Multiple frequency EMFs and capacitors

  • #1
mymodded
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TL;DR Summary
for time-varrying EMFs with multiply frequencies, how does a capacitor "charge" and "not charge" at the same time (which is why it simultaneously blocks low frequencies and not high ones)
In an RC low-pass filter, low frequencies pass through the filter, and only signals with high frequencies pass through the capacitor (where they are filitered out), and that happens because for low frequencies, the capacitor is charging, so they are blocked, while high frequencies don't allow charge to build up on the capacitor, so they are allowed to pass through.

My question is, how exactly does the capacitor charge and "not charge" at the same time, thus allowing certain frequencies to pass through?
 
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  • #2
Are you familiar with the differential equation that defines the relationship between voltage and current for a capacitor?
$$i(t) = C \frac{dv(t)}{dt} $$
 
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  • #3
mymodded said:
TL;DR Summary: for time-varrying EMFs with multiply frequencies, how does a capacitor "charge" and "not charge" at the same time (which is why it simultaneously blocks low frequencies and not high ones)

In an RC low-pass filter, low frequencies pass through the filter, and only signals with high frequencies pass through the capacitor (where they are filitered out), and that happens because for low frequencies, the capacitor is charging, so they are blocked, while high frequencies don't allow charge to build up on the capacitor, so they are allowed to pass through.

My question is, how exactly does the capacitor charge and "not charge" at the same time, thus allowing certain frequencies to pass through?
I don't think your understanding is quite correct here. The capacitor is carrying AC. It is not a one-way device like a diode, so it can pass current on every half cycle. Each frequency that is present can act independently, the higher ones passing more easily.
 
  • #4
berkeman said:
Are you familiar with the differential equation that defines the relationship between voltage and current for a capacitor?
$$i(t) = C \frac{dv(t)}{dt} $$
yes, and even with this, dv/dt is different for each frequency.

edit: I mean that since dv/dt is different frequencies, I don't think the capacitor can use dv/dt for each one of them at once.
 
  • #5
mymodded said:
I mean that since dv/dt is different frequencies, I don't think the capacitor can use dv/dt for each one of them at once.
I have two words for you: "Superposition" :wink:
 
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  • #6
berkeman said:
I have two words for you: "Superposition" :wink:
Sorry if I'm sounding too dumb, but wouldn't that mean the superpositioned frequency cannot be filtered out since there is only "one" frequency (the superpositioned frequency)?
 
  • #7
mymodded said:
since there is only "one" frequency (the superpositioned frequency)?
No, the superposition of all of the frequency components is most certainly not a single frequency. Have you learned about Fourier Analysis yet?

https://en.wikipedia.org/wiki/Fourier_series
 
  • #8
berkeman said:
Have you learned about Fourier Analysis yet?
No I haven't
 
  • #9
Read through that Wikipedia article to start to get a flavor for Fourier Series, then you can look through other references. Basically, any periodic waveform is composed of a number of sinusoidal "components" of different frequencies.
 
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  • #10
berkeman said:
Read through that Wikipedia article to start to get a flavor for Fourier Series, then you can look through other references. Basically, any periodic waveform is composed of a number of sinusoidal "components" of different frequencies.
alright thanks.
 
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