# why first order filter have slope 6db/octave

Started by January 28, 2010
```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.

Regards,
John
```
```>Rune Allnor wrote:
>> On 1 Feb, 00:52, Jerry Avins <j...@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

```
```On Feb 2, 12:58&#2013266080;am, John Monro <johnmo...@optusnet.com.au> wrote:
> rickman wrote:
> > On Jan 28, 7:57 am, "rammya.tv" <rammya...@ymail.com> wrote:
> >> hi all
> >> &#2013266080; 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. &#2013266080;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 &#2013266080; &#2013266080; &#2013266080; R
> >> -------\/\/\------+---------> output
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;|
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;|
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; ---
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; --- C
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;|
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;|
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; ---
> > &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;V
>
> > The output voltage is just Vout = Vin * Xc/(R+Xc). &#2013266080;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. &#2013266080;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. &#2013266080;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. &#2013266080;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. &#2013266080;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. &#2013266080;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, &#2013266080;you get the correect expression:
> &#2013266080; &#2013266080;Vout = Vin * jXc/(R+jXc)
> At the frequency where R = Xc the expression becomes:
> &#2013266080; &#2013266080;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
```
```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.
>
> 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
```