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GaryK [48]
3 years ago
12

)A cable has a bandwidth of 3000 Hz assigned for data communication. The SNR is 3162. How many signal levels do we need?

Computers and Technology
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

L=56.2

Explanation:

We need to use two theoretical formulas: Shannon Capacity and Nyquist Bit Rate.

From Shannon:

C=B*log_2(SNR+1)

From Nyquist:

C=2B*log_2(L)

Where:

C= Channel\hspace{3}capacity\\B=Bandwidth\hspace{3} of\hspace{3} the\hspace{3} channel\\SNR=Signal\hspace{3}to\hspace{3}noise\hspace{3}ratio\\L=Number\hspace{3}of\hspace{3}signal\hspace{3}levels\hspace{3}used\hspace{3}to\hspace{3}represent\hspace{3}data

Using the data provided by the problem and the equations above, let's find the signal levels we need. Using the Shannon equation:

C=B*log_2(SNR+1)

Where:

B=3000Hz\\SNR=3162

C=3000*log_2(3162+1)\\\\C=3000*log_2(3163)\\\\C=34881.23352\hspace{3}bps

Now, using Nyquist equation:

C=2B*log_2(L)

Where:

C=34881.23352\hspace{3}bps\\B=3000Hz

34881.23352=2*3000*log_2(L)\\\\34881.23352=6000*log_2(L)

Solving for L :

\frac{34881.23352}{6000} =log_2(L)\\\\5.81353892=log_2(L)\\\\2^{5.81353892} =2^{log_2(L)}\\ \\56.24055476=L\\\\L\approx56.2

Therefore, we need approximately 56.2 signal levels.

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  1. public static long calcTotal(long [][] arr2D){
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  3.        
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  5.        {
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  12.        return total;
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Explanation:

Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

Line 2: Declare a variable, total, and initialize it with zero.

Line 4: An outer for-loop to iterate through every rows of the two-dimensional array

Line 6: An inner  for-loop to iterate though every columns within a particular row.

Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

Line 12: Return the final total of all the element values as an output

Answer:

2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
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  17.        return average;
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Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

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  1. public static double calcRowAverage(long [][] arr2D, int row){
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  8.            {
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  11.                    total = total + arr2D[i][j];
  12.                    count++;
  13.                }
  14.            }
  15.            
  16.        }
  17.        
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  19.        
  20.        return average;
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Explanation:

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Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

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