Filter Banks and FFT, are they similar conceptually?
The answer is yes, the discrete Fourier transform (DFT), of which the fast Fourier transform (FFT) is an efficient implementation, can be interpreted quite literally as a filter bank. Recall the formula for the DFT:

$$
X[k] = \sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi k n}{N}}
$$

The "filterbank" interpretation for the above equation goes something like this:


In order to get the $k$-th DFT output $X[k]$, we start with the input signal $x[n]$.
Multiply $x[n]$ by a complex exponential function $e^{\frac{-j2\pi k n}{N}}$. By the frequency-shifting property of the DFT, this has the effect of shifting the content at frequency $\frac{2\pi k}{N}$ down to zero frequency.
Convolve the result with a filter that has the impulse response:
$$
h[n] = \begin{cases}
1, &amp;&amp; 0 \le n \le N-1 \\
0, &amp;&amp; \text{otherwise}
\end{cases}
$$
This filter (known as a moving average or a "boxcar") has a lowpass characteristic.


Thus, we can view the DFT as placing a bank of uniformly-spaced critically-sampled filters across the signal's spectrum. Each filter has the same frequency response as an $N$-sample moving average filter, centered at frequencies $e^{\frac{j2\pi k n}{N}}, k = 0, \ldots, N-1$. I described this similarly in an answer to a previous question here.