# 1.3: RF Power Calculations

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## 1.3.1 RF Propagation

As an RF signal propagates away from a transmitter the power density reduces conserving the power in the EM wave. In the absence of obstacles and without atmospheric attenuation the total power passing through the surface of a sphere centered on a transmitter is equal to the power transmitted. Since the area of the sphere of radius $$r$$ is $$4\pi r^{2}$$, the power density, e.g. in $$\text{W/m}^{2}$$, at a distance $$r$$ drops off as $$1/r^{2}$$. With obstacles the EM wave can be further attenuated.

Example $$\PageIndex{1}$$: Signal Propagation

A signal is received at a distance $$r$$ from a transmitter and the received power drops off as $$1/r^{2}$$. When $$r = 1\text{ km}$$, $$100\text{ nW}$$ is received. What is $$r$$ when the received power is $$100\text{ fW}$$?

Solution

The signal collected by the receiver is proportional to the power density of the EM signal. The received signal power $$P_{r} = k/r^{2}$$ where $$k$$ is a constant. This leads to

$\label{eq:1}\frac{P_{r}(1\text{ km})}{P_{r}(r)}=\frac{100\text{ nW}}{100\text{ fW}}=10^{6}=\frac{kr^{2}}{k(1\text{ km})^{2}}=\frac{r^{2}}{(10^{3}\text{ m})^{2}};\quad r=\sqrt{10^{12}\text{ m}^{2}}=1000\text{ km}$

## 1.3.2 Logarithm

A cellular phone can reliably receive a signal as small as $$100\text{ fW}$$ and the signal to be transmitted could be $$1\text{ W}$$. So the same circuitry can encounter signals differing in power by a factor of $$10^{13}$$. To handle such a large range of signals a logarithmic scale is used.

Logarithms are used in RF engineering to express the ratio of powers using reasonable numbers. Logarithms are taken with respect to a base $$b$$ such that if $$x = b^{y}$$, then $$y = \log_{b}(x)$$. In engineering, $$\log(x)$$ is the same as $$\log_{10}(x)$$, and $$\ln (x)$$ is the same as $$\log_{e}(x)$$ and is called the natural logarithm ($$e = 2.71828\ldots$$). In physics and mathematics $$\log x$$ (and programs such as MATLAB) means $$\ln x$$, so be careful. Common formulas involving logarithms are given in Table $$\PageIndex{1}$$.

Description Formula Example
Equivalence $$y=\log _{b}(x)\longleftrightarrow x=b^{y}$$ $$\log (1000) =3\text{ and } 10^{3}=1000$$
Product $$\log_{b}(xy)=\log_{b}(x)+\log_{b}(y)$$ $$\log (0.13\cdot 978) = \log (0.13) + \log (978) = −0.8861 + 2.990=2.104$$
Ratio $$\log_{b}(x/y)=\log_{b}(x)-\log_{b}(y)$$ $$\ln (8/2) = \ln (8) − \ln (2) = 3 − 1=2$$
Power $$\log_{b}(x^{p})=p\log_{b}(x)$$ $$\ln (3^{2})= 2 \ln (3) = 2\cdot 1.0986 = 2.197$$
Root $$\log_{b}(\sqrt[p]{x} )=\frac{1}{p}\log_{b} (x)$$ $$\log (\sqrt[3]{20})=\frac{1}{3}\log (20)=0.4337$$
Change of base $$\log_{b}(x)=\frac{\log_{k}(x)}{\log_{k}(b)}$$ $$\ln (100)=\frac{\log (100)}{\log (2)}=\frac{2}{0.30103}=6.644$$

Table $$\PageIndex{1}$$: Common logarithm formulas. In engineering $$\log x ≡ \log_{10} x$$ and $$\ln x ≡ log_{2} x$$.

## 1.3.3 Decibels

RF signal levels are expressed in terms of the power of a signal. While power can be expressed in absolute terms, e.g. watts ($$\text{W}$$), it is more useful to use a logarithmic scale. The ratio of two power levels $$P$$ and $$P_{\text{REF}}$$ in bels ($$\text{B}$$) is

$\label{eq:2}P(B)=\log\left(\frac{P}{P_{\text{REF}}}\right)$

where $$P_{\text{REF}}$$ is a reference power. Here $$\log x$$ is the same as $$\log_{10} x$$. Human senses have a logarithmic response and the minimum resolution tends to be about $$0.1\text{ B}$$, so it is most common to use decibels ($$\text{dB}$$); $$1\text{ B} = 10\text{ dB}$$. Common designations are shown in Table $$\PageIndex{2}$$. Also, $$1\text{ mW} = 0\text{ dBm}$$ is a very common power level in RF and microwave power circuits where the $$\text{m}$$ in $$\text{dBm}$$ refers to the $$1\text{ mW}$$ reference. As well, $$\text{dBW}$$ is used, and this is the power ratio with respect to $$1\text{ W}$$ with $$1\text{ W} = 0\text{ dBW} = 30\text{ dBm}$$.

$$P_{\text{REF}}$$ Bell units Decibel units
$$1\text{ W}$$ $$\text{BW}$$ $$\text{dBW}$$
$$1\text{ mW} = 10^{-3}\text{ W}$$ $$\text{Bm}$$ $$\text{dBm}$$
$$1\text{ fW} = 10^{-15}\text{ W}$$ $$\text{Bf}$$ $$\text{dBf}$$

Table $$\PageIndex{2}$$a: Common power designations (a) Reference powers, $$P_{\text{REF}}$$

Power ratio in $$\text{dB}$$
$$10^{-6}$$ $$-60$$
$$0.001$$ $$-30$$
$$0.1$$ $$-20$$
$$1$$ $$0$$
$$10$$ $$10$$
$$1000$$ $$30$$
$$10^{6}$$ $$60$$

Table $$\PageIndex{2}$$b: Common power designations (b) Power ratios in decibels ($$\text{dB}$$)

Power Absolute power
$$-120\text{ dBM}$$ $$10^{-12}\text{ mW} = 10^{-15}\text{ W} = 1\text{ fW}$$
$$0\text{ dBm}$$ $$1\text{ mW}$$
$$10\text{ dBm}$$ $$10\text{ mW}$$
$$20\text{ dBm}$$ $$100\text{ mW} = 0.1\text{ W}$$
$$30\text{ dBm}$$ $$1000\text{ mW} = 1\text{ W}$$
$$40\text{ dBm}$$ $$10^{4}\text{ mW} = 10\text{ W}$$
$$50\text{ dBm}$$ $$10^{5}\text{ mW} = 100\text{ W}$$
$$-90\text{ dBm}$$ $$10^{-9}\text{ mW} = 10^{-12}\text{ W} = 1\text{ pW}$$
$$-60\text{ dBm}$$ $$10^{-6}\text{ mW} = 10^{-9}\text{ W} = 1\text{ nW}$$
$$-30\text{ dBm}$$ $$0.001\text{ mW} = 1\:\mu\text{W}$$
$$-20\text{ dBm}$$ $$0.01\text{ mW} = 10\:\mu\text{W}$$
$$-10\text{ dBm}$$ $$0.1\text{ mW} = 100\:\mu\text{W}$$

Table $$\PageIndex{2}$$c: Common power designations (c) Powers in $$\text{dBm}$$ and watts

Example $$\PageIndex{2}$$: Power Gain

An amplifier has a power gain of $$1200$$. What is the power gain in decibels? If the input power is $$5\text{ dBm}$$, what is the output power in $$\text{dBm}$$?

Solution

Power gain in decibels, $$G_{\text{dB}} = 10 \log 1200 = 30.79\text{ dB}$$.

The output power is $$P_{\text{out|dBm}} = P_{\text{dB}} + P_{\text{in|dBm}} = 30.79 + 5 = 35.79\text{ dBm}$$.

Example $$\PageIndex{3}$$: Gain Calculations

A signal with a power of $$2\text{ mW}$$ is applied to the input of an amplifier that increases the power of the signal by a factor of $$20$$.

Figure $$\PageIndex{1}$$

1. What is the input power in $$\text{dBm}$$?
$\label{eq:3}P_{\text{in}}=2\text{ mW} = 10\cdot\log\left(\frac{2\text{ mW}}{1\text{ mW}}\right) = 10\cdot\log (2) = 3.010\text{ dBm}\approx 3.0\text{ dBm}$
2. What is the gain, $$G$$, of the amplifier in $$\text{dB}$$?
The amplifier gain (by default this is power gain) is
$\label{eq:4}G=20=10\cdot\log (20)\text{ dB}=10\cdot 1.301\text{ dB}=13.0\text{ dB}$
3. What is the output power of the amplifier?
$\label{eq:5}G=\frac{P_{\text{out}}}{P_{\text{in}}},\quad\text{and in decibels }G|_{\text{dB}}=P_{\text{out}}|_{\text{dBm}}-P_{\text{in}}|_{\text{dBm}}$
Thus the output power in $$\text{dBm}$$ is
$\label{eq:6}P_{\text{out}}|_{\text{dBm}}=G|_{\text{dB}}+P_{\text{in}}|_{\text{dBm}}=13.0\text{ dB} +3.0\text{ dBm} =16.0\text{ dBm}$
Note that $$\text{dB}$$ and $$\text{dBm}$$ are dimensionless but they do have meaning; $$\text{dB}$$ indicates a power ratio but $$\text{dBm}$$ refers to a power. Quantities in $$\text{dB}$$ and one quantity in $$\text{dBm}$$ can be added or subtracted to yield $$\text{dBm}$$, and the difference of two quantities in $$\text{dBm}$$ yields a power ratio in $$\text{dB}$$.

In Examples $$\PageIndex{2}$$ and $$\PageIndex{3}$$ two digits following the decimal point were used for the output power expressed in $$\text{dBm}$$. This corresponds to an implied accuracy of about $$0.01\%$$ or $$4$$ significant digits of the absolute number. This level of precision is typical for the result of an engineering calculation.

Example $$\PageIndex{4}$$: Power Calculations

The output stage of an RF front end consists of an amplifier followed by a filter and then an antenna. The amplifier has a gain of $$33\text{ dB}$$, the filter has a loss of $$2.2\text{ dB}$$, and of the power input to the antenna, $$45\%$$ is lost as heat due to resistive losses. If the power input to the amplifier is $$1\text{ W}$$, then:

Figure $$\PageIndex{2}$$

1. What is the power input to the amplifier expressed in $$\text{dBm}$$?
$$P_{\text{in}}=1\text{ W}=1000\text{ mW},\quad P_{\text{dBm}}=10\log (1000/1)=30\text{ dBm}$$
2. Express the loss of the antenna in $$\text{dB}$$.
$$45\%$$ of the power input to the antenna is dissipated as heat.
The antenna has an efficiency, $$\eta$$, of $$55\%$$ and so $$P_{2} = 0.55P_{1}$$.
$$\text{Loss} = P_{1}/P_{2} = 1/0.55 = 1.818 = 2.60\text{ dB}$$.
3. What is the total gain of the RF front end (amplifier + filter + antenna)?
$\label{eq:7}\text{Total gain }= (\text{amplifier gain})_{\text{dB}}+(\text{filter gain})_{\text{dB}}-(\text{loss of antenna})_{\text{dB}}=(33-2.2-2.6)\text{ dB}=28.2\text{ dB}$
4. What is the total power radiated by the antenna in $$\text{dBm}$$?
$\label{eq:8}P_{r}=P_{\text{in|dBm}}+(\text{amplifier gain})_{\text{dB}}+(\text{filter gain})_{\text{dB}}-(\text{loss of antenna})_{\text{dB}}=30\text{ dBm}+(33-2.2-2.6)\text{ dB}=58.2\text{ dBm}$
5. What is the total power radiated by the antenna?
$\label{eq:9}P_{R}=10^{58.2/10}=(661\times 10^{3})\text{ mW}=661\text{ W}$

This page titled 1.3: RF Power Calculations is shared under a CC BY-NC license and was authored, remixed, and/or curated by Michael Steer.