# Does using the Goertzel algorithm actually give better frequency resolution?

I am reading this article, and I am getting a little confused by the author's liberal use of 'frequency resolution' regarding the Goertzel Algorithm.

Basic question: Does using the Goertzel algorithm actually give you more frequency resolution over a specific band of interest, or does it simply efficiently compute the FFT over only the specified band of interest, but at the same frequency resolution specified by sampling frequency divided by number of samples?

For example, lets say the $$F_s$$ is 100 KHz, (fixed) and the number of data samples $$N$$ is 10000. (Also fixed). If I compute a normal FFT, where FFT length is also $$N$$, my frequency resolution is $$\frac{F_s}{N}$$ as to be expected, and it will be equal to 10 Hz. This means my bins are separated by 10 Hz, from -50,000 Hz to 50,000 Hz.

Now let us say I want to use the Geortzel Algorithm to only look at frequencies in the range of say, 20,000-21,000 Hz. If I use the same $$N$$ for number of samples, and use the same $$N$$ for my FFT size, then what is my frequency resolution? Still 10 Hz? Or is it $$\frac{21,000-20,000}{10000} = 0.1$$Hz?

I have a feeling that I am not really increasing my frequency resolution, as much as simply interpolating points on the main lobe, by using the same $$N$$ in to evaluate the frequencies from 21,000 to 20,000 as I did from 0 to 50,000.

Is this a correct understanding?