《信号与系统教学资料》信号与系统奥本海姆英文版课后答案chapter9.pdfVIP

Chapter 9 Answers 9.1 （a）The given integral may be written as ∫∞0 e?(5+σ )t e jωt dt If σ -5 ,then the function e?(5+σ )t 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 as ∫ 0 ?∞ e?(5+σ )t e jωt d t If σ -5 ,then the function e?(5+σ )t grows towards ∞ as t decreases towards - ∞ and the given integral does not converge .but if σ -5,then the integral does converge (c) The given integral may be written as ∫5?5 e?(5+σ )t e jωt d t Clearly this integral has a finite value for all finite values of σ . (d) The given integral may be written as ∫∞?∞ e?(5+σ )t e jωt d t If σ -5 ,then the function e?(5+σ )t grows towards ∞ as t decreases towards - ∞ and the given integral does not converge If σ -5, ,then function e?(5+σ )t 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 as ∫ 0 ?∞ e?(5+σ )t e jωt d t+ ∫ ∞ 0 e?(5+σ )t e jωt d t The first integral converges for σ -5, the second internal converges if σ -5,therefore, the given internal converges when σ 5. (f) The given integral may be written as ∫ 0 ?∞ e?(5+σ )t e jωt d t If σ 5 ,then the function e?(?5+σ )t 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)= ∫ ∞ ?∞ e ?5t u (t ? 1)e? dt dt = ∫ ∞ 0 e?(5+s)t dt = e?(5+s) s+5 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 e?5t u(-t- t0 ) has the Laplace transform G(s)= Ae(s+5)t0 s+5 The ROC is specified as Re{s} -5 . Therefore ,A=1 and t0 =-1 9.3 Using an analysis similar to that used in Example 9.3 we known that given signal has a Laplace transform of the form X(s) s 1 + 5 + s 1 + β The corresponding ROC is Re{s} max(-5,Re{ β }). Since we are giv