信号与系统奥本海姆英文版课后答案chapter1

1、Signals & Systems(Second Edition)Learning Instructions (Exercises Answers)Department of Computer Engineering 2005.12ContentsChapter 1 ···························

2、···························· 2Chapter 2 ····················

3、··································· 17Chapter 3 ·············

4、;·········································· 53Chapter 4 ·····

5、3;················································· 80C

6、hapter 5 ················································

7、83;······ 101Chapter 6 ·········································&

8、#183;············· 127Chapter 7 ··································

9、;····················· 137Chapter 8 ··························

10、83;···························· 150Chapter 9 ···················&

11、#183;··································· 158Chapter 10 ···········

12、3;···········································178Chapter 1 Answers1.1 Converting from

13、polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coordinates:, , , , , 1.3. (a) =, =0, because (b) , .Therefore, =, = (c) =cos(t). Therefore, =,= (d) , . Therefore, = =0，because <. (e) =, =1. therefore, =, =. (f) =. Therefore, =, =1.4. (a) The signal xn is shifted by 3 to th

14、e right. The shifted signal will be zero for n<1, And n>7. (b) The signal xn is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal xn is flipped signal will be zero for n<-1 and n>2. (d) The signal xn is flipped and the flipped signal is shi

15、fted by 2 to the right. The new Signal will be zero for n<-2 and n>4. (e) The signal xn is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n<-6 and n>0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signa

16、l by 1 to the right. Therefore, x (1-t) will be zero for t>-2. (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1,Therefore, x (1-t) +x(2-t) will be zero for t>-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) wi

17、ll bezero for t<1. (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will bezero for t<9.1.6 (a) x1(t) is not periodic because it is zero for t<0. (b) x2n=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x3n is as shown in the F

18、igure S1.6.-3-141-10-4111-1n5x3nTherefore, it is periodic with a fundamental period of 4.1.7. (a)=Therefore, is zero for >3. (b) Since x1(t) is an odd signal, is zero for all values of t. (c) Therefore, is zero when <3 and when . (d) Therefore, is zero only when .1.8. (a) (b) (c) (d) 1.9. (a)

19、is a periodic complex exponential. (b) is a complex exponential multiplied by a decaying exponential. Therefore, is not periodic. （c） is a periodic signal. =. is a complex exponential with a fundamental period of . (d) is a periodic signal. The fundamental period is given by N=m() = By choosing m=3.

20、 We obtain the fundamental period to be 10.(e) is not periodic. is a complex exponential with =3/5. We cannot find any integer m such that m( ) is also an integer. Therefore, is not periodic.1.10.x(t)=2cos(10t1)-sin(4t-1)Period of first term in the RHS =.Period of first term in the RHS = .Therefore,

21、 the overall signal is periodic with a period which the least common multiple of the periods of the first and second terms. This is equal to .1.11. xn = 1+Period of first term in the RHS =1.Period of second term in the RHS =7 （when m=2）Period of second term in the RHS =5 (when m=1)Therefore, the ove

22、rall signal xn is periodic with a period which is the least commonMultiple of the periods of the three terms inn xn.This is equal to 35.1.12. The signal xn is as shown in figure S1.12. xn can be obtained by flipping un and thenShifting the flipped signal by 3 to the right. Therefore, xn=u-n+3. This

23、implies that M=-1 and no=-3.0-1-2-3123XnnFigure S 1.121.13 y(t)= =Therefore 1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14.10-1210-1t1-2g(t)2-3-3tFigure S 1.14x(t)Therefore )This implies that A=3, t=0, A=-3, and t=1.1.15 (a) The signal xn, which is the input to S, is the same

24、 as yn.Therefore , yn= xn-2+ xn-3 = yn-2+ yn-3 =2xn-2 +4xn-3 +( 2xn-3+ 4xn-4) =2xn-2+ 5xn-3 + 2xn-4The input-output relationship for S is yn=2xn-2+ 5x n-3 + 2x n-4(b) The input-output relationship does not change if the order in which Sand S are connected series reversed. . We can easily prove this

25、assuming that S follows S. In this case , the signal xn, which is the input to S is the same as yn.Therefore yn =2xn+ 4xn-1= 2yn+4 yn-1 =2( xn-2+ xn-3 )+4(xn-3+ xn-4) =2 xn-2+5xn-3+ 2 xn-4The input-output relationship for S is once again yn=2xn-2+ 5x n-3 + 2x n-41.16 (a)The system is not memory less

26、 because yn depends on past values of xn.(b)The output of the system will be yn= =0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form , k . Therefore , the system is not invertible .1.17 (a) The system is not causal because the output y(t) at

27、 some time may depend on future values of x(t). For instance , y(-)=x(0).(b) Consider two arbitrary inputs x(t)and x(t).x(t) y(t)= x(sin(t)x(t) y(t)= x(sin(t)Let x(t) be a linear combination of x(t) and x(t).That is , x(t)=a x(t)+b x(t)Where a and b are arbitrary scalars .If x(t) is the input to the

28、 given system ,then the corresponding output y(t) is y(t)= x( sin(t) =a x(sin(t)+ x(sin(t)=a y(t)+ by(t)Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs xnand xn.xn yn =xn yn =Let xn be a linear combination of xn and xn. That is :xn= axn+b xnwhere a and b are arbitrary scalars

29、. If xn is the input to the given system, then the corresponding output yn is yn= =a+b = ayn+b ynTherefore the system is linear.(b) Consider an arbitrary input xn.Let yn =be the corresponding output .Consider a second input xn obtained by shifting xn in time:xn= xn-nThe output corresponding to this

30、input is yn= = = Also note that yn- n= .Therefore , yn= yn- nThis implies that the system is time-invariant.(c) If <B, then yn(2 n+1)B.Therefore ,C(2 n+1)B.1.19 (a) (i) Consider two arbitrary inputs x(t) and x(t). x(t) y(t)= tx(t-1) x(t) y(t)= tx(t-1)Let x(t) be a linear combination of x(t) and x

31、(t).That is x(t)=a x(t)+b x(t)where a and b are arbitrary scalars. If x(t) is the input to the given system, then the corresponding output y(t) is y(t)= tx (t-1)= t(ax(t-1)+b x(t-1)= ay(t)+b y(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x(t).Let y(t)= tx(t-1)be the correspon

32、ding output .Consider a second input x(t) obtained by shifting x(t) in time:x(t)= x(t-t)The output corresponding to this input is y(t)= tx(t-1)= tx(t- 1- t)Also note that y(t-t)= (t-t)x(t- 1- t) y(t)Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs xnand xn. xn yn = xn

33、-2 xn yn = xn-2.Let x(t) be a linear combination of xnand xn.That is xn= axn+b xnwhere a and b are arbitrary scalars. If xn is the input to the given system, then the corresponding output yn is yn = xn-2=(a xn-2 +b xn-2) =axn-2+bxn-2+2ab xn-2 xn-2 ayn+b ynTherefore the system is not linear.(ii) Cons

34、ider an arbitrary input xn. Let yn = xn-2 be the corresponding output .Consider a second input xn obtained by shifting xn in time:xn= xn- nThe output corresponding to this input is yn = xn-2.= xn-2- nAlso note that yn- n= xn-2- nTherefore , yn= yn- nThis implies that the system is time-invariant.(c)

35、 (i) Consider two arbitrary inputs xnand xn.xn yn = xn+1- xn-1xn yn = xn+1 - xn -1Let xn be a linear combination of xn and xn. That is :xn= axn+b xnwhere a and b are arbitrary scalars. If xn is the input to the given system, then the corresponding output yn is yn= xn+1- xn-1=a xn+1+b xn +1-a xn-1-b

36、xn -1 =a(xn+1- xn-1)+b(xn +1- xn -1) = ayn+b ynTherefore the system is linear.(ii) Consider an arbitrary input xn.Let yn= xn+1- xn-1be the corresponding output .Consider a second input xn obtained by shifting xn in time: xn= xn-nThe output corresponding to this input is yn= xn +1- xn -1= xn+1- n- xn

37、-1- nAlso note that yn-n= xn+1- n- xn-1- nTherefore , yn= yn-nThis implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x(t) and x(t).x(t) y(t)= x(t) y(t)= Let x(t) be a linear combination of x(t) and x(t).That is x(t)=a x(t)+b x(t)where a and b are arbitrary scalars. If x

38、(t) is the input to the given system, then the corresponding output y(t) is y(t)= = =a+b= ay(t)+b y(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x(t).Lety(t)= =be the corresponding output .Consider a second input x(t) obtained by shifting x(t) in time:x(t)= x(t-t)The output cor

39、responding to this input is y(t)= =Also note that y(t-t)= y(t)Therefore the system is not time-invariant.1.20 (a) Given x= y(t)=x= y(t)=Since the system liner ) =1/2(+)Therefore (t)=cos(2t)=cos(3t)(b) we know that (t)=cos(2(t-1/2)= (+)/2 Using the linearity property, we may once again write(t)=( +)

40、=(+)= cos(3t-1)Therefore,(t)=cos(2(t-1/2) =cos(3t-1)1.21.The signals are sketched in figure S1.21.t-10-112213x(t-1)10-112tx(2t+1)t4-1321012x(2-t)0.50.5t3/2-3/2t102x(4-t/2)t1012618412Figure S1.211.22 The signals are sketched in figure S1.22a11/2-1/2-1n73210x3- n1.23 The even and odd parts are sketche

41、d in Figure S1.2311/2-1/2-1n73210xn-4(b)1-12n0x3n+1(d)-111/2-1/2n210x3n(c)2112n0xnun-3=xn(f)-4-1-211/2n20x3- n/2 +(-1)nxn/2(g)011n2(h)Figure S1.22-1/2x0(t)1/2-1-2021t-201/22tx0(t)(a)t-1/2x0(t)1/2-1-2021t1-1/2x0(t)1/2-1-2021(b)x0(t)-t/20tt3t/2-3t/20x0(t)(c)Figure S1.237n10-7xon10-1/2-77-1/2nxn(a)1/2n

42、1/2-771n(n)31/2071/2-1nxon1(b)03/2-3/2-1/24n1/2xon3/251-5nxen(c)Figure S1.241.24 The even and odd parts are sketched in Figure S1.241.25 (a) periodic period=2/(4)= /2(b) periodic period=2/(4)= 2 (c) x(t)=1+cos(4t-2/3)/2. periodic period=2/(4)= /2 (d) x(t)=cos(4t)/2. periodic period=2/(4)= 1/2 (e) x(

43、t)=sin(4t)u(t)-sin(4t)u(-t)/2. Not period.(f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) xn=(1/2)cos(3n/4+cos(n/4). periodic, period=8.(e) periodic, period=16.1.27 (a) Linear, stable(b) Not period.(c) Linear(d) Linear, causal, stable(e) Time invariant, linear, c

44、ausal, stable(f) Linear, stable(g) Time invariant, linear, causal1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable(c)Memoryless, linear, causal(d) Linear, stable(e) Linear, stable(f) Memoryless, linear, causal, stable(g) Linear, stable1.29 (a) Consider two inputs to the system such t

45、hatand Now consider a third input n= n+n. The corresponding system outputWill be therefore, we may conclude that the system is additiveLet us now assume that inputs to the system such thatandNow consider a third input x3 n= x2 n+ x1 n. The corresponding system outputWill betherefore, we may conclude

46、 that the system is additive(b) (i) Consider two inputs to the system such that and Now consider a third input t= t+t. The corresponding system outputWill betherefore, we may conclude that the system is not additiveNow consider a third input x4 t= a x1 t. The corresponding system outputWill beTheref

47、ore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let n=2n+2+2n+1+2n and n= n+1+ 2n+1+ 3n. The corresponding outputs evaluated at n=0 areNow consider a third input x3 n= x2 n+ x1 n.= 3n+2+4n+1+5nThe corresponding outputs evaluated at n=0 is y30=15/4. Gnar

48、ly, y30 .This Therefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x1(t)=x(t)+2give the same output(c) n and 2n give the same outputd) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n0 and

49、yn=xn for n<0(f) Non invertible. x(n) and x(n) give the same result(g)Invertible. Inverse system y(n)=x(1-n)(h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system y(n) = x(n)-(1/2)xn-1(j) Non invertible. If x(t) is any constant, then y(t)=0(k) n and 2n result in yn=0(l) Invert

50、ible. Inverse system: y(t)=x(t/2)(m) Non invertible x1 n= n+ n-1and x2 n= n give yn= n(n) Invertible. Inverse system: yn=x2n1.31 (a) Note that x2t= x1 t- x1 t-2. Therefore, using linearity we get y2 (t)=y1 (t)- y1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 t+ x1 t+1. .Therefore, usin

51、g linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is shown in Figure S1.31y2 (t)y3 (t)02-1t-22024tFigure S1.311.32 All statements are true(1) x(t) periodic with period T; y1 (t) periodic, period T/2(2) y1 (t) periodic, period T; bx(t) periodic, period2T(3) x(t) periodic, period T; y2 (t) periodic, p

52、eriod2T(4) y2(t) periodic, period T; x(t) periodic, period T/2;1.33(1) True xn=xn+N; y1 (n)= y1 (n+ N0)i.e. periodic with N0=n/2if N is even and with period N0=n if N is odd.(2)False. y1 n periodic does no imply xn is periodic i.e. Let xn = gn+hn where Then y1 n = x 2n is periodic but xn is clearly

53、not periodic.(3)True. x n+N =xn; y2 n+N0 =y2 n where N0=2N(4) True. y2 n+N =y2 n; y2 n+N0 =y2 n where N0=N/21.34. (a) Consider If xn is odd, xn +x -n =0. Therefore, the given summation evaluates to zero.(b) Let yn =x1nx2n .Then y -n =x1-n x2-n =-x1nx2n =-yn.This implies that yn is odd.(c)Consider Us

54、ing the result of part (b), we know that xenxon is an odd signal .Therefore, using the result of part (a) we may conclude that Therefore,(d)Consider Again, since xe (t) xo (t) is odd,Therefore,1.35. We want to find the smallest N0 such that m(2 /N) N0 =2k or N0 =kN/m, where k is an integer, then N m

55、ust be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N0, then m/k should be the GCD of m and N. Therefore, N0=N/gcd(m,N).1.36(a)If xn is periodic This implies that a rational number . (b)T/T0 =p/q then xn = ,T

56、he fundamental period is q/gcd(p,q) and the fundmental frequency is (c) p/gcd(p,q) periods of x(t) are needed .1.37.(a) From the definition of We have (b) Note from part(a) that This implies that is even .Therefore,the odd part of is zero.(c) Here, and1.38.(a) We know that ThereforeThis implies that

57、 (b)The plot are as shown in Figure s3.18.1.39 We have Also, u'（t）11/2/2-/2t0tu'（t）12t0tu'（t）11/2-t0tu'（t）11/2-t0t11-1/2e-t/u'（t）1-0tu'（t）11/2-01/2e-t/tFigure s3.18We have Therefore,1.40.(a) If a system is additive ,then also, if a system is homogeneous,then(b) y(t)=x2(t) is such a systerm .(c) No.For example,consider y(t) with Then x(t)=0for t>1,but y(t)=1 for t>1.1.41. (a) yn=2xn.Therefore, the system is time invariant.