# Fft time axis for a signal that is a function of frequency

1. May 7, 2014

### wattacatta

Hello!
I am attempting to do an FFT on a signal e^(jω+nδω)t, where t is constant and for n=0:N-1. This signal starts off at ω when n =0 and then increments in steps of δω until it reaches the final frequency ω+(N-1)δω.

Performing an FFT on this signal will put me in the time domain so my x-axis is time. I have been told that the waveform will have a total duration of T = 1/Δω and Δt will be 1/(N*Δω). I asked my instructor why that is the case and he told me that it is all the same as if I were working in the time domain, the FFT works the same whether I am working with time, frequency, or apples. However, I cannot see this. I have been searching online to try to find an explanation but all I can find is on going from time to frequency, not from frequency back to time.
The closest thing I found on physics forums:

however, I can't post an entry to that thread. Also that thread confirms that T = 1/Δf and Δt = T/N = 1/(N*Δf), but doesn't really provide an explanation.

The way I see it, we get different results depending on what domain we are in:
When doing an FFT of a sampled time signal, you have a sampling frequency (fs) from which you can calculate the Sample Period(Ts) by just taking the inverse. We also know that in the frequency domain we will have frequency components in the range -fs/2 to fs/2.
But if you have a signal that we are essentially sampling in the frequency domain for a particular instant of time, "fs" is now possibly our Δω, so to get Ts, we can do 1/fs. But, this does not calculate Ts, rather it is Ttotal_observation_time (which in the time domain would be calculated as Ttot = (N-1)*Δt). I would jump for joy if the correct answer were that Δt = 1/Δf but it is not so, rather this is equal to the overall duration of the time domain signal.

So my confusion is that somehow, 1/fs yields the total observation time in the frequency domain, whereas in the time domain, it yields the sample period; Thus I think it does matter what domain I am in. Please someone help me see otherwise!If there is someone who can understand what I cannot see please provide me with some help that might make me understand. Thank you in advance!

2. May 7, 2014

### wattacatta

Furthermore, the range in the time domain is now 0 to 1/Δf, which does not parallell the -fs/2 to fs/2 range in frequency in any way.