# RC Filter Circuit Question

Given the image below, we know that the second filter circuit is a Passive Low Pass Filter circuit. I'm assuming it is still a Passive Low Pass Filter with a different corner frequency. Am I correct? So my transfer function for the top circuit is just a voltage divider, which is-

$$\frac{V_o}{V_s} = \frac{Z_c||R_2} {(Z_c||R_2)+R_1} \approx \frac{.67jw} {jw + .33}$$

But my corner frequency should be $\dfrac{1}{RC}$. Do I include $Z_c||R_2$ with $R$ or ignore the capacitor?

So given this circuit, which is a high-pass filter. I get my transfer function to equal,

$$\frac{V_o}{V_s} = \frac{R_2} {(R_2)+Z} \approx \frac{10K} {15k + \frac{1} {jw.1E-6}}$$

Now to find the corner frequency, which occurs at, $\dfrac{1}{RC}$, we set the magnitude of $\dfrac{V_o}{V_s}$ equal to $\dfrac{1}{\sqrt{2}}$. Included is a screenshot because I'm too lazy to type this in mathjax. I feel like I'm missing some key part of knowledge which is netting me a wrong answer, or I suck at math. Shouldn't the corner frequency occur at 667 rad/s.

$$V_o = V_s * \frac{Z}{Z+R1}$$ where $$Z = \frac{R_2}{j\omega R_2C +1}$$ Thus

$$\frac{V_o}{V_s} = \frac{Z}{Z+R_1}$$

$$\frac{V_o}{V_s} = \frac{R_2}{j\omega R_1 R_2C+R_1+R_2}$$

• Makes sense, now the corner frequency is just 1/R2*R1*C? Nov 30, 2014 at 3:33
• No - to derive the corner frequency from the denominator you must rewrite the transfer function to get the so-called "normal" form: H(s)=K/(1+w/wc). In this case, you must divide numerator and denominator by (R1+R2). This gives wc=1/RpC with Rp=R1||R2.
– LvW
Nov 30, 2014 at 10:28
• exactly, remember you are trying to make the Vo/Vs equation equal 1/sqrt(2) or 0.707. For a simple RC network it is easy to see that when wCR = 1 (note the complex conjugate denominator requires resolving to the real domain) you have have this. For more "complex" circuits re.Equations a bit more thought is needed but the principal is the same
– user16222
Nov 30, 2014 at 11:53
• Updated question. Nov 30, 2014 at 22:12