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Data Communication

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Data Communication

Lecture # 05

Course Instructor:

Engr. Sana Ziafat

- The conversion involves three techniques:line coding,block coding, andscrambling. Line coding is always needed; block coding and scrambling may or may not be needed.

- Data element is defined as smallest entity to represent piece of information
- Where as signal element is the shortest unit of digital signal.
- Data element is actually what we need to end and signal element is what we can send

“r” is the ratio of data element to signal element

- Data rate:
-Number of data elements sent in one second

-The unit is bits per second(bps)

- Signal Rate:
-Number of signals elements sent in one second.

-The unit is baud

- Increase the data rate while decreasing the signal rate.
- Increasing the data rate increase the speed of data transmission
- Decreasing the signal rate decrease the bandwidth requirement.

- Increasing the data rate increases the speed of transmission where as increasing the signal rate increases the bandwidth requirement
- HOW???

- S= c* N * 1/r

Example

A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:

Pulse Rate = 1/ 10-3= 1000 pulses/s

Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps

Example

A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1?

Solution

We assume that the average value of c is 1/2 . The baud rate is then

- No of signal levels
- Bit rate verses baud rate
- DC components
- Noise immunity
- Error detection
- Synchronization
- Cost of implementation

Example

In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Kbps? How many if the data rate is 1 Mbps?

Solution

At 1 Kbps:

1000 bits sent 1001 bits received1 extra bps

At 1 Mbps:

1,000,000 bits sent 1,001,000 bits received1000 extra bps

Note

In unipolar encoding, we use only one voltage level.

- It uses only one polarity of voltage level
- Bit rate same as data rate
- Dc component present
- Lack of synchronization for long sequence of 0’s & 1’s.
- Simple but obsolete

Note

In polar encoding, we use two voltage levels: positive & negative

Note

In NRZ-L the level of the voltage determines the value of the bit.

In NRZ-I the inversion or the lack of inversion determines the value of the bit.

Both have average signal rate of N/2 baud.

- Synchronization problem
- Dc component problem
- Sudden change of polarity in system (for case of NRZ-L)

Note

In Manchester and differential Manchester encoding, the transition

at the middle of the bit is used for synchronization.

Note

In bipolar encoding, we use three levels: positive, zero, and negative.

- AMI: Alternate Mark Inversion
- It uses three voltage levels
- Unlike RZ 0 level is used to represent a 0
- Binary 1’s are represented by alternating positive and negative.

- Pseudoternary
- variation of AMI encoding in which 1 bit is encoded as zero voltage level and 0 bit is encoded as alternating positive and negative voltages

Chapter 4 (B. A Forouzan)

Section 4.1

- Draw the graphs for all encoding techniques learnt in this lecture, using each of following data stream.
a.11111111

b.00000000

c.00110011

d.01010101

- What is the Nyquist sampling rate for each of the following signals?
a. A low pass signal with bandwidth of 300 KHz?

b. A band pass signal with bandwidth of 300KHz if the lowest frequency is 100 KHz?