# Signal Resolution

I have a question.

Suppose we have a signal $x(n)$, length (samples) $N=400$ which have been sampled with $f_s=8000 \mathtt{Hz}$. Also suppose $X(k)$ - the DFT transform of this signal.

How many zeros we must add at the end of $x(n)$ in order to change the frequency resolution of the DFT to $1 \mathtt{Hz}$?

My question: Is $7600$ zeros the right answer? Because $\Delta f = \frac{f_s}{N}$, so: $$1\mathtt{Hz} = \frac{8000}{x+400} \Rightarrow x=7600$$

Thanks, I appreciate.

One very important thing - zero padding does not increase your spectral resolution. What does matter is how many meaningful samples is in your signal ($N=400$). This is defining your frequency resolution, which is in fact 20Hz. Adding extra zeros to your signal is only interpolating values between your frequency bins - what you will observe is the side-lobes of your window.
• 7600 is OK if you want to have frequency vector with resolution of 1Hz. Although values of amplitude for most of the frequencies are interpolated and do not contain any extra information. spectral resolution $\ne$ resolution of frequency vector. – jojek Jun 27 '14 at 13:33