Calculate the convolution between these functions

[Telemachus]

Hi there. I must calculate the convolution between these functions
$$f(t)= e^{-t} H(t)$$
$$g(t)=e^t H(t)$$ H(t) the unit step Heaviside function.
So I have to find: $$f \star g$$

This is what I did:
$$f \star g=\displaystyle\int_{-\infty}^{\infty}e^{-\lambda}e^{t-\lambda}H(\lambda)H(t-\lambda)d\lambda=\displaystyle\int_{-\infty}^{\infty}e^{t-2\lambda} \left [ H(\lambda)-H(\lambda-t) \right ] d\lambda=$$
$$=\displaystyle\int_{-\infty}^{\infty}e^{t-2\lambda}H(\lambda)d\lambda-\displaystyle\int_{-\infty}^{\infty}e^{t-2\lambda} H(\lambda-t)d\lambda=\displaystyle\int_{0}^{\infty}e^{t-2\lambda}d\lambda-\displaystyle\int_{t}^{\infty}e^{t-2\lambda} d\lambda=\displaystyle\int_{0}^{t}e^{t-2\lambda} d\lambda=-2e^{-t}+2e^{t}$$

Is this fine?

 

[hunt_mat]

Looks good to me.

 

[I like Serena]

Hi Telemachus! How did you get from:

$H(\lambda)H(t-\lambda)$

to:

$[ H(\lambda)-H(\lambda-t)]$That just doesn't seem right.

 

[hunt_mat]

I think that the product $H(\lambda )H(t-\lambda )$ is zero if $\lambda <0$ and also 0 if $t<\lambda$, and so the integral would become:
$$\int_{0}^{t}e^{t-2\lambda}d\lambda$$
Sorry, I made a mistake.

 

[Telemachus]

Hi I like Serena. I've realized that $H(\lambda)H(t-\lambda)$ its equal to 0 for all lambda outside the interval [0,t], and then its easy to get the same result using a sum with the Heaviside function, I get the $H(\lambda)$ and subtracted all what I didn't want using another Heaviside function, but as hunt_mat said it can be reasoned on that other way, which is the same, using the definition for the function. Maybe that mid step was redundant.

 

[I like Serena]

Oh yes, I see now.
But what if t < 0?

And btw, I believe your last integration step is not quite right...
I get a different factor in the final answer. ;)

 

[Telemachus]

Oh, I haven't think about what happens for t<0. But I've seen that always appear an Heaviside factor for this kind of integrals, but I don't know why.

The answer you get is $$(-2e^{-t}+2e^{t})H(t)$$?

I see, I have another mistake, I've made as if I was differentiating :P
So the answer is: $$(\frac{-e^{-t}+e^{t}}{2})H(t)$$?

Explain me how it is to be reasoned to get the Heaviside function as a factor there.

 

[I like Serena]

Right! That took care of the integration step.

And yes, you'd get a Heaviside function in your answer as you surmised.

(Btw, ever hear of the sinh function? It matches rather nicely.)

As for negative t.
You considered the interval [0, t] and checked the Heaviside product for different values of lambda, which is good.
But if t < 0, you'll get the intervall [t, 0].
You should check that as well for different values of lambda.

If necessary you can try it for instance with t = -2 and lambda = -3, -1, +1 (that's what I did).
You'll see!


Edit: on a more intuitive note, your function f and g are zero for negative t.
This means that their convolution will be zero as well.

 

[Telemachus]

Alright, for negative t I have
$$H(-t-\lambda)=1,\lambda\leq{-t}$$
$$H(-t-\lambda)=0,\lambda\geq{-t}$$
And by the other hand I have
$$H(\lambda)=1,\lambda\geq{0}$$
$$H(\lambda)=0,\lambda<0$$

So its zero for all negative t?

I've seen your edit note now, thank you very much Serena :)

 

[hunt_mat]

Yes.

 

[Telemachus]

Thanks.

 

[I like Serena]

Cheers!