Convolution of two delta functions in frequency domain

In summary, when convolving in the frequency domain, the result is obtained by using the distributivity property to expand the equation. Laplace transforms can also be used to compute the convolution, but this may not be the expected method for all courses. It is important to understand the underlying principles and concepts in order to fully comprehend the result.
  • #1
atrus_ovis
101
0
Apparently, when convolving, for example:

[δ(ω-π) - δ(ω+π)] * (δ(ω+50π)-δ(ω-50π))

the result is
δ(ω+49π)-δ(ω-51π)-δ(ω+51π)+δ(ω-49π)

where δ() is the Dirac delta function, * the convolution operator and ω the frequency variable

How do we get to this? Can you help me on the intuition in this example and/or general in convolution in the frequency domain?

thank you.edit: i think i understand that we use the distributivity property to expand it.When we have , i.e. δ(ω+π) * δ(ω-50π) how do we continue?
 
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  • #2
atrus_ovis said:
Apparently, when convolving, for example:

[δ(ω-π) - δ(ω+π)] * (δ(ω+50π)-δ(ω-50π))

the result is
δ(ω+49π)-δ(ω-51π)-δ(ω+51π)+δ(ω-49π)

where δ() is the Dirac delta function, * the convolution operator and ω the frequency variable

How do we get to this? Can you help me on the intuition in this example and/or general in convolution in the frequency domain?

thank you.


edit: i think i understand that we use the distributivity property to expand it.When we have , i.e. δ(ω+π) * δ(ω-50π) how do we continue?

In terms of actually computing the convolution, you can use Laplace transforms to get the "frequency information" of both terms, and then use the inverse Laplace Transform to get the actual result. With convolutions you just work out the Laplace transforms of T1, and T2 (in T1 * T2) and then use it in the way that was said above.

You could alternatively do it just with the actual definition though, but that would require a different approach. The Laplace way is a lot easier.

Do you know about Laplace transforms?

What have you tried so far? Maybe you should show some working or any ideas you had or tried that failed.
 
  • #3
The course I'm on doesn't include Laplace transforms, so i guess it's not the expected way.

The equation I've written is part of a solved problem set, and I'm trying to actually understand why this result is correct because I'm preparing for my exams.
I know the rule about homework.I have the solution, I'm only asking for tips to understand it.

In the time domain, it's ok. f(x) convolved with δ(x-k) yields f(k) , as the delta function is 0 everywhere but k.
In the frequency domain, it seems that the frequency offset of the delta function, is simply added as an offset to the convolving pair.
 

Related to Convolution of two delta functions in frequency domain

1. What is the definition of convolution in the frequency domain?

In the frequency domain, convolution is a mathematical operation that combines two functions to create a third function. It is commonly used to analyze signals and systems, and can be thought of as a way to measure the overlapping areas between two functions.

2. How is convolution of two delta functions represented in the frequency domain?

In the frequency domain, the convolution of two delta functions is represented by the product of their respective Fourier transforms. This means that the resulting function will have spikes at the same frequencies as the individual delta functions, but with the amplitudes multiplied together.

3. What is the significance of convolving two delta functions in the frequency domain?

The convolution of two delta functions in the frequency domain is significant because it allows us to analyze the behavior of a system when it is excited by a single frequency signal. This is useful in signal processing and communication systems, as well as in other areas of engineering and science.

4. How does the convolution of two delta functions relate to the concept of filtering?

In signal processing, filtering is the process of removing unwanted frequencies from a signal. Convolution of two delta functions in the frequency domain can be seen as a form of filtering, as it results in a new function that only contains the frequencies present in both of the original functions.

5. Can the convolution of two delta functions be used to solve real-world problems?

Yes, the convolution of two delta functions has many practical applications in fields such as signal processing, communication systems, and image processing. It is a powerful tool for analyzing systems and can be used to solve a wide range of real-world problems.

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