Signals & Systems信号与系统奥本海姆英文版 课后全解答案.doc

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1、Signals & Systems(Second Edition)Learning InstructionsContentsChapter 1 2Chapter 2 17Chapter 3 53Chapter 4 80Chapter 5 101Chapter 6 127Chapter 7 137Chapter 8 150Chapter 9 158Chapter 10 178Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coor

2、dinates:, , , , , 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 the right. The shifted signal will be zero for n7. (b) The signal xn is shifted by

3、4 to the left. The shifted signal will be zero for n0. (c) The signal xn is flipped signal will be zero for n2. (d) The signal xn is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n4. (e) The signal xn is flipped and the flipped and the flipped signal is

4、 shifted by 2 to the left. This new signal will be zero for n0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal 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

5、-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) will bezero for t1. (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will bezero for t9.1.6 (a) x1(t) is not periodic because it is zero for t

6、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) is a periodic complex exponential. (b) is a complex exponential multiplied by a decaying exponential. Therefore, is not periodic.

7、 （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. 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

8、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, the overall signal is periodic with a period which the least common multiple of the periods of the first and second terms. This

9、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 overall signal xn is periodic with a period which is the least commonMultiple of the periods of the three terms inn xn.This is equal

10、 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 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 show

11、n 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 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

12、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 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+

13、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 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 conclud

14、e 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 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)

15、= 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 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

16、 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. 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.(

17、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 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

18、+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(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 o

19、utput 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 corresponding 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

20、)= 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-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 in

21、put 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) Consider an arbitrary input xn. Let yn = xn-2 be the corresponding output .Consider a second input xn obtained by shifting xn in time:xn=

22、 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) (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 i

23、s :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 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-1b

24、e 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-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 arb

25、itrary 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(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.

26、(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 corresponding 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(

27、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)=( +) =(+)= 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)t4-1

28、321012x(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 sketched 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

29、)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)1/2n1/2-771n(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

30、.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(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/

31、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, causal, stable(f) Linear, stable(g) Time invariant, linear, causal1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable(c)

32、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 thatand Now consider a third input n= n+n. The corresponding system outputWill be therefore, we may conclude that the system is additi

33、veLet 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 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 correspond

34、ing 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 beTherefore, 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+ 3

35、n. 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. Gnarly, 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)

36、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 yn=xn for n0(f) Non invertible. x(n) and x(n) give the same result(g)Invertible. Inverse system y(n)=x(1-n)(h) Invertible. Inverse sy

37、stem 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) Invertible. 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, using linearity we get Y3 (t)