Third Order High Pass Filter (Report)

  • #26
rude man
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The fact that the gain is always positive means that dG/dw is always negative, even though that is not explicitly stated.
Not explicitly stated because wrong.
Looky here: see the bump on the Butterworth frequency response? Have to look carefully maybe ...
 

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  • #27
NascentOxygen
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Not explicitly stated because wrong.
Looky here: see the bump on the Butterworth frequency response? Have to look carefully maybe ...
What is the significance of the small rectangles at points along each of your plots?
 
  • #28
rude man
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What is the significance of the small rectangles at points along each of your plots?
Your guess is as good as mine. Look like data points on empirically derived data but everything in this 144-page app note is seemingly computer-derived. And I see them only on the Chebyshev, not the Butterworth or Bessel curves.

I know my link wouldn't open so I suggest googling "analog filters" and finding "Chapter 8: Analog Filters - Analog Devices" which is how I found the paper. All you ever wanted to know about analog filters and then some.
 
  • #29
rude man
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The term "ripple" always refers to the frequency response (Butterworth is maximally flat without any ripple), whereas the term "overshoot" must be used for the step response only (time domain).
There is also "overshoot" on frequency plots. Let's not get pedantic. See my post 26.
As I said already, all this is of no relevance to the OP who simply wondered about the relationship between pulse distortion and dφ/df flatness (cf. his last posts, ##7 and 11).
 
  • #30
rude man
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You are listing low-pass characteristics, but is any of this relevant to the high-pass filters shown?
It is.
Consider simple illustration: a rectangular pulse of duration T into 1st order low-pass and high-pass networks, cutoff radian frequency = a:
Input is (1/s)(1 - exp(-sT).
Low-pass is a/(s+a) and hi-pass is s/(s+a).
Group delay is the same for both: dφ/dω = -τ/(ω2τ2 +1) with τ = 1/a as I derived earlier.
Output for the low-pass is 1 - exp(-at) - [1 - U(t-T)exp(-a(t-T))]. This is a distorted pulse with slow rise and fall times but a flat top (for T >> 1/a).
Output for the high-pass is exp(-at) - U(t-T)exp(-a(t-T)) which is also a distorted pulse with sharp rise and fall times but a drooping top (for T >> 1/a).
Totally complementary situation. Both filters distort the pulse, but in different ways.

Further: dφ/dω of the input pulse = -T/2.
So the total delay of this pulse into a network with linear phase over all ω (in which φ = -kω so dφ/dω = -k) would be -T/2 - k
which proves that a network with linear phase over all ω would not distort the pulse anywhere.
 
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  • #31
The Electrician
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Not explicitly stated because wrong.
Looky here: see the bump on the Butterworth frequency response? Have to look carefully maybe ...

The pdf attached to post #26 is page 8.25 from the Chapter 8 document found here:http://www.analog.com/media/en/trai...-Linear-Design/Chapter8.pdf?doc=ADA4661-2.pdf

However, on page 8.21 of that same Chapter 8 document under the heading "Butterworth" is found this: "The Butterworth filter is the best compromise between attenuation and phase response. It has no ripple in the pass band or the stop band, and because of this is sometimes called a maximally flat filter."

How do you reconcile the statement that the Butterworth filter has no ripple in the passband with the slight bump shown in the image on page 8.25?

The Wikipedia page also says that the Butterworth filter has no ripple.

Calculating and plotting the response of an 8 pole Butterworth filter to the same scale as the left hand image of figure 8.13 gives this result:

Butterworth.png


There's not the slightest ripple. The Butterworth response shown in the left hand image of figure 8.13 is in error. If it were truly a Butterworth response there could be no ripple.

dG/dω IS always negative because n is always positive, G is always positive and ω is always positive.
 
  • #32
rude man
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How do you reconcile the statement that the Butterworth filter has no ripple in the passband with the slight bump shown in the image on page 8.25?
Pls see my post #18: "Butterworth is not necessarily montonically non-increasing with frequency (for a low-pass). whereas Bessel is. I considered Butterworth "ripple" but better is "over-shoot". Semantic quibble.
The Butterworth response shown in the left hand image of figure 8.13 is in error. If it were truly a Butterworth response there could be no ripple.
Your word against analog devices'? Hmm ...
 
  • #33
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Your word against analog devices'? Hmm ...

It seems to be obvious that the various responses (as shown in the document under discussion) are measured curves and NOT idealized functions.
Hence, it cannot be a surprise that the curve named "Butterworth" does in fact deviate from the ideal response.
It is a well-known effect the - besides any parts tolerances - the real phase deviations of the used opamp (due to the limited GBW) cause a slight Q enhancement (which is responsible for the observed slight gain increase at the pole frequency.
 
  • #34
rude man
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It seems to be obvious that the various responses (as shown in the document under discussion) are measured curves and NOT idealized functions.
Hence, it cannot be a surprise that the curve named "Butterworth" does in fact deviate from the ideal response.
It is a well-known effect the - besides any parts tolerances - the real phase deviations of the used opamp (due to the limited GBW) cause a slight Q enhancement (which is responsible for the observed slight gain increase at the pole frequency.
I agree that's possible.
 
  • #35
The Electrician
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Your word against analog devices'? Hmm ...

No. Wikipedia's word against the image on page 8.25.

Analog Devices' word against itself. On page 8.21 of that same Chapter 8 document under the heading "Butterworth" is found this: "The Butterworth filter is the best compromise between attenuation and phase response. It has no ripple in the pass band or the stop band, and because of this is sometimes called a maximally flat filter."

I ask you again: How do you reconcile the statement on page 8.21 that the Butterworth filter has no ripple in the passband with the slight bump shown in the image on page 8.25?
"
 
  • #36
NascentOxygen
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Calculating and plotting the response of an 8 pole Butterworth filter to the same scale as the left hand image of figure 8.13 gives this result
Electrician, are you able to get your software to replicate Fig 8.14 of that document to show a comparison of the step response of your Butterworth LPF with that of a comparable Bessel?
 
  • #37
The Electrician
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Electrician, are you able to get your software to replicate Fig 8.14 of that document to show a comparison of the step response of your Butterworth LPF with that of a comparable Bessel?

I can do this, but in the process of doing so I noticed something a little off about the Bessel frequency response in figure 8.13 (left image). The response shown apparently isn't for an 8 pole Bessel filter. Here's an image showing the response of a 1 pole (red), 2 pole (magenta), 3 pole (blue) and 8 pole (Green) Bessel filter. What is shown in figure 8.13 is either a 1 pole or a 2 pole filter; it seems clear that it isn't an 8 pole filter.

Bessel1.png


Here's the step response of an 8 pole Butterworth filter:

Butter8step.png


This is the theoretically perfect step response of an 8 pole Butterworth filter. It doesn't exactly match the step response shown in figure 8.14, but that is apparently a real hardware filter and it doesn't have an accurate 8 pole Butterworth frequency response, so we shouldn't expect the step response to be exactly accurate either.

Here is the 8 pole Butterworth step response and the 2 pole Bessel step response together
ButterBesselStep.png


. I show the 2 pole Bessel because that more or less matches the frequency response of the Bessel in figure 8.13:
 
  • #38
NascentOxygen
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Electrician, if you scaled your 8 pole Bessel's frequency so its -3dB point co-incided with f=1.0 then you'd have a curve matching that in Fig 8.13, I think. It's difficult to estimate filter order in the passband without reference to the high-frequency gain rolloff or output phase range. Remarkably, it looks like a scale factor of exactly ⅓ may be what's needed, that will certainly be close enough for my purposes, anyway. The Analog Devices filters all co-incide at -3dB and all appear to rolloff at similar rates where f >> fo, so I'd say they seem fine.
 
  • #39
The Electrician
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OK. I tracked down the problem with the Bessel response. I was using Bessel poles normalized to have unit group delay.

Here is the response of the 8 pole Butterworth and 8 pole Bessel using standard Bessel poles:

ButterBesselResp.png


And here is the step response for the two:

ButterBesselStep2.png


These are the theoretically ideal curves. The frequency response of the Butterworth filter shows no ripple as stated on page 8.21, and the step response of the Bessel shows only a tiny bit of overshoot as the ideal should have. The Gaussian filter is the one with no step response overshoot.

Figure 8.15 on page 8.31 shows the Butterworth response and it appears to be calculated rather than measured. The Butterworth responses have no ripple.
 
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  • #40
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Regarding overshoot of a second-order function, here is the formula (in %):

γ=100*exp[-π/SQRT(4Qp²-1)]

As you can see, the overshoot depends on the pole-Q and is zero for Qp=o.5 only.
 
  • #41
NascentOxygen
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OK. I tracked down the problem with the Bessel response. I was using Bessel poles normalized to have unit group delay.

Here is the response of the 8 pole Butterworth and 8 pole Bessel using standard Bessel poles:

butterbesselstep2-png.png


And here is the step response for the two:

butterbesselstep2-png.png

These are the theoretically ideal curves. The frequency response of the Butterworth filter shows no ripple as stated on page 8.21, and the step response of the Bessel shows only a tiny bit of overshoot as the ideal should have. The Gaussian filter is the one with no step response overshoot.
Thank you for showing the plots are all in agreement with the Analog Devices plots, apart from their misleading Butterworth. It is remarkable to see the alacrity with which the more heavily damped 8 pole Bessell responds to a step input, in comparison with the ponderous though less damped 8 pole Butterworth. This is in stark contrast with the performance of second-order filters where it's the Butterworth that responds smartly to a step input but the Bessell is seen to become progressively slower.
 
  • #42
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Thank you for showing the plots are all in agreement with the Analog Devices plots, apart from their misleading Butterworth. .

Some days ago, I mentioned my opinion that the "misleading" Butterworth response could be a measured one. Meanwhile, I do not think this is true. We should not overlook the fact that we speak about an 8th-order (simulated) response.
There are two possible sources of error:
* Perhaps the simulation was based on real opam models (limited GBW)
* For an 8th-order filter just a small deviation of the filter coefficients from their ideal values may lead to the observed deviation.
(a1....a8: 5.126...13.138... 21.848... 25.691... 21.848... 13.138... 5.126... 1).

By the way, at the beginning of chapter 8.4 (Analog Devices) they confirm that Butterworth is identical to max. flat.
 
  • #43
NascentOxygen
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Analog Devices are publishers of great tutorial-style material, as evidenced by the helpful detail in the chapter we have been referred to by rude man. I said their graph is misleading simply because they include a plot which shows a departure from theory but they fail to acknowledge this discrepancy and don't offer an explanation. Possibly it is the dominant pole of an amplifier, but they need to explain this.
 
  • #44
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Possibly it is the dominant pole of an amplifier, but they need to explain this.
Yes - i am of the same opinion.
 

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