# Why is the bandwidth for a matched filter twice the normal nyquist theorem one?

Why is the needed channel bandwidth for a matched filter higher than the normal bandwidth derived by the nyquist theorem (2B=R, assuming L=2)?

After some digging I found this source that states the following:

Overcome the practical difficulties encountered with the ideal Nyquist channel by extending the bandwidth from the minimum value W = R_b/2 to an adjustable value between W and 2W.

What are these practical difficulties and how does increasing the channel bandwidth overcome them? Why do these difficulties only occur when using a matched filter?

edit 1: In response to "not specific to matched filter" comments, I realize that in practice 2.2 (or more) times the highest f is used instead of 2 to overcome imperfections. However, in my handbook B=R is used for matched filters (both in theory as in exercises) and 2B=R is used in every other context.

F.e. my handbook states the following for a matched filter: $$BER = Q(\sqrt{\frac{(A_1 - A_0)^2 T_s}{4N_0/2}}) = Q(\sqrt{\frac{(A_1 - A_0)^2}{2N_0 B}})$$ which implies R=B, does it not?

edit 2: My question was inherently flawed. Unluckily, in my handbook it just happened to be so that both the relevant exercise and the theory section about matched filters used a (flawed) sampling method at the middle of the bit puls merely as an example. The usage of B=R stemmed from this coincidence, not from the fact that a matched filter was used.

• This has got nothing to do with matched filters. Jan 16 '18 at 11:35
• @Andyaka I updated my question to reflect this. Jan 16 '18 at 12:07