About Fourier transform of periodic signal

Question

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 $\delta(\omega)$

Answer 1

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.

Answer 2

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.

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2. 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.

Answer 3

I have found this proof, i am not sure it is correct or not. So correct if I am wrong.

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 $$X(\omega)=\alpha \delta(\omega)$$ $$ x(t) = \frac {1}{2\pi} \int _{-\infty}^{+\infty} \alpha \delta(\omega) e^{j \omega t}$$ $$ 1= \frac {\alpha}{2\pi} \int _{-\infty}^{+\infty} \delta(\omega) e^{j \omega t} $$ $$ 1= \frac {\alpha}{2\pi} \int _{-\infty}^{+\infty} \delta(0) e^{0} $$ $$ 1= \frac {\alpha}{2\pi} . 1 $$ $$ \alpha=2\pi $$

Answer 4

• 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.