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20 May, 22:51

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.

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  1. 20 May, 23:03
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    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
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