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# delete - approx corresponds to his R and C again Now with...

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( 29 2 10 log 1 2 : ( ) ( ) ( ) ( ) A f RC where A attenuation in units of decibels dB Negative removed because it is a decrease f frequency in units of kilohertz kHz R resistance in units of kiloohms k C capacitan π = + = = = = The equation of interest is : 0.1 0.1(14) 4 0.01 ( ) : 10 1 2 #1 : 14 ; 20 , 10 1 2 (20) 0.0391... . ! #2 : 24 ; 62.5 , 1 A ce in units of microfarads F now solving for RC RC f Example A dB f kHz now RC which if you notice approx corresponds to his R and C Example A dB f kHz now RC μ π π - = = = - = = = = = = = 0.1(24) 4 0.01 40 20 0 1 2 (62.5) 0.0403 ...
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Unformatted text preview: approx corresponds to his R and C again Now with this formula one can plug in any values for A and f and thereby get the product RC desired Let s say you want A attenuation at frequency of-= = = = 0.1(40) 10 1 2 (20) 0.01 . .. 1 0.01 ; : z RC could correspond perhaps to R k and C F-= = Ω = I believe this is the situation your friend was looking for . .....
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