# How to calculate the FFT period

I have a question about finding the FFT duration for 802.11a preamble.

According to the standards, when the bandwith is $20 \textrm{MHz}$ and for $N=64$ we have $\Delta f = 20\textrm{MHz}/64 =312.5 \textrm{kHz}$ and it says FFT duration should be $1/\Delta f=1/(312.5 \textrm{kHz}) = 3.2 \textrm{us}$.

My question is that why FFT duration should be "1 over subcarrier spacing"?

• Why negative vote? it does not make sense anyone can downvote the post. – user59419 Jul 3 '17 at 4:38
• it does make sense, that's how we, as a community, assess the quality of questions, with respect to things like diligence and sufficient own research. – Marcus Müller Jul 3 '17 at 18:42
• No my friend . What I asked is really confusing to me I can find the answer by myself but I can never be sure about that unless somebody with better expertise can explain it. Now can you explain why it makes sense ? – user59419 Jul 3 '17 at 19:14
• At least tell me why downvote ? So anyone with certain reputation can click on downvote ? – user59419 Jul 3 '17 at 19:15
• @user59419 yes anyone with certain reputation can downvote. It is democratic. However, it would be nice if they left comments to improve your question. – AlexTP Jul 4 '17 at 8:44

OFDM can be thought as FDM with sinc pulse whose delay-$T$-shifted versions form an orthonormal basis. In frequency domain, they are seperated by $1/T$ which is denoted $\Delta f$ , i.e. subcarrier spacing. You fix a $T$, then $\Delta f = 1/T$; you fix $\Delta f$, then $T = 1/\Delta f$.
Look in frequency domain, because data is modulated in seperated subcarriers, the final signal is the sum of these subcarriers. Because the Fourier transform preserves the sum operation, in time domain, the final signal is also the sum of inverse-Fourier-transformed subcarrier signals that have the same duration $T$. Thus the final signal, which is sum of several signals having the same duration $T$, has the same duration $T$ that is called OFDM symbol duration or FFT duration or FFT period.