Convert from 'LIN MAG' to 'LOG MAG'

[Chest]

lin mag

Hi all,

Anyone know how to convert from 'LIN MAG' to 'LOG MAG'?

Thanks in advance.

 

[fanshuo]

convert linmag to logmag

use the exponential property of bipolar devices

 

[biff44]

+20 log 10

Depends. If you are measuring voltages or phase noises, use 'Log mag' = 20 Log ('Lin Mag').

If you are measuring power, 'Log Mag' = 10 Log ('Lin Mag')

Ex: 0.1 mW is the same as 10 Log (0.1 mw/1 mw) = -10 dBm

or
0.2 volts is the same as 20 Log (0.2v/1 v) = -13.9 dbv

 

[kspalla]

convert lin mag to logmag

Have you tried dB() function?

 

[yours]

convert linear mag to log mag

Depends. If you are measuring voltages or phase noises, use 'Log mag' = 20 Log ('Lin Mag').

If you are measuring power, 'Log Mag' = 10 Log ('Lin Mag')

Ex: 0.1 mW is the same as 10 Log (0.1 mw/1 mw) = -10 dBm

or
0.2 volts is the same as 20 Log (0.2v/1 v) = -13.9 dbv

I think there is something wrong. Measuring power is 20log().
Measuring voltages is 10log().

 

[radiohead]

phase to logmag

Depends. If you are measuring voltages or phase noises, use 'Log mag' = 20 Log ('Lin Mag').

If you are measuring power, 'Log Mag' = 10 Log ('Lin Mag')

Ex: 0.1 mW is the same as 10 Log (0.1 mw/1 mw) = -10 dBm

or
0.2 volts is the same as 20 Log (0.2v/1 v) = -13.9 dbv

I think there is something wrong. Measuring power is 20log().
Measuring voltages is 10log().

Nah. Power is 10log, voltage is 20log.

 

[Azulykit]

lin mag to log mag

Just think of voltage squared is proportional to power. That is why the 20 or 10 log is applied depending on the argument of the log function (voltage or power units).