This is the simplified transfer function of a bandpass (with only one single resistor).(Ro*s)/(Ro*s^2*C*L+s*L+Ro), {this is the transfer function}
Are we talking here about lowpass or bandpass?
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This is the simplified transfer function of a bandpass (with only one single resistor).(Ro*s)/(Ro*s^2*C*L+s*L+Ro), {this is the transfer function}
Hi there,This is the simplified transfer function of a bandpass (with only one single resistor).
Are we talking here about lowpass or bandpass?
He wants an input filter (either antialiasing or interference removing) for an A/D. It took 124 posts to establish that.Question to Ian0: Do you know what the questioner (pinkyponky) really wants?
In this thread, I have seen many different circuits - with and without opamp.
PP, the transfer function you have derived here is the transfer function of the circuit in post #150 if you let R=R1, L=L1, C=C1, and Rload =R2+R3.Hi MrAl!
Thank you for your support.
I already said the application. This filter will be used to filter the input signal of the ADC and output of the DAC.
You asked me to derive the transfer function. Eventually, Done!. Ok. How can I plot the response from this transfer function?.
How to design the R, L, and C values?. These values will be used to design a filter that attenuate the signal above 100kHz.
S=jw, w=2*pi*f, and j2=-1. After substitution of these values, the final Transfer Function is derived and attached below. Please have a look and tell me how to plot the response from the below transfer function.
Hello Electrician!PP, the transfer function you have derived here is the transfer function of the circuit in post #150 if you let R=R1, L=L1, C=C1, and Rload =R2+R3.
The response in post #150 is the response of your transfer function; that's what LTSpice does--it plots transfer function responses, among other things.
Are you required to plot the response from the mathematical expression of the transfer function, or will the plot from LTSpice be enough?
Hello LvW,Hi pinkyponky,
as I have told you already before, it is not necessary (and very involved and time consuming) to do all the math with complex figures.
Nobody is doing this (because you would start at "zero".)
This was already done long time ago - and the results are summarized in tables which allow to design a second order filter (or even higher order) for the desired approximation.
In post#177 I gave you the transfer function for your second-order RLC filter.
For your convenience, I repeat it again: The part names are according to your hand written figure in your post#174.
The parallel connection is Rp=R||RL.
H(s)=Ao / {1 + s[(RpC + L/(R+RL)] + s²AoLC}
Writing the transfer function in this form (so called "standard form") has the advantage that we immediately can identify the three main parameters which define the lowpass response:
* DC gain is Ao
* Pole frequency is wp=SQRT(1/AoLC)
* Quality factor (pole-Q) can be found by 1/Qp=wp[(RpC + L/(R+RL)]
* For DC the gain Ao can immediately found to be RL/(R+RL). You are free to select R undf RL according to your requirements.
* The pole frequency wp is in rather vicinity to the wanted 3dB-cutoff frequency wc (for a maximum flat magnitude, Butterworth approximation, it is identical wp=wc). Because Ao now is known you can compute the values for L and C together with the required pole-Q (next point). This results in a set of two equations for the two unknown values L and C.
* The required Q-values can be found in corresponding tables: Q=1/SQRT(2)=0.7071 for max. flat (Butterworth) response. Larger Q-values give a Chebyshev response with a slight "peak" in the vicinity of the pole frequency (tabulated for a peak ("ripple") of 0.05dB, 0.1dB, 0.3dB.........3dB).
Based on these information you immediately can design your filter.
I hope this helps.
Hi Ian0,IF I remember correctly, this filter precedes an ADC, so therefore must be an anti-aliasing filter.
Therefore the cutoff frequency must be less than half the sampling frequency.
The Q is a matter of choice, depending on many factors, such as:
1) how much the amplitude response can vary within the passband. If it is important, choose Butterworth (Q=0.7)
2) how important it is to have constant group delay (complex signals retain their shape). If this is important, choose Bessel (Q=0.577)
3) if there is energy in the signal close to the sampling frequency. If so, choose Chebyshev.
I need small size components since we have physical dimension problem on the PCB. If I go with high inductor value then it has high physical dimensions.A value of inductance in the mH region is correct.
Why? what happen?Abandon the idea of a filter that uses inductors.
Not possible Unless you reduce the load resistance.Why? what happen?
The experts like you its not a difficult design. It's very simple for you. I have given all my requirements and also I put my effort lot to design RLC fiter, such as deriving the tranfer function and also I have calculated the L and C values as per the post #22. Even-though I'm not able to complete this work. I'm requesting you that how to design a RLC filter with low inductance values.
Hi Bob,What would your response be if I said I was having trouble designing a 5 passenger car that can go from 0-60 in 5 sec, has a 500 mile range and the fuel tank can hold only 1 gallon?
Here's the relevant page from Zverev.
View attachment 247173Hi Ian0,
Could you please provide me the name of the this text book?.
by Jake Hertz
by Lianne Frith
by Jeff Child