Hi all, I'm a newbie when it comes to filters. I've recived a schematic at work which has an LC Pi filter as part of a video circuit. How do I go about analyzing that filter? How can i derive, from the given L & C values, what's the filter transfer function and behavior? Any input is welcome, Thanks! Ran
Analog filter analysis
Jul 05, 2006
7 Replies
Hello, You can express these circuit elements as their frequency domain impedance (laplace transform) Your filter circuit may be expressed as a voltage divider - use the impedances from above step for this. analyize the behavior of this voltage divider equation, this is what your filter will do
I am an undergarduate student, so I am not sure how much I would trust my advice.
Ian Coolidge
Thanks for the quick replay, Ian. Since this is a typical Pi filter, is there anywhere I can find "ready-made" transfer functions, to which I can input the capacitor and inductor values and get a quick answer? I guess the same goes for T and L filter topologies. thanks, Ran
============================= Passive filters have a large body of info.... 3 elements... might be a
3rd order LP filter.. if it was 2nd order, the freq=1/2*pi*sqrt(LC)..
It can be done analytically, but it's a lot of work. I'd download LT Spice and simulate it.
John
If you label the impedances like this: z1, z2, z3 for the left leg, center, and right leg of the PI network, and z0, zL for the input and output impedances, respectively, the formula for Vout / Vin will be z1*z3*zL / (z0 *z3*zL + z1*z3*zL + z0*z1*z2*z3 + z0*z1*z2*zL + z0*z1*z3*zL), (reduced from a voltage divider expression by CAS).
Given the formula, plug in the values for z0 and zL (for example, 50 ohms, or whatever), then put in the values for z1, z2 and z3 in terms of the variable s. For an inductor, use the inductance times s, so for a 0.1H inductor, use just 0.1s. For a capacitor, plug in 1/(s*C), so for a 1000 uF capacitor, it would be 1/(1e-3s) or 1000/s. That gives you a transfer function, and if you use a CAS, it will simplify a lot. For example, the formula above, for 50 ohm input and output impedance, with z1=z3=0.1H, z2=1000uF, reduces to
0.5s^2 / (25.5s^2 + 750s + 2.5e5 ).At this point you can plot it or evaluate it in the frequency domain by substituting jw for s, where j = sqrt(-1) and w = radian frequency (or
2pi*f ). When you plot it, you should take the absolute value of the result for the magnitude, or its angle if you want to plot phase. For the values given above, the filter magnitude response shows a rise from-infinity to a resonant peak at 100 rad/sec (15.91Hz), and then levels off to + infinity at -34db.
If you have Matlab, you could put the above transfer function in by myTF = tf ( [.5, 0, 0] , [25.5, 750, 2.5e5] ), just putting the numerator and denominator in, in descending powers of s. At that point you can use Matlab to plot frequency response or time domain response quite easily: bode(myTF), step(myTF).
-- john
Sorry, I made a mistake in the transfer function- it should have ended up at -3db for high frequency, since the L's have infinite impedance, the C has zero, and the input and output impedances are equal. The correct expression for the transfer function is z1*z3*zL / (z0*z1*z3 + z0*z2*z3 + z1*z2*z3 + z0*z1*zL + z0*z2*zL
+z0*z3*zL + z1*z2*zL + z1*z3*zL). The evaluation with 50 ohm in and out, 0.1H, 1000uF would be s^3 / (2s^3 + 1020s^2 +2e4s + 5e6) and the resonant peak would be about 71 rad/sec or 11.3Hz.-- john
Thank you very much for your help, John! Excellent work! Best wishes, Ran
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