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What is the bilateral Laplace transform of the following function:
f(t)=1 and t belongs to R
f(t)=1 and t belongs to R
Laplace transform of f(t)=1?
- Thread starter logicman112
- Start date
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 \(\mathcal{L}(c)=\mathcal{L}(c\cdot u(t))=\frac{c}{s}\)
Take your pick.
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 \(\mathcal{L}(c)=\mathcal{L}(c\cdot u(t))=\frac{c}{s}\)
Take your pick.
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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)
limits are the same from -infinity to +infinity
krishna chaitanya
- Joined Nov 8, 2010
- 5
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..
f(t) = u(t)+u(-t)
F(s) = 2/s... for practical needs.. u can use it..
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)?
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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!!!!
This way f(t)=1 is not u(t)+u(-t) for every t!!!!
Can you write down the integral you have problem with and your two ways of integrating it?
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!!!!
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!!!!
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.
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.
