RC Low Pass Filters

The R-C filter, incredibly simple in design, can be used as a low-pass or a high-pass filter. If the output is taken off the capacitor, as shown in Fig. 1, it will respond as a low-pass filter. If the positions of the resistor and capacitor are interchanged and the output is taken off the resistor, the response will be that of a high-pass filter.
Fig. 1: Low-pass filter.
Fig. 2: Low-pass filter.
A glance at Fig. 2 reveals that the circuit should behave in a manner that will result in a high-level output for low frequencies and a declining level for frequencies above the critical value. Let us first examine the network at the frequency extremes of f = 0 Hz and very high frequencies to test the response of the circuit.
At f = 0 Hz,
$$X_{C}=\frac{1}{2 \pi fC}=\infty \Omega$$
and the open-circuit equivalent can be substituted for the capacitor, as shown in Fig. $3$, resulting in $\mathrm{V}_{o}=\mathrm{V}_{i}.$ At very High frequencies, the reactance is
$$X_{C}=\frac{1}{2\pi fC} \equiv 0\Omega$$
and the short-circuit equivalent can be substituted for the capacitor, as shown in Fig. $4$, resulting in $\mathrm{V}_{o}=0$ V.
Fig. 3: R-C low-pass filter at low frequencies.
Fig. 4: R-C low-pass filter at high frequencies.
A plot of the magnitude of $V_{o}$ versus frequency will result in the curve of Fig. 5. Our next goal is now clearly defined: Find the frequency at which the transition takes place from a pass-band to a stop-band.
﻿For filters, a normalized plot is employed more often than the plot of $V_{o}$ versus frequency of Fig. 5.
Normalization is a process whereby a quantity such as voltage, current, or impedance is divided by a quantity of the same unit of measure to establish a dimensionless level of a specific value or range.
Fig. 5: Vo versus frequency for a low-pass R-C filter
Fig. 6: Normalized plot of Fig. 4.
A normalized plot in the filter domain can be obtained by dividing the plotted quantity such as $V_{o}$ of Fig. 5 with the applied voltage $V_{i}$ for the frequency range of interest. Since the maximum value of $V_{o}$ for the low-pass filter of Fig. 1 is $V_{i}$, each level of $V_{o}$ in Fig. 5 is divided by the level of $V_{i}$. The result is the plot of $A_{V}=V_{o} / V_{i}$ of Fig. 6. Note that the maximum value is 1 and the cutoff frequency is defined at the $0.707$ level.
At any intermediate frequency, the output voltage $\mathbf{V}_{o}$ of Fig. 1 can be determined using the voltage divider rule:
$$\mathbf{V}_{o}=\frac{X_{C} \angle-90^{\circ} \mathbf{V}_{i}}{R-j X_{C}}$$
or
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{X_{C} \angle -90^{\circ}}{R-j X_{C}}$$
$$=\frac{X_{C} \angle -90^{\circ}} {\sqrt{R^{2}+X_{C}^{2}} \angle -tan^{-1}(X_{C} / R)}$$
and
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle-90^{\circ}+tan ^{-1}\left(\frac{X_{C}}{R}\right)$$
The magnitude of the ratio $V_{o} / V_{i}$ is therefore determined by
$$A_{V}=\frac{V_{o}}{V_{i}}=\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}}=\frac{1}{\sqrt{\left(\frac{R}{X_{C}}\right)^{2}+1}}$$
and the phase angle is determined by
$$\theta=-90^{\circ}+\tan ^{-1} \frac{X_{C}}{R}=-\tan ^{-1} \frac{R}{X_{C}}$$
For the special frequency at which $X_{C}=R$, the magnitude becomes
$$A_{V}=\frac{V_{o}}{V_{i}}=\frac{1}{\sqrt{\left(\frac{R}{X_{C}}\right)^{2}+1}}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}=0.707$$
which defines the critical or cutoff frequency of Fig. 6.The frequency at which $X_{C}=R$ is determined by
$$\frac{1}{2 \pi f_{c} C}=R$$
and
$$f_{c}=\frac{1}{2 \pi R C} \tag{1}$$
The impact of Eq. (1) extends beyond its relative simplicity. For any low-pass filter, the application of any frequency less than $f_{c}$ will result in an output voltage $V_{o}$ that is at least $70.7 %$ of the maximum. For any frequency above $f_{c}$, the output is less than $70.7 \%$ of the applied signal. Solving for $\mathbf{V}_{o}$ and substituting $\mathbf{V}_{i}=V_{i} \angle 0^{\circ}$ gives
$$\mathbf{V}_{o}=\left[\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta\right] \mathbf{V}_{i}=\left[\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta\right] V_{i} \angle 0^{\circ}$$
and
$$\mathbf{V}_{o}=\frac{X_{C} V_{i}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta$$
The angle $\theta$ is, therefore, the angle by which $\mathbf{V}_{o}$ leads $\mathbf{V}_{i}$. Since $\theta=$ $-\tan ^{-1} R / X_{C}$ is always negative (except at $f=0 \mathrm{~Hz}$ ), it is clear that $\mathbf{V}_{o}$ will always lag $\mathbf{V}_{i}$, leading to the label lagging network for the network of Fig. 1.
At high frequencies, $X_{C}$ is very small and $R / X_{C}$ is quite large, resulting in $\theta=-\tan ^{-1} R / X_{C}$ approaching $-90^{\circ}$. At low frequencies, $X_{C}$ is quite large and $R / X_{C}$ is very small, resulting in $\theta$ approaching $0^{\circ}$.
At $X_{C}=R$, or $f=f_{c},-\tan ^{-1} R / X_{C}=-\tan ^{-1} 1=-45^{\circ}$. A plot of $\theta$ versus frequency results in the phase plot of Fig. 7. The plot is of $\mathbf{V}_{o}$ leading $\mathbf{V}_{i}$, but since the phase angle is always negative, the phase plot of Fig. 8 $V_{o}$ lagging $\left.\mathbf{V}_{i}\right)$ is more appropriate.
Note that a change in sign requires that the vertical axis be changed to the angle by which $\mathbf{V}_{o}$ lags $\mathbf{V}_{i}$. In particular, note that the phase angle between $\mathbf{V}_{o}$ and $\mathbf{V}_{i}$ is less than $45^{\circ}$ in the pass-band and approaches $0^{\circ}$ ﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿at lower frequencies.
Fig. 7: Angle by which Vo leads Vi.
Fig. 8: Angle by which Vo lags Vi.
In summary, for the low-pass $R-C$ filter of Fig. 1:
$$\begin{split} f_c &= { 1 \over 2 \pi RC}\\ \text{For} f &< fc, Vo > 0.707V_i\\ \text{whereas for} f &> fc, Vo < 0.707V_i\\ & \text{At fc, V_o lags V_i by} 45^\circ \end{split}$$
The low-pass filter response of Fig. 2 can also be obtained using the $R-L$ combination of Fig. 9 with
$$f_{c}=\frac{R}{2 \pi L}$$
In general, however, the $R-C$ combination is more popular due to the smaller size of capacitive elements and the nonlinearities associated with inductive elements. The details of the analysis of the low-pass $R-L$ will be left as an exercise for the reader.
Fig. 9: Low-pass R-L filter.
Example 1:
a. Sketch the output voltage $V_{o}$ versus frequency for the low-pass $R-C$ filter of Fig. $10$.
b. Determine the voltage $V_{o}$ at $f=100 \mathrm{kHz}$ and $1 \mathrm{MHz}$, and compare the results to the results obtained from the curve of part (a).
c. Sketch the normalized gain $A_{V}=V_{o} / V_{i}$.
Fig. 10: Example 1.
Solution: ﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿a. Eq. (1):
$$f_{c}=\frac{1}{2 \pi R C}=\frac{1}{2 \pi(1 \mathrm{k} \Omega)(500 \mathrm{pF})}=\mathbf{318.31 ~} \mathbf{~ k H z}$$
At $f_{c}, V_{o}=0.707(20 \mathrm{~V})=14.14 \mathrm{~V}$. See Fig. 11.
Fig. 11: Frequency response for the low-pass R-C network of Fig. 9.
b.
$$V_{o}=\frac{V_{i}}{\sqrt{\left(\frac{R}{X_{C}}\right)^{2}+1}}$$
At $f=100 \mathrm{kHz}$:
$$X_{C}=\frac{1}{2 \pi f C}=\frac{1}{2 \pi(100 \mathrm{kHz})(500 \mathrm{pF})}=3.18 \mathrm{k} \Omega$$
and
$$V_{o}=\frac{20 \mathrm{~V}}{\sqrt{\left(\frac{1 \mathrm{k} \Omega}{3.18 \mathrm{k} \Omega}\right)^{2}+1}}=\mathbf{1 9 . 0 8} \mathbf{V}$$
At $f=1 \mathrm{MHz}$ :
$$X_{C}=\frac{1}{2 \pi f C}=\frac{1}{2 \pi(1 \mathrm{MHz})(500 \mathrm{pF})}=0.32 \mathrm{k} \Omega$$
and $$V_{o}=\frac{20 \mathrm{~V}}{\sqrt{\left(\frac{1 \mathrm{k} \Omega}{0.32 \mathrm{k} \Omega}\right)^{2}+1}}=\mathbf{6 . 1} \mathbf{V}$$ Both levels are verified by Fig. $11$.
c. Dividing every level of Fig. $11$ by $V_{i}=20 \mathrm{~V}$ will result in the normalized plot of Fig. 12.
Fig. 12: Normalized plot of Fig. 10.