Receiving power from antenna (Homework)

The question:

"At a carrier frequency of 2 GHz, assuming a transmit antenna gain of 3 dBi and a receive antenna gain of 0 dBi, calculate the line-of-sight received power at a distance of 100 m from the basestation (in units of dBm). Assume 100 mW enters the transmit antenna port."

My attempt: The equation to use $$P_{r}=\frac{P_{t}G_{t}G_{r}\lambda^2}{(4\pi R)^2}$$

I am not sure what to do next, I think that $$\G_r=0\$$, hence the $$\P_r\$$ will be 0 since the numerator is.

EDIT:

Thanks for the help, I iwll try to solve it now:

$$\\lambda=\frac{c}{f}\$$, Therefore $$\\lambda=0.15\$$ m

$$\R=100\$$ m

$$\P_{t}=0.1\$$ W

$$\G_t=10^{0.3}=1.996\$$

$$\G_r=10^0=1 \$$

Hence $$\P_{r}=2.844*10^{-9}\$$ W $$\=-55.5\$$ dbm

Is this correct now?

• Why do you think it is 0? – PlasmaHH Oct 23 '18 at 10:41
• Because it is 0dbi. In general I struggle with what units should I use where. – Scavenger23 Oct 23 '18 at 10:48
• dB values are just factors of some -- sadly mostly context specific -- baseline values, like an antenna with a gain of 0dB has a factor of 1 for its received power – PlasmaHH Oct 23 '18 at 10:50
• So whenever I see dbi I should convert it to dB? What about the $\lambda$? How do I find it in this question? – Scavenger23 Oct 23 '18 at 10:53
• no, you should understand what it means in the specific context. – PlasmaHH Oct 23 '18 at 10:53

You are confusing dB with "linear" units.

To convert from mW (a linear unit) to dBm (decibels with respect to 1mW), use the equation:

Power_dBm = 10*log10(Power_mW)

As a general rule of thumb, you should always be very suspicious if you're ever multiplying two numbers in dB. Note the following property of logarithms:

log10(uv) = log10(u) + log10(v) (Note: Log base 10 shown here, but this is true for any base)

This means that if u and v are initially in linear units, and you're multiplying them together, that's equivalent to taking their logarithms and adding them together (or, converting to dB and then adding them together). Multiplying two numbers in dB has no obvious meaning that translates to linear units.

Ok, back to the original question. The equation you presented is only valid for linear units. In this case, your powers (Pr and Pt) are in mW, and your gains (Gt and Gr) are unitless - or, perhaps more precisely, they are mW per mW.

So, to use the equation correctly, you need to convert any numbers which are given in dB or dBm back to linear units, plug them in to the equation to find Pr, and then convert Pr back to dBm.

As analogsystemsrf pointed out, there is an equivalent version of the Friis formula which uses inputs in dB - note that now the terms are added together, rather than multiplied.

You have an exercise in PathLoss. That formula is

PathLoss = 22dB + 10 * log10 [ (distance / wavelength)^2 ] (1)

where the 22dB comes from 4*pi in the spherical spreading of energy.

To compute the received power, you need to know the transmitted power, the TX antenna gain, the RX antenna gain, and the PathLoss. Here the pathloss assumes only the losses as computed in equation (1), and no other losses: no rain, no reflections, no tree leaves, no diffraction over the top of hills, etc.