Which filters response is maximally flat at origin?
It seems to me that both High Pass Filter and Band Pass Filter have maximum flat response at origin, although it is almost zero in case of Bandpass Filter. Am I right?
This is a question that I came across in my friends exam yesterday.
The solution is given is that it is the Butterworth Filter. Is it true and if so, how? Are both HP and BP filters wrong. Please explain. Thanks
The classification of filters into general High-pass, Band-pass, Low-pass filters categories essentially indicates which frequency range is predominantly unattenuated. It has little to do with the type of filters such as Butterworth, which are defined with specific transfer function and imply a specific shapes in the frequency domain and correspondingly specific characteristics (such as ripples, flatness, etc).
Analog Butterworth filters are referred to as "maximally flat" because all the derivatives (up to the 2nth) of the filter gain function as a function of frequency are 0 at $\omega=0$ (see wikipedia). Note that this definition is given in the context of a low-pass Butterworth filter.
let's begin with a normalized-frequency lowpass prototype, where we want $|H_N(0)|=1$ and $|H_N(j\omega)|<1$ for $|\omega|>0$ and where we want the gain to essentially decrease as $|\omega|$ increases.
for normalized frequency, the $N$th-order prototype Butterworth frequency response (magnitude-squared) is
$$ |H_N(j\omega)|^2 \ = \ \frac{1}{1+\omega^{2N}} $$
if you compute the derivatives
$$ \frac{d |H_N(j\omega)|^2}{d \omega} $$
and evalutate it at $\omega=0$, you will see that all of these derivatives are zero up to the $2N$th derivative (the odd-order derivatives would be zero anyway).
this is what makes it maximally flat at $\omega=0$.
for the lowpass (LPF), highpass (HPF), and bandpass (BPF), tuned to a specific frequency, $\omega_0$, this same normalized-frequency prototype (once $H_N(s)$ is figured out) is used with these mappings:
$$ H_{LPF}(s) = H_N\left( \frac{s}{\omega_0} \right) $$
$$ H_{HPF}(s) = H_N\left( \frac{\omega_0}{s} \right) $$
$$ H_{BPF}(s) = H_N\left( Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) \right) $$
