help with 2 pole low pass filter/attenuator

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Thanks, Now for the high pass...

On my Sure model AA-AB32186 board, I noticed that the 22k shunt resistor is placed before the series input coupling cap [first image]. The resistance from shown test point to ground is 76k ohms.

BTW, I've replaced the damaged 22k shunt resistor with an 18k 1/8w 1%metal film resistor, since it was the nearest value small resistor I had on hand).

In most input circuits the shunt resistor is after the series cap, making input filter frequency calculation easy[second image].

How would one determine the low frequency cutoff in this case, by using the 76k or the 18k as the input impedance?

Thanks.

p.s. I'm using a first order high pass (series cap) to be 400Hz (near the -3db point of the midrange sealed box response).
 

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the resistance before the DC blocking cap has virtually no effect on the way the amp or the filtering operates.

When passing a signal into the amp the source is connected.
The source impedance appears in parallel to the 22k/18k and since the source impedance is usually very much smaller than 18k, the effect of the 18k is effectively nil.

The 18k is put in there to give the input terminal a reference to Signal Ground. It dissipates any leakage through the DC blocking capacitors to Signal Ground.
Generally this resistor is very high in value 100k to 2M2 is commonly seen.
Some amplifiers even omit it.

Omitting it gives problems if you mistakenly disconnect ot reconnect the interconnect cable while the amplifier is ON.
A DC voltage can build up on the input terminal and connecting the cable shorts this voltage to ground. This creates an instantaneous change in voltage at the input and results in an amplified version (an almighty "crack") appears at the output. This can damage speakers".
 
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