Deriving Corner Frequency for High Pass Filter

Discussion in 'Homework Help' started by jegues, Feb 2, 2011.

1. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Hello all,

I've derived a transfer function, $T(s) = \frac{\frac{-R_{2}}{R_{1}}}{1 + \frac{1}{sCR_{1}}}$, but I can't seem to derive the corner frequency.

I am suppose to show that it is, $\omega_{o} = \frac{1}{CR_{1}}$

The only way I know how to obtain the corner frequency was how I obtained it using a low pass filter. I set the magnitude of my transfer function at the corner frequency equal to my DC gain divided by √(2).

But in a high pass filter my DC gain is 0. (This is what you would expect, it allows high frequencies to pass and attenuates low frequencies, and at DC $\omega = 0$)

So how do I solve it? Can someone nudge me in the right direction?

Thanks again!

2. hgmjr Moderator

Jan 28, 2005
9,030
214
Does this help at all?

$\frac{-R_2}{R_1}\Large \left(\frac{1}{1+\frac{1}{sR_1C}}\right)\ =\ \frac{-R_2}{R_1}\Large \left(\frac{s}{s+\frac{1}{R_1C}}\right)$

hgmjr

3. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I'm still not seeing how I am supposed to derive the corner frequency from this.

Is this supposed to be some sort of canonical form like the following,

$\frac{-R_2}{R_1}\Large \left(\frac{s}{s+\omega_{o} \right)$

?

Can you give me another hint?

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Start by considering the gain at frequencies where ω>>1/R1C.

The gain under these conditions is a constant -R2/R1.

As the frequency decreases, the first corner frequency occurs when ω=1/R1C.

So at ω=1/R1C the gain starts to drop at a nominal 20dB per decade. It will continue to drop with decreasing frequency, and at that rate down to DC or ω=0.

The DC gain - based on your TF at s=0, would be -R2C. Which seems somewhat strange, in that the DC gain depends upon a C value.

Are you sure your TF is correct?

5. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I'm fairly certain it is the same as the one given in the solution. They list it as,

$\frac{-(\frac{R_{2}}{R_{1}})s}{s + \frac{1}{R_{1}C}}$

Which is the same as what I have, correct?

How did you compute that DC gain?

I see when s=0 giving a 0 in the numerator or ∞ in the denominator either way I look at it.

Can you clarfiy some more?

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Yeah - my brain "fart" on that one. The DC gain is zero.

7. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
So is there any reasonable way to derive the corner frequency from this transfer function?

8. Georacer Moderator

Nov 25, 2009
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I think you can approach is as a generic TF and notice that you have a pole on the frequency $|{s_0}|$, where $s_0=-\frac{1}{RC}$.

9. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I've thought of this as well, but how do I justify that this pole is indeed the corner frequency ?

10. Georacer Moderator

Nov 25, 2009
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1,266
Because there is no other pole or zero to skew the graph. There is only the DC initial gain and then the downwards slope of -20dB/decade caused by the pole.
Typical Low-Pass filter.

<edit>The aforementioned TF represents a HIGH-PASS filter, not a LOW-PASS one. Read further for more.
Excuse me for the misinformation.

Last edited: Feb 4, 2011
11. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
But this is a High-Pass Filter, we would want to achieve a 20dB/decade increase by means of a zero, provided and initial DC gain of 0.

12. Georacer Moderator

Nov 25, 2009
5,142
1,266
I have to stop being lazy and sloppy with things I think I know by heart.

Let's recap. The pole is calculated with the correct methodology and is in a positive frequency. There is only one of it, so the system must be either a low-pass filter, or a high-pass one (If you come up with other combinations, please say so). By looking at your TF, you can see that in s=0, the gain is 0. For s=$\infty$ the gain is -R2/R1.
Therefore we have a High-Pass filter in our hands.

13. jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Unity gain for a High-Pass filter occurs as $s \rightarrow \infty$.

So what I need to do to derive the corner frequency is,

$|T(s)| = \frac{1}{\sqrt{2}} \cdot T(s)|_{s \rightarrow \infty}$

14. Georacer Moderator

Nov 25, 2009
5,142
1,266
Your approach is a bit problematic.

For one, High Pass filters don't necessarily have unity gain on high frequencies, just as Low Pass filters don't have unity DC gain. That gain depends on the circuit that makes them.

Also, with that equation, you will find the corner frequency gain, not the frequency itself. You will then need to solve the TF for the frequency in order to calculate it.

Why don't you like the method with the calculation of the pole? It 's quite simple and painless.