Problem 3 a(t) u(-t)+2e-2a(t) Using Laplace transforms of elementary functions(table 9. 2), we find 1 R1=Re{s}<-1 R2= ReIsh s+2 Using the linearity property, ROC containing R1∩f2 Resh The pole-zero plot is shown below p-z map of X(sProblem 3 (a) −2t x(t) = e−t u(−t)+2e u(t) Using Laplace transforms of elementary functions (table 9.2), we find, −t e u(−t) −1 ←→ , R1 = Re{s} < −1 s + 1 − 1 2t e u(t) ←→ R2 = Re{s} > −2 s + 2, Using the linearity property, 2 � X(s) = −1 + , ROC containing R1 R2 s + 1 s + 2 s = (s + 1)(s + 2) ROC = −2 < Re{s} < −1 The pole-zero plot is shown below: × −2 × −1 e m p-z map of X(s) 9