# Analysis of a high-pass filter

I'm doing some analysis of op-amp circuitry including the high-pass filter shown below.

I was given three questions to try further my understanding on the analysis of such system.

However, I am not quite sure where to start on this one. I have listed the three questions below, the derivation of the transfer function is where I'm currently stuck.

1. Derive a general expression for the transfer function of the filter shown.

2. Let R1 = 3.9 kΩ, R2 = 1.8 kΩ, and C = 0.33 μF. What is the filter frequency magnitude response at very low frequencies?

3. Calculate the frequency at which the magnitude response will be 3 dB lower than it is at high frequencies.

Any help would be greatly appreciated.

Well, we are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_1&=\text{I}_2\\ \\ \text{I}_3&=\text{I}_4 \end{alignat*} \end{cases}\tag1

When we use and apply KVL, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\text{V}_\text{p}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2&=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3&=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4&=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{alignat*} \end{cases}\tag2

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

\begin{cases} \begin{alignat*}{1} \frac{\text{V}_\text{p}-\text{V}_1}{\text{R}_1}&=\frac{\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_2}{\text{R}_3}&=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{alignat*} \end{cases}\tag3

Now, when we have an ideal opamp we know that $$\\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_1=\text{V}_2\$$. So we can rewrite equation $$\(3)\$$ as follows:

\begin{cases} \begin{alignat*}{1} \frac{\text{V}_\text{p}-\text{V}_x}{\text{R}_1}&=\frac{\text{V}_x}{\text{R}_2}\\ \\ \frac{\text{V}_x}{\text{R}_3}&=\frac{\text{V}_3-\text{V}_x}{\text{R}_4} \end{alignat*} \end{cases}\tag4

Now, we can solve for $$\\text{V}_x\$$ and $$\\text{V}_3\$$:

• $$\text{V}_x=\frac{\text{V}_\text{p}\text{R}_2}{\text{R}_1+\text{R}_2}\tag5$$
• $$\text{V}_3=\frac{\text{V}_\text{p}\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag6$$

So, the transfer function is:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{p}}=\frac{1}{\text{V}_\text{p}}\cdot\frac{\text{V}_\text{p}\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag7$$

Now, applying this to your circuit we need to use (from now on I use the lowercase letters for the function in the 'complex' s-domain where I used Laplace transform):

$$\text{R}_1=\frac{1}{\text{sC}}\tag8$$

So, the transfer function will be:

$$\mathscr{H}\left(\text{s}\right)=\frac{\displaystyle\text{v}_3\left(\text{s}\right)}{\displaystyle\text{v}_\text{p}\left(\text{s}\right)}=\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\displaystyle\text{R}_3\left(\frac{1}{\text{sC}}+\text{R}_2\right)}\tag9$$

So, when we use the transformation $$\\text{s}=\text{j}\omega\$$ (where $$\\text{j}^2=-1\$$), we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\displaystyle\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}\tag{10}$$

The magnitude response can be found by solving for the absolute value of equation $$\(10)\$$:

$$$$\begin{split} \left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\left|\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\displaystyle\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)}\right|\\ \\ &=\frac{\displaystyle\left|\text{R}_2\left(\text{R}_3+\text{R}_4\right)\right|}{\displaystyle\left|\text{R}_3\left(\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right)\right|}\\ \\ &=\frac{\displaystyle\left|\text{R}_2\left(\text{R}_3+\text{R}_4\right)\right|}{\displaystyle\left|\text{R}_3\right|\cdot\left|\frac{1}{\text{j}\omega\text{C}}+\text{R}_2\right|}\\ \\ &=\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\displaystyle\text{R}_3\sqrt{\text{R}_2^2+\left(\frac{1}{\omega\text{C}}\right)^2}} \end{split}\tag{11}$$$$

The magnitude response will be at a maximum when $$\\omega\to\infty\$$, at that point the response is:

$$\lim_{\omega\space\to\space\infty}\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\sqrt{\text{R}_2^2+\left(0\right)^2}}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\text{R}_2}=\frac{\text{R}_3+\text{R}_4}{\text{R}_3}=1+\frac{\text{R}_4}{\text{R}_3}\tag{12}$$

So, we can solve for the cutoff frequency as follows:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\left(1+\frac{\text{R}_4}{\text{R}_3}\right)\space\Longrightarrow\space\omega=\frac{1}{\text{CR}_2}\tag{13}$$

Brandon, the transfer function of the whole circuit is nothing else than the product of the passive C-R2 block multiplied with the gain of the non-inverting opamp stage. Just basics!