This is probably basic but as I am new to the field it confuses me a bit. While looking at some solutions provided to a problem in the final step following happens:

$$\begin{aligned} \frac{1}{10}\sum_{l=-\infty}^{\infty}x[l]\sum_{k=0}^9 e^{-i2\pi(n-l)k\frac{1}{10}} &=\frac{1}{10}\sum_{l=-\infty}^{\infty} x[l]10\delta_{10}[n-l]\\ &=\sum_{s=-\infty}^{\infty}x[n-10s]\end{aligned}$$

Now, according to the DFT table we have provided for the course, the transformation in the 1st step is defined like this

$$e^{2i\pi(k-k_0)/N}\longmapsto N\delta[k-k_0]$$

The delta with subscript was never defined before appearing in this task so searching online I found this:

$$\delta_N[k]=\sum_{s=-\infty}^{\infty}\delta[k-sN]$$ My questions are: What exactly describes the $\delta_N[k]$?

In the 1st step regarding the delta, is one of the 2 sources wrong or could it be that in the transformation table it should be ''understood'' that the delta has to have the subscript depending on its period? Is the last step just the result of applying the properties of the delta(with subscript) function on the x signal?


1 Answer 1


You have to realize that all sequences in DFT relations are inherently periodic. This is usually understood, but it can be made explicit, for instance by that subscript. So both sources are correct, but one explicitly notes the periodicity of the sequence.

If you write down the DFT relation

$$\delta[n]=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi nk/N}\tag{1}$$

then you see that the expression on the right-hand side of $(1)$ is periodic in $n$ with period $N$. So $\delta[n]$ on the left-hand side is actually continued periodically. If we just consider one period, we don't care about that periodic continuation. However, if we are concerned about the behavior for all $n$, then we do need to take that periodicity into account, and we could have written $\delta_N[n]$ on the left-hand side of $(1)$ to be more explicit.

And in the given example we do care what happens outside that period because the index $l$ in the first sum extends from $-\infty$ to $\infty$.

The last step in the given equation is just a convolution of the sequence $x[n]$ with a unit impulse comb, so $x[n]$ is shifted by integer multiples of $10$ and all those shifted versions are summed up (hoping that the sum converges).


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