Excellent question !! pass band will add extra complexity and extra hardware requirements in-terms of circuit components, and also pass band means we are dealing with higher frequencies than low-pass and hence it is more Susceptible to noise element which we already suffer from by the thermal noise in and outside the receiver system. noise-power = k T B, with K=Boltzmann constant 1.38064852 × 10-23 m2 kg s-2 K-1; T=temp and finally (B = bandwidth = passband ).
Excellent question !! pass band will add extra complexity and extra hardware requirements in-terms of circuit components, and also pass band means we are dealing with higher frequencies than low-pass and hence it is more Susceptible to noise element which we already suffer from by the thermal noise in and outside the receiver system. noise-power = k T B, with K=Boltzmann constant 1.38064852 × 10-23 m2 kg s-2 K-1; T=temp and finally (B = bandwidth = passband ).