Saturday, 15 October 2016

proof - Equation for impulse train as sum of complex exponentials


Could someone please break down what's going on in this equation for me? I understand what the left side looks like, but not so much how the right side is the same thing.


Impulse train:



$$\sum_{m=-\infty}^\infty \delta[n-Nm]=\frac 1N\sum_{k=k_0}^{k_0+N-1}e^{ \ j 2\pi kn/N} \\ \\ n, m, N,k_0 \in \mathbb{Z}$$


where $\delta[n]$ is the Kronecker delta:


$$ \delta[n] = \begin{cases} 1 \quad n = 0 \\ 0 \quad n \ne 0 \\ \end{cases} \\ \\ n \in \mathbb{Z}$$



Answer



The summation on the left side of your equation represents a time domain discrete time periodic signal $x[n]$ whose period is N. $$ x[n] = \sum_{k=-\infty}^{\infty} \delta[n-kN] $$


And the summation on the right side of your equation is just the same (periodic) signal $x[n]$ represented via its Discrete Fourier Series DFS synthesis form which is an alternative representation for periodic signals $$ x[n] = \sum_{k=0}^{N-1} a_k e^{j \frac{2 \pi}{N} k n} $$


where the DFS coefficients $a_k$ are all $1/N$ found via DFS analysis equation: $$ a_k = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j \frac{2 \pi}{N} n k} $$


I hope you can see why all $a_k$ are $1/N$ (it can be easily seen from the properties of impulse inside the summation...)


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