信号与系统奥本海姆英文版课后答案chapter5

1、Chapter 5 Answers5.1 (a) let xn= un-1.Using the Fourier transform analysis equation (5.9).the Fourier transform of this signal is = = =(b) Let xn=.Using the Fourier transform analysis equation (5.9).The Fourier transform of signal is=The second summation in the righthand side of the above equation i

2、s exactly the same as result of part (a).Now ,=(1/2)Therefore=(1/2)+ =5.2 (a) let . Using the Fourier transform analysis equation (5.9).the Fourier transform of this signal is=+= (b) Let .using the Fourier transform analysis equation (5.9). the Fourier transform of this signal is=-=5.3 We note from

3、section 5.2 that a periodic signal with Fourier series representation xn= has a Fourier transform =( a ) Consider the signal .We note that the fundamental period of the signal is N=6. The signal may be written as = = Form this , we obtain the non-zero Fourier series coefficients of the range as Ther

4、efore , in the range ,we obtain (b) consider the signal .we note that the fundamental period of the signal is N=12.the signal maybe written as Form this ,we obtain the non-zero Fourier series coefficients of in the range as Therefore ,in the range ,we obtain 5.4 (a)Using the Fourier transform synthe

5、sis equation (5.8) (b)Using the transform synthesis equation (5.8)5.5 From the given information =The signal xnis zero when is a nonzero integer multiple of or when |n| .the value of can never be such that it is a nonzero integer multiple of .Therefore .xn=0 only for n=5.6 Throughout this problem, w

6、e assume thatXn (a) Using the time reversal property (Sec.5.3.6),we haveUsing the time shift property (Sec.5.3.3) on this .we have and Therefore (b) Using the time reversal property (Sec.5.3.6) ,we haveUsing the same conjugation property on this ,we have Therefore (c) Using the differentiation frequ

7、ency property (Sec.5.3.8),we haveUsing the same property second time ,Therefore5.7 (a) Consider the signal with Fourier transform We see that is real and odd .From Table 5.1 , we know that the Fourier transform of a real and odd signal is purely imaginary and odd. Therefore ,we may say that the Four

8、ier transform of a purely imaginary and odd signal is real and odd. Using this observation, we conclude that is purely imaginary and oddNote now thatTherefore , .therefore , is also purely imaginary .but is neither even nor odd(b)We note that is purely imaginary and odd. Therefore, has to be real an

9、d odd.(d) Consider a signal whose magnitude of the Fourier transform is and whose phase of the Fourier transform is .since and ,we may conclude that the signal is real (see Table 5.1,property5.3.4).Now, consider the signal with Fourier transform .Using the result from previous paragraph and the line

10、arity property of the Fourier transform .we may conclude that has to real .since the Fourier transform ,we may conclude that has to real . since the Fourier transform is neither purely imaginary nor purely real .the signal is neither even nor odd5.8 Consider the signal From the table 5.2, we know th

11、atUsing the accumulation property (Table 5.1，Property 5.3.5),we haveTherefore , in the range ,Also, in the range ,Therefore , in the range ,The signal xn has the desired Fourier transform .We may express xn mathematically as = 5.9 From property 5.3.4 in Table 5.1 , we know that for a real signal xn,

12、From the given information Therefore,We also know thatAnd that xn=0 for n0. therefore for n0.we conclude that x0=1Therefore 5.10 From table 5.2 we know that Using property 5.3.8 in table 5.1,Therefore , 5.11 We know from the time expansion property (Table 5.1, Property 5.3.7) that Therefore, is obta

13、ined by compressing by a factor of 2. Since we know that is periodic with a periodic of , we my conclude that has a periodic which is (1/2)=. Therefore,.5.12 Consider the signal For Table 5.2, we obtain the Fourier transform of x1n to be The plot of is as shown in Figure S5.12. Now consider the sign

14、al x1n=( x2n)2. Using the multiplication property (Table 5.1, Property 5.5), we obtain the Fourier transform of x2n to be This is plotted in Figure S5.12. From Figure S5.12. It is clear that is zero for . By using the convolution property (Table 5.1, Property 5.4), we know thatThe plot of is shown i

15、n Figure S5.12. It is clear that of then .5.13 When two LTI systems connected in parallel, the impulse response of the overall system is the sum of the impulse response of the individual. Therefore, hn= h1n+ h2nusing the linearity property (Table 5.1, Property 5.3.2) Given that h1n=(1/2)nun, we obta

16、inTherefore, Taking the inverse Fourier transform, h2n=-2(1/4)nun.5.14 From the given information, we have the Fourier transformof gn to be Also, when the input to the system is xn=(1/4)nun, the output is gn. Therefore,For Table 5.2, we obtainTherefore,Clearly, hn is a three point sequence.We have a

17、ndWe see that only if h1=0. We also haveSince we are also given that , we have h0-h2=1 (S5.14-1)Now not thatEvaluating this equation at n=2, we haveSince h1=0, (S5.14-2)Solving equation (S5.14-1) and (S5.14-2), we obtainTherefore,5.15 Consider xn=sin(wcn)/(n), the Fourier transformof xn is also show

18、n in Figure S5.15. We note that the given signal yn=xnxn. Therefore, the Fourier transformof yn is Employing the approach used in Example 5.15, we can convert the above periodic convolution into an aperiodic signal by definingThen we may writeThis is the aperiodic convolution of the rectangular puls

19、e shown in Figure S5.15 with the periodic square wave . The result of this convolution is also shown in Figure S5.15.From the figure, it is clear that we require 1+(2wc/) to be 1/2. therefore, wc=3/4.5.16 We may writewhere * denotes aperiodic convolution. We may also rewrite this as a periodic convo

20、lutionwhereand(a) Taking the inverse Fourier transform of , we get gn=(1/4)nun. therefore, a=1/4.(b) Taking the inverse Fourier transform of , we getThis signal is periodic with a fundamental periodic N=4.(c) We can easily show that is not conjugate symmetric. Therefore, xn is not real.5.17 Using th

21、e duality property, we have5.18 Knowing thatwe may use the Fourier transform analysis equation to writePutting =-2t in this equation, and replacing the variable n by the variable kBy comparing this with the continuous-time Fourier series synthesis equation, it is immediately apparent that ak=(1/3)(1

22、/2)|k| are the Fourier series coefficients of the signal 1/(5-4cos(2t).5.19 (a) Taking the Fourier transform of both sides of the difference equation, we haveTherefore,(b) Using Partial faction expansion,Using Table 5.2, and taking the inverse Fourier transform, we obtain5.20 (a) Since the LTI syste

23、m is causal and stable, a signal input-output pair is sufficient to determine the frequency response of the system. In this case, the input is xn=(4/5)nun and output is yn=n(4/5)nun. The frequency response is given by=/,Where and are the Fourier transforms of xn and yn respectively. Using Table 5.2,

24、 we haveUsing the differentiation in frequency property (Table 5.1, Property 5.3.8), we haveTherefore,(b) Since =/, we may write Taking the inverse Fourier transform of both sides5.21 (a) The given signal is xn=un-2- un-6=n-2+n-3+ n-4+n-5Using the Fourier transform analysis eq. (5.9), we obtain(b) U

25、sing the Fourier transform analysis eq. (5.9), we obtain（c）Using the Fourier transform analysis (5.9),we obtainX(ejw)=e2jw (d ) using the Fourier transform analysis eq.(5.9),we obtainX(ejw)= =-=(e) using the Fourier transform analysis eq(5.9),we obtain X(ejw)= =+（f）the given signal is xn=-3Using the

26、 Fourier transform analysis eq(5.9),we obtain(g) the given signal is therefore= -+ + 0|w| (h) the given signal is Xn= = + + 0|w|(i) xn is periodic with periodic 6. the Fourier series coefficients of xn are give byTherefore, form the result of section 5.2=)(j) using the Fourier transform analysis eq.

27、(5.9) we obtainUsing the differentiation in frequency property of the Fourier transformThereforeXn= (k)we haveIn the range , therefore, if ,then=periodic convolution of and Using the mechanics of periodic convolution demonstrated in example 5.15 ,we obtain In the range 5.22 (a) Using the Fourier tra

28、nsform analysis eq(5.8),we obtain（b）comparing the given Fourier transform analysis eq(5.8),we obtainxn= (c) Using the Fourier transform analysis eq(5.8),we obtain(d) the given Fourier transform is =Comparing the given Fourier transform with the analysis eq(5.8),we obtain（e）this is the Fourier transf

29、orm of a periodic signal with fundamental frequency Therefore its fundamental periodic is 4. also, the Fourier series coefficient of this Signal are . Therefore, the signal is given by(f) the given Fourier transform may be written as =Compare each of two terms in the right-hand side of the above equ

30、ation with the Fourier transform analysis eq. (5.9) we obtain (g) the given Fourier transform may be written as Therefore(h) the given Fourier transform may be written asCompare the given Fourier transform with the analysis eq. (5.8), we obtain5.23 (a) we have form eq.(5.9)(b) note that yn=xn+2 is a

31、n even signal. Therefore , is real and even . This Implies that .furthermore , form the time shifting property of the Fourier Transform we have .therefore, (c) we have form eq. (5.8)Therefore (d) we have form eq.(5.9)(e) from table 5.1, we have 11/2-1/2-4-71111/2-1/2-10147EvxnFigure s 5.232Therefore

32、, the desired signal is .this is as shown in figure (f) (i) from Parsevals theorem we have (ii) using the differentiation in frequency property of the Fourier transform we obtainAgain using Parsevals theorem, we obtain5.24 (1) for to be zero, the signal must be real and odd. Only signal (b) and (i)A

33、re real and odd.(2) to be zero , the signal must be real and even Only signal (b) and (h)Are real and even (3) Assume . Using the time shifting property of the Fourier transformWe have yn=xn+a , if is real, then yn is real and even (assuming that xn is real).Therefore, xn has to be symmetric about a

34、/ this is true only for signal (a) , (b), (d) , (e) , (f) , and (h).(4) since ,the given condition is satisfied only is xn=0. this is True for signal (b), (e) , (f) , (h) , and (f).(5) is always periodic with period . Therefore ,all signal satisfy this condition (6) since ,the given condition is sat

35、isfied only the samples of the signal Add up to zero. This is true for signal (b), (g) , and (i). 5.25. if the inverse Fourier transform of is xn, then AndTherefore , the inverse Fourier transform of B(w) is . Also, the inverse Fourier transform of is. Therefore, the time function corresponding to t

36、he inverse Fourier transform of will be . this is as shown in the 1-1/2-4111-1/2-101figure s 5.25 Figure s 5.255.26 (a) we may express as Therefore (b) We may express X3(ej)as . Therefore, .(c)We may express as =. (d)Using the fact H(ej) is the frequency of an idea lowpass filter with cutoff frequen

37、cy /6,we may draw X4(ej) as show in figure s5.26-/6 - 1/60-/6/6 Figure S5.265.27.(a) W(ej) will be the periodic convolution of X(ej) with P(ej).The Fourier transforms are sketched in figure S5.27. (b) The Fourier transform of Y(ej) of yn is Y(ej)= P(ej) H(ej). The LTI system with unit sample respons

38、e hn is an idea lowpass filter with cutoff Frequency /2 .Therefore, Y(ej) for each choice of pn are as shown in Figure S5.27.Therefore,yn in each case is :(i) yn=0(ii) yn= (iii) yn=(iv) yn=(v)5.28.Let . Taking the inverse Fourier transform of the above equation ,we obtain gnxn= n+n-1=yn.(a-ii)1/20-1

39、/2-0(a-i)1/20-(a-iv)0j/2-(a-iii)/2-/20(b-ii)0(a-iv)-/2/20(a-v)/2-/2(b-v)/2-/20(b-iv)/20j/2-/2(b-iii) Figure S5.27(a) If xn=(-1)n, gn=n-n-1.(b)If xn=(1/2)nun,gn has to be chosen such that gn= Therefore, there are many possible chosen for gn.5.29.(a)Let the output of the system be yn.We known that X(e

40、j)= X(ej) H(ej). In this part of the problem H(ej)= .（i）we haveTherefore Taking the inverse Fourier transform, we obtain(ii) we have Taking the inverse Fourier transform, we obtain(iii) We have ThereforeTaking the inverse Fourier transform, we obtain (b) Givenwe obtain(i)We have ThereforeWhere A=-j/

41、2(1-j), B=1/2, and C= 1/2(1+j), therefore (ii)In this case (c) Here +Therefore，+5.30 (a) the frequency response of the system is as shown in figure s5.30(b) the Fourier transform in figure s5.30 of xn is as shown in figure s5.30(i) the frequency response is as shown in figure s5.30. therefore , yn=s

42、in(n/8)(ii)the frequency response is as shown in figure s5.30. therefore , yn=2sin(n/8)-cos(n/4)(iii) the frequency response is as shown in figure s5.30. therefore , yn=1/6sin(n/8)-1/4 cos(n/4)-/61b-i-wn/j-/8-/4/4w/4w/4/8w-n/j-/3/311/2jb-iv-/2/6/21/bb-iii-/6/2/212-/2wFigure S5.30 (iv)The frequency r

43、esponse H(ej) is as show in Figure S5.30. Therefore, yn=-sin(n/4).(c) The frequency response H(ej) is as show in Figure S5.30.(i) The signal xn is periodic with period 8.The Fourier series coefficients of the signal are ak= The Fourier transform of this signal is X(ej)=. The Fourier transform Y(ej)

44、of the output is Y(ej)=X(ej) H(ej).Therefore,Y(ej)= In the range . Therefore, yn=.(ii) The signal xn is periodic with periodic 8.The Fourier series coefficients of the signal are The Fourier transform of this signal is The Fourier transform Y(ej) of the output is Y(ej) =X(ej) H(ej).Therefore, Y(ej)=

45、In the range .Therefore, yn=.(iii) Again in this case, the Fourier transform X(ej) of the signal xn is of the form show in part(i).Therefore,yn=.(iv) in this case, the output is yn=hn*xn=5.31.(a) Form the given information, it is clear that when the input to the system is a complex exponential of fr

46、equency 0 ,the output is a complex exponential of the same frequency but scaled by the |0|.Therefore,the frequency response of the system is H(ej)= |, for .(b)Taking the inverse Fourier transform of the frequency response ,we obtain hn= 5.32 From the synthesis equation (5.8) we have Also, since we h

47、ave therefore ,the question here amounts to asking whether it is true that since and are causal , this is indeed true.5.33 (a) Taking the Fourier transform of the given difference equation we have (b) The Fourier transform of the output will be .(i) In this case . Therefore Taking the inverse transform, we obtain . (ii) In this case .Therefore , Taking the inverse Fourier transform , we obtain , (iii) In this case .Therefore .Taking the inverse Fourier transform , we have . (iv) In this case .The