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

[Winterstone]

Anyway I will leave this and concentrate on designing it from scratch. I know I need a 3 pole high pass filter. I will choose a cut off frequency. Does anyone have an idea as to in which phase will Matlab be of use in the design process?

gauthamtechie,
what you did up to now was to ANALYZE a filter circuit with given values (using Wolfram-alpha service).
In contrast, your above cited question concerns the reciprocal process - that is: DESIGN of a filter. That is a complete different approach.
What exactly is your question regarding DESIGN?

 

[gauthamtechie]

Yes I wasn't clear enough. Earlier,before all the posts, I thought Analysis and design were nearly complementary, but now I understand that a lot of factors need to be taken care of in design of a similar filter for another cut off.

 

[Winterstone]

Yes I wasn't clear enough. Earlier,before all the posts, I thought Analysis and design were nearly complementary, but now I understand that a lot of factors need to be taken care of in design of a similar filter for another cut off.[/QUOTE

To say it with simple words:
*To analyze a given circuit yields one single solution
*To design a circuit (based on specific requirements) allows - in principle - an infinite number of solutions (different circuits with different dimensioning).

 

[The Electrician]

Hi Electrician,

Regarding your matrix formation, I have one clarification. What if I had a voltage source Vnew in the Leg with the inductor as well? What will I equate this to - (Vx - V1)*(s C1) + Vx*(1/(s L1)) + (Vx - Va)*(s C2) = ??? Should I look into per unit values and stuff?

And this term: Vx*(1/(s L1)) will change to (Vx-Vnew)*(1/(s L1)) right ?

Are you saying you want a transfer function involving the old input (which was V1) plus a new Vnew at the same time? If so, this complicates things somewhat.

The best thing might be to use a linear solver instead of inverting matrices. This technique could also be used to get the transfer function of your original problem like this. The original Y matrix is changed by removing the first row and column. This is equivalent to grounding what was the V1 node. Let's use Vin to represent the old input voltage. We inject a current into the Vx node equal to the voltage Vin divided by the impedance of C1; that current will be Vin/(1/(s C1)), or Vin*s*c1. If we do this, we get the same transfer function as before. The desired result is the (2,1) element of the result vector; I've added some red to help identify it.

Now, to have both inputs at the same time we add another current to the injected current. This current will be a current equal to Vnew divided by the impedance of L1; this will be Vnew/(s L1). Then we get this result:

 

[The Electrician]

And I finally got f=3333.83 !

Check back to early in the thread. I got 3334.86. Your answer is slightly different; probably the result of not carrying enough digits in the computation.