Program in C or C++ for JK flip-flop

JK Flip-Flop

JK flip-flop is capable of storing 1 bit. It has three inputs, labeled S(for set), R(for reset) and C(for clock).

image source-google | image by-Wikimedia Commons

Circuit Diagram:

image source-google | image by-Wikimedia Commons

Q(t+1)=J.Q(t)' + K'.Q(t)

Program in C:

---

---

 #include <stdio.h>
  typedef char bit;

  bit jk(bit J,bit K,bit Q)
  {
 
 return J&~(Q)|Q&~(K);
  
   }

  
int main()
{
 int i,j,k,a,res=0;
 int x,y,z;
 char b;
 printf("1.Truth Table of JK flip flop");
 printf("\n2.I/O operation\n");
 printf("Enter your choice: ");
 scanf("%d",&a);
 switch(a) 
 {
  case 1:
  printf("J  K  Q(t) | Q(t+1)\n");
  for(i=0;i<2;i++)
  {
   for(j=0;j<2;j++)
   {
    for(k=0;k<2;k++)
    {
    res=jk(i,j,k);
    printf("%d  %d  %d    |",i,j,k);
    printf("  %d\n",res);
   
    }
   }
  }
 break;
  
case 2:
 printf("Enter the value of J,K,Q(t):");
 scanf("%d%d%d",&x,&y,&z);
 
  res=jk(x,y,z);
  printf("Q(t+1)= %d\n\n",res);

  break;

 default:
  printf("invalid input");   
         
  }
 printf("do you want to continue(Y/N):");
 scanf(" %c",&b);
  if(b=='Y'||b=='y')
    {
     main();
 }
 else{
  return 0;
    }
}  
      

Output:

 1. Truth Table of JK flip flop
2.I/O operation
Enter your choice: 1
J  K  Q(t) | Q(t+1)
0  0  0    | 0
0  0  1    | 1
0  1  0    | 0
0  1  1    | 0
1  0  0    | 1
1  0  1    | 1
1  1  0    | 1
1  1  1    | 0
do you want to continue(Y/N): y
1. Truth Table of JK flip flop
2.I/O operation
Enter your choice: 2
Enter the value of J, K & Q(t): 1 1 1
Q(t+1)= 0

do you want to continue(Y/N): n

--------------------------------
Process exited after 25.18 seconds with return value 0
Press any key to continue . . .

Program in C++:

---

---

 #include <iostream>
 using namespace std;
  typedef char bit;

  bit jk(bit J,bit K,bit Q)
  {
 
 return J&~(Q)|Q&~(K);
  
   }

  
int main()
{
 int i,j,k,a,res=0;
 char b;
 cout<<"1.Truth Table of JK flip flop\n";
 cout<<"2.I/O operation\n";
 cout<<"Enter your choice: ";
 cin>>a;
 switch(a) 
  {
  case 1:
  cout<<"J  K  Q(t) | Q(t+1)"<<endl;
  for(i=0;i<2;i++)
  {
   for(j=0;j<2;j++)
   {
    for(k=0;k<2;k++)
    {
    res=jk(i,j,k);
   cout<<i<<"  "<<j<<"  "<<k<<"  ";
   cout<<"  | "<<res<<endl;
    }
   }
  }
 break;
  
 case 2:
  int j,k,l;
 cout<<"Enter the value of J,K & Q(t):";
 cin>>j>>k>>l;
 
  res=jk(j,k,l);
  
  cout<<"Q(t+1)= "<<res<<"\n\n";
  break;
  default:
  cout<<"invalid input";   
         
  }
  cout<<"do you want to continue(Y/N): ";
  cin>>b;
  if(b=='Y'||b=='y')
    {
     main();
 }
 else{
  return 0;
    }
}  
      

---

Output:

 1. Truth Table of JK flip flop
2.I/O operation
Enter your choice: 1
J  K  Q(t) | Q(t+1)
0  0  0    | 0
0  0  1    | 1
0  1  0    | 0
0  1  1    | 0
1  0  0    | 1
1  0  1    | 1
1  1  0    | 1
1  1  1    | 0
do you want to continue(Y/N): y
1. Truth Table of JK flip flop
2.I/O operation
Enter your choice: 2
Enter the value of J, K & Q(t): 1 1 1
Q(t+1)= 0

do you want to continue(Y/N): n

--------------------------------
Process exited after 25.18 seconds with return value 0
Press any key to continue . . .

Algorithm:

STEP 1: Start

STEP 2: Define 'jk' function

STEP 3: Perform the operation (J&~(Q)|Q&~(K)) where J, Q, K are one bit binary number

STEP 4: Invoke main function

STEP 5: Initialize variables

STEP 6: Print "Enter your choice"

STEP 7: Input choice 1 or 2

STEP 8: If the user enters 1 or 2 then follow the below steps else flow goes to default case & exit the program Switch(operator)

STEP 8.a case 1:
STEP 8.a.1: Check for condition (i<2), if Condition is true, go to step 7 otherwise go to step 12

STEP 8.a.2: Check for condition (j<2), if Condition is true, go to further step otherwise go to step 6

STEP 8.a.3: Check for condition (k<2), if Condition is true, go to further steps otherwise go to step 7

STEP 8.a.4: Call 'jk' function for value of Q(t+1)

STEP 8.a.5: Print i,j,k and Q(t+1) for truth table

STEP 8.b case 2:
STEP 8.b.1: Input three one bit binary number

STEP 8.b.2: Call 'jk' function for value of Q(t+1)

STEP 8.a.3: Print Q(t+1)

STEP 9: Stop