Short/Open Circuit time constant method: b1

[kougou]

Homework Statement

The short circuit time constant method for determining 3dB point for high pass filter says

b1= the sum of short circuit time constant seen by each individual capacitor with the all other capacitor replaced by short circuit.

b1 is also = the sum of pole1, pole2...


I am not very convince by this statement. How should I see that
[the sum of poles] = b1 = [the sum of short circuit time constant seen by each capacitor] ?

Thank you

 

[member 553083]

. Homework EquationsThe equations are given in the statement. The Attempt at a SolutionI am not sure how to conceptually prove this statement. I have read some articles and it says that it is due to the fact that all other capacitor are short circuited, hence resulting in b1 = [the sum of short circuit time constant seen by each capacitor]. However, I am not sure how to interpret this conceptually. For example, if the high pass filter consists of two capacitors, C1 and C2 with resistance R1 and R2 respectively, then the sum of the poles [C1/(R1 + R2)] + [C2/(R1 + R2)] = b1 [C1/R1 + C2/R2] ? I am not sure how to explain this. Any help would be appreciated. Thank you.