# What can we understand when we increase the $N$-point DFT for a given signal?

I have an input sequence $$x(n)=\cos(0.48\pi n)+\cos(0.52\pi n).$$ I am determining its spectrum (amplitude of DFT values) based on the finite number of samples.

For Example :

1. Taking the first $10$ samples of $x(n)$, $n=0, \ldots, 9$, calculating the $10$-point DFT of $x(n)$

2. Using the $10$ samples from part $1$, calculating and plotting the $100$-point DFT.

3. Taking the first $100$ samples of $x(n)$, $n=0, \ldots, 99$, ploting the $100$-point DFT of $x(n)$.

Questions:

• Why are we going for a $100$-point DFT/ $200$-oint DFT ?
• And what happens when we take $10$ samples and calculate $100$-point DFT ?
• Increase by zero-padding, increasing the sample rate, or sampling for a longer length of signal? – hotpaw2 Jun 21 '16 at 19:23
• Increase the sampling for a longer length of the signal – Ross Jun 21 '16 at 20:09

If the signal length $M$ remains the same but only number of points $N$ in DFT is increased (assuming $M \leq N$) then this will give you the finely spaced samples $X[k]$ of the same true spectrum $X(e^{j\omega})$ as given by the DFT frequency sampling relation: also known as the zero-padding result $$X[k] = X(e^{j\frac{2 \pi}{N}k}) = \sum_{n=0}^{N-1} {x[n]e^{-j\frac{2 \pi}{N}k}} = \sum_{n=0}^{M-1} {x[n]e^{-j\frac{2 \pi}{N}k}} ~~~ , ~~~~ k=0,1,2,...,N-1$$
On the other hand, if both the analysed signal length $M$ and the DFT length $N$ are increased (assuming $M=N$) by using a longer analysis window (considering $x[n]$ defined for a very long, possibly infinite length duration) then the windowed true spectrum of the underlying signal whose samples are investigated, is also enhanced, thereby providing samples $X[k]$ of a higher frequency resolution spectrum of the very long length signal $x[n]$