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What is the cut off frequency of this LC and RC combination?

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I left it out because I judged that the OP knew already, and there was no need to waste his time and mine explaining something he already knew.

Hi there Electrician,

Oh ok, well i think you did a good job of showing how to do the circuit, but now that we know that he doesnt know this yet (where the 1/sqrt(2) comes from) perhaps you would want to explain that to him. Just a thought :)
 
Thanks, I've learnt a lot of things.

Now I have to modify this filter to give me higher cut off frequency. Now, is it better I look at how to design this filter from scratch, which has a butterworth response?

Hello, gauthamtechie!

In a former post you have mentioned that your task is not homework.
Thus, I conclude that you plan to use the filter for a certain task.
Therefore, I like to point to the fact that this passive filter has a finite (complex) output resistance.
That means, any loading will destroy the designed filter response (Butterworth or whatever it is).
Thus, you have to use a buffer for decoupling the output.
As an alternative you could use from the beginning an active filter stage.
However, in this case, the performance depends on the requirements in the higher frequency region.
Do you have specific damping requirements ?
 
Hi there Electrician,

Oh ok, well i think you did a good job of showing how to do the circuit, but now that we know that he doesnt know this yet (where the 1/sqrt(2) comes from) perhaps you would want to explain that to him. Just a thought :)

What on earth are you talking about? He explained in post #15 that he knows exactly where it comes from.
 
What on earth are you talking about? He explained in post #15 that he knows exactly where it comes from.

Oh hi again Electrician, nice to see you are up at this time too, it's very early here i happen to be awake.


Well lets look at the time line here...

I said maybe you or i should explain it to him. You said he might already know, and asked him. He gave an answer. I then said maybe we should tell him because he doesnt yet know. So guess what i am saying on this Earth :)

In other words, ever after his reply, it appears he still doesnt know so maybe we should give the complete answer?
Maybe you should read his reply again more carefully? Not sure here, but whatever you want to do is fine with me here. I'll wait for your decision.

I'll post his reply here for your convenience:

Hi Electrician, I knew why you took 0.707 or rather SquareRoot(1/2).. And no it isn't a homework problem..

Vout/Vin is my transfer function. I need the cut off frequency. The cut off frequency is the -3 dB point; In terms of voltage that is the point where the output is 70.7% of the input which means Vout = 0.707 Vin ;And 0.707= Squareroot(1/2)
So I can equate the transfer function like this Vout/Vin = Squareroot(1/2)

Apparently The time difference is +10.5 hours between your place and mine; so i couldn't get back on your question in time!

Note i am posting this because it was not yet shown here or mentioned exactly why we want to use 1/sqrt(2).
Unless you think perhaps it is too obvious here to bother mentioning?
 
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It appears that nobody can figure out what you're getting at, so you'll just have to come right out with it.

Hi again,

Oh ok, sorry about that. He said that he understood that we needed the -3db point, and that is the point that the output is 70.7 percent of the input. But that's not always true. It depends on the type of filter, and it is very hard to guess sometimes what kind of filter it is and if it even works the way it is supposed to work.
So what i like to suggest is immediately after determining the response from the analysis, to do a plot of the response over a frequency range that is comparable to the fastest estimated time constant (or more). Then that will tell what value we can use for the comparative constant, which obviously will not always be 1/sqrt(2) times the input.

Again i apologize if i sounded a little too demanding earlier for this information.
 
Oh ok, sorry about that. He said that he understood that we needed the -3db point, and that is the point that the output is 70.7 percent of the input. But that's not always true.
When is the -3dB point not 70.7% of the input?
 
Collect terms and you will have:

V1*(s C1) + Vx*(-s C1) + Va*(0) = 1 The 1 here is the test current.

V1*(-s C1) + Vx*(s(C1+C2)+1/(s L1)) + Va*(-s C2) = 0

V1*(0) + Vx*(-s C2) + Va*(s C2 + 1/R1) = 0

Solve these equations for V1, Vx and Va using whatever method you like. I inverted the matrix of coefficients and the voltages V1, Vx and Va are the first column of the inverse; columns 2 and 3 are not needed. Form the expression Va/V1. Substitute numerical values and the rest follows.

I wasn't able to download mathematica, but wolfram alpha does have some features, so does anyone have an idea how I can input this matrix and find its inverse in wolfram alpha; if that is possible? I can't input the matrix like a normal one because it doesn't understand the term 's'.
 
You could try replacing the term 's' with 'jw' (i.e. j = i = sqrt(-1) and w = 2.pi.f)
 
Hi dougy,

thanks ! I replaced the s with j*ω*f and got the solution! A small clarification though: The transfer function is taken as: Z[3,1] /Z[1,1] .
The Electrician mentioned this after he presented the equations:

Solve these equations for V1, Vx and Va using whatever method you like. I inverted the matrix of coefficients and the voltages V1, Vx and Va are the first column of the inverse; columns 2 and 3 are not needed. Form the expression Va/V1. Substitute numerical values and the rest follows.

Is this another way of saying the Transfer function is the output/input wherein my output is across [3,1] and input is across [1,1] ? I didn't get this part: "V1, Vx and Va are the first column of the inverse".

Also How to go about further? I got my transfer function from the inverse. Then how do I equate to 0.707 and ask wolfram alpha to solve for f ?
 
Hi gauthamtechie!
If you want to make an active or passive filter, I recommend reading the book.

Electronic Filter Design Handbook
Arthur B. Williams
McGraw-Hill Book Company

It has all you need to quickly make your filter.
You will need only a pocket calculator.
MOR_AL
 

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Is this another way of saying the Transfer function is the output/input wherein my output is across [3,1] and input is across [1,1] ? I didn't get this part: "V1, Vx and Va are the first column of the inverse".

Remember how I ended up with the equations in this form:

If the 3 nodes along the top of the circuit, left to right, are labeled V1, Vx, Va, you can write the 3 node equations with a TestCurrent of 1 amp applied to the input:

(V1-Vx)*(s C1) = TestCurrent

(Vx - V1)*(s C1) + Vx*(1/(s L1)) + (Vx - Va)*(s C2) = 0

(Va-Vx)*(s C2) + Va*(1/R1) = 0

Collect terms and you will have:

V1*(s C1) + Vx*(-s C1) + Va*(0) = 1 The 1 here is the test current.

V1*(-s C1) + Vx*(s(C1+C2)+1/(s L1)) + Va*(-s C2) = 0

V1*(0) + Vx*(-s C2) + Va*(s C2 + 1/R1) = 0

Solve these equations for V1, Vx and Va using whatever method you like. I inverted the matrix of coefficients and the voltages V1, Vx and Va are the first column of the inverse; columns 2 and 3 are not needed. Form the expression Va/V1. Substitute numerical values and the rest follows.

They can be put into matrix form and then they look like this:

example1-png.74197


These equations could be solved by a linear equation solver, or by using Cramer's rule, or by brute force substitution and elimination; it's the student's choice. The result is then a column vector of the voltages at the three nodes when a current of 1 amp is injected at node V1:

result-png.74198


However you could just as well inject the current at node 2, and solving would give the voltages at node V1, Vx and Va. The setup would look like this:

example2-png.74199


If you were to do that you would get some voltages that are irrelevant to the problem at hand, but it could be done.

Now, with this next setup you are solving for all the voltages at the three nodes, with a 1 amp current injected into node V1, into node Vx and into node Va, all in one setup. The result is a 3x3 matrix; it's the same result you get if you just invert the admittance matrix:

example4-png.74200


The first column is the result of the injection of 1 amp into V1. The second column gives the result of the 1 amp applied to Vx, and the 3rd column gives the result of 1 amp into node Va. The second and third columns are superfluous to the problem at hand, and we ignore them. It's that first column that you want.

Now the (1,1) element of the inverse matrix is the voltage at V1 when 1 amp is injected at V1, and the (3,1) element of the inverse is the voltage at Va. The inverse is a Z matrix and the ratio Va/V1, the desired transfer function, is given by Z(3,1)/Z(1,1).

Some may argue that taking the inverse and getting two extra unneeded columns is overkill. This is the 21st century and computing power is everywhere. Even my HP50G calculator can invert the matrix in a second, using only a few milliwatt-seconds of energy. Formulating the problem as a matrix and taking the inverse to get the desired Z(1,1) and Z(3,1) elements allows a very compact, quick and elegant solution to the problem.
 

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Hello again Electrician,


Does that new circuit shed more light on this now (post 33) ?
 
After trying much to get this data fit into wolfram; I got this f~= -3.30566-1.90852i and more solutions in this wolfram link:

Have I made any errors and how do i proceed to get my frequency 3300 Hz??

You don't want to do that. You want a numerical solution not involving the roots of polynomials.

If you go to Wolfram Alpha home and type in "find roots", then click on where it says "use as a general topic instead", you'll get some description of what you have to do to get a numerical solution.

I tried to make it work with Newton's method using this:

"using Newton's method solve |((8*i*C1*C2*f^3*L1*Pi^3*R1)/(1 - 4*C1*f^2*L1*Pi^2 - 4*C2*f^2*L1*Pi^2 + 2*i*C2*f*Pi*R1 - 8*i*C1*C2*f^3*L1*Pi^3*R1))| = .707"

which becomes "using Newton's method solve |(363*f^3*i* Pi^3)/(625000000000000*(1 + (363*f*i*Pi)/2500000 - (f^2*Pi^2)/ 78125000 - (363*f^3*i*Pi^3)/625000000000000))| = .707"

when numerical values are substituted.

This should work, and it did find a solution of f = 1954.22266 + 1670.835 i

This is indeed a solution, but we want a solution where f is a pure real number; no imaginary part allowed!

I'm going to leave it up to you to flog Wolfram Alpha and see if you can make it work.
 
Hello Electrician,

A simple yes or no would have sufficed :)
 
Hello again Electrician,


Does that new circuit shed more light on this now (post 33) ?

It's superfluous to the purpose of this, the OP's thread.

The OP's problem is not a homework problem. He asked how to calculate the 3 dB point for his circuit. He does in fact know full well why the sqrt(1/2) value is relevant to the solution of his circuit.

There was no need to explain to him what he already knows as it applies to his circuit.

Your example circuit cannot be built in the real world, of course, and I think the OP is concerned with a real world problem. You could start another thread to philosophize about mathematical singularities.
 
It's superfluous to the purpose of this, the OP's thread.

The OP's problem is not a homework problem. He asked how to calculate the 3 dB point for his circuit. He does in fact know full well why the sqrt(1/2) value is relevant to the solution of his circuit.

There was no need to explain to him what he already knows as it applies to his circuit.

Your example circuit cannot be built in the real world, of course, and I think the OP is concerned with a real world problem. You could start another thread to philosophize about mathematical singularities.

Hello again Electrician,

Ok, if you dont like that circuit, which after all is a circuit isnt it (?) that can be analyzed like any other circuit, then simply take his circuit that he started with, but take the output from the 3.33k resistor, tapped at half it's value.
So 3.33k/2=1.665k, so that gives a resistor of 1.665k upper and 1.665k lower which is a simple voltage divider.
So now we have the output voltage at 1/2 of what it was before.

Now try to apply the same factor, 1/sqrt(2), and see why it does not work anymore, at least not the way he described it :)
 
Hello again Electrician,

Ok, if you dont like that circuit, which after all is a circuit isnt it (?) that can be analyzed like any other circuit, then simply take his circuit that he started with, but take the output from the 3.33k resistor, tapped at half it's value.
So 3.33k/2=1.665k, so that gives a resistor of 1.665k upper and 1.665k lower which is a simple voltage divider.
So now we have the output voltage at 1/2 of what it was before.

Now try to apply the same factor, 1/sqrt(2), and see why it does not work anymore, at least not the way he described it :)

I know why it doesn't work anymore, and so does the OP. It's not the same circuit he was asking about; that's why it doesn't "work" anymore. The method I described was for his circuit, not for any circuit in general.

He already knew, as did I, that the response at very high frequencies was unity. He even included a plot in the first post. Therefore, it was clear that the 3 dB down point was where the voltage was .707*unity = .707

For your new example of a voltage divider on the output, then clearly we would need to multiply the gain at very high frequencies (which is not unity anymore, of course) by .707 to get the cutoff gain. But that's not what the OP had.

I don't deny that there are circuits where the method to find the cutoff would not be to solve the equation Vout/Vin = .707. In the general case of a simple high pass like the OP has, but where the gain at very high frequencies is not unity, but something else, call it A[sub]∞[/sub], then the equation to be solved would be Vout/Vin = .707*A[sub]∞[/sub].

You keep bringing up examples that are different from the circuit the OP wanted to analyze. For his circuit, the reason for and use of the factor sqrt(1/2) was understood by the OP, and I knew this and saw no need to digress from the answer to his specific question.
 
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