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Circuit Analysis - Breakpoints of Passive, First Order Filters in a Circuit

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Acestuff

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I am having a great deal of problems trying to work out the corner frequencies of 2 filters (one high pass, one low pass) which are part of a circuit I need to analyze.

The problem is that these filters have 2 resistors in them; all the information on the internet refers to 1 resistor in the filter! I am using this equation:
Code:
f_c = 1/(2πRC)


With the low pass filter:
Intuitively I know that (excuse the poor terminology!) if I used the equation with just the 10k resistor, the cut-off frequency should be lower than calculated with the equation I stated because there is voltage "leaking" through the 100k resistor.

The use of Thevenin's theorem to solve this problem comes to mind. I'd then get a 9.09...kohm resistor in series with the output which would fit into the equation.

However, this means the denominator of the f_c equation is lower thus a higher frequency will be calculated. This goes against my intuition that the frequency should be lower!


I don't know where I'm going wrong so any help you can give would be amazing! The amount of time I've wasted thinking about these problems...


Low Pass:

vd05tz.jpg



High Pass:

1tx9nr.jpg


(Input from top, output to right)
 

LvW

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I recommend not to simply use one formula found elsewhere but, instead, to calculate by yourself.
That is the only way to understand what you are doing and if a particular formula is applicable or not!
Simply use the voltage divider formulas to calculate the frequency dependent output-input-ratio that is identical to the transfer function. Then you can see what happens when one particulöar resistor is increased resp. decreased.
 

Acestuff

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Thanks for your response. However, after just a couple of terrible lectures on the topic of amplifiers, my knowledge of them is very poor and I find nothing about them "simple"! Despite trying for hours to read about it, I barely understand anything about the transfer function or voltage dividers (I do computer science not electronics after all...)
 

LvW

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You need nothing more than to apply OHM`s law for impedances two times!
 

Acestuff

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You need nothing more than to apply OHM`s law for impedances two times!

I'm sorry but I'm not too sure what that means! Is there any chance you can supply an example for one of the filters?

In the meantime I'll do some reading on Ohm's law cause all I know of the law at the moment is V=IR!
 

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