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UNO [17]
3 years ago
8

Assume a flash memory can have 6 levels {0, 1, 2, 3, 4, 5} and the errors can be of limited value of magnitude 1 (i.e. a digit a

t level i can change to level (i 1) (mod 6)). With one digit we can choose {0, 2, 4} as the codewords to have a perfect error correcting code of limited magnitude 1. With two digits, what is the maximum number codewords in a limited mag- nitude 1 perfect error correcting code
Engineering
1 answer:
Elena-2011 [213]3 years ago
6 0

Answer:

Explanation:

/* General Code for achieving perfect error for any level and any

digit of codewords*/

#include<iostream>

using namespace std;

int main(){

  int level;

  //Input memory levels.

  cout <<" Enter number of Flash Memory levels : ";

  cin>>level;

  //construct array store levels.

  int lvl[level];

  for(int i = 0; i < level; i++){

      //assign levels name

      lvl[i] = i;

  }

  int digit;

  cout << "Enter number of digit codewords :";

  //Taking input of codeword digits

  cin >> digit;

  for(int i = 0; i < level;){

      //output Codeword

      cout << lvl[i];

      //Skipping for n number digit codeword in level

      i+=(digit+1);

     

  }

 

}

/*

a) i from 0 to (i = i+(2+1))<levels

b) for length (1) -> 3 codewords

  codewords -> {0,2,4}

for length (2) -> 2 codewords

codewords -> {0,3}

*/

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5 0
3 years ago
Given the system of equations−3x2+7x3=2,x1+2x2–x3=3,5x1−2x2=2(a). Compute the determinant. (b). Use Cramer’s rule to solve for t
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5 0
3 years ago
The solid spindle AB is connected to the hollow sleeve CD by a rigid plate at C. The spindle is composed of steel (Gs = 11.2 x 1
dalvyx [7]

Answer:

T_max = 12.63 kip.in

Ф_a = 1.093°

Explanation:

Given:

- The modulus of rigidity of solid spindle G_ab = 11.2 * 10^6 psi

- The diameter of solid spindle d_ab = 1.75 in

- The allowable stress in solid spindle τ_ab = 12 ksi

- The modulus of rigidity of sleeve G_cd = 5.6 * 10^6 psi

- The outer diameter of sleeve d_cd = 3 in

- The thickness of sleeve t = 0.25

- The allowable stress in sleeve τ_cd = 7 ksi

Find:

- The largest torque T that can be applied to end A that does not exceed allowable stresses and sleeve angle of twist 0.375°

- The corresponding angle through which end A rotates.

Solution:

- Calculate the polar moment of inertia of both spindle AB and sleeve CD.

   Spindle AB:    c_ab = 0.5*d_ab = 0.5(1.75) = 0.875 in

                           J_ab = pi/2 c^4 = pi/2 0.875^4 = 0.92077 in^4

   Sleeve CD:  c_cd1 = 0.5*d_cd = 0.5(3) = 1.5 in , c_cd2 = c_cd1 - t = 1.25

                     J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

- The stress criteria for maximum allowable torque in spindle AB:

                    T_ab = J_ab*τ_ab / c_ab

                    T_ab = 0.92077*12 / 0.875

                    T_ab = 12.63 kip.in

- The stress criteria for maximum allowable torque in sleeve CD:

                    T_cd = J_cd*τ_cd / c_cd1

                    T_cd = 4.1172*7 / 1.5

                    T_cd = 19.21 kip.in

- The angle of twist criteria for point D:

                    T_d = J_cd*G_cd*Ф / L

                    T_d = 4.1172*5.6*10^6*0.006545 / 8

                    T_d = 18.86 kip.in

- The maximum allowable Torque for the structure is:

                    T_max = min ( 12.63 , 19.21 , 18.86 )

                    T_max = 12.63 kip.in

- The angle of twist of end A:

                    Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

                    T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

                    12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

                    0.01908 rads = 1.093°

3 0
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An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin
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Answer:

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8 0
3 years ago
The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False
Alex

Answer:

(B) FALSE

Explanation:

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3 years ago
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