LC Filter Query

[Techman_7]

I have a interesting question:

What is difference between low pass LC filter cut off and resonant frequency.
Resonant frequency = 1/(2*pi*sqroot(L*C))

Now, Cut off frequency = 3dB frequency, so how to find the cut off frequency?

 

[KlausST]

Hi,

it depends on Q .. or the "R" involved in the circuit.

Klaus

 

[danadakk]

You have a transfer function for LC filter = Vout/Vin. Solve for Vout = -3 db.
Note result can exhibit two values if peaking/Q is present.

If Butterworth flat there will be one solution value.

How to obtain 3 dB frequency from transfer function?

How can I calculate the 3 dB frequency of the following transfer function? $$ H(j\omega)=\frac{1}{1-j\frac{250}{\omega}} $$ I have thought of doing the inverse Fourier transform of \$H(j\omega)\$ ...

electronics.stackexchange.com

 

[Techman_7]

Hi,

it depends on Q .. or the "R" involved in the circuit.

Klaus

Suppose L and C are ideal components, then how to determine cut off frequency? I tried deriving it by equating the transfer function to 1/2.
But, final equation has sqrt(-1) term in it.

Can anyone pls suggest the formula or derivation?

 

[KlausST]

Hi

Ideal components make no sense.
As low pass it would:
* f < fres: a = 1
* f = fres: chaos, short circuit. Infinite High voltage
* f > fres: 0

Klaus

 

[Akanimo]

The resonant frequency is where the inductive reactance and the capacitive reactance equal eah other. If you plot the capacitive reactance and the inductive reactance on a log-log graph, then you find that the frequency at which they equal is the frequency where the two poles of the filter are.

--- Updated Saturday at 7:56 PM ---

Suppose L and C are ideal components, then how to determine cut off frequency? I tried deriving it by equating the transfer function to 1/2.
But, final equation has sqrt(-1) term in it.

Can anyone pls suggest the formula or derivation?

The poles are actually complex for an underdamped LC filter.

 

[danadakk]

From chatGPT -

To design an LC low-pass filter, the following equations are commonly used for determining the cutoff frequency, which is where the filter attenuates the signal by 3 dB. The 3 dB point is the frequency at which the power of the signal is reduced to half of its maximum value.

1. Cutoff Frequency (f_c):​

The cutoff frequency fcf_cfc is determined by the inductance LLL and capacitance CCC as:

fc=12πLCf_c = \frac{1}{2\pi\sqrt{LC}}fc=2πLC1
Where:

• fcf_cfc is the cutoff frequency in Hz
• LLL is the inductance in Henries (H)
• CCC is the capacitance in Farads (F)

2. Impedance at Cutoff Frequency (Z_c):​

The impedance at the cutoff frequency for an LC low-pass filter is given by:

Zc=L/CZ_c = \sqrt{L / C}Zc=L/C

3. Transfer Function (H(s)):​

The transfer function H(s)H(s)H(s) for an LC low-pass filter (where s=jωs = j\omegas=jω, and ω=2πf\omega = 2\pi fω=2πf) can be expressed as:

H(s)=11+sLCH(s) = \frac{1}{1 + s \sqrt{LC}}H(s)=1+sLC1
At s=jωs = j\omegas=jω, the transfer function becomes:

H(jω)=11+jωLCH(j\omega) = \frac{1}{1 + j\omega \sqrt{LC}}H(jω)=1+jωLC1

4. Attenuation at a given frequency:​

For frequencies above the cutoff, the attenuation can be calculated by considering the magnitude of the transfer function:

∣H(jω)∣=11+(ωLC)2|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega \sqrt{LC})^2}}∣H(jω)∣=1+(ωLC)21
At the cutoff frequency fcf_cfc, the magnitude is:

∣H(jfc)∣=12≈0.707|H(jf_c)| = \frac{1}{\sqrt{2}} \approx 0.707∣H(jfc)∣=21≈0.707
This corresponds to a 3 dB drop.

Summary:​

• The 3 dB cutoff frequency is given by fc=12πLCf_c = \frac{1}{2\pi\sqrt{LC}}fc=2πLC1.
• The filter has a maximum passband gain of 0 dB and a 3 dB attenuation at the cutoff frequency.

 

[BradtheRad]

low pass LC filter cut off and resonant frequency.

Shall we take low pass filter as the keyword? This sounds like the second order LC filter generally used at the output of switched-coil converters. Load R is connected across capacitor.

R at low Ω results in a rolloff curve which creates a usable low pass filter (see scope trace below for load 1). However when R is high Ω (or absent), then conditions are ripe to permit a resonant (or near-resonant) waveform to instigate oscillations which dominate, grow in amplitude and destroy components. (See scope traces for load 2 & 3.)

For the same reason amplifier designers warn against operating a hi-fi amplifier without a load yet with an LC second-order low-pass filter.

--- Updated Sunday at 5:12 AM ---

Link below runs above schematic in Falstad's animated interactive simulator:

tinyurl.com/2c4gduc8

 

[Techman_7]

I got below reference, it seems useful but I am not able to derive the same equation for Fc at -3dB :

LC Filter Cut-off

 

[danadakk]

As R rises the filter exhibits less and less resonant effects (red curve 10 ohms, the others 100 and 1000 ohms)
because increasing R "effectively" lowers L. Essentially filter becomes RC at high values of R relative to
X_L :

The series R case vs Brads case for load effects, both relevant. Note useful way of looking
at R effects on L,C elements is use the transforms :

https://www.eetimes.com/convert-parallel-impedances-to-series-impedances/

 

[FvM]

Post #9 (infinite Q resonator) is no useful low pass filter. Low pass filter has defined Q according to implemented filter prototype, e.g Bessel, Butterworth or Chebyshev. Relation between resonance cut-off frequency is different. Also filter order plays a role, apparently you refer to 2nd order (one complex pole pair).

Basic design procedure:
1. choose cut-off frequency, filter type and order
2. get L, R and C values from filter design tables or equations

 

[D.A.(Tony)Stewart]

I have a interesting question:

What is difference between low pass LC filter cut off and resonant frequency.
Resonant frequency = 1/(2*pi*sqroot(L*C))

Now, Cut off frequency = 3dB frequency, so how to find the cut off frequency?

Cutoff or BW or BW(-3dB) for voltage is defined by the "half-power" breakpoint.


Q=Xo/R , the resonant peak rises above the baseline and slowly shifts to where the f-3dB Butterworth was located as the maximum Q possible matches the previous BW. But now the -3dB point for a 2nd order filter has shifted to the right by 1.414.

Resonant frequency, fo is defined by the the peak voltage frequency which shifts to the right with rising Q.
When raising the Q=0.707 for a Butterworth filter which by definition is critically damped with 2nd order angle of 45 degrees exactly matches a Chebyshev filter with 0 dB amplitude ripple.

Link

To design any filter you first must have a list of specs;

• specify type , (LP, HP, BP, BS, AP etc) , or conjugate matching
• Passive or Active,
• BW ,
• gain,
• order or bandstop points and bandpass points
• with optional passband options for ripple
• or group delay maximally flat
• or maximally steep skirts. Generally filter errors from tolerance sensitivity always increases with Q.
• Filters by name: Butterworth, Elliptical Cauer, Raised Cosine, Hilbert, etc.

 

[Akanimo]

Your circuit shows an ideal LC filter with no resistance. This can still be analysed though. Let's lay out some characteristics of this circuit.

Vin, L and C are in series, so the same current, I, flows through them.

Vout = I*1/(sC); Vin = I*sL + I*1/(sC).
Solving for transfer function for Vout/Vin yields poles at +-sqrt(-1/(L*C)).

Intuitively, because the filter has no resistance, it's poles have no real part. Taking the -1 out of the square root gives us a j so we have the pole frequencies at +-j/(2*π*sqrt(L*C)).

I got below reference, it seems useful but I am not able to derive the same equation for Fc at -3dB :

LC Filter Cut-off

View attachment 198744

--- Updated Sunday at 8:10 PM ---

In reality, just as have been pointed out. You need resistance to damp the resonance at the corner frequency.