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dBc
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<h2 id="example">Example</h2> <p>If a carrier (reference signal) has a power of C = 0 . 1 m W {\textstyle \mathrm {C} =0{.}1\,\mathrm {mW} } , and noise signal has power of S = 10 μ W {\textstyle \mathrm {S} =10\,\mathrm {\mu W} } . </p><p>Power of reference signal expressed in decibel is : </p> P C = 10 log 10 ( 0 . 1 m W 1 m W ) = − 10 d B m {\displaystyle P_{\mathrm {C} }=10\log _{10}\left({\frac {0{.}1\,\mathrm {mW} }{1\,\mathrm {mW} }}\right)=-10\,\mathrm {dBm} } <p>Power of noise expressed in decibel is : </p> P S = 10 log 10 ( 10 μ W 1 m W ) = − 20 d B m {\displaystyle P_{\mathrm {S} }=10\log _{10}\left({\frac {10\,\mathrm {\mu W} }{1\,\mathrm {mW} }}\right)=-20\,\mathrm {dBm} } <p>The calculation of dBc difference between noise signal and reference signal is then as follows: </p> P S − P C = − 20 d B m − ( − 10 d B m ) = − 10 d B c {\displaystyle P_{\mathrm {S} }-P_{\mathrm {C} }=-20\,\mathrm {dBm} -(-10\,\mathrm {dBm} )=-10\,\mathrm {dBc} } <p>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: </p> P S − P C = 10 log 10 ( S C ) = 10 log 10 ( 10 μ W 0 . 1 m W ) = − 10 d B c {\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} } . <h2 id="external-links">External links</h2> <ul><li><a href="http://www.rp-photonics.com/decibel.html">Encyclopedia of Laser Physics and Technology</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Taylor 1995, Guide for the Use of the International System of Units (SI), NIST Special Publication SP811 <a href="http://physics.nist.gov/cuu/pdf/sp811.pdf" target="_blank">http://physics.nist.gov/cuu/pdf/sp811.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>