通信系统习题new

1、Tut21Express the pulses shown in the figure by means of the unit step function u(t), and calculate their energy. (1)The pulse can be Mathematically described as: Using the unit step function notation, this can be expressed as: Without using the separate interval definitions, butrelying on the proper

2、ties of the unit step function, this can be written as: The pulse energy is defined as: To find the value of this integral, we can divide it into 3 separate integrals with limits T, 3T, 3T,4T, 4T, 6T, respectively. Hence, the energy is given by: (2)The pulse can be expressed as: In terms of the unit

3、 step function, this can be written as: The pulse energy is equal to: (3)The pulse can be expressed as: In terms of the unit step function, this can be written as: And the energy is equal to 2T.2. Evaluate the following integrals3.Find the output of the time-invariant linear system having the impuls

4、e response h(t) shown in the Figure(a), in response to the input signal x(t) presented in the Figure (b). In order to calculate the convolution x(t)*h(t), we need first to find the mirrored version of the signal x(t), change the variable to , i.e. find x(-k), as presentedThat mirrored version of the

5、 input signal is then shifted and signal x(t-) is considered.It is clearly visible that for t < 3 the signal x(t-k) and the impulse response h(t) do not overlap for any . Hence the integral of their product is equal to zero.The different overlapping possibilities are illustrated in the figure, an

6、d finally for t > 7, x(t-) and h(t) stop overlapping for any Therefore, the output signal y(t) can be expressed as: After performing integration, we getThe obtained output signal y(t) is plotted below. Itis worth to notice that the signal y(t) is smoother than the input signal x(t), and that the

7、time limits property holds. 4.Determine the auto-correlation function of the following signal (where A>0): Solution: The waveform is shown as follows Firstly note that the autocorrelation of a real function is always symmetric - thus you only need to determine the shape of the autocorrelation for

8、 r < 0 or r > 0. Secondly, note that the time-shifted image of the triangular pulse does not overlap the original function at all if r > T or r>-T Therefore the autocorrelation is zero in those regions. We can just work out the autocorrelation function for 0 < r < T and we have all

9、 the information required to solve the problem. When the image is shifted to the right by t = t units, the overlapping region goes from t to T. Therefore, the autocorrelation is This integral is trivial to evaluate，which gives youThe left-hand side is just a mirror-image of the right： you can substi

10、tute t=-t into Equation 3 to derive this： Thus the full expression for the autocorrelation isThis is shown graphically in Figure 2. 5.Determine the auto-correlation function of the signal , given the auto-correlation function of a periodic signal x(t) is defined as: where T is the period of the sign

11、al x(t)Solution:Tut31.Find the Fourier series expansion of the triangular waveformThe cyclically repeated pulse can be expressed as: with the boundaries of the interval【a，b】being-0.570 and 0.5T0, respectively.The coefficients are determined by: The pulse x(t) is an even function of time, so for anyv

12、alues of t and n within the considered intervalTherefore,which results in all coefficients bn being equal to 0; n1, 2, . .The coefficient a0 equals toThis result follows directly from the fact that a0 represents the d.c. term, which is clearly equal to 0 in the considered case.For other values of n,

13、 n = 1, 2, . , we haveThe first integral within the brackets equals to zero for any value of n, as we integrate there cos(x) function over an integer number of periods. The second and the third terms are of the same form, and can be computed using the formulaHence, we get The latest means thatExampl

14、e plots of thetruncated Fourier seriesexpansion for thetriangular wave for(a) 3 terms,(b) 5 terms,(c) 50 terms2.Find the Fourier transform of the rectangular pulseand draw its magnitude and phase spectra. From the definition of Fourier transform, we haveHence, the W(f) is a real function of frequenc

15、y. This is an expected result, because the w(t) is an evenfunction of time.Thus the magnitude spectrum | W(f)| is expressed as: To find the phase spectrum(f), let us first consider the fact that for it to be and odd function, we must haveTherefore, 0(f) = 0 if W(f) > 0 and 0(f) = ±knotherwis

16、e; k is an oddinteger.In addition, since w(t) is a real signal, 0(f) = -0(-f).3,Find the Fourier transform of the so called 'radio pulses' w1(t) and w2(t) defined as: The Fourier transform W,(f) can be found substituting w(t) = n(t) and 0= 0 into the modulation theorem. This substitution yie

17、lds: For w2(t), we need to take 0 = -0.5, which results in:4.Find the Fourier transform of the rectangular wave w(t) shown in the figure below and draw the magnitude spectrum | W(f)| in the frequency range - 300 Hz to 300 Hz. To find the Fourier transform of the wave w(t) we will use the formulaTher

18、efore, we need first to find the Fourier transform of the pulse h(t) which is repeated at regular intervals to generate the wave w(t). The pulse h(t) is a rectangular pulse with a width T =10 msThe pulses h(t) are repeated every T0 = 30 ms. Hence, assuming that time t is in seconds, the wave w(t) ca

19、n be expressed as: and its Fourier transform W(f) is given by: where the fundamental frequency f0 = 1/T0 = 33.333 Hz, and /-/(f) is the Fourier transform of a single pulse From the table of Fourier transforms Substituting this into the formula for W(f) yields: Because the different terms of the sum

20、in the formula for W(f) do not overlap, we can take the magnitude of |W(f)| as equal to the sum of the magnitudes of different terms. The weights of impulses for f = nf0; n = 0, ±1, ±2,can be presented in the form of a tableWeights of spectral lines for the magnitude spectrum |W(f)| Plot o

21、f the magnitude spectrum of the wave w(t) Tut51.A mixer is used to multiply two signals x(i) and y(i). Plot the magnitude spectrum of the output of the mixer ifwhere A = 2mV, fm = 2kHz, B = 1mV and fc = 200kHz. Repeat your considerations if the magnitude spectrum of x(i) is as given in Fig. 1, and y

22、(i) is the same as previously. Solution：In the first case, the mixer's output z(i) is given by: Utilizing the trigonometric identity: we get: Becausethe latest can be rewritten in the formSubstituting the numerical values for the constants yields: and the magnitude spectrum of z(i) is given in F

23、ig.2. Figure 2: Magnitude spectrum of waveform at the mixer's output.To find the magnitude spectrum in the second case, we need to utilize the frequency translation property of Fourier transform, which states In the considered case, 0 = 0. The magnitude spectrum of thez(i) is given in Fig.3.Figu

24、re 3: Magnitude spectrum of the z(i).2. Draw the magnitude spectrum of the signal xout(t) at the output of a non-linear device, having a characteristic: Assume that the input signal consists of a sum of a baseband signal and a carrier wave having its frequency much higher than the maximum frequency

25、in the baseband component. Repeat your analysis, assuming that the baseband signals of the form: SolutionIn the first case, the considered input signal xin(i) is given by the equation: where m(i) is a baseband signal. Substituting this to the formula for the output voltage of the non-linear device y

26、ields: Utilizing the trigonometric identity: we get: The plot of the magnitude spectrum of xoui(i) is given in Fig.4. Figure 4: Plot of the magnitude spectrum at the output of a non-linear device if the input signal is a superposition of a sinusoidal wave of frequency fc and the signal m(t) having t

27、hemagnitude spectrum |M(f)|.In the second part of the problem the xin(t) is of the form: Without the loss of generality, let us assume here that this corresponds to the situation where the signal m(t) considered in the previous case has a magnitude spectrum as shown in Fig.5. Figure 5: Magnitude spe

28、ctrum of the signal m(t) composed oftwo sinusoids.All of the derivations performed for the first part of the problem are exactly the same in this part. However, the signal m(t) is now of the form: Therefore, to find all the spectral components of the signal atthe output of the non-linear device, we

29、need to workout thespectral components ofSubstituting the previous equation for a2m方(t) we get: The plot of the magnitude spectrum of the xout(t) is given in Fig. 6, and the magnitudes of the spectral components are listed in Table 1. Figure 6: Magnitude spectrum of the xout(t).Table 1: Magnitudes o

30、f spectral components for xout(t).3. Consider a pulse w(t) given in Fig.l, and multiply it by a sinusoidIs the resulting waveform a base-band or a band-pass waveform? If the resulting waveform is a band-pass one, find its band-pass representation in terms of: the complex envelope g(t), the real and

31、imaginary parts of the envelope x(t) and y(t), the magnitude of the envelope R(t) and the phase (t). SolutionThe resulting waveform is given by: To determine if it is a base-band or band-pass waveform, one needs to check its magnitude spectrum. It can be computed using the real signal frequency tran

32、slation theorem Couch.Application of that theorem yields: where W(f) denotes the Fourier transform of the pulse w(t) derived in the previous example. As it is visible in Fig.3, the magnitude of Wf) decays rapidly with increase of f, and we can consider with a very good accuracy that W(6000) = 0. The

33、refore, we can consider the waveform w1(t) as having its power concentrated around the frequency 6000 Hz, and equalto zero at 0 Hz. Thus the waveform w1(t) is a band-pass waveform. In the time domain, it can be expressed as: Therefore, the complex envelope g(t), the real and imaginary parts of the e

34、nvelope x(t) and y(t), the magnitude of the envelope R(t) and the phase 6(t) are given by: 4.For the AM signal shown in Fig.9, calculate its modulation index. What is the carrier amplitude Ac and the tone signal amplitude Am in the above AM signal? Figure 9: An AM waveform with a sinusoidal message.

35、SolutionFrom the plot presented in Fig. 9, it is visible that themaximum value taken by the AM envelope is l00, and theminimum value is 30. By definition, the modulation index for an AM wave when a single tone message is applied, is equal to: where A is the maximum value and B is the minimum value t

36、he envelope takes. Hence, we have The carrier wave amplitude is equal in this case to the average between A and B: and the message signal amplitude is equal to tut61. A 250 kHz carrier signal is modulated by a 3.5 kHz tone signal with peak amplitude of 1.5 V. Draw the corresponding double-sided magn

37、itude spectrum for the following types of amplitude modulation:(a)DSB-FC (with index of modulation m=0.5) (b)DSB-SC (c)SSB-SC (upper sideband) Solution(a)DSB-FC, m=0.5 The modulated signal is described by the following formula: The carrier wave amplitude Ac can be calculated from the definition of t

38、he modulation index, i.e. (b)DSB-SC The modulated signal is described by the following formula: The corresponding double-sided spectra for the cases a) and b), as well as c) are givenin the following figure. 2. A DSB-SC wave is demodulated by applying it to a coherent detector. Evaluate the effect o

39、f a frequency error f in the local carrier frequency of the detector, measured with respect to the carrier frequency of the incoming DSB- SC wave. Repeat your considerations, assuming that the frequency of a local oscillator is correct, but there is a phase error Solution In general, a DSB-SC signal

40、 is given by: Demodulation in a product detector involves multiplication of the incoming modulated wave with a locally generated carrier wave, described in general by: For a proper demodulation, the following conditions should be maintained: However, when = 0 and f=0, the carrier waveand the product

41、 of w(t) and the incoming signal is given by: Because we have: thenAfter low-pass filtering, the demodulated signal isinstead of a signal 1/2Am(t) . The obtained signal y(t) is a DSB-SC signal, with a carrier frequency f. Hence, even for low f, serious distortions will arise due to overlapping of th

42、e spectra at positive and negative frequencies around DC, as is shown in Fig. 2. Figure 2: Overlapping of spectra at positive and negative frequencies.Assuming that f = 0 but /= 0, results inAfter passing through the low-pass filter, the demodulated signal becomesBecause cos() is just a constant, th

43、e demodulated signal is not distorted. The only problem occurs when = 0.5n ± kn, k = 0, 1, 2, . In such cases, the output of the low-pass filter is equal to zero.3. A sinusoidal signal of peak amplitude of 10 volts and a frequency of 12 kHz is applied to a VCO with sensitivity factor of 20Hz/vo

44、lt. Find the peak frequency deviation of the resulting FM signal.SolutionA VCO can be considered as a simple FM modulator. The instantaneous frequency at the VCO's output is given by: and the input signal, in the considered case, is: Because the sensitivity factor k0 of the VCO is equal to 20 Hz

45、/V, the maximum frequency deviation F equals 4. Draw a double-sided magnitude spectrum for the following narrowband FM signal:SolutionIn the considered case, the modulation factor mf = 0.2. Hence this is a narrow-band FM waveform, and it can be simplified as follows: This is a DSB-FC waveform having

46、 a double-sided magnitude spectrum given in Fig3.5. For the FM signal: determine the amplitudes of its major spectral components and draw the corresponding double-sided magnitude spectrum.SolutionIn the considered case, we have: Using an expansion for a wide-band FM signal where the message is a sin

47、gle tone sinusoid, we can find the significant spectral components. After expanding, the modulated signal is expressed as: Utilising the tables of Bessel functions of the First Kind, Jn(mf), we get:J 0(2) = 0.22 J1 (2) = 0.58 J 2(2) = 0.35 J3 (2) = 0.13 J4(2) = 0.03Therefore, there are 9 major spect

48、ral components in the magnitude spectrum of the considered signal. 6. Calculate the bandwidth requirements for the FM signal described by:In your calculations use:A）analysis of the magnitude spectrum, B）Carson's ruleSolutionThe modulated signal is a wide-band FM signal, with the modulation index

49、 equal toUtilising the tables of Bessel functions of the First Kind, Jn(mf), we get: From the listed values of Jn(5), we can see that there are 8 side-bands on each side of the carrier, and that the bandwidth required to transmit the signal isB = 2 x 8 x fm = 16 x 104 = 160 kHzUtilising Carson's

50、 rule we have:the upper bound on the bandwidth Bu = 2(mf +1) fm = 120 kHzthe lower bound on the bandwidth Bi = 2AF = 2mffm = 100 kHzFrom the results, we can see that there is a discrepancy between even the upper bound obtained from Carson's rule and the result obtained by analysing the magnitude

51、 spectrum. One, however, needs to take to account that Carson's rule caters for the signal power, not magnitude. Hence, the two last sidebands are really insignificant.7. An FM signal has sinusoidal modulation with a frequency fm = 15kHz and peak frequency deviation of 22.5 kHz. The carrier freq

52、uency is fc = 80 MHz and carrier amplitude is A = 5V. Determine the amplitudes of all the spectral components of the FM signal and draw its double-sided magnitude spectrum.SolutionThe same method as in case of Problem 5 is to be followed. Becausethe FM signal is a wide-band FM signal, and the tables

53、 of Bessel functions must beused.Tut71.A source has an alphabet ab a2, a3, a4, a5, with corresponding probabilities 0.1, 0.2, 0.3, 0.05, 0.15, 0.2. Find the entropy of this source. Compare this entropy with the entropy of a uniformly distributed source with the same alphabet.Solution：1. The entropy

54、for this source is If the source symbols are equi probable, then pi =1/6 and2.For a binary source:（a）Show that the entropy H is a maximum when the probability of sending a binary 1 is equal to the probability of sending a binary 0.（b）Fid the value of maximum entropy. Solution：(a)f(x)=-xlog2X-(1-x)lo

55、g2 (1-x) y(x)=ln2f(x)=-xlnx-( 1 -x)ln( 1 -x)y' (x)=-lnx-1+ln( 1-x)+1 =ln( 1 -x)/x=0 (1-x)/x=1 x=0.5 p1=p2=0.5(b) Hmax=1 bit.3. A single-digit, seven-segment LCD emits a 0 with probability 0.25; a 1 and 2 with probability of 0.15 each; 3, 4, 5, 6, 7 and 8 with probability 0.07 each; and a 9 with

56、a probability of 0.03. Find the average information of the source. Solution：H=-0.25log20.25-2x0.15log20.15-6x0.07log20.07-0.03log20.03=3.084 bits.4. A numerical keypad has the digits 0 to 9. Assume that the probability of sending any one digit is the same as that for sending any of the other digits.

57、 Calculate how often the buttons must be pressed in order to send out information at a rate of 2 bits/s.Solution：T=log210/2=1.661 s5.An analog telephone line has an SNR of 45 dB and passes audio frequencies over the range 300 to 3200 Hz. A modem is to be designed to transmit and receive data simultaneously (i.e. full duplex) over this line without errors.(a)If the frequency range 300 to 1200 Hz is to be used for the transmitted signal, what is the maximum transmitted data rate?（b）If the frequency rang