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Let me have an impulse train function as below,

$$ x[n] = \sum_{m=-\infty}^{\infty} {\delta[n-f_0 m]} $$

where, $f_0 \in \textbf{Z}$.

Now, I am trying to calculate its DTFT, so I put it into DTFT equation as below,

$$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} {\sum_{m=-\infty}^{\infty} {\delta[n-f_0 m]} e^{-j\omega n}} . $$

I substitute $n-f_0 m = p$, the equation can be re-written as below,

$$ X(e^{j\omega}) = \sum_{m=-\infty}^{\infty} { \left(\sum_{p=-\infty}^{\infty} {\delta[p]} e^{-j\omega p} \right) e^{-j\omega f_0 m}}. $$

In the parenthesis equation goes to 1, $X(e^{j\omega})$ only has the term as below,

$$ X(e^{j\omega}) = \sum_{m=-\infty}^{\infty} {e^{-j\omega f_0 m}}. $$

I divided this equation as two parts.

$$ X(e^{j\omega}) = \sum_{m=0}^{\infty} {e^{-j\omega f_0 m}}+\sum_{m=0}^{\infty} {e^{j\omega f_0 m}}-1. $$

And I calculated this and got the answer which was equal to 0.

$$ X(e^{j\omega}) = \frac{1}{1-e^{-j\omega f_0}} + \frac{1}{1-e^{j\omega f_0}} -1 = 0.$$

At this moment, I was stuck at here, because I could not find where is incorrect.

Please give me

1) some comments for fixing this issue

2) How can I get the DTFT of impulse train? Is this approach is correct?

Thank you very much in advance!


@Matt L. The infinite two sums can be considered as infinite series each of them.

So let me consider left part of above equation i.e. $ \sum_{m=0}^{\infty} {e^{-j\omega f_0 m}} $.

Its common ratio is $e^{-j\omega f_0}$, and its first term is 1.

We can get above fraction, and we also can calculate the 2nd part as the same manner.


From your answer, I tried to calculate the infinite series as below.

If you don't mind, this procedure is correct or not, please let me know.

$$c_k = \frac{1}{N} \sum_{n=0}^{N-1} {\sum_{m=-\infty}^{\infty} {\delta[n-mN] e^{-j \frac{2\pi}{N} nk}} }.$$

Above equation can be written as below, and except the case $m=0$ all $m$'s condition are 0 because the definition of dirac delta sequence.

$$\sum_{n=0}^{N-1} {\sum_{m=-\infty}^{\infty} {\delta[n-mN] e^{-j \frac{2\pi}{N} nk}} } = \sum_{m=-\infty}^{\infty} {\left( \delta[0-mN] e^{-j \frac{2\pi}{N} 0k}+\delta[1-mN] e^{-j \frac{2\pi}{N} 1k} +...+ \delta[(N-1)-mN] e^{-j \frac{2\pi}{N} (N-1)k}\right)}.$$

In other words, $ \delta[1-mN]=\delta[2-mN]=...=\delta[(N-1)-mN]$ equals to 0 whatever $m$ is.

So, $c_k=1/N$.

I understood up to here, and from now on I should understand DTFT of $x[n]$, i.e. $X(e^{j\omega})$.

Thank you for your comment.


@Matt L. Thank you always for your comment. Now I am struggling with the equation from $x[n]$ to $X(e^{j\omega})$.

What I am doing is as below, I am stuck at the infinite sum.

Because, $x[n]$ is

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

the DTFT can be written as below,

$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} { \sum_{n=-\infty}^{\infty}{e^{-j \left( \frac{2\pi}{N}k - \omega \right)n} } }.$$

I think this can be converted as below,

$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} {2\pi \delta \left(\omega-\frac{2\pi}{N}k \right)}. $$

Since the DTFT of $e^{j2\pi kn/N}$ is $2\pi\delta(\omega-2\pi k /N)$ as you mentioned previously.

Is the DTFT means that as below? I think the definition is DTFT is as below. If it is, I have no idea why the DTFT of impulse train is $k$ from $-\infty$ to $\infty$.

$$\sum_{n=-\infty}^{\infty}{e^{-j \left( \frac{2\pi}{N}k - \omega \right)n} }.$$

So, the above $X(e^{j\omega})$ could be re-written as

$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} {DTFT(e^{j2\pi kn/N})}.$$

Please give me some comment for fixing this, and understanding the DTFT of impulse train.

Thank you always!

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  • $\begingroup$ Your $x[n]$ does not make sense. The argument of $\delta[n]$ must be an integer, which is not the case if $f_0$ can be any real number. $\endgroup$
    – Matt L.
    Feb 13, 2015 at 10:02
  • $\begingroup$ Another problem is the formula for the geometric series, which only works if the magnitude of the complex number in the sum is less than $1$. This is not the case in your example. $\endgroup$
    – Matt L.
    Feb 13, 2015 at 10:05
  • $\begingroup$ Thanks! I've changed R to Z, but I have no idea about second comment. Could you give me the comment more in detail? Thank you! $\endgroup$
    – actlee
    Feb 13, 2015 at 10:15
  • $\begingroup$ In your last equation you replaced the infinite sums by two fractions. How did you get these expressions? $\endgroup$
    – Matt L.
    Feb 13, 2015 at 10:18

1 Answer 1

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A simple approach to show that the DTFT of an impulse train is an impulse train in the frequency domain, is to represent the periodic impulse train by its 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)

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

Since the DTFT of $e^{j2\pi kn/N}$ is $2\pi\delta(\omega-2\pi k /N)$ (periodically continued with period $2\pi$), the DTFT of $x[n]$ is

$$X(e^{j\omega})=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta(\omega-2\pi k/N)$$

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  • $\begingroup$ I attached some equations and question. If you may, please check them. :) thank you! $\endgroup$
    – actlee
    Feb 16, 2015 at 7:59
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    $\begingroup$ @AlbertRee: It looks like your calculation of $c_k$ is correct. But what is not correct, as pointed out before in one of my comments, is the application of the formula for the infinite geometric series, because the formula is only valid if the common ratio has a magnitude of less than one, but $|e^{j\omega f_0}|=1$, so the series does not converge and the formula can't be applied. $\endgroup$
    – Matt L.
    Feb 16, 2015 at 17:21
  • $\begingroup$ Yeah, those equations are only valid under the absolute value of common ratio is smaller than 1. Thank you. Now I figured out why those equations did not make sense. $\endgroup$
    – actlee
    Feb 17, 2015 at 1:44
  • $\begingroup$ L: Thank you. I also attached some lines, if you have some time, please have a look and give me some comment. Always I appreciate you! Thank you so much. $\endgroup$
    – actlee
    Feb 17, 2015 at 4:42
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    $\begingroup$ @AlbertRee: I see your problem. The solution is given in the last sentence of my answer above. Note the 'periodically continued with period $2\pi$'. So you have a sum of N shifted and periodically continued Dirac impulses, which can be written as an infinite sum of 'normal' (i.e. non-periodic) Dirac impulses. $\endgroup$
    – Matt L.
    Feb 17, 2015 at 7:53

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