Radio power and range – recap

From Wikipedia –

“In electromagnetic and antenna theory, antenna aperture, effective area, or receiving cross section, is a measure of how effective an antenna is at receiving the power of electromagnetic radiation (such as radio waves).

The power received by an antenna (in watts) is equal to the power density of the electromagnetic energy (in watts per square meter), multiplied by its aperture (in square meters). The larger an antenna’s aperture, the more power it can collect from a given electromagnetic field.”

Power density at a distance “d”  from the transmitting antenna (with output power  Pt) is given by

Pd =  Pt / (4* (pi) * d*d)

Where  pi  is 3.1414.  The denominator is the surface area of a sphere of radius “d”.

As the distance increases, the power density reduces as square of “d”.

Assuming receiving antenna has an aperture “Ar” sq-m, then power received is  given by:

Pr = pd * Ar = (Pt * Ar)/ (4* (pi) * d*d)

Let us  assume that the receiver is d0 meters away in line of sight from the transmitter and there are no obstructions in between (completely ideal scenario).  Assume Pt0 is  the  transmitter’s  power output.  Then Pr0 is theoretically given by:

Pr0  =  (Pt0*  Ar)/(4*pi*d0*d0)

Let us  calculate the power required if we  want  to double  the  distance “d”  but  still receive  the  same power Pr0 at the receiver.

Pr1 =  (Pt1 * Ar)/(4*pi*d1*d1)

Here d1 = 2*d0   and Pr1  = Pr0

d0*d0 = (Pt0 *  Ar)/(Pr0 * 4 *  pi)  –      [1]

d1*d1 = (Pt1 *  Ar)/(Pr1 * 4 *  pi)     – or –    4*d0*d0 = (Pt1 *  Ar)/(Pr0 * 4 *  pi)  – [2]

Using [1] and [2]

4  * (Pt0 *  Ar)/(Pr0 * 4 *  pi) =  4*d0*d0 = 4 * (Pt1 *  Ar)/(Pr0 * 4 *  pi)

– or –

Pt0 =  Pt1 / 4

– or –

Pt1 = 4 * Pt0

The transmitter has to increase it’s output power 4 times to reach twice the distance such that the receiver sees the same power as before.

How much is this increase in “dB” ?

10 * log (P1/P0)  = 10 * log (4) ~ 6 dB

So for every 6 dB increase in transmitted power, the range should double (under ideal conditions).

Let us take a real example. Assume transmitter power is 14 dBm and the receiver is around 1000 meters away and receiving the signal at -106 dBm.

The path loss here is 14 – (-106) = 120 dB.

120 = 10*log(F) =>     12 = log(F)   =>    F =  10^12

The received signal is 10^12 times weaker than the transmitted signal.


Let us say, we use a bad RF cable to connect the radio PCB to an external omni-directional antenna on the transmitter and the RF cable has a loss of say 1 dB. How much range loss does this translate to ?

Let Pt0 be the power (in watts) out of the transmitter’s antenna  when using an RF cable of say 0 dB (ideal) loss. Let Pt1 (In watts) be the power out of  the transmitter’s antenna when using an RF cable of  1 dB loss.

10 log (Pt1/Pt0)  = -1 dB

Pt1/Pt0  =10^(-0.1)

Pt1 = Pt0 * (.795)    [3]

The output power at the transmitter’s antenna is now 0.795 times the previous  power.

Pr0  =  (Pt0 * Ar)/(4 * pi * d0 * d0)         [4]  When using cable with 0 dB loss

Pr1  =  (Pt1 * Ar)/(4 * pi * d1 * d1)         [5]  When using cable with 1 dB loss

We  are interesting in finding the new distance  “d1” at which Pr1 is same as  Pr0.

Pr1 = Pr0       [6]

From [3], [4], [5] and [6]

(Pt0 *  Ar)/(4 * pi * d0 * d0) = ( 0.795 * Pt0 *  Ar)/(4 * pi * d1 *d1)

We  get,

d1*d1  = 0.795 * d0 * d0    =>     d1 = 0.891 *  d0

The range has reduced by around 11 %.

If power loss is 2 dB,  range loss is 20.5 %.  For 3 dB loss in power, the range loss is 29.2 % and so on.  We think 1 dB loss is not a big deal since we are not used to thinking in dB. As you can see now, small losses here and there (low quality cables and connectors, non optimum RF layout,  mismatched antenna etc) and add up to considerable loss in radio range.




Posted on May 13, 2018, in Uncategorized. Bookmark the permalink. Leave a comment.

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