1
$\begingroup$

I have a (presumably simple) Laplace Transform problem which I'm having trouble with:

$$\mathcal L\big\{t \sinh(4t)\big\} = ?$$

How would I go about solving this? Would you please show working if possible, or alternatively point me in the right direction regarding how to go about solving this?

Also, I'm studying Electrical Engineering at University, and in my ElectEng lectures, my lecturers are referring to Laplace transforms and the Laplace domain (with regard to the frequency response of circuits) in a way which I haven't been exposed to in my mathematics courses (as we've only briefly covered Laplace). Do you know any good resources which I could look at (either on the web or in the library) to get a better understanding of Laplace transforms (and, in particular, their application to circuit analysis)?

Sorry if the last part of this question is out of the scope of this website.

Thanks in advance.

$\endgroup$

4 Answers 4

3
$\begingroup$

These notes here look good.

The definition of the Laplace transform is $\mathcal{L}(f)(s) := \int_0^\infty e^{-st} f(t) dt$ and for $\sinh$ the following holds: $\sinh x = \frac12 (e^x -e^{-x})$. Now you put these two things together and compute

$$\frac12 \int_0^\infty xe^{4x} e^{-xs} dx - \frac12 \int_0^\infty xe^{-4x} e^{-xs} dx$$

Hope this helps.

$\endgroup$
2
  • $\begingroup$ Thanks bro! I don't remember ever begin taught hyperbolic trig functions, but now that I refer to my formula book I see them there so I'll take note. Thanks again! $\endgroup$
    – JonaGik
    Commented Mar 18, 2012 at 9:00
  • $\begingroup$ @JonaGik Glad I could help : ) $\endgroup$ Commented Mar 18, 2012 at 9:02
2
$\begingroup$

You got $t\operatorname{sinh}{4t}=t\frac{e^{4t}-e^{-4t}}{2}$ , so: $$\mathcal{L}\big\{t\operatorname{sinh}4t\big\}=\tfrac{1}{2}\left(\mathcal{L}\left\{{te^{4t}}\right\}+\mathcal{L}\left\{te^{-4t}\right\}\right)$$

We know that $\mathcal{L}\big\{e^{at}f(t)\big\}=F(s-a)$. Furthermore $\mathcal{L}\left\{t\right\}=\frac{1}{s^2}$. So in our case $\mathcal{L}\left\{te^{4t}\right\}=F(s-4)=\frac{1}{(s-4)^2}$. Finally:$$\mathcal{L}\big\{t \text{sinh} 4t\big\}=\frac{0.5}{(s-4)^2}+\frac{0.5}{(s+4)^2}$$

$\endgroup$
1
  • $\begingroup$ you're welcome..i'm an engineer too and I believe that this book (it's a bible for Electrical Engineers) will help you understand Fourier and Laplace transforms with specific applications and examples in your field: amazon.com/Signals-Systems-2nd-Alan-Oppenheim/dp/0138147574 $\endgroup$
    – chemeng
    Commented Mar 18, 2012 at 10:29
1
$\begingroup$
  1. There is a generic formula to write $G(s)=\mathcal{L}\{tf(t)\}(s)$ in terms of $F(s)=\mathcal{L}\{f(t)\}(s)$; seeing it involves integration by parts. Have you covered this rule?
  2. Hyperbolic sine is a difference of exponentials; can you find the Laplace transform of these?
$\endgroup$
2
  • $\begingroup$ Regarding #1, I don't believe I have covered that rule. Could you please point me to it and its derivation, if you don't mind? Regarding #2, that was the main thing I needed to see to solve it. Thanks. $\endgroup$
    – JonaGik
    Commented Mar 18, 2012 at 9:24
  • $\begingroup$ @Jona: It's an easy derivation the way Bitrex has it; just note that $te^{-st}=-D_s e^{-st}$ and the derivative can be interchanged with the integral (b/c of sufficient regularity). $\endgroup$
    – anon
    Commented Mar 18, 2012 at 9:53
0
$\begingroup$

$\mathcal{L}\{f(t)\}' = F'(s)= \int_0^\infty\frac{d}{ds}[e^{-st} f(t)] dt = \int_0^\infty e^{-st}[-t f(t)]dt$,

so $\mathcal{L}\{tf(t)\} = -F'(s).$

Since $\mathcal{L}\{\sinh(4t)\} = \frac{4}{s^2 - 8}$, (I "cheated" and checked a table!)

Then $\mathcal{L}\{t*\sinh(4t)\} = -\mathcal{L}\{\sinh(4t)\}' = -\frac{d}{ds}(\frac{4}{s^2 - 16})$.

In general, $\mathcal{L}\{t^nf(t)\} = (-1)^n\frac{d}{ds}^nF(s)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .