0, 2]. Suppose the power spectral density of X(t) is SX(f), the power spectral density of Y(t) is SY(f) = 3. In a Frequency Modulation (FM) system, the modulating signal is m(t) = 2 cos (6t) V, the

frequency sensitivity is kf = 0.3 Hz/V. Using Carson’s rule, bandwidth of the FM signal is approximately 4. An analog signal is first encoded into a binary Pulse-Code Modulation (PCM) wave.

Sampling rate of the PCM system is 8 kHz, number of representation levels is 64. The binary PCM wave is transmitted over a baseband channel using a 4-ary Pulse-Amplitude Modulation (PAM) (that is, a PAM with 4 amplitude levels). The minimum bandwidth requirement for transmitting the PAM wave is kHz.

5. Basic operations performed in the transmitter of a PCM system

include , and 6. Bandwidth efficiency of 4-ary Quadrature Amplitude Modulation (QAM) is bandwidth efficiency of 8-ary Phase-Shift Keying (8PSK) is 7. In a Delta Modulation (DM) system, sampling rate is fs = 8 kHz and step size is  = 0.1 V.

If the input to the DM system is a 1 kHz sinusoidal signal, then to avoid slope overload, the maximum amplitude of this input signal is V.

8. 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 Single-Sideband Modulation (SSB), respectively, then the minimum bandwidth required is kHz; if the multiplexing and modulation methods are Time-Division Multiplexing (TDM) and PAM, then the minimum bandwidth required is kHz;

9. The binary data sequence {bk} = 1010110 is applied to a duobinary (class I partial

response) system with precoding, as shown in Fig. P1-9 (see next page), where dk = bk  dk-1,  denotes modulo-two addition, and ak = 2×dk – 1. The initial value of dk is 1. The output of the duobinary encoder is {ck} = If at the receiving

ˆ} end, the first received digit is 1.25 due to noise, then the receiver output is {bk

共 10 页第 2 页

Fig. P1-9

10. A communication system has the signal-space diagram shown in Fig. P1-10, where

message points have equal probability of transmitting. Assume communication channel in this system is Additive White Gaussian Noise (AWGN) channel, and E/N0 is 12, where N0 is single-sideband power spectral density of the AWGN. Using union band, the

average probability of symbol error is overbounded as Pe  complementary error function erfc().)



1(t)

Fig. P1-10

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Section C(40%): Calculations (8% for each question)

1. In a coherent Binary Phase-Shift Keying (BPSK) system, symbols 1 and 0 are represented

by signals s1(t) and s2(t), respectively.

The signals are defined by

s(t)fct)1



0tTb 

s(t)ft)

c2



where Eb is the transmitted signal energy per bit, and Tb is one bit duration.

a) Determine the basis function(s) of coherent BPSK signal constellation; b) Plot the signal-space diagram of coherent BPSK system; c) Determine the error probability of BPSK;

d) If there is a phase error  between the phase references of the transmitter and

receiver, determine the error probability of BPSK in this condition again.

2. Spectrum of a message signal m(t) is shown in Fig. P3-2. This message signal is Double

Sideband-Suppressed Carrier (DSB-SC) modulated with a carrier wave Accos(2fct). a) If fc = 2 kHz, plot the spectrum of the modulated signal s(t); b) What is the baseband bandwidth W of the message signal m(t)?

c) What is the transmission bandwidth BT of the DSB-SC modulated signal?

d) What is the lowest value of fc that keeps the DSB-SC modulation from sideband

overlap?

-1 kHz

01 kHzf

Fig. P3-2

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3. Consider a Quadriphase-Shift Keying (QPSK) system. The transmitted signal set is

defined as:

cos[2fct(2i1)/4],0tT

si(t)

0, otherwise

where i = 1, 2, 3, 4. Every two input bits select one of the signals in transmitted signal

set to transmit. The rule of mapping is 10  s1(t), 00  s2(t), 01  s3(t), 11  s4(t). a) Determine the basis function(s) of QPSK signal constellation; b) Express si(t) using the basis function(s);

c) If the input binary sequence is 01101000, and suppose fc = 2/T, plot the QPSK

waveform;

d) Is the QPSK waveform continuous phase?

4. Consider the signal s(t) shown in Fig. P3-4,

a) Assuming h(t) is the matched filter of s (t), plot the impulse response of h (t); b) When s (t) is applied to h (t), plot the matched filter output in the time domain.

s(t)A

2T/3

0-A

Fig. P3-4

5. Suppose X(t) = Acos(2ft – ), where A is a constant, and f and  are independent.  is

uniformly distributed over the interval [0, 2]. Determine the power spectrum density of X(t) in terms of the probability density function of the frequency f.

T/3

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通信原理期终考试参考答案和评分标准

考试学期：04-05-3

Section A: (30分)

1.  The mobile radio channel is typically time variant. 2.  3. 

4.  Stationary is not necessary ergodic. 5.  FM is not a linear modulation scheme 6.  Figure of merit of DSB-SC is 1, not 1/3. 7.  A-law is also a nonlinear compression law. 8.  The matched filter is optimal with AWGN channel.

9.  Correlative-level coding is to use ISI to achieve 2W signaling rate in a

bandwidth of W Hz. 10.  FDM is used in ADSL. 11. 

12.  The BER is usually not Pe/log2M. 13.  16-QAM is not constant envelop. 14.  15. 

Section B: （30分） 1.







SX(f)exp(j2f)df

共 10 页第 6 页

[SX(ffc)SX(ffc)] 4

3. 7.2 4. 12

5. Sampling, Quantizing, and Encoding 6. 1, 1.5 7. 1.27 8. 240, 240

9. 0 -2 0 2 0 0 2, 0010110 10. erfc(3)

Section C:（40分）

(a) The basis function is:

1(t)

fct),0tTb. (2分) (b) The signal-space diagram is: (2分)

1(t

)

(d) With phase error 

, the error probability becomes Peerfc. 2

(2分)

2. (a) The spectrum is: (2分)

共 10 页第 7 页

-2 kHz

02 kHzf

(b) W = 1 kHz; (2分) (c) BT = 2 kHz; (2分)

(d) To avoid sideband overlap, the lowest frequency is 1 kHz. (2分)

3. (a) The basis functions are: (2分)

fct),0tT1(t)





(t)ft),0tT2c

(b)

si(t)

(2i1)1(t)(2i1)2(t),i1,2,3,4 (244

分)

(d) The QPSK waveform is not continuous phase. (2分)

4. (a) h(t) is shown below, where k is a positive constant. (4分)

共 10 页第 8 页

h(t)kA

0-kA

(b) The filter output is shown below. (4分)

Response of h(t) to s(t)

-kA2T

-kA2T/3

T4T/3

5T/3

5. The autocorrelation function of X(t) is: (2分)

RX()E[X(t)X(t)]

A2E[cos(2Ft2F)cos(2Ft)] A2E[cos(4Ft2F)cos(2F)]2

Since  is uniformly distributed over the interval [0, 2], we get: (2分)

共 10 页第 9 页

RX()E[cos(2F)]

AfF(f)cos(2f)df2

Since X(t) is a real-valued random process, SX(f) is an even function of frequency, we also have: (2分)

RX()SX(f)exp(j2f)df



SX(f)cos(j2f)df



where Sx(f) is the power spectrum density of X(t). Therefore, we get: (2分)

SX(f)fF(f)

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