# Fourier transform periodic signal

I have to find the Fourier transform is this signal $$\sum_{n= - \infty }^{+ \infty } e^{- 2 \pi | \frac{t-nT_0}{T_0} |}$$

I know that the Fourier transform of $$\sum_{n= - \infty }^{+ \infty } x_0 (t-nT)$$ Is $$X_0(kf_0) f_0 \delta (f- kf_0)$$ . Now I have to find $$X_0(kf_0)$$ and $$X_0 (f)$$

I know that the Fourier transform of $$e^{-a|t|}$$ Is $$\frac{2a}{a^{2} + w^{2} }$$ So the Fourier transform Of $$e^{- 2 \pi | \frac{t-nT_0}{T_0} |}$$ , also written as $$e^{-2\pi |\frac{t}{T_0} - n| }$$ , should be $$\frac{ 4 T_0 \pi }{4 (\pi )^{2} + w^{2} }$$

Now I Find $$X(k f_0)$$ with $$f= kf_0$$ , and I obtain $$X(k f_0) = \frac {T_0}{\pi + \pi k^{2} f_0^{2} }$$

Finally , I obtained that $$X(f) = \frac {1}{\pi} \frac {1}{1 + k^{2} f_0 ^{2} } \delta(f- kf_0 )$$ But the result on my book is $$X(f) = \frac {1}{\pi} \frac {1}{1 + k^{2} } \delta(f- kf_0 )$$

• Could you define the symbols $k$, $f$, and $f_0$ by writing the Fourier transform formula you have supposed? If not, could you provide a link where this identity is written? Jun 18, 2020 at 11:41
• This formula is written on my book , “segnali analogici e sistemi lineari “ by A. Vannucci Jun 18, 2020 at 12:17
• Could you provide the relationship between $k$, $f$, $f_0$ and $\omega$, $T_0$, $n$? Jun 18, 2020 at 14:39
• $T_0$ is the period , $f_0 = \frac {1}{period} = \frac {1}{f_0}$ so $f_0 T_0 = 1$ this is what I always studied on my book. n , is written , it’s a generic index. f is the Fourier transform of t but if a signal is periodic I have to find $X_0 (k f_0)$ so I have $f = f_0 k$. Jun 18, 2020 at 14:57
• @ElenaMartini, notation used in a book is not necessarily used anywhere else, so one must always explain it. Jun 18, 2020 at 20:20

Prove that $$x(t)=\sum_{n=-\infty}^{+\infty}e^{-2\pi\left|\frac{t-nT_0}{T_0}\right|}\xrightarrow{\text{FT}}X(f)=\frac{1}{\pi}\left(\frac1{1+k^2}\right)\delta(f-kf_0)$$ , given that $$k=f/f_0$$, $$T_0f_0=1$$, where $$T_0>0$$, $$n\in \mathbb{Z}$$.

Solution:

Let Fourier transform of $$x(t)=e^{\frac{-2\pi}{|T_0|}|t-nT_0|}$$ be $$X(f)$$ and that of $$x^\prime(t)=e^{\frac{-2\pi}{|T_0|}|t|}$$ be $$X^\prime(f)$$.

Time-shifting property of Fourier transform \begin{align*} y(t\pm t_0)&\xrightarrow{\text{FT}} Y(f)e^{\pm j2\pi ft_0}\\ \text{if}\quad y(t)&\xrightarrow{\text{FT}}Y(f) \end{align*} So, $$e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} X^\prime(f)e^{-j2\pi fnT_0}\tag{1}$$ You know that $$e^{-a|t|}\xrightarrow{\text{FT}} \frac{2a}{a^2+(2\pi f)^2}$$ So, $$e^{\frac{-2\pi}{|T_0|}|t|}\xrightarrow{\text{FT}}\frac{\frac{4\pi}{|T_0|}}{\left(\frac{2\pi}{|T_0|}\right)^2+(2\pi f)^2}\equiv \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)\tag{2}$$ From equation $$(1)$$ and $$(2)$$, we get $$e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)e^{-j2\pi fnT_0}\tag{3}$$

On summating equation $$(3)$$ over all integers $$n$$, we get $$\sum_{n=-\infty}^{+\infty}e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac1{\pi |T_0|}\left(\frac1{\frac1{(T_0)^2}+f^2}\right)\sum_{n=-\infty}^{+\infty}e^{-j2\pi fnT_0}=\frac{f_0}{\pi}\left(\frac1{f_0^2+f^2}\right)\sum_{n=-\infty}^{+\infty}e^{+j2\pi fnT_0}\equiv \frac{1}{\pi f_0}\left(\frac1{1+\left(\frac{f}{f_0}\right)^2}\right)\sum_{n=-\infty}^{+\infty}e^{+j2\pi fnT_0}\quad\quad(\because n\in (-\infty,+\infty)\text{ and }T_0f_0=1) \tag{4}$$

We know that the Fourier series of an impulse train is $$\sum_{n=-\infty}^{+\infty}\delta(t-nT)=\frac1{T}\sum_{n=-\infty}^{+\infty}e^{j2\pi nt/T}\tag{5}$$ Finding an impulse train that could divert the RHS of equation $$(4)$$ towards the answer by replacing $$t\rightarrow f$$ and $$T\rightarrow f_0$$: $$\sum_{n=-\infty}^{+\infty}\delta(f-nf_0)=\frac1{f_0}\sum_{n=-\infty}^{+\infty}e^{j2\pi nf/f_0}\tag{6}$$ Replacing $$k\rightarrow f/f_0$$ and substituting equation $$(6)$$ in equation $$(4)$$, we get $$\sum_{n=-\infty}^{+\infty}e^{\frac{-2\pi}{|T_0|}|t-nT_0|}\xrightarrow{\text{FT}} \frac{1}{\pi}\left(\frac1{1+\left(\frac{f}{f_0}\right)^2}\right)\sum_{n=-\infty}^{+\infty}\delta(f-nf_0)=\frac{1}{\pi}\left(\frac1{1+k^2}\right)\delta(f-kf_0)\quad\quad(\because \delta(f-nf_0)=0\ \forall\ n \neq k)$$ (Hence proved :))

• Thank you so much ! Unfortunately I’m Italian and my English is not perfect so I’m in trouble when I have to explain the notation on my book. And I’ve always studied only on it so This is the only notation I know and sometimes I forgot to write things ( as f = 1\T ) I take for granted Jun 19, 2020 at 13:41