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Considering a signal $f(t)=\sin(2\pi t)$ with frequency $10\textrm{ Hz}$:
Middle and bottom plots show the results of sampling at different rates.
I'm trying to understand the case shown at the bottom.
The name of this effect is Spectral Leakage. Remember the relation $$\frac{k}{N} = \frac{F}{F_S}$$ where $k$ is the bin number, $N$ is the FFT size, $F$ is the continuous frequency in Hz and $F_S$ is the sample rate in Hz. It can be seen that $k$ varies from $-N/2$ to $N/2-1$ (or from $0$ to $N-1$). So there are only $N$ continuous frequencies $F$ for which you will get a single impulse in discrete frequency domain.
So in the first figure, the frequency of the signal contains an exact integer number of periods within the FFT duration and hence the correlation of other DFT sinusoids with the signal is zero. In the second figure, the signal span is not equal to the integer number of periods that violates the orthogonality requirement. Hence, the correlation of different DFT sinusoids appears as leakage around the true frequency.
When discussing DFT, you have to remember two things:
So now, let's discuss your specific example. The real Fourier transform $X(e^{j\Omega})$ of your signal $x[n]$ is composed of two delta's at $\pm 1$Hz or $\Omega = \pm \frac{2\pi}{10}$ in normalized pulsation (i.e. $\Omega = 2\pi f/f_s$ where $f_s$ is your sampling frequency). This is illustrated in the figure below.
Then, in the time domain, you're applying a rectangular window $w[n]$ of length $N$ to $x[n]$, this is equivalent to convolving the real spectrum $X(e^{j\Omega})$ with the Fourier transform $W(e^{j\Omega})$ of your rectangular window, which is a sinc function $$X'(e^{j\Omega}) = X(e^{j\Omega}) \circledast W(e^{j\Omega})$$ where $\circledast$ represents the convolution.
Those sinc functions are null at multiples of $\Omega = \frac{2\pi}{N}$. Convolving with a delta is equivalent to moving the signal centered on this delta and so you end up with two sinc functions centered around $\Omega = \pm \frac{2\pi}{10}$. The Fourier Transform of $x'[n]$ is illustrated in the figure below (for the case of $N = 100$). To obtain such a plot showing what is "hidden" in $X'(e^{j\Omega})$, you can use zero-padding on $x'[n]$.
Finally, you're sampling the spectrum $X'(e^{j\Omega})$ of your truncated version of $x[n]$ at $\Omega _k = \frac{2\pi k}{N}$ to obtain the DFT $X[k]$. For $N = 100$ (i.e. an integer multiple of the period of $x[n]$), you will thus exactly sample at the places where
And so you go back to the two delta's at $k = \pm 10$ you observed in the first figure.
For $N = 128$ however, the sinc functions are no longer well positionned (the places where the sinc are null do no longer coincides with the pulsation $\Omega_k$ where you will sample). This leads to the second figure you have.
Think of it this way: You have a 10Hz signal. You try to make a 10Hz signals with some building blocks.
1) Your building blocks are this signals: 1Hz, 2Hz,.. 10Hz, .. 20Hz.
Quite easy right? You just make your 10Hz signals using the 10Hz "building block".
2) Your building blocks are: 0.6Hz, 1.2Hz, .. 9.6Hz, 10.2Hz, ... Now what can we do? Well.. the signal looks almost like a 10.2Hz so we use a big block of 10.2. But it's not exactly that so we add a little bit of 9.6 and some others.
That's exactly spectral leakage. You're trying to describe a signal with elements you don't have so you project them and use a component of each.
