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Find Laplace Transform of $\frac{cosbt - cosat}{t}$
1 Answer
written 7.9 years ago by | modified 20 months ago by |
Here f(t) = cosbt - cosat
By division of 't' property $L[\frac{1}{t} f(t)] * \int\limits_{s}^{\infty} f(s) d(s)$ ........... (1)
$\therefore L[f(t)] = L[cosbt - cosat] = F(s) = \frac{s}{s^2 + b^2} - \frac{s}{s^2 + a^2}$
Now,
$L[\frac{[cosbt - cosat]}{t}] = \int\limits_{s}^{\infty} \frac{s}{s^2 + b^2} - \frac{s}{s^2 + a^2} ds$ .......[from (1)]
$L[\frac{[cosbt - cosat]}{t}] = \bigg[\frac{1}{2} log(s^2 + b^2) - \frac{1}{2} log(s^2 + a^2)\bigg ]_{s}^{\infty}$
$\therefore L[\frac{[cosbt - cosat]}{t}] = \bigg[ -\frac{1}{2} log(s^2 + b^2) + \frac{1}{2} log(s^2 + a^2) \bigg]$
$\therefore L[\frac{[cosbt - cosat]}{t}] = \frac{1}{2} log \bigg ( \frac{(s^2 + a^2)}{s^2 + b^2} \bigg)$