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Assume we have communication system between transmit and receive over wireless channel.

The Friis path loss model is given by the following equation

$$P_r = P_t G_t G_r (\frac{\lambda}{4\pi D})^2$$

Where $P_{r(t)}$ is the received (transmit) power, $G_{t(r)}$ is the gain of the antennas at transmitter (receive) and $\lambda $ is the signal wavelength while $D$ is the distance between transmit receive ends.

In theoretical (academia) and simulation scenarios (for example MATLAB)I have come across examples wireless channel where a power delay profile is defined, that is time of arrival (TOA) of rays versus amplitude of those rays. For example, assume I have five rays arriving and delay spread $40 ns$ with amplitude

$$channel= [ 0.5+i0.5, 0.3+i0.3, 0.2+i0.2, 0.1+i 0.1, 0.3 ] $$

$$TOA= [0, 10*10^{-9} , 20*10^{-9}, 30*10^{-9} , 40*10^{-9}]$$

My simple question is: How are the two concepts related??

Thanks

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  • $\begingroup$ If your last ray has an amplitude of 0, does it really arrive? $\endgroup$ – Peter K. Nov 16 '15 at 19:20
  • $\begingroup$ Henry, I am trying to understand your question. The typo stopped me understanding your question. Relax! $\endgroup$ – Peter K. Nov 16 '15 at 19:31
  • $\begingroup$ both are different. One is short term fading and other is the long term fading. The gains obtained by the short term fades are not captured in the path loss equation. The gains arise from shadowing, diffraction,dispersion,reflection,etc. $\endgroup$ – phanitej Jul 13 '17 at 9:43
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Basically, multipath propagation causes time dispersion (hence causing a delay spread ($\sigma_{\tau}$)) and that dispersion influences the fading characteristics of the wireless channel.So delay spread is related to the fading characteristics of a channel and not really to the path loss. Particularly, if $\sigma_{\tau}$ is greater than the symbol period ($T_s$) then the wireless channel is said to be frequency selective, meaning that a signal will interfere with itself causing additional fading.

You should also note that the rms delay spread is not the maximum delay that your signal experiences as I think you may asssume, $\sigma_{\tau}$ can be calculated as follows:

$\sigma_{\tau} = \sqrt{\overline{\tau ^2} - \overline{\tau} ^2}$

where

$$\overline{\tau} = \frac{\Sigma P(\tau _i) \tau _i}{\Sigma P(\tau _i)},$$ $$\overline{\tau ^2} = \frac{\Sigma P(\tau _i) \tau _i ^2}{\Sigma P(\tau _i)}$$ and $P(\tau _i)$ is the power of the component with delay $\tau _i$

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So your channel is (rephrasing your numbers) : $$ h(t) = (0.5 + i0.5)\delta(t) + (0.3+i0.3)\delta(t-10) + (0.2+i0.2) \delta(t-20) + (0.1+i0.1) \delta(t-30) + 0.3\delta(t-40) $$ where $t$ is the time in nanoseconds.

This says that there are five ways to get from the transmitter to the receiver: a direct path (I'm assuming the $t=0$ version is direct), and four longer paths.

The Friis path loss model says that, given the transmit power, the Tx and Rx antenna gains, the wavelength, and the distance then the attenuation due to just the transmission path will be given by the equation.

How does this relate to your channel?

All it means is that each coefficient of the various delay elements comprises partly of an attenuation due to the path loss. Each coefficient will have a different attenuation because each coefficient corresponds to a different path distance --- so the value of $D$ for each coefficient will be different.

The coefficient with the smallest path loss component will be the $0.5 + i0.5$ first one ("zero" delay), and the coefficient with the largest path loss component will be the $0.3$ one (delay of 40ns).

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  • $\begingroup$ thank you. Do you mean thag given the power delay profile I can compute the receid power of each ray according to friis law? $\endgroup$ – Henry Nov 16 '15 at 19:58
  • $\begingroup$ There are many possible causes of path loss. Friis just tries to capture a simplified version related to a few parameters. For a simplification, you could assume the first raw is just due to Friis, but the others will include a component of reflection attenuations... which will be hard to find. $\endgroup$ – Peter K. Nov 16 '15 at 20:03

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