# Bandwidth computation for NRZ baseband transmission scheme

In Andrew Tanenbaum's book on Computer Networks, while explaining about bandwidth of a NRZ (Non Return to Zero) baseband transmission scheme, he says the following:

With NRZ, the signal may cycle between the positive and negative levels up to every 2 bits (in the case of alternating 1s and 0s). This means that we need a bandwidth of at least B/2 Hz when the bit rate is B bits/sec. This relation comes from the Nyquist rate [Eq. (2-2)].

I have 2 questions here.

1. What does the first statement mean exactly? Does the author imply that for a signal that switches between 1s and 0s alternately, if a particular bit is a 1, then the next time we get a 1 again is after 2 bits?
2. How does the first statement suggest the second statement?

The following excerpt from the book explains Nyquist's theorem and is followed by Eqn. (2-2).

Nyquist proved that if an arbitrary signal has been run through a low-pass filter of bandwidth B, the filtered signal can be completely reconstructed by making only 2B (exact) samples per second. Sampling the line faster than 2B times per second is pointless because the higher-frequency components that such sampling could recover have already been filtered out. If the signal consists of V discrete levels, Nyquist’s theorem states:

$$maximum\; data\; rate = 2B \log_2 V \;bits/sec$$

The first statement says that for the given binary sequence of $$a_k = \{ 1,0,1,0,1,0,...\}$$ the transmitted NRZ line-signal would oscillate at the maximum rate : $$V = \{+A,-A,+A,-A,...\}.$$ which is alternating at a rate half of the binary data transmission bit-rate. This maximum frequency of line voltage oscillation should be capable of passing through the channel filter. Hence, the channel bandwidth should be at least 500 Hz in that case.