# Double integral of product of Diracs

1. Aug 16, 2010

### juriguen

Hi there!

I am having a bit of a trouble when I try to work out a demonstration involving Dirac delta functions. I know, they are not real functions, and all that, but it only makes my life more difficult :)

Lets begin by the beginning to see if anyone can help. The first equation I will write I think comes straight from the definition of the Dirac distribution:
$$\int_{-\infty}^{\infty} f(t) \delta(t-nT) \mathrm{d}t = f(nT)$$

Ok, so far so good. But now I want to evaluate a more complicated expression:
$$\sum_{n=0}^{N-1} f(nT) g(nT)$$

I guess, there should be no problem to rewrite each sampled function by means of the integral involving the Dirac distribution, so that the equation becomes:
$$\sum_{n=0}^{N-1} f(nT) g(nT) = \sum_{n=0}^{N-1} \int_{-\infty}^{\infty} f(t) \delta(t-nT) \mathrm{d}t \int_{-\infty}^{\infty} g(\tau) \delta(\tau-nT) \mathrm{d}\tau$$

But now it comes when I don't know how to continue. For me the following demonstration would be just right, but the result is quite surprising:
$$\sum_{n=0}^{N-1} f(nT) g(nT) = \int_{-\infty}^{\infty} f(t) \int_{-\infty}^{\infty} g(\tau) \sum_{n=0}^{N-1} \delta(t-nT) \delta(\tau-nT) \mathrm{d}t \mathrm{d}\tau = \int_{-\infty}^{\infty} f(t) \int_{-\infty}^{\infty} g(\tau) \sum_{n=0}^{N-1} \delta(t-\tau) \mathrm{d}t \mathrm{d}\tau = N \int_{-\infty}^{\infty} f(t) g(t) \mathrm{d}t$$

There's almost definitely something wrong there, since I believe the above result should only hold in the limit when N tends to infinity. I say this because the equality basically resembles the interpretation of the integral as a Riemann sum.

Any help would be really appreciated!

Jose

2. Aug 17, 2010

### ross_tang

The key step that's incorrect should be you replace:

$$F_1=\delta (t-n T)\delta (\tau -n T)$$

with

$$F_2=\delta (t-\tau )$$

Clearly, $$F_1$$ is non-zero when both $$t = n T$$ and $$\tau = n T$$. Though it is true that $$t = \tau$$, but you missed $$t = n T$$.

There is a second problem as well. A product of delta functions won't be equal to a single delta function.

3. Aug 17, 2010

### juriguen

Thanks for the answer ross_tang. I suspected that was the wrong step. However, how would the demonstration continue otherwise?

Ok, I found my answer. Anyone who may want to continue posting is welcome, since the problem is interesting on its own (at least for me). But the above demo is an intermediate step for another proof which, in fact, is not needed. The one I did have to prove was the following:

$$\sum_{n=0}^{N-1} f(nT) g^{*}(nT) = \sum_{n=0}^{N-1} \left( \sum_{k} \hat{f}(k) \mathrm{e}^{j2\pi kn\frac{T}{\tau}} \right) \left( \sum_{k'} \hat{g}^{*}(k') \mathrm{e}^{-j2\pi k'n\frac{T}{\tau}} \right) = \sum_{k} \hat{f}(k) \sum_{k} \hat{g}^{*}(k') N\delta_{k,k'} = N \sum_{k} \hat{f}(k)\hat{g}^{*}(k)$$

where the first step holds only for periodic functions (which is my case), and the last is true when tau=NT, which is my case too!

Last edited: Aug 17, 2010