# Maximum frequency of a high-pass filter

simulate this circuit – Schematic created using CircuitLab

A) Derive the expression for $$\ \mathcal{H}(s)\$$ where $$\\mathcal{H}(s) = \frac{V_o}{V_i}\$$

I started by using the voltage divider at the $$\V_{out}\$$ node.

$$\V_o = \frac{R}{R_c+\frac{1}{sc}+R}\ \cdot V_i\tag1\$$

B) At what frequency will the magnitude of $$\H(j\omega)\$$ be maximum?

The frequency response of the system is as follows

$$\ \mathcal{H}(s) = \frac{CRS}{CRS+CR_CS+1} = \frac{RCS}{(R+R_C)CS+1} = \frac{\frac{R}{R+R_C}S}{S+\frac{1}{(R+R_C)C}} \tag2\$$

$$\ \mathcal{H}(j\omega) = \frac{\frac{R}{R+R_C} j\omega}{j\omega+\frac{1}{(R+R_C)C}}\tag3\$$

The magnitude will be maximum when the frequency is infinity. This is because when the frequency of the voltage source is infinite ($$\\omega = \infty \$$), the capacitor behaves as a short circuit, and thus there is no voltage across the capacitor.

$$\ \mathcal{H}(j\omega) = \frac{\frac{R}{R+R_C} \infty}{\infty+\frac{1}{(R+R_C)C}} \tag3\$$

However, why wouldn't $$\ \omega = 0\$$ work? Because the capacitor will act as an open circuit when the frequency is zero thus $$\V_{out}\$$ will receive no voltage. Therefore, $$\ \omega = 0\$$ is minimum.

C) What is the maximum value of the magnitude of $$\ \mathcal{H}(j \omega)\$$?

$$\ |\mathcal{H}(j\omega)|_{max}=\frac{R}{R+R_C}\tag4\$$

D) At what frequency will the magnitude of $$\\mathcal{H}(j\omega) \$$ equal its maximum value divided by $$\ \sqrt{2}\$$

We start off finding the magnitude of the transfer function

$$\ | \mathcal{H}(j \omega)| = \frac{\frac{R\omega}{R+R_C}}{\sqrt{\omega^2+(\frac{1}{(R+R_C) \cdot C}})^2} \tag5\$$

Now we can use the equation $$\ | \mathcal{H}(j \omega_c)| = \frac{ | \mathcal{H}(\omega)|_{max}}{\sqrt{2}} \$$ to solve for the cutoff frequencies

$$\ \frac{\frac{R\omega}{R+R_C}}{\sqrt{\omega^2+(\frac{1}{(R+R_C) \cdot C}})^2} = \frac{\frac{R}{R+R_C}}{\sqrt{2}} \tag6\$$

Solving for $$\ \omega \$$ in this equation I get

$$\ \omega^2 = \frac{1}{C^2(R+R_C)^2} \rightarrow \omega = \frac{1}{C(R+R_C)} \tag7\$$

E) Assume a resistance of 5 ohms is connected in series with the 80 $$\\mu F\$$ capacitor in the circuit. $$\R_C\$$ is 20 ohms. Calculate $$\\omega_c\$$

$$\ \omega_C = \frac{1}{C(R_C+R)} = \frac{1}{(80 \times 10^{-6})(5+20)} = 500 rad/sec\$$

Plugging this back into the transfer function:

$$\\frac{\frac{20}{20+5}j\times 500}{j \times 500+\frac{1}{(20+5)\cdot (80 \times 10^{-6})}} = 0.5656 \angle 45\tag8\$$

• For B: When you "simplified" the equation your frequency ended up in many terms. It is easier to work with the not simplified expression. There only 1/jwC depends on the freq. It goes to zero for max freq which gives the max output voltage. – Oldfart Nov 11 '18 at 1:00
• @Oldfart Why does removing the frequency (letting it go to zero) give the max output voltage? Is it because the capacitor behaves like an open circuit when its frequency is = 0? Because of this all the voltage will be across the open and $V_{out} = V_{in}$? – Arthur Green Nov 11 '18 at 1:10
• I apologize I can't edit my previous comment. It should be because when the frequency of the source is infinity then the capacitor will act as a short circuit. This means that $V_{out} = V_{in}$ – Arthur Green Nov 11 '18 at 1:26