PI Filter Calculation

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You need to know the input resistance. Since it is not shown, 50 ohms is the value it seems to be the custom to assume.

You can get an idea how the filter will behave, by figuring the rolloff frequencies for the various components.

The rightmost end is labelled 'VIN'. The signal goes through 6.8 uH, which passes lows.

C10 diverts highs. Etc.

If there is an output resistance, and it is a low enough ohm value to load the network, then it must be factored into the calculations as well.
 
I´m not sure if we could define a bandwidth curve for the circuit employing a standard analysis in the same way would be done for filters designed to attenuate frequency ranges. Although an intrinsic inductance is taken on the Bead as its main characteristic, its value is not precisely determined by manufacturer in term of a nominal value, but can occur in a certain range. The purpose of the circuit at all is supposed to suppress noise inside a bandwidth by dissipating it inside the Bead device.

Anyway, in normal operation, once you can measure the exact value of the inductance, you can performs a ladder analysis as suggested and quickly know the main behavior, but a simulation would give you a more precise answer.
 
Sorry.

The above two circuits are not related to each other. They are separate circuits.

I am very basics in electronics and I was not able to understand your reply.

can I please know if this below website can give me the cutoff value for the above filters ?

http://www.calculatoredge.com/electronics/bw pi low pass.htm

can i add 0.1uF, 4.7uF, 10uF and FB1 and L2 to get the cut off frequency using the above link ?

Will this help me ?
 


The calculator works the opposite way. It gives you component values.

Your bottom schematic is easier to understand, so I drew it in a simulator (only the C-L-C Pi section).

It demonstrates the fact that we need to know what ohmic resistance is involved, because difference resistance values yield different rolloff frequencies.



Furthermore the last plot reveals something not specifically brought out in primers. A filter needs to be driven by sufficient current, in order for it to perform correctly.

When resistance is high enough, the CLC tank circuit becomes isolated to a degree. An instigating waveform allows its resonant action to kick in unexpectedly. As a result this CLC tank circuit resonates at a frequency of 35 kHz.
 
As this seems to be power supply decoupling, I am not sure that regarding it as a filter is the correct way to go. I think the theory is to raise the source impedance of any interference, with the series inductance then attenuate the interference with a capacitor. So it depends where the interference is coming from. So if the supply is feeding a RF power amplifier, the set up would be series choke from the amplifier, then a RF decoupling capacitor, which will attenuate the basic RF. Now there might be a residue of modulation (video/audio/data), which then need a much higher inductance and a much bigger capacitor. The effect being that the RF amp will then not put any spurious signals on to the supply line. Conversely if the amplifier is a wideband low level amplifier, then it will not put any rubbish on the supply line, but would need the filtering (wired the other way around) to stop noise that is picked up on the supply line from affecting its performance.
Simulation is only so good, many capacitors have undesireable series inductance, which normally stops them from working over 1MHZ, so the component construction will affect the decoupling as much as the value on of it.
Frank
 
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