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davors
Joined: 16 Oct 2004
Posts: 5
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 18 Oct 2004 17:50 hmmm... let me think... |
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hmmm...
as what i know. for a LOW PASS filter, when the capacitor is an open cct, no current flows via any of the elements since they rall in series with the capacitor... the current through the resistances is zero, the voltage across them is zero... so Vi = Vo right? when the capacitor is shorted then, frequency is arbitrarily big, so the Vo = 0...
for high pass case... is it just the reverse thing? hmmm... any idea? 
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gzif
Joined: 26 Sep 2004
Posts: 6
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| 18 Oct 2004 18:12 Re: hmmm... let me think... |
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For High pass put series C and shunt R
for Low pass put series R and shunt C
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adel_48
Joined: 17 Dec 2003
Posts: 333
Helped: 37
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| 19 Oct 2004 16:22 Re: hmmm... let me think... |
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the current through a capacitor = C dv/dt which when solved for a sinusoidal wave Magnitude(V) = Magnitude(i)/ (2*pi*f*C) thus for very low frequencies applying a voltage across the capacitor yields very small current which makes the capacitor acts as a very high impedance (or open circuit at d.c). Similarly for very high frequencies applying a voltage across the capacitor yields very high current which makes the capacitor act as a short circuit at high frequencies.
thus for high pass filter at low frequencies the capacitor isolates the input voltage and the resistance has no current in it thus producing an output voltage =0
at high requencies the capacitor acts as a short ciruit connecting the input and the output voltages
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