quick and easy integral question

count_volta

Joined Feb 4, 2009
435
Hey, I am doing signal processing. Calculating the Bilateral Laplace of something.

Here is my question.

Why does this integral equal this?

I mean here is my solution.

e^-(s+a)t/ (s+a) (minus signs cancel)

evaluated at 0 (top limit) and - ∞ (lower limit)

so for the top limit you get 1/(s+a) and for the second limit you end up with e^-(s+a) (-∞) /(s+a)

As you see the minus sign on -(s+a) and the minus sign on -∞ should cancel right, and if thats true then this does not converge at all, but goes to ∞.

So how the heck does this work? What am I doing or understand wrong?

ELECTRONERD

Joined May 26, 2009
1,147
I have NO idea! But, I would like to know what the "∫" kind of thing means before your problem?

count_volta

Joined Feb 4, 2009
435
LOL Austin. The fancy $$\int$$ means integral.

The main idea of calculus is derivatives and the opposite of derivatives which are integrals. Derivatives are rates of change like we talked about before, and integrals basically sum up the area underneath a graph (function) Like 90% of signal processing is integrals, so pay attention to them in Calculus.

Anyway does nobody here know the answer to such a simple integral problem? Or does nobody want to help. Come on now people. This is important. I am having trouble understanding anti-causal functions because of this.

Ratch

Joined Mar 20, 2007
1,070
count_volta,

Why does this integral equal this?
It doesn't. That integral equals minus infinity.

Ratch

count_volta

Joined Feb 4, 2009
435
Ratch I'm not sure how familiar you are with Bilateral Laplace transform. Oppenheim Signals and Systems says that

And the region of convergence is Re (s) <-a

This is supposed to be an anti-causal function x(t)= -e^-at*u(-t) and that is its Laplace transform.

I think I will just ask my professor and then post what he said here.

Joined Jul 7, 2009
1,583
I did the integral in the attached file and it's clear the thing only converges for s < -a.

Attachments

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gregcoll

Joined Oct 11, 2009
21
a straightforward (and you might kick yourself after seeing the solution) is:
1. make the substitution -> x = -(s+a)*t, then -> -1/(s+a)*dx = dt
2. substitute all of this back in and you get: -int[(-1/(s+a)*exp(x)*dx] =
1/(s+a)*int[exp(x)*dx]
3. the solution of the exponential function is itself of course and so you get
1/(s+a)*exp(x).
4. substitute -(s+a) = x back into your solution and then you can evaluate at the original limits of -inf to 0 and you should get 1/(s+a) - 0 or simply -> 1/(s+a)
5. remember substitution!!! (especially trig substitution) for methods to solve many types of integrals

count_volta

Joined Feb 4, 2009
435
Thank you someonesdad and gregcoll that helped. someonesdad explanation did it, I finally get it.

My main problem was not doing the integral itself, but the limits of integration with infinity after you plug them in.

Actually this is the whole point. If s < -a, then and only then the integral converges and equals to 1/(s+a). If s > a, then the integral diverges and the Laplace transform is not defined.

We did not need to worry about this type of stuff for the unilateral Laplace. Blasted annoying anti-causal functions.

Seriously real circuits are always causal, why torture people with this? What signal is defined from time = 0 and goes backwards in time? Okay perhaps if you invert a signal or whatever, I would still rather look at it with time going forward not backward.

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notxjack

Joined Sep 7, 2009
20
Yeah, you have to actually do the integral limit out like someone's dad. You can't slacker through it with a CAS and just say it diverges.

Funny what a silly sign change will do to a problem =X

LKjell

Joined Oct 17, 2009
6
Something is indeed wrong because the inverse transformation of
$$\frac{1}{s+a} => e^{-at}$$
Hence

$$\int_0^\infty e^{-at}e^{-st}dt$$

What you have wrong is the limit. It is from 0 to infinity. If it was Fourier transformation then it is from minus infinity to plus infinity. And you set $$\sigma = 0$$

jabar

Joined Oct 13, 2009
14
i think s+a must be minus in this case.

count_volta

Joined Feb 4, 2009
435
Something is indeed wrong because the inverse transformation of
$$\frac{1}{s+a} => e^{-at}$$
Hence

$$\int_0^\infty e^{-at}e^{-st}dt$$

What you have wrong is the limit. It is from 0 to infinity. If it was Fourier transformation then it is from minus infinity to plus infinity. And you set $$\sigma = 0$$
This is directly out of an example in the book. They work it out and get the answer. I just posted what they wrote.

I really don't think its wrong. Remember the above is only true if Re{s} < -a

LKjell

Joined Oct 17, 2009
6
$$\int e^{at}dt = \frac{1}{a}e^{at}$$

$$-\int_a^b = \int_b^a$$

$$\int_0^{-\infty} e^{-(s+a)t}dt = -\frac{1}{s+a}[e^{-(s+a)t}]_0^{-\infty}$$

$$\lim_{t\to -\infty}e^{-(s+a)t} = \infty, \qquad \sigma, a >0$$

$$\lim_{t\to 0}e^{-(s+a)t} = 1$$

notxjack

Joined Sep 7, 2009
20
$$\lim_{t\to -\infty}e^{-(s+a)t} = \infty, \qquad \sigma, a >0$$

that integral has a domain (values of s+a) of convergence, so we can analyze it using a radius (values of s+a) of convergence, which is SOP for most bilateral laplace transforms.

Eduard Munteanu

Joined Sep 1, 2007
86
The usual (unilateral) Laplace transform goes from 0 to +inf, while the bilateral Laplace transform goes from -inf to +inf.