Laplace transform of two functions

Question

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) $?

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

Mhenni Benghorbal · Accepted Answer · 2013-11-24T23:35:59.913

1

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

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

edited Nov 24 '13 at 23:35

answered Nov 24 '13 at 23:15

Mhenni Benghorbal

48,099

Is $F'$ the transform of $f(t)$ or the derivative of the transform of $f(t)$? – Bob Shannon Nov 24 '13 at 23:16
It's the derivative of the transform. – Sudarsan Nov 24 '13 at 23:17
@Bob: It is the derivative of the Laplace transform of the function which is in your case $f(t)=\frac{\cos(5t)}{\sqrt{\pi t}}$. – Mhenni Benghorbal Nov 24 '13 at 23:21
So what would $t$ be in this case? – Bob Shannon Nov 24 '13 at 23:25
@Bob: I'll add more material. – Mhenni Benghorbal Nov 24 '13 at 23:30
I see. So if t was raised to the second power we'd do $(-1)^2F''(s)$ instead? Can we use $L{f(t)g(t)} = L{f(t)}L{g(t)}$ instead? – Bob Shannon Nov 24 '13 at 23:38
@Bob: Convolution is different: $L(f*g)=L(f)L(g)$. See here. – Mhenni Benghorbal Nov 24 '13 at 23:41
It is? We couldn't take $f=\sqrt{\frac{\pi}{t}}$ and $g=cos(5t)$ and then compute the transforms manually without any tables? – Bob Shannon Nov 24 '13 at 23:44
@Bob: You should take your time to understand these concepts and distinguish between them. You should solve more problems. – Mhenni Benghorbal Nov 24 '13 at 23:48

Laplace transform of two functions

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