why first order filter have slope 6db/octave
in [DSP]
|
From: HardySpicer on 1 Feb 2010 16:27 On Feb 2, 9:01 am, Jerry Avins <j...(a)ieee.org> wrote: > HardySpicer wrote: > > ... > > >> So if the shelf I had intended to be level is a bit low on one end, it > >> ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms. > > > I don't think you are right there. although the use of the word ripple > > factor is clearly a miss-noma I still don't understand how you go > > about designing a Butterworth filter with say a 1dB attenuation in the > > passband at 1kHz and a 40 dB attenuation in the stopband at 20kHz? Do > > you define only 3dB attenuations and 'work back"? > > All low-pass Butterworth filters of the same order have the same shape > when plotted in normalized co-ordinates, f/fc and log gain. What is > usually specifies is maximum attenuation in the passband and rate of > attenuation in the stopband. The stopband rate in dB/decade is always a > multiple of 20, and the multiple is the order. (A third-order filter > rolls off at 60 dB/decade.) > > Given the maximum attenuation allowed in the passband, i.e., the minimum > A(f) in the passband, and the order n, it is easy to solve A(f) = > 1/sqrt(1+(f/fc)^2n) for fc. > > For your filter (1dB attenuation in the passband at 1kHz and a 40 dB > attenuation in the stopband at 20kHz) the rules of thumb indicate that a > first-order filter with fc=2 KHz meets the spec. (The response will be 1 > dB down at fc/2, 3 dB down at fc, 5 dB down at 2fc, and 20 dB down at > 10fc. You picked the kind of easy numbers I would use on a quiz.) > > An alternate solution that I would rarely use associates 2 values of > A(f) with assigned values of f and solves the simultaneous equations for > n and fc. Of course, one rounds n up to an integer. > > Jerry > > P.S. A first-order Butterworth (and Chebychev!) is a simple R-C. > -- > Engineering is the art of making what you want from things you can get. > ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Ok so you are applying the 3dB rules and working back in freq. The other method gives you the direct solution. I am not saying that the slopes won't be multiples of +/- -20dB/decade - for LTI systems they have to be of course. ok so here is what is happening. For a nomalised freq v the magnitude- squared Butetrworth is given by |H(v)|^2=1/(1+eps^2.v^2n) what you have is |H(v)|^2=1/(1+v^2n) so you are absorbing the constand eps^2 into the term v^2n. The problem with that is that I don't think the calculations are as simple as you need to 'work back" so to speak. Hardy
From: John Monro on 2 Feb 2010 00:58 rickman wrote: > On Jan 28, 7:57 am, "rammya.tv" <rammya...(a)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. Regards, John
From: steveu on 2 Feb 2010 01:40 >Rune Allnor wrote: >> On 1 Feb, 00:52, Jerry Avins <j...(a)ieee.org> wrote: >> >>> We speak different languages. On this side of the Atlantic, Butterworth >>> means maximally flat, >> ... >>> Because A(f) falls monotonically as f >>> increases, there is no ripple. There are many filters with no ripple, >>> but only the Butterworth is maximally flat in the sense I defined above. >> >> I think the problem is the definition of the term 'ripple'. >> >> If one interprets it literally, as something like 'fluctuating >> back and forth', then yes, the Butterworth has no ripple. >> >> If, on the other hand, one interprets the term as 'deviation >> from desired value', then it is immediately clear that the >> Butterworth lowpass filter has a non-zero ripple everywhere >> except at DC. > >It does the art no good to debase clear terms. if "ripple" is >interpreted to mean "deviation from desired value", then nearly every >filter, calculated transfer function, and approximation will exhibit >ripple. We will need to invent a new term. "Actual ripple" perhaps? > >> I wasn't there to see what actually happened, but I wouldn't be >> surprised if the term 'ripple' was preferred over 'deviations >> from desired value' for rather pragmatic reasons... > >A lot of sloppy language gets used, most of it informally. We will lose >the ability to communicate if we adopt every illogical neologism. Just as much sloppy language gets used quite formally. The two of us can't communicate unless the comms channel is operating above capacity. That's completely backwards from proper English. It makes ripple for droop seem a minor distortion of the English language. Names are generally chosen to obscure rather than illuminate. Professional jargon isn't much different. Steve
From: rickman on 2 Feb 2010 20:16 On Feb 2, 12:58 am, John Monro <johnmo...(a)optusnet.com.au> wrote: > rickman wrote: > > On Jan 28, 7:57 am, "rammya.tv" <rammya...(a)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. Rick
From: John Monro on 2 Feb 2010 22:31
rickman wrote: > On Feb 2, 12:58 am, John Monro <johnmo...(a)optusnet.com.au> wrote: >> rickman wrote: >>> On Jan 28, 7:57 am, "rammya.tv" <rammya...(a)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. > > Rick I 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 |