Nov 22, 2013 #1 A adnan012 Advanced Member level 1 Joined Oct 6, 2006 Messages 468 Helped 2 Reputation 4 Reaction score 2 Trophy points 1,298 Activity points 4,923 hi, the second equation comes after solving the first equation. I need the steps involved. Can anyone help me? Attachments integral.JPG 14.2 KB · Views: 91
hi, the second equation comes after solving the first equation. I need the steps involved. Can anyone help me?
Nov 22, 2013 #2 A albbg Advanced Member level 4 Joined Nov 7, 2009 Messages 1,316 Helped 448 Reputation 898 Reaction score 409 Trophy points 1,363 Location Italy Activity points 10,049 First of all remember that ln(1/x)=-log(x). Second \[\Gamma\] doesn't depends from frequecy then it can be brought out from the integral. Thus the integral result is simply \[\omega\]b-\[\omega\]a. This means the equation can be rewritten as: -ln|gamma|*(wb-wa)=pi/(R0*C1) that means: ln|gamma|=-pi/[(wb-wa)*(R0*C1)] that is: |gamma|=exp{-pi/[(wb-wa)*(R0*C1)]}
First of all remember that ln(1/x)=-log(x). Second \[\Gamma\] doesn't depends from frequecy then it can be brought out from the integral. Thus the integral result is simply \[\omega\]b-\[\omega\]a. This means the equation can be rewritten as: -ln|gamma|*(wb-wa)=pi/(R0*C1) that means: ln|gamma|=-pi/[(wb-wa)*(R0*C1)] that is: |gamma|=exp{-pi/[(wb-wa)*(R0*C1)]}
Nov 23, 2013 #3 A adnan012 Advanced Member level 1 Joined Oct 6, 2006 Messages 468 Helped 2 Reputation 4 Reaction score 2 Trophy points 1,298 Activity points 4,923 Thank you very much.