Question

A 1 MB digital file needs to transmit a channel with bandwidth of 10 MHz and the SNR is 10 dB. What is the minimum amount of time required for the file to be completely transferred to the destination?

Answer 1

Answer:

A 1 MB digital file needs 0.23 seconds to transfer over a channel with bandwidth 10 MHz and SNR 10 dB.

Explanation:

We can calculate the channel capacity using Shannon's Capacity formula:

C = B + log₂ (1 + SNR)

Where C = Channel Capacity

           B = Bandwidth of the Channel

           SNR = Signal to Noise Ratio

We are given SNR in dB so we need to convert it into a ratio.

$SNR_{dB}$ = 10log₁₀ (SNR)

10 = 10log₁₀ (SNR)

1 = log₁₀ (SNR)

SNR = 10¹

SNR = 10

So, using Shannon Channel Capacity formula:

C = 10 x 10⁶ log₂ (1 + 10)

C = 34.5 MHz

Total amount of time required to transmit a 1MB file:

1MB = 1 x 8 Mbytes = 8Mb

C = 34.5 MHz = 34.5 Mb/s

Time required = 8Mb/34.5Mb/s = 0.23 seconds

A 1 MB digital file needs 0.23 seconds to transfer over a channel with bandwidth 10 MHz and SNR 10 dB.