Question

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, \(P_R=-90\text{ dBm}\)?

• A. \(\dfrac{15}{4\pi}\)
• B. \(\dfrac{15}{2\pi}\)
• C. \(\dfrac{75}{2\pi}\)
• D. \(\dfrac{3}{4\pi}\)

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.

Answer 1

The Correct Option is B

Step 1: Convert the power levels to a linear ratio.
Both \(P_T\) and \(P_R\) are given in dBm, so the ratio between them in dB is
\[ P_R(\text{dB})-P_T(\text{dB})=-90-10=-100\text{ dB} \]
Converting this decibel difference to a plain linear power ratio,
\[ \frac{P_R}{P_T}=10^{-100/10}=10^{-10} \]

Step 2: Write the Friis equation as a power ratio.
Since \(G_T=G_R=1.0\), the Friis equation simplifies to
\[ \frac{P_R}{P_T}=\left(\frac{\lambda}{4\pi D}\right)^2 \]

Step 3: Equate the two expressions for the ratio.
\[ \left(\frac{\lambda}{4\pi D}\right)^2=10^{-10} \]
Taking the square root of both sides,
\[ \frac{\lambda}{4\pi D}=10^{-5} \]

Step 4: Solve for D.
\[ D=\frac{\lambda}{4\pi\times10^{-5}}=\frac{0.30}{4\pi\times10^{-5}}\text{ m} \]
\[ D=\frac{0.30\times10^{5}}{4\pi}\text{ m}=\frac{30000}{4\pi}\text{ m} \]

Step 5: Convert D from metres to kilometres.
\[ D=\frac{30000}{4\pi\times1000}\text{ km}=\frac{30}{4\pi}\text{ km}=\frac{15}{2\pi}\text{ km} \]

Step 6: Rule out the other options.
Option (A), \(15/(4\pi)\), is exactly half of the correct value, the kind of error that appears if the square root step in Step 3 is skipped incorrectly. Option (C), \(75/(2\pi)\), is five times too large. Option (D), \(3/(4\pi)\), is too small by a factor of \(10\), consistent with a slip in the power-of-ten conversion.

Final Answer:
\[ \boxed{D=\frac{15}{2\pi}\text{ km}} \]