# cascading two single pole low pass filters

Discussion in 'Homework Help' started by PG1995, Jun 2, 2012.

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1. ### PG1995Active Member

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Hi

Could you please help me with the query included in the attachment? Thanks a lot for the help.

Regards
PG

2. ### ChrisP58Well-Known Member

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No. The roll of the two will add. The first filter has -10db/decade, the second will attenuate that signal -10db/decade again, for a total of -20db/decade.

3. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

When you connect two circuits like this where the impedance of the first stage does not interfere with the impedance of the second stage (output impedance of first stage is very low) then the two filter functions simply multiply.

For a general low pass filter we have a transfer function like this:
Hs1=1/(s*A+1)

and if we put two of them together like your circuit we have two of the same responses multiplied together so we get:
Hs2=Hs1*Hs1

and so we have for the combined response:
Hs2=1/(s*A+1)^2

The simplest cases are with A=1 so Hs1 becomes:
Hs1=1/(s+1)

and Hs2 becomes:
Hs2=1/(s+1)^2

The amplitude of Hs1 is:
Ampl1=1/sqrt(w^2+1)

and the amplitude of Hs2 is:
Ampl2=1/(w^2+1)

So we see we have Ampl1 is the square root of Ampl2, which means the slope is less for Ampl1.
Also, a plot of these two would reveal the different response slopes.

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

When you connect two circuits like this where the impedance of the first stage does not interfere with the impedance of the second stage (output impedance of first stage is very low) then the two filter functions simply multiply.

For a general low pass filter we have a transfer function like this:
Hs1=1/(s*A+1)

and if we put two of them together like your circuit we have two of the same responses multiplied together so we get:
Hs2=Hs1*Hs1

and so we have for the combined response:
Hs2=1/(s*A+1)^2

The simplest cases are with A=1 so Hs1 becomes:
Hs1=1/(s+1)

and Hs2 becomes:
Hs2=1/(s+1)^2

The amplitude of Hs1 is:
Ampl1=1/sqrt(w^2+1)

and the amplitude of Hs2 is:
Ampl2=1/(w^2+1)

So we see we have Ampl1 is the square root of Ampl2, which means the slope is less for Ampl1.
Also, a plot of these two would reveal the different response slopes.

In the attachment blue is Ampl1 and red is Ampl2...

Last edited: Jun 3, 2012
6. ### WinterstoneBanned

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Just a small correction: A simple first order lowpass always has a slope of -20 dB/dec.
(That means: 10 dB/dec are impossible).

A comment to cascading two first order sections:
You get a second-order lowpass with two real poles which has a very bad selectivity.
With one opamp and 2 RC combinations you can realize a classical 2nd order filter with complex pole pair and good selectivity.

7. ### PG1995Active Member

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Thank you, Chris, MrAl, Winterstone.

@MrAl: Thanks a lot for the detailed reply.

Hi

Please have a look here and kindly help me with the query. Thanks.

Regards
PG

8. ### WinterstoneBanned

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PG1995,
as mentioned before, the single opamp circuit allows for a complex pole pair - which means (as you have mentioned) that all known responses (Butterworth, Chebyshev,...) are possible.
Nobody would use the single pole cascade as shown in Fig. 2. because it has no advantages.

Remark: The 2nd order circuit is rather sensitive to opamp gain variations (a known disadvantage of this Sallen-Key topology). Thus, I recommend to use a gain of unity or a gain of two with two identical resistors.
Design formulas are available for both alternatives.

Last edited: Jun 3, 2012
9. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I think what you meant was that circuit 1 has the sharpest response and circuit 2 is not as good.

The circuit with the op amp has the ability to force a response which is sharper than just two low pass filters cascaded. The drawback is sometimes there is overshoot where the response goes up high before it starts to drop down as a normal low pass would. We can take a look at the two responses so you can see by direct examination of the responses why they are different.

To put it more simply, the Q of circuit 1 can be made higher than the Q of circuit 2.

With R1=R2=R and C1=C2=C we get the same response for circuit 1 as for circuit 2.
With C1>C2 we should see a greater slope in circuit 1 as compared to circuit 2.
So you can see that the values must be properly set to see a better response in circuit 1.

Last edited: Jun 3, 2012
10. ### WinterstoneBanned

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Quote MrAl: The drawback is sometimes there is overshoot where the response goes up high before it starts to drop down as a normal low pass would.

Perhaps it sounds "academical" only - but I like to point out that for my opinion the magnitude peaking of some higher-Q responses (chebyshev approximations) cannot be considered as a "drawback". It is nothing else than the logical consequence of a pole Q>0.707.
More than that, for higher-order lowpass filters (n>3) consisting of a cascade of 2nd order stages this effect even is wanted and necessary! Example: A 4th order Butterworth filter must contain one stage having this property (gain peaking with Q>0.707).

11. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

Yes it's only a drawback when you dont want an overshoot, ie when you want a flat response up to the cut point. I brought this up because sometimes the overshoot is way high and the circuit almost acts like a bandpass instead of low pass. Sure, if you want a bandpass that's great but if you want a low pass that's not good

12. ### WinterstoneBanned

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Yes, I understood the background of your remark. However, if I cannot accept such a gain peaking (by the way: I believe the term "overshoot" applies to the time domain) I do not choose a pole Q>0.707 (Butterworth).
For my opinion, more interesting is the question: Which topology? (Sallen-Key, multi-loop feedback, or something else?).

13. ### PG1995Active Member

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Thank you, Winterstone, MrAl.

Regards
PG

Last edited: Jun 3, 2012
14. ### WinterstoneBanned

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Hi PG1995,

Q1: The answer is simple: A 1st order lowpass is a 1st order lowpass. That`s all. There are no alternatives like Butterworth, Chebyshev,....The only differeces are different gain values. However, they do not influence the filter response at all (amplifier is not part of the RC network - it only amplifies).

15. ### PG1995Active Member

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Thanks a lot, Winterstone.

I don't know why you weren't able to open Q2. Could you please try again? Please let me know if the problem persists. Thanks.

Regards
PG

16. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG,

Q1: Look at that statement "Sallen-Key Butterworth configuration" as a specification rather than a misnomer. Thus, he is specifying what kind of filter they *will* be rather than what kind they *could* be. In other words, think of those two filters with the right values to make them Butterworth. With this in mind, read it over again.

Q2: It is probably that the Butterworth factor can vary while the Chebyshev must be of a certain value.

The word "overshoot" is a generic English word that can be used in various contexts to indicate something that has gone beyond some limit.

Last edited: Jun 3, 2012
17. ### WinterstoneBanned

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Quote MrAl: ....... In other words, think of those two filters with the right values to make them Butterworth. With this in mind, read it over again.

*MrAl, do you really believe that two first order lowpass sections in cascade (each with a real pole) can form a 2nd order Butterworth response with a complex pole pair? I don`t think so.

*PG1995, as to Q2: You are right that - based on the shown circuit diagrams - one cannot speak about a specific 2nd order response (like Butterworth). However, in connection with the shown transfer curve (pass region of a bandpass) it is clear - because of the maximum flat characteristic - that it is a Butterworth bandpass.
Any further question?

18. ### lebevtiMember

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what is meant by "selectivity"?

19. ### WinterstoneBanned

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I think, the purpose o a filter is to select a certain frequency band - either to let it pass or to provide attenuation.
In reality, there is always a certain transition band between the pass band and the stop band region.
The selectivity of a filter is high (low) in case of a small (broad) transition region.
These regions are defined for each application by the user based on certain damping requirements.

*Example: Pass band of a low pass from 0 to 1 kHz. Attenuation in the stopband (starting at 2 kHz) at least 40 dB.
A filter with the same pass band but with a stop band attenuation of at least 60 dB would have a higher (better) selectivity. OK ?

20. ### lebevtiMember

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ok, gotcha...i know the concept u speak of, just never heard of the term 'selectivity'

21. ### WinterstoneBanned

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In most cases this term ("selectivity") is used to describe the bandwidth of a bandpass.
Selectivity is proportional to the quality factor Q of the bandpass and invers proportional to its bandwidth..

Additional remark: Search in "Google" or Wikipedia for "selectivity". You will find a lot of explanations.

Last edited: Jun 3, 2012