How can we calculate the transfer function from this filter?

Question

I'm trying to calculate the transfer function of this high pass filter :

So basically I'm trying to find how I can find \$V_{\text{out}}\$.

Due to the amplifier we know that $$V_{in} = V_{\text{out}}$$ I know how to calculate \$V_{\text{in}}\$ but I can't seem to find \$V_{\text{out}}\$ because I am supposed to find this transfer function (since it is a Butterworth high pass filter):

$$|H_{ph}(j \omega)| = \frac{1}{\sqrt{1+ \left(\frac{\omega_c}{\omega}\right)^{2n}}} = \frac{\left(\frac{\omega}{\omega_c}\right)^{n}}{\sqrt{1+ \left(\frac{\omega}{\omega_c}\right)^{2n}}}$$

here \$n=1\$ because it's a first order filter

So I was wondering if any of you could help me and find this transfer function knowing that:

$$\underline{H}(j\omega) = \frac{\underline{V}_{\text{out}}}{\underline{V}_{\text{in}}}$$

Answer 1

Most of you reasonning concerning the Vout and Vin is correct. To find the transfer function, you need to do the Voltage divider.

$$V_{out} = V_{in}\frac{R_{1}}{{R_{1} + \frac{1}{j\omega C_{1}}}} $$

if we manipulate the equation:

$$\frac{V_{out}}{V_{in}}= \frac{j\omega}{{j\omega +\frac{1}{C_{1}R_{1}}}} $$

the cuf off frequency in this problem is:

$$\omega_{c}= \frac{1}{CR} $$

so the equation become:

$$H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{j\omega}{j\omega + \omega_{c}} $$

So there you have the general transfer equation. The op-amp is just a follower with unity Gain.

You can rearrange the equation to fit your format

$$H(j\omega) = \frac{\frac{j\omega}{\omega_{c}}}{\frac{j\omega}{\omega_{c}}+ 1} $$

Answer 2

John's method will certainly work, and works without you needing to know some of the formulas that you've given.

However, if you want to use your equation for \$|H(j\omega)|\$, then you need \$\omega_c\$.

$$\omega_c=\frac{1}{RC}=1000$$

$$\omega=2\pi 1000$$

$$|H(j\omega)|=\frac{1}{\sqrt{1+\left(\frac{1000}{2\pi 1000}\right)^2}}$$

Either will give you about 0.9876.