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Shannon Theorem Statement: - In information theory, the Shannon–Hartley theorem tells the maximum rate at which information can be transmitted over a communications channel of a specified bandwidth in the presence of noise. C = B log 2 (1 + SNR) Prepared By : Abdul Samad Abb

Shanon theorem and Free space Path loss theory

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This slide is about shanon theorem ,its applications and its limitations. it also gives an overview of Free Space Path Loss(fspl)

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Page 1: Shanon theorem and Free space Path loss theory

Shannon Theorem

Statement:- In information theory, the Shannon–Hartley

theorem tells the maximum rate at which information can be

transmitted over a communications channel of a specified

bandwidth in the presence of noise.

C = B log2(1 + SNR)

Prepared By : Abdul Samad Abbasi

Page 2: Shanon theorem and Free space Path loss theory

3.2

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 C is calculated as:

Example 1

This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.

Page 3: Shanon theorem and Free space Path loss theory

3.3

We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is

usually 3162. For this channel the capacity is calculated as

Example 2

This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or

improve the signal-to-noise ratio.

Page 4: Shanon theorem and Free space Path loss theory

3.4

The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be

calculated as:

Example 3

Page 5: Shanon theorem and Free space Path loss theory

If it is required to transmit at 50 kbit/s, and a bandwidth of 1 MHz is used, then the minimum SNR required is given by: C = B log2(1 + SNR) 50 = 1000 log2(1+S/N) So S/N= 0.035 corresponding to an SNR of -14.5 dB. This shows that it is possible to transmit using signals which are actually much weaker than the background noise level.

Example 4

Page 6: Shanon theorem and Free space Path loss theory

Why 64-QAM Is Better And What Are Its Disadvantages?

64-QAM is better because it transfer 6 bits per symbol and this is higher than the values of BPSK,QPSK,8-PSK,16QAM.

Page 7: Shanon theorem and Free space Path loss theory

Disadvantages

For Binary we require simple circuitry but for 64-QAM we require complex DSP.

Increase in transfer of bits per symbols also increases error rate.

Page 8: Shanon theorem and Free space Path loss theory

Free Space Path Loss

Statement:-Free-space path loss (FSPL) is the loss in signal strength of an electromagnetic wave that would result from a line-of-sight path through free space (usually air), with no obstacles nearby

to cause reflection or diffraction.

Page 9: Shanon theorem and Free space Path loss theory

Derivation

For free propagation waves in radio channel, the path loss model is:

Where

=

=

=

=

Page 10: Shanon theorem and Free space Path loss theory

What is Digital Communication

Digital Communication is information transmitted electronically and is encoded

digitally. It is passed through digital devices such as computer. Digital communication

are fast efficient and reliable.

Examples of digital communication are Email, faces and phone text.

Page 11: Shanon theorem and Free space Path loss theory

Why Is Modulation Done?

We can send signals of different frequencies through same channel without interference.

Height of antenna depends on the frequency of signal

λ=c/f Height of Antenna= λ/4For Example if 1Hz signal is transmitted Then Height of antenna should be equal to 75x10⁶But if it is modulated with carrier wave of 88MHz then height of antenna would be equal to 0.85m.