GATE 2026 EC Q.21: solution and answer

GATE 2026 · ECQ.21 · 1 mark · MCQ

Consider the Friis' transmission equation $P_{R} = (P_{T} G_{T} G_{R} λ^{2}) / (4 π D)^{2}$ , 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 $λ$ is the wavelength in free space.

Given: $G_{T} = G_{R} = 1.0$ , $λ = 0.30 m$ and $P_{T} = + 10 dBm$ .

Choose the distance ( $D$ ), in km, from the following options at which the received power, $P_{R} = - 90 dBm$ ?

(A)
$\frac{15}{4 π}$
(B)
$\frac{15}{2 π}$
(C)
$\frac{75}{2 π}$
(D)
$\frac{3}{4 π}$

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Answer and solution

ANSWER

(B) $\frac{15}{2 π}$ km, about 2.39 km

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

With $G_{T} = G_{R} = 1$ , the Friis equation becomes $\frac{P _{R}}{P _{T}} = (\frac{λ}{4 π D})^{2}$
Both powers are given in dBm, so their difference is the power ratio in dB: $P_{R} - P_{T} = - 90 - (+ 10) = - 100 dB$ . That gives $\frac{P _{R}}{P _{T}} = 1 0^{- 100/10} = 1 0^{- 10}$ . (Check: $P_{T} = 10 mW$ and $P_{R} = 1 0^{- 9} mW = 1 0^{- 12} W$ , and their ratio is $1 0^{- 10}$ .)
Take the square root: $\frac{λ}{4 π D} = 1 0^{- 10} = 1 0^{- 5}$ . In dB terms, the free-space path loss is $20 lo g_{10} (\frac{4 π D}{λ}) = 100 dB$ , so $\frac{4 π D}{λ} = 1 0^{5}$ .
Solve for the distance: $D = \frac{λ}{4 π \times 1 0 ^{- 5}} = \frac{0.30 \times 1 0 ^{5}}{4 π} m = \frac{30000}{4 π} m$
Convert to km: $D = \frac{30}{4 π} km = \frac{15}{2 π} km \approx 2.39 km$ . This is option (B).

KEY IDEA

Subtracting one dBm value from another gives a dB ratio. Here $P_{R} - P_{T} = - 100 dB$ is the free-space path loss with unity-gain antennas, and $(\frac{λ}{4 π D})^{2} = 1 0^{- 10}$ . Because Friis loss depends on the square of distance, 100 dB of power loss means $\frac{4 π D}{λ} = 1 0^{5}$ , not $1 0^{10}$ .

WATCH OUT FOR

The usual mistakes are about the square. One is forgetting the square root, which leaves $\frac{4 π D}{λ} = 1 0^{10}$ and a distance far larger than any option. Another is converting the 100 dB power ratio with $1 0^{- 100/20}$ and then taking a square root as well, which applies the halving twice. A third is leaving $D$ in metres ( $\frac{30000}{4 π}$ m) and not turning it into km before comparing with the options. It also helps to remember that dBm minus dBm is a plain dB ratio, so there is no need to convert to watts.

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 2 Antennas questions, with notes