Beginning Low-Pass Filter Questions

[baltimorebest]

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.

 

[gneill]

I think you want 1/(2*\piRC) 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" .

 

[baltimorebest]

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?

 

[Mindscrape]

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)

 

[baltimorebest]

I tried to read it, and I have a slightly better idea. So perhaps the cutoff frequency divided by the original frequency?

 

[Mindscrape]

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.

 

[baltimorebest]

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?

 

[Mindscrape]

They are neither in series or parallel. You can use the equation on the website for getting the right impedance.

 

[baltimorebest]

But that impedance uses the value of 'R' which I don't have. I just have 'RC'

 

[gneill]

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!).

 

[gneill]

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.

 

[Mindscrape]

Use

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

where s is 2πf. Or do what gneill suggests.

 

[gneill]

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.

 

[baltimorebest]

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?

 

[gneill]

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.

 

[baltimorebest]

I see. Thank you very much for all of your help. Same to Mindscrape.