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Beginning Low-Pass Filter Questions

  • #1

Homework Statement


Consider a low-pass filter with RC=2x10-4 s.
a) Calculate the -3db frequency for the filter.
b) By what factor will the amplitude of a 10 kHz input sine wave be reduced by the filter?
c) By how many db will the power be reduced by the filter at 10 kHz?


Homework Equations





The Attempt at a Solution



I believe I got part a. I just said that the -3db frequency is equal to 1/RC which is 50000 1/s. I need help with parts b and c though. Thank you.
 

Answers and Replies

  • #2
gneill
Mentor
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I think you want 1/(2*[tex]\pi[/tex]RC) for the "cutoff" frequency.

You might want to ponder how the slope of the low pass filter response curve in its Bode plot can help you.

Here's a pretty good http://www.electronics-tutorials.ws/filter/filter_2.html" [Broken].
 
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  • #3
Thanks for responding. I see what you are saying, but doesn't that cutoff frequency determine where the remaining amount is 70% of the original? Does that help for part b?
 
  • #4
1,860
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You should read that site that gneill pointed you to, it's got all the basics and even has a couple examples of exactly what part b asks you to do.

For part c use the fact that dB=20log(Vout/Vin)
 
  • #5
I tried to read it, and I have a slightly better idea. So perhaps the cutoff frequency divided by the original frequency?
 
  • #6
1,860
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Look at the part of the website that says

Vout=Vin Xc/Z

then look at examples 1 and 2 below it. The only difference being that your question wants the ratio of Vout/Vin, which is even less work for you.
 
  • #7
Oh I think I was misunderstanding the question. I think got it now. My only remaining question for part b is that how do I calculate the impedance? I don't know if my resistor and capacitor are in series or parallel? Do I assume series?
 
  • #8
1,860
0
They are neither in series or parallel. You can use the equation on the website for getting the right impedance.
 
  • #9
But that impedance uses the value of 'R' which I don't have. I just have 'RC'
 
  • #10
gneill
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20,793
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The simple RC low pass filter is a voltage divider, where one circuit element is reactive (either an inductor or a capacitor). In this case it's a capacitor.

To remember how the resistor and capacitor connect, remember that the impedance of capacitors goes down with increasing frequency; They block DC and let high frequencies pass. So a low pass filter would have the capacitor placed so that it would shunt to ground high frequencies, while presenting a high impedance to ground for low frequencies.

Attached is a Bode plot and typical circuit for a low pass RC filter that has your filter's corner frequency (-3dB frequency). The red curve is the actual Vout/Vin curve, while the black line denotes the salient characteristics of the filter (and is often used for design work, because it's very easy to draw!).
 

Attachments

  • #11
gneill
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But that impedance uses the value of 'R' which I don't have. I just have 'RC'
So improvise! Choose a resistance value, say 1000 Ohms, and find the capacitance from your time constant.
 
  • #12
1,860
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Use

Vout/Vin=1/(1+R*C*s)

where s is 2πf. Or do what gneill suggests.
 
  • #13
gneill
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Use

Vout/Vin=1/(1+R*C*s)

where s is 2πf. Or do what gneill suggests.
That's the elegant way, of course. Make s = j2πf, since the expression encapsulates both the magnitude and phase responses (it's a complex value). The absolute value will give the amplitude response.
 
  • #14
Ah ok so Vout/Vin = 0.079577. So it's by a factor of 0.079577.
So for part c, do I just plug that number into dB=20log(Vout/Vin) and get -21.98. So that means it decreases by 21.98 db?
 
  • #15
gneill
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Yup. Round to -22dB and call it a day!

Note that you could read that directly off of the Bode plot, even the "stick figure" version that's straight lines.
 
  • #16
I see. Thank you very much for all of your help. Same to Mindscrape.
 

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