Inverse DFT and the DFT Filter Bank Sum

The Length $N$ inverse DFT is given by [264]

$$x(n) = \frac{1}{N}\sum_{k=0}^{N-1} X(k) e^{j2\pi nk/N}, \quad n=0,1,2,\ldots,N-1.$$ (10.16)

This suggests that the DFT Filter Bank can be inverted by simply remodulating the baseband filter-bank signals $y_k(n)$ , summing over $k$ , and dividing by $N$ for proper normalization. That is, we are led to conjecture that

$$x(n-N+1) = \frac{1}{N}\sum_{k=0}^{N-1} y_k(n) e^{j2\pi nk/N}, \quad n=0,1,2,\ldots\,.$$ (10.17)

This is in fact true, as we will later see. (It is straightforward to show as an exercise.)