# RC filter

Discussion in 'Homework Help' started by skusku, Aug 28, 2012.

1. ### skusku Thread Starter Active Member

Aug 9, 2009
70
1
Hi, this ain't homework though but i suspect something similar for a test.

All i want to know is how to calculate the cut off freq of this 2stage filter with an Rload resistor.

circuit - http://i49.tinypic.com/rbwsoz.jpg

3db cut off - http://i49.tinypic.com/2em0qb4.jpg

Above are the practical results.

So the default equation goes Fc=1/(2∏RC)

I understand that Rs and R1 is in series but my equation and practical doesn't match up.

Fc=1/[(2∏(R1+Rs)//R2//R3)*(C1+C2)]
= 212 Hz

Thank you

Last edited: Aug 28, 2012
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I obtain a transfer function as follows

$G(s)=\frac{V_{out}}{V_2}=\frac{1}{a$$s^2+bs+c$$}$

where

$a={(R_s+R_1)R_2C_1C_2}$

$b=\frac{1}{R_3C_2}+\frac{R_s+R_1+R_2}{(R_s+R_1)R_2C_1}+\frac{1}{R_2C_2}$

and

$c=\frac{R_s+R_1+R_2+R_3}{(R_s+R_1)R_2R_3C_1C_2}$

The -3dB point would be found for ω when

$\frac{c}{\sqr{{$$c-\omega^2$$}^2+b^2\omega^2}}=\frac{1}{\sqr2}$

For which I obtain ω=762.691 radians/sec or f=121.386Hz

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Can you tell me how you get this equation ?

4. ### Audioguru New Member

Dec 20, 2007
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The second RC filter on the right overloads the first RC filter on the left so they make a very poor filter. There should be a buffer amplifier between the RC filters.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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If

$G_{(s)}=\frac{1}{a $$s^2 +bs +c$$}$

the DC gain is simply

$G_{(0)}=\frac{1}{ac}$

$G_{$$j \omega$$}=\frac{1}{a$$-\omega^2+j \omega b+c$$}$

From which we obtain the magnitude

$|G_{$$\omega$$}|=\frac{1}{a sqrt{ {$$c-\omega^2$$ }^2 +\omega^2 b^2}}$

The -3dB point on the DC gain has a gain value

$|G_{-3dB}|=\frac{1}{\sqrt{2}} \times \frac{1}{ac}$

The value of ω which satisfies this condition is found by equating

$|G_{$$\omega$$}|=|G_{-3dB}|$

or

$\frac{1}{a sqrt{ {$$c-\omega^2$$ }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}} \times \frac{1}{ac}$

which after multiplying through by ac gives

$\frac{c}{sqrt{ {$$c-\omega^2$$ }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}}$

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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OK I understand but I still have a question.
Why don't you use a classical form of a transfer function?

$G_{(s)}=\frac{a}{$$s^2 +bs +c$$}$

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I wasn't particularly concerned with standard forms. Variable a just happened to be the term I'd factored out in the denominator. No doubt I could have made one the inverse of the other to agree with convention.

I must pose the obvious question - Would it have made any difference to the final relationship?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
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I think that the answer look like this

$\frac{a}{ sqrt{ {$$c-\omega^2$$ }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}} \times \frac{a}{c}$

So if we multiplying through by c and divide by a the answer will look the same.

So this equation holds for any second-order filter ?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I guess it's definitely true of a low pass type without any pass-band ripple - such as the Butterworth filter.

What about the Chebyshev LPF with a 3dB passband ripple say...? I'd have to think about it some more.