[sdiy] Wanted: non mathematical description of the functionof RC-filters

[rburnett at richieburnett.co.uk]

I like Tom's intuitive and non-mathematical RC filter explanation.

It's worth noting that the circuits for the 1st-order RC low-pass and 
1st-order RC high-pass are actually the same.  The only difference is 
where you take the output from.  Take the output from across the 
capacitor and you get low-pass filter action.  Conversely, take the 
output from across the resistor and you get high-pass action.  (Add the 
two together and you get back to the complete input signal again.)

Another way of looking at this is to say that the high-pass output is 
equal to the difference between the original input and the low-pass 
output.  I think that's quite intuitive:-  If you remove all the 
high-frequencies from a signal by low-pass filtering it, then the 
difference from the original input is the high-frequencies that you just 
removed!  (This is only guaranteed to work for 1st-order filters 
though!)

To electrical engineers the 1st-order lowpass and highpass filter 
transfer functions are:

LP = 1/(s+1)
HP = s/(s+1)

and you can show that: HP = 1 - LP

...but that would be maths!!! ;-)

-Richie,

PS. You can also get from the low-pass transfer function to the 
high-pass transfer function by differentiating once, because 
differentiation "tilts" the low-pass frequency response by +6dB/oct 
turning it into the highpass filter response.  That's where the 
additional 's' in the numerator of the HP filter equation comes from.


On 2013-08-09 20:01, Tom Wiltshire wrote:
> Before I understood RC filters, I'd seen a capacitor charging up
> through a resistor. A capacitor is like a little battery, and the
> resistor changes how fast you charge it up. Smaller resistor equals
> faster charge. Pretty straightforward, especially if you can watch it
> on a scope or a voltmeter. Do a few experiments, try a few different
> caps and resistors, get a feel for it. Discharging a capacitor is
> similar. Short it with a wire and it discharges pretty much instantly.
> Use a big resistor and it takes a while. Watch that on the meter too.
> 
> The next step is feeding frequencies through it. Imagine we feed a
> sine wave in where we previous used DC. Obviously if the resistor is
> large, the capacitor won't have had time to charge up fully before the
> sine wave starts going back down again. So the output tries to follow
> the sine wave, but the amplitude of the sine wave is reduced. If the
> resistor is bigger, the capacitor charges less, and the output is
> quieter. If the frequency goes up, the capacitor doesn't have as much
> time to charge, so the output is quieter. By now you get the idea.
> 
> That was how I got it at first. Highpass took me a bit longer, but
> once I'd seen the lowpass, it wasn't hard to see it the same way, but
> with the voltage divider the other way up. But for high pass, you need
> the voltage divider concept too, which I didn't have initially.
> 
> There, that wasn't too bad, was it?
> 
> I was probably twelve years old or so, and didn't have (or need) any
> mathematical understanding of it.
> 
> HTH,
> Tom
> 
> 
> 
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