# Deriving corner frequency, Active Low Pass

Discussion in 'Homework Help' started by jegues, Jan 30, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Hello all, I'm working on the following question. (See figure attached)

Now I was able to complete the first question by deriving the transfer function,

$T(jw) = \frac{-R_{2}}{R_{1} + R_{1}R_{2}jwC}$

and for DC gain, $\omega \rightarrow 0$

so $T(jw) = \frac{-R_{2}}{R_{1}}$

I can't figure out how to derive the corner frequency for this circuit.

Where do I start?

I know that $\omega_{c} = \frac{1}{\tau}$

and $f_{c} = \frac{\omega_{c}}{2 \pi}$

But how do I make the actual derivation? Can I simply say that,

$\tau = R_{2}C$?

Once I get this out of the way I'll make an attempt at the remaining questions.

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Last edited: Jan 30, 2011
2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
There is a small problem. DC gain occurs when ω -> 0, not ∞
In any case the corner frequency is the frequency that reduces the DC gain by 3 dB
So we know that the magnitude of the gain must be 1 / √2 times the DC gain
So if you divide numerator and denominator by R2 you can set the denominator equal to √2 and solve for ω.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Alrighty I think I managed to get it with your help!

I tried answering the remaining questions as well.

Could someone quickly check my work?

One concern I have is when solving so that the DC gain is 20DB.

The DC gain is given as,

$\frac{ -R_{2}}{R_{1}}$

So when I'm solving for R2 to achieve this gain, how do I get rid of the negative? Surely we don't want a negative resistance, right?

Thanks again!

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4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Gain in dB is always relative to "something". The negative sign with the gain just means there is an inversion which you could also think of as a 180° phase shift. A gain of +20 dB means that regardless of phase shift, the MAGNITUDE of the output needs to be 10 times as large as the input because
Code ( (Unknown Language)):
1.
2. 20 dB = 20 * log (10 / 1)
3. So
4. | -R2 / R1 | = 10 will satisfy this requirement.
5.
This would be COMPLETELY independent of calling the output level at DC the 0 dB level and comparing that to the output at say 10 kHz for example. In order to use dB for anything there must be a comparison or a ratio of two things involved.

When computing MAGNITUDE we take the square root of a squared quantity. Squaring anything removes the negative sign, because taking the square root of a negative number has no real solution. Of course there is a complex solution, but that is a story for another thread.

5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Alright, thanks that definatly cleared that confusion up.

Is there any mistakes in my work answering the remaining questions? Or does everything look good?