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1、.Chapter 9 Answers9.1 （a）The given integral may be written as If -5,then the integral does converge (b) The given integral may be written as tIf -5 ,then the function grows towards as t decreases towards -and the given integral does not converge .but if -5 ,then the function grows towards as t decre
2、ases towards -and the given integral does not converge If -5, ,then function grows towards with increasing t and the given integral does not converge If =5, then the integral stilldoes not have a finite value. therefore, the integral does not converge for any value of .(e) The given integral may be
3、written as t+ tThe first integral converges for -5,therefore, the given internal converges when 5 ,then the function grows towards as t decrease towards - and the given integral does not converge .but if -5.(b) By using eg.(9.3), we can easily show that g(t)=Au(-t-) has the Laplace transform G(s)= T
4、he ROC is specified asmax(-5,Re). Since we are given that the ROC is Res-3, we know that Re=3 . there are no constraints on the imaginary part of .9.4 We know form Table 9.2 that , Res-1We also know form Table 9.1 that x(t)= X(s)= The ROC of X(s) is such that if was in the ROC of , then - will be in
5、 the ROC of X(s). Putting the two above equations together ,we have x(t)= (-t) =X(s)= =-, 1the denominator of the form -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 =.Clearly ,X(s) has a zero at s=-2 .since in X(s) the order of the denomi
6、nator 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)= Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denomina
7、tor polynomial exceeds the 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 as Clearly ，X（s）has a zero at s=1.since in X(s) the order of the numerator polynomial exceeds th
8、e 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-plan
9、e 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 -axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal would be Res . From property 5 in section 9.2 we knew that x(t)
10、can not be a right-side signal (d) Yes . Since the signal is absolutely integrable, The ROC must include , the -axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal could be Res2 such that - (ii)-2 Res- (iii)-3Res-2(iv)Res-3Therefore ,we may find four different signals
11、 the given Laplace transform.9.8 From Table 9.1,we know that G(t)= .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() exists , we may infer that -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 s
