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In Reply to: RE: First Watt B4 arrives today - setup questions for Tymp I-Ds posted by Davey on April 21, 2017 at 06:49:37
I know the formula for creating a 1st-order high-pass filter via a single capacitor, plugging in the power amps input impedance and the desired x/o frequency. What is the formula for the corresponding 1st-order low-pass filter? Can the low-pass be accomplished just as easily, with a single capacitor on the power amps input jacks, or are more parts required, like a resistor as well as a cap?
Edits: 04/21/17Follow Ups:
Same formula. No, you can't just a single (shunt) capacitor at the input jacks because you need a series element (resistor) to create a low-pass filter.
At speaker-level you would use an inductor, but you can't do that at line-level because the L value would be very large.You need to fully characterize both power amps input R and the preamp output R before even attempting this.
Dave.
Edit: Sorry for the duplicate information, Neo beat me to it and posted while I was typing.
Edits: 04/22/17
You need to use a series resistor and shunt cap for the LP, creating a voltage divider. The resistor will introduce a insertion loss. You can match the insertion loss on the HP by also using a resistor but as you said a single series cap is sufficient.
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