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- Nov 17, 2011
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The frequency response of a high pass filter shows a low output at low frequencies and a high output at high frequencies. It lets high frequencies pass, as the name says. This is achieved by a combination of frequency dependent components, e.g. capacitor or inductors.
How this is done in an actual filter depends on the specific circuit used. There are thousands if not milllions of ways for constructing such a circuit. You can read this tutorial for starters.
How this is done in an actual filter depends on the specific circuit used. There are thousands if not milllions of ways for constructing such a circuit. You can read this tutorial for starters.
- Joined
- Nov 28, 2011
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A high-pass filter is a frequency-dependent attenuator. A simple one-pole high-pass filter consists of a series capacitor, followed by a resistor to ground. This configuration forms a kind of voltage divider, where the output amplitude is determined by the ratio between the "top" resistance (the reactance of the capacitor) and the "bottom" resistance (the resistor to ground).
For very high frequencies, where the capacitor's reactance is low, the output will be almost the same amplitude as the input, i.e. a tiny bit lower. As the frequency drops, the capacitor's reactance increases, and the circuit attenuates the signal to a greater extent, because the top resistance in the voltage divider gets higher with reducing frequency.
The output amplitude can never be quite as high as the input amplitude, because some signal is always lost in the capacitor (even if it's as little as 0.001 dB).
The voltage divider analogy is not perfectly accurate because reactance does not behave exactly like resistance; the current and voltage in a capacitor are 90° out of phase, instead of perfectly in phase as they are in a resistor. But that's a general explanation.
For very high frequencies, where the capacitor's reactance is low, the output will be almost the same amplitude as the input, i.e. a tiny bit lower. As the frequency drops, the capacitor's reactance increases, and the circuit attenuates the signal to a greater extent, because the top resistance in the voltage divider gets higher with reducing frequency.
The output amplitude can never be quite as high as the input amplitude, because some signal is always lost in the capacitor (even if it's as little as 0.001 dB).
The voltage divider analogy is not perfectly accurate because reactance does not behave exactly like resistance; the current and voltage in a capacitor are 90° out of phase, instead of perfectly in phase as they are in a resistor. But that's a general explanation.
