# Laplace transform of f(t)=1?

Discussion in 'Math' started by logicman112, Nov 10, 2010.

1. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
What is the bilateral Laplace transform of the following function:
f(t)=1 and t belongs to R

2. ### Georacer Moderator

Nov 25, 2009
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1,266
Are you sure you want a bilateral and not a unilateral transformation?

In the bilateral Laplace transformation you have $\int_{-\infty}^\infty \exp^{-st}dt = c(\frac{1}{s}\exp^{\infty \cdot t})=\infty$, which isn't very useful.

On the other hand, the unilateral transformation gives $\mathcal{L}(c)=\mathcal{L}(c\cdot u(t))=\frac{c}{s}$

Last edited: Nov 10, 2010
3. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
Think about the Fourier transform of f(t) =1, the integral is the same, but the result of the integral is 2*pi*impulse(w)

4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
The integral might be the same, but the limits of integration differ. It makes all the difference.

5. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
limits are the same from -infinity to +infinity

6. ### krishna chaitanya New Member

Nov 8, 2010
5
0
theoretically ... it doesn't exist..if u apply bilateral Laplace transform...... but u can take it as
f(t) = u(t)+u(-t)
F(s) = 2/s... for practical needs.. u can use it..

7. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
Why the integral from -infinity to +infinity seems to be infinity if we integrate directly but seems to be 2/s if we integrate by u(t) and u(-t)?

Last edited: Feb 21, 2011
8. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
How do you define u(t)? for t>0 it is +1 , for t<0, it is 0 but I take it +1 for t=0.
This way f(t)=1 is not u(t)+u(-t) for every t!!!!

9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Can you write down the integral you have problem with and your two ways of integrating it?

10. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
I can not type integral symbols and so on. As You wrote before and you may have forgot!!it, the result of the integral is infinity. But another way is the separation of 1*e^(-s*t)*dt to (u(t)+u(-t))*e^(-s*t)*dt. If you integrate from -infinity to +infinity from the latter expression, you will have 1/s+1/s.
It is very simple, please do it yourself.
my question remains: Why the two different solutions by you and the other friend leads to completely different answers!!!!

11. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I had some inaccuracies in the first post, but I don't see anything different in the result.
However, you have to define very well what it is that you want to transform. I will integrate the function f(t)=1 from -infinity to infinity. I emphasise that this is different from f(t)=u(t).

So here it is. With one integral:
$\int_{-\infty}^\infty e^{-st}dt=-\frac1s\cdot(e^{-s\infty}-e^{s\infty})=\\
-\frac1s(-\infty)=\\
\infty$

for s>0. For s<0 the result is $-\infty$, useless either way.

With the integral broken in two:
$\int_{-\infty}^{\infty} e^{-st}dt=\\
\int_{-\infty}^0 e^{-st}dt + \int_0^{\infty} e^{-st}dt=\\
-\frac1s(e^0-e^{s\infty})-\frac1s(e^{-s\infty}-e^0}=\\
\infty+\frac1s=\\
\infty$

You see that the result is the same either way.