rickman wrote:> On Feb 2, 12:58 am, John Monro <johnmo...@optusnet.com.au> wrote: >> rickman wrote: >>> On Jan 28, 7:57 am, "rammya.tv" <rammya...@ymail.com> wrote: >>>> hi all >>>> i would like to know the technical description or derivation about >>>> the slope of a filter >>>> ie >>>> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave) >>>> slope. >>>> with regards >>>> rammya >>> I see you got a lot of replies, but not an answer. I have always had >>> to "understand" things like this on a physical basis rather than a >>> purely mathematical basis, so I think I can answer your question so >>> you will understand it too. >>> A 1st order filter would use a single reactive component such as a low >>> pass filter with one capacitor and one resistor. >>> Vin R >>>> -------\/\/\------+---------> output >>> | >>> | >>> --- >>> --- C >>> | >>> | >>> --- >>> V >>> The output voltage is just Vout = Vin * Xc/(R+Xc). Xc is 1/2pi*f*C. >>> When f is large (well above the corner frequency where Xc = R), Xc is >>> small and R+Xc is approximately R. So Vout/Vin = Xc/R or >>> Vout/Vin = 1/(2pi*R*C*f) >>> So as F changes by a factor of two, the output voltage changes by a >>> factor of two. A voltage change of a factor of two results in 3 dB >>> voltage, but dB is conventionally expressed as a power ratio, since P >>> = V**2/R the result is 6 dB power per octave. >>> Does that help? >>> I think a lot of newbies have trouble with decibels more so than the >>> applications of them. For example, it is common to talk about a -3 dB >>> point in filters and other applications, but that does not mean the >>> voltage is half of the 0 dB point. The *power* is half and R = Xc in >>> the example above, but the reactive component is not in phase with the >>> resistive component so that the voltages do not add up. Each voltage >>> is sqrt(2) times the 0dB value. >>> Maybe this was just a problem for me when I was learning this stuff. >>> But I find a lot of people don't really get decibels. >>> Rick >> Rick, >> The expression you give, Vout = Vin * Xc/(R+Xc) is fine in >> the region you are talking about where Xc << R, but it is >> not very accurate otherwise. >> >> In particular, at a frequency where R = Xc the above >> expression becomes: Vout = Vin * 1/2 >> This gives a gain (Vout/Vin) of 0.5 ( -6.0 dB), which is >> incorrect. >> >> If instead of Xc, the reactance of C, you use jXc, the >> impedance of C, you get the correect expression: >> Vout = Vin * jXc/(R+jXc) >> At the frequency where R = Xc the expression becomes: >> Vout = Vin * 1/sqrt(2) >> This gives a gain of 0.707 (-3.0 dB) which is correct. > > > Did you read my post??? I clearly stated the region where the > approximation was reasonable and in the later part I discuss why the > gain is not as it would appear in the region where the reactive > component is out of phase with the resistive component. You merely > repeated what I posted with complex math which may well be beyond the > grasp of the OP. As I stated, I was trying to explain it in simpler > terms which someone can understand without resorting to complex > arithmetic. > > RickI did not 'merely' repeat your post; I corrected the expression you used. I am sorry that you have taken offence. I hope my post was useful to the OP. Regards, John