The normalized frequency response of a N points bandpass digital filter is:
$\frac{sin(\omega_{c2}(n-M))}{\pi(n-M)}-\frac{sin(\omega_{c1}(n-M))}{\pi(n-M)}$ where $\omega_{c1}$ is the normalized lower cutoff angular frequency , where $\omega_{c2}$ is the normalized upper cutoff angular frequency and $M = floor(N/2)$.
However the normalized frequency response of a N points bandstop digital filter is:
$\frac{sin(\omega_{c1}(n-M))}{\pi(n-M)}-\frac{sin(\omega_{c2}(n-M))}{\pi(n-M)}$ where $\omega_{c1}$ is the normalized lower cutoff angular frequency and where $\omega_{c2}$ is the normalized upper cutoff angular frequency .
But obviously $H_{bp}(e^{j\omega}) =-H_{bs}(e^{j\omega})$ Does this mean that a band pass filter has the same frequency response to a band stop filter if you switch the angular cutoff frequencies?
And what implications does this have in digital filter design?