Flip Flops

Basic Flip Flops

This article deals with the basic flip flop circuits like S-R Flip Flop, J-K Flip Flop, D Flip Flop, and T Flip Flop along with truth tables and their corresponding circuit symbols.

Before going to the topic it is important that you get knowledge of its basics. Click on the links below for more information.

TAKE A LOOK : BOOLEAN LOGIC

TAKE A LOOK : LOGIC GATES

TAKE A LOOK : HALF ADDER AND FULL ADDER

Flip flops are actually an application of logic gates. With the help of Boolean logic you can create memory with them. Flip flops can also be considered as the most basic idea of a Random Access Memory [RAM]. When a certain input value is given to them, they will be remembered and executed, if the logic gates are designed correctly. A higher application of flip flops is helpful in designing better electronic circuits.

The most commonly used application of flip flops is in the implementation of a feedback circuit. As a memory relies on the feedback concept, flip flops can be used to design it.

There are mainly four types of flip flops that are used in electronic circuits. They are

1. The basic Flip Flop or S-R Flip Flop
2. Delay Flip Flop [D Flip Flop]
3. J-K Flip Flop
4. T Flip Flop

1. S-R Flip Flop

The SET-RESET flip flop is designed with the help of two NOR gates and also two NAND gates. These flip flops are also called S-R Latch.

• S-R Flip Flop using NOR Gate

The design of such a flip flop includes two inputs, called the SET [S] and RESET [R]. There are also two outputs, Q and Q’. The diagram and truth table is shown below.

S-R Flip Flop using NOR Gate

From the diagram it is evident that the flip flop has mainly four states. They are

S=1, R=0—Q=1, Q’=0

This state is also called the SET state.

S=0, R=1—Q=0, Q’=1

This state is known as the RESET state.

In both the states you can see that the outputs are just compliments of each other and that the value of Q follows the value of S.

S=0, R=0—Q & Q’ = Remember

If both the values of S and R are switched to 0, then the circuit remembers the value of S and R in their previous state.

S=1, R=1—Q=0, Q’=0 [Invalid]

This is an invalid state because the values of both Q and Q’ are 0. They are supposed to be compliments of each other. Normally, this state must be avoided.

• S-R Flip Flop using NAND Gate

The circuit of the S-R flip flop using NAND Gate and its truth table is shown below.

S-R Flip Flop using NAND Gate

Like the NOR Gate S-R flip flop, this one also has four states. They are

S=1, R=0—Q=0, Q’=1

This state is also called the SET state.

S=0, R=1—Q=1, Q’=0

This state is known as the RESET state.

In both the states you can see that the outputs are just compliments of each other and that the value of Q follows the compliment value of S.

S=0, R=0—Q=1, & Q’ =1 [Invalid]

If both the values of S and R are switched to 0 it is an invalid state because the values of both Q and Q’ are 1. They are supposed to be compliments of each other. Normally, this state must be avoided.

S=1, R=1—Q & Q’= Remember

If both the values of S and R are switched to 1, then the circuit remembers the value of S and R in their previous state.

• Clocked S-R Flip Flop

It is also called a Gated S-R flip flop.

The problems with S-R flip flops using NOR and NAND gate is the invalid state. This problem can be overcome by using a bistable SR flip-flop that can change outputs when certain invalid states are met, regardless of the condition of either the Set or the Reset inputs. For this, a clocked S-R flip flop is designed by adding two AND gates to a basic NOR Gate flip flop. The circuit diagram and truth table is shown below.

Clocked S-R Flip Flop

A clock pulse [CP] is given to the inputs of the AND Gate. When the value of the clock pulse is ’0′, the outputs of both the AND Gates remain ’0′. As soon as a pulse is given the value of CP turns ’1′. This makes the values at S and R to pass through the NOR Gate flip flop. But when the values of both S and R values turn ’1′, the HIGH value of CP causes both of them to turn to ’0′ for a short moment. As soon as the pulse is removed, the flip flop state becomes intermediate. Thus either of the two states may be caused, and it depends on whether the set or reset input of the flip-flop remains a ’1′ longer than the transition to ’0′ at the end of the pulse. Thus the invalid states can be eliminated.

2. D Flip Flop

The circuit diagram and truth table is given below.

D Flip Flop

D flip flop is actually a slight modification of the above explained clocked SR flip-flop. From the figure you can see that the D input is connected to the S input and the complement of the D input is connected to the R input. The D input is passed on to the flip flop when the value of CP is ’1′. When CP is HIGH, the flip flop moves to the SET state. If it is ’0′, the flip flop switches to the CLEAR state.

To know more about the triggering of flip flop click on the link below.

TAKE A LOOK : TRIGGERING OF FLIP FLOPS

TAKE A LOOK : MASTER-SLAVE FLIP FLOP CIRCUIT

3. J-K Flip Flop

The circuit diagram and truth-table of a J-K flip flop is shown below.

J-K Flip Flop

A J-K flip flop can also be defined as a modification of the S-R flip flop. The only difference is that the intermediate state is more refined and precise  than that of  a S-R flip flop.

The behavior of  inputs J and K is same as the S and R inputs of the S-R flip flop. The letter J stands for SET and the letter K stands for CLEAR.

When both the inputs J and K have a HIGH state, the flip-flop switch to the complement state. So, for a value of Q = 1, it switches to Q=0 and for a value of Q = 0, it switches to Q=1.

The circuit includes two 3-input AND gates. The output Q of the flip flop is returned back as a feedback to the input of the AND along with other inputs like K and clock pulse [CP]. So,  if the value of CP is ’1′, the flip flop gets a CLEAR signal and with the condition that the value of Q was earlier 1. Similarly output Q’ of the flip flop is given as a feedback to the input of the AND along with other inputs like J and clock pulse [CP]. So the output becomes SET when the value of CP is 1 only if the value of Q’ was earlier 1.

The output may be repeated in transitions once they have been complimented for J=K=1 because of the feedback connection in the JK flip-flop. This can be avoided by setting a time duration lesser than the propagation delay through the flip-flop. The restriction on the pulse width can be eliminated with a master-slave or edge-triggered construction.

4. T Flip Flop

This is a much simpler version of the J-K flip flop. Both the J and K inputs are connected together and thus are also called a single input J-K flip flop. When clock pulse is given to the flip flop, the output begins to toggle. Here also the restriction on the pulse width can be eliminated with a master-slave or edge-triggered construction. Take a look at the circuit and truth table below.

• Complements • Subtraction using complements.

• Complements
• Subtraction using complements.

By the end of this class you should be able to:

• Obtain the r’s and (r-1)’s complements in decimal, binary,
and any other number system
• Perform subtraction by addition of the complements


Decimal number complements:

9’s complement of the decimal number N = (10n
 – 1) – N
= n (9’s) – N
i.e. {subtract each digit from 9}

Example Æ 9’s complement of 134795 is 865204
Similarly

1’s complement of the binary number N = (2n
 -1) – N = n (1’s) – N
Example Æ 1’s complement of 110100101 is 001011010
which can be obtained by replacing each one by a zero and each
zero by one.

r’s complement: EE200(class 2-2) Prof. M.M. Dawoud 2of 5
10’s complement of the decimal number N = 10n
 – N = (r-1)’s
complement + 1

Example Æ 10’s complement of 134795 is 865205


Example Æ find the 9’s and 10’s complements of 314700.

Answer Æ 9’s complement = 685299
 10’s complement=685300

Rule: To find the 10’s complement of a decimal number leave all
leading zeros unchanged. Then subtract the first non-zero digit
from 10 and all the remaining digits from 9’s.

The 2’s complement of a binary number is defined in a similar
way.

Example: Find the 1’s and 2’s complements of the binary
number 1101001101

Answer Æ 1’s complement is 0010110010
 2’s complement is 0010110011

Example: Find the 1’s and 2’s complements of 100010100

Answer Æ 1’s complement is 011101011
 2’s complement is 011101100

Subtraction using r’s complement:

To find M-N in base r, we add M + r’s complement of N

Result is M + (rn
 – N)

1) If M > N then result is M – N + rn (rn is an end carry and
can be neglected.
 EE200(class 2-2) Prof. M.M. Dawoud 3of 5
2) If M < N then result is rn
 –(N-M) which is the r’s complement
of the result.

Example: Subtract (76425 – 28321) using 10’s complements.


Answer Æ 10’s complement of 28321 is 71679

 Then add Æ 7 6 4 2 5
 + 7 1 6 7 9

 1 4 8 1 0 4

Therefore the difference is 48104 after discarding the end carry.


Example: subtract (28531 – 345920)

Answer Æ It is obvious that the difference is negative. We
also have to work with the same number of digits, when dealing
with complements.

 10’s complement of 345920 is 654080

 Then add Æ 0 2 8 5 3 1
 + 6 5 4 0 8 0

 6 8 2 6 1 1
Therefore the difference is negative and is equal to the 10’s
complement of the answer.
 Difference is Æ - 317389

The same rules apply to binary.

Example: subtract (11010011 – 10001100)

Discard
No end carry EE200(class 2-2) Prof. M.M. Dawoud 4of 5
Answer Æ 2’s complement of 10001100 is 01110100

Then add Æ 1 1 0 1 0 0 1 1
 + 0 1 1 1 0 1 0 0

 1 0 1 0 0 0 1 1 1
The difference is positive and is equal to 01000111

The same rules apply to subtraction using the (r-1)’s
complements. The only difference is that when an end carry is
generated, it is not discarded but added to the least significant
digit of the result. Also, if no end carry is generated, then the
answer is negative and in the (r-1)’s complement form.


Example: Subtract (76425 – 28321) using 9’s complements.
Answer Æ 9’s complement of 28321 is 71678

 Then add Æ 7 6 4 2 5
 + 7 1 6 7 8

 1 4 8 1 0 3
 1
 4 8 1 0 4



Example: subtract (11010011 – 10001100) using 1’s complement.

Answer Æ 1’s complement of 10001100 is 01110011

Then add Æ 1 1 0 1 0 0 1 1
 + 0 1 1 1 0 0 1 1

 1 0 1 0 0 0 1 1 0
 1

Discard
Difference
Difference EE200(class 2-2) Prof. M.M. Dawoud 5of 5
 1 0 1 0 0 0 1 1 1 

Make a simple currency converter for 3 currencies

Question :  You are required to write a program for Currency Exchange rates. The basic idea is that user/reader will be able to interchange different currencies using our program. User will have three options i.e. Pakistani Rupees, US Dollars, and Euro. In the end of the program, user should be asked if he/she wants to make another conversion. If user presses y then the whole program should start again. If user presses n then the program should terminate.

 #include <iostream>  
 using namespace std;  
 void main()  
 {  
      int amount=0;  
      char choice1, choice2, choice3='Y';  
      while (choice3=='Y')  
      {  
           cout<<"Select the currencies that you want to exchange\n";  
           cout<<" - Enter R for Pakistani Rupees\n";  
           cout<<" - Enter E for Euro\n";  
           cout<<" - Enter D for Dollar\n"<<endl;  
           cout<<"Please select the currency that you want to convert"<<endl;  
           cin>>choice1;  
           cout<<"Please enter the currency that you want to convert into"<<endl;  
           cin>>choice2;  
           cout<<"Enter Amount: ";  
           cin>>amount;  
           if ( choice1=='R' && choice2=='E')  
           {  
                //division of input rupees by 112 to convert into euros  
                cout<<amount <<" Rupees = "<<amount/112<<" Euros"<<endl;    
           }  
           if ( choice1=='R' && choice2=='D')  
           {  
                //division of input rupees by 84 to convert into dollars  
                cout<<amount <<" Rupees = "<<amount/84<<" Dollars"<<endl;   
           }  
           if ( choice1=='E' && choice2=='R')  
           {  
                //multiplication of input Euros by 112 to convert into rupees  
                cout<<amount <<" Euros = "<<amount*112<<" Rupees"<<endl;   
           }  
           if ( choice1=='E' && choice2=='D')  
           {  
                //multiplication of input Euros by 1.33 to convert into dollars  
                cout<<amount <<" Euros = "<<amount*1.33<<" Dollars"<<endl;   
           }  
           if ( choice1=='D' && choice2=='R')  
           {  
                //multiplication of input dollars by 84 to convert into rupees  
                cout<<amount <<" Dollars = "<<amount*84<<" Rupees"<<endl;   
           }  
           if ( choice1=='D' && choice2=='E')  
           {  
                //division of input dollars by 1.33 to convert into euros  
                cout<<amount <<" Dollars = "<<amount/1.33<<" Euros"<<endl;   
           }  
           cout<<"Do you want to make another conversion? ( Y/N ) :";  
           cin>>choice3;  
      }  
      cout<<"Goodbye :) "<<endl;  
 }  

Dangling point.

 What is a dangling pointer?

   
A dangling pointer arises when you use 
the address of an object after
its lifetime is over. This may occur 
in situations like returning
addresses of the automatic variables
from a function or using the
address of the memory block after 
it is freed. The following
code snippet shows this:

          class Sample
          {
          public:
                  int *ptr;
                  Sample(int i)
                  {
                   ptr = new int(i);
                  }

                  ~Sample()
                  {
                   delete ptr;
                  }
                  void PrintVal()
                  {
cout << "The value is " << *ptr;
                  }
          };

          void SomeFunc(Sample x)
          {
cout << "Say i am in someFunc " << endl;
          }

          int main()
          {
           Sample s1 = 10;
           SomeFunc(s1);
           s1.PrintVal();
          }

In the above example when PrintVal() function is
called it is called by the pointer that has been
freed by the       destructor in SomeFunc.

void pointer

void *

Sometimes we know we want a pointer, but we don't necessarily know or care what it points to. The C/C++ language provides a special pointer, the void pointer, that allows us to create a pointer that is not type specific, meaning that it can be forced to point to anything.Why is this useful? One common application of void pointers is creating functions which take any kind of pointer as an argument and perform some operation on the data which doesn't depend on the data contained. A function to "zero-out" memory (meaning to turn off all of the bits in the memory, setting each byte to the value 0) is a perfect example of this.

void memzero(void *ptr, size_t len)
{
 for(; len>0; len--) {
  *(char *)ptr = 0;
 }
}

This function takes a pointer to any piece of memory, meaning that we can pass in any kind of pointer we want, and the number of bytes to zero out. It then walks along the memory zeroing out each byte. Without void pointers, it would be more difficult to write a generic function like this.

Other example

void increaseChar (char* charData)
{
    ++(*charData);
}

void increaseInt (int* intData)
{
    ++(*intData);
}

int main ()
{
  char a = 'x';
  int b = 1602;
  increaseChar (&a);
  increaseInt (&b);
  cout << a << ", " << b << endl;
  string str;
  cin >> str;
  return 0;
}

string

C-Strings (Character Arrays)

STRING: It is an array of type char.

Syntax for declaration

char <array/string name> [max. number of characters to be stored +1];

The number of elements that can be stored in a string is always n-1, if the size of the array specified is n. This is because 1 byte is reserved for the NULL character '\0' i.e. backslash zero. A string is always terminated with the NULL character.

Example:
char str[80]; 
In the above example, str can be used to store a string with 79 characters.

Initializing a string

A string can be initialized to a constant value when it is declared.

char str[ ] = "Good";
    Or
char str[]={'G','o','o','d','\0'};

Here. 'G' will be stored in str[0], 'o' in str[1] and so on.

Note: When the value is assigned to the complete string at once, the computer automatically inserts the NULL character at the end of the string. But, if it is done character by character, then we have to insert it at the end of the string.

Reading strings with/without embedded blanks

To read a string without blanks cin can be used
cin>>str;
To read a string with blanks cin.getline() or gets() can be used.
cin.getline(str,80);
   -Or-
gets(str);

Printing strings

cout and puts() can be used to print a string.
cout<<str:
   Or
puts(str);

Note: For gets( ) and puts(), the header file stdio.h has to be included. puts() can be used to display only strings. It takes a line feed after printing the string.

cin	gets()
It can be used to take input of a value of any data type.	It can be used to take input of a string.
It takes the white space i.e. a blank, a tab, or a new line character as a string terminator.	It does not take the white space i.e. a blank, a tab, or a new line character, as a string terminator.
It requires header file iostream.h	It requires the header file stdio.h
Example: char S[80]; cout<<"Enter a string:”; cin>>S;	Example: char S[80]; cout<<"Enter a string:"; gets(S);

cout	puts()
It can be used to display the value of any data type.	It can be used to display the value of a string.
It does not take a line feed after displaying the string.	It takes a line feed after displaying the string.
It requires the header file iostream.h	It requires the header file stdio.h
Example: char S[80]="Computers"; cout<<S<<S; Output: ComputersComputers	Example: char S[80]="Computers"; puts(S); puts(S); Output: Computers Computers

Counting the number of characters in a string and printing it backwards

#include<iostream.h>
#include<stdio.h>
int main( )
{
  char str[80];
  cout<<"Enter a string:";
  gets(str);
  for(int l=0; str[l]!='\0';l++);  //Loop to find length
    cout<<"The length of the string is : "<<l<<endl ;
  for(int i=l-1;i>=0;i--)    //Loop to display the string backwards
    cout<<str[i];
  return 0;
}

Function to count the number of words in a string

void count(char S[])
{
  int words=0;
  for(int i=0;S[i]!='\0';i++)
  {
    if (S[i]==' ')
      words++;                   //Checking for spaces
  }
  cout<<"The number of words="<<words+1<<endl;
}

Function to find the length of a string

int length(char S[ ])
{
  for(int i=0;S[i]!='\0';i++);
     return i;
}

Function to copy the contents of string S2 to S1

void copy(char S1[ ], char S2[ ])
{
   for(int i=0;S2[i]!='\0';i++)
      S1[i]=S2[i];
   S1[i]='\0';
}

Function to concatenate the contents of string S2 to S1

void concat(char S1[ ], char S2[ ])
{
   for(int l=0;S1[l]!='\0';l++);
     for(int i=0;S2[i]!='\0';i++)
         S1[l++]=S2[i];
     S1[l]='\0';
}

Function to compare strings STR1 to STR2.

The function returns a value>0 if //STR1>STR2, a value<0 if STR1<STR2, and value 0 if STR1=STR2

int compare(char STR1[ ],char STR2[])
{
  for(int I=0;STR1[I]==STR2[I] && STR1[I]!='\0'&&STR2[I]!='\0'; I++);
     return STR1[I]-STR2[I];
}

To reverse the contents of string S and store it in string Rev

void Reverse(char S[], char Rev[])
{
   for(int C1=0; S[C1]!='\0'; C1++);
      C1--;
   for(int C2=0;C1>=0;C2++,C1--)
      Rev[C2]=S[C1];
   Rev[C2]='\0';
}

Function to check whether a string S is a palindrome or not

int Palin(char S[])
{
   for(int L=0;S[L]!='\0';L++);    //To find length
      for(int C=0;(C<L/2) && (S[C]==S[L-C-1]);C++);
         return (C==L/2)?1:0; //Returns 1 if Palindrome else 0
}

Function to change the case of string S to uppercase

void Upper(char S[])
{
   for(int i=0;S[i]!='\0';i++)
      S[i] = (S[i]>='a' && S[i]<='z')?(S[i]-32):S[i];
}

Function to change the case of string S to lower case

void Lower(char S[])
{
   for(int i=0;S[i]!='\0';i++)
       S[i] = (S[i]>='A' && S[i]<='Z')?(S[i]+32):S[i];
}

Function to extract n characters from left side of the string and store it in a different string.

Example: 4 characters from ENVIRONMENT=ENVI

int SLeft(char S[ ], int n, char result[ ])
{
   for(int l=0;S[l]!='\0';l++);
     if(n<=I)    //characters extracted should be <=length
     {
        for(int i=0;i<n;i++)
            result[i]=S[i];
        result[i]='\0';
        return 1;
     }
     else
        return 0;
}

Function to extract n characters from right side of the string and store it in a different string.

Example: 4 characters from ENVIRONMENT=MENT

int SRight(char S[ ], int n, char result[ ])
{
   for(int l=0;S[l]!='\0';l++);
     if(n<=I)     //characters extracted should be <=length
     {
        for(int j=0;i=l-n;S[i]!=’/0’;i++,j++)
            result[j]=S[i];
        result[j]='\0';
        return 1;
     }
     else
        return 0;
}

Function to extract n characters from specified location loc of the string and store it in a different  string.

Example: 4 characters from third location in string ENVIRONMENT= VIRO

int substring(char S[ ], int n, int loc, char result[ ])
{
  for(int l=0;S[l]!='\0';l++);
     if(n<=I)      //characters extracted should be <=length
     {
        for(int j=0;i=l-n;S[i]!=’/0’;i++,j++)
           result[j]=S[i];
        result[j]='\0';
        return 1;
     }
     else
        return 0;
}

Two Dimensional Array

Two Dimensional Array

It is a collection of data elements of same data type arranged in rows and columns (that is, in two dimensions).

Declaration of Two-Dimensional Array

Type arrayName[numberOfRows][numberOfColumn];

For example,
int Sales[3][5];

Initialization of Two-Dimensional Array

An two-dimensional array can be initialized along with declaration. For two-dimensional array initialization, elements of each row are enclosed within curly braces and separated
by commas. All rows are enclosed within curly braces.
int A[4][3] = {{22, 23, 10},
              {15, 25, 13},
              {20, 74, 67},
              {11, 18, 14}};

Referring to Array Elements

To access the elements of a two-dimensional array, we need a pair of indices: one for
the row position and one for the column position. The format is as simple as:
name[rowIndex][columnIndex]

Examples:
cout<<A[1][2];      //print an array element
A[1][2]=13;         // assign value to an array element
cin>>A[1][2];       //input element

Using Loop to input an Two-Dimensional Array from user

int mat[3][5], row, col ;
for (row = 0; row < 3; row++)
  for (col = 0; col < 5; col++)
    cin >> mat[row][col];

Arrays as Parameters

Two-dimensional arrays can be passed as parameters to a function, and they are passed by reference. When declaring a two-dimensional array as a formal parameter, we can omit the size of the first dimension, but not the second; that is, we must specify the number of columns. For example:
   void print(int A[][3],int N, int M)
In order to pass to this function an array declared as:
   int arr[4][3];
we need to write a call like this:
   print(arr);
Here is a complete example: 

#include <iostream>
using namespace std;
void print(int A[][3],int N, int M)
{
  for (R = 0; R < N; R++)
    for (C = 0; C < M; C++)
       cout << A[R][C];
}

int main ()
{
  int arr[4][3] ={{12, 29, 11},
                  {25, 25, 13},
                  {24, 64, 67},
                  {11, 18, 14}};
  print(arr,4,3);
  return 0;
}

Function to read the array A

void Read(int A[][20], int N, int M)
{
  for(int R=0;R<N;R++)
    for(int C=0;C<M;C++)
    {
      cout<<"(R<<','<<")?";
      cin>>A[R][C];
     }
}

Function to display content of a two dimensional array A

void Display(int A[][20],int N, int M)
{
  for(int R=0;R<N;R++)
  {
     for(int C=0;C<M;C++)
        cout<<setw(10)<<A[R][C];
     cout<<endl;
   }
}

Function to find the sum of two dimensional arrays A and B

void Addition(int A[][20], int B[][20],int N, int M)
{
  for(int R=0;R<N;R++)
    for(int C=0;C<M;C++)
      C[R][C]=A[R][C]+B[R][C];
}

Function to multiply two dimensional arrays A and B of order NxL and LxM

void Multiply(int A[][20], int B[][20], int C[][20],int N, int L, int M)
{
  for(int R=0;R<N;R++)
   for(int C=0;C<M;C++)
   {
      C[R][C]=0;
      for(int T=0;T<L;T++)
        C[R][C]+=A[R][T]*B[T][C];
    }
}

Function to find & display sum of rows & sum of cols. of a 2 dim. array A

void SumRowCol(int A[][20], int N, int M)
{
  for(int R=0;R<N;R++)
  {
     int SumR=0;
     for(int C=0;C<M;C++)
       SumR+=A[R][C];
     cout<<"Row("<<R<<")="<<SumR<<endl;
   }
  for(int R=0;R<N;R++)
  {
    int SumR=0;
    for(int C=0;C<M;C++)
      SumR+=A[R][C];
    cout<<"Row("<<R<<")="<<SumR<<endl;
   }
}

Function to find sum of diagonal elements of a square matrix A

void Diagonal(int A[][20], int N, int &Rdiag, int &LDiag)
{
  for(int I=0,Rdiag=0;I<N;I++)
    Rdiag+=A[I][I];
  for(int I=0,Ldiag=0;I<N;I++)
    Ldiag+=A[N-I-1][I];
}

Function to find out transpose of a two dimensional array A

void Transpose(int A[][20], int B[][20],int N, int M)
{
  for(int R=0;R<N;R++)
    for(int C=0;C<M;C++)
       B[R][C]=A[C][R];