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What is the inverse LaPlace Transform of 7[e^(-7s)]/(s+7) I suspect it is the result of two functions one of whome is right time shifted by 7 sec.

2007-08-02 21:08:44 · 1 answers · asked by 037 G 6 in Science & Mathematics Mathematics

1 answers

the e^-7s tells you it is a step function
7e^-7s /(s+7)
= 7e^-7s L{ e^-7t }(s)
= 7 L{ u(t-7)*e^-7(t-7) }(s)
= L{ 7u(t-7)*e^-7(t-7) }(s)
so
L^-1{ 7e^-7s /(s+7) }(t)
= 7u(t-7)*e^-7(t-7)
=
{ 0 .......................t<7
{ 7e^-7(t-7)...........t≥7
is the answer
I assumed that you were familiar with the unit step function
u(t-7)
=
{ 0 ...........t<7
{ 1............t≥7

the end
.

2007-08-03 02:16:10 · answer #1 · answered by The Wolf 6 · 1 0

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