电子科技大学《通信原理》课程试题A卷含答案

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电子科技大学《通信原理》课程试题（A卷）

(120分钟) 一 二 三 四 五 六 七 八 九 十 总分 评卷教师

|f|,11. An analog signal with the spectrum M(f)25000,Assume that 50 dB.

|f|2000Hzothers is to be transmitted over a PCM system.

Pe0 and that there is no ISI. The peak signal-to-noise ratio at the receiver output needs to be at least

(a) find the maximum allowable time interval between sample values

TS.

(b) What will be the number of quantizing steps needed?

(c) Determine the minimum bit rate required in the PCM signal.

(d) What is the first null bandwidth of the PCM signal for the polar NRZ signaling case？

Solution

a. fs=2B=2*2000=4000Hz，Ts=1/fs=1/4000=0.25ms b. 50=6.02n+4.77，n=7.5，n=8，M=2n=256 c. R=nfs=8*4000=32000bit/s=32 Kbit/s d. Bnull=R=32 KHz

2. A binary line code that has a rectangular pulse shape is sent over a channel without ISI. Assume the impulse response

of the overall system is

sin(f0t)he(t)=f0t, where

f0103Hz.

(a) Find the zero points in the system impulse response and the minimum symbol period Ts without introducing ISI. (b) Find the maximum bit rate that can be supported by this system. (c) Find the absolute bandwidth B for this transmission system.

3ftk(k0)tkfk10s(k0),Ts1ms Solution: (a)00 (b)R (c)B

1Ts1kb/s

R20.5kHz

3. Assume that an AM transmitter is modulated with a video testing signal given by

m(t) 0.30.5sin1t, where f1=4 MHz. Let Ac =100.

(a) Sketch the AM waveform.

(b) What are the percentages of positive and negative modulation? (c) Evaluate the total average power. (d) Evaluate the peak envelope power.

Solution:

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(a) m(t)= 0.3+0.5 sin 1t fm= f1 =4 MHz AC100

S(t)100(0.70.5sin1t)cosct

AmaxAC12010020% (b) %pos.mod.AC100ACAmin10020%neg.mod.80%

AC100 (c)

Pnorm12Ac[1m(t)]221002(0.70.5sinw1t)2 210020.250.49307522 (d)

max[m(t)]0.30.50.2 1[Ac(10.2)]27200 2

PPEP4. Consider a random data pattern consisting of binary 1’s and 0’s, where the probability of obtaining either a binary 1 or 0 is 1/2 and the duration of one symbol is values of

Tb1 ms. Calculate the PSD for polar RZ signaling assuming peak

1 V.

Solution：

3Tbsin(fTb/2)sin(0.510f)F(f)0.51032fTb/20.5103fTbfTbSa()0.5103Sa(0.5103f) 22sin(0.5103f)f12R(0)E[a][1(1)2]122n

1R(k)(11)0 where k02 |F(f)|2j2kfTb(f)R(k)eTbkTb2fTbSa()2.5104Sa2(0.5103f)422

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5,0tT mv is a known signal, whereT5. Assume that s(t)0,otherwise(a) (b) (c) (d)

Solution:

（a）The impulse response of the matched filter is

Find the impulse response of the matched filter. Find the optimal sampling time t0.

Find the output of the matched filter when s(t) is sent. Plot the output of the matched filter.

8s. For the case of white noise,

5,0t8106h(t)Ks(t0t) t0T0,otherwiseK1

h(t)=s(t0-t) s(t)/mv 5 s(-t) t/μs 8 -8 0 8 5 t/μs 0 b. The minimum time that we have to wait before the maximum signal level occurs at the filter output is

t0T8s

And t0 is the optimal sampling time. c. The output of the matched filter is

so(t)h(t)s(t)s()h(t)d-60, t<0 or t>1610t -625d25t, 0t<8100825d40010-625t, 810-6t1610-6t8e. The output of the matched filter is shown as in the following figure.

so(t) 2×10-4 0 t0=8 16 t/μs 6. For a binary baseband system , the received signal is

5V,s(t)5V,0tT,for a binary 10tT,for a binary 0 and T1ms

N0106W/Hz. The block diagram of 2Assume the channel noise is White Gaussian Noise with the PSD of Pn(f)receiver is as follow

Let LPF has an equivalent bandwidth of B=4R and a unity gain. (a) Evaluate the optimum threshold VT.

(b) Evaluate the error bit probability for the data at the receiver output. ( Pe can be represented by Q[.] ) Solution:

so15Vso25V

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1VToptPeminso1so2550(V)

22(so1so2)24o2Qso122o

2

Q25Q2106141103Q255

7. Assume that the input signal power of LSSB receiver is 2W, as shown in the following Fig., and carrier frequency is 100MHz. If the bandwidth of modulating signal m(t) is 4MHz, and the PSD of noise for channel

9N/2210W/Hz,and coherent demodulation is used, please determine: is0Product detectorModulated LSSBsignal plus noiseIFfilter2cosctLow-passfilter~(t)m

(a) the transfer function of the ideal IF filter; (b) the input ratio of signal power and noise power (c) the ouput ratio of signal power and noise power Solution:

(a) in order to assure the signal to pass the IF filter, the bandwidth of IF filter is equal to that of the modulated signal,

That is to say, BIF=Bm=4MHz. thus

(S/N)inof the detector;

(S/N)outof the detector.

K(costant)(96MHzf100MHz) H()

otherwise0(b) the power of input signal being

Si2W, and the power of input noise being

Ni2BTPn(f)2*(2*109)*(4*106)32*103Wthus the input ratio of signal power and noise power (S/N)in125 (c) because the G=1 for SSB, so

(S/N)out125

8. A binary baseband signal is passed through a multilevel modulated bandpass system. The wave shape of the binary

baseband signal is rectangular, and the data rate is 50Mb/s. Evaluate

(a) Assume that a QPSK signal is used to send the data. What is the absolute bandwidth of the QPSK signal? What is

the null to null bandwidth of the QPSK signal? Is there AM component on the QPSK signal?

(b) Assume the baseband signal is pass through a raised cosine-rolloff filter with a 50% rolloff factor first, and then be

transmitted using QPSK signal. What is the absolute bandwidth of the QPSK signal? What is the transmission bandwidth of the QPSK signal? Is there AM component on the QPSK signal?

(c) Assume the available channel bandwidth is 20MHz, and the rolloff factor is 50%. What is the minimum multilevel

for the multilevel modulated signal? Solution

1) The absolute bandwidth of the QPSK signal is ∞.

For QPSK signal,

L2l4l2

R50MHz 2The null to null bandwidth of the QPSK signal is Bnull2D2There isn’t AM component on the QPSK signal.

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2) BTR501r10.537.5MHz l2The absolute bandwidth and the transmission bandwidth of the QPSK signal are same, 37.5MHz.

There is AM component on the QPSK signal.

R501r10.53.75lmin4 3) lBT20 The minimum multilevel is L=24=16

9. Assume that the input data stream is {+1, -1, +1, +1, -1, -1, +1, -1, +1, +1, -1, +1, -1, +1, +1}, Ac=1.

(a) Plot the MSK type II modulation waveforms x(t) and y(t). (b) Let the period of carrier Tc = Tb/2, plot the BPSK signal s(t).

MSK IIxm(t)yxyxyxyxyxyxyxyData onx(t)Data ony(t)x(t)y(t)BPSK

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