For T_{2  Flip flop,}

For T₁ Flip flop,

Step 4: Lastly according to the equation got from K map create the design for 2 bit synchronous up down counter.

Related posts:

Design a 2 bit Synchronous down counter using T Flip flop?
Design a 2 bit Synchronous up counter using T Flip flop? 

These are the following steps to Design a 3 bit synchronous up counter using T Flip flop:

Step 1: To design a synchronous up counter, first we need to know what number of flip flops are required. we can find out by considering a number of bits mentioned in the question. So, in this, we required to make 3 bit counter so the number of flip flops required is 3 [2ⁿ where n is a number of bits].

Step 2: After that, we need to construct state table with excitation table.
Note: To construct excitation table from state table you should know the excitation table of respective flip flop, in this case, it is T flip flop. So check the excitation table for T flip flop Which is:

T Flip Flop Excitation Table

Present state	Next State	T
0	0	0
0	1	1
1	0	1
1	1	0

So, the above table is the excitation table for T Flip Flop.

State Table with  excitation  table

Present State			Next State			Flip Flop
Q3	Q2	Q1	Q'3	Q'2	Q'1	T3	T2	T1
0	0	0	0	0	1	0	0	1
0	0	1	0	1	0	0	1	1
0	1	0	0	1	1	0	0	1
0	1	1	1	0	0	1	1	1
1	0	0	1	0	1	0	0	1
1	0	1	1	1	0	0	1	1
1	1	0	1	1	1	0	0	1
1	1	1	0	0	0	1	1	1

Above  table is created as per follow :

When Q₃ =0 which is present state and Q₃‘=0 which is next state then T₃ become 0 [As per excitation table, have a look ]
Similarly, if Q₃ is 0 and Q₃‘ is 1 then T₃  become 1.
In similar way it goes on .

Step 3: After making the excitation table the next thing  to do is dig out the equation from the boolean algebra or K map for the design of the counter. So, for T₁ , T₂ and T₃ we got 1,  Q₁ and Q₁.Q₂

K-Map

 

For T₃ Flip flop,

 

T₃= Q₁.Q₂

For T_{2  Flip flop,}

For T₁ Flip flop,

Step 4: Lastly according to the equation got from K map create the design for 3 bit synchronous up counter.

In above design, T₁ is getting input 1 and T₂ is getting input from the output of the T₁ flip flop and lastly, T₃ is getting input from the output of T₁  and T₂ . A clock is attached to it which is in blue colour.

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What is Sum of Product form?

‘Sum of Product’ is commonly abbreviated as SOP form. When an expression is expressed in a sum of binary terms ( A term refers to a set of binary variables, where each binary variable is combined with an operation) called Minterms then it is said to be Sum of Products.

In other words, An expression consisting only of minterms is called Sum of the Product. For example, A.B+A’B is an SOP expression.
[other concept Maxterm, (A+B).(A’+B’)]

SOP can be categorized into two forms :

– Canonical or Standard SOP form: In this, each and every binary variable must have in each term. For example- A.B+A’B’

– Minimal SOP form: In this, the standard SOP expression is reduced up to the minimum possible expression.

A Sum of Products question can be asked in three ways:-

1) In the form Truth table
2) In the form of non-canonical Expression
3) In the form of Boolean function

Case 1: For the given table, simplify it in SOP expression

Points

• Always consider high output (1)
• In SOP, every term in the expression is referred to as Minterm
• A Minterm is represented as m (small m)

How do you find the SOP expression from truth table?

Solution :
Step 1. Since there are 2 variables, so, therefore, there would be 2ⁿ combinations which are 2² =4.
We consider a high output as minterm. a minterm is denoted as m.
Y= m₂+m₃
Y = A.B’ +A.B                            :- It is in Canonical SOP form

In Sum of Product each term is combined with OR operation and within each term, every variable combined with AND operation.

Step 2. Now narrow the founded expression down to minimal SOP form. In this, you should know the rules of Boolean expression or K-map

Y= A.( B’ + B)
Y= A.1
Y= A

OR

Considering, A= 1, A’=0

Y= A

A is your answer.

Case 2: Simplify any expression to SOP or Sum of product form, when Boolean function is given

Simplify the following Boolean function in SOP form
F(X,Y,Z) = Σm(1,4,5,6,7)

Points to Remember

• First, check the type of m, if m is small then it would Minterm, meaning we have to solve the boolean function in respect to Sum of Product form, whereas if m is capital then it would be Maxterm, means we need to solve it in respect to Product of Sum.
• On the left side, See how many binary variables are there, in the above case there are 3 binary variables, that’s 2³ possible combinations.

How do you find the SOP expression from the boolean function?

Step 1. Make the truth table, if you want to see the actual picture or skip the step.

X	Y	Z	F
0	0	0	0 (m0)
0	0	1	1 (m1)
0	1	0	0 (m2)
0	1	1	0 (m3)
1	0	0	1 (m4)
1	0	1	1 (m5)
1	1	0	1 (m6)
1	1	1	1 (m7)

In the above table, we put output as 1 or high at the mentioned position in the above Boolean function.
For example, in Boolean function  first value is 1, so we put the output high to respective place in the truth table as output 1 at  row 001  ( represents 1)
Similarly, we have 4 , so at row 100 (represents 4)  in the truth table, we put high output.

Step 2. Now make the expression as per the above table,

F= m₁ + m₄ + m₅ + m₆ + m₇
F=  X’Y’Z + XY’Z’ + XY’Z+ XYZ’ + XYZ                :- Canonical SOP form

Step 3. Now Solve the function to minimal SOP form using K map or Boolean rules
K-map is always considered the easiest and fastest way.

 

F= X + Y’Z                   :- Minimal SOP form

This your answer

 

Case 3: Simplify any expression in SOP form, when a is given

Simplify the following expression in SOP form
F= x’z’+y’z’+yz’+xy

 OR

Simplify the following non-canonical expression in SOP form

F= x’z’+y’z’+yz’+xy

Point to remember

• Put 1 in place of missing input and convert that 1 to the appropriate Boolean rule. Suppose in above term x’z’, y is missing ( replace 1 with (y+y’))
• Make sure the expression is in Canonical form

How do you find the SOP expression from the non-canonical expression?

Step 1.   First, check the given expression is in Canonical form or not, if yes simply make K-map (or jump to 4th step ) and simplify it, otherwise follow the steps.

Step 2. Now, the above expression isn’t in canonical form as there are three binary variables or inputs, x , y and z which can’t be together found in each and every term. So to do this, make it canonical form by doing the following:-

F =  (x’.y’. 1) + (y’z’ .1) + (y z’ .1) + ( x. y.1 )

Since, in each term, a variable is missing, to bring it, in each term we have to put 1 to get the expression.

Step 3.  Put relevant input in place of 1 on the basis of Boolean rules

There is a rule, where

x + x’ =1 in Boolean algebra, we will apply it.

In the first term, y is missing  (x’y’), similarly in the second one, x is missing (y’z’) and so one.

F =  x’z'( y +y’)  +  y’z'(x+x’) + yz'(x+x’)+ xy(z+z’)
F=  x’z’y+x’z’y’+y’z’x+y’z’x’+yz’x+yz’x’+xyz+xyz’

Now, Arranging in it alphabetical order for ease of understanding

F=x’yz’+x’y’z’+xy’z’+x’y’z’+xyz’+x’yz’+xyz+xyz’

It is now in Canonical SOP or Sum of Product form

Step 3. This is the last step, shrink the expression down to minimal SOP form using K-map
or Boolean algebra

Which gives, F = z’ + xy

 

The main difference between them is, Go-Back-N protocol retransmits all the frames starting from the damaged or corrupted frame, whereas Selective Repeat protocol only retransmits frames that are damaged.

Difference between Selective Repeat and Go-Back- N sliding window protocol:

Selective Repeat sliding window protocol:

Selective repeat is one of the sliding window protocols which is essentially responsible for detecting and correcting the error caused in the data link layer.

The Selective repeat protocol basically only retransmits the damaged or lost frame.

The retransmitted frame in the selective repeat protocol is always received out of sequence.

Go-back-N sliding window protocol:

Go-back-N is also one of the sliding window protocols. This protocol mechanism is essentially used to control and detect the errors occurring in the data link layer.

During the data transmission between the sender and the receiver; if the acknowledgement is lost or the frame has been damaged then the sender has to resend all the frames starting from the damaged or corrupted frame.

There are two types of modes, User Mode and Kernel Mode. A kernel mode is also known as supervisor mode.

User mode

It has restricted access to the resources. CPU has restrictions, therefore, it can have only access to limited instructions and memory. Utility applications such as text editor, media player are run in this mode.

When an application or program is executed, its initial state and operation mode are loaded on stack. At this point, CPU starts executing the program in this particular mode. Interrupts and traps are used to signal  or flag the CPU to switch back to Kernel mode and do the essential tasks by storing the current state of the user program to the stack again. After completing it, CPU resumes the program from where it left.

Kernel Mode

It has full access to memory, I/O and other resources. In this, CPU can execute any instructions and have full access to underlying hardware. The core functionalities of the operating system always run in Kernel mode.

Having these authorities, processes using this mode have full rights to access resources, allowing them read/write to the storage media, enable and disable hardware and more. Difference between Kernel Mode and User Mode :

Key Differences:

1. The mode in which there is an unconditional, unrestricted and full permission to access the system’s hardware by the current executing piece of code is known as the kernel mode. The mode in which there is no means of accessing the system’s hardware directly by the current piece of code is also known as the user mode.
2. The kernel mode is a very powerful and impactful mode which means that it can refer to any memory block of the system and can also orders to execute an instruction to the CPU. The user mode is standardized and normal viewing mode which means that it cannot reference to any memory block or execute any information on its own as it requires some Application Protocol Interface (API) to do these tasks.

3. The kernel mode is the most unrestricted and candid mode that a system can have which means that only the trustable sources can be allowed to operate in this very mode. The user mode is more like a generic mode that can have, which means that any user can access this mode for a particular system and is allowed to operate in this particular mode.

4. As the kernel mode is very important and candid this means that if the system crashes in this mode then it could be very fatal and increases the complexity. As the user mode is quite general and user-friendly this means that even if the system crashes in this mode then it is not fatal at all and can simply restart the session.

5. The kernel mode is very crucial to the system as well as the system programmer so everything that runs easily runs on your PC might not run here because its purpose is to employ hardware for a particular task. The user mode is user friendly space where all the general code will definitely execute.

Simplify the following expressions using Boolean Algebra :-

•  A + AB

= A.1+AB
= A(1+B)                            :-  OR (A+1 = 1)
= A.1
= A

• AB+AB’

= A(B+B’)                            :- AB+AC = A(B+C)
= A(B+B’)                           :-  A + A’ =1
= A(1)
= A

• A’BC + AC

= C(A’B+A)                         :- A+BC =  (A+B).(A+C)
= C((A’+A).(A+B))             :- A’+A=1
= C(1.(A+B))
= C(A+B)

• A’B+ABC’+ABC

= A’B+AB(C’+C)                       :- A’+A=1
= A’B+AB
= B(A’+A)                                   :- A’+A=1
= B

• A’ B’ C + A’ B C’ + A B’ C’ + A B C