Homework Help: High pass filter and low pass filter question?

1. Feb 26, 2012

Grendelle

You have a series circuit consisting of a ac power supply, a 1150.0 ohm resistor and a 144.0 nF capacitor. If the circuit is configured as a low pass filter, what frequency will cause the gain to be 0.25?

Formula
Gain = Vout / Vin = R / [R^2 + (1/ωC)^2]^1/2

Solution
Plugging in R=1150 ohms
C=144*10^-9 F

to find a value for ω

And then using ω=2 π f
I got f= 0.05223Hz which is wrong.

Also what will be the formula for high pass filter?

2. Feb 26, 2012

vela

Staff Emeritus
Show your calculations for finding $\omega$.

3. Feb 26, 2012

rude man

1. Formula for gain is wrong.

2. Gain expression for high-pass would be ωRC/√(1 + ω2R2C2)

4. Feb 26, 2012

Grendelle

Calculations for ω

ω=2∏f

The formula by prof has given me is Gain = Vout / Vin = R / [R^2 + (1/ωC)^2]^1/2 for a low pass filter.

0.25 = 1150 / [1150^2 + (1/ω*144*10^-9)^2]^1/2

My bad. I had a calculation mistake previously. I did it again and this time I solved it to get ω=1.5592*10^3
Using ω=2∏f, f=248.15 Hz.

Which is still wrong... :(

I'm confused as to what frequency they are asking me for. Is the this frequency just plain ω? Am I making a mistake by calculating f=ω/2∏?

5. Feb 26, 2012

Staff: Mentor

That is not a low pass filter. You seem to have dived into this problem without even sketching the circuit you are dealing with. Can you draw the circuit diagram for the low pass filter using the 2 specified components? Can you derive the transfer function Vout / Vin for it, rather than placing your faith in something without knowing whether it's the right formula or not?

They are one and the same. But if in doubt, you can always write both.

6. Feb 26, 2012

Staff: Mentor

You are asking the wrong question. What you should be asking of yourself at this point is, "What is the circuit for my high pass filter?"

Then determine for yourself its transfer function, and sketch it. That way, you can't go wrong.

7. Feb 27, 2012

rude man

No, you're right, f in Hz = ω/2π.

But the formula your prof gave you, if he really did, is still wrong.

Sketch the diagram for the low-pass circuit, write a summation of currents into the node at the capacitor. Then you can determine the right formula yourself. Don't just accept a formula from someone else, derive it for yourself, or you really won't learn what you need to.