# Beginning Low-Pass Filter Questions

## 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?

## 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

gneill
Mentor
I think you want 1/(2*$$\pi$$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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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?

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)

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

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.

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?

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

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

gneill
Mentor
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

• Low Pass.pdf
10.8 KB · Views: 135
gneill
Mentor
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.

Use

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

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

gneill
Mentor
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.

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
Mentor
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.

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