# Math Help - second shifting theorem

1. ## second shifting theorem

Hi all,

I am having trouble gettin the answer to the attached question. I need to find f(t) and then find L[f(t)] using the t -shifting theorem.

I tried working it out...

f(t) = e^t - e^t u(t-2)

But when i take the Laplace transform i can't get the answer. The answer is

(1 - e^(2 - 2s))/ (s -1)

ArTiCk

2. Applying the defintion of LT we have...

$\mathcal{L}\{f(t)\} = \int_{0}^{2} e^{-(s-1)t}\cdot dt = - \frac{1}{s-1} |e^{-(s-1)t}|_{0}^{2} = \frac{1-e^{-2(s-1)}}{s-1}$

Kind regards

$\chi$ $\sigma$

3. Originally Posted by ArTiCK
Hi all,

I am having trouble gettin the answer to the attached question. I need to find f(t) and then find L[f(t)] using the t -shifting theorem.

I tried working it out...

f(t) = e^t - e^t u(t-2)

But when i take the Laplace transform i can't get the answer. The answer is

(1 - e^(2 - 2s))/ (s -1)

ArTiCk
To use the theorem you mention, you should note that $f(t) = e^t [H(t) - H(t - 2)]$.

Then LT[f(t)] = G(s - 1) where G(s) = LT[g(t)] and $g(t) = H(t) - H(t - 2)$.

The answer so obtained is the same as the answer found by the previous poster.

4. Originally Posted by mr fantastic
To use the theorem you mention, you should note that $f(t) = e^t [H(t) - H(t - 2)]$.

Then LT[f(t)] = G(s - 1) where G(s) = LT[g(t)] and $g(t) = H(t) - H(t - 2)$.

The answer so obtained is the same as the answer found by the previous poster.
Hi,

I don't quite understand your notation. I found f(t) using step functions and that gave f(t) = e^t - e^t u (t -2). From my lectures the t shifiting theorem is:

If LT[f(t)] = F (s), then

LT[f(t-a) u(t-a)] = e^(-as)F(s)

Could you please explain what H(t) is?

Thanks once again,
ArTiCk

5. Originally Posted by ArTiCK
Hi,

I don't quite understand your notation. I found f(t) using step functions and that gave f(t) = e^t - e^t u (t -2). Mr F says: It should be clear from my previous reply (go back and read it) that this is wrong. It does not give a 'pulse' between t = 0 and t = 2.

From my lectures the t shifiting theorem is:

If LT[f(t)] = F (s), then

LT[f(t-a) u(t-a)] = e^(-as)F(s)

Could you please explain what H(t) is?

Thanks once again,
ArTiCk
The notation H(t) (the Heaviside step function) means the same as u(t) (the unit step function).