# Low-pass filter + signal inversion

Discussion in 'Homework Help' started by halfknot, Apr 22, 2013.

1. ### halfknot Thread Starter New Member

Apr 21, 2013
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Hello,

I'm pretty bad and inexperienced at this, so I get easily confused. I am trying to sketch and design a 1st order op-amp low-pass filter circuit that gives signal inversion. Does that mean just a simple op-amp low-pass filter with a NOT gate on the input/output?
This is my sketch for an op-amp low pass filter circuit.

I also want to perform an analysis to find the voltage transfer function. To do this, can I combine C1 and R2 in R2/(1+R2*SC1) where S = j*ω? Is the voltage transfer function the Vo/Vi from which I get the gain?

Thank you

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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This simply op-amp low-pass filter provides "signal inversion" in passband.
And, yes when you find Vo/Vi you will find the "gain".

3. ### halfknot Thread Starter New Member

Apr 21, 2013
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So I got Vo/Vi = (-R2/R1)*[1/(1+S*R2*C1)]. Assuming this is right; I'm told that the bandwidth is 1kHz, and the gain magnitude is 15.
I can use the frequency to find out S, but I still have to find suitable values for the capacitors and the 2 resistors. Do I just assign values that give the expected result? How would I go around this?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,590
1,285
The voltage gain in the passband is equal to −R2/R1 = 15
And you need a pole at 1Khz.
So simple you can assume R2 = 15K. And now you can easily find the remaining component values.

5. ### halfknot Thread Starter New Member

Apr 21, 2013
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Sorry for being dense, but I'm not quite how to get the value of the capacitor?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Pole is located at Fp = 1/(2 * pi * R2 * C2) ---> C2 = 1/(2 * pi * R2 * Fp) ≈ 0.16/(R2 * Fp)

7. ### halfknot Thread Starter New Member

Apr 21, 2013
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Cheers, guys.