0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.