Question:medium
Consider the Friis' transmission equation \(P_R=\dfrac{P_TG_TG_R\lambda^2}{(4\pi D)^2}\), where \(P_R\) and \(P_T\) are the received and the transmitted powers, respectively. \(G_T\) and \(G_R\) are the gain of transmitting and receiving antennas, respectively, \(D\) is the distance between the transmitting and receiving antennas, and \(\lambda\) is the wavelength in free space.
Given: \(G_T=G_R=1.0\), \(\lambda=0.30\text{ m}\) and \(P_T=+10\text{ dBm}\).
Choose the distance \((D)\), in km, from the following options at which the received power, \(P_R=-90\text{ dBm}\)?
Consider the Friis' transmission equation \(P_R=\dfrac{P_TG_TG_R\lambda^2}{(4\pi D)^2}\), where \(P_R\) and \(P_T\) are the received and the transmitted powers, respectively. \(G_T\) and \(G_R\) are the gain of transmitting and receiving antennas, respectively, \(D\) is the distance between the transmitting and receiving antennas, and \(\lambda\) is the wavelength in free space.
Given: \(G_T=G_R=1.0\), \(\lambda=0.30\text{ m}\) and \(P_T=+10\text{ dBm}\).
Choose the distance \((D)\), in km, from the following options at which the received power, \(P_R=-90\text{ dBm}\)?
Given: \(G_T=G_R=1.0\), \(\lambda=0.30\text{ m}\) and \(P_T=+10\text{ dBm}\).
Choose the distance \((D)\), in km, from the following options at which the received power, \(P_R=-90\text{ dBm}\)?
Show Hint
Convert the 100 dB power difference to a linear ratio first, then use \(P_R/P_T=(\lambda/4\pi D)^2\) to solve for D, remembering to take a square root.
Updated On: Jul 20, 2026
- \(\dfrac{15}{4\pi}\)
- \(\dfrac{15}{2\pi}\)
- \(\dfrac{75}{2\pi}\)
- \(\dfrac{3}{4\pi}\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Express the Friis equation in decibels.
Taking $10\log_{10}$ of both sides of $P_R=P_TG_TG_R\left(\dfrac{\lambda}{4\pi D}\right)^2$ turns the squared ratio into a $20\log_{10}$ term:
\[ P_R(\text{dBm})=P_T(\text{dBm})+10\log_{10}G_T+10\log_{10}G_R+20\log_{10}\left(\frac{\lambda}{4\pi D}\right) \]
Step 2: Drop the zero-gain terms.
Since $G_T=G_R=1.0$, a linear gain of exactly $1$, we get $10\log_{10}(1)=0$ dB for each, so those terms vanish.
Step 3: Plug in the known powers.
\[ -90=10+20\log_{10}\left(\frac{\lambda}{4\pi D}\right) \] \[ 20\log_{10}\left(\frac{\lambda}{4\pi D}\right)=-100 \] \[ \log_{10}\left(\frac{\lambda}{4\pi D}\right)=-5 \]
Step 4: Remove the logarithm.
\[ \frac{\lambda}{4\pi D}=10^{-5} \]
Step 5: Solve for D in metres, then convert to km.
\[ D=\frac{\lambda\times10^{5}}{4\pi}=\frac{0.30\times10^{5}}{4\pi}\text{ m}=\frac{30000}{4\pi}\text{ m} \] Dividing by $1000$ to get kilometres,
\[ D=\frac{30}{4\pi}\text{ km}=\frac{15}{2\pi}\text{ km} \] \[ \boxed{D=\dfrac{15}{2\pi}\text{ km}} \]
Taking $10\log_{10}$ of both sides of $P_R=P_TG_TG_R\left(\dfrac{\lambda}{4\pi D}\right)^2$ turns the squared ratio into a $20\log_{10}$ term:
\[ P_R(\text{dBm})=P_T(\text{dBm})+10\log_{10}G_T+10\log_{10}G_R+20\log_{10}\left(\frac{\lambda}{4\pi D}\right) \]
Step 2: Drop the zero-gain terms.
Since $G_T=G_R=1.0$, a linear gain of exactly $1$, we get $10\log_{10}(1)=0$ dB for each, so those terms vanish.
Step 3: Plug in the known powers.
\[ -90=10+20\log_{10}\left(\frac{\lambda}{4\pi D}\right) \] \[ 20\log_{10}\left(\frac{\lambda}{4\pi D}\right)=-100 \] \[ \log_{10}\left(\frac{\lambda}{4\pi D}\right)=-5 \]
Step 4: Remove the logarithm.
\[ \frac{\lambda}{4\pi D}=10^{-5} \]
Step 5: Solve for D in metres, then convert to km.
\[ D=\frac{\lambda\times10^{5}}{4\pi}=\frac{0.30\times10^{5}}{4\pi}\text{ m}=\frac{30000}{4\pi}\text{ m} \] Dividing by $1000$ to get kilometres,
\[ D=\frac{30}{4\pi}\text{ km}=\frac{15}{2\pi}\text{ km} \] \[ \boxed{D=\dfrac{15}{2\pi}\text{ km}} \]
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