Explain Fading Techniques. Explain Hamming code with Example.
Fading Techniques:
1) Flat Fading:
Such types of fading occur when the bandwidth of the
transmitted signal is less than the coherence
bandwidth of the channel. Equivalently if the symbol
period of the signal is more than the rms delay spread of the channel,
then the fading is at fading.
So we can say that at fading occurs
when;
BS << BC
Where BS is the signal bandwidth and BC is the coherence bandwidth. Also,
TS >>στ
Where TS is the symbol period and στ is the rms delay spread. And in such
a case, mobile channel has a constant gain and linear phase response over its
bandwidth.
2) Frequency
Selective Fading:
Frequency selective fading occurs when the signal
bandwidth is more than the coherence bandwidth of the mobile radio channel or
equivalently the symbols duration of the signal is less than the rms delay spread.
BS >> BC and TS <<στ
· At the receiver, we obtain multiple copies of the
transmitted signal, all attenuated and delayed in time. The channel introduces
inter symbol interference.
A rule of thumb for a channel to have at fading is if στ /TS<= 0.1
3) Fast Fading:
In a fast fading channel, the channel impulse response changes rapidly within the symbol duration of the signal. Due to Doppler spreading, signal undergoes frequency dispersion leading to distortion. Therefore a signal undergoes fast fading if
TS >> TC
Where TC is the coherence time and
BS >> BD
Where BD is the Doppler spread. Transmission involving very low data rates suffers from fast fading.
4) Slow Fading:
In such a channel, the rate of the change of the channel impulse response is much less than the transmitted signal.
We can consider a slow faded channel a channel in which channel is almost constant over at least one symbol duration. Hence
TS << TC and BS >> BD
We observe that the velocity of the user plays an important role in deciding whether the signal experiences fast or slow fading.
Hamming code :
Hamming code is a set of error-correction codes that can be used to detect and correct the errors that can occur when the data is moved or stored from the sender to the receiver.
Redundant bits are extra binary bits that are generated and added to the carrying bits of data transfer.
The number of redundant bits can be calculated by:
𝟐𝒓> 𝒎 + 𝒓 + 𝟏
where, r = redundant bit, m = data bit
Example, the number of data bits is 7, then the number of redundant bits can be calculated using:
𝟐𝟒> 𝟕 + 𝟒 + 𝟏
Thus, the number of redundant bits= 4
Parity bits:
A parity bit is a bit appended to a data of binary bits to ensure that the total number of 1’s in the data is even or odd. Parity bits are used for error detection. There are two types of parity bits:
1) Even parity bit:
In the case of even parity, for a given set of bits, the number of 1’s are counted. If that count is odd, the parity bit value is set to 1, making the total count of occurrences of 1’s an even number. If the total number of 1’s in a given set of bits is already even, the parity bit’s value is 0.
2) Odd Parity bit:
In the case of odd parity, for a given set of bits, the number of 1’s are counted. If that count is even, the parity bit value is set to 1, making the total count of occurrences of 1’s an odd number. If the total number of 1’s in a given set of bits is already odd, the parity bit’s value is 0.
Determining the position of redundant bits:
These redundancy bits are placed at the positions which correspond to the power of 2.
Example : Data = 1011001
As in the above example:
The number of data bits = 7
The number of redundant bits = 4
The total number of bits = 11
The redundant bits are placed at positions corresponding to power of 2- 1, 2, 4, and 8.
Suppose the data to be transmitted is 1011001, the bits will be placed as follows:Determining the Parity bits:
1. R1 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the least significant position.
R1: bits 1, 3, 5, 7, 9, 11
To find the redundant bit R1, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R1 is an even number the value of R1 (parity bit’s value) = 0
2. R2 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the second position from the least significant bit.R2: bits 2,3,6,7,10,11
To find the redundant bit R2, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R2 is odd the value of R2(parity bit’s value)=1
3. R4 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the third position from the least significant bit.
R4: bits 4, 5, 6, 7
To find the redundant bit R4, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R4 is odd the value of R4(parity bit’s value) = 1
4. R8 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the fourth position from the least significant bit.
R8: bit 8,9,10,11
To find the redundant bit R8, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R8 is an even number the value of R8(parity bit’s value)=0.
Thus, the data transferred is:
Error detection and correction: Suppose in the above example the 6th bit is changed from 0 to 1 during data transmission, then it gives new parity values in the binary number:
The bits give the binary number as 0110 whose decimal representation is 6. Thus, the bit 6 contains an error. To correct the error the 6th bit is changed from 1 to 0.
Follow Us