# Frequency Resolution Problem

Im trying to tackle the following problem while still not having a firm idea on what "frequency resolution" means :

Suppose we sample a continuous time signal with sampling period Ts = 1/2000, and then use a window of length 1000 on the resulting discrete time signal. If we transform it using a 2000 point DFT what would its frequency resolution be ?

Can anyone help me figure this out ?

• Do you want potential plot resolution with interpolation, peak location estimation resolution given an S/N, result bin separation, or peak separation resolution with a separation criteria? All these produce different frequency resolutions for the same length DFT. – hotpaw2 Feb 1 at 21:33

Depends a bit on what you are trying to achieve.

If you do an FFT of length $$N$$ of a signal sampled at sampled at a rate of $$F_s$$, then many people would say your frequency resolution is $$\frac{F_s}{N}$$. Whether that's correct or not, really depends on how exactly you define frequency resolution and what you are planning to do with it.

What's really happening is that you sample a frequency domain function with a sampling interval of $$\frac{F_s}{N}$$. As soon as you pick an FFT size, you are sampling in both domains with the sampling intervals being $$\frac{1}{F_s}$$ in time and $$\frac{F_s}{N}$$ in frequency.

Frequency domain sampling has all the same properties, requirements and problems as time domain sampling, you can get aliasing, you can interpolate, there is assumed periodicity in the other domain, etc.

By simply applying the sampling theorem we could argue that the frequency resolution required to fully characterize a signal is simply the inverse of the length in the time domain. This works well for signals that are inherently time bound, such as the impulse response of an LTI system.

However it's not practical for long continuous signals. In this case you need to pick a frequency resolution that's "good enough" for your application, and that really depends on the requirements and goal of your specific application.

The Sampling is given by $${T}_{s} = \frac{1}{2000}$$ [Sec].
The Window length is 1000 Samples.
Since the Window length must be equal to the data length we infer the data length is 1000 samples which means the sampling time is $$0.5$$ [Sec].

The Bin resolution in DFT is the ration between the sampling interval to the number of DFT Samples, which in this case is 2000. Hence the bin resolution is $$\frac{1}{4000}$$ [Hz].