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For $a > 0$, is there any known representation of the Laplace transform of $f(t+a)$ in terms of the Laplace Transform of $f(t) $

Note: In my application, $f(t)$ is not a periodic function and the functional form of $f(t)$ is not actually known a-priori, because I have to couple it to another set of equations.

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Let $s = \sigma + j\omega$, the inverse Laplace transform of $f(t+a)$ is given by $$f(t+a) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{s(t+a)} \mathrm{d}s = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{sa}e^{st} \mathrm{d}s.$$

Hence the bilateral Laplace transform of $f(t+a)$ is $F(s)e^{sa}$ where $F(s)$ is the Laplace transform of $f(t)$. For the unilateral case, see Matt L. answer.

This is sometimes called the shifting theorem. See Theorem 12.16 here.

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  • $\begingroup$ Thanks! I actually derived that same answer by using a very squirrely interpretation of the Second Shifting Theorem with a negative shift, wasn't sure of my result! Appreciate your help. $\endgroup$ – Sharat V Chandrasekhar May 24 '17 at 18:36
  • $\begingroup$ Unless I'm missing something, the result doesn't seem to check out against a simple example where $f(t)= t^2$ $$ L[f(t)] =L[t^2]=\frac{2}{s^3} $$ and $$ L[f(t+a)] = L[(t+a)^2] =\frac{2}{s^3}+\frac{2a}{s^2}+\frac{a^2}{s} $$ which does not equal $2\frac{e^{sa}}{s^3}$ $\endgroup$ – Sharat V Chandrasekhar May 24 '17 at 19:09
  • $\begingroup$ The last line should read: which does not (at least algebraically) equal $2\frac{e^{sa}}{s^3}$ for $a \ne 0$ The 5-min edit rule is a bit inconvenient. $\endgroup$ – Sharat V Chandrasekhar May 24 '17 at 19:18
  • $\begingroup$ I would note that: $$ L[f(t+a)] = L[(t+a)^2] =\frac{2}{s^3}+\frac{2a}{s^2}+\frac{a^2}{s} =\frac{2}{s^3}(1+as+\frac{{as}^2}{2}) $$ where the term in parentheses is the Taylor's series expansion up to the second order of $e^{sa}$ ! $\endgroup$ – Sharat V Chandrasekhar May 24 '17 at 19:25
  • $\begingroup$ I'm not clear as to which inverse transform you're talking about. I was only comparing the functional forms of the unilateral Laplace transform. $\endgroup$ – Sharat V Chandrasekhar May 25 '17 at 2:39
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You have to be clear about which flavor of the Laplace transform you're talking about. If you consider the bilateral Laplace transform

$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt\tag{1}$$

then the relationship

$$\mathcal{L}\{f(t+a)\}=e^{as}F(s)\tag{2}$$

clearly holds, also for $a>0$.

However, if you consider the unilateral Laplace transform

$$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt\tag{3}$$

then for $a>0$

$$\mathcal{L}\{f(t+a)\}=e^{as}\int_a^{\infty}f(t)e^{-st}dt\tag{4}$$

which is generally not equal to $e^{as}F(s)$ (unless $f(t)=0$ in $[0,a]$).

In your example $f(t)=t^2$ you implicitly use the unilateral Laplace transform, so $(2)$ doesn't hold.

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  • $\begingroup$ Thanks. Yes. I was talking about the unilateral transform. I guess that for this case, there may not be a more convenient representation of $ L{f(t+a)}$ $\endgroup$ – Sharat V Chandrasekhar May 25 '17 at 2:43
  • $\begingroup$ @SharatVChandrasekhar: Yes, for the unilateral transform you just get Eq. (4) in my answer, and generally there is no more convenient expression. This is also clear because for $a>0$ the unilateral Laplace transform of $f(t+a)$ generally cuts off a part of the function (the part inside the interval $[0,a]$). $\endgroup$ – Matt L. May 25 '17 at 8:22
  • $\begingroup$ Many thanks to everyone for their input. In the event, I was able to reformulate the underlying problem more elegantly, so that I don not have to worry about transforming f(t+a) anymore. $\endgroup$ – Sharat V Chandrasekhar May 25 '17 at 15:09

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