# Calculate the convolution between these functions

• Telemachus
In summary, the conversation discussed the process of calculating the convolution between two functions using the unit step Heaviside function. The final result was found to be (-2e^(-t) + 2e^t)H(t), with the Heaviside function accounting for the zero values of the functions for negative t values. There was also a mention of the sinh function as a possible alternative.
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?

Looks good to me.

Hi Telemachus! How did you get from:

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

to:

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

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.

Last edited:

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.

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. ;)

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.

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.

Last edited:

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 :)

Yes.

Thanks.

Cheers!

## 1. What is the purpose of calculating convolution between functions?

The purpose of calculating convolution between functions is to determine how two functions interact with each other. It can help us understand how the output of one function is affected by the input of another function.

## 2. How is convolution calculated?

Convolution is calculated by taking the integral of the product of two functions as one function is shifted over the other. This can be represented mathematically as (f * g)(t) = ∫f(τ)g(t-τ)dτ.

## 3. What is the difference between convolution and correlation?

The main difference between convolution and correlation is the order in which the functions are multiplied. In convolution, one function is inverted and shifted over the other, while in correlation, the functions are multiplied without any changes in order or direction.

## 4. Can convolution be used for any type of functions?

Yes, convolution can be used for any type of functions, as long as they are integrable. This includes continuous, discrete, and even complex functions.

## 5. What are some real-world applications of convolution?

Convolution has many real-world applications in various fields, such as signal processing, image processing, and physics. It is used to analyze and manipulate signals, filter noise, and model complex systems.

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