How to graph amplitude vs. frequency for a Low-Pass Filter?

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To graph the amplitude of a low-pass filter's output versus frequency, the correct equation to use is Vout = Vin * (Xc / sqrt(R^2 + Xc^2), where Xc represents capacitive reactance, defined as 1/(2πfC). The confusion arose from incorrectly using "Rc" instead of "Xc" for capacitive reactance. It's important to ensure that the frequency range plotted is appropriate to avoid a straight line output. Clarifying these terms and using the correct equations will lead to a successful graph.
eroc1002
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Homework Statement


For my lab class my professor wants us to graph the theoretical amplitude of the output from a low-pass filter vs. frequency. I have the resistance and capacitance that I chose for the RC circuit and a V(in). Is there an equation that I should be able to use to relate these and produce a graph?

Homework Equations


In my notes I have an equation Vout = Vin*(Rc/sqrt(R^2 + Rc^2) and Rc = 2πfε but I have no idea what epsilon represents because I forgot to ask my professor when I was taking the notes.

3. Attempt at a solution

After doing some research, I tried to use 1/sqrt(1+(2πfRC)^2)to get a ratio of Vo/Vin but all I got was a straight line.
 
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I think the epsilon should be C and RC is the capacitive reactance 2πfC.

As to why you got a straight line, it probably has to do with not plotting in the correct frequency range.
 
eroc1002 said:
In my notes I have an equation Vout = Vin*(Rc/sqrt(R^2 + Rc^2) and Rc = 2πfε
First, don't use "Rc" for capacitive reactance. Use Xc. "R" is always resistance in this context.
Second, capacitive reactance is not 2πfC. It's the inverse.
 
rude man said:
... capacitive reactance is not 2πfC. It's the inverse.
Right you are. I missed that. :oops:
 
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