hi W,
Did you check the link I added to post #18.?

Using your resistor values I chose the preferred value Caps to suit.
E

Ah i did not saw the link update when i was posting, sorry. I had checked the link but is it ok to ignore the output impedance R3,50 ohm? To add on as to why i was wondering on how to choose the component value. I am required to justify why i chose this capacitor or resistor value before purchasing the component which is why i was trying to find the component value from the transfer function.

 

Wow thank you! That is a very complicated function . Is Mathematica able to help derive the complex conjugate or all the intermediate steps must be done manually?

Hi Wompalamba, you have sent me the pdf-file which describes the method you are using. Now I know what you are doing - it is the classical way of finding the 3dB- points - however, in a somewhat unusual form.
Because I am not sure if you really know what you are doing (and WHY) - I will roughly describe the principle as follows:

* You must find the MAGNITUDE of the transfer function. For this purpose, you need the conjugate complex form (for multiplying numerator and denominator with this form).

* Now you can find the magnitude of the function because the numerator can be splitted into a real and an imaginary part - and the denominator is a real expression.

* The magnitude of the whole function is SQRT(Re²+Im²) and you have to find the frequency where the denominator (which is a real function) equals SQRT(2). When the denominator has the form SQRT(1 + aw + bw²...) we arrive at the equation
1 + aw + bw²...=2 which can be written as (aw + bw²...-1=0).

Note: In the above derivation I have assumed that theDC gain is k=1.

 

Wow thank you! That is a very complicated function . Is Mathematica able to help derive the complex conjugate or all the intermediate steps must be done manually?

Mathematica can derive the complex conjugate, but that is the easiest part of the problem to do by hand.

 

Hello,

Here is what i get...

Amplitude function A(w):
Vout=Vin/sqrt((w*T3+w*T2+w*R23*T1+w*T1)^2+(-w^2*T1*T2+R23+R13+1)^2)
where
T1=R1*C1
T2=R2*C2
T3=R1*C2
R23=R2/R3
R13=R1/R3

Then find the passband gain A1 (look at the graph of the function and decide where that will be, and for this example it will be at w=0), so A1=A(0).
Divide that by sqrt(2) to get:
G=A1/sqrt(2)
Equate the amplitude A(w) to G:
A(w)=G
Solve that for w, the only valid result will be the angular cutoff frequency.

To check, insert your solution for w back into A(w) and check to see that you get G which was found earlier.

The transfer function can be found using various methods such as Nodal Analysis or "network collapse and expansion" or Norton and Thevenin theorems.

 

hi W,
As you use LTS, this is a asc file.
E

Update:
Check this link.

http://sim.okawa-denshi.jp/en/CRCRkeisan.htm

The calculated value for Fo with these component values is:



not quite as good as the result with component values shown in post #15

 

The calculated value for Fo with these component values is:

View attachment 227327

not quite as good as the result with component values shown in post #15

Hi,

Just a suggestion, see if you can reduce it by asserting time constants in place of resistor and capacitor values such as T1=R1*C1 and T2=R2*C2. That often makes it more readable also.
Sometimes we use T11=R1*C1, T22=R2*C2, T21 for R2*C1, etc.

 

hi Elect'.
For reference only, using your calc values in LTS.
E

 

For reference only, using your calc values in LTS.

Can you show it at -3.01dB because 20log(1/√2) = -3.01029995664dB ≈ -3.01dB

 

hi jony,
Do you mean raise the lower reference frequency dB to say 0dB, so that -3dB point is referenced to the actual filter.?
E

 

hi jony,
This is a 'fudge' by using 2.78Vac as the Analysis value, rather than 1V, is it what you are asking.
E

 

Do you mean raise the lower reference frequency dB to say 0dB, so that -3dB point is referenced to the actual filter.?

No, do exactly the same as you did, but this time use -3.01dB instead of -3dB. So that you can get closer results to the one get by The Electrician.
The reference lies at 20log(50/139) ≈ - 8.88dB

 

This is a 'fudge' by using 2.78Vac as the Analysis value, rather than 1V, is it what you are asking.

This way will also work. THX

 

hi jony
Ref post #31.
This is this closest I can get using, using 10,000 sampling points

Is this satisfactory.?

E

Wait one second.!

 

hi jony.
Noted a problem , that Post #25 equation has been changed.! R2 was 39Ω, it now 50Ω.

So I will modify the LTS sim to suit.
E

Update:
This image is what I see, using the values in post #25.
The LTS image in Post #23 was using the initial posted value of 39R