# Stupidly simple RC filter question

Discussion in 'Homework Help' started by donbenjy, Apr 15, 2011.

1. ### donbenjy Thread Starter New Member

Apr 15, 2011
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Hey,

So I'm going though an analysis for this guitar effects circuit: http://www.geofex.com/article_folders/tstech/tsxtech.gif

Looking at the input buffer section, I'm trying to reverse engineer the corner frequency for the high pass filter. I realise that spice/qucs will tell me in about 10 seconds that the Fc is about 15.5Hz (and also that the 1k resistor is small enough to ignore and use 1/2∏RC), but I'm trying to do it by hand, because I feel like it would help me understand things better, and be a more efficient designer.

Could anybody help me through where I'm missing something (or maybe I just need to push through the calculus to solve it, but still..) in order to calculate the Fc for the information given? I think the reason I'm getting stuck is because I know that I need a magnitude of 0.707 V (assuming a 1V input), but I don't know what the phase angle will be, so I can't solve it that way!

I know that the total impedance (in kohms) Zt= 511 + Xc j, but I don't know how to get this into polar form (or at least a workable polar form).

I'm thinking that there is probably a really obvious way to solve this that I'm gonna kick myself for, but I can't work it out!

Ben

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Assuming that the input resistance of transistor Q1 is very big (and thus omitted) and that the value of the input capacitor is 0.02mF, then we can calculate the corner frequency of the RC input circuit as follows:

$f_c=\frac{1}{2 \cdot \pi \cdot R \cdot C}=15.57Hz$

3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
659
I think you meant to write 0.02μF instead of 0.02mF.

4. ### donbenjy Thread Starter New Member

Apr 15, 2011
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I know that equation, but does it ALWAYS apply to any network of RC? I assumed that it only worked in cases of a series/parallel divider pair!

(Also, I just realised my simulation data was complicating things - I was taking an absolute value of 0.707 for my Fc, rather than 0.707 of the output signal!

Which raises another question - in an RC filter, is the 0.707 point ALWAYS 45 degrees out of phase? (assuming a perfect cap) With that knowledge, it would be easier to solve unknown networks

Is there a simple enough proof for why the formula for Fc works? The website does allude to it, but it would be great to read an explanation specifically applied to filter networks.

Thanks for the help btw!

5. ### Audioguru AAC Fanatic!

Dec 20, 2007
9,418
903
The phase change caused by a filter cannot be heard. A reduction in level of 0.707 times is just a little but is noticeable by most people. 0.707 times the voltage also causes the load current to be 0.707 times which causes the output power to be half (-3dB).

In my calculations for the cutoff frequency I don't bother with pi since 1/2pi= 0.16.
So my formula is 0.16/RC. It is very simple.

6. ### donbenjy Thread Starter New Member

Apr 15, 2011
5
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Yeah, I know that it won't be heard, but obviously when comparing -3dB to the 0dB point, is that taken as just the magnitude of the two total voltages (i.e. ignoring the phase angle)? I guess so, although couldn't that lead to some ambiguity depending on what's happening at the load? If there was some kind of feedback to a filter stage (even this simple one), then it wouldn't respond linearly across all frequencies?

Is this why 1/2piFC works? Rearranging to get

R = 1/2piFC
then substituting for Xc
R = 1/2piFC = Xc
R = Xc

When R = Xc, why does this give 1/2 power?

Thanks

7. ### Audioguru AAC Fanatic!

Dec 20, 2007
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903
At the frequency where the reactance of the capacitor equals the resistor value then the phase shift causes the output to be 0.707 times but not half.

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8. ### donbenjy Thread Starter New Member

Apr 15, 2011
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Ah ok, so it doesn't matter what order the components are in, the formula works for any network of Rs and Cs? The reason I was getting stuck is because I failed to realise the importance of the phase shift! Thanks

9. ### Audioguru AAC Fanatic!

Dec 20, 2007
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Of course it matters what order the R and C are in.
A capacitor feeding a resistor to ground is a highpass filter that cuts low frequencies.
A resistor feeding a capacitor to ground is a lowpass filter that cuts high frequencies.

A single RC lowpass filter has an output that is 0.707 at its calculated cutoff frequency and drops at 6dB (half the level) for each higher octave.

10. ### donbenjy Thread Starter New Member

Apr 15, 2011
5
0
Well yeah, but regardless of the network (i.e. something more than a potential divider of one resistor and one capacitor), the formula will always work? I was only concerned about phase angle because I can now see that's where my attempts to solve in reverse failed.

How, for example, would I solve the decoupling in the power supply network in the top left of the schematic above (47u cap and 2 x 10k resistors)? I know from software analysis that Fc is 0.707Hz, but that doesn't come out using the formula for various values of Rtotal.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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As for virtual ground circuit (47uF cap and 2 x 10k resistors).
If you want to find Fc you need to find equivalent resistance seen by 47uF capacitor.
So capacitor see 10K resistors connect in parallel plus parallel connect the load resistance of a virtual ground circuit.
Fc≈1/ (2 *pi * R*C) = 0.16/(R*C) = 0.16/ (5K*47uF) ≈ 0.68Hz

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