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HACTEHA [7]
4 years ago
10

Given: an 802.11 wlan transmitter that emits a 50 mw signal is connected to a cable with 3 db loss. the cable is connected to an

antenna with 16 dbi gain. what is the eirp power output?
Physics
1 answer:
CaHeK987 [17]4 years ago
8 0

-- A transmitter has 50 mW output power.  

50 mW is equivalent to +17 dBm.

(+17 is a magic number.  It tells us that the transmitter could very well be based on a single modulated Gunn diode oscillator, which, after resonating and filtering to remove the unwanted puree, hash, and garbage, typically delivers right around +17 dBm at the output.  

-- The power passes through a piece of lossy cable, where it loses 3 dB.

+17 dBm went into the cable.   +14 dBm came out of the other end.  

(The lost 3 dBm warmed the cable.)  

-- The power was then coupled (losslessly) to an antenna with +16 dB "gain".

+14 dBm went into the antenna.  It was shaped and focused so that coming out of the antenna in a certain direction, it sounded as loud as a source that's radiating (+14 + 16) = <em>+30 dBm = 1 watt</em> .

This is NOT 1 watt of real power output.  The antenna has no batteries, it isn't plugged into a wall outlet, and it has no actual 'gain'.  

That 1 watt is "eirp" . . . <em>"</em><em>E</em><em>ffective </em><em>I</em><em>sotropic </em><em>R</em><em>adiated </em><em>P</em><em>ower"</em>.  The antenna focuses most of its power in one certain narrow direction, and then, <u><em>in that direction</em></u>, it sounds as loud as an antenna would that took 1 watt and spread it equally in <em>all</em> directions.

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Nitrogen fixing bacteria changes dead plants and animals into ammonia compounds.

<h2>What is nitrogen fixation?</h2>

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<h3>Characteristics of Nitrogen fixing bacteria</h3>

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7 0
2 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
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Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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