Nyquist Theorem to calculate maximum data rate is -

l o g_{2}

bits/sec.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

l o g_{2}

V bits/sec

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l o g_{2}

(1 + S/N)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l o g_{2}

(1 + S/N)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 2H $l o g_{2}$ bits/sec.
Nyquist theorem to calculate maximum data rate is 2H log2log2 bits/sec.

Nyquist bit rate formuladefines the theoretical maximum bit rate. r = 2 X B X log2 L. In this formula, B is the bandwidth of the channel, L is the number of signal levels used to represent data, and r is the bit rate in bits per second.

The Nyquist rate, is twice the bandwidth of a band limited function or a band limited channel.

Data rate governs the speed of data transmission. A very important consideration in data communication is how fast we can send data, in bits per second, over a channel.

Suggest Corrections

Similar questions

{log}_{2} [{log}_{3} ({log}_{2} x)] = 1

\int_{0}^{1} \frac{x}{1 + \sqrt{x}} d x =

a^{x} = N

{log}_{a} N = x

{log}_{2} a b = {log}_{2} a + {log}_{2} b

{log}_{2} (x y) = 5

{log}_{1 / 2} (\frac{x}{y}) = 1

A = | ({log}_{2} | cos 24^{o} |) + ({log}_{2} | cos 48^{o} |) + ({log}_{2} | cos 96^{o} |) + ({log}_{2} | cos 192^{o} |) |

{log}_{2} (3 x - 1) - {log}_{2} (x - 2) = 3

Join BYJU'S Learning Program