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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.
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.
= 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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Answer:
What do you want me to do
Explanation:
<span>#include
#include
using namespace std;
int main() {
const int SCORES_SIZE = 4;
vector bonusScores(SCORES_SIZE);
int i = 0;
bonusScores.at(0) = 10;
bonusScores.at(1) = 20;
bonusScores.at(2) = 30;
bonusScores.at(3) = 40;
for (i = 0; i < SCORES_SIZE; ++i) {
cout << bonusScores.at(i) << " ";
}
cout << endl;
return 0;
}</span>