# DFT Signal DFT Length N , FFT

If We sample an Signal let say sine(2 * pi * f) with f=1Hz and a sampling Frequecy of Fs = 8Hz, is it right that the length of the data schoul be N = Fs/f or multiple of Fs/f like N= d*(Fs/f) with d=1,2,3,.., is that right ? i read often, that it is better to increase N, my question : we can only increase N with Nnew= d * Nold, right ? and not like Nnew= 1,2 * Nold?

for my example , with f=1HZ, Fs= 8Hz, T=1s, Ts=0,125s, N=8

we can increase to f=1HZ, Fs= 8Hz, T=2s, Ts=0,125s, Nnew=16,

but it would not be right to increase ist to f=1HZ, Fs= 8Hz, T=1,5s, Ts=0,125s, Nnew=12, wich is not right because x[n]=x[n+N] doesnt fit, right ?

...... if it is right above, then where do i know how to get the right lenght N if i down know my signal ? do i have to Look the x[n] Series of the Signal one by one to see where N (full Ciycle) of the signal ist ?

Thanks

• Does the provided answer below gives you what you want? – Gilles Jan 6 at 9:35

## 1 Answer

For a time domain signal the length or the number of samples in the sampled signal can be computed as shown in Equation $$(1)$$ $$T = NT_s \equiv T = \frac NF_s \implies N = TF_s\tag{1}$$ Where $$T$$ is the signal total duration in [seconds] and $$F_s$$ is the sampling frequency in [Hertz], and $$N$$ the total number of samples.

The length of the DFT depends on the number of point taken in computing the DFT. For a $$N$$-point DFT on a signal of $$N$$ total samples you have $$N$$ as the length of your DFT.