# Guidance for digital Communication

Discussion in 'Wireless & RF Design' started by guru200773, Apr 30, 2010.

1. ### guru200773 Thread Starter Member

Apr 26, 2010
94
1
hello sir I have undergone a cell phone trouble shooting course.... I want to know the basics of digital communication techniques... Along with mobile phone communication working of mobiles and their hardware block diagram how it is working and abt software... they doesnt teach this so pls help me... If u suggest free Ebook that will be useful for my career pls help me..... THanking u in advance

2. ### bertus Administrator

Apr 5, 2008
15,649
2,348
guru200773 likes this.
3. ### guru200773 Thread Starter Member

Apr 26, 2010
94
1
thank u so much dude.....

4. ### bluedraco20 New Member

May 6, 2010
5
0
Dude,

I thought you were in Avionics dude.

Peace dude.

5. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
As you probably know, digital is basic 1's and 0's, ON and OFF, TRUE and FALSE, etc. Digital communications relies on a reconcilation between the sender and the receiver, otherwise they won't be able to transfer any intelligent information through data. Consider the example below:

Suppose hypothetically I wanted to create a digital system in which a HIGH ($V_O_H$) is equal to 2V to 5V ($2V < V_O_H < 5V$). Likewise, let's suppose a LOW ($V_O_L$) is equal to 0V to 2V ($0V < V_O_L < 2V$). An important concept for digital communication is noise margin. If some noise gets into my signal, which is more than likely, what if it bumps my signal up to exactly 2V? How is the receiver supposed to interpret that? Unfortunately, the receiver can't do anything about those types of situations. Therefore, some noise margin must be implemented in order to compensate for data loss and noise.

Instead, let's propose a new setup in which $0V < V_O_L < 1V$ and $4V < V_O_H < 5V$. Now we have some noise margin; 3V to be exact. Nevertheless, noise could bump my signal into the "No Man's Land" and my receiver wouldn't be able to interpret it. Therefore we have to go to yet another proposition.

Now let's suppose that the sender has $0V < V_O_L < 2V$, and $4V < V_O_H < 6V$. Rather than have the receiver have the same conditions, let's try something slightly different: $0V < V_O_L < 1V$ and $5V < V_O_H < 6V$. We now have some noise margin for our data communications.

If you look on the datasheets for digital gates and other components, you will find those exact HIGH and LOW conditions.

Austin

Last edited: May 8, 2010
6. ### guru200773 Thread Starter Member

Apr 26, 2010
94
1
ya im belongs to avionics but there also digital communications are used... So i wanna to bulid up a fundamental dude