Hello to all,
Let's consider that a signal is sampled at every dt, creating a discrete signal of N samples. The frequency step in the fft should be 1/dt/N. But I expect that by increasing the number of samples and consequently decreasing the dt, the frequency step will decrease, leading to a more accurate FFT. But I understand that the only way to decrease the frequency step df is by increasing the overall time of the signal (length of the window). Where do I do wrong with this?
Hi,
df = 1/(dt*(N-1)) = 1/T where T is the total duration of your signal and N the number of sampling points.
I think that 1/dt/N is ambiguous and leads you to this mistake.
As you said, increase the signal duration is the only way to decrease df.
hope it helps,
Bustillo julien
I think the first thing you need to think over is what does "more accurate FFT" mean to you, and what you want to achieve by running the FFT?
Ultimately it depends on the application how wide frequency buckets you want, and that will determine the number of samples and the sampling frequency. Usually if its a real signal, sampling frequency is determined by your hardware, obviously if you generate it with a code, you can do pretty much what you want.
Check this topic, it explains things nicely:
https://electronics.stackexchange.com/questions/12407/what-is-the-relation-between-fft-length-and-frequency-resolution
Hope this helps!
Best regards,
David
Dear Dimitrios,
welcome,
Julien is right and i second him,
If the sampling frequency is fs and the number of samples is N, then the lowest frequency resolved will be fs/N. Assuming the sampling time Ts =dt, then
The frequency resolution= 1/ dt N. Ndt is the overall time window T. So
the frequency resolution= 1/ T, so there is no contradiction. If you want to increase the resolution by decreasing the minimum resolved frequency you must increase the the time window length T. In creasing fs keeping the window constant will increase the highest observed frequency fmax=fs.
Best wishes
Thank you all for the answers. What happened if I can't increase the window? I know that I can increase the sampling frequency but let's say double the sampling frequency, the length of the window will be doubled and again, df will remain the same. What is happening in such cases?
Dear Dimitrious,
welcome,
I answered your new question in my first post. If you increase fs keeping the width of the time window T the same, then yo can observe the higher frequency components in the waveform.The frequency resolution will remain the same as
fmin= 1/T
Best wishes
What happened if I can't increase the window?
In this case it depends on your signal...
If your signal doesnt last a lot, you can use zero padding (take attention to avoid the step at the beginning of the zero padding). This method doesn't increase the accuracy by itself but is equivalent to upsample your FFT.
If you want to work on complex signals, such as ulstrasonic codas, it becomes more tricky...