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In Fourier transform for periodic signal, I checked different books and I found a different explanation in each book. Let's take the explanation in Signals and Systems by Rajeshwari & Rao:

The resulting Fourier transform for a periodic signal consist of a train of impulses in frequency, with areas of impulses proportional to the Fourier series coefficients.

To suggest the general result, let us consider $x(t)$ with Fourier transform $X(\omega)$ which is a single impulse of area $2\pi$ at $\omega=\omega_0$, that is, $$X(\omega)=2\pi\delta(\omega-\omega_0)$$ To determine $x(t)$ for which this is Fourier transform we can apply the inverse Fourier transform to obtain $$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} 2\pi \delta(\omega-\omega_0)e^{j\omega t}d\omega$$

The things I want to ask is:

  1. If we have Fourier series of a periodic signal which will be one impulse, then the Fourier transform of that impulse will be the same single impulse? That's what it is explained above?
  2. Why we used shifted impulse? Why we can't take $\delta(\omega)$
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    $\begingroup$ Personally, whenever impulses are involved, I just choose to trust the mathematicians and use the results blindly (and I try to use them as little as possible). Impulses are not even functions, and there is so much hand-waving in how textbooks use them, it's not even funny. $\endgroup$ – MBaz Nov 5 '15 at 13:54
  • $\begingroup$ @MBaz " impulse are not even function" you mean delta impulse is not even function? do you have any idea about the answer i wrote, is it correct? $\endgroup$ – Aadnan Farooq A Nov 5 '15 at 14:01
  • $\begingroup$ Yes, the dirac delta is not a function, it's a "distribution". And I didn't check in detail but I believe your answer is "engineering textbook correct", which is what most of us here need. $\endgroup$ – MBaz Nov 5 '15 at 14:23
  • $\begingroup$ Impulses are not proper functions but they are defined with the property $\int_{-\infty}^\infty \delta(x) f(x) dx = f(0)$ and with that the math isn't that hard. $\endgroup$ – clay Nov 5 '15 at 15:48
  • $\begingroup$ Take a look at this answer over at math.SE showing the relation between the Fourier series and the Fourier transform of a periodic function. $\endgroup$ – Matt L. Nov 6 '15 at 10:37
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We can figure out what's going on if we first understand a simple identity and then just compute the Fourier transform of the periodic function.

A useful identity

First let's prove that $$D(\omega - \omega') \equiv \int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} = 2\pi \, \delta(\omega - \omega')\,. $$ We just use a test function $\tilde{g}(\omega)$: \begin{align} \int_{-\infty}^\infty \frac{d\omega}{2\pi} \underbrace{ \left(\int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} \right)}_{D(\omega - \omega')} \tilde{g}(\omega) &= \int_{-\infty}^\infty dt \, e^{-i \omega' t}\left( \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{g}(\omega) e^{i \omega t}\right) \\ &= \int_{-\infty}^\infty dt \, g(t) e^{-i \omega' t} \\ &\equiv \tilde{g}(\omega') \, . \\ \end{align} This shows that by definition $D(\omega-\omega') = 2\pi \, \delta(\omega - \omega')$. We didn't have to say anything weird about distributions or use any magic.

Fourier transform of periodic function

Now consider a periodic function $x(t)$ with period $T$. Since $x$ is periodic we can write it as a Fourier series $$x(t) = \sum_{n=-\infty}^\infty \hat{x}_n e^{i 2\pi n t/T} \, .$$

Now let's compute the Fourier transform, \begin{align} \tilde{x}(\omega) &\equiv \int_{-\infty}^\infty x(t)e^{-i\omega t} \, dt\\ \text{plug in Fourier series} \qquad &= \sum_{n=-\infty}^\infty \hat{x}_n \underbrace{\int_{-\infty}^\infty dt \, e^{i(2\pi n/T -\omega) t}}_{D(2\pi n / T - \omega)} \\ &= \sum_{n=-\infty}^\infty 2\pi \, \hat{x}_n \delta(\omega - 2\pi n / T) \, . \end{align} So there we go: the Fourier transform of a periodic function $x(t)$ is a sum of delta functions appearing at $x(t)$'s Fourier series frequencies (i.e. the frequencies $2 \pi n / T$). The weight of each delta function is $2\pi$ times the Fourier series coefficient $\hat{x}_n$. The factor of $2\pi$ is there simply because we used the Fourier transform convention with angular frequency $\omega$ but the Fourier series convention with regular frequency.

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  • $\begingroup$ What is $D$ here? Is it defined as the integral next to it? I would expect so, but I've never seen $\equiv$ used as a shorthand for definition. $\endgroup$ – John P Nov 13 at 21:43
  • $\begingroup$ @JohnP Yes, I'm using $\equiv$ as a definition. Perhaps that's incorrect use of the symbol. $\endgroup$ – DanielSank Nov 13 at 23:42
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1.The FT of a periodic signal is not one, but (potentially) infinite impulses.

Assume an arbitrary periodic function $f_T(t)$ with period $T$ and consider its Fourier series representation in which $\omega_0=\frac{2\pi}{T}$: $$f_T(t)=\sum_{n=-\infty}^{+\infty}c_n e^{j n \omega_0 t}$$ Take the Fourier transform of the sides: \begin{align} \mathcal{F}\{f_T(t)\}=&\mathcal{F}\{\sum_{n=-\infty}^{+\infty}c_n e^{j n \omega_0 t}\}\\ =&c_n\sum_{n=-\infty}^{+\infty}\mathcal{F}\{ e^{j n \omega_0 t}\}\\ =&2\pi c_n\sum_{n=-\infty}^{+\infty}\delta(\omega-n\omega_0) \end{align} This means that the Fourier transform of a periodic signal is an impulse train where the impulse amplitudes are $2\pi$ times the Fourier coefficients of that signal.


  1. We need shifted impulses since they correspond to different frequency components in the Fourier series representation. More clearly, complex exponentials in the frequency domain with different coefficients of $\omega_0$ in their exponent translate to different shifted versions of impulse.
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  • $\begingroup$ For whatever it's worth, the proof of the last line of this answer is given in the first part of my answer. $\endgroup$ – DanielSank Sep 2 '16 at 9:34
  • $\begingroup$ I think it would be good to explicitly write the Fourier transform convention used in this answer. In other words, define $\mathcal{F}$. $\endgroup$ – DanielSank Sep 2 '16 at 9:42
  • $\begingroup$ There is no single definition of Fourier transform. We can use $\mathcal{F}\{f\}(\omega) = (1/2\pi) \int f(t) \exp(-i \omega t) \, dt$, or $\int f(t) \exp(-i 2\pi \nu t) \, dt$, or $(1/\sqrt{2\pi}) \int f(t) \exp(-i \omega t) \, dt$... We can switch the sign of $i$. There are a rather large number of conventions used in practice. By not being specific we are contributing to the pile of confusion people constantly go through when learning about Fourier analysis. $\endgroup$ – DanielSank Sep 2 '16 at 23:36
  • $\begingroup$ The $2\pi$ in the final equation of this answer depends crucially on the fact that the Fourier series and Fourier transform conventions being used are different in that one uses angular frequency while the other does not. Some readers (i.e. you and I) can just figure that out because we're experienced. However, when writing technical documentation which will be read by a wide variety of people, I think relying on that kind of experience is a grave error. Have a good day :) $\endgroup$ – DanielSank Sep 3 '16 at 0:50
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  • The Fourier transform of the time domain impulse $\delta(t)$ is constant $1$, not another impulse. Analogously, the Fourier series coefficient of a periodic impulse train is a constant.
  • Fourier transform applies to finite (non-periodic) signals. Fourier series representations with coefficients apply to infinitely periodic signals. In your first question you are mixing these up.
  • The book is using a shifted impulse to demonstrate the frequency shift property.
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  • $\begingroup$ what will be case if we take the Fourier transform of 1? it will $\delta(\omega)$ but how we can get the $2\pi$ factor with that? $\endgroup$ – Aadnan Farooq A Nov 5 '15 at 7:37
  • $\begingroup$ The Fourier transform of 1 should be $2\pi \delta(\omega)$. Just substitute $0$ in for $\omega_0$ in the textbook example to see that. $\endgroup$ – clay Nov 5 '15 at 7:51
  • $\begingroup$ Yes it is correct $2\pi \delta(\omega)$, but in the text book it was just written "single impulse of area $2\pi$" I am confuse to understand that.. is there any mathematical proof? $\endgroup$ – Aadnan Farooq A Nov 5 '15 at 7:53
  • $\begingroup$ $2\pi \delta(\omega)$ is an impulse of area $2\pi$. $\delta(\omega)$ is defined to have an area of $1$. You just showed the proof for $e^{j\omega_0 t} \Leftrightarrow 2\pi \delta(\omega - \omega_0)$. Substitute $\omega_0=0$ and you have the proof of $1 \Leftrightarrow 2\pi \delta(\omega)$ $\endgroup$ – clay Nov 5 '15 at 8:18
  • $\begingroup$ Yes even for that how the $2\pi$ factor comes with $2\pi \delta(\omega - \omega_0)$?? $\endgroup$ – Aadnan Farooq A Nov 5 '15 at 9:22
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I have found this proof, i am not sure it is correct or not. So correct if I am wrong.

enter image description here Proof:
$$x(t) = \delta(t)$$ $$ X(\omega) = \int _{-\infty}^{+\infty}\delta(t) e^{-j \omega t}dt$$ $$ X(\omega) = \int _{-\infty}^{+\infty}\delta(0) e^{0} dt=1$$

What If $x(t) = 1$ then $X(\omega) = \delta(\omega)$. Since all the energy is concentrated at 0. Validate by inverse fourier transform.
lets take $$X(\omega)=\alpha \delta(\omega)$$ $$ x(t) = \frac {1}{2\pi} \int _{-\infty}^{+\infty} \alpha \delta(\omega) e^{j \omega t}$$ $$ 1= \frac {\alpha}{2\pi} \int _{-\infty}^{+\infty} \delta(\omega) e^{j \omega t} $$ $$ 1= \frac {\alpha}{2\pi} \int _{-\infty}^{+\infty} \delta(0) e^{0} $$ $$ 1= \frac {\alpha}{2\pi} . 1 $$ $$ \alpha=2\pi $$ enter image description here

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  • $\begingroup$ This looks completely correct. $\endgroup$ – clay Nov 5 '15 at 15:45
  • $\begingroup$ There is a minor notation inconsistency, but apart from that this is correct. The inconsistency: Writing $\delta(0)$ in the integrals is not ok. You have $\delta(t)$ (or some other variable instead of $t$) there, and by the definition of the $\delta$-distribution the integration process over the term with the $\delta$ "picks" the value associated with $\delta(0)$. $\endgroup$ – M529 Jun 4 '16 at 8:49

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