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Topic Title: 2 Pole High Pass Filter
Topic Summary: Is there a simple equation to calculate how much my signal will be attenuated by a 2 pole filter?
Created On: 27 March 2012 09:02 AM
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 27 March 2012 09:02 AM
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safortune

Posts: 8
Joined: 27 October 2010

Hi all,

I have a circuit that has 2 high pass filters and low pass filter all cascading. I can work out there individual cutoff frequencies to be HPF1 = 1.44Hz, LPF1 = 160kHZ and HPF2 = 1.6Hz.

My first question is, if I have a 8Hz sin wave how much will my signal be attenuated by after passing through the three filters? Is there a simple calculation for this?

It is the cascading effects that are confusing me here I think: I would say that the two high pass filters should be multiplied together so that 0.707 and 0.707 would give me an attenuation of 0.5 of the voltage is this correct?

Any and all advice would be much appreciated. Thanks in advance.

regards,

Steve
 27 March 2012 01:50 PM
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ANFierman

Posts: 138
Joined: 25 July 2008

Hi Steve,

Depends if your filters are buffered from each other or not.

Calculating buffered filter stages is relatively simple. Calculate effect of each stage separately then multiply them all together taking account of any gain in the buffer stages.

Calculating buffered filter stages is horrendous. There are whole textbooks on it. Realistically no-one does it by hand any more. There are various online tools and bits of s/w you can use to good effect but to really see what's happening (and especially at HF & VHF) you need to use a SPICE or similar simulation tool.

Above around VHF, SPICE can be used but RF and Microwave simulation tools offer more ways to account for parasitics etc., that matter less at lower RF. SPICE can deal with s-parameters but there are better tools for that.

QUCS is a good if rather quirky tool for general simulation but is particularly good for RF:
http://qucs.sourceforge.net/

For what you're doing , you might like to have a look at:

https://www.circuitlab.com/circuit/hgw6zf/simple-cascaded-filters-01/

which I just put together in CircuitLab.
https://www.circuitlab.com/

You can play with the circuit in your browser (note that IE is not supported, so use a real browser) but to save any changes you will need to sign up for a (free) account.

Been using CircuitLab a lot recently. As a learning/teaching aid it's excellent. As a serious simulation tool it's a long way behind the free, unlimited, mixed mode, use any vendor's models, industrial strength LTspiceIV from:

http://www.linear.com/designtools/software/#LTspice

-------------------------
Andy Fierman

---------------------------
http://signality.co.uk
---------------------------
 27 March 2012 02:41 PM
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safortune

Posts: 8
Joined: 27 October 2010

Hi,

Thanks for getting back to me. The filters are buffered by op amps (of gains 10 and 4.75 respectively).

I just had a look at the circuitlab you kindly created for me. It does like a useful tool but I wonder if you can help me understand the theory better?

I read that to calculate the cutoff of a 2nd order HPF, I should use the following equation:

1/[2(pi)sqrt(R1R2C1C2)]

where the Rs and Cs are the respective components for each of the high pass filters. Since my signal will between 1 and 8Hz i thought it safe to assume there will be no effect from the 160KHz LPF.

Am I correct to assume there will be no effect on the frequency by the op amp buffers? Hence the "taking into account the gains" refers to the amplitude of the signal being passed through?

So if I have understood correctly to this point: I have a 2nd order high pass filter with an Fc of 1.5Hz and I pass a 500mV pk 2 pk signal through the circuit @ 8Hz is there a calculation I can use to find the level of attenuation at 8Hz?

Your help is very much appreciated.

Steve
 27 March 2012 11:09 PM
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ANFierman

Posts: 138
Joined: 25 July 2008

Can't think clearly enough just now about the maths of cascading filters ...

"the cutoff of a 2nd order HPF, ... 1/[2(pi)sqrt(R1R2C1C2)] ... " is too much of a simplification.

Imagine what would happen with two lpfs where R1C1 >> R2C2.

You get one rolloff then another much further away. The cutoff frequency would be set by R1C1 leading into a 20dB/decade (6dB/Octave) slope with R2C2 steepening the the slope to 40dB/Decade (12dB/Octave).

Try it in CircuitLab.

This might help:

https://www.circuitlab.com/circuit/3dfhet/simple-cascaded-filters-02/

Basically if the transfer function of a stage is H(s) and the transfer function of a 2nd stage is G(s) and there's a linear gain K in between them then the overall transfer function is:

H(s)*K*G(s)

The simulation shows the three stages as Laplace transfer function blocks buffered by unity gain between each (i.e. K1 = 1, K2 = 1)

So the overall transfer function is:

1/(1+(2*pi*1.44)/s) * 1/(1+(2*pi*1.44)/s) * 1/(1+s/(2*pi*160k))

If you ignore the 160kHz lpf stage this simplifies to

1/(1+(2*pi*1.44)/s) * 1/(1+(2*pi*1.44)/s)

Which if you do the algebra will give you the classic 2nd order transfer function.

You can then work out the magnitude and the phase by substituting s = j*2*pi*f and grinding through more algebra.

This is a reasonable introduction:

http://www.ti.com/lit/an/snoa224a/snoa224a.pdf


There's much better around but haven't got time to find it out for you.

If you get more involved in trying stuff like this out I recommend you get LTspice and learn that. It's far more well developed and although has maybe a steeper learning curve, is ultimately a much more useful and powerful tool.

Some examples here:

https://docs.google.com/open?id=0B2wonnsWWfTXOTVmNDU2Y2MtYzJlZi00M2MyLWExZDAtM2JhY2NiNGI4ODBi



-------------------------
Andy Fierman

---------------------------
http://signality.co.uk
---------------------------

Edited: 27 March 2012 at 11:59 PM by ANFierman
 28 March 2012 10:54 PM
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ANFierman

Posts: 138
Joined: 25 July 2008

Found some better search terms ...

gain and phase in simple passive filters

and turned this up:

http://www.electronics-tutoria.../filter/filter_2.html

This has the

fc = 1/[2(pi)sqrt(R1R2C1C2)]

equation, which is a bit dubious for reasons already explained.

The stuff starting at the heading "The Low Pass Filter" and ending at the heading "Second-order Low Pass Filter" is OK for understanding how to work out the attenuation of a 1st order RC lpf. The section under "Second-order Low Pass Filter" is poorly explained and makes no mention of the fact that the two 1st order stages are unbuffered and so interact.

The summary section and the bit about time constant is OK.

These pages are OK too:
http://en.wikipedia.org/wiki/RC_circuit
http://www.learnabout-electron..._theory/filters82.php
http://zone.ni.com/devzone/cda/tut/p/id/13236#toc0
http://www.analog.com/library/.../EDCh%208%20filter.pdf

-------------------------
Andy Fierman

---------------------------
http://signality.co.uk
---------------------------
 29 March 2012 09:27 AM
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safortune

Posts: 8
Joined: 27 October 2010

I can't say I fully understand all the theory yet but I will delve more deeply when I have more time. For now, I know what I need to, to progress with my project, so thanks for that.

I will look at your links when I have more time as I could definitely benefit from understanding it better.

Your help is much appreciated.

Cheers,

Steve
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