1
$\begingroup$

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:

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

$\endgroup$
3
$\begingroup$

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...)

$\endgroup$
2
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ I don't think the Geometric series can directly derive the LHS. The magnitude of common ratio here is one but it should be less than one for the series to converge. $\endgroup$ – msm Nov 15 '16 at 23:33
  • 2
    $\begingroup$ It's not a geometric series, it's a geometric sum (finite number of addends). Hence, there are no convergence problems. $\endgroup$ – Maximilian Matthé Nov 16 '16 at 8:39
  • $\begingroup$ See my answer for the details concerning the geometric sum interpretation. $\endgroup$ – Matt L. Nov 16 '16 at 8:52
  • $\begingroup$ Oh OK it makes sense (the number of terms is finite in the sum). Thanks for clarification. $\endgroup$ – msm Nov 16 '16 at 9:35
  • $\begingroup$ the fact that it is a finite sum is not sufficient alone to warrant convergence of the series, as it also depends on the summand being finite throughout the range of summation, which is the case in this example. On the other hand I find the DFT approach of explaning the equality a little redundant albeit being perfectly valid... $\endgroup$ – Fat32 Nov 16 '16 at 11:24
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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)

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.