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=--=--.。
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