1
$\begingroup$

I know how to do a basic laplace transform, but how does one deal with transforming complex combination of functions?

For example, how would we handle:

$$\mathcal{L}\left( \ \sqrt{\frac{t}{\pi}}cos(5t) \right) = ... $$

From a table of laplace transforms it is known that: $$\mathcal{L}\left( \ \frac{cos(5t)}{\sqrt{\pi t}} \right) = \frac{e^{-5/s}}{\sqrt{s}}$$

This table value must be of some use to solve this problem, but how?

EDIT: Can we use $\mathcal{L}\left( f(t) *g(t) \right) = \mathcal{L}\left( f(t)\right) * \mathcal{L}\left( g(t)\right) $?

$\endgroup$
  • 3
    $\begingroup$ There's a relation between $\dfrac{d}{ds} \mathcal{L}[f](s)$ and $t\cdot f(t)$. Which? $\endgroup$ – Daniel Fischer Nov 24 '13 at 23:13
  • $\begingroup$ There should be an identity for the Laplace transform of $t.f(t)$ if $\mathcal{L}(f(t))$ is known. $\endgroup$ – Sudarsan Nov 24 '13 at 23:14
  • $\begingroup$ Hrm -- there's a known identity for taking the transform of $g(t)f(t)$ where the transform of $f(t)$ is known? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:15
1
$\begingroup$

Hint: Use the property

$$ L(t f(t)) = -F'(s). $$

Added Note that,

$$\frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} = t \frac{\cos(5t)}{\sqrt{\pi t}}.$$

Now, take Laplace transform of both sides of the above equality

$$ \mathcal{L}\left\{ \frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} \right\}= \mathcal{L}\left\{ t \frac{\cos(5t)}{\sqrt{\pi t}} \right\}=-\frac{d}{ds}\frac{e^{-5/s}}{\sqrt{s}}=\dot\,. $$

$\endgroup$
  • $\begingroup$ Is $F'$ the transform of $f(t)$ or the derivative of the transform of $f(t)$? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:16
  • $\begingroup$ It's the derivative of the transform. $\endgroup$ – Sudarsan Nov 24 '13 at 23:17
  • $\begingroup$ @Bob: It is the derivative of the Laplace transform of the function which is in your case $f(t)=\frac{\cos(5t)}{\sqrt{\pi t}}$. $\endgroup$ – Mhenni Benghorbal Nov 24 '13 at 23:21
  • $\begingroup$ So what would $t$ be in this case? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:25
  • $\begingroup$ @Bob: I'll add more material. $\endgroup$ – Mhenni Benghorbal Nov 24 '13 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.