chapter-9-作业参考答案

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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=--=--.。

(含答案)九年级英语人教全一册课后作业Unit 9 I like music that I can

答卷时应注意事项１、拿到试卷，要认真仔细的先填好自己的考生信息。

２、拿到试卷不要提笔就写，先大致的浏览一遍，有多少大题，每个大题里有几个小题，有什么题型，哪些容易，哪些难，做到心里有底；３、审题，每个题目都要多读几遍，不仅要读大题，还要读小题，不放过每一个字，遇到暂时弄不懂题意的题目，手指点读，多读几遍题目，就能理解题意了；容易混乱的地方也应该多读几遍，比如从小到大，从左到右这样的题；４、每个题目做完了以后，把自己的手从试卷上完全移开，好好的看看有没有被自己的手臂挡住而遗漏的题；试卷第１页和第２页上下衔接的地方一定要注意，仔细看看有没有遗漏的小题；５、中途遇到真的解决不了的难题，注意安排好时间，先把后面会做的做完，再来重新读题，结合平时课堂上所学的知识，解答难题；一定要镇定，不能因此慌了手脚，影响下面的答题；６、卷面要清洁，字迹要清工整，非常重要；７、做完的试卷要检查，这样可以发现刚才可能留下的错误或是可以检查是否有漏题，检查的时候，用手指点读题目，不要管自己的答案，重新分析题意，所有计算题重新计算，判断题重新判断，填空题重新填空，之后把检查的结果与先前做的结果进行对比分析。

亲爱的小朋友，你们好!经过两个月的学习，你们一定有不小的收获吧，用你的自信和智慧，认真答题，相信你一定会闯关成功。

相信你是最棒的！Unit9I like music that I can dance toSection AI.根据句意和汉语提示写出所缺的单词1.The play begins with a short___________(对话)between father and son.2.They want to know more about___________(澳大利亚的)history.3.In my mind,Feng Xiaogang is an excellent___________(导演).4.Sometimes we___________(闭上)off our eyes to the facts.refusing to face the truth.5.This picture wasn't true because it was___________(粘贴)from the Internet.6.We love peace.but we are not afraid of___________(战争).7.I can’t___________(抽出)the time for a holiday at the moment.8.The trousers are very expensive.The cloth feels___________(平滑的).9.When I am feeling_________(沮丧)，he will always listen to me and try to help me.10.This dictionary is available in___________(电子的)form.Ⅱ.根据句意和首字母提示写出所缺的单词11.The Internet is very useful and it can offer p___________of information to us.12.He used to like treating friends to dinner in restaurants,but now he p___________to invite them to his home.13.You can't make a noise in class.in that c___________you'll be punished.14.I s___________we'll go to the park to have a picnic next week.15.We went to see him o___________in a while when he was in Beijing.Ⅲ.从方框中选出合适的短语，并用其正确形式填空16.I don’t___________walking very much today.17.The crops grow well this year because there is___________rain.18.He was so tired that he___________his brain to have a rest.19.I want to get home___________to watch the football match.20.-Are you going climbing tomorrow,Jack?-I’m not sure.It___________the weat her.Ⅳ.单项选择()21.-What makes you___________that I'm against the rule?-Look!You're late again.A.supposeB.imagineC.predictD.accept()22.Please remember to_________the electricity and water before you leave t he laboratory.A.take offB.shut offC.go offD.put off()23.You can’t come here soon.___________,we decide not to wait for you an y longer.A.HoweverB.In that caseC.So farD.After that()24.-Which invention do you like best?-Wechat.It is an invention__________can help us communicate with others online fre ely.A.whoB.thatC.what()25.Some people prefer tea__________milk.However,I like drinking tea wit hout anything in it.A.toB.withC.ofV.根据汉语意思完成句子。

Chapter9反译法课后练习答案

Chapter9反译法课后练习答案Chapter 9 反译法Drills 9.1.14. 我认为，安理会不能也不应对这些极其严重的侵犯人权的行为漠不关心（袖手旁观）。

5．我们认为，他今天上午发言的要旨和口气都无助于在中东建立持久和平。

6．我觉得再也不能忍受下去了。

7．我看你可不必担心。

8．我想不是这样。

（我想并非如此）Drills 9.1.26.那天他碰巧不在那里。

7.她不是乘公共汽车来上学的。

8.我来这里不是要逗弄你。

9.我们战斗了这么多年，不是为了在最后向敌人的走狗屈服。

10.她把信留在这里，并不是有意想让你看。

11.人民摆脱了殖民主义者的奴役，绝不是为了套上霸权主义者的枷锁。

12.你如果拒绝应邀参加会议，那就不礼貌了。

13.我们要到那里去，并不是代表我国政府去进行谈判。

14.不要因为我使唤了她就认为我不好。

15.他这样做，并不是因为没有校对员，而是他不愿在报纸上无意中出现任何差错。

16.我离开了家，但并不是因为我怕我父亲。

/我不是怕我父亲才离开家的。

/ cf. 我没有离开家，因为我怕我父亲。

Drills 9.1.36.她没有回答我。

7.禁止招贴。

8.滚石不生苔，专业不聚财。

/ 见异思迁，一事无成。

9.在任何情况下，我们都不会放弃原则。

10.她发现她确实不能指望谁了。

11.当然，林登·约翰逊在这方面是别人望尘莫及的。

12.我走开了，不去参观名胜了。

Drills 9.2a1.焦急不安2.不出所料3.不及格、不履行、没做到4.无上权威、至高权威5.不失体面、不失尊严6.一大片不毛之地7.困惑不解、茫然不知所措8.心不在焉9.不负所望10.安然无恙11.一点也不差12.得不治之症13.一股不可小看的力量、一股力量，决不能掉以轻心14.不迟、不晚15.油漆未干16.无期徒刑17.放心不下18.不为过、没有扩大化、没有超出范围、没有越过雷池一步Drills 9.2b12.不要客气。

13.她临处死刑而不屈。

Chapter 9参考译文及答案

Chapter 9Learning Aims学完本章，学生应能：1．解释客户投诉索赔的一般原因；2．学会如何进行投诉和索赔；3．学会如何对投诉和索赔作出答复；4．掌握投诉理赔信件的撰写原则、结构等有关知识；5．熟悉一些化工产品的名称。

Background Information在国际贸易中，一家公司向另一家公司提出投诉或索赔是很常见的。

在执行合同的过程中，买卖双方都应遵守合同条款并严格履行各自的义务。

任何一方如果不能严格履约，就会给另一方带来麻烦，有时还会使另一方遭受巨大损失。

在这种情况下，受损失的一方有权根据合同规定要求责任方保证不再发生此类事件，这种行为叫“投诉”，或要求责任方赔偿损失，这种行为叫“索赔”。

相关的信件就是投诉信或索赔信。

通常有两种情况促使买方提出投诉或索赔：一是由于货物错发、质量不良、货物损坏、短重或迟发而引起的真正的投诉或索赔；二则是因为买方不想收到货物或找到了更低价的货源，所以吹毛求疵以此为借口解除合同。

撰写投诉索赔信件需要注意以下几点原则：①第一时间迅速、清楚告诉对方问题所在；②在信件中注明发票号、产品信息等情况，叙述问题清晰有条理；③提供足够的依据；④明确说明投诉或索赔想要得到的具体要求，如果未得到满意回应则礼貌地加强措辞；⑤结尾语气友好但必须坚决。

投诉信往往包含以下要点：①信的开头应对提出的抱怨表示遗憾；②详尽陈述事实，说明出现了什么差错；③说明不满的原因，并要求予以解释；④提及由此而造成的不便；⑤提出如何纠正的建议。

每一项投诉不管看起来多么微不足道，对写信方来讲都是很重要的。

所以收到这类信后，应立即作答。

答复信应说明事实，礼貌而公平，并遵循以下要点：①对来信表示感谢。

②对给对方造成的不便表示歉意。

③说明你将采取什么行动。

④期待进一步合作。

写索赔信应遵循以下要点：①告知收信人有关索赔货物的详细情况。

②清楚地表达你的索赔要求。

③要求对方采取什么行动：换货，维修，退款或赔偿。

chapter9投资作业-chapter15习题

习题Chapter 93. You are a consultant to a large manufacturing corporation that is considering a project with the following net after-tax cash flows (in millions of dollars):The project’s beta is 1.8. Assuming that, what is the net present value of the project? What is the highest possible beta estimate for the project before its NPV becomes negative?4. Are the following true or false? Explain.a. Stocks with a beta of zero offer an expected rate of return of zero.b. The CAPM implies that investors require a higher return to hold highly volatile securities.c. You can construct a portfolio with beta of .75 by investing .75 of the investment budget in T-bills and the remainder in the market portfolio.In problems 13 to 15 assume that the risk-free rate of interest is 6% and the expected rate of return on the market is 16%.13. Ashare of stock sells for $50 today. It will pay a dividend of $6 per share at the end of the year. Its beta is 1.2. What do investors expect the stock to sell for at the end of the year?14. I am buying a firm with an expected perpetual cash flow of $1,000 but am unsure of its risk. If I think the beta of the firm is .5, when in fact the beta is really 1, how much more will I offer for the firm than it is truly worth?15. A stock has an expected rate of return of 4%. What is its beta?17. Suppose the rate of return on short-term government securities (perceived to be riskfree) is about 5%. Suppose also that the expected rate of return required by the market for a portfolio with a beta of 1 is 12%. According to the capital asset pricing model (security market line):a. What is the expected rate of return on the market portfolio?b. What would be the expected rate of return on a stock with beta＝0?c. Suppose you consider buying a share of stock at $40. The stock is expected to pay $3 dividends next year and you expect it to sell then for $41. The stock risk has been evaluated at beta＝－0.5. Is the stock overpriced or underpriced?21. The security market line depicts:a. A security’s expected return as a function of its systematic risk.b. The market portfolio as the optimal portfolio of risky securities.c. The relationship between a security’s return and the return on an index.d. The complete portfolio as a combination of the market portfolio and the risk-free asset.22. Within the context of the capital asset pricing model (CAPM), assume:• Expected return on the market ＝15%.• Risk-free rate _ 8%.• Expected rate of return on XYZ security ＝17%.• Beta of XYZ security ＝1.25.Which one of the following is correct?a. XYZ is overpriced.b. XYZ is fairly priced.c. XYZ’s alpha is＝-0.25%.d. XYZ’s alpha is=0 .25%.The following table shows risk and return measures for two portfolios. answer question 26 and 2726. When plotting portfolio R on the preceding table relative to the SML, portfolio R lies:a. On the SML.b. Below the SML.c. Above the SML.d. Insufficient data given.27. When plotting portfolio R relative to the capital market line, portfolio R lies:a. On the CML.b. Below the CML.c. Above the CML.d. Insufficient data given.31. Karen Kay, a portfolio manager at Collins Asset Management, is using the capital asset pricing model for making recommendations to her clients. Her research department has developed the information shown in the following exhibit.a. Calculate expected return and alpha for each stock.b. Identify and justify which stock would be more appropriate for an investor who wants toi. add this stock to a well-diversified equity portfolio.ii. hold this stock as a single-stock portfolio.Chapter 111. A portfolio management organization analyzes 60 stocks and constructs a meanvariance efficient portfolio using only these 60 securities.a. How many estimates of expected returns, variances, and covariances are needed to optimize this portfolio?b. If one could safely assume that stock market returns closely resemble a single index structure, how many estimates would be needed?2. The following are estimates for two of the stocks in problem 1.The market index has a standard deviation of 22% and the risk-free rate is 8%.a. What is the standard deviation of stocks A and B?b. Suppose that we were to construct a portfolio with proportions:Stock A: .30Stock B: .45T-bills: .25Compute the expected return, standard deviation, beta, and nonsystematic standard deviation of the portfolio.4. Consider the two (excess return) index model regression results for A and B:a. Which stock has more firm-specific risk?b. Which has greater market risk?c. For which stock does market movement explain a greater fraction of return variability?d. Which stock had an average return in excess of that predicted by the CAPM?e. If rf were constant at 6% and the regression had been run using total rather than excess returns, what would have been the regression intercept for stock A ?Use the following data for problems 5 through 9. Suppose that the index model for stocks A and B is estimated from excess returns with the following results:5. What is the standard deviation of each stock?6. Break down the variance of each stock to the systematic and firm-specific components.7. What are the covariance and correlation coefficient between the two stocks?8. What is the covariance between each stock and the market index?9. Are the intercepts of the two regressions consistent with the CAPM? Interpret their values.15. Based on current dividend yields and expected growth rates, the expected rates of return on stocks A and B are 11% and 14%, respectively. The beta of stock A is .8, while that of stock B is 1.5. The T-bill rate is currently 6%, while the expected rate of return on the S&P 500 index is 12%. The standard deviation of stock A is 10% annually, while that of stock B is 11%.a. If you currently hold a well-diversified portfolio, would you choose to add either of these stocks to your holdings?b. If instead you could invest only in bills and one of these stocks, which stock would you choose? Explain your answer using either a graph or a quantitative measure of the attractiveness of the stocks.16. Assume the correlation coefficient between Baker Fund and the S&P 500 Stock Index is .70. What percentage of B aker Fund’s total risk is specific (i.e., nonsystematic)?a. 35%b. 49%c. 51%d. 70%17. The correlation between the Charlottesville International Fund and the EAFE Market Index is 1.0. The expected return on the EAFE Index is 11%, the expected return on Charlottesville International Fund is 9%, and the risk-free return in EAFE countries is 3%. Based on this analysis, the implied beta of Charlottesville International is:a. Negativeb. .75c. .82d. 1.00chapter 122. Which of the following most appears to contradict the proposition that the stock market is weakly efficient? Explain.a. Over 25% of mutual funds outperform the market on average.b. Insiders earn abnormal trading profits.c. Every January, the stock market earns abnormal returns.3. Suppose that, after conducting an analysis of past stock prices, you come up with the following observations. Which would appear to contradict the weak form of the efficient market hypothesis? Explain.a. The average rate of return is significantly greater than zero.b. The correlation between the return during a given week and the return during the following week is zero.c. One could have made superior returns by buying stock after a 10% rise in price and selling after a 10% fall.d. One could have made higher-than-average capital gains by holding stocks with low dividend yields.4. Which of the following statements are true if the efficient market hypothesis holds?a. It implies that future events can be forecast with perfect accuracy.b. It implies that prices reflect all available information.c. It implies that security prices change for no discernible reason.d. It implies that prices do not fluctuate.5. Which of the following observations would provide evidence against the semistrong form of the efficient market theory? Explain.a. Mutual fund managers do not on average make superior returns.b. You cannot make superior profits by buying (or selling) stocks after the announcement of an abnormal rise in dividends.c. Low P/E stocks tend to have positive abnormal returns.d. In any year approximately 50% of pension funds outperform the market.6. The semistrong form of the efficient market hypothesis asserts that stock prices:a. Fully reflect all historical price information.b. Fully reflect all publicly available information.c. Fully reflect all relevant information including insider information.d. May be predictable.7. Assume that a company announces an unexpectedly large cash dividend to its shareholders.In an efficient market without information leakage, one might expect:a. An abnormal price change at the announcement.b. An abnormal price increase before the announcement.c. An abnormal price decrease after the announcement.d. No abnormal price change before or after the announcement.8. Which one of the following would provide evidence against the semistrong form of the efficient market theory?a. About 50% of pension funds outperform the market in any year.b. All investors have learned to exploit signals about future performance.c. Trend analysis is worthless in determining stock prices.d. Low P/E stocks tend to have positive abnormal returns over the long run.Chapter 141. Which security has a higher effective annual interest rate?a. A 3-month T-bill selling at $97,645 with par value $100,000.b. A coupon bond selling at par and paying a 10% coupon semiannually.2. Treasury bonds paying an 8% coupon rate with semiannual payments currently sell at par value. What coupon rate would they have to pay in order to sell at par if they paid their coupons annually? (Hint: what is the effective annual yield on the bond?)3. Two bonds have identical times to maturity and coupon rates. One is callable at 105, the other at 110. Which should have the higher yield to maturity? Why?4. Consider a bond with a 10% coupon and with yield to maturity _ 8%. If the bond’s yield to maturity remains constant, then in 1 year, will the bond price be higher, lower, or unchanged? Why?5. Consider an 8% coupon bond selling for $953.10 with 3 years until maturity making annual coupon payments. The interest rates in the next 3 years will be, withcertainty, . Calculate the yield to maturity and realized compound yield of the bond.6. Philip Morris may issue a 10-year maturity fixed-income security, which might include a sinking fund provision and either refunding or call protection.a. Describe a sinking fund provision.b. Explain the impact of a sinking-fund provision on:i. The expected average life of the proposed security.ii. Total principal and interest payments over the life of the proposed security.c. From the investor’s point of view, explain the rationale for demanding a sinking fund provision.7. Bonds of Zello Corporation with a par value of $1,000 sell for $960, mature in 5 years, and have a 7% annual coupon rate paid semiannually.a. Calculate the:i. Current yield.ii. Yield to maturity (to the nearest whole percent, i.e., 3%, 4%, 5%, etc.).iii. Realized compound yield for an investor with a 3-year holding period and a reinvestment rate of 6% over the period. At the end of 3 years the 7% coupon bonds with 2 years remaining will sell to yield 7%.b. Cite one major shortcoming for each of the following fixed-income yield measures:i. Current yield.ii. Yield to maturity.iii. Realized compound yield.9. A 20-year maturity bond with par value of $1,000 makes semiannual coupon payments at a coupon rate of 8%. Find the bond equivalent and effective annual yield to maturity of the bond if the bond price is:a. $950.b. $1,000.c. $1,050.12. Consider a bond paying a coupon rate of 10% per year semiannually when the market interest rate is only 4% per half year. The bond has 3 years until maturity.a. Find the bond’s price today and 6 months from now after the next coupon is paid.b. What is the total (6 month) rate of return on the bond?14. A bond with a coupon rate of 7% makes semiannual coupon payments on January15 and July 15 of each year. The Wall Street Journal reports the asked price for the bond on January 30 at 100:02. What is the invoice price of the bond? The coupon period has 182 days.20. A 30-year maturity, 8% coupon bond paying coupons semiannually is callable in 5 years at a call price of $1,100. The bond currently sells at a yield to maturity of 7% (3.5% per half-year).a. What is the yield to call?b. What is the yield to call if the call price is only $1,050?c. What is the yield to call if the call price is $1,100, but the bond can be called in 2 years instead of 5 years?22. A 2-year bond with par value $1,000 making annual coupon payments of $100 is priced at $1,000. What is the yield to maturity of the bond? What will be the realized compound yield to maturity if the 1-year interest rate next year turns out to be (a) 8%,(b) 10%26. Alarge corporation issued both fixed and floating-rate notes 5 years ago, with terms given in the following table:a. Why is the price range greater for the 9% coupon bond than the floatingrate note?b. What factors could explain why the floating-rate note is not always sold at par value?c. Why is the call price for the floating-rate note not of great importance to investors?d. Is the probability of call for the fixed-rate note high or low?e. If the firm were to issue a fixed-rate note with a 15-year maturity, what coupon rate would it need to offer to issue the bond at par value?f. Why is an entry for yield to maturity for the floating-rate note not appropriate?27. On May 30, 1999, Janice Kerr is considering one of the newly issued 10-year AAA corporate bonds shown in the following exhibit.a. Suppose that market interest rates decline by 100 basis points (i.e., 1%). Contrast the effect of this decline on the price of each bond.b. Should Kerr prefer the Colina over the Sentinal bond when rates are expected to rise or to fall?c. What would be the effect, if any, of an increase in the volatility of interest rates on the prices of each bond?Chapter 152. Which one of the following statements about the term structure of interest rates is true?a. The expectations hypothesis indicates a flat yield curve if anticipated futureshort-term rates exceed current short-term rates.b. The expectations hypothesis contends that the long-term rate is equal to the anticipated short-term rate.c. The liquidity premium theory indicates that, all else being equal, longer maturities will have lower yields.d. The liquidity preference theory contends that lenders prefer to buy securities at the short end of the yield curve.8. Suppose the following table shows yields to maturity of zero coupon U.S. Treasury securities as of January 1, 1996:a. Based on the data in the table, calculate the implied forward 1-year rate of interest at January 1, 1999.b. Describe the conditions under which the calculated forward rate would be an unbiased estimate of the 1-year spot rate of interest at January 1, 1999.c. Assume that 1 year earlier, at January 1, 1995, the prevailing term structure for U.S. Treasury securities was such that the implied forward 1-year rate of interest at January 1, 1999, was significantly higher than the corresponding rate implied by the term structure at January 1, 1996. On the basis of the pure expectations theory of the term structure, briefly discuss two factors that could account for such a decline in the implied forward rate.16. Below is a list of prices for zero-coupon bonds of various maturities.a. An 8.5% coupon $1,000 par bond pays an annual coupon and will mature in 3years. What should the yield to maturity on the bond be?b. If at the end of the first year the yield curve flattens out at 8%, what will be the1-year holding-period return on the coupon bond?21. The yield to maturity (YTM) on 1-year zero-coupon bonds is 5% and the YTM on 2-year zeros is 6%. The yield to maturity on 2-year-maturity coupon bonds with coupon rates of 12% (paid annually) is 5.8%. What arbitrage opportunity is available for an investment banking firm? What is the profit on the activity?22. Suppose that a 1-year zero-coupon bond with face value $100 currently sells at $94.34, while a 2-year zero sells at $84.99. You are considering the purchase of a2-year-maturity bond making annual coupon payments. The face value of the bond is $100, and the coupon rate is 12% per year.a. What is the yield to maturity of the 2-year zero? The 2-year coupon bond?b. What is the forward rate for the second year?c. If the expectations hypothesis is accepted, what are (1) the expected price of the coupon bond at the end of the first year and (2) the expected holding-period return on the coupon bond over the first year?d. Will the expected rate of return be higher or lower if you accept the liquidity preference hypothesis?。

操作系统第九版部分课后作业习题答案解析

CHAPTER 9 Virtual Memory Practice Exercises9.1 Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.Answer:A page fault occurs when an access to a page that has not beenbrought into main memory takes place. The operating system veriﬁesthe memory access, aborting the program if it is invalid. If it is valid, a free frame is located and I/O is requested to read the needed page into the free frame. Upon completion of I/O, the process table and page table are updated and the instruction is restarted.9.2 Assume that you have a page-reference string for a process with m frames (initially all empty). The page-reference string has length p;n distinct page numbers occur in it. Answer these questions for any page-replacement algorithms:a. What is a lower bound on the number of page faults?b. What is an upper bound on the number of page faults?Answer:a. nb. p9.3 Consider the page table shown in Figure 9.30 for a system with 12-bit virtual and physical addresses and with 256-byte pages. The list of freepage frames is D, E, F (that is, D is at the head of the list, E is second, and F is last).Convert the following virtual addresses to their equivalent physical addresses in hexadecimal. All numbers are given in hexadecimal. (A dash for a page frame indicates that the page is not in memory.)• 9EF• 1112930 Chapter 9 Virtual Memory• 700• 0FFAnswer:• 9E F - 0E F• 111 - 211• 700 - D00• 0F F - EFF9.4 Consider the following page-replacement algorithms. Rank these algorithms on a ﬁve-point scale from “bad” to “perfect” according to their page-fault rate. Separate those algorithms that suffer from Belady’s anomaly from those that do not.a. LRU replacementb. FIFO replacementc. Optimal replacementd. Second-chance replacementAnswer:Rank Algorithm Suffer from Belady’s anomaly1 Optimal no2 LRU no3 Second-chance yes4 FIFO yes9.5 Discuss the hardware support required to support demand paging. Answer:For every memory-access operation, the page table needs to be consulted to check whether the corresponding page is resident or not and whether the program has read or write privileges for accessing the page. These checks have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.9.6 An operating system supports a paged virtual memory, using a central processor with a cycle time of 1 microsecond. It costs an additional 1 microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers 1 million words per second. The following statistical measurements were obtained from the system:• 1 percent of all instructions executed accessed a page other than the current page.•Of the instructions that accessed another page, 80 percent accesseda page already in memory.Practice Exercises 31•When a new page was required, the replaced page was modiﬁed 50 percent of the time.Calculate the effective instruction time on this system, assuming that the system is running one process only and that the processor is idle during drum transfers.Answer:effective access time = 0.99 × (1 sec + 0.008 × (2 sec)+ 0.002 × (10,000 sec + 1,000 sec)+ 0.001 × (10,000 sec + 1,000 sec)= (0.99 + 0.016 + 22.0 + 11.0) sec= 34.0 sec9.7 Consider the two-dimensional array A:int A[][] = new int[100][100];where A[0][0] is at location 200 in a paged memory system with pages of size 200. A small process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated bythe following array-initialization loops, using LRU replacement andassuming that page frame 1 contains the process and the other twoare initially empty?a. for (int j = 0; j < 100; j++)for (int i = 0; i < 100; i++)A[i][j] = 0;b. for (int i = 0; i < 100; i++)for (int j = 0; j < 100; j++)A[i][j] = 0;Answer:a. 5,000b. 509.8 Consider the following page reference string:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, ﬁve, six, or seven frames? Remember all frames are initially empty, so your ﬁrst unique pages will all cost one fault each.•LRU replacement• FIFO replacement•Optimal replacement32 Chapter 9 Virtual MemoryAnswer:Number of frames LRU FIFO Optimal1 20 20 202 18 18 153 15 16 114 10 14 85 8 10 76 7 10 77 77 79.9 Suppose that you want to use a paging algorithm that requires a referencebit (such as second-chance replacement or working-set model), butthe hardware does not provide one. Sketch how you could simulate a reference bit even if one were not provided by the hardware, or explain why it is not possible to do so. If it is possible, calculate what the cost would be.Answer:You can use the valid/invalid bit supported in hardware to simulate the reference bit. Initially set the bit to invalid. O n ﬁrst reference a trap to the operating system is generated. The operating system will set a software bit to 1 and reset the valid/invalid bit to valid.9.10 You have devised a new page-replacement algorithm that you thinkmaybe optimal. In some contorte d test cases, Belady’s anomaly occurs. Is the new algorithm optimal? Explain your answer.Answer:No. An optimal algorithm will not suffer from Belady’s anomaly because —by deﬁnition—an optimal algorithm replaces the page that will notbe used for the long est time. Belady’s anomaly occurs when a pagereplacement algorithm evicts a page that will be needed in the immediatefuture. An optimal algorithm would not have selected such a page.9.11 Segmentation is similar to paging but uses variable-sized“pages.”Deﬁnetwo segment-replacement algorithms based on FIFO and LRU pagereplacement schemes. Remember that since segments are not the samesize, the segment that is chosen to be replaced may not be big enoughto leave enough consecutive locations for the needed segment. Consider strategies for systems where segments cannot be relocated, and thosefor systems where they can.Answer:a. FIFO. Find the ﬁrst segment large enough to accommodate the incoming segment. If relocation is not possible and no one segmentis large enough, select a combination of segments whose memoriesare contiguous, which are “closest to the ﬁrst of the list” andwhich can accommodate the new segment. If relocation is possible, rearrange the memory so that the ﬁrstNsegments large enough forthe incoming segment are contiguous in memory. Add any leftover space to the free-space list in both cases.Practice Exercises 33b. LRU. Select the segment that has not been used for the longestperiod of time and that is large enough, adding any leftover spaceto the free space list. If no one segment is large enough, selecta combination of the “oldest” segments that are contiguous inmemory (if relocation is not available) and that are large enough.If relocation is available, rearrange the oldest N segments to be contiguous in memory and replace those with the new segment.9.12 Consider a demand-paged computer system where the degree of multiprogramming is currently ﬁxed at four. The system was recently measured to determine utilization of CPU and the paging disk. The results are one of the following alternatives. For each case, what is happening? Can the degree of multiprogramming be increased to increase the CPU utilization? Is the paging helping?a. CPU utilization 13 percent; disk utilization 97 percentb. CPU utilization 87 percent; disk utilization 3 percentc. CPU utilization 13 percent; disk utilization 3 percentAnswer:a. Thrashing is occurring.b. CPU utilization is sufﬁciently high to leave things alone, and increase degree of multiprogramming.c. Increase the degree of multiprogramming.9.13 We have an operating system for a machine that uses base and limit registers, but we have modiﬁed the ma chine to provide a page table.Can the page tables be set up to simulate base and limit registers? How can they be, or why can they not be?Answer:The page table can be set up to simulate base and limit registers provided that the memory is allocated in ﬁxed-size segments. In this way, the base of a segment can be entered into the page table and the valid/invalid bit used to indicate that portion of the segment as resident in the memory. There will be some problem with internal fragmentation.9.27.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which (if any) of the following will (probably) improve CPU utilization? Explain your answer.a. Install a faster CPU.b. Install a bigger paging disk.c. Increase the degree of multiprogramming.d. Decrease the degree of multiprogramming.e. Install more main memory.f. Install a faster hard disk or multiple controllers with multiple hard disks.g. Add prepaging to the page fetch algorithms.h. Increase the page size.Answer: The system obviously is spending most of its time paging, indicating over-allocationof memory. If the level of multiprogramming is reduced resident processeswould page fault less frequently and the CPU utilization would improve. Another way toimprove performance would be to get more physical memory or a faster paging drum.a. Get a faster CPU—No.b. Get a bigger paging drum—No.c. Increase the degree of multiprogramming—No.d. Decrease the degree of multiprogramming—Yes.e. Install more main memory—Likely to improve CPU utilization as more pages canremain resident and not require paging to or from the disks.f. Install a faster hard disk, or multiple controllers with multiple hard disks—Also animprovement, for as the disk bottleneck is removed by faster response and morethroughput to the disks, the CPU will get more data more quickly.g. Add prepaging to the page fetch algorithms—Again, the CPU will get more datafaster, so it will be more in use. This is only the case if the paging action is amenableto prefetching (i.e., some of the access is sequential).h. Increase the page size—Increasing the page size will result in fewer page faults ifdata is being accessed sequentially. If data access is more or less random, morepaging action could ensue because fewer pages can be kept in memory and moredata is transferred per page fault. So this change is as likely to decrease utilizationas it is to increase it.10.1、Is disk scheduling, other than FCFS scheduling, useful in asingle-userenvironment? Explain your answer.Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O, such as when aWeb browser retrieves data in the background while the operating system is paging and another application is active in the foreground.10.2.Explain why SSTF scheduling tends to favor middle cylindersover theinnermost and outermost cylinders.The center of the disk is the location having the smallest average distance to all other tracks.Thus the disk head tends to move away from the edges of the disk.Here is another way to think of it.The current location of the head divides the cylinders into two groups.If the head is not in the center of the disk and a new request arrives,the new request is more likely to be in the group that includes the center of the disk;thus,the head is more likely to move in that direction.10.11、Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer:a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The total seek distance is 1745.c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The total seek distance is 9769.d. The LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek distance is 3319.e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The total seek distance is 9813.f. (Bonus.) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The total seek distance is 3363.12CHAPTERFile-SystemImplementationPractice Exercises12.1 Consider a ﬁle currently consisting of 100 blocks. Assume that the ﬁlecontrol block (and the index block, in the case of indexed allocation)is already in memory. Calculate how many disk I/O operations are required for contiguous, linked, and indexed (single-level) allocation strategies, if, for one block, the following conditions hold. In the contiguous-allocation case, assume that there is no room to grow atthe beginning but there is room to grow at the end. Also assume thatthe block information to be added is stored in memory.a. The block is added at the beginning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the beginning.e. The block is removed from the middle.f. The block is removed from the end.Answer:The results are:Contiguous Linked Indexeda. 201 1 1b. 101 52 1c. 1 3 1d. 198 1 0e. 98 52 0f. 0 100 012.2 What problems could occur if a system allowed a ﬁle system to be mounted simultaneously at more than one location?Answer:4344 Chapter 12 File-System ImplementationThere would be multiple paths to the same ﬁle, which could confuse users or encourage mistakes (deleting a ﬁle with one path deletes theﬁle in all the other paths).12.3 Why must the bit map for ﬁle allocation be kept on mass storage, ratherthan in main memory?Answer:In case of system crash (memory failure) the free-space list would notbe lost as it would be if the bit map had been stored in main memory.12.4 Consider a system that supports the strategies of contiguous, linked, and indexed allocation. What criteria should be used in deciding which strategy is best utilized for a particular ﬁle?Answer:•Contiguous—if ﬁle is usually accessed sequentially, if ﬁle isrelatively small.•Linked—if ﬁle is large and usually accessed sequentially.• Indexed—if ﬁle is large and usually accessed randomly.12.5 One problem with contiguous allocation is that the user must preallocate enough space for each ﬁle. If the ﬁle grows to be larger than thespace allocated for it, special actions must be taken. One solution to this problem is to deﬁne a ﬁle structure consisting of an initial contiguous area (of a speciﬁed size). If this area is ﬁlled, the operating system automatically deﬁnes an overﬂow area that is linked to the initialc ontiguous area. If the overﬂow area is ﬁlled, another overﬂow areais allocated. Compare this implementation of a ﬁle with the standard contiguous and linked implementations.Answer:This method requires more overhead then the standard contiguousallocation. It requires less overheadthan the standard linked allocation. 12.6 How do caches help improve performance? Why do systems not use more or larger caches if they are so useful?Answer:Caches allow components of differing speeds to communicate moreefﬁcie ntly by storing data from the slower device, temporarily, ina faster device (the cache). Caches are, almost by deﬁnition, more expensive than the device they are caching for, so increasing the number or size of caches would increase system cost.12.7 Why is it advantageous for the user for an operating system to dynamically allocate its internal tables? What are the penalties to the operating system for doing so?Answer:Dynamic tables allow more ﬂexibility in system use growth — tablesare never exceeded, avoiding artiﬁcial use limits. Unfortunately, kernel structures and code are more complicated, so there is more potentialfor bugs. The use of one resource can take away more system resources (by growing to accommodate the requests) than with static tables.Practice Exercises 4512.8 Explain how the VFS layer allows an operating system to support multiple types of ﬁle systems easily.Answer:VFS introduces a layer of indirection in the ﬁle system implementation. In many ways, it is similar to object-oriented programming techniques. System calls can be made generically (independent of ﬁle system type). Each ﬁle system type provides its function calls and data structuresto the VFS layer. A system call is translated into the proper speciﬁc functions for the ta rget ﬁle system at the VFS layer. The calling program has no ﬁle-system-speciﬁc code, and the upper levels of the system call structures likewise are ﬁle system-independent. The translation at the VFS layer turns these generic calls into ﬁle-system-speciﬁc operations.。

Chapter-9_NP完全问题

南京理工大学
9 NP完全问题 NP Complete Problem
南京理工大学
本章内容提要
• • • • • • • 易解问题与难解问题 P类问题和类问题类问题和NP类问题类问题和 NP完全问题完全问题 co-NP类问题类问题 NPI类问题类问题 P、NP、co-NP、NPI类之间的区别与联系、、、类之间的区别与联系 NP完全问题的计算机处理技术简介完全问题的计算机处理技术简介
南京理工大学
9.1.2 易解问题与难解问题的主要区别
在学术界已达成这样的共识：在学术界已达成这样的共识：把多项式时间复杂性作为易解问题与难解问题的分界线，主要原因有：易解问题与难解问题的分界线，主要原因有： 1) 多项式函数与指数函数的增长率有本质差别
问题规模 n logn n
1 10 20 50 100 0 3.3 4.3 5.6 6.6 1 10 20 50 100
特点：对同一输入实例，运行算法，所得结果是一样的。特点：对同一输入实例，运行算法A，所得结果是一样的。
• 定义定义9.2(P类问题如果对于某个判定问题，存在一个非类问题): 类问题如果对于某个判定问题∏，负整数k，对于输入规模为n的实例能够以O(nk)的时间运的实例，负整数，对于输入规模为的实例，能够以的时间运行一个确定性算法，得到yes或no的答案，则称该判定问题的答案，行一个确定性算法，得到或的答案 ∏是一个是一个P(Polynomial)类问题。类问题。是一个类问题
3) 多项式时间复杂性忽略了系数，不影响易解问题与难解多项式时间复杂性忽略了系数，
问题的划分
问题规模n 问题规模 5 10 100 1000 多项式函数 n8 390625 108 1016 1024 108n 5×108 × 109 1010 1011 n1000 51000 101000 102000 103000 1.1n 1.611 2.594 13780.6 指数函数 20.01n 1.035 1.072 2

(完整word版)国际财务管理课后习题答案chapter9

CHAPTER 9 MANAGEMENT OF ECONOMIC EXPOSURESUGGESTED ANSWERS AND SOLUTIONS TO END-OF-CHAPTERQUESTIONS AND PROBLEMSQUESTIONS1. How would you define economic exposure to exchange risk?Answer: Economic exposure can be defined as the possibility that the firm’s cash flows and thus its market value may be affected by the unexpected exchange rate changes.2. Explain the following statement: “Exposure is the regression coefficient.”Answer: Exposure to currency risk can be appropriately measured by th e sensitivity of the firm’s future cash flows and the market value to random changes in exchange rates. Statistically, this sensitivity can be estimated by the regression coefficient. Thus, exposure can be said to be the regression coefficient.3. Suppose that your company has an equity position in a French firm. Discuss the condition under which the dollar/franc exchange rate uncertainty does not constitute exchange exposure for your company.Answer: Mere changes in exchange rates do not necessarily constitute currency exposure. If the French franc value of the equity moves in the opposite direction as much as the dollar value of the franc changes, then the dollar value of the equity position will be insensitive to exchange rate movements. As a result, your company will not be exposed to currency risk.4. Explain the competitive and conversion effects of exchange rate changes on the firm’s operating cash flow.Answer: The competitive effect: exchange rate changes may affect operating cash flows by altering the firm’s competitive position.The conversion effect: A given operating cash flows in terms of a foreign currency will be converted into higher or lower dollar (home currency)amounts as the exchange rate changes.5. Discuss the determinants of operating exposure.Answer: The main determinants of a firm’s operating exposure are (1) the structure of the markets in which the firm sources its inputs, such as labor and materials, and sells its products, and (2) the firm’s ability to mitigate the effect of exchange rate changes by adjusting its markets, product mix, and sourcing.6. Discuss the implications of purchasing power parity for operating exposure.Answer: If the exchange rate changes are matched by the inflation rate differential between countries, firms’ competitive positions will not be altered by exchange rate changes. Firms are not subject to operating exposure.7. General Motors exports cars to Spain but the strong dollar against the peseta hurts sales of GM cars in Spain. In the Spanish market, GM faces competition from the Italian and French car makers, such as Fiat and Renault, whose currencies remain stable relative to the peseta. What kind of measures would you recommend so that GM can maintain its market share in Spain.Answer: Possible measures that GM can take include: (1) diversify the market; try to market the cars not just in Spain and other European countries but also in, say, Asia; (2) locate production facilities in Spain and source inputs locally; (3) locate production facilities, say, in Mexico where production costs are low and export to Spain from Mexico.8. What are the advantages and disadvantages of financial hedging of the firm’s operating exposure vis-à-vis operational hedges (such as relocating manufacturing site)?Answer: Financial hedging can be implemented quickly with relatively low costs, but it is difficult to hedge against long-term, real exposure with financial contracts. On the other hand, operational hedges are costly, time-consuming, and not easily reversible.9. Discuss the advantages and disadvantages of maintaining multiple manufacturing sites as a hedge against exchange rate exposure.Answer: To establish multiple manufacturing sites can be effective in managing exchange risk exposure, but it can be costly because the firm may not be able to take advantage of the economy of scale.10. Evaluate the following statement: “A firm can reduce its currency exposure by diversifying across different business lines.”Answer: Conglomerate expansion may be too costly as a means of hedging exchange risk exposure. Investment in a different line of business must be made based on its own merit.11. The exchange rate uncertainty may not necessarily mean that firms face exchange risk exposure. Explain why this may be the case.Answer: A firm can have a natural hedging position due to, for example, diversified markets, flexible sourcing capabilities, etc. In addition, to the extent that the PPP holds, nominal exchange rate changes do not influenc e firms’ competitive positions. Under these circumstances, firms do not need to worry about exchange risk exposure.PROBLEMS1. Suppose that you hold a piece of land in the City of London that you may want to sell in one year. As a U.S. resident, you are concerned with the dollar value of the land. Assume that, if the British economy booms in the future, the land will be worth £2,000 and one British pound will be worth $1.40. If the British economy slows down, on the other hand, the land will be worth less, i.e., £1,500, but the pound will be stronger, i.e., $1.50/£. You feel that the British economy will experience a boom with a 60% probability and a slow-down with a 40% probability.(a) Estimate your exposure b to the exchange risk.(b) Compute the variance of the dollar value of your property that is attributable to the exchange rate uncertainty.(c) Discuss how you can hedge your exchange risk exposure and also examine the consequences of hedging.Solution: (a) Let us compute the necessary parameter values:E(P) = (.6)($2800)+(.4)($2250) = $1680+$900 = $2,580E(S) = (.6)(1.40)+(.4)(1.5) = 0.84+0.60 = $1.44Var(S) = (.6)(1.40-1.44)2 + (.4)(1.50-1.44)2= .00096+.00144 = .0024.Cov(P,S) = (.6)(2800-2580)(1.4-1.44)+(.4)(2250-2580)(1.5-1.44)= -5.28-7.92 = -13.20b = Cov(P,S)/Var(S) = -13.20/.0024 = -£5,500.You have a negative exposure! As the pound gets stronger (weaker) against the dollar, the dollar value of your British holding goes down (up).(b) b2Var(S) = (-5500)2(.0024) =72,600($)2(c) Buy £5,500 forward. By doing so, you can eliminate the volatility of the dollar value of your British asset that is due to the exchange rate volatility.2. A U.S. firm holds an asset in France and faces the following scenario:In the above table, P* is the euro price of the asset held by the U.S. firm and P is the dollar price of the asset.(a) Compute the exchange exposure faced by the U.S. firm.(b) What is the variance of the dollar price of this asset if the U.S. firm remains unhedged against thisexposure?(c) If the U.S. firm hedges against this exposure using the forward contract, what is the variance of thedollar value of the hedged position?Solution: (a)E(S) = .25(1.20 +1.10+1.00+0.90) = $1.05/€E(P) = .25(1,800+1,540+1,300 +1,080) = $1,430Var(S) = .25[(1.20-1.05)2 +(1.10-1.05)2+(1.00-1.05)2+(0.90-1.05)2]= .0125Cov(P,S) = .25[(1,800-1,430)(1.20-1.05) + (1,540-1,430)(1.10-1.05)(1,300-1,430)(1.00-1.05) + (1,080-1,430)(0.90-1.05)]= 30b = Cov(P,S)/Var(S) = 30/0.0125 = €2,400.(b) Var(P) = .25[(1,800-1,430)2+(1,540-1,430)2+(1,300-1,430)2+(1,080-1,430)2]= 72,100($)2.(c) Var(P) - b2Var(S) = 72,100 - (2,400)2(0.0125) = 100($)2.This means that most of the volatility of the dollar value of the French asset can be removed by hedging exchange risk. The hedging can be achieved by selling €2,400 forward.MINI CASE: ECONOMIC EXPOSURE OF ALBION COMPUTERS PLCConsider Case 3 of Albion Computers PLC discussed in the chapter. Now, assume that the pound is expected to depreciate to $1.50 from the current level of $1.60 per pound. This implies that the pound cost of the imported part, i.e., Intel’s microprocessors, is £341 (=$512/$1.50). Other variables, such as the unit sales volume and the U.K. inflation rate, remain the same as in Case 3.(a) Compute the projected annual cash flow in dollars.(b) Compute the projected operating gains/losses over the four-year horizon as the discounted present value of change in cash flows, which is due to the pound depreciation, from the benchmark case presented in Exhibit 12.4.(c) What actions, if any, can Albion take to mitigate the projected operating losses due to the pound depreciation?Suggested Solution to Economic Exposure of Albion Computers PLCa) The projected annual cash flow can be computed as follows:______________________________________________________Sales (40,000 units at £1,080/unit) £43,200,000Variable costs (40,000 units at £697/unit) £27,880,000Fixed overhead costs 4,000,000Depreciation allowances 1,000,000Net profit before tax £15,315,000Income tax (50%) 7,657,500Profit after tax 7,657,500Add back depreciation 1,000,000Operating cash flow in pounds £8,657,500Operating cash flow in dollars $12,986,250______________________________________________________b) ______________________________________________________Benchmark CurrentVariables Case Case______________________________________________________Exchange rate ($/£) 1.60 1.50Unit variable cost (£) 650 697Unit sales price (£) 1,000 1,080Sales volume (units) 50,000 40,000Annual cash flow (£) 7,250,000 8,657,500Annual cash flow ($) 11,600,000 12,986,250Four-year present value ($) 33,118,000 37,076,946Operating gains/losses ($) 3,958,946______________________________________________________c) In this case, Albion actually can expect to realize exchange gains, rather than losses. This is mainly due to the fact that while the selling price appreciates by 8% in the U.K. market, the variable cost of imported input increased by about 6.25%. Albion may choose not to do anything.。

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第9章作业参考答案P233 4. 25℃时，下列电池的电动势为0.518 V（忽略液接电位）：
Pt | H2(100 kPa)，HA(0.01 mol·L-1)，A-(0.01 mol·L-1) || SCE
计算弱酸HA的K a值。

解电池组成为
Pt | H2(100 kPa)，HA(0.01 mol·L-1)，A-(0.01 mol·L-1) || SCE
E = φ(+) - φ(-)= φSCE–
而= 0.059lg[H+]
故E - φSCE = -0.059lg[H+]，0.518 – 0.244 = -0.059lg[H+]
[H+] = 2.29×10-5 mol·L-1
根据pH = p K a + lg
有pH = p K a，即K a = [H+] = 2.29×10-5 mol·L-1
6. 测得下列电池电动势为0.873 V(25 ℃)
Cd | Cd(CN)42- (0.08 mol·L-1) CN- (0.100 mol·L-1) || SHE
试计算Cd(CN)42-的稳定常数。

解电池组成为
Cd | Cd(CN)42- (0.08 mol·L-1) CN- (0.100 mol·L-1) || SHE
E = φ(+) - φ(-)= φSHE–
而＝＋
根据K稳＝
有0.873 V = 0.403 V –
K稳＝7.1×1018
11. 20.00 mL 0.1000 mol·L-1 Fe2+溶液在1 mol·L-1 H2SO4溶液中，用0.1000 mol·L-1 Ce4+溶液滴定，用Pt(+)，SCE(-)组成电池，测得电池电动势为0.50 V。

此时已加入多少毫升滴定剂？
解E = φ(+) - φ(-) = - φSCE
= + - φSCE
0.50 = 0.68 + – 0.244
= 12
根据反应：Fe2+ + Ce4+ = Fe3+ + Ce3+，1 mol Fe2+与1 mol Ce4+反应
根据[Fe3+]/[Fe2+] = 12和nFe2+ + nFe3+ = 20 mL×0.100 mol·L-1
得：nFe3+ = 18.46×10－4 mol即有相当量的Ce4+加入
则加入Ce4+的体积为18.46×10-4 mol/0.100 mol·L-1 = 18.46 mL
12. 下列电池
Ag | Ag2CrO4，CrO42-(x·mol·L-1) || SCE
测得E＝－0.285 V，计算CrO42-的浓度（忽略液接电位）。

已知K sp(Ag2CrO4) = 9.0×10－12。

解E = φ
- φ(-)= φSCE–
E = φSCE–- 0.059lg[Ag+]
而[Ag+] ＝= 0.799V
所以-0.285 = 0.244 – 0.799 -
[CrO42-] = 1.3×10-2 mol·L-1
14. 某种钠敏感电极的选择性系数约为30（说明H +存在将严重干扰Na＋的测定）。

如用这种点击测定pNa ＝3的Na＋溶液，并要求测定误差小于3％，则试液的pH必须大于几？
解
为被测离子活度；为干扰离子活度；K ij为电位选择性系数。

即30×[H+]/10-33%
得[H+] 10-6即pH 6。

17. 在25℃时用标准加入法测定Cu2＋浓度，于100mL铜盐溶液中添加0.1 mol·L-1 Cu(NO3)2溶液1mL，电动势增加4mV。

求原溶液中总铜离子浓度。

解根据题意有
E1＝K’ ＋
E2＝K’ ＋
其中：c x为待测液的浓度，c s = 0.1mol·L-1为加入标准液的浓度。

V0＝100mL为待测液的体积，V s＝1mL为加入的标准物的体积。

电动势增量
ΔE＝E2－E1＝－
则c x= 2.64×10-3 mol·L-1。