奥本海姆《信号与系统》第二版信号与系统答案.pdf

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1、 Signals & Systems (Second Edition) Learning Instructions (Exercises Answers)(Exercises Answers) Department of Computer Engineering 2005.12 1 Contents Chapter 1 2 Chapter 2 17 Chapter 3 35 Chapter 4 62 Chapter 5 83 Chapter 6 109 Chapter 7 119 Chapter 8 132 Chapter 9 140 Chapter 10 160 2 Chapter 1 An

2、swers 1.1 Converting from polar to Cartesian coordinates: 111cos222je 111c o s ()222je 2cos()sin()22jjje 2c o s ()s i n ()22jjje 522jjjee 42 ( c o s ()s i n () )1442jjje 944122jjjee 944122jjjee 412jje 1.2 converting from Cartesian to polar coordinates: 055je, 22je , 233jje 21322jje, 412jje, 2221jje

3、4(1)jje, 411jje 122213jje 1.3. (a) E=4014tdte, P=0, because E (b) (2)42( )jttxe, 2( )1tx.Therefore, E=22( )dttx=dt=, P=211limlim222( )TTTTTTdtdtTTtxlim1 1T (c) 2( ) tx=cos(t). Therefore, E=23( )dttx=2cos( )dtt=, P=2111(2 )1limlim2222cos( )TTTTTTCOStdtdtTTt (d) 1 12nnu nx, 2 11 4nu nnx. Therefore, E=

4、204131 4nnnx P=0，because E. (e) 2 nx=()28nje, 22 nx=1. therefore, E=22 nx=, P=211limlim1 122121 NNNNnNnNNNnx. (f) 3 nx=cos4n. Therefore, E=23 nx=2cos()4n=2cos()4n, P=1limcos214nNNnNN1 cos()112lim()2122NNnNnN 1.4. (a) The signal xn is shifted by 3 to the right. The shifted signal will be zero for n7.

5、 (b) The signal xn is shifted by 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

6、flipped and the flipped signal is 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-

7、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) will be zero for t1. 3 (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be zero for t9. 1.6 (a) x1(t)

8、is not periodic because it is zero for t3. (b) Since x1(t) is an odd signal, 2 vnx is zero for all values of t. (c) 11311 33221122vnnnnnu nunxxx Therefore, 3 vnx is zero when n3 and when n. (d) 1554411( )( )()(2)(2)22vtttttu tutxxxee Therefore, 4( )vtxis zero only when t. 1.8. (a) 01( )22cos(0)tttxe

9、 (b) 02( )2cos()cos(32 )cos(3 )cos(30)4tttttxe (c) 3( )sin(3)sin(3)2tttttxee (d) 224( )sin(100 )sin(100)cos(100)2tttttttxeee 1.9. (a) 1( ) tx is a periodic complex exponential. 101021( )jtjttjxee (b) 2( ) tx is a complex exponential multiplied by a decaying exponential. Therefore, 2( ) tx is not per

10、iodic. （c）3 nx is a periodic signal. 3 nx=7jne=j ne. 3 nx is a complex exponential with a fundamental period of 22. (d) 4 nx is a periodic signal. The fundamental period is given by N=m(23 /5) =10().3m By choosing m=3. We obtain the fundamental period to be 10. (e) 5 nx is not periodic. 5 nx is a co

11、mplex exponential with 0w=3/5. We cannot find any integer m such that m(02w ) is also an integer. Therefore, 5 nx is not periodic. 1.10. x(t)=2cos(10t1)-sin(4t-1) Period of first term in the RHS =2105. Period of first term in the RHS =242 . Therefore, the overall signal is periodic with a period whi

12、ch the least common multiple of the periods of the first and second terms. This is equal to . -3 -1 41-1 0-41 1 1 -1 n 5 x3n 4 0 -1 -2 -3 1 2 3 Xn n Figure S 1.12 1 0 -1 2 1 0 -1 t 1 -2 g(t) 2 -3 -3 t Figure S 1.14 x(t) 1.11. xn = 1+74jne25jne Period of first term in the RHS =1. Period of second ter

13、m in the RHS =7/42=7 （when m=2） Period of second term in the RHS =5/22=5 (when m=1) Therefore, the overall signal xn is periodic with a period which is the least common Multiple 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 o

14、btained by flipping un and then Shifting the flipped signal by 3 to the right. Therefore, xn=u-n+3. This implies that M=-1 and no=-3. 1.13 y(t)= tdtx)( =dtt)2()2(=2, 022, 12, 0,ttt Therefore 224dtE 1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14. Therefore kkktkttg12(3)2(3)()

15、This implies that A1=3, t1=0, A2=-3, and t2=1. 1.15 (a) The signal x2n, which is the input to S2, is the same as y1n.Therefore , y2n= x2n-2+21 x2n-3 = y1n-2+ 21 y1n-3 =2x1n-2 +4x1n-3 +21( 2x1n-3+ 4x1n-4) =2x1n-2+ 5x1n-3 + 2x1n-4 The input-output relationship for S is yn=2xn-2+ 5x n-3 + 2x n-4 5 (b)

16、The input-output relationship does not change if the order in which S1and S2 are connected series reversed. . We can easily prove this assuming that S1 follows S2. In this case , the signal x1n, which is the input to S1 is the same as y2n. Therefore y1n =2x1n+ 4x1n-1 = 2y2n+4 y2n-1 =2( x2n-2+21 x2n-

17、3 )+4(x2n-3+21 x2n-4) =2 x2n-2+5x2n-3+ 2 x2n-4 The input-output relationship for S is once again yn=2xn-2+ 5x n-3 + 2x n-4 1.16 (a)The system is not memory less because yn depends on past values of xn. (b)The output of the system will be yn= 2nn=0 (c)From the result of part (b), we may conclude that

18、 the system output is always zero for inputs of the form kn, 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 x1(t)and x2(t). x1(t) y1

19、(t)= x1(sin(t) x2(t) y2(t)= x2(sin(t) Let x3(t) be a linear combination of x1(t) and x2(t).That is , x3(t)=a x1(t)+b x2(t) Where a and b are arbitrary scalars .If x3(t) is the input to the given system ,then the corresponding output y3(t) is y3(t)= x3( sin(t) =a x1(sin(t)+ x2(sin(t) =a y1(t)+ by2(t)

20、 Therefore , the system is linear. 1.18.(a) Consider two arbitrary inputs x1nand x2n. x1n y1n =001kxnnnnk x2n y2n =002kxnnnnk Let x3n be a linear combination of x1n and x2n. That is : x3n= ax1n+b x2n where a and b are arbitrary scalars. If x3n is the input to the given system, then the corresponding

21、 output y3n is y3n= 003kxnnnnk =)(2100kbxkaxnnnnk=a001kxnnnnk+b002kxnnnnk = ay1n+b y2n Therefore the system is linear. (b) Consider an arbitrary input x1n.Let 6 y1n =001kxnnnnk be the corresponding output .Consider a second input x2n obtained by shifting x1n in time: x2n= x1n-n1 The output correspon

22、ding to this input is y2n= 002kxnnnnk= n 1100kxnnnnk= 01011kxnnnnnnk Also note that y1n- n1= 01011kxnnnnnnk. Therefore , y2n= y1n- n1 This implies that the system is time-invariant. (c) If nxB, then yn(2 n0+1)B. Therefore ,C(2 n0+1)B. 1.19 (a) (i) Consider two arbitrary inputs x1(t) and x2(t). x1(t)

23、 y1(t)= t2x1(t-1) x2(t) y2(t)= t2x2(t-1) Let x3(t) be a linear combination of x1(t) and x2(t).That is x3(t)=a x1(t)+b x2(t) where a and b are arbitrary scalars. If x3(t) is the input to the given system, then the corresponding output y3(t) is y3(t)= t2x3 (t-1) = t2(ax1(t-1)+b x2(t-1) = ay1(t)+b y2(t

24、) Therefore , the system is linear. (ii) Consider an arbitrary inputs x1(t).Let y1(t)= t2x1(t-1) be the corresponding output .Consider a second input x2(t) obtained by shifting x1(t) in time: x2(t)= x1(t-t0) The output corresponding to this input is y2(t)= t2x2(t-1)= t2x1(t- 1- t0) Also note that y1

25、(t-t0)= (t-t0)2x1(t- 1- t0) y2(t) Therefore the system is not time-invariant. (b) (i) Consider two arbitrary inputs x1nand x2n. x1n y1n = x12n-2 x2n y2n = x22n-2. Let x3(t) be a linear combination of x1nand x2n.That is x3n= ax1n+b x2n where a and b are arbitrary scalars. If x3n is the input to the g

26、iven system, then the corresponding output y3n is y3n = x32n-2 =(a x1n-2 +b x2n-2)2 =a2x12n-2+b2x22n-2+2ab x1n-2 x2n-2 ay1n+b y2n Therefore the system is not linear. (ii) Consider an arbitrary input x1n. Let y1n = x12n-2 be the corresponding output .Consider a second input x2n obtained by shifting x

27、1n in time: x2n= x1n- n0 The output corresponding to this input is y2n = x22n-2.= x12n-2- n0 7 Also note that y1n- n0= x12n-2- n0 Therefore , y2n= y1n- n0 This implies that the system is time-invariant. (c) (i) Consider two arbitrary inputs x1nand x2n. x1n y1n = x1n+1- x1n-1 x2n y2n = x2n+1 - x2n -1

28、 Let x3n be a linear combination of x1n and x2n. That is : x3n= ax1n+b x2n where a and b are arbitrary scalars. If x3n is the input to the given system, then the corresponding output y3n is y3n= x3n+1- x3n-1 =a x1n+1+b x2n +1-a x1n-1-b x2n -1 =a(x1n+1- x1n-1)+b(x2n +1- x2n -1) = ay1n+b y2n Therefore

29、 the system is linear. (ii) Consider an arbitrary input x1n.Let y1n= x1n+1- x1n-1 be the corresponding output .Consider a second input x2n obtained by shifting x1n in time: x2n= x1n-n0 The output corresponding to this input is y2n= x2n +1- x2n -1= x1n+1- n0- x1n-1- n0 Also note that y1n-n0= x1n+1- n

30、0- x1n-1- n0 Therefore , y2n= y1n-n0 This implies that the system is time-invariant. (d) (i) Consider two arbitrary inputs x1(t) and x2(t). x1(t) y1(t)= d(t) x1 x2(t) y2(t)= (t) x2d Let x3(t) be a linear combination of x1(t) and x2(t).That is x3(t)=a x1(t)+b x2(t) where a and b are arbitrary scalars

31、. If x3(t) is the input to the given system, then the corresponding output y3(t) is y3(t)= d(t) x3 =(t) xb+(t) ax21d =ad(t) x1+b(t) x2d= ay1(t)+b y2(t) Therefore the system is linear. (ii) Consider an arbitrary inputs x1(t).Let y1(t)= d(t) x1=2)(x-(t) x11t be the corresponding output .Consider a sec

32、ond input x2(t) obtained by shifting x1(t) in time: x2(t)= x1(t-t0) The output corresponding to this input is y2(t)= (t) x2d=2)(x-(t) x22t =2)(x-)t-(t x0101tt Also note that y1(t-t0)= 2)(x-)t-(t x0101tt y2(t) Therefore the system is not time-invariant. 8 1.20 (a) Given x)(t=jte2 y(t)=tje3 x)(t=jte2

33、y(t)=tje3 Since the system liner tjetx21(2/1)(jte2) )(1ty=1/2(tje3+tje3) Therefore x1(t)=cos(2t) )(1ty=cos(3t) (b) we know that x2(t)=cos(2(t-1/2)= (jejte2+jejte2)/2 Using the linearity property, we may once again write x1(t)=21( jejte2+jejte2) )(1ty=(jejte3+jejte3)= cos(3t-1) Therefore, x1(t)=cos(2

34、(t-1/2) )(1ty=cos(3t-1) 1.21.The signals are sketched in figure S1.21. Figure S1.21 1.22 The signals are sketched in figure S1.22 1.23 The even and odd parts are sketched in Figure S1.23 t -1 0 -1 1 2 2 1 3 x(t-1) a 1 1/2 -1/2 -1 n 7 3 2 1 0 x3- n 0.5 0.5 t 3/2 -3/2 t 4 -1 3 2 1 0 1 2 x(2-t) 1 0 -1

35、1 2 t x(2t+1) x(4-t/2) t 10 12 6 1 8 4 1 2 )()()(tutxtx t 1 0 2 1 1/2 -1/2 -1 n 7 3 2 1 0 xn-4 (b) -1 1 1/2 -1/2 n 2 1 0 x3n (c) 1 -1 2 n 0 x3n+1 (d) 2 1 1 2 n 0 xnun-3=xn (f) 9 t -1/2 x0(t) 1/2 -1 -2 0 2 1 t 1 -1/2 x0(t) 1/2 -1 -2 0 2 1 -1/2 x0(t) 1/2 -1 -2 0 2 1 t Figure S1.22 0 1 1 n 2 (h) -4 -1

36、-2 1 1/2 n 2 0 x3- n/2 +(-1)nxn/2 (g) 7 n 1 0 -7 xon x0(t) -t/2 0 t x0(t) t 3t/2 -3t/2 0 (a) 1 0 -1/2 -7 7 -1/2 n xn 1 (c) Figure S1.24 -2 0 1/2 2 t x0(t) (a) (b) (c) Figure S1.23 1/2 n 1/2 -7 7 1 n xe(n) 3 (b) 1/2 0 7 1/2 -1 n xon 3/2 5 1 -5 n xen 0 3/2 -3/2 -1/2 4 n 1/2 xon 10 1.24 The even and od

37、d parts are sketched in Figure S1.24 1.25 (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) No

38、t 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, causal, stable (f) Linear, stable (g) Time invariant, linear, causal 1.28 (a) Linear

39、, stable (b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable (f) Memoryless, linear, causal, stable (g) Linear, stable 1.29 (a) Consider two inputs to the system such that 111.Sex ny nx n and 221.Sexnynx n Now consider a third input x3n= x2n

40、+x1n. The corresponding system output Will be 33121212eeeeynxnx nxnx nxny nyn therefore, we may conclude that the system is additive Let us now assume that inputs to the system such that /4111.Sjex ny nex n and /4222.Sjexnynexn Now consider a third input x3 n= x2 n+ x1 n. The corresponding system ou

41、tput Will be /433331122/4/41212cos/4sin/4cos/4sin/4cos/4sin/4jemememejjeeynex nnx nnx nnx nnx nnxnnxnex nexny nyn therefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that 11 211111Sdx tx ty tx tdt and 222211Sdxtxtytx tdt Now consider a third input x3

42、t= x2t+x1t. The corresponding system output Will be 2333211111211dxtytxtdtd x tx tx tx tdty tyt therefore, we may conclude that the system is not additive Now consider a third input x4 t= a x1 t. The corresponding system output Will be 2444211211111dxtytxtdtd ax tax tdtdx tax tdtay t Therefore, the

43、system is homogeneous. (ii) This system is not additive. Consider the fowling example .Let n=2n+2+ 2n+1+2n and x2n= n+1+ 2n+1+ 3n. The corresponding outputs evaluated at n=0 are 120203/2yandy Now consider a third input x3 n= x2 n+ x1 n.= 3n+2+4n+1+5n The corresponding outputs evaluated at n=0 is y30

44、=15/4. Gnarly, y30 021yyn .This 444442,1010,xn xnxnynxnotherwise 4445442,1010,xn xnaxnynaynxnotherwise 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 output d) In

45、vertible. Inverse system; y(t)=dx(t)/dt (e) Invertible. Inverse system y(n)=x(n+1) for n0 and yn=xn for n1,but y(t)=1 for t1. 1.41. (a) yn=2xn.Therefore, the system is time invariant. (b) yn=(2n-1)xn.This is not time-invariant because yn- N0(2n-1)2xn- N0. (c) yn=xn1+(-1)n+1+(-1)n-1=2xn.Therefore, th

46、e system is time invariant . 1.42.(a) Consider two system S1 and S2 connected in series .Assume that if x1(t) and x2(t) are the inputs to S1.then y1(t) and y2(t) are the outputs.respectively .Also,assume that if y1(t) and y2(t) are the input to S2 ,then z1(t) and z2(t) are the outputs, respectively

47、. Since S1 is linear ,we may write 11212,sax tbxtay tbyt where a and b are constants. Since S2 is also linear ,we may write 21212,say tbytaz tbzt We may therefore conclude that )()()()(212121tbtatbtazzxxss Therefore ,the series combination of S1 and S2 is linear. Since S1 is time invariant, we may w

48、rite 11010sxtTytT and 21010sytTztT Therefore, 1 21010s sx tTz tT Therefore, the series combination of S1 and S2 is time invariant. (b) False, Let y(t)=x(t)+1 and z(t)=y(t)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.

49、 00. ( )( ).00 x ty t0( )( )( )( )0 x tx ty ty t1 1-1/2e-t/ u（t） 1 - 0 t u（t） 1 1/2 - 0 1/2e-t/ t Figure s3.18 15 (c) Let us name the output of system 1 as wn and the output of system 2 as zn .Then 11 2 2 212224y nznw nw nw n 241121nxnxnx The overall system is linear and time-invariant. 1.43. (a) We

50、 have )(tytxs Since S is time-invariant. )(TtyTtxs Now if x (t) is periodic with period T. xt=x(t-T). Therefore, we may conclude that y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b) 1.44 (a) Assumption : If x(t)=0 for tt0 ,then y(t)=0