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The inverse Laplace transform ofX(s)=2+2se−2s+4e−4s(s2+4s+3),Re(s)>−1 is of the formx(t)=(e−t+e−3tu(t)+[−e−t−a−3e−3(t−a)]u(t−a)+2[e−t−2a−e−3(t−2a)]u(t−2a)value of a is ______2 |
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Answer» The inverse Laplace transform of X(s)=2+2se−2s+4e−4s(s2+4s+3),Re(s)>−1 is of the form x(t)=(e−t+e−3tu(t)+[−e−t−a−3e−3(t−a)]u(t−a)+2[e−t−2a−e−3(t−2a)]u(t−2a)
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