# Filter Banks and FFT, are they similar conceptually?

I'm using Filter Banks (Narrow Band Pass) created using the functions in Matlab. However, I wanted to know if there are fundamental limitations like that of FFT which also affect the Filter Banks. Is there any co-relation between the two?

Basically, I don't know if they share the same problem of time and frequency being inversely related. If they do share this problem, this means that we cannot have higher time and higher frequency resolution at the same time.

• Too broad a question. Please be little more specific May 7, 2014 at 4:41
• Time frequency uncertainty is universal to all linear time-frequency representations. Filter banks and short time fourier transforms all fall into this category. May 7, 2014 at 8:58
• In both cases you will have to deal with the uncertainty principle. See also this answer to a related question: dsp.stackexchange.com/questions/15607/… May 7, 2014 at 10:21

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: $$h[n] = \begin{cases} 1, && 0 \le n \le N-1 \\ 0, && \text{otherwise} \end{cases}$$ 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.

• I'm not sure what you mean by the "peak" of the convolution. The convolution is apparent by rewriting the DFT equation as $X[k] = \sum_{n=0}^{N-1} x_s[n]$, where $x_s[n]$ is the shifted version of $x[n]$. This is equivalent to convolving the shifted version with a boxcar filter. The "critically sampled" part of it is that each block of $N$ input samples yields only 1 sample in each of $N$ filters in the bank. May 9, 2014 at 16:11