RC filter

Thread Starter

skusku

Joined Aug 9, 2009
80
Hi, this ain't homework though but i suspect something similar for a test.

Last week i had a similar practical about an active filter...

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

With orcad I get 122Hz. Can someone please help me with the theory cut off freq?

Thank you
 
Last edited:

t_n_k

Joined Mar 6, 2009
5,455
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
 

Audioguru

Joined Dec 20, 2007
11,248
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.
 

t_n_k

Joined Mar 6, 2009
5,455
Can you tell me how you get this equation ?
If

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

the DC gain is simply

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

At frequency ω [rads/sec]

\(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}} \)
 

Jony130

Joined Feb 17, 2009
5,487
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}{\( s^2 +bs +c\)}\)
 

t_n_k

Joined Mar 6, 2009
5,455
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?
 

Jony130

Joined Feb 17, 2009
5,487
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 ?
 

t_n_k

Joined Mar 6, 2009
5,455
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.
 
Top