If a carrier (reference signal) has a power of ${\textstyle \mathrm {C} =0{.}1\,\mathrm {mW} }$ , and noise signal has power of ${\textstyle \mathrm {S} =10\,\mathrm {\mu W} }$ .

Power of reference signal expressed in decibel is :

${\displaystyle P_{\mathrm {C} }=10\log _{10}\left({\frac {0{.}1\,\mathrm {mW} }{1\,\mathrm {mW} }}\right)=-10\,\mathrm {dBm} }$ 

Power of noise expressed in decibel is :

${\displaystyle P_{\mathrm {S} }=10\log _{10}\left({\frac {10\,\mathrm {\mu W} }{1\,\mathrm {mW} }}\right)=-20\,\mathrm {dBm} }$ 

The calculation of dBc difference between noise signal and reference signal is then as follows:

${\displaystyle P_{\mathrm {S} }-P_{\mathrm {C} }=-20\,\mathrm {dBm} -(-10\,\mathrm {dBm} )=-10\,\mathrm {dBc} }$ 

It is also possible to compute the dBc power of noise signal with respect to reference signal directly as logarithm of their ratio as follows:

${\displaystyle P_{\mathrm {S} }-P_{\mathrm {C} }=10\log _{10}\left({\frac {\mathrm {S} }{\mathrm {C} }}\right)=10\log _{10}\left({\frac {10\,\mathrm {\mu W} }{0{.}1\,\mathrm {mW} }}\right)=-10\,\mathrm {dBc} }$ .

• Encyclopedia of Laser Physics and Technology