Data rate for an ASK modulation [closed]

Question

A telephone line with a bandwidth of 3 kHz and an SNR of 6 dB. I am asked to calculate the maximum bitrate on this line and the error probability, knowing that ASK modulation is used.

For ASK modulation:

• Bit 1 is represented by Acos(wt), and 0 by zero.
• Error probability: Q(sqrt(Ed/2N)), where Ed is the energy of the signal sent during one period after calculating Ed = ((A^2) * Tb)/2 with Tb as the time of one bit.
• The error probability is expressed by the function Q(sqrt((A^2) * Tb)/4N) with N as the power spectral density of AWGN noise.

The SNR (Signal-to-Noise Ratio) is expressed as ((A^2)/4)/(N*B) = 10^0.6 = 4 with B as the bandwidth.

Now, to find the value of the error probability, a relationship between the Q(x) function and the SNR needs to be established. The problem is finding the bitrate; the options are:

1. R = 2Blog(M)
2. R = B*log(1+SNR)
3. B = (1+d)*R

The given answer for this exercise is R = B/3, which I didn't understand were to comes from.

Answer 1

The maximum bitrate on a communication channel is typically calculated using the Nyquist formula for the maximum data rate and Shannon's channel capacity theorem. However, in your case, the given answer \$\displaystyle\text{R}=\frac{\text{B}}{3}\$ seems inconsistent with the standard formulas. Let's break it down step by step:

1. Bandwidth: \$\displaystyle\text{B}=3\space\text{kHz}\$;
2. Signal-to-Noise Ratio: \$\displaystyle\text{SNR}=6\space\text{dB}=4\$

Since \$\displaystyle\text{SNR}\$ in linear scale equals \$\displaystyle10^\frac{\text{SNR}\space\left[\text{dB}\right]}{10}\$.

The relationship between SNR and bandwidth for a channel is usually given by Shannon's channel capacity formula:

$$\text{C}=\text{B}\log_2\left(1+\text{SNR}\right)\tag1$$

Where \$\displaystyle\text{C}\space\left[\text{bits/second}\right]\$ is the channel capacity, \$\displaystyle\text{B}\space\left[\text{Hz}\right]\$ is the bandwidth and \$\displaystyle\text{SNR}\$ is the signal-to-noise ratio.

Using this formula, you can calculate the maximum bitrate \$\displaystyle\text{R}\$ given the bandwidth and \$\displaystyle\text{SNR}\$:

$$\text{R}=\text{C}=\text{B}\log_2\left(1+\text{SNR}\right)=3\space\text{kHz}\cdot\log_2\left(1+4\right)\approx6.96578\space\text{kbps}\tag2$$

So, according to Shannon's channel capacity theorem, the maximum bitrate should be approximately \$\displaystyle6.96578\space\text{kbps}\$ based on the given bandwidth and \$\displaystyle\text{SNR}\$.

Regarding the relationship \$\displaystyle\text{R}=\frac{\text{B}}{3}\$, it might be a specific condition or simplification for this particular exercise, but it doesn't align with the standard formulas of channel capacity in information theory. If you have additional information or constraints specific to the question, it might clarify why that particular relationship was used.