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Deriving Corner Frequency for High Pass Filter

  • Thread starter jegues
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  • #1
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Homework Statement



See Explanation below for details of my confusion.

Homework Equations





The Attempt at a Solution



Hello all,



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



I am suppose to derive the corner frequency and show that it is indeed, [tex]\omega_{o} = \frac{1}{CR_{1}}[/tex]



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 [tex]\omega = 0[/tex])



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



Thanks again!
 

Answers and Replies

  • #2
berkeman
Mentor
57,300
7,279

Homework Statement



See Explanation below for details of my confusion.

Homework Equations





The Attempt at a Solution



Hello all,



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



I am suppose to derive the corner frequency and show that it is indeed, [tex]\omega_{o} = \frac{1}{CR_{1}}[/tex]



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 [tex]\omega = 0[/tex])



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



Thanks again!
The HPF has unity gain at infinite frequency. What is s at infinite frequency?
 
  • #3
1,097
2
The HPF has unity gain at infinite frequency. What is s at infinite frequency?
Well for physical frequencies, [tex]s = j\omega[/tex]

[tex]\text{So if, } \omega \rightarrow \infty[/tex]

[tex]s \rightarrow \infty[/tex]
 
  • #4
berkeman
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Well for physical frequencies, [tex]s = j\omega[/tex]

[tex]\text{So if, } \omega \rightarrow \infty[/tex]

[tex]s \rightarrow \infty[/tex]
Yep. So what does that give you for your transfer function at infinite frequency?
 
  • #5
1,097
2
Yep. So what does that give you for your transfer function at infinite frequency?
[tex]T(s)| _{s \rightarrow \infty} = \frac{-R_{2}}{R_{1}}[/tex]

Are these questions working towards giving me my corner frequency, or how to derive it?

EDIT: Ahhh I think I see my confusion now! I was trying to solve it the same way I solved the low pass filter but the unity gain of the low pass and high pass filters are different! So I have to do the magnitude of the transfer function corresponds to 1/sqrt(2) times the gain as s approaches infinity!
 
Last edited:
  • #6
1,097
2
The HPF has unity gain at infinite frequency. What is s at infinite frequency?
berkeman, I have a quick question about the language you used here.

What do you mean when you say HPF has unity gain at infinite frequency?

Unity gain from what I understand is when I have a gain of 1.

I know that for derivations of corner frequencys (both high and low pass),

[tex]|T(jw)| = \frac{1}{\sqrt{2}} * T(jw)|_{w \rightarrow \text{Unity Gain}}[/tex]
 
  • #7
berkeman
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57,300
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berkeman, I have a quick question about the language you used here.

What do you mean when you say HPF has unity gain at infinite frequency?

Unity gain from what I understand is when I have a gain of 1.

I know that for derivations of corner frequencys (both high and low pass),

[tex]|T(jw)| = \frac{1}{\sqrt{2}} * T(jw)|_{w \rightarrow \text{Unity Gain}}[/tex]
Yes, maybe I should have used a different word. Filters can have whatever gain they are designed for, so the HPF has whatever gain at infinite frequency.

I should have said more like a "reference" gain, which is the gain that it stabilizes at as the frequency goes to infinity. Like a LPF has some "reference" gain at DC. The attenuated values as you go down the slope(s) of the filter are with respect to the reference gain where there is no attenuation.

Sorry for the confusion.
 
  • #8
1,097
2
Yes, maybe I should have used a different word. Filters can have whatever gain they are designed for, so the HPF has whatever gain at infinite frequency.

I should have said more like a "reference" gain, which is the gain that it stabilizes at as the frequency goes to infinity. Like a LPF has some "reference" gain at DC. The attenuated values as you go down the slope(s) of the filter are with respect to the reference gain where there is no attenuation.

Sorry for the confusion.
So there isn't an actual accepted term that electrical engineers use to describe this "reference" gain?

Or is it just referred to as the "stable" gain?
 
  • #9
berkeman
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  • #10
berkeman
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You could call it the "passband gain"....
 
  • #11
1,097
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You could call it the "passband gain"....
Hmph... You'd think they'd have a specific term in place to describe just this... Weird.
 

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