Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

cutoff frequency

Status
Not open for further replies.

SneaKSz

Member
Hi all,

I'm struggling with the following opamp circuit which acts like a filter :
Capture.JPG

H=1+ Z2/Z1

Z1 = R2+1/sC2 = (s.C2.R2+1)/sC2
Z2 = (R1/sC1)/(R1+1/sC1)= (R1/sC1)/((s.C1.R1+1)/sC1)= R1/(s.C1.R1+1)
H= 1+R1/(s.C1.R1+1) / (s.C2.R2+1)/sC2 =1+ (R1*sC2)/((s.C1.R1+1)*(s.C2.R2+1))

and now things are about to heat up!

H=1+ (R1*sC2)/((s.C1.R1+1)*(s.C2.R2+1)) = (((s.C1.R1+1)*(s.C2.R2+1)) + (R1*sC2))/((s.C1.R1+1)*(s.C2.R2+1))

So the denominator is fine now = (s.C1.R1+1)*(s.C2.R2+1)

now the numerator : (s.C1.R1+1)*(s.C2.R2+1) + (R1*sC2) = s.C1.R1.C2.R2+s.C1.R1+s.C2.R2+1+R1*sC2
now the options are :
: s.C2.(R1+R2)+s.C1.R1.C2.R2+s.C1.R1+1
: s.R1.(C1+C2)+s.C1.R1.C2.R2+s.C2.R2+1
:s.R1.C2(1+s.R2.C1)+s.C1.R1+s.C2.R2+1

none of them can lead to a result as = X( 1+s.CxRx)(1+s.CxxRxx)

Someone who sees something I'm unable to?

Cheers
 
It is a simple filter that cuts high frequencies.
At the frequency that the reactance of C1 equals the resistance of R1 then the gain of the circuit is 0.707 times (-3db) the gain at much lower frequencies. The output level drops at 6dB per octave.
 
Well I want to calculate those frequencies, that's the point. I know how the filter works but I'm unable to get those frequencies ( poles and zero's).

Kind regards
 
I hired Mathematica program to find the transfer function of your circuit.
And this is what I get

23.PNG

And the plot

Opamp4.png
 

Attachments

  • Opamp4.zip
    28.1 KB · Views: 133
Last edited:
I also analysis this circuit separately
For this circuit
Capture1.JPG

I got one Zero at

[LATEX]Fz = \frac{1}{2*\pi*C2*(R1+R2)}[/LATEX]

And one Pole at

[LATEX]Fp = \frac{1}{2*\pi*C2*R2}[/LATEX]

And for this circuit

Capture.JPG

One zero at

[LATEX]Fz = \frac{1}{2*\pi*C1*(R1||R2)}[/LATEX]

and one Pole

[LATEX]Fp = \frac{1}{2*\pi*C1*R1}[/LATEX]
 
Last edited:
Hi all,

I'm struggling with the following opamp circuit which acts like a filter :
View attachment 55993

H=1+ Z2/Z1

Z1 = R2+1/sC2 = (s.C2.R2+1)/sC2
Z2 = (R1/sC1)/(R1+1/sC1)= (R1/sC1)/((s.C1.R1+1)/sC1)= R1/(s.C1.R1+1)
H= 1+R1/(s.C1.R1+1) / (s.C2.R2+1)/sC2 =1+ (R1*sC2)/((s.C1.R1+1)*(s.C2.R2+1))

and now things are about to heat up!

H=1+ (R1*sC2)/((s.C1.R1+1)*(s.C2.R2+1)) = (((s.C1.R1+1)*(s.C2.R2+1)) + (R1*sC2))/((s.C1.R1+1)*(s.C2.R2+1))

So the denominator is fine now = (s.C1.R1+1)*(s.C2.R2+1)

now the numerator : (s.C1.R1+1)*(s.C2.R2+1) + (R1*sC2) = s.C1.R1.C2.R2+s.C1.R1+s.C2.R2+1+R1*sC2
now the options are :
: s.C2.(R1+R2)+s.C1.R1.C2.R2+s.C1.R1+1
: s.R1.(C1+C2)+s.C1.R1.C2.R2+s.C2.R2+1
:s.R1.C2(1+s.R2.C1)+s.C1.R1+s.C2.R2+1

none of them can lead to a result as = X( 1+s.CxRx)(1+s.CxxRxx)

Someone who sees something I'm unable to?

Cheers


Hi there,

First of all i see you dont have the numerator written correctly. It contains at least one term with s to the power of 2 in it making this a second order equation.

As you probably know, you can solve for the roots and therefore factor the numerator using the quadratic formula. This will lead to a little more complicated solution than for the denominator of course, but that's one way to do it. Example:
Code:
s=(sqrt(C2^2*R2^2+(2*C2^2-2*C1*C2)*R1*R2+(C2^2+2*C1*C2+C1^2)*R1^2)+C2*R2+(C2+C1)*R1)/(2*C1*C2*R1*R2)
That's not as simple as we would like to see but that's part of a solution in factoring the numerator:
s^2*C1*C2*R1*R2+s*C2*R2+s*C2*R1+s*C1*R1+1

which of course can be written as:
s^2*C1*C2*R1*R2+s*(C2*R2+C2*R1+C1*R1)+1

You might also however take note that all the component values are given in this exercise, so you dont really have to work out the problem symbolically unless of course you want to do that anyway. Numerically it's just a matter of finding the two roots of the numerator and that will give you the required numerical values to use to factor it.

Either way though you can use the quadratic formula to get the results you are after. You're lucky this is only second order as these are much simpler than third or higher order transfer functions.
 
Last edited:
Hello Jony and mrAl,

I have to thank you guys for the work and effort!
I understand the formulas that you guys are using!

I use the program from texas ( TI nspire) to get the zero's of this second order function :
Capture.JPG

and then a bode plot:
bode.JPG

so the 2 zero's have to be : 133965 Hz and 4654.51 Hz
and the 2 poles : 10261 Hz and 1539 Hz

so then I marked the points of these cut off frequencies:

My question is: do these frequencies even make sense? Because the - 3dB points are not really that correctly?

Or is it because I'm working with a region between 0 dB and 5 dB so it's hard to accomplish these dB's and frequencies?

Thanks for the effort!

kind regards
 

Attachments

  • bode21.jpg
    bode21.jpg
    139.8 KB · Views: 182
Form your transfer function we can find the Center frequency (mid-band frequency)

s² + s (ωo/Q) + ωo²

01-png.56054


[LATEX]Fo = \frac{1}{2*\pi\sqrt{C1C2R1R2}} = 3.974238KHz [/LATEX]

And we also can find the Q factor

The numerical transfer function look like this

[LATEX]
H(s) = \frac{s^2 + 138620.24500s + 6.23545*10^8}{s^2 + 74145.71244s + 6.23545*10^8}[/LATEX]

And from this we can find two Poles at

P1 = 1.53921665825KHz
P2 = 10.2614399856KHz

And we also have two Zeros

Z1 = 740.788454832Hz
Z2 = 21.3213087495KHz

And how to find 3dB cut off frequencies ?
I don't know. Maybe form Q factor
 

Attachments

  • 01.PNG
    01.PNG
    6 KB · Views: 286
Last edited:
Reply for post 7 and part of 8...

Hi there,

Sneak:
Not sure what you are talking about here, and i see you're representing some frequencies as regular frequencies and some as angular frequencies. Maybe take a look at that.

The center frequency occurs when w=sqrt(1/(C1*C2*R1*R2)) which is a frequency of 3974Hz, with a peak amplitude of (C2*R2+(C2+C1)*R1)/(C2*R2+C1*R1)=1.8695652=5.4348db, and the two -3db points are at F1=836.6344Hz and F2=18878.7054Hz.

Sneak, Jony:
You can find the two -3db points by setting the amplitude equal to the max amplitude at the center frequency divided by the square root of 2:
MaxAmpl/sqrt(2)=Ampl(Hjw)

and then finding the two solutions, and the MaxAmpl is just Ampl(Hjw(wc)) so we solve:
Ampl(Hjw(wc))/sqrt(2)=Ampl(Hjw)

and there will be two real solutions for w.

The above is taken using the theoretical response of a perfect op amp. The real life op amp will probably roll off the frequency somewhere after the second -3db point, so there would be another point to consider if the circuit was to be used with higher frequencies.
 
Last edited:
MrAl
Can you tell me how you find the peak amplitude from the transfer function ?


Hi Jony,

Yes sure...


First solve for the amplitude of the transfer function, then solve for the center frequency. Then insert the center frequency into the equation for the amplitude you found. The amplitude is an equation in all reals and the center frequency is a real so you end up with an equation that is in all reals.

The amplitude above can be found by finding the amplitude of the numerator and dividing by the amplitude of the denominator. The amplitude is the square root of the sum of squares of the real and imaginary parts.

Note that we are dealing with a bandpass type response here. If the response was bandstop for example, we would have to find the minimum response first not the maximum which we are calling the 'peak' here.
 
Last edited:
Oki I find the maximum amplitude

[LATEX]Amax = 1 + \frac{(C2 R1)}{(C1 R1 + C2 R2)}[/LATEX]

I find Amax by substituting S = jωo

But can you describe in more detail the way how you find 3dB frequencies.
 
Last edited:
Oki I find the maximum amplitude

[LATEX]Amax = 1 + \frac{(C2 R1)}{(C1 R1 + C2 R2)}[/LATEX]

I find Amax by substituting S = jωo

But can you describe in more detail the way how you find 3dB frequencies.


Hi again,



Starting with the original transfer function:
Hs=(s^2*C1*C2*R1*R2+s*C2*R2+s*C2*R1+s*C1*R1+1)/((s*C1*R1+1)*(s*C2*R2+1))

in the numerator we subst s=j*w and we get:

HjwN=j^2*w^2*C1*C2*R1*R2+j*w*C2*R2+j*w*C2*R1+j*w*C1*R1+1

and after simplifying we get:
HjwN=-w^2*C1*C2*R1*R2+j*w*C2*R2+j*w*C2*R1+j*w*C1*R1+1

so the real part in the numerator is:
RPN=1-w^2*C1*C2*R1*R2

and the imag part in the numerator is:
IPN=w*C2*R2+w*C2*R1+w*C1*R1

and so the amplitude of the numerator is:
AmplN(w)=sqrt(RPN^2+IPN^2)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C2*R1+w*C1*R1)^2)

The denominator is:
(s*C1*R1+1)*(s*C2*R2+1)

and after subst s=j*w and finding the amplitude the same way we did the numerator we get:
AmplD(w)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C1*R1)^2)

Now we have an expression for the amplitude:
Ampl(w)=AmplN(w)/AmplD(w)

Now here we would subst w=w0 and calculate the amplitude at the center frequency to get the
max amplitude, but since you already have that we take that amplitude:
1+C2*R1/(C1*R1+C2*R2)

and divide that by the square root of 2 thus:
(1+C2*R1/(C1*R1+C2*R2))/sqrt(2)

and then set that equal to the amplitude we computed above AmplN(w)/AmplD(w):
(1+C2*R1/(C1*R1+C2*R2))/sqrt(2)=AmplN(w)/AmplD(w)

Written out this looks like this:
((C2*R1)/(C2*R2+C1*R1)+1)/sqrt(2)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C2*R1+w*C1*R1)^2)/sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C1*R1)^2)

Now we simplify that expression and at this point it gets much simpler to substitute the actual values for the
variables C1, C2, R1, and R2. We could work symbolicially, but it would take a lot more work.

Doing that, we get a much more simple equation:
3.8457892807561354e-18*w^4-5.4217779933648401e-8*w^2+1.495274102079395=0

We then solve this for w, and reject negative values, leaving us with two positive values for w.
We then divide by 2*pi and get the two -3db frequencies F1 and F2. One is below F0 and one is above F0.

Of course sometimes we dont know the center frequency w0 yet, so we would have to solve for that first. To do that, we would compute the Amplitude as we did above and take the first derivative with respect to w and then set that equal to zero:
d(Ampl(w))/dw=0

and solve this for w and test each solution for validity in the amplitude equation, and that would give us w0. Again, doing this numerically is a *lot* easier than symbolically.
 
Last edited:
I thought that with finding the poles and zeros of a transfer function, I would be able to know at which frequency the output will be 20dB/decade or -20dB/decade. But I thought that when I would find the poles, I would find the cutoff frequencies like a simple low pass filter, I'm clearly wrong.

Thanks for the whole explanation mr al ! and of course Jony.
 
Hi again Sneak,

It depends on how close the poles and zeros are to each other, but you can get a general idea of the response by looking at the Bode asymptotic plot. It's never going to be as exact as a true curve plot though.

In the following diagram, you can see the Bode asymp plot as the bold thick black line, and the true curve in blue, and the -3db horizontal green line cuts the true curve at the -3db points. You can see the Bode plot isnt too far from the true 3db points but is quite a bit higher than the true curve.
You might want to try this yourself and verify that i drew the Bode plot correctly as i dont draw Bode plots any more because the computer software can draw the true curve much faster and of course it is more accurate.

The horizontal dashed lines are spaced 1db apart, with the lowest line being 0db.
 

Attachments

  • Bode-01.gif
    Bode-01.gif
    14.5 KB · Views: 173
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top