# Task 9. Data Measurement

Text of the task. Between two working sections of the deposit, it is planned to install an additional channel for exchanging stereo audio signals (messages) for daily reporting communication sessions. Determine the required bandwidth of this channel in kilobits based on the calculation that the sessions will last no more than 51 minutes. The requirements for the mono signal per second are listed below:

- Sampling frequency: 63 Hz
- Bit depth: 17 bits
- Metadata volume: 47 bytes for every 5 Kilobits of audio
The solution to the problem. The bandwidth of a channel is a measure of the ratio of the volume of transmitted data to the transmission time, i.e., simply put - how much data can be transmitted over a set time period. It is also worth noting that the indicators provided are for a mono signal, and the calculation needs to be made for a stereo signal, i.e., the resulting indicators should be doubled. First, let's convert all the data to a unified scale - seconds and bits.

Time for data transmission $=51 * 60=3060$ seconds

To determine the amount of data transmitted, we need to multiply the sampling frequency by its depth over the allocated period

$$
\text { Data Volume }=63 * 17 * 3060=3277260 \text { bits }
$$

Now let's calculate the volume of metadata

$$
\text { Data Volume }=\frac{47 * 8 * 3277260}{5 * 1024}=240674 \text { bits }
$$

Thus, the total volume of data that needs to be transmitted will be Total data volume $=(3277260+240674) * 2=7035868$ bits

And finally, let's determine the bandwidth in kilobits per second:

$$
\text { Bandwidth }=\frac{7035868}{1024 * 3060}=2.25 \text { Kbit } / \text { sec }
$$