信号与系统奥本海姆英文版课后答案chapter8.pdf

Chapter 8 Answers 8 1 Using Table 4 1 take the inverse Fourier transform of c Y j This gives 2 c jw t y tx t e Therefore 2 c jw t m te 8 2 a The Fourier transform Y jw of y t is given by c Y jwXt j Clearly is just a shifted version of Y jw X jw Therefore x t may be recovered from y t simply by multiplying y t by There is no constraint that needs to be placed on c jw t e c to ensure that x t is recoverable from y t b We know that 1 Re cos c y ty tx tt The Fourier transform of 1 Y jw 1 y t is as shown in Figure S8 2 1 11 22 cc Y jwX jX j 1000 c 1000 c 0 c c c X j c X j Y j Figure S8 2 If we want to prevent the two shifted replicas of from multiplied by cos Y jw2000 t the output will be 1 1 cos 2000 sin 2000 cos 2000 sin 4000 2 x tg ttx tttx tt The Fourier transform of this signal is 1 11 4000 4000 44 XjwX jX j jj This implies that 1 Xjw is zero for 2000 When is passed through a lowpass fiter with cutoff frequency y t 2000 the output will clearly be zero Therefore y t 0 8 4 Consider the signal 3 sin 400 2sin 400 y tg ttt 23 sin 200 sin 400 2sin 400 sin 200 1 cos 800 22sin 400 1 cos 800 2 1 2 sin 200 1 4 sin 1000 sin 600 sin 400 1 2 sin 1200 sin 400 ttt tttt ttt ttt If this signal is passed through a lowpass filter with cutoff frequency 400 then the output will be 1 sin 200 yt 8 5 The signal is as shown in Figure S8 5 x t x t tt Envelop of A t 0 150 Figure S8 5 The envelope of the signal t is as shown in Figure S8 5 Clearly is we want to use asynchronous demodulation to recover the signal x t we need to ensure that A is greater than the height of the highest sidelobe see Figure S8 5 Let us now determine the height of the highest sidelobe The first zero crossing of the signal h x t occurs at time such that o t 00 1000 1 1000 tt Similarly the second zero crossing happens at time such that 1 t 11 10002 2 1000 tt The highest sidelobe occurs at time 01 tt2 that is at time 2 3 2000t At this time the amplitude of x tis 2 sin 3 2 2000 3 20003 x t Therefore A should at least be 2000 3 The modulation index corresponding to the smallest permissible value of A is Maxvalue m Min possible of value x t of 1 A 10003 2000 32 8 6 Let us denote the Fourier transform of sin c t by This will be rectangular pulse which is nonzero only in the range H jw c Taking The Fourier transform of the first equation given in the problem we have cos cos cos 1 1 2 1 cc c cc G jFT x ttFT x ttH jw FT x ttH jw X jX jH jw G j is as shown in Figure S8 6 The Fourier transform of cos c g tt is also shown in Figure S8 6 Clearly if we want to recover x tfrom cos c g tt then we have to pass cos c g tt through an ideal lowpass filter with gain 4 and cutoff frequency M Therefore A 4 8 7 In Figure S8 7 we show X j G j and Q j We also show a polt of The Fourier transform of cos c g tt then we need to ensure that 1 0 2 c and 2 an ideal lowpass filter with passband gain of 2 and a cutoff frequency of c is used to filter 0 cos g tt c 1 2 Q j 0 1 2 c c 0 c M 0 1 2 1 2 M X j 1 G j 2 c Figure S8 6 M 2 c c cosFT g t 1 4 1 4 1 4 M 0 X j G j 1 2 1 2 1 0 M 0 151 0 cosFT g tt 0 0c 0c 0 0c 0 0c Figure S8 7 8 8 a From Figure S8 8 it is clear that Y j is conjugate symmetric Therefore y t is real b This part of the problem explores the demodulation of SSB signals through synchronous demodulation This idea is explored in more detail in problem 8 29 Let us assume that we use the synchronous demodulation system shown in the Figure S8 8 The Fourier Transform 1 Y j of the signal 1 y t c is shown in the Figure S8 8 Clearly if we use an ideal lowpass filter with cutoff frequency and passhand gain of 2 we should recover the original signal x t Therefore 2sin sin c c t x ty tt t X j A 8 9 Let the signal 1 x t and 2 x thave Fourier Transform 1 Xj and 2 Xj as shown in the Figure S8 9 When SSB modulation is performed on the signals 1 x t and 2 x t we would obtain the signal 1 y t and 2 y t respectively The Fourier Transform 1 Y j and 2 Yj of these signals would be as shown in the Figure S8 9 see Section 8 4 for details a From the figure it is clear that the signal 12 y ty ty t w ould have a Fourier Transform Y j which is as shown in the Figure S8 9 Figure S8 8 2Wc2Wc M M Aj Wc Wc Aj 2 Aj 2 Wc Aj 2Aj 2 c Wc M M 0 1 2 0 M 0 M X jH j FT x t h t coswc 0 Wc Aj X jw A 2 FT y t sinwct 152 From this figure it is obvious that Y j is zero for 2 c b In order to obtain 1 x t from y t we have to first remove any contribution in y t from 2 x t From the previously draw figures it is clear that we can remove all contribution to y t from 2 x t by first lowpass filtering y t using a lowpass filter with cutoff frequency c We may then follow this by a synchronous demodulation system This idea is illustrated in the Figure S8 9 Therefore 10 sinsin cos 2 cc tA x ty tt t t In order to determine the value of the gain A we first plot the Fourier Transform of 10 sin cos ct x ty t t t From this it is clear that A 4 8 10 a From Section 8 5 we know that in order to avoid aliasing 2 2 M T where M is the Maximum frequency in the original signal and T is the period of In this case Therefore c t 3 10T 1000 M Therefore 0X j for 1000 b For Figure 8 24 we know that the Fourier Transform Y j of the signal consists of shifted replicas of y t X j The replica of X j centered around 0 is scaled by where T is the width of each pulse of By using a lowpass filter we may recover c t X j from Y j The lowpass filter needs to have a passband gain of T In this case this evaluates to 33 10 10 0 254 8 11This signal is c t 22 1122 cccc jw tjw tj w tj w t c ta ea ea ea e L Since is real The Fourier Transform c t k aa k Y j of the signal y tx t c t is 1122 2 2 ccc Y ja X ja X ja X ja X j L c This is plotted in Figure S8 11 FDM Signal FT 1 2 2 c 2 c c c 1 2 1 2 c c c 2 c 2 c 1 4 2 c c 2 c c sin cos ct ct FTy t t Demodulator x t c c c c 0 0 cos ct y t 1 1 2 zj c c c c 1 zj 153 z j z j a The Fourier transform G jw of g t is 11 Cc G Ja X Jaj This is as shown in Figure s8 11 clearly by comparing G J and Y JW we know That g t may be obtained from y t by passing Y t through an ideal bandpass filter Which has a passband gain of unity in the range c 2 3 2 b if 11 j a aaethe n 11 j a aaeAlso 11 11 11 1 11 2 cos 2 cc j tj t c jtajta cc g ta ea eta aex tae atata therefore Aaanda in this case 2000 m Therefore 3 0 5 10sec T we need to how 10 different pulses within a duration of T Therefore each pulled can be at most 4 0 5 10secwide 8 12 a we know that 1 0 2 pp jd Therefore 1 1 2 1 2 11 0 cos 222 T T T pd b since p j statistics eq 8 28 we know that it must have zero crossings every 1 T Therefore 1 8 13 given 0 1 2 p kTfork cos cos cm y ttmt cos cos cos sin sin cos cmc tmttmt m But since cmand 2 m 0 22 ccmc mm y jw jj m z j zj 0c Figure s8 11 1 2 c zja 0 c 2 2 c 154 8 14 when a signal x t is amplitude modulated with 0 jn e then the Fourier transform OF the result is 0 1 jj y ex e 8 15 when a signal x t is amplitude modulated with cos 0n then the Fourier transform OF the result is 00 1 11 22 jjj y ex ex e 12 j y ey e j only when 0 is either 0 or 8 16 we know that c n 5 sin sin 22 nn 22 11 22 jj y jx ex e jj This is as shown in the figure s8 16 From the figure it is obvious that 35 0 0 88 y jfo rand 8 17 the Fourier transforms jjj x eG eandy eare shown in figure s8 17 Figure s8 16 Figure s8 17 8 j z e 0 8 1 2 j 1 2j j z e 2 0 2 3 8 5 8 0 1 j z e 0 2 2 j a e 0 1 4 4 j Q e 2 2 0 j z e 2 4 j x e 0 4 2 155 2 0 2 j z e 1 1 2 2 1 2 cos 2 n FT z n 0 1 2 sin 2 n FT z nh n 8 18 the Fourier transforms j x e and j y e are as shown in figure s8 18 from this figures it is clear that we wish to accomplish single sideband modulation using the system in particular we are interest in retaining the upper sidebands of the signal note that in figure 8 21 of section8 4 is shown a continues time signal sideband system for retaining the lower sidebands In this section it was also mentioned see eq 8 21 that in order to retain the upper sidebands the frequency response of the filter used in the system had to be charge to 0 0 j j H j let us now show that this does indeed give us the desired output we redraw the system give in the problem with appropriate labels for the intermediate outputs The fourier transforms of this intermediate outputs are shown in figure s8 18 from figure s8 18 it is clear that the choice of j H e was appropriate 8 19 since 10 different signals have to be squeezed in within a bandwidth of 2 each signal is allowed to occupy a bandwidth of 2 105 after sinusoidal modulation therefore before from the figure it is obvious that 0 0 2 y jfor sinusoidal modulation each signal can occupy only a bandwidth of 10 the Fourier transform j i y e of the signal obtained by upsampling i x n by a factor of N can be nonzero in the range only for 20 therefore n has to be at least 20 8 20 note that by choosing p n we would be eble to getand at the output of the multipliers furthermore note that 2 k nk 1 v e 1 v n 2 1v n 2 2 jj v e and 2 22 jj e j v ev e this is illustrated in figure s8 20 therefore the output of the two branches will be as shown in figure s 8 20