R-C High-Pass Filters

 As noted early in Section (R-C low pass filters), a high-pass $ R $ - $ C $ filter can be constructed by simply reversing the positions of the capacitor and resistor, as shown in Fig. 1.
High-pass filter.
Fig. 1: High-pass filter.
At very high frequencies the reactance of the capacitor is very small, and the short-circuit equivalent can be substituted, as shown in Fig. 2. The result is that $ \mathbf{V}_{o}=\mathbf{V}_{i} $.
R-C high-pass filter at very high frequencies.
Fig. 2: R-C high-pass filter at very high frequencies.
R-C high-pass filter at f = 0 Hz.
Fig. 3: R-C high-pass filter at f = 0 Hz.
At $ f=0 mathrm{~Hz} $, the reactance of the capacitor is quite high, and the open-circuit equivalent can be substituted, as shown in Fig. 3. In this case, $ \mathbf{V}_{o}=0 \mathrm{~V} $. A plot of the magnitude versus frequency is provided in Fig. 4, with the normalized plot in Fig. 5.
Vo
versus frequency for a high-pass R-C filter.
Fig. 4: Vo versus frequency for a high-pass R-C filter.
Fig. 5: Normalized plot of Fig. 4.
At any intermediate frequency, the output voltage can be determined using the voltage divider rule:
$$\mathbf{V}_{o}=\frac{R \angle 0^{\circ} \mathbf{V}_{i}}{R-j X_{C}}$$
or
$$\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{R \angle 0^{\circ}}{R-j X_{C}}=\frac{R \angle 0^{\circ}}{\sqrt{R^{2}+X_{C}^{2}} \angle-tan ^{-1}\left(X_{C} / R \right)}$$
and
$$ \frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{R}{\sqrt{R^{2}+X_{C}^{2}}} \angle tan ^{-1}\left(X_{C} / R \right) $$
The magnitude of the ratio $ \mathbf{V}_{o} / \mathbf{V}_{i} $ is therefore determined by
$$A_{V}=\frac{V_{o}}{V_{i}}=\frac{R}{\sqrt{R^{2}+X_{C}^{2}}}=\frac{1}{\sqrt{1+\left(\frac{X_{C}}{R} \right)^{2}}}$$
and the phase angle $ \theta $ by
$$\theta=tan ^{-1} \frac{X_{C}}{R}$$
For the frequency at which $ X_{C}=R $, the magnitude becomes
$$\frac{V_{o}}{V_{i}}=\frac{1}{\sqrt{1+\left(\frac{X_{C}}{R} \right)^{2}}}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}=0.707$$
as shown in Fig. 5.
The frequency at which $ X_{C}=R $ is determined by
$$X_{C}=\frac{1}{2 \pi f_{c} C}=R$$
and
$$ f_{c}=\frac{1}{2 \pi R C} $$
For the high-pass $ R-C $ filter, the application of any frequency greater than $ f_{c} $ will result in an output voltage $ V_{o} $ that is at least $ 70.7 \% $ of the magnitude of the input signal. For any frequency below $ f_{c} $, the output is less than $ 70.7 \% $ of the applied signal.
Phase-angle response for the high-pass R-C filter.
Fig. 6: Phase-angle response for the high-pass R-C filter.
For the phase angle, high frequencies result in small values of $ X_{C} $, and the ratio $ X_{C} / R $ will approach zero with $ tan^{-1}\left(X_{C} / R \right) $ approaching $ 0^{\circ} $, as shown in Fig. 6. At low frequencies, the ratio $ X_{C} / R $ becomes quite large, and $ tan ^{-1}\left(X_{C} / R \right) $ approaches $ 90^{\circ} $.
For the case $ X_{C}=R $, $ tan ^{-1}\left(X_{C} / R \right)= tan^{-1} 1=45^{\circ} $. Assigning a phase angle of $ 0^{\circ} $ to $ \mathbf{V}_{i} $ such that $ \mathbf{V}_{i}=V_{i} \angle 0^{\circ} $, the phase angle associated with $ \mathbf{V}_{o} $ is $ \theta $, resulting in $ \mathbf{V}_{o}=V_{o} \angle \theta $ and revealing that $ \theta $ is the angle by which $ \mathbf{V}_{o} $ leads $ \mathbf{V}_{i} $.
Since the angle $ \theta $ is the angle by which $ \mathbf{V}_{o} $ leads $ \mathbf{V}_{i} $ throughout the frequency range of Fig. 6, the high-pass $ R $-$C$ filter is referred to as a leading network.
In summary, for the high-pass $ R$-$C $ filter:
$$f_{c}=\frac{1}{2 \pi R C}$$
For $f < fc$, $V_o < 0.707V_i$
whereas for $f > fc$, $V_o > 0.707V_i$
At $f_c$, $V_o$ leads $V_i$ by $45^\circ$
The high-pass filter response of Fig. $ 5 $ can also be obtained by interchanging their positions, as shown in Fig. 7.
High-pass R-L filter.
Fig. 7: High-pass R-L filter.
Example 1: Given $ R=20 \mathrm{k} \Omega $ and $ C=1200 \mathrm{pF} $ :
a. Sketch the normalized plot if the filter is used as both a high-pass and a low-pass filter.
b. Sketch the phase plot for both filters of part (a).
c. Determine the magnitude and phase of $ \mathbf{A}_{V}=\mathbf{V}_{o} / \mathbf{V}_{i} $ at $ f=\frac{1}{2} f_{c} $ for the high-pass filter.
Solution:
a.
$$ f_{c}=\frac{1}{2 pi R C}=\frac{1}{(2 \pi)(20 \mathrm{k} \Omega)(1200 \mathrm{pF})} =6631.46 \mathrm{~Hz} $$
The normalized plots appear in Fig. $ 8 $.
Fig. 8: Normalized plots for a low-pass and a high-pass filter using the same elements.
Fig. 9: Phase plots for a low-pass and a high-pass filter using the same elements.
b. The phase plots appear in Fig. $ 9 $.
c.
$$ f=\frac{1}{2} f_{c}=\frac{1}{2}(6631.46 \mathrm{~Hz})=3315.73 \mathrm{~Hz} $$
$$\begin{aligned} X_{C} &=\frac{1}{2 \pi f C}=\frac{1}{(2 \pi)(3315.73 \mathrm{~Hz})(1200 \mathrm{pF})} \\ & \cong 40 \mathrm{k} \Omega \\ A_{V}&=\frac{V_{o}}{V_{i}}=\frac{1}{\sqrt{1+\left(\frac{X_{C}}{R}\right)^{2}}}\\ &=\frac{1}{\sqrt{1+\left(\frac{40 \mathrm{k} \Omega}{20 \mathrm{k} \Omega} \right)^{2}}}\\ &=\frac{1}{\sqrt{1+(2)^{2}}}=\frac{1}{\sqrt{5}}=0.4472 \\ \theta&=tan^{-1} \frac{X_{C}}{R}=tan ^{-1} \frac{40 \mathrm{k} \Omega}{20 \mathrm{k} \Omega}\\ &=tan ^{-1} 2=63.43^{\circ} \end{aligned}$$
and
$$\mathrm{A}_{V}=\frac{\mathrm{V}_{o}}{\mathrm{~V}_{i}}=\mathbf{0 . 4 4 7 2} \angle 63.43^{\circ}$$