Cut-off Frequency (Electrical Engineering)

[jghlee]

So this is a pretty general question regarding the cut-off frequency for any filter. In my study of filters, I've come across certain equations that the book gives for wc.

For example, Low/High Pass filters have wc = 1/(RC) or R/L, and Bandpass/stop filters have the center frequency wo^2 = 1/(LC). I'm able to derive these equations from some pretty generic L/H Pass Filter examples so I know where they come from.

Now my question is, do these equations apply to any filters of its kind? Meaning, can I always find the wc for a L/H pass filter by simply plugging in the 1/(RC) or R/L values? I'm terrible confused because all of my course homework seems to point this out but I'm not too sure in situations for when there's multiple L/C/R's or when I have RLC filters in which case a simple 1/(RC) doesn't seem to accurately depict the wc since it's missing the L value..

If somebody could clarify the meaning of those wc and wo equations, I'd appreciate it very much.

 

[Antiphon]

They don't apply to all filters but often apply.

The most general statement you can make, that always applies (by convention, not axiom) is that the cutoff wavelength is where the transmission is 3 dB down. The exact (to maybe 7 figures) cutoff wavelength is at 3.0103 dB down. This seems arcane but it's very simple.

If you have a simple voltage divider Z1/(Z1+Z2), the 3dB point is where |Z1|=|Z2|. Put in any element for Z1 and 2 to make any simple first order filter. The higher order filters also follow the 3dB convention even though it is less obvious that it would make sense.

 

[jghlee]

First off, thanks for reply!

So basically the sure way of knowing what the cut-off frequency is essentially using the cutoff frequency definition where H(jw) = Hmax / (root 2).

This leads me to another question. What are the purposes of these "general" equations like H(s) = (R/L) / [s + (R/L)] or H(s) = (1/RC) / [s + (1/RC)] in the case of LPF? Is there a special meaning in working/deriving our frequency response to look like this? Seems like for LPF/HPF we always want to get that s by itself in the denominator. My hunch is that, whenever you get a freq response equation that fits this mold, you're able to use those generic wc and wo values...

 

[MisterX]

The magnitude of frequency response of the system with transfer function H(s) is equal to |H(jω)|.

if one takes

$$H(s) = \frac{1/RC}{s + 1/RC}$$

and substitutes jω for s we have

$$H(jω) = \frac{1/RC}{jω + 1/RC}$$

When ω = 1/RC, the denominator is a complex number with the real part equal in magnitude to the imaginary part.

$$H(jω) = \frac{1/RC}{ (1/RC)(j + 1)}$$The magnitude of the denominator with ω = 1/RC is sqrt(2) times what it is when ω = 0.

 

[tedbradly]

Not sure if this helps, but a quick rule you can apply to any circuit network (I think) is an LC in parallel has infinite impedance at w = 1/sqrt(lc) and one in series as zero impedance at that w. So for simple networks, you can figure out the passband/stopband instantly by replacing the LCs with shorts/opens in your mind and asking yourself if the signal fully passes or fully stops at w = 1/sqrt(LC).