Equation for impulse train as sum of complex exponentials

Question

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 1

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). $$ x[n] = \sum_{k=0}^{N-1} a_k e^{j \frac{2 \pi}{N} k n} $$

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...)

Answer 2

It can be explained by two ideas:

1. Consider the geometric sum. There you directly get to it.
2. The right-hand side can be understood as the N-point DFT of $g[n]=1/N$. The DFT of a constant is an impulse. Since we are have the discrete Fourier Transform, the spectrum is considered periodic adn hence the Dirac on the left-hand side repeats every N points. Note that the complex exponentials on the right-hand side are invariant to $m$ in $(n+mK)$, hence for each m, the same right-hand side is calculated.

Answer 3

All the above answers are mathematically correct. Just to visualize the kind of shapes intermediate signals take, you can read my recent article on DSPrelated.

Answer 4

In this answer I will elaborate on the geometric sum interpretation mentioned in Maximilan Matthé's answer.

Consider the term on the right-hand side:

$$S=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi kn/N}\tag{1}$$

For $n=lN$ (with integer $l$) we get

$$S=\frac{1}{N}\sum_{k=0}^{N-1}1=1\tag{2}$$

For other values of $n$ we can use the formula of the geometric sum to get

$$S=\frac{1}{N}\frac{1-e^{j2\pi n}}{1-e^{j2\pi n/N}}=\frac{1}{N}\frac{1-1}{1-e^{j2\pi n/N}}=0\tag{3}$$

Note that the denominator is not zero when $n \ne lN$.

Consequently, we have

$$S=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi kn/N}=\begin{cases}1,&n=lN \quad\quad l \in \mathbb{Z}\\0,&\text{otherwise}\end{cases}\tag{4}$$

which can also be written in terms of the discrete-time unit impulse $\delta[n]$ as in the OP's equation:

$$S=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi kn/N}=\sum_{m=-\infty}^{\infty}\delta[n-mN]\tag{5}$$

Answer 5

Periodic impulse train can be expressed by discrete time fourier series, $$x[n]=\sum_{m=-\infty}^{\infty}\delta[n-mN]=\sum_{k=0}^{N-1}c_ke^{j2\pi kn/N}\tag{1}$$

where the Fourier coefficients $c_k$ are given by

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

For the given $x[n]$, we have $c_k=1/N$ and from (1)

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi kn/N}\tag{2}$$