Suppose a transmitter produces 50 W of power. (a) Express the transmit power in units of dBm and dBW. (b) If the transmitter’s power is applied to a unity gain antenna with a 900-MHz carrier frequency, what is the received power in dBm at a free space distance of 100 m? (c) Repeat (b) for a distance of 10 km. (d) Repeat (c) but assume a receiver antenna gain of 2.

a. 46.99dBm, 16.99 dBW

b. 3.5 x 10^-6 W, -24.56 dBm, -54.56 dBw

c. -3.5 * 10^-6W , -64.56dBm, -94.56dBW

d. -62.56 dBm

Explanation:

a.

50 watts to dBm:

P(dBm) = 10 ⋅ log(1000⋅50W)

= 46.98970004336018

= 46.99 dBm --- Approximated

50 watts to dBW:

P(dBW) = 10 ⋅ log( 50W / 1W)

= 10 . log(50)

= 16.98970004336018

= 16.99 dBW

b.

Given

f = 900MHz,

λ = c/f = 3e^8/9e^8 = ⅓

Pr(d) = (Pt * Gt * Gr * λ²) / ((4 π)² d² L)

Gr, Gt are assumed to be 1

So

Pr(100m) = (50 * 1 * 1 * (1/3)²) / ((4 π)² * 100² * 1)

= 3.5 x 10^-6 W

convert to dBm

= 10 log (1000 * 3.5 * 10^-6)

= −24.5593195564972

= -24.56 dBm

To dBW

= 10 log (3.5 * 10^-6W/1W)

= 10 * log (3.5 * 10^-6)

= −54.5593195564972

= -54.56 dBw

c.

10km = 10000m

Pr(100m) = (50 * 1 * 1 * (1/3)²) / ((4 π)² * 10000² * 1)

3.5 x 10^-10 W

convert to dBm

= 10 log (1000 * 3.5 * 10^-10)

= -64.56 dBm

To dBW

= 10 log (3.5 * 10^-10W/1W)

= 10 * log (3.5 * 10^-10)

= -94.56 dBw

d. A gain of 2

Power = -64.56 dBM

Gain = -64.56 + 2

= -62.56dBM

explanation:

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answer:

31

explanation: