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I have a simple question that confuses me.

Assume that we have the following results of a measurement campaign, distance (km) and the corresponding path loss $L_{e}$ (dB). I want to estimate the path loss exponent of the simplified path loss model

$$L_{m,\text{dB}} = K_\text{dB} - 10 n \log_{10}\left(\frac{d}{d_0}\right)\text ,$$

where $K$ is a constant that depends on antenna characteristics and average channel attenuation at reference distance $d_0$ and is given by $K_\text{dB} = 20 \log_{10}\left(\frac{\lambda}{4 \pi d_0}\right)$ or $K_\text{dB} = 20 \log_{10}\left(\frac{\lambda \sqrt{G_t G_r}}{4 \pi d_0}\right)$, where $G_t$ and $G_r$ are the transmit and receive antenna gains, respectively, and $\lambda$ is the wavelength.

To estimate the path loss exponent $n$, I am finding $n$ that minimizes the MMSE error for the dB power measurement

$$\begin{align} \text{MMSE}(n) &= \sum_i^N \left[L_e(d_i) - L_m(d_i)\right]^2\\ &= \sum_i^N{ \left[L_e(d_i) - K + 10 n \log_{10}\left(\frac{d_i}{d_0}\right)\right]^2}\text , \end{align}$$

where $N$ is the total number of measurements.

  • It seems the exact value of $n$ depends on the reference distance $d_0$ and the antenna gains $G_t$ and $G_r$, right?
  • In other words, for different $d_0, G_t, G_r$, I will have different value for $n$. Is this in conflict with the fact that on a log-log scale, the path loss is represented by a straight line with a slope of $10 n$ per decade?
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  • $\begingroup$ I believe the path loss exponent depends only on the propagation environment; it shouldn't depend on the distance or the antennas. I'm not an expert on this, but why don't you try a simple curve fit instead of applying MMSE? $\endgroup$ – MBaz Feb 25 '17 at 18:35
  • $\begingroup$ But curve fitting techniques minimizes the MMSE, right? So, it should be the same thing. $\endgroup$ – Noor Feb 25 '17 at 18:39
  • $\begingroup$ Well, there are many fitting techniques, the simpler ones do not minimize the MMSE as far as I know, but I don't know much :) en.wikipedia.org/wiki/Curve_fitting $\endgroup$ – MBaz Feb 25 '17 at 19:48
  • $\begingroup$ What I'm sure of is that $n$ does not depend on the distance or the antennas. $\endgroup$ – MBaz Feb 25 '17 at 19:49
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The path loss exponent does not depend on the reference distance. There are two mistakes in your path loss formula. First, path loss does not include the gains of transmit or receive antennas. Second, the path loss exponent must also be part of the expression for the path loss at the reference distance.

According to the Friis transmission equation the ratio between the received power $P_r$ and the transmitted power $P_t$ is given by

$$\frac{P_r}{P_t}=G_tG_r\left(\frac{\lambda}{4\pi d}\right)^2\tag{1}$$

where $G_t$ and $G_r$ are the gains of the transmitter and receiver antennas, respectively, $\lambda$ is the wavelength, and $d$ is the distance between the transmit and receive antennas. The squared term in parentheses (or its inverse) is the free-space path loss. Note that it does not include the antenna gains. In order to account for other environments than free-space, a path loss exponent $n$ is introduced:

$$\frac{P_r}{P_t}=G_tG_r\left(\frac{\lambda}{4\pi}\right)^2\frac{1}{d^n}\tag{2}$$

In free space we have $n=2$; in lossy environments, $n$ can be greater than $2$.

If we introduce a reference distance $d_0$, Eq. $(2)$ becomes

$$\frac{P_r}{P_t}=G_tG_r\left(\frac{\lambda}{4\pi}\right)^2\frac{1}{d_0^n}\left(\frac{d_0}{d}\right)^n\tag{3}$$

In dB we get

$$\begin{align}10\log_{10}\left(\frac{P_r}{P_t}\right)&=10\log_{10}(G_t)+10\log_{10}(G_r)+\ldots\\&\ldots +20\log_{10}\left(\frac{\lambda}{4\pi}\right)-10n\log_{10}(d_0)-10n\log_{10}\left(\frac{d}{d_0}\right)\end{align}\tag{4}$$

where

$$PL_0=20\log_{10}\left(\frac{\lambda}{4\pi}\right)-10n\log_{10}(d_0)\tag{5}$$

is the path loss at the reference distance $d_0$. Note that in $(4)$ the reference distance $d_0$ will cancel out, so the estimate of the path loss exponent $n$ will not depend on $d_0$. On the other hand, the estimate of $n$ will have a (weak) dependence on the antenna gains. This is however not in conflict with the fact that the path loss as a function of distance is a straight line on a log-log plot.

Note that in practice, the path loss at reference distance $d_0$ is often just set to the free-space path loss, i.e using $n=2$ in Eq. $(5)$ $[1]$. In that case the reference distance $d_0$ will no longer cancel out in $(4)$ and the estimate of the path loss exponent $n$ will depend on the choice of $d_0$. Alternatively, $PL_0$ can also be measured at distance $d_0$, or its can be optimized together with $n$ to obtain an optimal fit to measured data $[1]$.

[1] Wireless Communications, A. Goldsmith, 2005 Cambridge University Press, p. 41.

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  • $\begingroup$ Can you please provide a reference for Eqn. (3) or (5) or the statement "the path loss exponent must also be part of the expression for the path loss at the reference distance"? I am asking this as the books I have does not have this dependency. For instance, in the book of "wireless communications" by Rapport published in 2002, Section 3.9.1 page 104, it is mentioned "the reference path loss is calculated using the free space path loss formula" which is for n = 2. In other words, the free space path loss formula is not a function of n. The same discussion is also in Goldsmith book in ch. 2. $\endgroup$ – Noor Feb 26 '17 at 17:23
  • $\begingroup$ @Noor: Eq. (3) is just the same as Eq. (2) with a reference distance included. Eq. (5) is what you get if you write everything in dB and leave out the gain terms and the contribution of the path loss depending on $d$. This is how it should be done, in which case there's no dependence on $d_0$. I've added a paragraph on how it sometimes is done, in which case you are right that there is a dependence on $d_0$ (which shouldn't be the case if one does things right). $\endgroup$ – Matt L. Feb 27 '17 at 8:18
  • $\begingroup$ Thanks for clarifications. My understanding is, by definition, $PL_0$ is calculated using the free space path loss formula. What you are saying (although it makes sense), it does not follow the definition of $PL_0$ (i.e., reference distance close to the transmitter and hence follow free space path loss). I am still a bit confused. $\endgroup$ – Noor Feb 27 '17 at 18:46
  • $\begingroup$ @Noor: I'm not sure if everybody defines $PL_0$ like that; as mentioned by Goldsmith, $PL_0$ can also be measured or even optimized to get the best fit. I think using the free-space path loss formula is just a very easy and straightforward method, that's why it's probably used. $\endgroup$ – Matt L. Feb 27 '17 at 18:50

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