Transmitter Power-Distance Relation?

[CMOS]

transmitter output power distance

What is the formula to calculate approximate distance to which a transmitter will radiate if transmitter output power is known.

Or what will be the power output required to transmit upto a distance of 30 meters?

 

[XNOX_Rambo]

You can transmit to the LMC if you wish - depending on the receivers sensitivity.

I.e. you have to rephrase your question.

 

[FANT]

An handy formula to know the attenuation vs distace is the following:

Att ( dB ) = 20 dog(10 ) (41,87 * F ( MHz ) * d ( Km ))

But in any case it depends on the sensibility of the receiver and from the transmitter power.

Also take care about attenuation due to the cables and antennas gain.


Mandi

 

[CMOS]

Yes I know I had to specify sensitivity of my receiver. But I am going to use standard commercial FM receiver. I don't have any idea about its sensistivity. Nothing has been mentioned in its manual.

 

[vfone]

A typical sensitivity of an FM tuner (measured in dBm) is –100dBm for a SNR of 60dB.
The free space path loss at 30m will be ~ 42dB, without accounting for absorption or dispersion by the atmosphere. Assuming that you are using two vertical dipoles (one for TX and one for RX), and adding some cable loss, the theoretical TX power shall be approximately -60dBm.
To have a stable link connection I would go with more power than this (depend by your application). Probably for voice would be ok a TX level greater than -30dBm.
To convert dBm in mW use the formula: P_mW = 10^(P_dBm/10)

 

[CMOS]

An handy formula to know the attenuation vs distace is the following:
Att ( dB ) = 20 dog(10 ) (41,87 * F ( MHz ) * d ( Km ))
Mandi

Is this formula correct? I am asking this because "vfone" told that at 30m attenuation will be 42dB but using above formula I am getting 82.6dB at 108MHz. But if I use
Att ( dB ) = 10 log(10 ) (41,87 * F ( MHz ) * d ( Km ))
I get 42dB.

Where am I going wrong?

 

[vfone]

Path loss formula that I used for free space is:

Ploss = 20*log((4*Pi*d)/Lambda)

where:
d = distance in meters
Lambda = wavelength in meters
Pi = 3.14

 

[XNOX_Rambo]

10 or 20 depends on if you are using power (a quadratic variable) or e.g. voltage
(a linear variable), in a log expression.

Back to basics: **broken link removed**

/Rambo

 

[FANT]

From the site:

**broken link removed**


Attenuation of the electromagnetic wave while propagating through space. This attenuation is calculated using the following formula:

Free space loss =32.4 + 20xLog F(MHz) + 20xLog R(Km)

F is the RF frequency expressed in MHz. R is the distance between the transmitting and receiving antennas. At 2.4 Ghz, this formula is: 100+20xLog R(Km).

As you can see is everything expressed as 20 log (10)

Mandi

 

[GeekWizard]

Free space loss =32.4 + 20xLog F(MHz) + 20xLog R(Km)

Where does this 32.4 factor come from?

 

[vfone]

Free space path loss formulas:
**broken link removed**
**broken link removed**