# Convolution of two Sinc functions

• roam
In summary: I think you are there. The combined result is a product of two rectangular F.T.'s. You do not need to integrate at all. How about your inverse F.T.?Those aren't equal. What they are is the transform pair.Yes, you will get the narrower of the two transform functions, and therefore the wider of the two sinc functions as the convolution. Of course there may be a re-scaling factor.Sorry I meant that we get the FT pair:$$sinc(at) * sinc(bt) \leftrightarrow \frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left #### roam ## Homework Statement Calculate the convolution of ##sinc(at)## and ##sinc(bt),## where ##a## and ##b## are positive real numbers and ##a>b.## ## Homework Equations Convolution integral ## The Attempt at a Solution The fact that ##a>b## tells us that the graph of ##sinc(at)## is ##a-b## times more "compressed" than that of ##sinc(bt).## So from the definition of the convolution integral:$$sinc(at) * sinc(bt) = \int^\infty_{-\infty} sinc(a \tau).sinc(bt-\tau) d \tau = \int^\infty_{-\infty} \frac{\sin(a\tau)}{a\tau} \frac{\sin(bt-\tau)}{bt-\tau} d\tau \tag{1}$$I've read that the convolution of two sinc functions at two different points is itself a sinc function located at the point of the difference between the two. So how exactly do I proceed from equation (1) to arrive at this result? Delta2 This one I was able to google and find something that should work: It gave a somewhat simple answer of taking the F.T. of the convolution function which is a product of the individual F.T.'s. Each of the F.T.'s is a rectangular function so the product should be a narrowed rectangular function. You then take the inverse F.T. of the narrowed rectangular function to get the result. There may be other ways to solve it directly, (e.g. complex variable residue theorem), but the method that showed up in a google should be useful. roam Thank you for the suggestion. So, since we have the Fourier transform pair ##\Pi (t) \leftrightarrow sinc(\nu),## we must also have ##sinc(t) \leftrightarrow \Pi (-\nu).## And by the scaling property the convolution becomes:$$sinc(at) * sinc(bt)= \frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right)$$Is this the right? Personally, I don't think this is correct since apparently the convolution of two sinc functions has to be just a larger sinc function. Suggestion: Google the F.T. of a single sinc function. The result after a lengthy derivation is a rectangular function. It will probably require a little algebra working with the product of the two rectangular functions. Your "a" and "b" are frequency terms. These are proportional to the frequency limits that are found in the rectangular F.T. solution. If my calculations are correct, it is simply the lower frequency "b" that will determine the limit in the product of the rectangular functions. (Perhaps my calculations are incorrect, but I don't get an average and/or difference of the frequencies.) One case they showed in the google was when the two frequencies are the same, the result is simply a sinc function (at that frequency), and not at the difference of the two frequencies. Last edited: Please read my edited post #4. roam said: Thank you for the suggestion. So, since we have the Fourier transform pair ##\Pi (t) \leftrightarrow sinc(\nu),## we must also have ##sinc(t) \leftrightarrow \Pi (-\nu).## And by the scaling property the convolution becomes:$$sinc(at) * sinc(bt)= \frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right)$$Is this the right? Personally, I don't think this is correct since apparently the convolution of two sinc functions has to be just a larger sinc function. Those aren't equal. What they are is the transform pair. Yes, you will get the narrower of the two transform functions, and therefore the wider of the two sinc functions as the convolution. Of course there may be a re-scaling factor. roam SammyS said: Those aren't equal. What they are is the transform pair. Yes, you will get the narrower of the two transform functions, and therefore the wider of the two sinc functions as the convolution. Of course there may be a re-scaling factor. @SammyS I question what the function ## \Pi(\nu) ## above represents. When I googled the F.T. of the sinc function, they gave a result proportional to ## (1/2)(sgn(\omega+\omega_o)-sgn(\omega-\omega_o)) ## which is basically a rectangle for the F.T. that is bounded by ## -\omega_o ## and ## +\omega_o ##. In this case with f(t)=sinc(at), the F.T. rectangle would be bounded by ## -a ## and ## +a ## (in frequency space). With the definitions for F.T. (and frequency) that the OP is using, he might get a factor of ## 1/(2 \pi) ## on his ## -a ## and ##+a ##, etc. Any help is appreciated. I am on a learning curve for this calculation as well. Charles Link said: @SammyS I question what the function ## \Pi(\nu) ## above represents. When I googled the F.T. of the sinc function, they gave a result proportional to ## (1/2)(sgn(\omega+\omega_o)-sgn(\omega-\omega_o)) ## which is basically a rectangle for the F.T. that is bounded by ## -\omega_o ## and ## +\omega_o ##. In this case with f(t)=sinc(at), the F.T. rectangle would be bounded by ## -a ## and ## +a ## (in frequency space). With the definitions for F.T. (and frequency) that the OP is using, he might get a factor of ## 1/(2 \pi) ## on his ## -a ## and ##+a ##, etc. Any help is appreciated. I am on a learning curve for this calculation as well. I assumed that this is what OP meant by the ##\ \Pi\ ## function. The way you expressed it using the sign function is much clearer. Charles Link SammyS said: Those aren't equal. What they are is the transform pair. Yes, you will get the narrower of the two transform functions, and therefore the wider of the two sinc functions as the convolution. Of course there may be a re-scaling factor. Sorry I meant that we get the FT pair:$$sinc(at) * sinc(bt) \leftrightarrow \frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right)$$Do I simplify this and return to time domain? I think the combined scalar would be ##1/|ab|,## but what would be the argument of the resulting ##\Pi(...)## function? roam said: Sorry I meant that we get the FT pair:$$sinc(at) * sinc(bt) \leftrightarrow \frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right)$$Do I simplify this and return to time domain? I think the combined scalar would be ##1/|ab|,## but what would be the argument of the resulting ##\Pi(...)## function? See my post #7. The ## \Pi(\nu) ## function is one that I'm not familiar with. It appears SammyS has seen it previously (see his answer to my question), and apparently ## \Pi(\nu) = 1 ## for ## -1<\nu<1 ## and is zero everywhere else. In the product of your ## \Pi ## functions, for a>b, the "b" will determine the rectangle, i.e. the result of the product will be ## \Pi(\nu/b) ##. Recommend a google of the ## \Pi ## function, but I think I have it correct. It is mathematically very simple and very easy to multiply two of them together. Also ## \Pi(\nu/b)=1 ## for ## -b<\nu<b ## and is zero everywhere else. Last edited: Charles Link said: See my post #7. The ## \Pi(\nu) ## function is one that I'm not familiar with. It appears SammyS has seen it previously (see his answer to my question), and apparently ## \Pi(\nu) = 1 ## for ## -1<\nu<1 ## and is zero everywhere else. Well ## \Pi(\nu) ## is the rectangular pulse, defined in my notes as$$A\Pi \left( \frac{t}{T} \right) = \left\{\begin{matrix}A, \ |t|<T/2\\ 0, \ else\end{matrix}\right.$$Its FT is ##AT sinc(\nu T),## and by duality we have the pair:$$sinc(\nu T) \leftrightarrow \Pi \left( \frac{-\nu}{T} \right).$$So how can we combine the two ##\Pi## functions in my post #9 to return to time domain? roam said: Well ## \Pi(\nu) ## is the rectangular pulse, defined in my notes as$$A\Pi \left( \frac{t}{T} \right) = \left\{\begin{matrix}A, \ |t|<T/2\\ 0, \ else\end{matrix}\right.$$Its FT is ##AT sinc(\nu T),## and by duality we have the pair:$$sinc(\nu T) \leftrightarrow \Pi \left( \frac{-\nu}{T} \right).$$So how can we combine the two ##\Pi## functions in my post #9 to return to time domain? You need to do an inverse F.T. = Remember, we took the F.T. of the convolution integral to then process the functions, so we need to take an inverse F.T . of our result. The resulting ## \Pi ## function (which is the product of the two ## \Pi ## functions and is simply the narrower of the two ## \Pi ## functions) has a sinc function as its inverse transform=a very simple result. The only complication here is that you are using ## \nu ## in your F.T.'s and much of the literature and googled results use ## \omega=2 \pi \nu ##. In addition, your definition of the ## \Pi ## function may introduce a factor of 2 in its argument. (For the F.T. of sinc(bt), instead of just ## \Pi(\nu/b) ## there may be a ## \pi ## or ## 2 \pi ## in the ## \nu/b ##. Note the "b" in sinc(bt) is like an ## \omega_o ##, so there's likely to be a ## \pi ## or ## 2\pi ## multiplying the ##\nu/b ##.) Last edited: That is right. We want to process the following equation in Fourier domain before taking its inverse transform:$$\frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right) \tag{1}$$The result would be something of the form ##\frac{1}{|a.b|} \Pi \left( ... \right).## My question is how do you exactly work out the argument of the resulting rectangular pulse? roam said: That is right. We want to process the following equation in Fourier domain before taking its inverse transform:$$\frac{1}{|a|} \Pi \left( \frac{-\nu}{a} \right) . \frac{1}{|b|} \Pi \left( \frac{-\nu}{b} \right) \tag{1}

The result would be something of the form ##\frac{1}{|a.b|} \Pi \left( ... \right).##

My question is how do you exactly work out the argument of the resulting rectangular pulse?
With a graph. Graph each ## \Pi ## function (by hand), you'll see the result. The one problem here is I question the ## \nu/a ## and ## \nu/b ##. I think there should be a ## \pi ## or ## 2 \pi ## in there. Incidentally, the derivation of computing the F.T. of the sync function is somewhat lengthy, even though the result is simple. (I googled the derivation earlier. The calculation was done using ## \omega ## for the F.T. and the result was a positive rectangle F.T. for ## -b<\omega<b ## with an additional scaling factor multiplying the rectangle function and the F.T. was zero outside this rectangle interval.) You really need a precise answer for the F.T. of the sinc(bt) using the ## \nu ## frequency notation. We know it's going to be some kind of ## \Pi ## function, but we need both the scaling factor and the precise argument of the function.

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roam
Please see my edited additions to post #14. You're very much on the right track, but the exact form of the F.T. of sinc(bt) needs to be worked out precisely.

I tried to graph each ##\Pi##. The factors of 1/a and 1/b in the argument of each ##\Pi## tell us that they are stretched by a and b respectively. And the heights are also altered by the factor in front of each function. If we superimpose the two pulses it would look something like this (compared to the original unscaled ##\Pi##):

So, how exactly does this help me find the expression for the multiplication in equation (1)?

roam said:
I tried to graph each ##\Pi##. The factors of 1/a and 1/b in the argument of each ##\Pi## tell us that they are stretched by a and b respectively. And the heights are also altered by the factor in front of each function. If we superimpose the two pulses it would look something like this (compared to the original unscaled ##\Pi##):

So, how exactly does this help me find the expression for the multiplication in equation (1)?
That part is simple. To make it easier, have both rectangles be height 1. In any case, how much is "0" times "0" ? , how much is "0" times "1"? How much is "1" times "1"? You are missing the obvious. The shape of the product of the two functions is the narrower of the two functions.

By narrower of the two functions do you mean ##a-b## (i.e. ##\frac{1}{|a.b|} \Pi \left( \frac{-\nu}{a-b} \right)##)?

No. Simply ## (1/|ab|)\Pi(-\nu/b) ## for b<a.

TracerBullit and roam
Now from this we need to take the inverse transform. The best I can determine from the googled literature is the exact scaling factor that is picked up from each sinc function that got Fourier transformed into the rectangle is ## \pi/a ## and ## \pi/b ##. (Remember the F.T. of a convolution ## h(\omega)=f(\omega)g(\omega) ##). When doing the inverse F.T., (of ## h(\omega) ## ),the scaling factors of the one function will be used in the inverse F.T. and get canceled and we'll get H(t)=sinc(bt) multiplied by the scaling factors from the other function which is ## \pi/a ##. (Be sure and see post #19 also).(In taking the product of ## f(\omega) ## and ## g(\omega) ## , the ## f(\omega) ## basically disappeared, but it still has its scaling factors.)

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TracerBullit and roam

## 1. What is the definition of Convolution of two Sinc functions?

The Convolution of two Sinc functions is a mathematical operation that combines two functions to create a new function. It can be thought of as a way to measure the overlap between two functions and is often used in signal processing and image filtering.

## 2. How is the Convolution of two Sinc functions calculated?

The Convolution of two Sinc functions is calculated by taking one function and flipping it horizontally, shifting it across the other function, multiplying the values at each point, and then summing the results.

## 3. What is the significance of Convolution of two Sinc functions in signal processing?

In signal processing, Convolution of two Sinc functions is used to filter signals by removing certain frequencies. It is also used to convolve a signal with a system's impulse response to determine the system's output.

## 4. Can the Convolution of two Sinc functions be calculated analytically?

Yes, the Convolution of two Sinc functions can be calculated analytically using the Fourier transform of the two functions. This method is often used to simplify the calculation process.

## 5. What are some real-world applications of Convolution of two Sinc functions?

The Convolution of two Sinc functions has various applications in signal processing, image processing, and audio processing. It is used in edge detection, blurring, noise reduction, and other image filtering techniques. It is also used in audio equalization and system identification.