# Calculate balance between signal and noise to allow 32 kbit/s transfer

I have to calculate proper balance/ratio between signal and noise ( S/N ) in decibels ( dB ), to be able to transfer data over the channel at 32 kbit/s. Channels bandwidth is 100 kHz.

I was given no formula or additional explanation to have as a start off, the textbook we are using is horrible and I can't reach my professor for explanation. I have to prepare for a exam so thanks for any help in advance.

I did however pick up a formula used to solve a similar problem where the signal/noise ratio has been provided in dB and the formula goes: $W \cdot log_2 \bigg ( 1 + \frac{S}{N} \bigg)$ This problem stated that the channel allows bandwidth between 400Hz and 3600Hz so $W = 3600 - 400 = 3200 Hz$

But I have no clue how to proceed from here.

The signal-to-noise ratio needed for $\mathrm{32\ kbit/s}$ with a bandwith of $\mathrm{100\ kHz}$ is calculated in the following way:

$\underbrace{C=B\cdot \log_2{\left(1+\dfrac SN\right)}}_{\textrm{Shannon's equation}}\longrightarrow \dfrac CB=\log_2{\left(1+\dfrac SN\right)}\longrightarrow 2^\frac{C}{B}=1+\frac SN\longrightarrow \frac SN=2^\frac CB - 1$

If we substitute the values in the equation we get:

$\dfrac SN=2^\frac{32\cdot 10^{3}\ \textrm{bits/s}}{10^5\ \textrm{Hz}} - 1=2^{0.32} - 1=0.2483\longrightarrow 10\cdot\log_{10}{(0.2483)}=-6.05\textrm{ dB}$.

In Shannon's equation:

• C is the channel capacity in bits per second;
• B is the bandwidth of the channel in hertz;
• S is the average signal power received over the bandwidth in watts;
• N is the average noise power received over the bandwidth in watts;
• S/N is the signal-to-noise-ratio as a linear power ratio (not as decibels)

Sources: