0
23kviews
Types of Small Scale Fading
1 Answer
2
502views

The type of fading experienced by a signal propagating through a mobile radio channel depends on the nature of the transmitted signal with respect to the characteristics of the channel. Depending on the relation between the signal parameters (such as bandwidth, symbol period, etc.) and the channel parameters (such as rms delay spread and Doppler spread), different transmitted signals will undergo different types of fading. The time dispersion and frequency dispersion mechanisms in a mobile radio channel lead to four possible distinct effects, which are manifested depending on the nature of the transmitted signal, the channel, and the velocity. While multipath delay spread leads to time dispersion and frequency selective fading, Doppler spread leads to frequency dispersion and time selective fading. The two propagation mechanisms are independent of one another. Figure 13 shows a tree of the two different types of fading.

enter image description here

Figure 13: Classification of small scale fading based on time delay spread

1. Fading Effects Due to Multipath time Delay Spread

Time dispersion due to multipath causes the transmitted signal to undergo either flat or frequency selective fading.

i) Flat fading

If the mobile radio channel has a constant gain and linear phase response over a bandwidth which is greater than the bandwidth of the transmitted signal, then the received signal will undergo flat fading. This type of fading is the most common type of fading. In flat fading, the multipath structure of the channel is such that the spectral characteristics of the transmitted signal are preserved at the receiver. The response of a flat fading channel for a pulse of duration $T_s$ is illustrated in Figure 14.

enter image description here

Figure 14: Response of a flat fading channel

It can be seen from Figure 14, that if the channel gain changes over time, a change of amplitude occurs in the received signal. Over time, the received signal r (t) varies in gain, but the spectrum of the transmission is preserved. The delay spread $\tau$ introduced by the channels much smaller than the symbol period Ts. Hence the total symbol is received in an interval of $\mathrm{T} \mathrm{s}+\tau$ which is similar to the transmitted symbol. Hence chances if Intersymbol interference is minimum. Similarly , the signal spectrum S(f) is much smaller than the coherence bandwidth of the channel. Hence , on the receiver side the signal spectrum is undistorted.

Flat fading channels are also known as amplitude varying channels and are sometimes referred to as narrowband channels, since the bandwidth of the applied signal is narrow as compared to the channel flat fading bandwidth. Typical flat fading channels cause deep fades, and thus may require 20 or 30 dB more transmitter power to achieve low bit error rates during times of deep fades as compared to systems operating over non-fading channels. The distribution of the instantaneous gain of flat fading channels is important for designing radio links, and the most common amplitude distribution is the Rayleigh distribution.

enter image description here

Figure 15:Response of a frequency selective fading channel

ii) Frequency Selective Fading

If the channel possesses a constant-gain and linear phase response over a bandwidth that is smaller than the bandwidth of transmitted signal, then the channel creates frequency selective fading on the received signal. Under such conditions the channel impulse response has a multipath delay spread which is greater than the reciprocal bandwidth of the transmitted message waveform. When this occurs, the received signal includes multiple versions of the transmitted waveform which are attenuated (faded) and delayed in time, and hence the received signal is distorted. Frequency selective fading is due to time dispersion of the transmitted symbols within the channel. Thus the channel induces intersymbol interference (ISI). Frequency selective fading channels are much more difficult to model than flat fading channels since each multipath signal must be modeled and the channel must be considered to be a linear filter. Summarizing, the following two conditions should be true:

$B_{S} \gt B_{C}$

$T_{S}\lt\sigma_{\tau}$

2. Fading Effects Due to Doppler Spread

enter image description here

Figure 16: Small scale fading based on Doppler spread

Depending on how rapidly the transmitted baseband signal changes as compared to the rate of change of the channel, a channel may be classified either as a fast fading or slow fading channel.

i) Fast fading

In a fast fading channel,the coherence time of the channel is smaller than the symbol period of the transmitted signal. This causes frequency dispersion (also called time selective fading) due to Doppler spreading, which leads to signal distortion. Viewed in the frequency domain, signal distortion due to fast fading increases with increasing Doppler spread relative to the bandwidth of the transmitted signal. Therefore, a signal undergoes fast fading if

$T_{s} \gt T_{c}$

$B_{S} \lt B_{D}$

ii) Slow fading

In a slow fading channel, the channel impulse response changes at a rate much slower than the transmitted baseband signal s(t). In this case, the channel may be assumed to be static over one or several reciprocal bandwidth intervals. In the frequency domain, this implies that the Doppler spread of the channel is much less than the bandwidth of the baseband signal. Therefore, a signal undergoes slow fading if

$T_{S} \gt \gt T_{C}$

$B_{S} \lt \lt B_{D}$

It should be noted that when a channel is specified as a fast or slow fading channel, it does not specify whether the channel is flat fading or frequency selective in nature. Fast fading only deals with the rate of change of the channel due to motion. The relation between the various multipath parameters and the type of fading experienced by the signal are summarized in figure 17.

enter image description here

Figure 17

Examples:

1) Calculate the mean excess delay, rms delay spread, and the maximum excess delay (10 dB) for the multipath profile given in the figure below. Estimate the 50 $\%$ coherence bandwidth of the channel. Would this channel be suitable for AMPS or GSM service without the use of an equalizer?

$\rightarrow$ enter image description here

$\bar{\tau} = \frac{0.01 (0) + 0.1 (1) + (0.2) + 1 (5)}{0.01 + 0.1 \times 2 + 1}$

$\bar{\tau} = \frac{5.3}{1.21} = 4.38 \mu s$

$ \bar{\tau^2} = \frac{0.01(0)^2 + 0.1 (1)^2 + 0.1 (2)^2 + 1(5)^5}{0.01 + 0.1 \times 2 +1 }$

$= \frac{25.5}{1.21} = 21.07 \mu s^2$

$\sigma_{\tau} = \sqrt{ \bar{\zeta^2} – ( \bar{\zeta)^2}}$

$\sigma_{\tau} = 1.373 \mu s$

Since $50 \%$ correlation is accepted

$B_c = \frac{1}{ 5 \ \sigma_{\tau}}$

= 145. 66 KHz.

For an AMPS signal $B_s$ = 30 KHz here $B_s \lt \lt B_c$ hence it will be a faded signal. Hence use of equalizer is not necessary.

For GSM signal $B_c$ = 200 KHz. Here $B_s \gt \gt B_c$ hence it will be a frequency selective faded signal. Hence equalizer is required.

2) Consider a transmitter which radiates a sinusoidal carrier frequency of 1850 MHz. For a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving

(a) directly towards the transmitter,

(b) directly away from the transmitter,

(c) in a direction which is perpendicular to the direction of arrival of the transmitted signal.

$\rightarrow$ Carrier frequency $=1850 \mathrm{MHz}$

Therefore, wavelength $\mathrm{\lambda}=\mathrm{c} / \mathrm{f}=0.162 \mathrm{m}$

=$1850 \times 10$

Vehicle speed $v=60 \mathrm{mph}=26.82 \mathrm{m} / \mathrm{s}$

(a) The vehicle is moving directly towards the transmitter.

The Doppler shift in this case is positive and the received frequency is given

by $\mathrm{f}=\mathrm{fc}+\mathrm{fd}$

$=1850.00016 \mathrm{MHz}$

(b) The vehicle is moving directly away from the transmitter.

The Doppler shift in this case is negative and hence the received frequency is given by

$f=f_{c}-f_{d}=1850 \times 10^{6}-\frac{26.82}{0.162}=1849.999834 \mathrm{MHz}$

The vehicle is moving perpendicular to the angle of arrival of the transmitted signal.

In this case, $\mathrm{S}=90^{\circ}, \cos \theta=0$ , and there is no Doppler shift.

The received signal frequency is the same as the transmitted frequency of 1850 MHz

Please log in to add an answer.