# Low Pass Filter

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#### Mechatronics Eng

##### New Member
Hi friends,

I have a Low Pass Filter circuit connected as shown in the attached file. Please, would help to let me know how to connect oscilloscope to check input and output voltages and filter cutoff frequency. Thanks

#### Winterstone

##### Banned
Hi Mech-Ing,

sorry, but I do not understand your problem.
When you know how to operate a scope you should know what it can measure - and how.
And when you have designed a filter for a specific application you should know what it does (resp. what it shall do).
So - what information do you need in addition?

#### MrAl

##### Well-Known Member
Most Helpful Member
Hi guys,

Winterstone:
He's probably just trying to test this filter to see if it does what he wants it to do, or something like that

Mech:
Connect channel A probe to Vin, connect channel B probe to Vo.
Using a DC input of 1v, drive the filter input Vin. After about a second measure Vo and make a note of that call it VoDC.
Now using a sine wave generator connected to Vin with frequency of 1Hz and with a peak of 1v do the same, measure the output, call it Vo1.
Now using the sine generator again set to 10Hz do the same again, measure the output, call it Vo10.
Next set the sine to 100Hz, do the same again, call the output Vo100.
Again for 1000Hz, call the output Vo1k.
Again for 10000Hz, call the output Vo10k
Now you have readings at DC,1Hz, 10Hz, 100Hz, 1kHz, and 10kHz.

The idea now is to find the frequency that caused an output that is 0.707 times the DC output level. That is the cutoff frequency.

Lets look at an example...
Say when you input 1 DC you get an output of 1v DC (you actually should see this with this filter).
Then at 1Hz you get an output of around 1v peak.
Then at 10Hz you get an output of around 1v peak.
Then at 100Hz you still get an output of around 1v peak, maybe a tiny tiny bit less.
Then at 1000Hz you notice you get an output close to 0.7v peak.
Then at 10kHz you get an output around 25mv peak or close to zero.
Now you examine all the values above and notice that when you multiply the reading at 1kHz by the DC level (which was 1) you get 0.7, so that means 1kHz is very close to what the cutoff frequency really is.
So now you set the input side to 900Hz and test again, you get an output of 0.75v peak.
Then you set the input side to 1100Hz and test again, you get an output of 0.68v peak.
Now you can see that 0.707 (the number we are looking for here) is between 900Hz and 1100Hz, so you adjust the frequency generator until you see an output of very close to 0.707v peak and that is the cutoff frequency.

Keep in mind the above was for a filter with an output of 1v DC for an input of 1v DC. If the filter put out 2v DC for an input of 1v DC you would be looking for a number that is twice 0.707 which is 1.41v instead.
Also, this procedure is for a low pass filter only which can pass DC as well as AC. If it only could pass AC we'd have to use a very low frequency like 1Hz as the reference instead of 1v DC.
If this wasnt a low pass filter we'd have to use a different procedure to find the reference output.

Just to note, your filter here should have a cutoff of just over 1kHz.

#### Winterstone

##### Banned
Hi guys,
Winterstone:
He's probably just trying to test this filter to see if it does what he wants it to do, or something like that

Yes - I could imagine.
However, have you ever designed a circuit without knowing how to test it?

#### MrAl

##### Well-Known Member
Most Helpful Member
Yes - I could imagine.
However, have you ever designed a circuit without knowing how to test it?

Hi there Winterstone,

Well he never said he designed it, he just said he wanted to connect the scope to it and check the input and output and also the cutoff frequency, right? Some people are just getting started in electronics here so are not seasoned like you are.

Original post:

Hi friends,

I have a Low Pass Filter circuit connected as shown in the attached file. Please, would help to let me know how to connect oscilloscope to check input and output voltages and filter cutoff frequency. Thanks

Last edited:

#### Mechatronics Eng

##### New Member
Winterstone ,
Thank you for your reply. As MrAl said, I am probably just trying to test this filter and see if the circuit works!

MrAl ,

Thank you very much for your prompt response and I am deeply grateful to you and for your detailed explanation.

I just want to tell you that I have set wrong values of the capacitors. Their values should be ( 0.01µF ) , not ( 0.01 mF ). Ok.

From your expression , I can conclude your idea. Please, see attached file. If it is right, Please, could you explain how did you reach that ( cut off frequency = 0.707 * VoDC ) ? is it from the formula : (ƒc=1\ 2π RC) ?

#### MrAl

##### Well-Known Member
Most Helpful Member
Hi,

Well actually the frequency itself is not 0.707 times VoDC, but rather 0.707 (sometimes approximated as 0.71 or even 0.7) times VoDC is the amplitude at which the cutoff frequency will produce at the output with a 1v input.
In other words, with 1v input AC you'll see 0.7v output AC when you reach the right frequency, which is the cutoff frequency.

I assumed your caps were 0.01uf.

#### Winterstone

##### Banned
Hi there Winterstone,
Well he never said he designed it, he just said he wanted to connect the scope to it and check the input and output and also the cutoff frequency, right? Some people are just getting started in electronics here so are not seasoned like you are.

Well, so it was a misunderstanding on my side (only now I realize the name of the questioner: mechatronics_eng) .
In this case, he will be happy to get such a detailed test procedure from you.

#### Winterstone

##### Banned
Hi mech_eng,

a short comment to your filter circuit.
It is an equal-component Sallen-Key topology with unity gain (the resistor in the feedback path has no effect) - and, thus, has a pole quality factor of only Qp=0.5.
This value is the lower limit for an active realization since this value can be realized also as a pure passive RC filter.
As a consequence, the magnitude response at the end of the passband (approx. 1 kHz) is not as "constant" as it could be.

Therefore, my suggestion: Increase the upper capacitor from 0.01uF to 0.02uF . This will give you a maximally flat response (Butterworth approximation) with Qp=0.707.
It is well known that such a lowpass response is a good trade-off between time and frequency domain. By the way: In this case, the 3-dB cut-off will be at 1.1 kHz.

#### MrAl

##### Well-Known Member
Most Helpful Member
Hi again,

Winterstone:
Oh you've talked to him before this?

Mech:
The calculation for the cutoff frequency is:
wo=(sqrt((sqrt(C1^2*R2^4+(4*C1^2-4*C1*C2)*R1*R2^3+(8*C2^2-8*C1*C2+6*C1^2)*R1^2*R2^2+(4*C1^2-4*C1*C2)*R1^3*R2+C1^2*R1^4)-C1*R2^2+(2*C2-2*C1)*R1*R2-C1*R1^2)/(C1)))/(sqrt(2)*C2*R1*R2)

but if the two capacitors are the same value so that C2=C1 then we have:
wo=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(sqrt(2)*C1*R1*R2)

then using either formula the cutoff frequency in Hertz is:
fo=wo/2/pi

but the calculation for the amplitude is simpler:
w=2*pi*f with f in Hertz, then:
Ampl=Vin/sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C1*R2+w*C1*R1)^2)

The phase shift is calculated from:
x=((1-w^2*C1*C2*R1*R2))/((1-w^2*C1*C2*R1*R2)^2+(w*C1*R2+w*C1*R1)^2)
y=((-w*C1*R2-w*C1*R1))/((1-w^2*C1*C2*R1*R2)^2+(w*C1*R2+w*C1*R1)^2)

then the actual phase shift in degrees is:
TH=atan2(y,x)*180/pi

and where atan2(y,x) is the two argument inverse tangent.

But if again C2=C1 then we have a simpler formula for the phase shift:
x=(1-w^2*C1^2*R1*R2)
y=(-w*C1*R2-w*C1*R1)
TH=atan2(y,x)*180/pi

where TH is again the phase shift in degrees.

The atan2(y,x) two argument function can be transformed into the atan(y/x) single argument form with a few changes and observation that the function changes for a certain threshold frequency but it's not worth the effort.

All these calculations of course assume a perfect op amp. With a real life op amp you may have to limit the input voltage level if it is a cheap op amp like the LM358 or go to a better quality op amp.

Audioguru has some interesting comments in the post that follows this one. The op amp will play an increasingly important role in the circuit as the frequency gets higher. It may be ok around 1kHz but up near 10kHz or so the input will have to be limited, but it's best to figure in the slew rate also to be sure.

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#### audioguru

##### Well-Known Member
Most Helpful Member
The opamp has no power supply so it will not work.
The input of the opamp is not shown to be biased at half the supply voltage so it will not work.
The lousy old LM324 quad opamp is too slow to make a half-decent filter.
The LM324 has crossover distortion and is noisy (hiss).

#### Winterstone

##### Banned
Hi again,
Winterstone:
Oh you've talked to him before this?

Why do you think so? I don`t understand.

Mech:
The calculation for the cutoff frequency is:
wo=(sqrt((sqrt(C1^2*R2^4+(4*C1^2-4*C1*C2)*R1*R2^3+(8*C2^2-8*C1*C2+6*C1^2)*R1^2*R2^2+(4*C1^2-4*C1*C2)*R1^3*R2+C1^2*R1^4)-C1*R2^2+(2*C2-2*C1)*R1*R2-C1*R1^2)/(C1)))/(sqrt(2)*C2*R1*R2)

but if the two capacitors are the same value so that C2=C1 then we have:
wo=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(sqrt(2)*C1*R1*R2)

With regard to the above set of formulas I like to place some comments.
I must confess that I have some problems reading and evaluating the very long expressions – nevertheless, I am not sure about the numerator in the shown formulas and if the whole expression can be rearranged to satisfy the required form w=1/T with T=RC time constant. More than that, I don’t know how to interprete the given expression for the denominator (SQRT(2)*C1*R1*R2).

Anyway, in order not to „frustrate“ beginners (and guys like the questioner) I point to the fact that in practice filter design is not so complicated as it seems (looking at the formulas from MrAl).
In contrary, it’s very simple as shown next:

1.) For the given circuit the so called pole frequency is
wp=1/SQRT(T1*T2) with T1=R1C1, T2=R2C2.

2.) The relation between wp and the cutoff frequency wc=a*wp is simply a fixed factor a, which depends on the selected lowpass approximation.
These factors can be derived from tables contained in all textbooks (and wikipedia, etc....) dealing with filter design.
Examples: Butterworth a=1, Bessel a= 0.577, Chebyshev (1 dB) a=0.952.

3.) After selection of a suitable approximation (a value) the corresponding pole quality factor Qp can be derived also from these tables.
Butterworth case: Qp=0.7071 (Bessel Qp=0.577, Chebyshev (1dB) Qp=0.9565).
For the given Sallen-Key circuit and R1=R2 we have

Qp=SQRT(C2/C1)/2 (with C2 shown as the upper capacitor in the diagram).

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#### MrAl

##### Well-Known Member
Most Helpful Member
Hi there Winterstone,

You're right in that there are other ways of looking at a filter such as in terms of it's poles, and you should write more about this which could be helpful. What i attempt to do is to present an exact formula for the required quantity and that's what i did here, but once in a while i'll throw in an approximation instead or in addition to. I also dont make too much attempt to reduce the formula to absolute simplest form as i often leave that up to the reader. The formulas i present do in fact work and if you would like to simplify that is entirely up to you to do so, and i think this of course has merit.

You will notice though that for this particular solution of wo i did simplify because most people make both caps the same value, so it came down to a relatively simple formula that needs no extra reference material to decipher:
wo=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(sqrt(2)*C1*R1*R2)

and you can see that's not really that complicated. Since you had a problem with the denominator for some reason i'll show this as:
wo=N/D

where
N=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)
and
D=sqrt(2)*C1*R1*R2

where you can see the denominator D is simply C1 times R1 times R2 times the square root of 2, not too hard to figure out

You see that by making use of this formula we dont need to reference anything else because it is complete in itself.

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#### cowboybob

##### Well-Known Member
Most Helpful Member
While I'm a big fan of bread boarding, sometimes a circuit simulator is just simpler.

Below is a TINA-TI3 (available free from TI.COM) circuit more or less similar to the one you provided:

It is a great aid, when analyzing signal filters, to employ a sweeping signal to the circuit to allow viewing the resultant effect(s) to the input signal.

This could also be seen using the TINA's on board oscilloscope. In this case, however, the signal analyzer gives a more definitive visual representation of the circuit's response to a variable frequency input to the circuit.

Other members on this forum prefer different simulators that are also free.

View attachment 66212

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#### Winterstone

##### Banned
Hi there MrAl,

thanks for replying. May I place some final comments?

Hi there Winterstone,
You're right in that there are other ways of looking at a filter such as in terms of it's poles, and you should write more about this which could be helpful. What i attempt to do is to present an exact formula for the required quantity and that's what i did here, but once in a while i'll throw in an approximation instead or in addition to. I also dont make too much attempt to reduce the formula to absolute simplest form as i often leave that up to the reader.
I absolutely agree - in principle, I prefer the same approach.

You will notice though that for this particular solution of wo i did simplify because most people make both caps the same value, so it came down to a relatively simple formula that needs no extra reference material to decipher:
wo=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(sqrt(2)*C1*R1*R2)

OK, I see that the frequency has the form wo=R/C*R^2=1/RC leading to the unit 1/sec.
Regarding both capacitors: Yes, very often both caps are made equal - provided this choice allows the wanted filter characteristic. And - as mentioned before - for the circuit under discussion this choice would allow a maximum pole Q of 0.5 only. For the shown circuit the choice R1=R2 (as already shown in the circuit diagram) allows more flexibility.

Since you had a problem with the denominator for some reason i'll show this as:
wo=N/D where
N=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)
and
D=sqrt(2)*C1*R1*R2
where you can see the denominator D is simply C1 times R1 times R2 times the square root of 2, .

As mentioned before, I didn't recalculate your formulas - and because in 99% of all similar formulas the denominator of a 2nd-order filter function has the form D=sqrt(R1R2C1C2)
I was not sure about the sqrt(2) in your expression. Therefore my question. But it's clear now to me. Thank you.
W.

#### Winterstone

##### Banned
While I'm a big fan of bread boarding, sometimes a circuit simulator is just simpler.

View attachment 66212

Hi cowboybob,

There are tenth of different lowpass topologies - each with its own advantages and disadvantages. This makes the decision to select one specific structure a typical trade-off process.

And your simulation results remind me to point to a big disadvantage of the Sallen-Key lowpass topology.

As can be seen in your diagram, for larger frequencies the magnitude of the transfer function is increasing again (poorer attenuation within the stopband).
Some people think that this would be a parasitic effect caused by the simulator. However, that's not true.
This effect can be observed also in reality (hardware). It is caused by the direct signal path between input and output through the feedback capacitor.
For increasing frequencies the output signal from the opamp becomes smaller and smaller - until the signal portion coming through the feedback C becomes dominant
because it creates an output voltage across the finite opamp output impedance. This effect sometimes is called "tail effect".
This leads to the recommendation to select for Sallen-Key filters an opamp with a very small output impedance (even for unity gain operation).

W.

#### MrAl

##### Well-Known Member
Most Helpful Member
Hi Winterstone,

Yes the units should work out ok as these formulas are checked before posting. Sure a mistake could occur however.

For the case where both resistors are the same R2=R1 but the caps are not the same, we can simplify the main formula again to include only R1 instead of both R1 and R2 with the assumption that R2=R1. This provides the following simpler formula:
wo=sqrt(sqrt(2)*sqrt(C2^2-2*C1*C2+2*C1^2)+C2-2*C1)/(sqrt(C1)*C2*R1)
fo=wo/2/pi

If we happen to have both resistors the same and both capacitors the same with both C2=C1 and R2=R1 we have a very very simple formula:
wo=sqrt(sqrt(2)-1)/(R1*C1)

#### cowboybob

##### Well-Known Member
Most Helpful Member
a big disadvantage of the Sallen-Key lowpass topology.

Hi W.,

To finish the quoted sentence fragment: "is that it ain't much good for nothin' ".

Just my opinion.

To be sure, you can get some deep attenuation (up to 100 dBs) but it can hardly be considered a "sharp-tailed" low pass configuration and is pretty much relegated to attenuating audio freqs above 100Hz.

So, with that in mind, as a mid-range/treble audio blocker, it works pretty well. Guess it might be, in fact, "good for something".

Anyway, just following the OPs choice of the topology. Got no idea, of course, what he's got planned for it.

#### audioguru

##### Well-Known Member
Most Helpful Member
The lousy old 741 opamp ruins the filter. A more modern and half-decent opamp should be used.

#### Mechatronics Eng

##### New Member
Well, so it was a misunderstanding on my side (only now I realize the name of the questioner: mechatronics_eng) .
In this case, he will be happy to get such a detailed test procedure from you.

Thank you Wint. that is true.

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