Is this a high-pass filter?
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The frequency where the reatances are equal is the resonance frequency. The cutoff frequency occurs when the magnitude of H(s) is equal to 1/√2 of H(max).The heuristic argument is that the cutoff frequency occurs where the reactances are equal. Set
jωL = (jωC)^-1
and solve for ω
then substitute that back into the transfer function.
You're missing a 'C' in the 3rd last line in the w^2 term. It should be '-2LC' instead of '-2L'.I am trying to calculate the cut-off freqency, wc, for the low-pass filter below. Could someone help review my calculation below and see if it makes sense? I am wondering if my approach makes sense as it looks too complicated to obtain the final answer.
Thanks for reviewing this. You are right - I missed a "c" in the equation. The math is difficult and I don't know how to solve the last equation.You're missing a 'C' in the 3rd last line in the w^2 term. It should be '-2LC' instead of '-2L'.
The tricky bit is that depending on the values of R, L, and C, there will be a different resonant frequency. With the fix mentioned above, you can go about finding that frequency but it wont be pretty as you are dealing with resonance.
Take a look at the image I provided. The peak shifts the -3dB frequency to a larger value. The math won't be pretty is all I am saying.
Cheers,
JP
You need to get in the habit of always (and I mean ALWAYS) checking your units. Most of the mistakes you make (most, not all) will affect the units. If the units don't work out, you KNOW the answer is wrong.I am trying to calculate the cut-off freqency, wc, for the low-pass filter below. Could someone help review my calculation below and see if it makes sense? I am wondering if my approach makes sense as it looks too complicated to obtain the final answer.