英文版《信号与系统》第九章习题解答
信号与系统奥本海姆英文版课后答案chapter9
Chapter 9 Answers9.1 （a ）The given integral may be written as(5)0t j t e e dt σω∞-+⎰If σ<-5 ,then the function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge .but if >-5,then the integral does converge (b) The given integral may be written as0(5)t j tee d σω-+-∞⎰t If σ>-5 ,then the function (5)te σ-+ grows towards ∞as t decreases towards -∞and the given integral doesnot converge .but if σ<-5,then the integral does converge (c) The given integral may be written as5(5)5t j t e e d σω-+-⎰t Clearly this integral has a finite value for all finite values of σ. (d) The given integral may be written as(5)t j t e e d σω∞-+-∞⎰tIfσ>-5 ,then the function (5)t e σ-+ grows towards ∞as t decreases towards -∞and the given integraldoes not converge If σ<-5, ,then function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge If σ=5, then the integral stilldoes not have a finite value. therefore, the integral does not converge for any value of σ. (e) The given integral may be written as0(5)t j tee d σω-+-∞⎰t+ (5)0t j t e e d σω∞-+⎰t The first integral converges for σ<-5, the second internal converges if σ>-5,therefore, the given internal converges whenσ<5.(f) The given integral may be written as0(5)t j t e e d σω-+-∞⎰tIf σ>5 ,then the function (5)te σ--+grows towards ∞ as t decrease towards -∞ and the given integral does not converge .but if σ<5,then the integral does converge. 9.2 (a)X(s)= 5(1)t dt eu t e dt ∞---∞⎰- =(5)0s tedt ∞-+⎰ =(5)5s es -++As shown in Example 9.1 the ROC will be {}Re s >-5. (b) By using eg.(9.3), we can easily show that g(t)=A 5te-u(-t-0t ) has the Laplace transformG(s)= 0(5)5s t Ae s ++The ROC is specified as {}Re s <-5 . Therefore ,A=1 and 0t =-19.3 Using an analysis similar to that used in Example 9.3 we known that given signal has a Laplace transform of the formX(s)115s s β+++The corresponding ROC is {}Re s >max(-5,Re{β}). Since we are given that the ROC isRe{s}>-3, we know that Re{β}=3 . there are no constraints on the imaginary part of β. 9.4 We know form Table 9.2 that111()sin(2)()()()Lt x t e t u t X s X s -=-←−→=-, Re{s}>-1 We also know form Table 9.1 thatx(t)= 1()Lx t -←−→X(s)= 1()X s -The ROC of X(s) is such that if 0s was in the ROC of 1()X s , then -0s will be in the ROC of X(s). Putting the two above equations together ,we havex(t)= 1x (-t) =sin(2)()t e t u t --X(s)= =-222(1)2s -+, {}Re s <1the denominator of the form 2s -2s+5. Therefore, the poles of X(s) are 1+2j and 1-2j.9.5 (a) the given Laplace transform may be written as ()X s =24(1)(3)s s s +++.Clearly ,X(s) has a zero at s=-2 .since in X(s) the order of the denominator polynomial exceeds the order of the numerator polynomial by 1 ,X(s) has a zero at ∞. Therefore ,X(s) has one zero in finite s-plane and one zero at infinity.(b) The given Laplance transform may be written asX(s)=1(1)(1)s s s +-+= 11s -Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceedsthe order the numerator polynomial by 1,X(s) has a zero at .therefore X(s) has no zero in the finite s-plane and one zero at infinity.(c) The given Laplace transform may be written as22(1)(1)()1(1)s s s X s s s s -++==-++ Clearly ，X （s ）has a zero at s=1.since in X(s) the order of the numerator polynomial exceeds the order of the denominator polynomial by 1,X(s) has zeros at ∞ .therefore , X(s) has one zero in the s-plane and no zero at infinity .9.6 (a) No. From property 3 in Section 9.2 we know that for a finite-length signal .the ROC is the entire s-plane .therefore .there can be no poles in the finite s-plane for a finite length signal . Clearly in this problem this not the case.(b) Yes. Since the signal is absolutely integrable, The ROC must include, the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal would be Re{s}<2. From property 5 in section 9.2 we know that this would correspond to a left-sided signal(C) No . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2. therefore ,we can never have an ROC of the form Re{s}> α. From property 5 in section 9.2 we knew that x(t) can not be a right-side signal(d) Yes . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal could be α<Re{s}<2 such that α<0 .From property 6 in section 9.2 ,we know that this would correspond to a two side signal9.7 We may find different signal with the given Laplace transform by choosing different regions of02s =- 13s =- 212s j =- 312s j =-Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 12(ii)-2< Re{s}<- 12(iii)-3<Re{s}<-2 (iv)Re{s}<-3Therefore ,we may find four different signals the given Laplace transform. 9.8 From Table 9.1,we know thatG(t)= 2()()(2)L te x t G s X s ←−→=-. The ROC of G(s) is the ROC of X(s) shifted to the right by 2We are also given that X(s) has exactly 2 poles at s=-1 and s=-3. since G(s)=X(s-2), G(s)also has exactly two poles ,located at s=-1+2=1 and s=-3+2=-1 since we are given G(j ω) exists , we may infer that j ω-axis lies in the ROC of G(s). Given this fact and the locations of the poles ,we may conclude that g(t) is a two sidesequence .Obviously x(t)= 2te g(t) will also be two sided9.9 Using partial fraction expansion X(s)= 4243s s -++Taking the inverse Laplace transform, X(t)=443()2()t t e u t e u t ---9.10 The pole-zero plots for each of the three Laplace transforms is as shown in Figure S9.10(a) form Section 9.4 we knew that the magnitude of the Fourier transform may be expressed aswe se that the right-hand side of the above expression is maximum for ω=0 and decreases as ω becomesincreasing more positive or more negative . Therefore 1()H j ω is approximately lowpass(b) From Section 9.4 we know that the magnitude of the Fourier transform may be express aswe see that the right-hand side of the above expression is zero for ω=0.It then increams withincreasing |ω| until |ω| reach 1/2. Then it starts decreasing as |ω| increase even further. Therefore | 2H (j )ω| is approximately bandpass.(c) From Section 9.4 we know that the magnitude of the Fourier transform may be express as2We see that the right-hand side of the above expression is zero for ω=0. It then increases withincreasing |ω| until |ω| reaches 21. Then |ω| increases,| 3()H j ω| decreases towards a value of1(because all the vector lengths became almost identical and the ratio become 1) .Therefore |3()H j ω| is approximately highpass.9.11 X(s) has poles ats=1-2and 1-2.X(s) has zeros ats=1212.From Section 9.4 we know that |X(j ω)| is11(Length of vector from toto ωωThe terms in the numerator and denominator of the right-band side of above expression cancel ourgiving us |X(j ω)|=1.9.12 (a) If X(s) has only one pole, then x(t) would be of the form A ate -.Clearly such a signal violates condition 2. Therefore , this statement is inconsistent with the given information.(b) If X(s) has only two poles, then x(t) would be of the form A 0sin()ate t ω- .Clearly such a signal could be made to satisfy all three conditions(Example:0ω=80π,α=19200). Therefore, this statement is consistent with the given information. (c) If X(s) has more than two poles (say 4 poles), then x(t) could be assumed to be of theform 00sin()sin()at btAe t Be t ωω--+. Clearly such a signal could still be made to satisfy all three conditions. Therefore, this statement is consistent with the given information. 9.13 We have1}Re{,1)(->+=s s s X β.Also,(Length of vector form ω to -1)(Length of vector form ω to 11}Re{1),()()(<<--+=s s X s X s G αTherefore, ].11[)(2s s s s G -++-=ααβComparing with the given equation for G(s), ,1-=α .21=β9.14. Since X(s) has 4 poles and no zero in the finite s-plane, we many assume that X(s) is of the form .))()()(()(d s c s b s a s As X ----=Since x(t) is real ,the poles of X(s) must occur in conjugate reciprocal pairs. Therefore, we mayassume that b=*a and d=*c . This result in .))()()(()(**c s c s a s a s As X ----=Since the signal x (t) is also even , the Laplace transform X(s) must also be even . This implies thatthe poles have to be symmetric about the j ω-axis. Therefore, we may assume that c=*a -. This results in .))()()(()(**a S a s a s a s As X ++--=We are given that the location of one of the poles is (1/2)4πj e . If we assume that this pole is a, we have 4444AX(s)=.1111(s-)(s-)(s+)(s+)2222j j j j e e e e ππππ--This gives us22().11()()44AX s s s s s =Also ,we are give that()(0)4x t dt X ∞-∞==⎰Substituting in the above expression for X(s), we have A=1/4. Therefore,221/4().11()()44X s s s s s =9.15. Taking the Laplace transform of both sides of the two differential equations, we haves X(s)=1)(2+-s Y and s Y(s)=2X(s) . Solving for X(s) and Y(s), we obtain4)(2+=s s s X and Y(s)= 22s 4+.The region of convergence for both X(s) and Y(s) is Re{s}>0 because both are right-hand signals. 9.16. Taking the Laplace transform of both sides of the given differential equations ,we obtain ).(])1()1()[(223s X s s s s Y =+++++αααα therefore,.)1()1(1)()()(223αααα+++++==s s s s X s Y s H(a) Taking the Laplace transform of both sides of the given equation, we haveG(s) = s H(s)+ H(s). Substituting for H(s) from above,.1)1()1()1()(22223αααααα++=++++++=s s s s s s s GTherefore, G(s) has 2 poles.(b) we know that H(s) =.))(1(122αα+++s s s Therefore, H(s) has poles at and j ),2321(,1+--α ).2321(j --α If the system has to be stable,then the real part of the poles has to be less than zero. For this to be true, we require that ,02/<-α i.e.,0>α.9.17 The overall system show in Figure 9.17 may be treated as two feedback system of the form shown in figure 9.31 connected in parallel. By carrying out an analysis similar to that described in Section 9.8.1, we find the system function of the upper feedback system to be.82)/2(41/2)(1+=+=s s s s HSimilarly, the system function of the lower feedback system is .21)2/1(21/1)(2+=+=s s s HThe system function of the overall system is now.1610123)()()(221+++=+=s s s s H s H s HSince H(s)=Y(s)/X(s), we may write]123)[(]1610)[(2+=++s s X s s s Y . Taking the inverse Laplace transform, we obtaindtt dx t x t y dt t dy dt t y d )(3)(12)(16)(10)(2+=++9.18. ( a) From problem 3.20, we know that differential equation relating the input and output of the RLC circuit is2()()()().d y t dy t y t x t dtdt++=Taking the Laplace transform of this (while nothing that the system is causal and stable), we obtain 2()[1]().Y s s s X s ++= Therefore ,2()1(),()1Y s H s X s s s ==++ 1{}.2e s ℜ>-(b) We note that H(s) has two poles at12s =--12s =-+From Section 9.4 we know that the magnitude of the Fourier transform may be expressed asWe see that the right hand side of the above expression Increases with increasing |ω| until |ω| reaches 12. Then it starts decreasing as |ω| increasing even further. It finally reaches 0 for |ω|=∞.Therefore 2|()|H j ω is approximately lowpass.(c) By repeating the analysis carried out in Problem 3.20 and part (a) of this problem with R =310-Ω,we can show that2()1(),()1Y s H s X s s s ==++ {}0.0005.e s ℜ>-(d) We haveWe see that when |ω| is in he vicinity 0.0005, the right-hand side of the above equation takes onextremely large value. On either side of this value of |ω| the value of |H (j ω)| rolls off rapidly. Therefore, H(s) may be considered to be approximately bandpass. 9.19. (a) The unilateral Laplace transform isX(s) = 20(1)t st e u t e dt -∞--+⎰= 20t st e e dt -∞--⎰=21+s {} 2.e s ℜ>-(b) The unilateral Laplace transform is2(3)0()[(1)()(1)]t st X s t t e u t e dt δδ-∞-+-=++++⎰2(3)0[()]t st t e e dt δ-∞-+-=+⎰612e s -=++ {} 2.e s ℜ>- (c) The unilateral Laplace transform is240()[()()]t t st X s e u t e u t e dt -∞---=⎰240[]t t st e e e dt -∞---=+⎰1124s s =+++ {} 2.e s ℜ>-9.20. In Problem 3.29, we know that the input of the RL circuit are related by ).()()(t x t y dtt dy =+Applying the unilateral Laplace transform to this equation, we have ).()()0()(s x s y y s sy =+--(a) For the zero-state response, set (0)0y -=.Also we have u s x =)(L{)(2t u et-}=21+s .Therefore,y(s)(s+1)=.21+sComputing the partial fraction expansion of the right-hand side of the above equation and then taking its inverse unilateral Laplace transform, we have ).()()(2t u e t u e t y t t ---=(b) For the zero-state response, assume that x(t) = 0.Since we are given that (0)1y -=,.11)(0)(1)(+=⇒=+-s s y s y s sy Taking the inverse unilateral Laplace transform, we have ()().t y t e u t -=Figure S9.212()2()().t t y t e u t e u t --=-9.21. The pole zero plots for all the subparts are shown in figure S9.21. (a) The Laplace transform of x(t) isX(s)= 230()t t st e e e dt ∞---+⎰= (2)(3)00[/(2)]|[/(3)]|s t s te s e s -+∞-+∞-++-+ =211252356s s s s s ++=++++(b) Using an approach similar to that show in part (a), we have41(),4L t e u t s -←−→+ {} 4.e s ℜ>-Also,551(),55L t j t e e u t s j -←−→+-and(){}551,555LT t j t e e u t e s s j --←−→ℜ>-++.From this we obtain()()()()55555215sin 52525LTt t j t t j t e t u t e e e e u t js ----⎡⎤=-←−→⎣⎦++ ,where {}5e s ℜ>- .Therefore,()()(){}245321570sin 5,51490100LTt t s s e u t e t u t e s s s s --+++←−→ℜ>-+++. R b Im(c)The Laplace transform of ()x t is ()()023t t st X s e e e dt --∞=+⎰()()()()2300/2|/3|s t s t e s e s ----∞-∞⎡⎤⎡⎤=--+--⎣⎦⎣⎦ 211252356s s s s s -=+=---+.The region of convergence (ROC) is {}2e s ℜ<.(d)Using an approach along the lines of part (a),we obtain(){}21,22LT t e u t e s s -←−→ℜ>-+. (S9.21-1) Using an approach along the lines of part (c) ,we obtain(){}21,22LT t e u t e s s -←−→ℜ<-. (S9.21-2)From these we obtain()()222224t LT t t s e e u t e u t s --=+-←−→-, {}22e s -<ℜ<. Using the differentiation in the s-domain property , we obtain(){}22222228,2244t LT d s s te e s ds s s -+⎡⎤←−→-=--<ℜ<⎢⎥-⎣⎦-. (e)Using the differentiation in the s-domain property on eq.(S9.21-1),we get()(){}2211,222LT t d te u t e s ds s s -⎡⎤←−→-=ℜ>-⎢⎥+⎣⎦+.Using the differentiation in the s-domain property on eq (S9.21-2),we get ()(){}2211,222LT t d te u t e s ds s s ⎡⎤--←−→=-ℜ<⎢⎥-⎣⎦-.Therefore,()()()(){}222224,2222t LT t t st e te u t te u t e s s s ---=--←−→-<ℜ<+-.(f)From the previous part ,we have ()()(){}2221,22LT t t t e u t te u t e s s -=--←−→-ℜ<-.(g)Note that the given signal may be written as ()()()1x t u t u t =-- .Note that (){}1,0LTu t e s s←−→ℜ>.Using the time shifting property ,we get(){}1,0s LT e u t e s s--←−→ℜ>.Therefore ,()1x t()()11,sLT e u t u t s----←−→ All s . Note that in this case ,since the signal is finite duration ,the ROC is the entire s-plane.(h)Consider the signal ()()()11x t t u t u t =--⎡⎤⎣⎦.Note that the signal ()x t may beexpressed as ()()()112x t x t x t =+-+ . We have from the previous part()()11sLT e u t u t s----←−→, All s . Using the differentiation in s-domain property ,we have()()()12111s s s LT d e se e x t t u t u t ds ss ---⎡⎤--+=--←−→=⎡⎤⎢⎥⎣⎦⎣⎦, All s . Using the time-scaling property ,we obtain()121s s LT se e x t s --+-←−→, All s .Then ,using the shift property ,we have()21212s sLT s se e x t es ---+-+←−→ ,All s . Therefore ,()()()21122112s s s sLT s se e se e x t x t x t e s s----+--+=+-+←−→+, All s. (i) The Laplace transform of ()()()x t t u t δ=+ is (){}11/,0X s s e s =+ℜ>.(j) Note that ()()()()33t u t t u t δδ+=+.Therefore ,the Laplace transform is the same as the result of the previous part.9.22 (a)From Table 9.2,we have()()()1sin 33x t t u t =.(b)From Table 9.2 we know that()(){}2cos 3,09LT st u t e s s ←−→ℜ>+. Using the time scaling property ,we obtain()(){}2cos 3,09LT s t u t e s s -←−→-ℜ<+Therefore ,the inverse Laplace transform of ()X s is()()()cos 3x t t u t =--.(c)From Table 9.2 we know that ()()(){}21cos 3,119LTt s e t u t e s s -←−→ℜ>-+. Using the time scaling property ,we obtain ()()(){}21cos 3,119LTt s e t u t e s s -+-←−→-ℜ<-++. Therefore ,the inverse Laplace transform of ()X s is ()()()cos 3t x t e t u t -=--.(d)Using partial fraction expansion on ()X s ,we obtain ()2143X s s s =-++ .From the given ROC ,we know that ()x t must be a two-sided signal .Therefore ()()()432t t x t e u t e u t --=+-.(e)Using partial fraction expansion on ()X s ,we obtain()2132X s s s =-++. From the given ROC ,we know that ()x t must be a two-sided signal ,Therefore,()()()332ttx t e u t e u t --=+-.(f)We may rewrite ()X s as ()2311s X s s s =+-+1=1=+Using Table 9.2 ,we obtain()())())()/2/23cos /2sin/2t t x t t e u t u t δ--=+.(g)We may rewrite ()X s as ()()2311s X s s =-+.From Table 9.2,we know that(){}21,0LT tu t e s s ←−→ℜ>.Using the shifting property ,we obtain()(){}21,11LT t e tu t e s s -←−→ℜ>-+.Using the differentiation property ,()()()(){}2,11LT t t t d s e tu t e u t te u t e s dt s ---⎡⎤=-←−→ℜ>-⎣⎦+. Therefore,()()()()33t t x t t e u t te u t δ--=--.9.23.The four pole-zero plots shown may have the following possible ROCs:·Plot (a): {}2e s ℜ<- or {}22e s -<ℜ< or {}2e s ℜ>.·Plot (b): {}2e s ℜ<- or {}2e s ℜ>-. ·Plot (c): {}2e s ℜ< or {}2e s ℜ>. ·Plot (d): Entire s-plane.Also, suppose that the signal ()x t has a Laplace transform ()X s with ROC R . (1).We know from Table 9.1 that()()33LT te x t X s -←−→+.The ROC 1R of this new Laplace transform is R shifted by 3 to the left .If ()3t x t e - is absolutely integrable, then 1R must include the jw -axis.·For plot (a), this is possible only if R was {}2e s ℜ> . ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ> . ·For plot (d),R is the entire s-plane. (2)We know from Table 9.2 that(){}1,11LT t e u t e s s -←−→ℜ>-+.Also ,from Table 9.1 we obtain()()(){}2,11LT t X s x t e u t R R e s s -⎡⎤*←−→=ℜ>-⎡⎤⎣⎦⎣⎦+I If ()()te u t x t -*is absolutely integrable, then 2R must include the jw -axis.·For plot (a), this is possible only if R was {}22e s -<ℜ<. ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(3)If ()0x t = for 1t > ,then the signal is a left-sided signal or a finite-duration signal . ·For plot (a), this is possible only if R was {}2e s ℜ<-. ·For plot (b), this is possible only if R was {}2e s ℜ<-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(4)If ()0x t =for 1t <-,then the signal is a right-sided signal or a finite-duration signal ·For plot (a), this is possible only if R was {}2e s ℜ>.·For plot (b), this is possible only if R was {}2e s ℜ>- . ·For plot (c), this is possible only if R was {}2e s ℜ>.·For plot (d),R is the entire s-plane.9.24.(a)The pole-zero diagram with the appropriate markings is shown Figure S9.24.(b)By inspecting the pole-zero diagram of part (a), it is clear that the pole-zero diagram shown in Figure S9.24 will also result in the same ()X jw .This would correspond to the Laplace transform()112X s s =-, {}12e s ℜ<.(c)≮()X jw π=-≮()1X jw .(d)()2X s with the pole-zero diagram shown below in Figure S9.24 would have the property that ≮()2X jw =≮()X jw .Here ,()211/2X s s -=-. (e) ()()21/X jw X jw =.(f)From the result of part (b),it is clear that ()1X s may be obtained by reflecting the poles and zeros in the right-half of the s-plane to the left-half of the s-plane .Therefore, ()11/22s X s s +=+.From part (d),it is clear that ()2X s may be obtained by reflecting the poles (zeros) in the right-half of the s-plane to the left-half and simultaneously changing them to zeros (poles).Therefore,()()()()2211/22s X s s s +=++9.25.The plots are as shown in Figure S9.25. 9.26.From Table 9.2 we have()()(){}2111,22LT t x t e u t X s e s s -=←−→=ℜ>-+and()()(){}3111,33LTt x t e u t X s e s s -=←−→=ℜ>-+.Using the time-shifting time-scaling properties from Table 9.1,we obtain()(){}22112,22s LT s e x t e X s e s s ---←−→=ℜ>-+and()(){}33223,33s LT s e x t e X s e s s---+←−→-=ℜ>--.Therefore, using the convolution property we obtain ()()()()23122323s s LTe e y t x t x t Y s s s --⎡⎤⎡⎤=-*-+←−→=⎢⎥⎢⎥+-⎣⎦⎣⎦. 9.27.From clues 1 and 2,we know that ()X s is of the form()()()AX s s a s b =++. Furthermore , we are given that one of the poles of ()X s is 1j -+.Since ()x t is real, the poles of ()X s must occur in conjugate reciprocal pairs .Therefore, 1a j =-and 1b j =+and ()()()11AH s s j s j =+-++. From clue 5,we know that ()08X =.Therefore, we may deduce that 16A = and ()21622H s s s =++ .Let R denote the ROC of ()X s .From the pole locations we know that there are two possible choices of R .R may either be {}1e s ℜ<-or {}1e s ℜ>-.We will now useclue 4 to pick one .Note that()()()()22LTt y t e x t Y s X s =←−→=-.The ROC of ()Y s is R shifted by 2 to the right .Since it is given that ()y t is not absolutely integrable ,the ROC of ()Y s should not include the jw axis -.This is possible only ofR is {}1e s ℜ>-.9.28.(a) The possible ROCs are(i) {}2e s ℜ<-.(ii) {}21e s -<ℜ<-. (iii) {}11e s -<ℜ<.( iv) {}1e s ℜ>.(b)(i)Unstable and anticausal. (ii) Unstable and non causal. （iii ）Stable and non causal. (iv) Unstable and causal. 9.29.(a)Using Table 9.2,we obtain (){}1,11X s e s s =ℜ>-+and(){}1, 2.2H s e s s =ℜ>-+(b) Since ()()()y t x t h t =*,we may use the convolution property to obtain()()()()()112Y s X s H s s s ==++.The ROC of ()Y s is {}1e s ℜ>-.(c) Performing partial fraction expansion on ()Y s ,we obtain . ()1112Y s s s =-++.Taking the inverse Laplace transform, we get()()()2t t y t e u t e u t --=-. (d)Explicit convolution of ()x t and ()h t gives us()()()y t h x t d τττ∞-∞=-⎰()()20t e e u t d ττττ∞---=-⎰t t e e d ττ--=⎰ for0t >()2.t t e e u t --⎡⎤=-⎣⎦ 9.30.For the input ()()x t u t =, the Laplace transform is (){}1,0.X s e s s=ℜ>The corresponding output ()()1t t y t e te u t --⎡⎤=--⎣⎦ has the Laplace transform()()(){}221111,0111Y s e s s s s s s =--=ℜ>+++. Therefore,()()()(){}21,0.1Y s H s e s X s s ==ℜ>+ Now ,the output ()()3123t t y t e e u t --⎡⎤=-+⎣⎦has the Laplace transform()()(){}12316,0.1313Y s e s s s s s s s =-+=ℜ>++++ Therefore , the Laplace transform of the corresponding input will be()()()()(){}1161,0.3Y s s X s e s H s s s +==ℜ>+ Taking the inverse Laplace transform of the partial fraction expansion of ()1,X s we obtain ()()()3124.t x t u t e u t -=+9.31.(a).Taking the Laplace transform of both sides of the given differential equation and simplifying, weobtain()()()212Y s H s X s s s ==--.b).The partial fraction expansion of ()H s is()1/31/321H s s s =--+. (i).If the system is stable ,the ROC for ()H s has to be {}12e s -<ℜ< . Therefore ()()()21133t t h t e u t e u t -=---.(ii).If the system is causal, the ROC for ()H s has to be {}2e s ℜ> .Therefore()()()21133t t h t e u t e u t -=-.(iii)If the system is neither stable nor causal ,the ROC for ()H s has to be {}1e s ℜ<-.Therefore ,()()()21133t t h t e u t e u t -=--+-9.32. If ()2t x t e =produces ()()21/6t y t e =,then ()()21/6H =. Also, by taking the Laplace transform of both sidesof the given differential equation we get ()()()()442s b s H s s s s ++=++.Since ()21/6H = ,we may deduce that 1b = .Therefore()()()()()222424s H s s s s s s +==+++. 9.33.Since ()()()t t t x t e e u t e u t --==+-,()()(){}112,111111X s e s s s s s -=-=-<ℜ<+-+-. We are also given that ()2122s H s s s +=++.Since the poles of ()H s are at 1j -±, and since ()h t is causal ,we may conclude that the ROC of()H s is {}1e s ℜ>-.Now()()()()()22221Y s H s X s s s s -==++-. The ROC of ()Y s will be the intersection of the ROCs of ()X s and ()H s .This is {}11e s -<ℜ<. We may obtain the following partial fraction expansion for ()Y s :()22/52/56/5122s Y s s s s +=-+-++. We may rewrite this as ()()()222/521411551111s Y s s s s ⎡⎤⎡⎤+=-++⎢⎥⎢⎥-++++⎢⎥⎢⎥⎣⎦⎣⎦.Nothing that the ROC of ()Y s is {}11e s -<ℜ<and using Table9.2,we obtain ()()()()224cos sin 555t t t y t e u t e tu t e tu t --=-++9.34.We know that()()(){}111,0LTx t u t X s e s s=←−→=ℜ> Therefore,()1X s has a pole at0s =.Now ,the Laplace transform of the output()1y t of the system with()1x t as the input is()()()11Y s H s X s =Since in clue 2, ()1Y s is given to be absolutely integrable ,()H s must have a zero at 0s =whichcancels out the pole of ()1X s at 0s =.We also know that()()(){}2221,0LT x t tu t X s e s s=←−→=ℜ> Therefore , ()2x s has two poles at 0s =.Now ,the Laplace transform of the output ()2y t of the system with ()2x t as the input is()()()22Y s H s X s =Since in clue 3, ()2Y s is given to be not absolutely integrable ,()H s does not have two zeros at0s =.Therefore ,we conclude that ()H s has exactly one zero at 0s =. From clue 4 we know that the signal ()()()()2222d h t dh t p t h t dt dt=++is finite duration .Taking the Laplace transform of both sides of the above equation ,we get ()()()()222P s s H s sH s H s =++. Therefore,()()222P s H s s s =++.Since ()p t is of finite duration, we know that ()P s will have no poles in the finite s-plane .Therefore, ()H s is of the form()()1222Ni i A s z H s s s =-=++∏,where i z ,1,2,....,i N =represent the zeros of ()P s .Here ,A is some constant.From clue 5 we know that the denominator polynomial of ()H s has to have a degree which is exactly one greater than the degree of the numerator polynomial .Therefore, ()()1222A s s H s s s -=++.Since we already know that ()H s has a zero at 0s = ,we may rewrite this as ()222As H s s s =++ From clue 1 we know that ()1H is0.2.From this ,we may easily show that 1A = .Therefore,()222s H s s s =++. Since the poles of ()H s are at 1j -± and since ()h t is causal and stable ,the ROC of ()H s is {}1e s ℜ>-.9.35.(a) We may redraw the given block diagram as shown in Figure S9.35. From the figure ,it is clear that()()1F s Y s s=. Therefore, ()()1/f t dy t dt =. Similarly, ()()/e t df t dt =.Therefore, ()()221/e t d y t dt =.From the block diagram it is clear that()()()()()()()21111266d y t dy t y t e t f t y t y t dtdt=--=--.。
信号系统习题解答khdaw
kh da
(c)
后
答 案
解 (a)、(c)、(d)为连续信号；(b)为离散信号；(d)为周期信号；其余为非周期信号；(a)、 (b)、(c)为有始（因果）信号。
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18 若有线性时不变系统的方程为 1-8
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若在非零 f( t )作用下其响应 y (t ) = 1 − e − t ，试求方程
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解 因为 f( t ) → y (t ) = 1 − e − t ，由线性关系，则 2 f (t ) → 2 y (t ) = 2(1 − e − t )
且
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x(t ) = y ′(t )
1-5 已知某系统的输入 f( t )与输出 y( t )的关系为 y( t ) = | f( t )|，试判定该系统是否为线 性时不变系统？
解 设 T 为系统的运算子，则可以表示为
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6
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uC du +C C R dt
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由于
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−2A1 + A2 = 2 所以
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故有
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信号与系统奥本海姆英文版课后答案chapter9
Chapter 9 Answers（a ）The given integral may be written as(5)0t j t e e dt σω∞-+⎰If σ<-5 ,then the function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge .but if >-5,then the integral does converge (b) The given integral may be written as0(5)t j tee d σω-+-∞⎰t If σ>-5 ,then the function (5)te σ-+ grows towards ∞as t decreases towards -∞and the given integral doesnot converge .but if σ<-5,then the integral does converge (c) The given integral may be written as5(5)5t j t e e d σω-+-⎰t Clearly this integral has a finite value for all finite values of σ. (d) The given integral may be written as(5)t j t e e d σω∞-+-∞⎰tIfσ>-5 ,then the function (5)t e σ-+ grows towards ∞as t decreases towards -∞and the given integraldoes not converge If σ<-5, ,then function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge If σ=5, then the integral stilldoes not have a finite value. therefore, the integral does not converge for any value of σ. (e) The given integral may be written as0(5)t j tee d σω-+-∞⎰t+ (5)0t j t e e d σω∞-+⎰t The first integral converges for σ<-5, the second internal converges if σ>-5,therefore, the given internal converges whenσ<5.(f) The given integral may be written as0(5)t j t e e d σω-+-∞⎰tIf σ>5 ,then the function (5)te σ--+grows towards ∞ as t decrease towards -∞ and the given integral does not converge .but if σ<5,then the integral does converge. (a)X(s)= 5(1)t dt eu t e dt ∞---∞⎰- =(5)0s tedt ∞-+⎰ =(5)5s es -++As shown in Example the ROC will be {}Re s >-5. (b) By using eg., we can easily show that g(t)=A 5te-u(-t-0t ) has the Laplace transformG(s)= 0(5)5s t Ae s ++The ROC is specified as {}Re s <-5 . Therefore ,A=1 and 0t =-1Using an analysis similar to that used in Example we known that given signal has a Laplace transform of the formX(s)115s s β+++The corresponding ROC is {}Re s >max(-5,Re{β}). Since we are given that the ROC isRe{s}>-3, we know that Re{β}=3 . there are no constraints on the imaginary part of β. We know form Table that111()sin(2)()()()Lt x t e t u t X s X s -=-←−→=-, Re{s}>-1 We also know form Table thatx(t)= 1()Lx t -←−→X(s)= 1()X s -The ROC of X(s) is such that if 0s was in the ROC of 1()X s , then -0s will be in the ROC of X(s). Putting the two above equations together ,we havex(t)= 1x (-t) =sin(2)()t e t u t --L ←−→X(s)= 1()X s -=-222(1)2s -+, {}Re s <1the denominator of the form 2s -2s+5. Therefore, the poles of X(s) are 1+2j and 1-2j.(a) the given Laplace transform may be written as ()X s =24(1)(3)s s s +++.Clearly ,X(s) has a zero at s=-2 .since in X(s) the order of the denominator polynomial exceeds the order of the numerator polynomial by 1 ,X(s) has a zero at ∞. Therefore ,X(s) has one zero in finite s-plane and one zero at infinity.(b) The given Laplance transform may be written asX(s)=1(1)(1)s s s +-+= 11s -Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceedsthe order the numerator polynomial by 1,X(s) has a zero at ∞.therefore X(s) has no zero in the finite s-plane and one zero at infinity.(c) The given Laplace transform may be written as22(1)(1)()1(1)s s s X s s s s -++==-++Clearly ，X （s ）has a zero at s= in X(s) the order of the numerator polynomial exceeds the order of the denominator polynomial by 1,X(s) has zeros at ∞ .therefore , X(s) has one zero in the s-plane and no zero at infinity .(a) No. From property 3 in Section we know that for a finite-length signal .the ROC is the entire s-plane .therefore .there can be no poles in the finite s-plane for a finite length signal . Clearly in this problem this not the case.(b) Yes. Since the signal is absolutely integrable, The ROC must include, the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal would be Re{s}<2. From property 5 in section we know that this would correspond to a left-sided signal(C) No . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2. therefore ,we can never have an ROC of the form Re{s}> α. From property 5 in section we knew that x(t) can not be a right-side signal(d) Yes . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal could be α<Re{s}<2 such that α<0 .From property 6 in section ,we know that this would correspond to a two side signalWe may find different signal with the given Laplace transform by choosing different regions of 02s =- 13s =- 2132s j =- 3132s j =-Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 12(ii)-2< Re{s}<- 12(iii)-3<Re{s}<-2 (iv)Re{s}<-3Therefore ,we may find four different signals the given Laplace transform. From Table ,we know thatG(t)= 2()()(2)L te x t G s X s ←−→=-. The ROC of G(s) is the ROC of X(s) shifted to the right by 2We are also given that X(s) has exactly 2 poles at s=-1 and s=-3. since G(s)=X(s-2), G(s)also has exactly two poles ,located at s=-1+2=1 and s=-3+2=-1 since we are given G(j ω) exists , we may infer that j ω-axis lies in the ROC of G(s). Given this fact and the locations of the poles ,we may conclude that g(t) is a two sidesequence .Obviously x(t)= 2te g(t) will also be two sidedUsing partial fraction expansion X(s)= 4243s s -++Taking the inverse Laplace transform, X(t)=443()2()t t e u t e u t ---The pole-zero plots for each of the three Laplace transforms is as shown in Figure(a) form Section we knew that the magnitude of the Fourier transform may be expressed aswe se that the right-hand side of the above expression is maximum for ω=0 and decreases as ω becomesincreasing more positive or more negative . Therefore 1()H j ω is approximately lowpass (b) From Section we know that the magnitude of the Fourier transform may be express as 1313(length of vector from to -+j )(Length of vector from to --j )ωωwe see that the right-hand side of the above expression is zero for ω= then increams withincreasing |ω| until |ω| reach 1/2. Then it starts decreasing as |ω| increase even further. Therefore | 2H (j )ω| is approximately bandpass.(c) From Section we know that the magnitude of the Fourier transform may be express as21313(length of vector from to -+j )(Length of vector from to --j ) ωωWe see that the right-hand side of the above expression is zero for ω=0. It then increases withincreasing |ω| until |ω| reaches 21. Then |ω| increases,| 3()H j ω| decreases towards a value of1(because all the vector lengths became almost identical and the ratio become 1) .Therefore |3()H j ω| is approximately highpass.X(s) has poles at s=13-213-2(s) has zeros at s=132and 132.From Section we know that |X(j ω)| is1313(Length of vector from toto 22221313(length of vector from to -+j )(Length of vector from to --j )ωωωωThe terms in the numerator and denominator of the right-band side of above expression cancel ourgiving us |X(j ω)|=1.(a) If X(s) has only one pole, then x(t) would be of the form ate - such a signal violates condition 2. Therefore , this statement is inconsistent with the given information.(b) If X(s) has only two poles, then x(t) would be of the form A 0sin()ate t ω- .Clearly such a signal could be made to satisfy all three conditions(Example:0ω=80π,α=19200). Therefore, this statement is consistent with the given information. (c) If X(s) has more than two poles (say 4 poles), then x(t) could be assumed to be of theform 00sin()sin()at btAe t Be t ωω--+. Clearly such a signal could still be made to satisfy all three conditions. Therefore, this statement is consistent with the given information. We have1}Re{,1)(->+=s s s X β.Also,(Length of vector form ω to -1)(Length of vector form ω to 1 I m-1 -3 Re -4 Re 1 Im Im Re1}Re{1),()()(<<--+=s s X s X s G αTherefore, ].11[)(2s s s s G -++-=ααβComparing with the given equation for G(s), ,1-=α .21=β. Since X(s) has 4 poles and no zero in the finite s-plane, we many assume that X(s) is of the form .))()()(()(d s c s b s a s As X ----=Since x(t) is real ,the poles of X(s) must occur in conjugate reciprocal pairs. Therefore, we mayassume that b=*a and d=*c . This result in .))()()(()(**c s c s a s a s As X ----=Since the signal x (t) is also even , the Laplace transform X(s) must also be even . This implies thatthe poles have to be symmetric about the j ω-axis. Therefore, we may assume that c=*a -. This results in .))()()(()(**a S a s a s a s As X ++--=We are given that the location of one of the poles is (1/2)4πj e . If we assume that this pole is a, we have 4444AX(s)=.1111(s-)(s-)(s+)(s+)2222j j j j e e e e ππππ-- This gives us22().11()()4422AX s s s s s =Also ,we are give that()(0)4x t dt X ∞-∞==⎰Substituting in the above expression for X(s), we have A=1/4. Therefore, 221/4().11()()4422X s s s s s =. Taking the Laplace transform of both sides of the two differential equations, we haves X(s)=1)(2+-s Y and s Y(s)=2X(s) . Solving for X(s) and Y(s), we obtain4)(2+=s s s X and Y(s)= 22s 4+.The region of convergence for both X(s) and Y(s) is Re{s}>0 because both are right-hand signals. . Taking the Laplace transform of both sides of the given differential equations ,we obtain ).(])1()1()[(223s X s s s s Y =+++++αααα therefore,.)1()1(1)()()(223αααα+++++==s s s s X s Y s H(a) Taking the Laplace transform of both sides of the given equation, we haveG(s) = s H(s)+ H(s). Substituting for H(s) from above,.1)1()1()1()(22223αααααα++=++++++=s s s s s s s GTherefore, G(s) has 2 poles.(b) we know that H(s) =.))(1(122αα+++s s s Therefore, H(s) has poles at and j ),2321(,1+--α ).2321(j --α If the system has to be stable,then the real part of the poles has to be less than zero. For this to be true, we require that ,02/<-α .,0>α.The overall system show in Figure may be treated as two feedback system of the form shown in figure connected in parallel. By carrying out an analysis similar to that described in Section we find the system function of the upper feedback system to be.82)/2(41/2)(1+=+=s s s s HSimilarly, the system function of the lower feedback system is .21)2/1(21/1)(2+=+=s s s HThe system function of the overall system is now.1610123)()()(221+++=+=s s s s H s H s HSince H(s)=Y(s)/X(s), we may write]123)[(]1610)[(2+=++s s X s s s Y . Taking the inverse Laplace transform, we obtaindtt dx t x t y dt t dy dt t y d )(3)(12)(16)(10)(2+=++. ( a) From problem , we know that differential equation relating the input and output of the RLC circuit is2()()()().d y t dy t y t x t dtdt++=Taking the Laplace transform of this (while nothing that the system is causal and stable), we obtain 2()[1]().Y s s s X s ++= Therefore ,2()1(),()1Y s H s X s s s ==++ 1{}.2e s ℜ>-(b) We note that H(s) has two poles at132s =--132s =-+From Section we know that the magnitude of the Fourier transform may be expressed as1313(Length of vector from to -+j )(Length of vector from to --j )ωω We see that the right hand side of the above expression Increases with increasing |ω| until |ω| reaches 12. Then it starts decreasing as |ω| increasing even further. It finally reaches 0 for |ω|=∞.Therefore 2|()|H j ω is approximately lowpass.(c) By repeating the analysis carried out in Problem and part (a) of this problem with R =310-Ω, wecan show that2()1(),()1Y s H s X s s s ==++ {}0.0005.e s ℜ>-(d) We have33(Vect.Len.from to -0.0005+j )(Vect.Len.from to -0.0005-j )ωωWe see that when |ω| is in he vicinity , the right-hand side of the above equation takes onextremely large value. On either side of this value of |ω| the value of |H (j ω)| rolls off rapidly. Therefore, H(s) may be considered to be approximately bandpass. . (a) The unilateral Laplace transform isX(s) = 20(1)t st e u t e dt -∞--+⎰= 20t st e e dt -∞--⎰=21+s {} 2.e s ℜ>-(b) The unilateral Laplace transform is2(3)0()[(1)()(1)]t st X s t t e u t e dt δδ-∞-+-=++++⎰2(3)0[()]t st t e e dt δ-∞-+-=+⎰612e s -=++ {} 2.e s ℜ>- (c) The unilateral Laplace transform is240()[()()]t t st X s e u t e u t e dt -∞---=⎰240[]t t st e e e dt -∞---=+⎰1124s s =+++ {} 2.e s ℜ>-. In Problem , we know that the input of the RL circuit are related by ).()()(t x t y dtt dy =+Applying the unilateral Laplace transform to this equation, we have ).()()0()(s x s y y s sy =+--(a) For the zero-state response, set (0)0y -=.Also we have u s x =)(L{)(2t u et-}=21+s .Therefore,y(s)(s+1)=.21+sComputing the partial fraction expansion of the right-hand side of the above equation and then taking its inverse unilateral Laplace transform, we have ).()()(2t u e t u e t y t t ---=(b) For the zero-state response, assume that x(t) = we are given that (0)1y -=,.11)(0)(1)(+=⇒=+-s s y s y s sy Taking the inverse unilateral Laplace transform, we have ()().t y t e u t -=Figure2()2()().t t y t e u t e u t --=- . The pole zero plots for all the subparts are shown in figure . (a) The Laplace transform of x(t) isX(s)= 230()t t st e e e dt ∞---+⎰= (2)(3)00[/(2)]|[/(3)]|s t s te s e s -+∞-+∞-++-+ =211252356s s s s s ++=++++(b) Using an approach similar to that show in part (a), we have41(),4L t e u t s -←−→+ {} 4.e s ℜ>-Also,551(),55L t j t e e u t s j -←−→+-and(){}551,555LT t j t e e u t e s s j --←−→ℜ>-++.From this we obtain()()()()55555215sin 52525LTt t j t t j t e t u t e e e e u t js ----⎡⎤=-←−→⎣⎦++ ,where {}5e s ℜ>- .Therefore,()()(){}245321570sin 5,51490100LTt t s s e u t e t u t e s s s s --+++←−→ℜ>-+++. -2 -3Im R a -2 R Im e Im R f Im Rg Im R h-2 2 4 -4 ImRd R b Im c R Im(c)The Laplace transform of ()x t is ()()023t t st X s e e e dt --∞=+⎰()()()()2300/2|/3|s t s t e s e s ----∞-∞⎡⎤⎡⎤=--+--⎣⎦⎣⎦ 211252356s s s s s -=+=---+.The region of convergence (ROC) is {}2e s ℜ<.(d)Using an approach along the lines of part (a),we obtain(){}21,22LT t e u t e s s -←−→ℜ>-+. Using an approach along the lines of part (c) ,we obtain(){}21,22LT t e u t e s s -←−→ℜ<-.From these we obtain()()222224t LT t t s e e u t e u t s --=+-←−→-, {}22e s -<ℜ<. Using the differentiation in the s-domain property , we obtain(){}22222228,2244t LT d s s te e s ds s s -+⎡⎤←−→-=--<ℜ<⎢⎥-⎣⎦-. (e)Using the differentiation in the s-domain property on eq.,we get()(){}2211,222LT t d te u t e s ds s s -⎡⎤←−→-=ℜ>-⎢⎥+⎣⎦+.Using the differentiation in the s-domain property on eq ,we get ()(){}2211,222LT t d te u t e s ds s s ⎡⎤--←−→=-ℜ<⎢⎥-⎣⎦-.Therefore,()()()(){}222224,2222t LT t t st e te u t te u t e s s s ---=--←−→-<ℜ<+-.(f)From the previous part ,we have ()()(){}2221,22LT t t t e u t te u t e s s -=--←−→-ℜ<-.(g)Note that the given signal may be written as ()()()1x t u t u t =-- .Note that (){}1,0LTu t e s s←−→ℜ>.Using the time shifting property ,we get(){}1,0s LT e u t e s s--←−→ℜ>.Therefore ,()1x t()()11,sLT e u t u t s----←−→ All s . Note that in this case ,since the signal is finite duration ,the ROC is the entire s-plane.(h)Consider the ()()()11x t t u t u t =--⎡⎤⎣⎦that the signal ()x t may beexpressed as ()()()112x t x t x t =+-+ . We have from the previous part()()11sLT e u t u t s----←−→, All s . Using the differentiation in s-domain property ,we have()()()12111s s s LT d e se e x t t u t u t ds ss ---⎡⎤--+=--←−→=⎡⎤⎢⎥⎣⎦⎣⎦, All s . Using the time-scaling property ,we obtain()121s s LT se e x t s --+-←−→, All s .Then ,using the shift property ,we have()21212s sLT s se e x t es ---+-+←−→ ,All s . Therefore ,()()()21122112s s s sLT s se e se e x t x t x t e s s----+--+=+-+←−→+, All s. (i) The Laplace transform of ()()()x t t u t δ=+ is (){}11/,0X s s e s =+ℜ>.(j) Note that ()()()()33t u t t u t δδ+=+.Therefore ,the Laplace transform is the same as the result of the previous part. (a)From Table ,we have()()()1sin 33x t t u t =.(b)From Table we know that()(){}2cos 3,09LT st u t e s s ←−→ℜ>+. Using the time scaling property ,we obtain()(){}2cos 3,09LT s t u t e s s -←−→-ℜ<+Therefore ,the inverse Laplace transform of ()X s is()()()cos 3x t t u t =--.(c)From Table we know that ()()(){}21cos 3,119LTt s e t u t e s s -←−→ℜ>-+.Using the time scaling property ,we obtain ()()(){}21cos 3,119LTt s e t u t e s s -+-←−→-ℜ<-++. Therefore ,the inverse Laplace transform of ()X s is()()()cos 3t x t e t u t -=--.(d)Using partial fraction expansion on ()X s ,we obtain ()2143X s s s =-++ .From the given ROC ,we know that ()x t must be a two-sided signal .Therefore ()()()432t t x t e u t e u t --=+-.(e)Using partial fraction expansion on ()X s ,we obtain()2132X s s s =-++. From the given ROC ,we know that ()x t must be a two-sided signal ,Therefore,()()()332ttx t e u t e u t --=+-.(f)We may rewrite ()X s as ()2311s X s s s =+-+()()2211/23/2s =-+()()()()222211/23/21/23/2s s =+-+-+Using Table ,we obtain()())())()/2/23cos 3/23sin3/2t t x t t e t u t e t u t δ--=+.(g)We may rewrite ()X s as ()()2311s X s s =-+.From Table ,we know that(){}21,0LT tu t e s s ←−→ℜ>.Using the shifting property ,we obtain()(){}21,11LT t e tu t e s s -←−→ℜ>-+.Using the differentiation property ,()()()(){}2,11LT t t t d s e tu t e u t te u t e s dt s ---⎡⎤=-←−→ℜ>-⎣⎦+. Therefore,()()()()33t t x t t e u t te u t δ--=--.four pole-zero plots shown may have the following possible ROCs:·Plot (a): {}2e s ℜ<- or {}22e s -<ℜ< or {}2e s ℜ>.·Plot (b): {}2e s ℜ<- or {}2e s ℜ>-. ·Plot (c): {}2e s ℜ< or {}2e s ℜ>. ·Plot (d): Entire s-plane.Also, suppose that the signal ()x t has a Laplace transform ()X s with ROC R . (1).We know from Table that()()33LT te x t X s -←−→+.The ROC 1R of this new Laplace transform is R shifted by 3 to the left .If ()3t x t e - is absolutely integrable, then 1R must include the jw -axis.·For plot (a), this is possible only if R was {}2e s ℜ> . ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ> . ·For plot (d),R is the entire s-plane. (2)We know from Table that(){}1,11LT t e u t e s s -←−→ℜ>-+.Also ,from Table we obtain()()(){}2,11LT t X s x t e u t R R e s s -⎡⎤*←−→=ℜ>-⎡⎤⎣⎦⎣⎦+ If ()()te u t x t -*is absolutely integrable, then 2R must include the jw -axis.·For plot (a), this is possible only if R was {}22e s -<ℜ<. ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(3)If ()0x t = for 1t > ,then the signal is a left-sided signal or a finite-duration signal . ·For plot (a), this is possible only if R was {}2e s ℜ<-. ·For plot (b), this is possible only if R was {}2e s ℜ<-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(4)If ()0x t =for 1t <-,then the signal is a right-sided signal or a finite-duration signal ·For plot (a), this is possible only if R was {}2e s ℜ>.·For plot (b), this is possible only if R was {}2e s ℜ>- . ·For plot (c), this is possible only if R was {}2e s ℜ>.·For plot (d),R is the entire s-plane..(a)The pole-zero diagram with the appropriate markings is shown Figure .(b)By inspecting the pole-zero diagram of part (a), it is clear that the pole-zero diagram shown in Figure will also result in the same ()X jw .This would correspond to the Laplace transform()112X s s =-, {}12e s ℜ<.(c)≮()X jw π=-≮()1X jw .(d)()2X s with the pole-zero diagram shown below in Figure would have the property that ≮()2X jw =≮()X jw .Here ,()211/2X s s -=-. (e) ()()21/X jw X jw =.(f)From the result of part (b),it is clear that ()1X s may be obtained by reflecting the poles and zeros in the right-half of the s-plane to the left-half of the s-plane .Therefore, ()11/22s X s s +=+.From part (d),it is clear that ()2X s may be obtained by reflecting the poles (zeros) in the right-half of the s-plane to the left-half and simultaneously changing them to zeros (poles).Therefore,()()()()2211/22s X s s s +=++plots are as shown in Figure . Table we have()()(){}2111,22LT t x t e u t X s e s s -=←−→=ℜ>-+and()()(){}3111,33LTt x t e u t X s e s s -=←−→=ℜ>-+.Using the time-shifting time-scaling properties from Table ,we obtain()(){}22112,22s LT s e x t e X s e s s ---←−→=ℜ>-+and()(){}33223,33s LT s e x t e X s e s s---+←−→-=ℜ>--.Therefore, using the convolution property we obtain ()()()()23122323s s LTe e y t x t x t Y s s s --⎡⎤⎡⎤=-*-+←−→=⎢⎥⎢⎥+-⎣⎦⎣⎦. clues 1 and 2,we know that ()X s is of the form()()()AX s s a s b =++. Furthermore , we are given that one of the poles of ()X s is 1j -+.Since ()x t is real, the poles of ()X s must occur in conjugate reciprocal pairs .Therefore, 1a j =-and 1b j =+and ()()()11AH s s j s j =+-++. From clue 5,we know that ()08X =.Therefore, we may deduce that 16A = and ()21622H s s s =++ .Let R denote the ROC of ()X s .From the pole locations we know that there are two possible choices of R .R may either be {}1e s ℜ<-or {}1e s ℜ>-.We will now useclue 4 to pick one .Note that()()()()22LTt y t e x t Y s X s =←−→=-.The ROC of ()Y s is R shifted by 2 to the right .Since it is given that ()y t is not absolutely integrable ,the ROC of ()Y s should not include the jw axis -.This is possible only ofR is {}1e s ℜ>-..(a) The possible ROCs are(i) {}2e s ℜ<-.(ii) {}21e s -<ℜ<-. (iii) {}11e s -<ℜ<.( iv) {}1e s ℜ>.(b)(i)Unstable and anticausal. (ii) Unstable and non causal. （iii ）Stable and non causal. (iv) Unstable and causal. .(a)Using Table ,we obtain(){}1,11X s e s s =ℜ>-+and(){}1, 2.2H s e s s =ℜ>-+(b) Since ()()()y t x t h t =*,we may use the convolution property to obtain()()()()()112Y s X s H s s s ==++.The ROC of ()Y s is {}1e s ℜ>-.(c) Performing partial fraction expansion on ()Y s ,we obtain . ()1112Y s s s =-++.Taking the inverse Laplace transform, we get()()()2t t y t e u t e u t --=-. (d)Explicit convolution of ()x t and ()h t gives us()()()y t h x t d τττ∞-∞=-⎰()()20t e e u t d ττττ∞---=-⎰t t e e d ττ--=⎰ for0t >()2.t t e e u t --⎡⎤=-⎣⎦ the input ()()x t u t =, the Laplace transform is (){}1,0.X s e s s=ℜ>The corresponding output ()()1t t y t e te u t --⎡⎤=--⎣⎦ has the Laplace transform()()(){}221111,0111Y s e s s s s s s =--=ℜ>+++. Therefore,()()()(){}21,0.1Y s H s e s X s s ==ℜ>+ Now ,the output ()()3123t t y t e e u t --⎡⎤=-+⎣⎦has the Laplace transform()()(){}12316,0.1313Y s e s s s s s s s =-+=ℜ>++++ Therefore , the Laplace transform of the corresponding input will be()()()()(){}1161,0.3Y s s X s e s H s s s +==ℜ>+ Taking the inverse Laplace transform of the partial fraction expansion of ()1,X s we obtain ()()()3124.t x t u t e u t -=+.(a).Taking the Laplace transform of both sides of the given differential equation and simplifying, we obtain ()()()212Y s H s X s s s ==--. The pole-zero plot for ()H s is as shown in figure .b).The partial fraction expansion of ()H s is()1/31/321H s s s =--+. (i).If the system is stable ,the ROC for ()H s has to be {}12e s -<ℜ< . Therefore ()()()21133t t h t e u t e u t -=---.(ii).If the system is causal, the ROC for ()H s has to be {}2e s ℜ> .Therefore()()()21133t t h t e u t e u t -=-.(iii)If the system is neither stable nor causal ,the ROC for ()H s has to be {}1e s ℜ<-.Therefore ,()()()21133t t h t e u t e u t -=--+-. If ()2t x t e =produces ()()21/6t y t e =,then ()()21/6H =. Also, by taking the Laplace transform of both sides ofthe given differential equation we get ()()()()442s b s H s s s s ++=++.Since ()21/6H = ,we may deduce that 1b = .Therefore()()()()()222424s H s s s s s s +==+++. ()()()t t t x t e e u t e u t --==+-,()()(){}112,111111X s e s s s s s -=-=-<ℜ<+-+-. We are also given that ()2122s H s s s +=++.Since the poles of ()H s are at 1j -±, and since ()h t is causal ,we may conclude that the ROC of()H s is {}1e s ℜ>-.Now()()()()()22221Y s H s X s s s s -==++-. The ROC of ()Y s will be the intersection of the ROCs of ()X s and ()H s .This is {}11e s -<ℜ<. We may obtain the following partial fraction expansion for ()Y s :()22/52/56/5122s Y s s s s +=-+-++. We may rewrite this as ()()()222/521411551111s Y s s s s ⎡⎤⎡⎤+=-++⎢⎥⎢⎥-++++⎢⎥⎢⎥⎣⎦⎣⎦.0 -1 2 ReImFigureNothing that the ROC of ()Y s is {}11e s -<ℜ<and using ,we obtain ()()()()224cos sin 555t t t y t e u t e tu t e tu t --=-++know that()()(){}111,0LTx t u t X s e s s=←−→=ℜ> Therefore,()1X s has a pole at0s =.Now ,the Laplace transform of the output()1y t of the system with()1x t as the input is()()()11Y s H s X s =Since in clue 2, ()1Y s is given to be absolutely integrable ,()H s must have a zero at 0s =whichcancels out the pole of ()1X s at 0s =.We also know that()()(){}2221,0LT x t tu t X s e s s=←−→=ℜ> Therefore , ()2x s has two poles at 0s =.Now ,the Laplace transform of the output ()2y t of the system with ()2x t as the input is()()()22Y s H s X s =Since in clue 3, ()2Y s is given to be not absolutely integrable ,()H s does not have two zeros at0s =.Therefore ,we conclude that ()H s has exactly one zero at 0s =. From clue 4 we know that the signal ()()()()2222d h t dh t p t h t dt dt=++is finite duration .Taking the Laplace transform of both sides of the above equation ,we get ()()()()222P s s H s sH s H s =++. Therefore,()()222P s H s s s =++.Since ()p t is of finite duration, we know that ()P s will have no poles in the finite s-plane .Therefore, ()H s is of the form()()1222Ni i A s z H s s s =-=++∏,where i z ,1,2,....,i N =represent the zeros of ()P s .Here ,A is some constant.From clue 5 we know that the denominator polynomial of ()H s has to have a degree which is exactly one greater than the degree of the numerator polynomial .Therefore, ()()1222A s s H s s s -=++.Since we already know that ()H s has a zero at 0s = ,we may rewrite this as ()222As H s s s =++ From clue 1 we know that ()1H 0.2 this ,we may easily show that 1A = .Therefore,()222s H s s s =++. Since the poles of ()H s are at 1j -± and since ()h t is causal and stable ,the ROC of ()H s is {}1e s ℜ>-..(a) We may redraw the given block diagram as shown in Figure . From the figure ,it is clear that()()1F s Y s s=. Therefore, ()()1/f t dy t dt =. Similarly, ()()/e t df t dt =.Therefore, ()()221/e t d y t dt =.From the block diagram it is clear that()()()()()()()21111266d y t dy t y t e t f t y t y t dtdt=--=--.。
信号处理导论(英文影印版) 习题答案_9-11章
9.12⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−−−−−−−−=0001221121221111111111111111111111111)1j j j j j j j j DFTdirect X 课后答案网ww w.kh da w.co m9.18或其它或或其它]2154300[]0154320[)2]154320[)1===x x x 本题考查两点基本概念：1）一序列先进行mod －N reduction 后再作N －point DFT 与其直接作N －point DFT 结果相等。
2）不同序列有可能有相同的mod －N reduction 结果课后答案网ww w.kh da w.co m9.30碟形图见黑板）)2]28682868[~]3101124211781744[1−−=−−−−−=∗=y h x y 课后答案网ww w.kh da w.co m9.41]1010101010120202020101[]1010101010120202020101[]00010201[~],10101212[~],01012121[~],02020202[~],20101323[~]00000011[]01111222[]22221113[]11333311[]31111000[333],1111[h 8815432154321−−−−−−=∗=−−−−−−=−=−−−=−−−=−−=−−=⎪⎪⎪⎩⎪⎪⎪⎨⎧=====∴=∴−−=∴−−h x y y y y y y y x x x x x M ns convolutio circular length methodsave overlap ，即每段间重叠阶数为每段长度取要做）Q Q 课后答案网ww w.kh da w.co m]1010101010120202020101[]01010101[~],20201111[~],10102303[~],32320101[~]01111[]22221[],11333[],31111[5x (888)241443322114321−−−−−−=−−==−−−==−−==−−−=======∴≤+≤∴−−y y y y y y y y y y y x x x x y 重叠相加得到以上取为滤波器阶数）的长度，为每子段即每段卷积输出长度要做N M N M N L ns convolutio circular length methodadd overlap i Q 课后答案网ww w.kh da w.co m)()()()()()3()()()()()()()()()()()()()()2()()()()()(ωωωωωωωωωωωωωωωω−===−∴−=−=−==−∴−=−=−==∑∑∑∑∑∑∑∑∞+−∞=−−∞+−∞=∗∞+−∞=−∞+−∞=−−=∞+−∞=∞+−∞=−∞+−∞=−−=∞+−∞=∗D ek d ek d D symmetrynot butvalue real only is k d oddis D D el d el d ek d D imaginaryis D D el d el d ek d D realandodd is k d k kj k kj k d l lj k d l lj l k k kj l lj k d l lj l k k kj k d 为实数为奇函数为奇函数为实数课后答案网 ww w.kh da w.co m。
信号与系统 Chapter9 SS-L3
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9 The Laplace Transform
9.9 The Unilateral LT
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Definition: X (s) x(t)est dt 0
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Example 9.28
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奥本海姆《信号与系统》(第2版)课后习题-第7章至第9章(下册)(圣才出品)
第二部分课后习题第7章采样基本题7.1已知实值信号x（t），当采样频率时，x（t）能用它的样本值唯一确定。
问在什么ω值下保证为零？解：对于因其为实函数，故是偶函数。
由题意及采样定理知的最大角频率即当时，7.2连续时间信号x（t）从一个截止频率为的理想低通滤波器的输出得到，如果对x（t）完成冲激串采样，那么下列采样周期中的哪一些可能保证x（t）在利用一个合适的低通滤波器后能从它的样本中得到恢复？解：因为x（t）是某个截止频率的理想低通滤波器的输出信号，所以x（t）的最大频率就为＝1000π，由采样定理知，若对其进行冲激采样且欲由其采样m点恢复出x（t），需采样频率即采样时间问隔从而有（a）和（c）两种采样时间间隔均能保证x（t）由其采样点恢复，而（b）不能。
7.3在采样定理中，采样频率必须要超过的那个频率称为奈奎斯特率。
试确定下列各信号的奈奎斯特率：解：（a）x（t）的频谱函数为由此可见故奈奎斯特频率为（b）x（t）的频谱函数为由此可见故奈奎斯特频率为（c）x（t）的频谱函数为由此可见，当故奈奎斯特频率为7.4设x（t）是一个奈奎斯特率为ω0的信号，试确定下列各信号的奈奎斯特率：解：（a）因为的傅里叶变换为可见x（t）的最大频率也是的最大频率，故的奈奎斯特频率为0 。
（b）因为的傅里叶变换为可见x （t）的最大频率也是的最大频率．故的奈奎斯特频率仍为。
（c）因为的傅里叶变换蔓可见的最大频率是x（t）的2倍。
从而知x 2（t）的奈奎斯特频率为2（d）因为的傅里叶变换为，x（t）的最大频率为，故的最大频率为，从而可推知其奈奎斯特频率为7.5设x（t）是一个奈奎斯特率为ω0的信号，同时设其中。
当某一滤波器以Y（t）为输入，x（t）为输出时，试给出该滤波器频率响应的模和相位特性上的限制。
解：p（t）是一冲激串，间隔对x（t）用p（t-1）进行冲激采样。
先分别求出P（t）和P（t-1）的频谱函数：注意0ω是x（t）的奈奎斯特频率，这意味着x（t）的最大频率为02ω，当以p（t-1）对x（t）进行采样时，频谱无混叠发生。
《信号与系统》奥本海姆英文版课后答案
2
2 (f) x [n] = . Therefore, E = = = , 3 cos x cos( n) cos( n) 3[n]
4
4
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2 )1 lim 1 ( 2 2 N 2 N 1 n N 1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal x[n] is flipped signal will be zero for n<-1 and n>2. (d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4. (e) The signal x[n] 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 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 t<1.
英文版《信号与系统》第九章习题解答
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Chapter 9
Problem Solution
数字信号处理英文版课后答案 (9)
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h 1 [n ]
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h 1 [n ] 0.5n sin 0.5n 0.5 sinc0.5n ) n 0.5n
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1 x[n] x[n 1] x[n 2] 3
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信号与系统-奥本海姆-中文答案-chapter-9
第九章9.6 解：(a) 若是有限持续期信号Roc 为整个s 平面，故存在极点不可能，故不可能为有限持续期。
(b) 可能是左边的。
(c) 不可能是右边的，若是右边信号，它并不是绝对可积的。
(d) x(t)可能为双边的。
9.8 解：因为te t x t g 2)()(=的傅氏变换，)(ωj G 收敛所以)(t x 绝对可积若)(t x 为左边或者右边信号，则)(t x 不绝对可积故)(t x 为双边信号9.10 解：(a) 低通(b) 带通(c) 高通9.14 解：dt e t x s X st ⎰∞∞--=)()(， 由)(t x 是偶函数可得)()()(t d e t x s X st --=⎰-∞∞ dt e t x t s ⎰∞∞----=)()(dt e t x t s ⎰∞∞---=)()( )(s X -=421πj e s =为极点，故421πj e s -=也为极点，由)(t x 是实信号可知其极点成对出现，故421πj e s -=与421πj e s --=也为极点。
)21)(21)(21)(21()(4444ππππj j j j e s e s e s e s Ms X --++--=由⎰∞∞-=4)(dt t x 得 4)0(=x所以，M ＝1/4 即，42}Re{42<<-s 9.21 解： (a) 3121)(+++=s s s X 2}Re{->s (b) 25)5(541)(2++++=s s s X 4}Re{->s (c) 3121)(----=s s s X 2}Re{<s (d) 22)2(1)2(1)(--+=s s s X 2}Re{2<<-s (e) 22)2(1)2(1)(-++-=s s s X 2}Re{2<<-s (f) 2)2(1)(-=s s X 2}Re{<s (g) )1(1)(s e ss X --= 0}Re{>s (h) 22)1()(se s X s -=- 0}Re{>s(i) ss X 11)(+= 0}Re{>s (j) ss X 131)(+= 0}Re{>s9.23 解：1． Roc 包括 Re{s}=32． Roc 包括 Re{s}=03． Roc 在最左边极点的左边4． Roc 在最右边极点的右边图1：1，2}Re{>s2，2}Re{2<<-s3，2}Re{-<s4，2}Re{>s图2： 1，2}Re{->s2，2}Re{->s3，2}Re{-<s4，2}Re{->s图3： 1，2}Re{>s2，2}Re{<s3，2}Re{<s4，2}Re{>s图4： 1，S 为整个平面2，S 为整个平面3，S 为整个平面4，S 为整个平面9.25 解：图略9.27 解：)(t x Θ为实信号，)(s X 有一个极点为j s +-=1)(s X ∴另一个极点为j s --=1)1)(1()(j s j s M s X ++-+=∴ 又Θ8)0(=X16=∴M 则，)1(8)1(8)(j s j j s j s X -+-++= 1}Re{->s 或者1}Re{-<s 之一使其成立又 )(2t x e tΘ不是绝对可积的 ∴ 对任一个s ，右移2，不一定在Roc 中因此，1}Re{-<s9.35 解： (a) )(1)(*)(s X st u t x L −→−Θ 那么方框图表示的方程为)(*)(*)(6)(*)()()(*)(*)()(*)(2)(t u t u t y t u t y t y t u t u t x t u t x t x --=++即 ⎰⎰⎰⎰⎰⎰∞-∞-∞-∞-∞-∞---=++t tt t t t dt d y d y t y dt d x d x t x ττττττττ)(6)()()()(2)( 对两边求导可得)(6)()()()()(2222t x dt t dx dtt x d t y dt t dy dt t y d --=++ (b) 126)(22++--=s s s s s H 121-==s s 是)(s H 的二重极点，由于系统是因果的所以 1}Re{->sRoc 包含虚轴，所以系统是稳定的。
英文版《信号与系统》第二章习题解答
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信号与系统答案 西北工业大学 段哲民 第九章 习题
第九章 习题9.1 图题9.1所示电路，已知)(1t x 与)(2t x 为状态变量，试证明以下各对变量是否都可以作为状态变量。
)(),()6();(),()5();(),()4();(),()3();(),()2();(),()1(2131t i t i t u t i t u t i t u t u t u t i y u t i R R R C L C L R C C L L 。
9.1图题++--3R u L u答案解 本题说明状态变量的选取不是唯一的。
若各组变量之间存在着非奇异线性变换关系，则这些变量组即可作为状态变量。
又因为研究的是电路的状态，故可采取无激励电路（即令电路中的激励均为零）。
（1） )()(2t x t i L =)()()(231t x R t x t u L -=即 ⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=⎥⎦⎤⎢⎣⎡)()(110)()(213t x t x R t u t i L L（2）=---=)()(1)(1)(21211t x t x R t x R t i C)()()11(2121t x t x R R -+-)()(1t x t u C =即 ⎥⎦⎤⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡+-=⎥⎦⎤⎢⎣⎡)()(011)11()()(2121t x t x R R t u t i C C（3） )(11t x u R -=)()()(231t x R t x t u L -=即 ⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=⎥⎦⎤⎢⎣⎡)()(101)()(2131t x t x R t u t u L R（4）)()()11()(2121t x t x R R t i C -+-=)()()(231t x R t x t u L -=即⎥⎦⎤⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡--+-=⎥⎦⎤⎢⎣⎡)()(11)11()()(21321t x t x R R R t u t i L C（5）)()()11()(2121t x t x R R t i C -+-=)()(233t x R t u R =即⎥⎦⎤⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡)()(01)11()()(213213t x t x R R R t u t i R C可见以上5对变量的变换矩阵，其行列式的值均不为零，即它们均为非奇异矩阵，故以上的五对变量组均可作为该电路的状态变量。
信号与系统第二版课后习题解答(6-7-9)奥本海姆
Chap 66.1 Consider a continuous-time LTI system with frequency response()()|()|H j H j H j e ωωω=and real impulse response h(t). Suppose that we apply an input 00()cos()x t t ωφ=+ to this system .The resulting output can be shown to be of the form0()()y t Ax t t =-Where A is a nonnegative real number representing anamplitude-scaling factor and 0t is a time delay.(a)Express A in terms of |()|H j ω.(b)Express 0t in terms of0()H j ω Solution:(a) For 0()()y t Ax t t =-So 0()()jt Y j AX j eωωω-= 0()()()j t Y j H j Ae X j ωωωω-== So |()|A H j ω=(b) for 0()H j t ωω=- So 0()H j t ωω=-6.3 Consider the following frequency response for a causal and stable LTI system:1()1j H j j ωωω-=+ (a) Show that |()|H j A ω=,and determine the values of A. (b)Determine which of the following statements is true about ()τω,the group delay of the system.(Note()(())/d H j d τωωω=-,where ()H j ωis expressed in aform that does not contain any discontinuities.)1.()0 0for τωω=>2.()0 0for τωω>>3 ()0 0for τωω<>Solution:(a) for |()|1H j ω== So A=1(b) for )(2)()()1()1()(ωωωωωωarctg arctg arctg j j j H -=--=+∠--∠=∠ 212)()(ωωωωτ+=∠-=d j H d So ()0 0for τωω>>6.5 Consider a continuous-time ideal bandpass filter whose frequency response is⎩⎨⎧≤≤=elsewherej H c c,03||,1)(ωωωω (a) If h(t) is the impulse response of this filter, determine a functiong(t) such that)(sin )(t g t t t h c πω=(b) As c ω is increased, dose the impulse response of the filter get more concentrated or less concentrated about the origin?Solution(a) Method 1. Let1()()()()()()2h t x t g t H j X j G j ωωωπ=↔=* They are shown in the figures,where1,sin ()(){0,c c ctx t X j t ωωωωωωπ<=↔=> So we can get()2cos(2)()2[(2)(2)]c c c g t t G j ωωπδωωδωω=↔=-++Method 2. Using the inverse FT definition,it is obtained331(){}2c c c cj t j t h t e d e d ωωωωωωωωπ--=+⎰⎰ 11{sin 3sin }{sin }{2cos 2}c c c c t t t t t tωωωωππ=-= (b) more concentrated.Chap 77.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is10,000s ωπ=.For what values ofω is ()X j ω guaranteed to be zero? Solution:According to the sampling theorem 2s M w w > That is 110000500022M s w w ππ<== So if 5000M w w π>=,0)(=jw X7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency 1,000c ωπ=.If impulse-train sampling is performed on x(t), which of the following sampling periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter?(a) 30.510T -=⨯(b) 3210T -=⨯(c) 410T -= Solution: π1000==c M w wFrom the sampling theorem,∴π20002=>M s w w ,that is 3102000222-==<πππM s w T ∴the conditions (a) and (c) are satisfied with the sampling theorem,(b) is not satisfied.7.3 The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:(a)()1cos(2,000)sin(4,000)x t t t ππ=++ (b)sin(4,000)()t x t tππ=(c) 2sin(4,000)()()t x t t ππ= Solution: (a) )4000sin()2000cos(1)(t t t x ππ++=max(0,2000,4000)4000M w πππ==∴ the Nyquist rate is 28000s M w w π>= (b) sin(4000)()t x t tππ= 4000M w π=∴ the Nyquist rate is 28000s M w w π>= (c) 2sin(4000)()t x t t ππ⎛⎫= ⎪⎝⎭ 2sin(4000)()t x tt ππ⎛⎫= ⎪⎝⎭221(1cos(8000))2t t ππ=- ∴8000M w π=∴the Nyquist rate is 216000s M w w π>=7.4 Let x(t) be a signal with Nyquist rate 0ω. Determine the Nyquist rate for each of the following signals:(a)()(1)x t x t +- (b)()dx t dt(c)2()x t(d)0()cos x t t ωSolution:(a) we let 1()()(1)y t x t x t =+-So 1()()()(1)()j j Y j X j e X j e X j ωωωωωω--=+=+ So the Nyquist rate of signal (a) is 0ω.(b) we let 2()()dx t y t dt= So 2()()Y j j X j ωωω=So the Nyquist rate of signal (b) is0ω. (c) we let 23()()y t x t = So 31()()*()2Y j X j X j ωωωπ= So the Nyquist rate of signal (c) is 20ω.(d) we let 40()()cos y t x t t ω=For 000cos [()()]FT t ωπδωωδωω→-++ So 4001()((()(())2Y j X j X j ωωωωω=-++ So the Nyquist rate of signal (d) is 03ω7.9 Consider the signal 2sin 50()()t x t tππ= Which we wish to sample with a sampling frequency of 150s ωπ= to obtain a signal g(t) with Fourier transform ()G j ω.Determine the maximum value of 0ω for which it is guaranteed that0()75() ||G j X j for ωωωω=≤Where ()X j ω is the Fourier transform of x(t).Solution: 2sin(50)()t x t t ππ⎛⎫= ⎪⎝⎭))100cos(1(2122t t ππ-= ∴100M w π=But π150=s wthe figure about before-sampling and after-sampling of )(jw H isWe can see that only when π500≤w , the before-sampling and after-sampling of )(jw H have the same figure.So if 0..)..(75)(w w for jw X jw G ≤=The maximum value of 0w is π50.Chap 99.2 Consider the signal 5()(1)t x t e u t -=- and denote its Laplace transform by X(s).(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and 0t such that the Laplace transform G(s) of 50()()t g t Ae u t t -=-- has the same algebraic form as X(s).what is the region of convergencecorresponding to G(s)?Solution:(a). According to eq.(9.3), we will getdt e t x s X st -∞∞-⎰=)()(dt e t u e st t --∞∞--=⎰)1(5dt e t s )5(1+-∞⎰=)5()5()5()5()5(1)5(+=+--=+-=+-+-∞+-s e s e s e s s t s ROC:Re{s}>-5 (b). )()(05t t u Ae t g t --=-−→←LT 0)5(5)(t s e s A s G ++-=, Re{s}<-5 ∴ If )()(s X s G =then it ’s obviously that A=-1, 10-=t , Re{s}<-5.9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity: (a)1113s s +++ (b) 211s s +- (c) 3211s s s -++ Solution :(a).1, 1)3)(1(423111+++=+++s s s s s ∴ it has a zero in the finite s-plane, that is 2-=sAnd because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(b). 0, 111)1)(1(1112-=-++=-+s s s s s s ∴ it has no zero in the finite s-plane.And because the order of the denominator exceeds the order of the numerator by 1∴ X(s) has 1 zero at infinity.(c). 1, 011)1)(1(112223-=++++-=++-s s s s s s s s s ∴ it has a zero in the finite s-plane, that is 1=sAnd because the order of the denominator equals to the order of the numerator∴ X(s) has no zero at infinity.9.7 How many signals have a Laplace transform that may be expressed as 2(1)(2)(3)(1)s s s s s -++++ in its region of convergence?Solution:There are 4 poles in the expression, but only 3 of them have different real part.∴ The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across.∴ There are 4 signals having the same Laplace transform expression.9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If2()() ()t g t e x t and G j ω=[ the Fourier transform of g(t)]converges, determine whether x(t) is left sided, right sided, or two sided.Solution:)()(2t x e t g t =∴)2()(-=s X s G ROC: R(x)+Re{2}And x(t) have three possible ROC strips:),1(),1,3(),3,(+∞-----∞∴g(t) have three possible ROC strips: ),1(),1,1(),1,(+∞---∞ IF jw s s G jw G ==|)()(Then the ROC of )(s G is (-1,1)∴)(t x is two sides. 9.9 Given that1(),{}Re{}sat e u t Re s a s a -↔>-+ Determine the inverse Laplace transform of22(2)(),Re{}3712s X s s s s +=>-++ Solution: It is obtained from the partial-fractional expansion:22(2)2(2)42()712(4)(3)43s s X s s s s s s s ++-===+++++++,Re{}3s >-We can get the inverse Laplace transform from given formula and linear property.43()4()2()t t x t e u t e u t --=-9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a): 1}Re{,.........)3)(1(1)(1->++=s s s s H (b): 221(),{}12s H s e s s s =ℜ>-++(c): 232(),{}121s H s e s s s =ℜ>-++ Solution:(a). 1}Re{,.........)3)(1(1)(1->++=s s s s H It ’s lowpass.(b).21}Re{,.........1)(22->++=s s s s s H It ’s bandpass.(c). 1}Re{., (1)2)(223->++=s s s s s H It ’s highpass.9.13 Let ()()()g t x t x t α=+- ,Where ()()t x t e u t β-=. Andthe Laplace transform of g(t) is 2(),1{}11s G s e s s =-<ℜ<-. Determine the values of the constantsαand βSolution: ()()()g t x t x t α=+-,and ()()t x t e u t β-=The Laplace transform : ()()()G s X s X s α=+- and()1X s s β=+,Re{}1s >- From the scale property of Laplace transform, ()1X s s β-=-+,Re{}1s < So 2(1)(1)()()()111s G s X s X s s s s βαββαβαα--+=+-=+=+-+-,1Re{}1s -<< From given 2()1s G s s =-,1Re{}1s -<< We can determine : 11,2αβ=-=。
信号与系统习题集英文版
信号与系统习题集第一部分：时域分析一、填空题1. =---)3()()2(t t u e t δ ( )。
2. The unit step response )(t g is the zero-state response when the input signal is( ).3. Given two continuous – time signals x(t) and h(t), if their convolution is denoted by y(t), then the convolution of )1(-t x and )1(+t h is ( ).4. The convolution =+-)(*)(21t t t t x δ( ).5. The unit impulse response )(t h is the zero-state response when the input signal is( ).6. A continuous – time LTI system is stable if its unit impulse response satisfies the condition: ( ) .7. A continuous – time LTI system can be completely determined by its ( ). 8. =⎰∞∞-(t)dt 2sin 2 δtt ( ). 9. Given two sequences }1,2,2,1{][=n x and }5,6,3{][=n h , their convolution =][*][n h n x ( ).10. Given three LTI systems S1, S2 and S3, their unit impulse responses are )(1t h , )(2t h and )(3t h respectively. Now, construct an LTI system S using these three systems: S1 parallel interconnected by S2, then series interconnected by S3. the unit impulse response of the system S is ( ).11. It is known that the zero-stat response of a system to the input signal x(t) is ⎰∞-=td x t y ττ)()(, then the unit impulse response h(t) is ( ). 12. The complete response of an LTI system can be expressed as a sum of its zero-state response and its ( ) response.13. It is known that the unit step response of an LTI system is )(2t u e t -, then the unit impulse response h(t) is ( ). 14. =++-=⎰∞dt t t t t x ))1()1((2sin )(0δδπ( ). 15. We can build a continuous-time LTI system using the following three basicoperations: ( ) , ( ), and ( ).16. The zero-state response of an LTI system to the input signal )1()()(--=t u t u t xis )1()(--t s t s , where s(t) is the unit step response of the system, then the unit impulse response h(t) is ( ).17. The block diagram of a continuous-time LTI system is illustrated in the following figure. The differential equation describing the input-output relationship of the system is ( )。
信号与系统奥本海姆第九章答案
s s s s H , Re{s}>-1 ∴ )(6)()()()(2)(2222t x dt t dx dt t x d t y dt t dy dt t y d ??=++ (b) It’s obviouse that this system is stable.
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信号与系统奥本海姆第九章答案
Chapter 9 9.21 Solution: (a). Q
)()()(32t u e t u e t x t t ??+= ∴ )3)(2(523121)(+++=+++=s s s s s s X , 2}Re{?>s (b). Q )()5(sin )()(54t u t e t u e t x t t ??+= ∴
13/123/1)1)(2(121)()()(2+?+?=+?=??==s s s s s s s X s Y s H (b). 1. The system is stable. ∴ ROC: (-1,2) ∴ )()()(31231t u e t u e t h t t ????= 2. The system is causal. ∴ ROC: ),2(+∞ ∴ )()()(31231t u e t u e t h t t ??= 3. The system is neither stable nor causal ∴ ROC: )1,(??∞ ∴ ) ()()(31231t u e t u e t h t t ?+??=? 9.32. Solution: from (1) Q t e t x 2)(=, for all t and x(t) is a eigen function ∴ t t s e e s H t y 2226 1|)()(=?== ∴ 61|)(2==s s H from (2) Q )()()()(2)(4t bu t u e t h dt t dh t +=+? ∴ s b s s h s sH ++=+41)(2)( ∴ ) 2)(4()4()(++++=s s s s b s s H when 2=s , 6146262)2(=××+= b h ∴ 862=+b ,1=b ∴ 0}Re{,........) 4(2)2)(4()2(2)(>+=+++=s s s s s s s s H 9.33. Solution: Q )()()(||t u e t u e e t x t t t ?+==?? ∴ ) 1)(1(21111)(?+?=??+=s s s s s X ∴ )()()(s H s X s Y =2 21)1)(1(22+++??+?= s s s s s )22)(1(22++??=s s s ) 22(5652)1(522++++??=s s s s 1)1(541)1()1(52)1(521)1(54)1(52)1(52222+++++++??=+++++??=s s s s s s s ∴ )(sin 5 4)(cos 52)(52)(t tu e t tu e t u e t h t t t ??++?= 9.35. Solution: According to the block-diagram, we will know (a) 126121611)(2222++??=++??=s s s s s
信号与系统(双语)_福州大学中国大学mooc课后章节答案期末考试题库2023年
信号与系统（双语）_福州大学中国大学mooc课后章节答案期末考试题库2023年1.People often refer to the sampling of discrete-time signal as ( )参考答案:resampling2.The situation in which convert a sequence to a higher equivalent samplingrate is often referred to as ( )参考答案:interpolation3.The response generated solely by the system's initial energy strorage iscalled as ( )参考答案:zero-input response4.The narrow band signal's central frequency is ( ) the origin.参考答案:far from5.Though the convolution plays an important role in the analyzing the outputof LTI systems, it can only be used to analyzing the ____________________of LTIsystems. ( )参考答案:Zero-State response6.Given a system y[n]=2x[n]+3, Please determine which of the followings isWRONG. ( )参考答案:Memoryless7.The response of a LTI system to a complex exponential is ( ) complexexponential with a change in ( )参考答案:the same, amplitude8.Which of the following actions or devices can restore or rconstruct adiscrete-time signal into a continuous-time signal by interpolation method? ( )参考答案:D/A converter。
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2 H (s ) = s (s + 4 )
Re{s} > 0
The system is unstable.
15
Chapter 9 9.33 A causal LTI system H (s ) =
x(t ) = e
t
Problem Solution
s +1 s 2 + 2s + 2
, ∞ < t < +∞
Determine and sketch the response y (t )
2 X (s ) = 2 s 1
1 < Re{s} < 1 Re{s} > 1
2(s + 1) (s + 1)(s 1) s 2 + 2s + 2
s +1 H (s ) = 2 s + 2s + 2 Y (s ) = X (s )H (s ) =
2 2 4 y (t ) = + e t u ( t ) + e t (cos t )u (t ) + e t (sin t )u (t ) 5 5 5
17
Chapter 9
Problem Solution
9.35 Consider a causal LTI system with the input x(t ) and output y (t ) . (b) Is this system stable?
2
②
1
③
1
④
2
σ
① Re{s} < 2 anticausal , unstable ② -2 < Re{s} < 1 noncausal , unstable
③ -1 < Re{s} < 1
noncausal , stable Causal , unstable
12
④ Re{s} > 1
Chapter 9 9.31 Consider a continuous-time LTI system (a) Determine H (s ) .
Determine the values of the constants αand β.
x(t ) = βe t u (t ) ← → x( t ) ← →
x(t ) + αx( t ) ← →
β
s +1
Re{s} > 1 Re{s} < 1
αβ
α = 1 1 β= 2
β
s +1
β
s +1 s +1
Re{s} < 1
1 1 h(t ) = + e t u ( t ) e 2t u ( t ) 3 3
13
Chapter 9 9.32 Consider a causal LTI system ,
1. x(t ) = e
2t
Problem Solution
dh(t ) 2. + 2h(t ) = e 4t u (t ) + bu (t ) dt
1 2t (-∞ < t < +∞ ) → y (t ) = e 6
(-∞ < t < +∞ )
b——unknown constant
Determine the system function H (s )and b. Solution sH (s ) + 2 H (s ) = 1 + b s+4 s
2
}ห้องสมุดไป่ตู้
Re{s} > 4
8
Chapter 9
Problem Solution
(i ) x(t ) = δ (t ) + u (t )
X (s ) = 1+ 1 s
Re{s} > 0
(j) x(t ) = δ (3t ) + u (3t )
1 x(t ) = δ (t ) + u (t ) 3
1 1 X (s ) = + 3 s
Homework: 9.2 9.5 9.7 9.8 9.9 9.13 9.21(a,b,i,j)
9.22(a,b,c,d) 9.28 9.31 9.32 9.33 9.35 9.45
1
Chapter 9 9.2 Consider the signal x(t ) = e 5t u (t 1) (a) Evaluate X (s )
14
Chapter 9
1 2t y (t ) = H (2)e = e 6
2t
Problem Solution
(-∞ < t < +∞ )
1 b(2 + 4) + 2 H (2) = = b =1 2(2 + 2)(2 + 4 ) 6
b (s + 4 ) + s H (s ) = s (s + 2 )(s + 4 )
(a )
(b )
1 1 2s + 4 + = s + 1 s + 3 (s + 1)(s + 3)
1 s +1 = 2 s 1 s 1
1 , 2
0, 1
(c)
s3 1 = s 1 2 s + s +1
1, 0
3
Chapter 9
Problem Solution
9.7 How many signals have a Laplace transform that may be s 1 expressed as
Problem Solution
(b) g (t ) = Ae 5t u ( t t0 ) G (s ) = X (s ) A, t0 , ROC = ?
L (t ) ←→ 1 e u
5t
s+5
s L e e 5(t 1)u (t 1) ←→
s+5
e (s +5 ) L e 5t u (t 1) ←→ Re{s} > 5 s+5 e (s +5 ) L e 5e 5(t 1)u ( t + 1) ←→ Re{s} < 5 s+5 g (t ) = e 5t u ( t + 1)
2 4 + X (s ) = Re{s} > 3 s+3 s+4
x(t ) = 2e 3t + 4e 4t u (t )
(
)
6
Chapter 9
Problem Solution
t 9.13 Let g (t ) = x(t ) + αx( t ) , where x(t ) = βe u (t ) s and G (s ) = 2 -1 < Re{s} < 1 s 1
+
(β αβ )s β αβ s = G (s ) = 2 s2 1 s 1
7
Chapter 9 9.21 Determine the Laplace transform.
Problem Solution
(a ) x(t ) = e 2t u (t ) + e 3t u (t )
X (s ) = 1 1 2s + 5 + = Re{s} > 2 s + 2 s + 3 (s + 2 )(s + 3)
(s + 2)(s + 3)(s 2 + s + 1)
in its region of convergence?
jω
Poles :
1 3 + j 2 2
s1 = 2, s1 = 3, s3, 4
1 3 = ± j 2 2
3 2
1 3 j 2 2
σ
There are four signals.
4
Chapter 9
Problem Solution
9.8 Let x(t ) be a signal that has a rational Laplace transform with exactly two poles,located at s = -1 and s = -3. If g (t ) = e 2t x(t ) and G ( jω ) convergence,determine whether x(t ) is left sided, right sided,or two sided.
b (s + 4 ) + s H (s ) = s (s + 2)(s + 4)
Causal Re{s} > 0
x(t ) = e 2t (-∞ < t < +∞ ) is the eigenfunction of system.
s = 2 Re{s} > 0 H (s ) s=2 is finite
A = 1 t 0 = 1
2
Chapter 9
Problem Solution
9.5 For each of the following algebraic expressions for the Laplace transform of a signal ,determine the number of zeros located in the finite s-plane and the number of zeros located at infinity:
s (cos 3t )u ( t ) ← → 2 s +9 Re{s} < 0