second shifting theorem

**ArTiCK** · October 2nd 2009, 05:22 AM

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)

Thanks in advance,
ArTiCk

**chisigma** · October 2nd 2009, 05:39 AM

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

$\chi$ $\sigma$

**mr fantastic** · October 2nd 2009, 05:50 AM

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)

Thanks in advance,
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.

**ArTiCK** · October 2nd 2009, 04:03 PM

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

**mr fantastic** · October 2nd 2009, 04:10 PM

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).

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