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.
Low-pass filter.
Fig. 1: Low-pass filter.
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.
R-C low-pass filter at low frequencies.
Fig. 3: R-C low-pass filter at low frequencies.
R-C low-pass filter at high 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.
versus frequency for a low-pass R-C filter
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:
$$\mathbf{V}_{o}=\frac{X_{C} \angle-90^{\circ} \mathbf{V}_{i}}{R-j X_{C}}$$
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{X_{C} \angle -90^{\circ}}{R-j X_{C}}$$
$$=\frac{X_{C} \angle -90^{\circ}} {\sqrt{R^{2}+X_{C}^{2}} \angle -tan^{-1}(X_{C} / R)} $$
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle-90^{\circ}+tan ^{-1}\left(\frac{X_{C}}{R}\right)$$
The magnitude of the ratio $ V_{o} / V_{i} $ is therefore determined by
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
which defines the critical or cutoff frequency of Fig. 6.The frequency at which $ X_{C}=R $ is determined by
$$\frac{1}{2 \pi f_{c} C}=R$$
$$ f_{c}=\frac{1}{2 \pi R C} \tag{1} $$
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
$$\mathbf{V}_{o}=\left[\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta\right] \mathbf{V}_{i}=\left[\frac{X_{C}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta\right] V_{i} \angle 0^{\circ}$$
$$\mathbf{V}_{o}=\frac{X_{C} V_{i}}{\sqrt{R^{2}+X_{C}^{2}}} \angle \theta$$
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.
Angle by which Vo leads Vi.
Fig. 7: Angle by which Vo leads Vi.
Angle by which Vo lags 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.
Low-pass R-L filter.
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.
At $f=100 \mathrm{kHz}$:
$$X_{C}=\frac{1}{2 \pi f C}=\frac{1}{2 \pi(100 \mathrm{kHz})(500 \mathrm{pF})}=3.18 \mathrm{k} \Omega$$
$$V_{o}=\frac{20 \mathrm{~V}}{\sqrt{\left(\frac{1 \mathrm{k} \Omega}{3.18 \mathrm{k} \Omega}\right)^{2}+1}}=\mathbf{1 9 . 0 8} \mathbf{V}$$
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.

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