Question

Construct a frequency response curve for a high-pass R C filter in which $R=300\text{k}\Omega$ and $C=100\text{pF}$$\left(1\text{pF}={10}^{-12}\text{F}\right)$ . Cover a range of ${\left(V_{\text{p}}\right)}_{\text{o}}/{\left(V_{\text{p}}\right)}_{\text{i}}$ of 0.001 to 0.9999. Plot ${\left(V_{\text{p}}\right)}_{\text{o}}/{\left(V_{\text{p}}\right)}_{\text{i}}$ versus $\ln f$.

Short answer

At the higher value of the ratio $\ln f$ the value of the ratio $\frac{{\left(V_{\text{p}}\right)}_{\text{o}}}{{\left(V_{\text{p}}\right)}_{\text{i}}}$ is constant but increasing the value of the $\ln f$ value of the $\frac{{\left(V_{\text{p}}\right)}_{\text{o}}}{{\left(V_{\text{p}}\right)}_{\text{i}}}$ decreases rapidly and after further increase in $\ln f$ the value of $\frac{{\left(V_{\text{p}}\right)}_{\text{o}}}{{\left(V_{\text{p}}\right)}_{\text{i}}}$ again becomes constant.

Step-by-step solution

Step 1. Introduction

The expression for the ratio of the peak voltages at output to the input.

$\frac{{\left(V_{\text{p}}\right)}_{\text{o}}}{{\left(V_{\text{p}}\right)}_{\text{i}}}=\frac{1}{\sqrt{{\left(2\pi fRC\right)}^2+1}}$

Consider the term for the ratio of the peak voltages at output to the input, $\frac{{\left(V_{\text{p}}\right)}_{\text{o}}}{{\left(V_{\text{p}}\right)}_{\text{i}}}$.

$\begin{array}{l}y=\frac{1}{\sqrt{{\left(2\pi fRC\right)}^2+1}}\\ y^2=\frac{1}{{\left(2\pi fRC\right)}^2+1}\\ y^2{\left(2\pi fRC\right)}^2+y^2=1\\ {\left(2\pi fRC\right)}^2=\frac{1-y^2}{y^2}\end{array}$

$\begin{array}{l}2\pi fRC=\sqrt{\frac{1-y^2}{y^2}}\\ f=\frac{1}{2\pi RC}\sqrt{\frac{1-y^2}{y^2}}\end{array}$

Step 2. Explanation

The frequency is

$f=\frac{1}{2\pi RC}\sqrt{\frac{1-y^2}{y^2}}$

Now substitute the given values and plot the frequency as required. Excel sheet has been used to calculate all the values and plot.