Find the Received Power in dBm and dBW at a Free Space

If a transmitter produces 70 watts of power, express the transmit power in units of (a) dBm, and (b) dBW. If 70 watts is applied to a unity gain antenna with an 1800 MHz carrier frequency, find the received power in dBm at a free space distance of 500 m from the antenna, what has received power P(15 km)? Assume unity gain for the receiver antenna.

Solution:

Given Data:

P_t = 70 W

a) Express Power in unit dBm
b) dBW

a) For Power in dBm:

P(dBm) = 10 log₁₀ (P(W)) + 30

Here in our case

P(W) = P_t = 70W

So,

=10 * log₁₀ (70) +30
=10 * (1.845) + 30
P(dBm) = 48.45

So,

P = 48.45 dBm

b) For Power in dBW:

P(dBW) = 10 log₁₀(P/1W)

Here again Power = P_t = 70W so,

P(dBW) = 10 log₁₀(70/1) = 10(1.845)

P(dBW) = 18.45

So,

P = 18.45 dBW

Now: If P_t = 70W

G_t = 1 dB
f = 1800 MHz
d = 500 m
P_r = ?
G_r 1 dB

So,

P_r = (P_t*G_t*G_r*λ²)/(4πd)²

Where

λ = c₀/f

=(3*10⁸)/(1800*10⁶)
=0.0016*10^8-6 = 0.0016*10²

λ = 0.16*10^2-2 = 0.16 m

So,

P_r = (70 * 1 * 1 * (0.16)²)/(4 * 3.14 * 500)²
=(70 * 0.0256)/(6283.18)²
=1.792 / 39478415.67

P_r = 0.0000000454 W

Now in dBm units:

P_r = 10 * log₁₀(P_r) + 30
= 10 * log₁₀(0.0000000454) + 30
= 10 * (-7.342) + 30

P_r = -43.42 dBm

Now again P_r = ? But d = 15 km = 15000 m

P_r(15 km) = P_r(500) + 20 log₁₀[500/15000]
=-43.42 + 20 log₁₀(0.03)
= -43.42 + (-29.542)

P_r(15 km) = -72.96 dBW

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