# Objective 3.5 — The Friis Equation | LO# | Description | |----------|----------| | 3.5 | I can calculate the maximum communication distance between a transmitter and receiver using the Friis and Line-of-Sight equations. | By the end of this reading you should be able to: 1. Write the **Friis equation** and explain where each term comes from. 2. Predict how received power changes with power, gain, frequency, or range. 3. Solve the Friis equation for the **maximum range** at which a receiver still gets the power it needs. 4. Work a complete **radio communication problem**, combining Friis with LOS. ## Friis Equation Introduction Previously, we discussed the four wireless RF propagation methods. We left off with a fairly simple equation to calculate the maximum line of sight (LOS) range for two communication systems. However, LOS is not the only factor that determines whether or not those two systems can communicate. In fact, just because two antennas are within the maximum LOS range doesn't mean they can actually communicate. Electrical signals attenuate (become weaker) as they propagate over a distance. If the received signal is too weak, the receiving radio will not be able to detect or process it. In fact, every radio has a certain threshold for received power, or Signal-to-Noise Ratio (SNR), which must be met in order to establish a communications link. The Friis equation provides the means to calculate received power, given a number of attributing factors. If the received power, or the SNR, is above the threshold, **and** the radios are within the maximum LOS range, they can effectively communicate. :::{admonition} Key Concept :class: key-concept The $1/R^2$ is nothing but the surface area of an expanding sphere. Range and wavelength enter **squared**; power and gain do not. Double the range and you quarter the received power. ::: ## Friis Equation The equation for determining how much power actually reaches the receiving antenna is called the Friis (rhymes with Greece) equation: $$P_{R} = \ P_{T}G_{T}G_{R}\frac{\lambda^{2}}{{(4\pi R)}^{2}}$$ where the variables are as follows: - *P­R­*: The power collected by the receiving antenna (in Watts) - *P­T­*: The power sent by the transmitting antenna (in Watts) - *G­T­*: The gain of the transmitting antenna (a linear ratio, no units) - *G­R­*: The gain of the receiving antenna (a linear ratio, no units) - *λ*: The wavelength of the radio signal (in meters) - *R*: The distance between the two antennas (in meters) Let's look at these variables and how they affect the strength of the signal that gets to our receiving antenna. **Power Transmitted, *P­T­*:** The power transmitted, *P­T­*, clearly affects received power. In fact, *P­T­* and *P­R­* are directly proportional, so as *P­T­* increases, *P­R­* also increases. **Distance (or Radius) between Antennas, *R*:** To understand the effect of distance on the amount of received power, consider a point source radiating *P­T­* Watts of power equally in all directions, which is effectively a sphere. At any point on a sphere of radius, *R*, away from the transmitting point source, we can define the power density, *S*, in W/m², as the total power radiated, *P­T­*, divided by the surface area of the sphere, 4π*R*². That is, $$S = \frac{P_{T}}{4\pi R^{2}}$$ From this, we see the power density is proportional to $\frac{1}{R^{2}}$. Soon, we will see the received power, *P­R­*, is proportional to the power density. As a result, received power is proportional to $\frac{1}{R^{2}}$. **Gain of the Transmitting Antenna, *G­T­*:** The gain of the transmitting antenna, *G­T­*, is its ability to focus energy in one direction; therefore, the gain acts as a multiplier of the transmitted power. As a result, the power density, *S*, on a sphere at a distance, *R*, in the direction of maximum gain, for any type of antenna is: $$S = P_{T}G_{T}\frac{1}{4\pi R^{2}}$$ where the gain, *G­T­*, is unitless. Therefore, when the receiving antenna is placed in the direction of maximum gain, the power density will also be maximized. We will often rely on this assumption to determine whether two radios can communicate. Since *P­R­* is proportional to *S*, we see *P­R­* is also proportional to *G­T­*. **Gain of Receiving Antenna:** The receiving antenna also has a gain, which describes how much energy it collects from the transmitted signal. In order to see how the gain of the receiving antenna affects how much power we receive, recall the transmitted signal propagates through free space in the shape of a sphere. As we get farther and farther from the transmitting antenna, the surface of the sphere starts to appear flat to the receiving antenna -- this is called a "plane wave". This idea is shown in the figure below, and is similar to the perception of a flat earth due to its large radius compared to the limited surface area that we perceive. ![](./ECE215_B3_Obj05_Reading_media/media/image1.png) Figure 1: The propagation of a signal from the transmitter to receiver. Note how the spherical wave appears nearly flat at the receiver. The amount of power the receiving antenna can capture from free space depends on the effective area of the antenna, *A­eff­* (the lesson slides write $A_e$). Although this can be thought of as the physical size of the antenna, they are not quite the same, as the effective area is actually the size of the antenna relative to the wavelength of the signal and proportional to the inherent gain of the antenna (G­R­). As an example, parabolic antennas have an effective area that is slightly smaller than their physical area. The power received then is the effective area of the receiving antenna times the power density. $$P_{R} = SA_{eff} = P_{T}G_{T}\frac{A_{eff}}{4\pi R^{2}}$$ As discussed, the effective area of the receiving antenna is related to the wavelength of the signal: $$A_{eff} = \frac{\lambda^{2}}{4\pi}G_{R}$$ Substituting this equation into the one above gives us the final Friis equation as introduced earlier, > **Key Concept -- Friis Equation** > > $$P_{R} = P_{T}G_{T}G_{R}\frac{\lambda^{2}}{(4\pi R)^{2}}$$ Here is a quick summary of how the variables affect the power received: - Doubling the transmitted power (P­T­), doubles the received power (P­R­). - Doubling the gain of either the transmitting or receiving antenna (G­T­ or G­R­) doubles the received power. If both are doubled, the received power quadruples. - Doubling the wavelength of the signal quadruples the received power! Doubling the wavelength is equivalent to halving the frequency. Although we have alluded to the fact that lower frequencies travel farther than higher frequencies, the Friis equation is the proof. For example, broadcasting a lower frequency means we transmit more power to the receiver. - Doubling the distance reduces the received power (P­R­) to one-fourth. The strength of an electromagnetic wave is inversely proportional to the square of the distance. This is intuitive, as the farther you are from a transmitter, the weaker the signal is. When using the Friis equation, your calculator is your friend, but because of the form of the equation, be sure to avoid the following errors: - Calculating λ incorrectly. If you neglect to include the speed of light, your answer will be off by a factor of $90\times10^{15}$. - Using the wrong distance. Remember that the distance is in meters, not kilometers. Therefore, in the problem below, R = 50,000 m. - Not squaring the 4 or the π in the denominator. ### Example Problem 1 A search and rescue helicopter is at 450' AGL because of clouds, while the airman they are trying to find is on a 50-foot hill. The airman's survival radio transmits 6 W of power at 121.5 kHz, and uses an antenna with a gain of 3.0. The helicopter's radio has a gain of 3.2 and requires a minimum of 50 μW of power to receive the transmission. If the helicopter is 50 km away from the airman, will the helicopter crew be able to hear the airman? **Understand**: To ensure the helicopter and the airman can communicate, they must have both LOS **and** sufficient power. **Identify Key Information:** - **Knowns:** We know the height of both the helicopter and airman and we know the specifications of each radio. Since these types of problems often have quite a few unknowns, it is useful to write them explicitly as follows: | Symbol | Value | |:--------------------:|:--------------------:| |h­helo­ | 450 feet | |h­airman­ | 50 feet | |f | 121.5 kHz | |P­T­ | 6W | |P­R, min­ | 50 μW | |G­T­ | 3 | |G­R­ | 3.2 | |R | 50 km | - **Unknowns:** The maximum LOS range and the maximum range based on the minimum power. Note that we are given the *minimum* power the receiver needs to process the signal. Since the power is at a minimum far away from the transmitter, this will give us the *maximum* distance the two radios can communicate across. - **Assumptions:** The communications link relies solely on LOS and not any of the other means (forward scatter, ground wave, or sky wave). Additionally, we can assume the beams of the transmitting antenna and receiving antenna are aligned such that their maximum gains overlap. **Plan:** There are three key steps to this problem: - First we must calculate the maximum LOS range, based on the heights of each radio. If LOS is established (i.e., the *actual* range is less than the LOS range), we can move on. Note, we calculate the line of sight in miles, but the distance is given in kilometers. We must therefore convert (use 1 mile = 1.61 km). The equation sheet calls this range $R_{LOS,mi}$: heights in feet, answer in miles. - Then, we calculate the received power, using the Friis equation, based on the *actual* range. - Finally, we compare the received power to the minimum power that the receiver needs to process the signal. **Solve**: First, we calculate the LOS of the helicopter and the airman by: $$r = \ \sqrt{2h_{helo}} + \sqrt{2h_{airman}} = \ \sqrt{2*450} + \ \sqrt{2*50} = \ \sqrt{900} + \ \sqrt{100} = 40\ miles$$ Converting this distance to kilometers: $$40\ miles*\ \frac{1.61\ km}{1\ mile}\ = \ \ 64.4\ km.$$ The actual separation of 50 km is well within the 64.4 km maximum LOS range, so LOS is established. Therefore, we can move on to solving for power received. Before using the Friis equation, we need to find the wavelength: $$\lambda = \ \frac{c}{f}\ = \ \ \frac{3*10^{8}\ m/s}{121.5*10^{3}\ Hz}\ = 2.469\ km$$ Notice this is a huge wavelength, more than 1.5 miles. Now, using the Friis equation: $$P_{R} = \ P_{T}G_{T}G_{R}\frac{\lambda^{2}}{{(4\pi R)}^{2}}\ = (6\ W)(3)(3.2)\frac{(2469\ m)^{2}}{(4\pi*50,000\ m)^{2}} = 889\ \mu W\ $$ Since the receiver only needs 50 μW of power, the actual signal is more than strong enough to allow the helicopter crew to hear the airman. **Answer**: Yes, the helicopter crew will be able to hear the airman. LOS is established and the power received is 889 μW, which is much greater than the required 50 μW. **Going further -- the maximum distance:** The problem asked only about 50 km. Solving the Friis equation for the range at which the received power falls to 50 μW (the method Example Problem 3 uses below) gives 210.9 km, far beyond the 64.4 km LOS range. So the farthest the helicopter can be and still hear the airman is 64.4 km. Here **line of sight** is the binding limit, not power. Either limit can bind, so always compute both. ### Example Problem 2 Two circular parabolic dishes, each with a radius of 250 mm, are used to send a microwave communication signal at 2.7 GHz. Both dishes have an efficiency of 91%. If the transmit power is 1.2 kW, how much power is received if the antennas are 120 km apart? **Understand**: We have two parabolic antennas trying to communicate. We need to first calculate the gain of the two antennas. **Identify Key Information**: - **Knowns:** We know the following quantities: | Symbol | Value | |:------------------------------------:|:------------------------------------:| | Antenna 1 Radius $(r_{1})$ | 250 mm | | Antenna 1 Efficiency $(η_{1})$ | 91% | | Antenna 2 Radius $(r_{2})$ | 250 mm | | Antenna 2 Efficiency $(η_{2})$ | 91% | | $(P_{T})$ | 1.2 kW | | Range $(R)$ | 120 km | | Frequency $(f)$ | 2.7 GHz | - **Unknowns:** We need to calculate the received power, given the constraints above. We also don't know the gains for the antennas. - **Assumptions:** The communications link relies solely on LOS and not any of the other means (forward scatter, ground wave, or sky wave). Additionally, we can assume the beams of the transmitting antenna and receiving antenna are aligned such that their maximum gains overlap. Lastly, we can assume LOS is established. **Plan**: First, we will calculate the wavelength of the signal, then calculate the gain of each antenna, and, finally, use the Friis equation to calculate the received power. **Solve**: The wavelength of the transmitted signal is $$\lambda = \ \frac{c}{f}\ = \ \ \frac{3*10^{8}\ m/s}{2.7*10^{9}\ Hz}\ = 0.1111\ m\ = \ 111.1\ mm$$ We can now calculate the gain of each antenna: $$G = \ \frac{{(2\pi r)}^{2}}{\lambda^{2}}*\eta\ = \ \frac{{(2\pi*.25\ m)}^{2}}{{(0.1111\ m)}^{2}}*0.91\ = \ \ 181.9$$ Since the gain is the same for both transmitting and receiving, this gain will work for both antennas. We can now use the Friis equation to calculate the received power. $$P_{R} = (1.2*10^{3}\ W)(181.9)(181.9)\frac{(0.1111\ m)^{2}}{(4\pi*120,000\ m)^{2}} = 215.5\ nW\ $$ **Answer**: The receiving antenna receives 215.5 nW of power. ### Example Problem 3 A control station uses a monopole antenna on top of a 15' tower to control an unmanned surveillance aircraft which will be orbiting at an altitude of 3000' AGL. The station broadcasts an 800 W control signal at a frequency of 300 MHz. The transmitting antenna has a gain of 3.2 (a practical monopole over an imperfect ground plane, so its gain is given rather than taken from the sheet's ideal row) and the receiving antenna on the aircraft has a gain of 3.0. If the aircraft needs a minimum of 3.3 nW of power to process the signal, what is the maximum distance the aircraft can be controlled by the ground station? **Understand:** The ground control station (GCS) is broadcasting to a UAS in the area and we need to know how far away the UAS can be and still remain under positive control. Note, we are using the *minimum* required power to solve for the *maximum* acceptable distance. **Identify Key Information**: - **Knowns:** We know the following quantities | Symbol | Value | |:---------------------------------:|:---------------------------------:| | P­T­ | 800 W | | h­UAS­ | 3000 feet | | h­GCS­ | 15 feet | | f | 300 MHz | | G­T­ | 3.2 | | G­R­ | 3 | | P­R,min­ | 3.3 nW | - **Unknowns:** We don't know the maximum range or the wavelength. - **Assumptions:** The communications link relies solely on LOS and not any of the other means (forward scatter, ground wave, or sky wave). Additionally, we can assume the beams of the transmitting antenna and receiving antenna are aligned such that their maximum gains overlap. **Plan:** We must solve for both maximum LOS range and the maximum range that the antennas can still communicate with. The second part is a Friis problem, but we need to solve for R, rather than P­R­. **Solve**: First, we calculate the maximum LOS range: $$r_{max,\ \ total} = \ \sqrt{2h_{station}} + \sqrt{2h_{UAS}}\ = \ \sqrt{2*15} + \ \sqrt{2*3000} = \ 82.94\ miles = 133.5\ km.$$ Now, we rearrange the Friis equation, solving for *R*: $$\mathbf{R} = \ \frac{\lambda}{4\pi}\sqrt{\frac{P_{T}G_{T}G_{R}}{P_{R}}}$$ Then, we use the other variables to find our answer. In this case, instead of explicitly solving for the wavelength, we can substitute $\lambda = \ \frac{c}{f}\ $ to find: $$\mathbf{R} = \ \ \frac{c}{4\pi f}\sqrt{\frac{P_{T}G_{T}G_{R}}{P_{R}}}$$ Now we can plug in all our numbers and solve for R: $$\mathbf{R} = \ \ \frac{c\sqrt{P_{T}G_{T}G_{R}}}{4\pi f\sqrt{P_{R}}} = \frac{(3 \times 10^{8}\frac{m}{s})\ \sqrt{800\ W*(3.2)(3.0)}}{(4\pi)(300\ MHz)\sqrt{3.3*\ 10^{- 9}\ W}}\ = \ \ 121.4\ km$$ So, now, we have calculated two ranges: a maximum LOS range of 133.5 km and a maximum power range of 121.4 km. The maximum range the UAS can remain under positive control is the smaller of the two distances, or 121.4 km. **Answer**: The maximum distance the aircraft can be controlled is 121.4 km. ::::{admonition} Type-along :class: type-along A UAV at **10,000 feet** relays to a ground station at **50 feet**. The UAV transmits $P_T = 5$ W at $f = 2$ GHz with $G_T = 4$; the ground antenna has $G_R = 100$ and sensitivity $P_{R,\min} = 1$ nW. 1. What is the maximum LOS distance? 2. What is the maximum range from the power budget alone? 3. Which limit binds? 4. The UAV also uplinks to a satellite **directly overhead** at an altitude of **400 km**, with the same transmitter and gain. The satellite has $G_R = 100$ and $P_{R,\min} = 1.5$ pW. What range goes into the Friis equation, and does the uplink close? :::{admonition} Solution :class: dropdown **1.** $R_{LOS,mi} = \sqrt{2(10000)} + \sqrt{2(50)} = 141.4 + 10.0 = 151.4$ miles, and $151.4 \times 1.61 = 243.8$ km. **2.** $\lambda = 0.15$ m. Setting $P_R = 1$ nW: $(4\pi R)^2 = (5)(4)(100)(0.15)^2/10^{-9} = 4.5\times10^{10}$, so $4\pi R = 2.121\times10^5$ and $R = 16.9$ km. **3.** **Power binds, by more than a factor of ten** — 16.9 km, not 243.8 km. Do not assume which limit binds: in Example Problem 1 it is line of sight, and here it is power. Either can bind, so compute both. **4.** Straight up there is no horizon in the way, so there is no LOS limit to compute. The range is the difference of the two altitudes, in meters: $R = 400{,}000 - (10{,}000)(0.305) = 396{,}950$ m. Then $P_R = (5)(4)(100)(0.15)^2/(4\pi \cdot 396{,}950)^2 = 1.81$ pW, which is at least 1.5 pW, so the uplink closes. ::: :::: ## Wrap-Up $P_R = P_T G_T G_R \lambda^2/(4\pi R)^2$ is power bookkeeping over a one-way path, with every gain a linear ratio. Range and wavelength enter squared, which is why doubling range quarters the received power and why lower frequencies reach farther. A link needs **both** line of sight and adequate power: the maximum range is the smaller of the two limits, and either one can bind, so compute both. Objective 3.6 asks what changes when the signal has to make the trip twice.