1. Disadvantages of Analog Transmission
Use amplifier to strengthen a weak signal but the amplifier itself introduces noise into the transmission medium
When you amplify (boost) the signal, you are also amplifying the distortion.
Cumulative effect, more amplifications result in more distortion.
NDSL

2. Advantage of Digital Signal
Use repeater to strengthen the signal.
Absence of a pulse means 0
Presence of pulse means 1. If 1, then strengthen the pulse using the repeater.
No cumulative distortion.
0 remains always 0
1 remains always 1
NDSL

3. Figure 4.1 Line coding and decoding
Encoder & decoder, not modem
NDSL

4. Figure 4.2 Signal element versus data element
# of bits / # of signal elements
NDSL

5. Data Rate versus Signal Rate
The number of data elements (bits) sent in 1s
The unit is bits per second (bps)
Called bit rate
Signal rate
The number of signal elements sent in 1s
The unit is the baud
Signal rate is sometimes called the pulse rate, the modulation rate, or the baud rate
Relationship between data rate and signal rate
S: signal rate in baud, c: the case factor, N: data rate (bps), r: data elements per signal elements
Worse case, c=1: 10101…
Best case, c=0: 000…
NDSL

6. Example 4.1
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
NDSL

7. Figure 4.3 Effect of lack of synchronization
Same square wave
Different synchronization, sampling
Receiver clock is faster.
The bit intervals of sender do not match.

8. Example 4.3
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, the receiver receives 1001 bps instead of 1000 bps.
At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps.
NDSL

9. Figure 4.4 Line coding schemes
NDSL

10. Non-Return-to-Zero (NRZ)
Figure 4.5 Unipolar: all signal levels are on one side of the horizontal time axis
Non-Return-to-Zero (NRZ)
Average power to send one bit
It is called NRZ because the signal does not return to zero at the middle of the bit.
NDSL

11. NRZ-L (NRZ-Level), NRZ-I (NRZ-Invert)
Figure 4.6 Polar NRZ-L and NRZ-I schemes
NRZ-L (NRZ-Level), NRZ-I (NRZ-Invert) N N
Bit rate = N bits/s
NRZ-L and NRZ-I both have an average signal rate of N/2 Bd, i.e., N/2 signal elements/s
High power at f=0, the DC component
Serious problem: power is not consumed evenly over the bandwidth.
At low frequencies, voltages tend to be high regardless of the data
At high frequencies, voltages tend to be low regardless of the data.
80% of the power is consumed from 0 to the average N/2.
NDSL

12. Example 4.4
A system is using NRZ-I to transfer 1-Mbps data. What are the average signal rate and minimum bandwidth?
Solution
The average signal rate is S = c x N x R = 1/2 x N x = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz.
Note c = 1/2 for the avg. case as worst case is 1 and best case is 0
What is the maximum bandwidth required?
c=1, S=1Mbaud

13. Figure 4.7 Polar RZ scheme Three voltage values: +, 0, -
Each symbol has a transition in the middle. Either from high to zero or from low to zero. Low to 0 means 0. High to 0 means 1.
Requires a wider bandwidth compared with NRZ
No DC components.
Self synchronization - transition indicates symbol value.
More complex as it uses three voltage level.

14. Figure 4.8 Manchester and differential Manchester schemes
Coming down is Going up is one.
Uses only two voltage levels. Always go through 0.
The transition at the middle of the bit is used for synchronization.
No inversion, going one way is 1. Level changes.
Inversion, going two ways is 0. Level remains the same.