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Let $f(t) = \begin{cases}t & \text{if}\,0<t<1 \\ 2-t & \text{if}\, 1<t < 2 \\ 0 & \text{otherwise} \end{cases}$

If you draw this function, it looks like an isosceles triangle with holes at the vertices at $(0,0)$, $(1,1)$, and $(2,0)$, so essentially, the $f(t) =t$ piece of this function is switched on at $t = 0$, and switched off at $t=1$, the $f(t) = 2-t$ piece of this function is switched on at $t = 1$ and switched off at $t=2$, and everywhere else the function is zero.

Earlier, I found the Laplace transform of $f(t)$ to be $\displaystyle \hat{f}(s) = \frac{1}{s^{2}} - \frac{2}{s^{2}}e^{-s} + \frac{1}{s^{2}}e^{-2s}$.

Therefore, $\displaystyle \hat{f}(s-1) = \frac{1}{(s-1)^{2}}- 2e\frac{2}{(s-1)^{2}}e^{-s}+e^{2}\frac{1}{(s-1)^{2}}e^{-2s}$, and $\displaystyle \mathbf{ s\hat{f}(s) = \frac{s}{(s-1)^2}- 2e \frac{s}{(s-1)^{2}}e^{-s}+e^{2}\frac{s}{(s-1)^{2}}e^{-2s}}$

Now, I am being asked to find $\mathbf{\mathcal{L}^{-1}[s \hat{f}(s-1)]}$, the inverse Laplace transform of $s \hat{f}(s-1)$.

The answer given in the back of the book is $\begin{cases} e^{t}(t+1) & \text{for}\,0<t<1 \\ e^{t}(1-t) & \text{for} \, 1 < t < 2 \\ 0 & \text{for}\, t>2 \end{cases}$

When approaching $\mathcal{L}^{-1}[s \hat{f}(s-1)]$ directly, none of the terms stood out to me as things with recognizable inverse Laplace transform forms. If the first term, for example, had been just $\displaystyle \frac{1}{(s-1)^{2}}$ and not $\displaystyle \frac{s}{(s-1)^{2}}$, I would know what to do. But that extra $s$ in the numerator is bothering me. I thought that maybe it was a convolution, where $f*g(t) = \int_{0}^{t}f(\tau)g(t - \tau)d\tau$, since $\mathcal{L}[f*g(t)] = \hat{f}(s)\hat{g}(s)$, but I'm not entirely sure what the inverse Laplace transform of $s$ is.

Finally, what I decided to do was work backwards. I started with the solution, and took the Laplace transform of it, hoping I might be able to reverse-engineer it, so to speak, and figure out how to do it that way. But, this isn't really practical - because unless you know what the answer is supposed to be ahead of time, you don't really know what mathematical tricks to use in order to rewrite it in terms of the functions you started with.

How should I approach this problem? In addition, can you give me any hints as to how to handle the extra $s$'s? And, if I do need to use convolution, could you work out one of the terms for me in full, so I could see how to do it? Then, I might be able to apply it to the other ones myself and get the answer I'm supposed to get for the whole thing. Thank you.

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Here is a sketch of a direct route. By definition, you have $$\hat{f}(s)=\int_0^\infty e^{-st}f(t)\,dt\implies s\hat{f}(s-1)=\int_0^\infty se^{-(s-1)t}f(t)\,dt=\int_0^\infty se^{-st}(e^t f(t))\,dt$$

The extra factor of $s$ is the only obstacle to this being the definition of a Laplace transform. To get rid of it, we may observe that $se^{-st}=-\frac{d}{dt}(e^{-st})$. This immediately suggests integration by parts. The remaining details are left to the interested reader, but 1) the boundary term of this partial integration should vanish, 2) the remaining integral is of the form of a Laplace transform. The inverse Laplace transform of $s\hat{f}(s-1)$ is then obtained by inspection of this last term.

Postscript

It seems worth emphasizing that a standard table of Laplace transforms include the following functional identities, of which the above are examples: \begin{array}{ccc} f(t) &\Leftrightarrow& \hat{f}(s)=\int_0^\infty e^{-st}f(t)\,dt &\text{(Definition of Laplace transform)}\\ e^{pt}f(t) &\Leftrightarrow& \hat{f}(s-p) &\text{(Shift in frequency domain)}\\ f(t-\tau)H(t-\tau) &\Leftrightarrow& e^{-s\tau}\hat{f}(s) &(\text{Shift in time domain for }\tau>0)\\ f'(t) &\Leftrightarrow& s\hat{f}(s)-f_0 &\text{(Differentiation in time domain)}\\ \end{array}

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  • $\begingroup$ I updated my question to include all my work so far, including the part where I feel like I've got my legs stuck in tar and can't get out or move anymore. $\endgroup$ – ALannister Jan 3 '17 at 23:09
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You can do partial fraction: $$\frac{s}{(s-1)^2}=\frac{1}{s-1}+\frac{1}{(s-1)^2}.$$ Then you can use the formulas $$L^{-1}(F(s-a))=e^{at}f(t)$$ and $$L^{-1}(e^{-as}F(s))=f(t-a)U(t-a)$$ to deal with the factor $e^{-(s-1)}$, where $U(t-a)$ is the unit step function.

Here is one example for the factor $e^{s-b}$ where $b$ is different from the shift in $F(s-a)$: $$L^{-1}\big(\frac{1}{s-1}\cdot e^{-2(s-1)}\big)e^t\cdot L^{-1}\big(\frac{1}{s}\cdot e^{-2s}\big)=U(t-2).$$ This used the second formula above.

Eventually you will have many terms with or without $U(t-1)$ and $U(t-2)$. Remember that $U(t-a)$ is 0 on $[0,a)$ and $1$ on $[a,\infty)$. You then just need to discuss three cases: $[0,1),[1,2),[2,\infty)$.

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    $\begingroup$ from one kitty to another ;) We cats need to stick together! I'll give this a try. $\endgroup$ – ALannister Jan 3 '17 at 20:52
  • $\begingroup$ how do you deal with $L^{-1}(e^{-as}F(s-b))$? I.e., the case where you want to find $\displaystyle L^{-1}\left[ e^{-2s}\left(\frac{1}{s-1} \right)\right]$? $\endgroup$ – ALannister Jan 3 '17 at 21:46
  • $\begingroup$ @Semiclassical perhaps you could answer the follow-up question I just asked Kitty. $\endgroup$ – ALannister Jan 3 '17 at 22:03
  • $\begingroup$ @JessyCat The only relevant difference in that case is the factor in front. But this is easily incorporated into the definition of the Laplace transform by a slight change of the integration variable $t$. $\endgroup$ – Semiclassical Jan 3 '17 at 22:08
  • $\begingroup$ @JessyCat: I added an example of that. $\endgroup$ – KittyL Jan 3 '17 at 23:04
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$L^{-1}\{e^{-as} F(s-b)\} = u_a(t)\cdot L^{-1}\{F(s-b)\}\big(t-a\big) = u_a(t)\cdot e^{b(t-a)} L^{-1}\{F(s)\}\big(t-a\big)$,

if this is what you are asking for.

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