2022年信号与系统奥本海姆英文版课后答案chapter .pdf

Signals & Systems (Second Edition) Learning Instructions (Exercises Answers) Department of Computer Engineering 2005.12 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页，共 17 页Contents Chapter 1 2Chapter 2 17Chapter 3 53Chapter 4 80Chapter 5 101Chapter 6 127Chapter 7 137Chapter 8 150Chapter 9 158Chapter 10 178精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页，共 17 页Chapter 1 Answers 1.1Converting from polar to Cartesian coordinates: 111cos222je111cos()222je2cos()sin()22jjje2cos()sin()22jjje522jjjee42(cos()sin()1442jjje944122jjjee944122jjjee412jje1.2converting from Cartesian to polar coordinates: 055je, 22je, 233jje21322jje, 412jje, 2221jje4(1)jje, 411jje122213jje1.3. (a) E=4014tdte, P=0, because E(b) (2)42( )jttxe, 2( )1tx.Therefore, E=22( )dttx=dt=, P=211limlim222( )TTTTTTdtdtTTtxlim11T(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=204131 4nnnxP=0，because E. (e) 2 nx=()28nje, 22 nx=1. therefore, E=22 nx=, P=211limlim1122121 NNNNnNnNNNnx. (f) 3 nx=cos4n. Therefore, E=23 nx=2cos()4n=2cos()4n, P=1limcos214nNNnNN1cos()112lim()2122NNnNnN1.4.(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 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 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-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. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 17 页(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) is not periodic because it is zero for t3. (b) Since x1(t) is an odd signal, 2 vnxis zero for all values of t. (c) 11311 33221122vnnnnnu nunxxxTherefore, 3 vnxis zero when n3 and when n. (d) 1554411( )( )()(2)(2)22vtttttu tutxxxeeTherefore, 4( )vtxis zero only when t. 1.8. (a) 01( )22cos(0)tttxe(b) 02( )2 cos()cos(32)cos(3 )cos(30)4tttttxe(c) 3( )sin(3)sin(3)2tttttxee(d) 224( )sin(100 )sin(100)cos(100)2tttttttxeee1.9. (a) 1( ) txis a periodic complex exponential. 101021( )jtjttjxee(b) 2( ) txis a complex exponential multiplied by a decaying exponential. Therefore, 2( ) txis not periodic. （c）3 nxis a periodic signal. 3 nx=7jne=jne. 3 nxis a complex exponential with a fundamental period of 22. (d) 4 nxis a periodic signal. The fundamental period is given by N=m(23/5) =10().3mBy choosing m=3. We obtain the fundamental period to be 10. (e) 5 nxis not periodic. 5 nxis a complex exponential with 0w=3/5. We cannot find any integer m such that m(02w) is also an integer. Therefore, 5 nxis 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 which 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精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 17 页0 -1 -2 -3 1 2 3 X n 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 +74jne-25jnePeriod of first term in the RHS =1. Period of second term 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 obtained 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,tttTherefore 224dtE1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14. Therefore kkktkttg12(3)2(3)() 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+21x2n-3 = y1n-2+21y1n-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 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 5 页，共 17 页(b) The input-output relationship does not change if the order in which S1and S2are connected series reversed. . We can easily prove this assuming that S1follows S2. In this case , the signal x1n, which is the input to S1is the same as y2n. Therefore y1n =2x1n+ 4x1n-1 =2y2n+4 y2n-1 =2( x2n-2+21x2n-3 )+4(x2n-3+21x2n-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 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(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) Therefore , the system is linear. 1.18.(a) Consider two arbitrary inputs x1nand x2n. x1n y1n=001kxnnnnkx2n y2n=002kxnnnnkLet 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=003kxnnnnk=)(2100kbxkaxnnnnk=a001kxnnnnk+b002kxnnnnk= ay1n+b y2n Therefore the system is linear. (b) Consider an arbitrary input x1n.Let 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页，共 17 页y1n=001kxnnnnkbe the corresponding output .Consider a second input x2n obtained by shifting x1n in time: x2n= x1n-n1 The output corresponding to this input is y2n=002kxnnnnk=n 1100kxnnnnk=01011kxnnnnnnkAlso 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) 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) 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(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 given 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 x1n in time: x2n= x1n- n0 The output corresponding to this input is y2n= x22n-2.= x12n-2- n0 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页，共 17 页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 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 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- n0- 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)x1x2(t)y2(t)=(t)x2dLet 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)=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)x11tbe 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)= (t)x2d=2)(x-(t)x22t=2)(x-)t -(tx0101ttAlso note that y1(t-t0)=2)(x- )t-(tx0101tty2(t) Therefore the system is not time-invariant. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 8 页，共 17 页1.20 (a) Given x)(t=jte2y(t)=tje3x)(t=jte2y(t)=tje3Since 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)/2Using the linearity property, we may once again write x1(t)=21(jejte2+jejte2) )(1ty=(jejte3+jejte3)= cos(3t-1) herefore, x1(t)=cos(2(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 1 2 t x(2t+1) x(4-t/2) t 10 12 6 1 8 4 1 2 )()()(tutxtxt 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) 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 9 页，共 17 页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 -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 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 10 页，共 17 页1.24 The even and odd 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) 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, causal, stable (f) Linear, stable (g)Time invariant, linear, causal 1.28 (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, stable 1.29 (a) Consider two inputs to the system such that 111.Sexnynx nand 221.SexnynxnNow consider a third input x3n=x2n+x1n. The corresponding system output Will be 33121212eeeeynxnxnxnxnxnynyntherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that /4111.Sjex nynex nand /4222.SjexnynexnNow consider a third input x3 n= x2 n+ x1 n. The corresponding system output Will be /433331122/4/ 41212cos/ 4sin/4cos/ 4sin/4cos/ 4sin/ 4jemememejjeeynexnnxnnxnnxnnxnnxnnxnex nexnynyntherefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 11 页，共 17 页211111Sdxtxtytxtdtand 222211SdxtxtytxtdtNow consider a third input x3t=x2t+x1t. The corresponding system output Will be 2333211111211dxtytxtdtdxtxtxtxtdtytyttherefore, 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 axtax tdtdxtaxtdtay tTherefore, the system is homogeneous. (ii) This system is not additive. Consider the fowlingexample .Let n=2n+2+ 2n+1+2n and x2n= n+1+ 2n+1+ 3n. The corresponding outputs evaluated at n=0 are 120203/ 2yandyNow consider a third input x3 n= x2 n+ x1 n.= 3n+2+4n+1+5n The corresponding outputs evaluated at n=0 is y30=15/4. Gnarly, y30021yyn.This 444442,1010,xn xnxnynxnotherwise4445442,1010,xn xnaxnynaynxnotherwiseTherefore, 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) Invertible. 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, the 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 . Since S1is linear ,we may write 11212,sax tbxtay tbytwhere a and b are constants. Since S2 is also linear ,we may write 21212,say tbytaz tbztWe may therefore conclude that )()()()(212121tbtatbtazzxxssTherefore ,the series combination of S1 and S2 is linear. Since S1 is time invariant, we may write 11010sxtTytTand 21010sytTztTTherefore, 1 21010s sxtTz tTTherefore, 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. 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 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 15 页，共 17 页(c) Let us name the output of system 1 as wn and the output of sy