同济大学通信系统原理(英文班)练习题附答案

Chapter 1

Homework:

Fill-in Questions:

log M) bits 1. In a M-ary communication system, each symbol contains ( 2

information content.

2. The purpose of communication is to ( transfer information ). Efficiency of the digital communication system can be measured by the specifications such as (R B) , (R b) , ( η). And reliability can be measured by ( P e) , (P b).

3. The basic factors for measuring the merit of a communication system are (efficiency ), and ( reliability ).

4. The main influence of constant parameter channel on signal transmission are usually described by their ( Amplitude-Frequency ) and ( Phase-Frequency ) characteristics.

5. Common characteristics of random parameter channels are (transmission attenuation of the signal is varying with time ), (transmission delay of the signal varies with time ), and (signal arrives at the receiver over several paths ).

Multiple-Choice Questions:

1.For analog and digital communications, which is (are) true? ( A B C D ).

A. digital communication typically uses more bandwidth.

B. analog communication cares more about fidelity

C. digital communication cares more about probability for correct decision

D. digital communication typically uses analog carrier to carry baseband signals

Single-Choice Questions:

1. The symbol rate of 16-ary digital signal is 1200Bd, then the corresponding information rate is ( D ); If with the same information rate, the symbol rate of 8-ary digital signal is ( D ).

A. 1600b/s, 1200Bd

B. 1600b/s, 3200Bd

C. 4800b/s, 2400B

D. 4800b/s, 1600B

2. Assume the delay difference of two paths in a random parameter channel is τ, the frequencies of maximum transmission loss are ( B ), the frequencies =

1

ms

of minimum transmission loss are ( B ).

A. n kHz, (n+0.5)kHz

B. (n+0.5)kHz, n kHz

C. n kHz, (n+0.2)kHz C. (n+0.2)kHz, n kHz

True or False

When there is no input signal, the additive interference doesn’t exist, but the multiplicative interference still exists. ( F )

Homework: 2.1, 2.4, 2.5, 2.6, 2.8, 2.11

Fill-in Questions:

1. AWGN noise is short term for ( Additive White Gaussian Noise ).

2. The power spectral density and ( the autocorrelation function ) of a stationary random process are a pair of Fourier transform.

3. If bandwidth is 10 MHz, signal-to-noise ratio is 20dB, the Shannon capacity is C=( 66.6 Mbps ).

Single-Choice Questions:

1. Assume X(t) is a generalized stationary random process with zero mathematical expectation, then the average power of X(t) is ( D ).

A. E[X (t )]

B. E 2[X (t )]

C. R(∞)

D. D[X (t )]

2. For a narrow band random process 00()()cos ()sin C S X t X t t X t t ωω=-, if ()X t is a Gaussian process, then ( A ). (A) ()C X t and ()S X t are also Gaussian processes; (B) Only ()C X t is also Gaussian process;

(C) ()C X t and ()S X t are not Gaussian processes; (D) Only ()S X t is also Gaussian processes.

https://www.360docs.net/doc/2b6022252.html,parison of AM, DSB, SSB and FM these four communication system,

which has the best efficiency? Which has the best reliability? Which have the same efficiency?

SSB has best efficiency. FM has the best reliability. AM and DSB have the same efficiency.

2.What is the requirement on the characteristics of the filter for producing the

VSB signal?

[H(f+f0)+H(f?f0)]=C

3.What kind of baseband signal is suitable for VSB modulation?

The analog baseband signal with D.C. component and low frequency component is suitable for VSB modulation.

4.Let the expression of a FM modulated signal be

10cos(2π×106t+5sin103πt)

Find: (1) the maximum frequency deviation of the modulated signal.

(2) the frequency of the carrier.

(3) the bandwidth and the average power of the modulated signal.

(4) if the FM circuit constant is k f=5kHz/V, then the expression of the baseband signal can be written as:

=5,∴?f=m f?f m=5×500=2500Hz

(1) ∵m f=?f

f m

(2) f0=106Hz

(3) B=2(m f+1)f m=2×(5+1)×500=6000Hz

P =

A 22

=

1022

=50 W

(4) ∵?f =k f ?A m ∴A m =

?f k f

=

25005000

=0.5 V

∴the baseband signal can be written as:

m (t )=0.5cos 103πt

5. Assume the amplitude of a FM modulated signal is 10V , the instantaneous frequency of the FM signal is:

f (t )=106+104cos (2π×103t ) (Hz)

Find: (1) the expression of the FM modulated signal.

(2) the maximum frequency deviation , the modulation index, and the bandwidth of the modulated signal.

(1) s (t )=10cos [2π×106t +10sin (2π×103t )]

(2) ?f =104, m f =10, B =2(m f +1)f m =2×(10+1)×1000=22 kHz

Fill-in Questions:

1. Assume the expression of a FM signal is )102sin 5102cos(536t t ?+?ππ, the carrier

frequency of this FM signal is (106)Hz, the maximum frequency deviation is (5000) Hz, the bandwidth of this FM signal is (12×103 ) Hz ，the average power is (12.5 )W.

Single-Choice Questions:

1. Which analog modulation has the highest spectrum efficiency?

( C )

(A) AM (B) DSB (C) SSB (D) VSB

Calculation Questions:

1. Let a baseband modulating signal be a sinusoidal wave with the frequency 10kHz, and the amplitude 1V . It modulates the phase of a carrier with frequency 10MHz, and the maximum phase deviation of modulation is 10rad. (1) Calculate the approximate bandwidth of the phase modulated signal. (2)If the frequency of the modulating signal is changed to 5kHz, calculate the bandwidth of the phase modulated signal.

Answer: (1) We has known that: m 10 kHz, A 1 V m f == , and maximum phase deviation

max 10 rad ?=.

The instantaneous phase of the carrier can be represented as:

()()

p t k m t ?=, where

10

p k =.

Then, the instantaneous frequency deviation of the carrier can be represented as:

d

()

sin p m m d t k t dt

?ωω=, and maximum frequency deviation is p m k ωω?=. Hence, the frequency modulation index is:

10

p m

f p m

m

k m k ωω

ωω?=

=

==

Thus, the approximate bandwidth of the phase modulated signal is

2(1)2(110)10220 kHz

f m B m f =+=+?=

(2) If the frequency of the modulating signal is changed to 5kHz, then the approximate bandwidth of the phase modulated signal is:2(1)2(110)5110 kHz f m B m f =+=+?=.

Chapter4

Fill-in Questions:

1. The basic process, from( Sampling the analog signal ), ( the quantization ), and ( converting to binary symbols ) , is usually called as pulse code modulation (PCM).

Single-Choice Questions:

1. The purpose of nonuniform quantization is ( B ).

A. Protect big signal

B. Protect small signal

C. remove quantization noise

D. Increase quantization levels

2. In Delta modulation, the best way to avoid overload quantization noise is ( A ).

A. Increase sampling frequency

B. Increase quantization step

C. Reduce sampling frequency

D. Reduce quantization step

3. The sampling theorem points out that if the highest frequency of a low-pass analog signal is

H f , then the signal can be represented by its samples when the sampling frequency no less

than ( B ).

(A) H f (B) 2H f (C) 3H f (D) 5H f 4. If the samples obey uniform distribution over [,]a a -, then the quantization noise power is determined by

( C )

(A) a (B) a and quantization interval ? (C) quantization interval ? (D) other parameters.

Multiple-Choice Questions:

1. Digitization of analog signals includes ( B 、D 、E )

(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). Sampling

Short-Answer Questions:

1. In Delta modulation system, why we increase sampling frequency s f rather than quantization step ? to avoid overload quantization noise?

Answer:

Given the sampling frequency and the quantization step ?, then the slope of a step is:

s k f =??

It is the maximum possible slope of a step-shaped wave, or it is called the maximum tracking slope of the decoder. When the slope of the input signal of the delta modulator exceeds this maximum, overload quantization noise will be generated. Therefore, in order to avoid overload quantization noise, it is necessary to make the product of ? and s f large enough, so that the slope of the signal can not exceed this product. One the other hand, the value of ? is directly related to the magnitude of the basic quantization noise. If the value of ? is large, then the basic quantization noise must also be large. Therefore, only the method of increasing s f to increase the product s k f =?? can ensure that the basic quantization noise and the overload quantization noise do not exceed the limit.

Fill-in Questions:

1.( 2 )Baud/Hz is the highest possible unit bandwidth rate, and is also called as the Nyquist

rate.

2.The time-domain equalizer is used to overcome (intersymbol interference (ISI)).

3.The basic process, from ( Sampling the analog signal), ( the quantization ), and

( converting to binary symbols ) , is usually called as pulse code modulation (PCM). Single-Choice Questions:

1. If the samples obey uniform distribution over [,]

a a

-, then the quantization noise power is determined by ( C)

(A) a(B) a and quantization interval ?(C) quantization interval ?(D) other parameters.

Multiple-Choice Questions:

1. Digitization of analog signals includes ( B, E )

(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). Sampling

2. Which are symbol code type for baseband digital signals? (A, B, C, D, E )

(A). HDB3 (B). AMI (C). CMI (D). Biphasic code (E). 5B6B (F). BPSK

(G).MSK

Short-Answer Questions:

1.Which are the design principles of code for digital baseband signal?

As we know, the performance of a practical baseband transmission system can be found in its eye pattern (refer to the figure below). Please find out the characteristics in the figure 1, and point out their impacts on transmission performance.

Answer:

(1) The location of the central perpendicular line is the optimum sampling instant.

(2) The middle horizontal line represents the optimum decision threshold level.

(3) The perpendicular height of the shadow region represents the distortion range of the

received signal.

(4) The slope of the bevel edge of the “eye” represents sensi tivity of the sampling instant

to the timing error.

(5) Under no noise situation, the degree of the opening of the “eye is the noise tolerance;

if noise at the sampling instant exceeds this tolerance, then error decision may happen.

Calculation Questions:

1.Assume there are 4 overall transfer characteristics of baseband transmission systems

shown in Figure 2. If symbol rate is 2000Bd, please illustrate whether they can transmit information without ISI? Which system has better transmission characteristic?

Solution: ∵symbol rate is 2000Bd,

∴ISI exists in system (a) and (c), ISI does not exist in system (b) and (d).

Since system (b) can be realized physically, and its η=1.33 Bd/Hz

system (d) can’t be realized physically, and its η=1 Bd/Hz

so system (b) has better transmission characteristic.

1000

-1000 (Hz) (Hz)

1500

-1500 500

-500

(Hz)

750

-750 (Hz)

2000

-2000

Chapter6

1. Assume there is a space-ground communication system, the symbols rate is 0.5MB, the bandwidth of receiver is 1MHz. Antenna gains of ground station and space station are respectively 40dB and 6dB ，the path loss is (60+10lg d )dB, where d is the distance (km). Given the transmitting power is 10W, the double-side power spectral density of white noise is 2×10?12W/Hz. Requiring the symbol error probability of the system is P e =10?5, try to find the maximum communication distance under these conditions as below.

(1) Adopting 2FSK modulation and coherent demodulation. (2) Adopting 2DPSK modulation and coherent demodulation. 解：（1）采用2FSK 方式传输，进行相干解调，其误码率为 5102

2

1-==r

erfc

P e 可计算出所需要的信噪比为r=18.3，噪声功率为 ()W B n n 661202104101104--?=???==σ 接收端信号幅度平方为

()W r a n 462210464.11043.1822--?=???==σ 传输中信号功率衰减为

()dB a A 35.5110464.110

2lg 1022lg 104

22=??=-

根据给定条件，51.35+40+6=60+10lgd ，可以算出传输距离为 ()km d 543210735.3==

(2) 采用2DPSK 方式传输，选择差分相干方式进行解调，其误码率

5

e 1

e 102r P --==，

可计算出所需要的信噪比r =10.82，接收端信号幅度平方为

22

65n 2210.824108.65610 W a r σ--==???=?

传输中信号功率衰减为 225

2210

10lg 10lg 53.63 dB 28.65610A a -?==?

根据给定条件，53.63+40+6=60+10lgd ，可以算出传输距离

3.963109183 km d ==。

Fill-in Questions:

1. Three basic digital modulations are ( ASK ),( FSK ), and( PSK ).

2. In the 2ASK, 2FSK, 2PSK systems, ( 2ASK ) is the worst in the fading channel; ( 2PSK ) is influenced by the phase variation of the fading channel; ( 2FSK) has the best robustness in the fading channel.

3. For 8FSK system, if symbol error rate Pe is 1?10-6, then bit error rate P b is ((4/7)?10-6 ).

Single-Choice Questions:

1. To obtain the same bit error probability, the M-ary signal needs to

( A )

A ．Occupy wider frequency band or use more power in exchange;

B ．Occupy wider frequency band or use little power in exchange;

C ．Occupy narrower frequency band or use more power in exchange;

D ．Occupy narrower frequency band or use little power in exchange.

2. In digital modulation system, if 8PSK is used for information transmission, then the maximum frequency band utilization ratio achieved can be ( C ). A. 1Hz s bit ?/

B. 2Hz s bit ?/

C. 3Hz s bit ?/

D. 4Hz s bit ?/

3. Which of the following statements is (are) true about QPSK: ( C ) A. it has 1 phase. B. it has 2 phases C. it has 4 phases

D. it has 2 amplitudes

4. In 2ASK, 2PSK, 2FSK systems, assume they have the same signal to noise ratio. Which one has the lowest bit error probability? ( B ) (A) 2ASK; (B) 2PSK; (C) 2FSK; (D) The three systems have the same bit error probability.

Multiple-Choice Questions:

2.Which of the following is (are) TRUE? ( A B C)

A. 4ASK and 16PSK use the same bandwidth as 2ASK.

B. For a communication system, higher SNR means better performance.

C. MFSK use more bandwidth than MPSK and MASK

D. Baseband signal is a sinusoidal wave of low frequency

3.Which of the following is (are) not TRUE? ( A B D)

A. In digital communication, the frequency bandwidth of the receiver must be larger than

that of received signal.

B. The frequency band utilization ratio of digital baseband system must be less than 2

/.

bit?

Hz

s

C. The more channel numbers in TDM system, the more information rate.

D. MSK signal is a kind of continuous phase FSK signal, so it can be demodulated by

envelope detection.

3. Let rb be the signal-to-noise ratio per bit, Eb be the energy per bit, E be the energy per symbol, N0 be the one-sided AWGN PSD, snr be the signal-to-noise ratio, which of the following is (are) true? ( A B ) (A). rb ≤ Eb / N0 (B). snr ≤ E / N0 (C). rb ≤ Eb ? N0 (D). snr ≤ E ? N0 (E). rb ≥ Eb / N0 (F). snr ≥ E / N0 (G). rb ≥ Eb ? N0 (H). snr ≥ E ? N0 Calculation Questions:

2.If the symbol rate of a baseband signal of 1?106 symbols/second. The baseband signal is

modulated using 2ASK. Received signal power is -100 dBm. One-sided AWGN PSD is -174 dBm/Hz.

(a)Determine the minimum bandwidth W for receiving without ISI at sampling time.

(b)Calculate the received signal-to-noise ratio SNR and Shannon capacity.

(c)Calculate the symbol error rate P e in 2ASK and 2FSK, for non-coherent

demodulation.

Answer: (a) Symbol duration is T=1/(1?106)=1 us. According to Nyquist criterion, the minimum baseband bandwidth for receiving without ISI is B=1/(2T)=0.5 MHz. 2ASK bandwidth W=2?B=1 MHz.

(b) SNR(dB)=P s(dB)-P n(dB). P n(dB)=-174+W(dB)=-174+10log(1M)=-114 dBm. So SNR(dB)=-100-(-114)=14 dB. We have SNR=10(14/10)=25.1.

The Shannon capacity is C=W?log(1+SNR)=4.71 Mbps.

(c) As both OOK and 2FSK are binary system, the bit error rate P b equals the symbol error rate P e.

2ASK: P b = P e =1/2?e(-SNR/4)=9.41?10-4.

2FSK: P b = P e =1/2?e(-SNR/2)=1.77?10-6.

3.If the symbol rate of a baseband signal of 1.0?106symbols/second. Two-sided AWGN

PSD is -207 dBW/Hz. The baseband signal is modulated using 8PSK. Received signal power is -133 dBW. (12%)

a)Determine the minimum bandwidth W for receiving without ISI at sampling time.

(2%)

b)Calculate the received signal-to-noise ratio SNR and Shannon capacity. (6%)

c)Calculate the bit error rate P b in 2ASK and 2DPSK for non-coherent demodulation.

(4%)

Answer: (a) Symbol duration is T=1/(1.0?106)=1 us. According to Nyquist criterion, the minimum baseband bandwidth for receiving without ISI is B=1/(2T)=0.5 MHz. 8PSK bandwidth W=2?B=1 MHz.

(b) SNR(dB)=P s(dB)-P n(dB). P n(dB)=-207+3+W(dB)=-204+10log(1M)=-144 dBW. So SNR(dB)=-133-(-144)=11 dB. We have SNR=10(11/10)=12.6.,

The Shannon capacity is C=W?log(1+SNR)=3.77 Mbps.

(c) As both 2ASK and 2DPSK are binary system, the bit error rate P b equals the symbol error rate P e.

2ASK: P b = P e =1/2?e(-SNR/4)=2.14?10-2.

2DPSK: P b = P e =1/2?e(-SNR)=1.69?10-6.

Chapter8

8. Assume a 2FSK signal can be expressed as:

??

?≤≤=≤≤=s s

T t t f A t s T t t f A t s 0,2cos )(0,2cos )(22

11ππ Where, s T f 31=，s

T f 5

2=，and the transmitting probabilities of s 1(t) and s 2(t) are equal.

(1) Please give the block diagrams of the correlation receiving and the matched filtering respectively.

(2) If the double-sided PSD of AWGN is n 0/2, find the error symbol probability.

解：（1）2FSK 的相关器形式的最佳接收机结构如图所示。

(3)

（2）2FSK 的匹配滤波器形式的最佳接收机结构如图所示

r 2

r

由题意知信号是等能量的，即

2

)()(200

220

2121s

T T b T A dt t s dt t s E E E s

s =====?

?

故系统的误码率为

200421

221n T A erfc n E erfc P s b e ==

Fill-in Questions:

1. For signal s(t) with duration of T, the match filter should have an impulse response of ?(t) = ( ks(t 0?t) ).

Multiple-Choice Questions:

1. Let rb be the signal-to-noise ratio per bit, Eb be the energy per bit, E be the energy per symbol, N0 be the one-sided AWGN PSD, snr be the signal-to-noise ratio, which of the following is (are) true? ( A B ) (A). rb ≤ Eb / N0 (B). snr ≤ E / N0

(C). rb ≤ Eb ? N0

(D). snr ≤ E ? N0 (E). rb ≥ Eb / N0

(F). snr ≥ E / N0 (G). rb ≥ Eb ? N0

(H). snr ≥ E ? N0

Chapter9

Fill-in Questions:

1. In the basic level E-1 of PCM digital telephone system, the time period of a frame is

( 125 )us, the output bit rate is ( 2.048 )Mb/s, the multiframe rate is ( 500 )frames/s; the transmission rate of each signaling is ( 2 )kb/s.

Single-Choice Questions:

1. The information rate of PCM30/32 system is ( ) kbit/s, there are ( ) time slots in a frame. A. 64, 30 B. 64, 32 C. 2048, 32 D. 2048, 30

2. The inverse polynomial of the primitive polynomial is also a primitive polynomial. Hence, if 41x x ++ is a primitive polynomial, then which one is its inverse polynomial ( D ) (A) 421x x ++; (B) 421x x x +++; (C) 4321x x x x ++++; (D) 431x x ++

3. Which operation mode is not included in the pure ALOHA system ( D ) (A) Transmission mode; (B) Listening mode; (C) Retransmission mode; (D) Reserved mode

4. Which property is not belong to m sequence ( B ) (A) Balance; (B) Length of the runs

(C)Shift addition characteristic; (D) Autocorrelation characteristic

True or False

2. The more channel numbers in TDM system, the more information rate. ( T )

3. In SDH system, the information rate of STM-4 is 4 times that of STM-1. ( T )

4. ITU constituted 2 recommendations of logarithm compression law for telephone signal,

they are A-law and μ-law respectively. A-law is used in China ’s mainland. ( T ) 5. Arbitrary two codewords used for CDM must be strictly orthogonal. ( F )

Short-Answer Questions:

Plot the block diagrams of 3 channel frequency division multiplexing (FDM) telephone communication system , including the block diagram of transmitter , the block diagram of receiver , and the frequency spectrum of FDM signal .

Answer :

(c) Frequency spectrum of FDM signal.

Calculation Questions:

The 8-ary digital expression of a primitive polynomial is 23, (1) Find its corresponding characteristic equation; (2) According to its characteristic equation, plot the block diagram of m sequence generator (linear feedback shift register); (3) Write down all status and find the period of this m sequence.

Answer:

(1) its characteristic equation f(x)=x4+x+1

(2) The block diagram of this linear feedback shift register is shown as below:

(3) a3a2a1a0

Initial status: 1 0 0 0

1 1 0 0

1 1 1 0

1 1 1 1

0 1 1 1

1 0 1 1

0 1 0 1

1 0 1 0

1 1 0 1

0 1 1 0

0 0 1 1

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

Period =24?1=15

Chapter11

Fill-in Questions:

1. If OFDM has 1024 sub-carrier and each symbol has a duration of 1ms, the total bandwidth is ( 1.025×106Hz).

Single-Choice Questions:

1. About the advance digital bandpass modulation system, the incorrect represents is ( C )

(A) MSK has constant envelope and rapid descent of the power spectral density outside the frequency band;

(B) The power spectral density of the GMSK signal is more concentrated than that of the MSK signal;

(C)OFDM signal are not suitable used in the radio channels with serve fading.

(D)The disadvantage of GMSK system is the intersymbol interference.

2. About the advanced digital bandpass modulation system, the incorrect represents is ( D)

(A) MSK and GMSK belong to improved FSK system.

(B) the power spectral density of the GMSK signal is more concentrated than that of the MSK signal.

(C)OFDM signals belong to a kind of frequency division modulation system with multi-frequencies.

(D)OFDM signal are not suitable used in the radio channels with serve fading.

3. Which of the following is not the characteristic of a MSK signal? ( D)

A. Continuous phase

B. Constant envelop

C. Minimum bandwidth

D. Minimum symbol error probability

Multiple-Choice Questions:

1. Which of the following is (are) not TRUE? ( A B D)

A. In digital communication, the frequency bandwidth of the receiver must be larger than

that of received signal.

B. The frequency band utilization ratio of digital baseband system must be less than 2

/.

bit

Hz

s

C. The more channel numbers in TDM system, the more information rate.