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I'm doing some homework for my circuitry class and I'm stuck.

I have to find the value \$\frac{V_{in}}{V_{out}}\$ of a low pass filter, the schematic of which I have posted below. enter image description here

I got the break frequency, which I have shown below, but I do not know how to find the value of \$V_{in}\$ or \$V_{out}\$. I am also confused by what an octave is, but I think it is adding the same frequency per octave.

enter image description here

Any help would be appreciated!

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  • \$\begingroup\$ * you know the voltage ratio of a 2 resistor divider? substitute the impedance of the cap(Zc=1/ωC) for R. The calc. method is exactly the same. \$\endgroup\$
    – D.A.S.
    Commented Dec 4, 2017 at 1:14

1 Answer 1

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An octave is twice (or half) the frequency away from a given point. Eg an octave higher than 100Hz is 200Hz.

You do not need to know the exact value of \$V_{out}\$ or \$V_{in}\$, just the ratio and the transfer function can determine this.

  1. Derive the transfer function.
    This is a simple 1st order filter \$\frac{V_{out}}{V_{in}} = \frac{\omega_o}{s + \omega_o} \$

  2. Determine the "break frequency" a.k.a. cutoff frequency.
    This is straight forward for simple RC low pass filters & you have posted this \$ f_{cuttoff} = \frac{1}{2 \pi R C}\$.

  3. determine the frequency of interest.
    This is "two octave's above the cuttoff". Remembering 1 octave is x2.

  4. Determine the output ratio at this frequency.
    Using the transfer function from #1. and replacing s with \$j\omega \$ & then determining the magnitude

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  • \$\begingroup\$ So does that mean that two octaves would be (Break Frequency x 3) or would it be (Break frequency + Break Frequency + Break Frequency)? \$\endgroup\$
    – John
    Commented Dec 3, 2017 at 20:50
  • \$\begingroup\$ it would be break x4. \$\endgroup\$
    – user16222
    Commented Dec 3, 2017 at 21:00
  • \$\begingroup\$ So, I've found that the interest frequency is about 23578.508 Khz and the break frequency is 5894.627 Khz. I can't find what the ratio is though, my book only tells me that the output voltage is 70.7% of the input. So would it just be 100/70.7? I really don't know what value I am looking for \$\endgroup\$
    – John
    Commented Dec 3, 2017 at 21:16
  • \$\begingroup\$ Think about the transfer function. If you replace "s" with "jw" \$\endgroup\$
    – user16222
    Commented Dec 3, 2017 at 21:44
  • \$\begingroup\$ ^^^ In other words, turn this circuit into the Laplace domain. This will help you with your bode plot. \$\endgroup\$
    – user103380
    Commented Dec 3, 2017 at 22:55

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