Question

I can't wrap my head around this nyquist theorem question.

According to Nyquist, what frequency is necessary to support a bit rate of 96,000 bits per second using 8 bits per signal component?

How the heck do you calculate this?

Answer 1

> According to Nyquist, what frequency is necessary to support a bit rate of 96,000 bits per second using 8 bits per signal component?

It has been awhile since I touched this stuff, so let it be known I cannot assure this being correct to any degree. (Last time I did touch this stuff was in middle school, so...)

Nyquist defines the data rate (bits per second) \(= 2 * f * n\) where \(f\) is maximum frequency, and \(n\) is bit per signal component.

Assuming I remembered that definition correctly, with some algebraic substitution:
\[96000 = 2 * f * 8 \Rightarrow f = 6000\]

So 6000 ㎐ is necessary, to support a bit rate of 96,000 bits per second using 8 bits per signal component.

Answer 2

Sweet, that's the answer I eventually got. Thank you