dbm()

Returns the decibel measure of a voltage referenced to a 1 milliwatt signal

Syntax
y = dbm(v, z)

Arguments

Name	Description	Range	Type	Default	Required
v	voltage (the peak voltage)	(-∞, ∞)	integer, real, complex		yes
z	impedance	(-∞, ∞)	integer, real, complex	50.0	no

Examples

y = dbm(100)
returns 50
y = dbm(8-6*j)
returns 30

Defined in

Built in

See Also

db(), pae()

Notes/Equations

The voltage is assumed to be a peak value. Signal voltages stored in the dataset from AC and harmonic balance simulations are in peak volts. However, noise voltages obtained from AC and HB simulations are in rms volts. Using the dbm() function with noise voltages will yield a result that is 3 dB too low unless the noise voltage is first converted to peak:
noise_power = dbm(vout.noise * sqrt(2));

Understanding the dbm() Function

Given a power Po in Watts, the power in dB is:

Po_dBW

=

10*log(mag(Po/(1 W)))

while the power in dBm is:

Po_dBm	=	10*log(mag(Po/(1 mW)))
=	10*log(mag(Po/(1 W)))+ 30
=	Po_dB + 30

Given a voltage Vo in Volts, the voltage in dB is:

V_dBV

=

20*log(mag((Vo/(1 V)))

This is the db() function - voltage in dB relative to 1V. Although dB is a dimensionless quantity, it is normal to attach dB to a value in order to differentiate it from the absolute value.

Given a real impedance Zo, the power-voltage relation is:

Po

=

(Vo) 2 /(2*Zo)

Using the above, Po in dBm is then:

Po_dBm	=	10*log(mag(Po/(1 W))) + 30
=	10*log(mag((Vo/(1 V)) 2 /(2*Zo/(1 Ohm)))) + 30
=	10*log(mag((Vo/(1 V)) 2 )) - 10*log(mag(2*Zo/(1 Ohm))) + 30
=	20*log(mag(Vo/(1 V))) - 10*log(mag(Zo/(50 Ohm))) + 10

This is the dbm() function - voltage in dBm in a Zo environment.