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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?

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Answer:

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

Step-by-step 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.

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User Rando Hinn
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