1. . Who is Most Merciful and Beneficial With the Name of Allah
PUCIT-BS(CS)F06

2. Nyquist and shanon
Noiseless Channel: Nyquist Bit Rate
Noisy Channel: Shannon Capacity
Using Both Limits

3. Nyquist Bit Rate
For a noiseless channel ,the Nyquist bit rate formula defines the theoretical maximum bit rate
Bit Rate =2*B*log2 L
B is the bandwidth of the channel ,L is the number of signal levels used.

4. Example
Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as
Bit Rate = 2  3000  log2 2 = 6000 bps

5. Example
Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as:
Bit Rate = 2 x 3000 x log2 4 = 12,000 bps

6. Shanon Capacity
To determine the theoretical highest data rate for a noisy channel shanon introduced a formula .
C = B log2 (1 + SNR)
B is the bandwidth .SNR is the signal to noise ratio, capacity is the capacity of the channel in bits per second.

7. Shanon Capacity
Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated as
C = B log2 (1 + SNR)
= B log2 (1 + 0)
= B log2 (1)
= B  0 = 0

8. Example
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually For this channel the capacity is calculated as
C = B log2 (1 + SNR) = 3000 log2 ( ) = 3000 log2 (3163)
C = 3000  = 34,860 bps

9. Example
We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?
First, we use the Shannon formula to find our upper limit.
C = B log2 (1 + SNR) = 106 log2 (1 + 63)
= 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
4 Mbps = 2  1 MHz  log2 L  L = 4

10. Transmission Impairment
Signal received may differ from signal transmitted [because of the imperfection of transmission media] causing:
Analog – degradation of signal quality
Digital – bit errors
Most significant impairments are
Attenuation
Distortion
Noise

11. Transmission Impairments

12. Attenuation Attenuation means loss of energy.
Amplifiers are used to amplify the signal.
Example – wire carrying electrical signals gets warm
Energy in the signal is converted to heat.
Attenuation is measured in Decibels.
Decibels – measures strength of signal
negative is signal has attenuated,
positive if signal is amplified
dB = 10 log10(P2 / P1 )  where P2 and P1 are powers at points 1,2

13. Attenuation

14. Example
Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as
10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB

15. Example
Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 * P1. In this case, the amplification (gain of power) can be calculated as
10 log10 (P2/P1) = 10 log10 (10P1/P1)
10 log10 (10) = 10 (1) = 10 dB

16. Attenuation
One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two.
In the above figure a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as

17. Attenuation
dB = –3 + 7 – 3 = +1

18. Distortion
Distortion means that signal changes its form or shape.
Because the velocity of propagation of a signal through medium varies with frequency.
Thus various frequency components of a signal will arrive at the receiver at different times, resulting in phase shifts between different frequencies.

19. Distortion(change form/shape)

20. Noise
Additional signals inserted between transmitter and receiver – external energy

21. Noise
Four types of noise .
Thermal noise.
Cross talk.
Impulse noise.

22. Thermal Noise
Thermal noise : thermal noise is due the thermal agitation of electrons .it is present in all thermal devices and transmission media. It is function of temperature

23. Cross talk
Crosstalk is the effect of one wire on another

24. Induced Noise
Induced noise – comes from sources such as motors or appliances – devices act as a sending antenna and transmission media acts as receiver

25. Impulse noise
Impulse noise – a spike – comes from power lines, lightning, etc..

26. Signals ……. Throughput Propagation speed Propagation time Wavelength
How fast data can pass through an entity [a point or a network]
Propagation speed
The distance a signal or a bit can travel through a medium in one second
In a vacuum, light is propagated with a speed of 3x108m/s
Propagation time
Time required for a signal or a bit to travel from one point of transmission medium to another
Calculated by dividing the distance by the propagation speed
Wavelength
The distance a simple signal can travel in one period
Propagation speed divided by the frequency

27. Throughput
Propagation speed
Propogation Time
Wavelength End of the Lecture