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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.

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  • \$\begingroup\$ This has got nothing to do with matched filters. \$\endgroup\$
    – Andy aka
    Commented Jan 16, 2018 at 11:35
  • \$\begingroup\$ @Andyaka I updated my question to reflect this. \$\endgroup\$
    – Runge Kutta
    Commented Jan 16, 2018 at 12:07

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One of these difficulties is,that you need a sampling rate that is higher than the double of the highst frequency in the signal. For that reason, signal paths for sampling include an low-pass filter to get rid of the higher spectrum of the signal that is not used. But the problem which come with the low-pass filter is, that they have an infinit spectrum and can not perfectly cut-off frequencies.

Example: If you would like to sample a signal and get a bandwith of 20kHz, you musst cut-off at 20kHz but you will still get signal components beyond 20kHz. Because of that, your sampling rate would not be > 40kHz, rather than 50kHz or more. It depends on the signal to noise ratio that your application needs.

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  • \$\begingroup\$ So in my opinion, it is not specific to matched filter design. \$\endgroup\$
    – Finkman
    Commented Jan 16, 2018 at 11:07
  • \$\begingroup\$ "But the problem which come with the low-pass filter is, that they have an infinit spectrum and can not perfectly cut-off frequencies". Could you elaborate on this? Do you mean that that the filter borders are not perfectly vertical or is there another effect at work? \$\endgroup\$
    – Runge Kutta
    Commented Jan 16, 2018 at 11:41
  • \$\begingroup\$ Yes, filter borders can't be perfectly vertical. Such filters only exist in therory and are called ideal filter. There is a nice explanation on wiki about ideal and real filters: en.wikipedia.org/wiki/Low-pass_filter#Ideal_and_real_filters \$\endgroup\$
    – Finkman
    Commented Jan 16, 2018 at 12:42
  • \$\begingroup\$ As an addition: a simple rc low-pass has 20dB attenuation per decate and is a first grade filter. A second grade filter would have 40dB per decate and so on. Now you might question: why don't use an (almost) infinit grade filter design? It's because it's easyer to simplye take a higher sample rate and to keep phase distortion small, when required. (Since every filtering action has an impact on the phase) \$\endgroup\$
    – Finkman
    Commented Jan 16, 2018 at 12:48

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