T Section Lowpass LC Filter Frequency Formula

[Laserray]

I have been trying to find out how the cutoff equation for the T section LC Lowpass filter is derived. I remember from college that f=1/(2pi√(LC)) for an LC circuit. But, with the T section, the cutoff frequency is given by f=1/(pi√(LC)), with L being the sum of the two inductors. My math has gotten very weak since college and I really could use some help figuring out how you get to that latter equation. I have searched the web for hours and can't find explanations, just calculators. I don't really want to spend the next several days poring over my old electronics textbooks trying to relearn the theory. Can anyone give me some pointers here (please keep it on the simple side, there are some serious ruts in this 65 year old brain). Thanks

 

[DaveE]

Try use Thevenin/Norton equivalent circuit transformations to reduce the complexity of the circuit. Remember to include source and load impedances in your model.

 

[Laserray]

I vaguely recall using Norton theorem, with resistive circuits. I'll give it a whirl. Thanks for the pointer Dave. You are really challenging these old brain cells (its in there, but hasn't been accessed in 50 some years).

 

[DaveE]

Same as for resistors, just use Zl = sL for inductors & Zc = 1/(sC) for capacitors, s = 2πfj.

 

[Laserray]

Thanks again Dave. Funny how seeing various letters arranged a certain way on the screen (or page) can trigger memory pathways. As soon as I saw the impedance equations in your reply, I said to myself "Yep! that rings a bell". Now, I have a job to get the rest of the brain cells firing. Let's see, I wonder if there are any threads on this forum to determine how thick dust would be on a book sitting on a shelf for 40 years (actually only 31, we moved in 1987). Have great day. Ray

 

[Tom.G]

Constant K filter. Z₁ ⋅ Z₂ = K²

Z₀ = √(Z₁Z₂) = √(L/C)
Z₁ = X_L1
Z₂ =C_C1

L = R/(π⋅f_c)
C = 1/(π⋅f_c⋅R)
f_c = 1/(π⋅√(L⋅C))

-----L1---+-----------
          |
          |  C1
         ----
         ----
          |
          |
----------+-----------

-----L1/2---+----L1/2-------
            |
            |  C1
           ----
           ----
            |
            |
------------+---------------

------+----L1-----+-----------
      |           |
      |  C1/2     |  C1/2
     ----        ----
     ----        ----
      |           |
      |           |
------+------------+-----------

Cheers,
Tom

 

[Laserray]

Thanks Tom. But darn,I was kind of looking forward to the challenge of doing the Norton analysis. Now I prob will have to make myself do it. That's ok, thanks for your help. Have great day. Ray