Convolution of a dirac delta function

In summary, the conversation discusses the convolution of a dirac delta function and a unit step function. The problem involves finding the inverse laplace of a function and using convolution to solve the final portion of the problem. The conversation also covers the notations and symbols used, as well as the confusion with the integral of the dirac delta function. In the end, the solution is found to be m(t) = h(t) * u(t - a), which can be simplified to 10 * m(t). The person asks for confirmation on their solution and for any additional help or resources.
  • #1
pka
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Alright...so I've got a question about the convolution of a dirac delta function (or unit step). So, I know what my final answer is supposed to be but I cannot understand how to solve the last portion of it which involves the convolution of a dirac/unit step function. It looks like this:

10 * Inverse laplace of [ H(s) * e ^ (-5s) ]

where H(s) = (1/20) * (1 - e ^ -20t).

---Note:
This is what I've done to lead me to the dirac/unit step. Btw, I'm calling it the dirac/unit step function because I get the dirac delta function in my answer whereas the answer has a unit step function. So, just for clarity's sake I will call the unit step function u(t - a) and the dirac delta function d(t - a).

Now, let's continue.

Saying L(s) = e ^ -5s. So that its inverse laplace, l(t) = d(t - 5).

Let's also say that M(s) = H(s) * L(s).

Convolution time! And I get m(t) = (1/20) * (1 - e ^ -20t) * Integral from 0 to t of d(tau - 5) d(tau).

I'm sorry about my notations, I don't know how to put in an integral sign or...any other fancies. =/

This is where my trouble starts. I thought the integral of a dirac delta function would be just h(t). But that's not right. In order for my answer to make any sense then the integral of d(tau - 5) should be just d(t - 5) to get a fairly simple answer of d(t - 5) * h(t).

Any help in this matter would be greatly appreciated. Links too! If I've posed a really simple question in too much writing then feel free to let me know or if I'm thinking about this way too hard then please...also let me know. But many thanks to any advice or help anyone can offer me.
 
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  • #2
Actually...I think I've solved my problem! In integral of the dirac delta should be just the unit step function...giving me what I need. And so...the convolution turns out to be m(t) = h(t) * u(t - a). So...then it's just 10 * m(t).

:D Can anyone tell me if my answer is correct in its thought and all that? :D
 

FAQ: Convolution of a dirac delta function

What is a Dirac delta function?

A Dirac delta function is a mathematical function that is used to represent an impulse or point source of energy. It is defined as zero everywhere except at the origin, where it is infinite, with an area under the curve of 1.

What is convolution?

Convolution is a mathematical operation that combines two functions to create a third function. It is often used in signal processing and image processing to describe the output of a linear time-invariant system given an input signal.

How is convolution of a Dirac delta function calculated?

The convolution of a Dirac delta function with another function is simply the other function evaluated at the point of the delta function. Mathematically, it can be represented as: f(x) * δ(x-a) = f(a). This means that the output of the convolution is a shifted version of the original function at the point where the delta function is located.

What are the applications of convolution of a Dirac delta function?

Convolution of a Dirac delta function is commonly used in signal processing to simulate the response of a system to a pulse input. It is also used in image processing for edge detection and noise reduction. Additionally, it has applications in physics and engineering for solving differential equations and modeling physical systems.

What are the limitations of using Dirac delta function in convolution?

One limitation of using Dirac delta function in convolution is that it is an idealized function and does not exist in the real world. In practical applications, it can be approximated by a very narrow and tall pulse, but this may introduce errors in the calculations. Additionally, the Dirac delta function is only defined for real numbers, so it cannot be used for discrete or complex systems.

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