High-Pass Filter with Limited Attenuation

The filter of Fig. $ 1 $ is designed to limit the low-frequency attenuation in much the same manner as described for the low-pass filter of the previous section.
At $ f=0 \mathrm{~Hz} $ the capacitor can assume its open-circuit equivalence, and $ \mathbf{V}_{o}=\left[R_{2} /\left(R_{1}+R_{2}\right)\right] \mathbf{V}_{i} $. At high frequencies the capacitor can be approximated by a short-circuit equivalence, and $ \mathbf{V}_{o}=\mathbf{V}_{i} $.
High-pass filter with limited attenuation.
Fig. 1: High-pass filter with limited attenuation.
Fig. 2: Determining R for the fc calculation for the filter of Fig. 1.
The resistance to be employed when determining $ f_{c} $ can be found by finding the Thevenin resistance for the capacitor $ C $, as shown in Fig. 2. A careful examination of the resulting configuration will reveal that $$ R_{T h}=R_{1} \| R_{2} $ and $ f_{c}=1 / 2 \pi\left(R_{1} \| R_{2}\right) C $$
Fig. 3: High-pass filter with limited attenuation.
A plot of $ V_{o} $ versus frequency is provided in Fig. 3(a), and a sketch of $ A_{V}=V_{o} / V_{i} $ appears in Fig. 3(b).
An equation for $ \mathbf{A}_{V}=\mathbf{V}_{o} / \mathbf{V}_{i} $ can be derived by first applying the voltage divider rule:
$$\mathbf{V}_{o}=\frac{R_{2} \mathbf{V}_{i}}{R_{2}+R_{1} \|-j X_{C}}$$
and
$$ \begin{split} \mathbf{A}_{V}&=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{R_{2}}{R_{2}+R_{1} \|-j X_{C}}\\ &=\frac{R_{2}}{R_{2}+\frac{R_{1}\left(-j X_{C}\right)}{R_{1}-j X_{C}}}=\frac{R_{2}\left(R_{1}-j X_{C}\right)}{R_{2}\left(R_{1}-j X_{C}\right)-j R_{1} X_{C}}\\ &=\frac{R_{1} R_{2}-j R_{2} X_{C}}{R_{1} R_{2}-j R_{2} X_{C}-j R_{1} X_{C}} =\frac{R_{1} R_{2}-j R_{2} X_{C}}{R_{1} R_{2}-j\left(R_{1}+R_{2}\right) X_{C}}\\ &=\frac{1-j \frac{R_{2} X_{C}}{R_{1} R_{2}}}{1-j \frac{\left(R_{1}+R_{2}\right)}{R_{1} R_{2}} X_{C}} \end{split}$$
so that
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{1-j\left(f_{1} / f\right)}{1-j\left(f_{c} / f\right)} \tag{1}$$
with $ \quad f_{1}=\frac{1}{2 \pi R_{1} C} \quad $ and $ \quad f_{c}=\frac{1}{2 \pi\left(R_{1} \| R_{2}\right) C} $
The denominator of Eq. (1) is simply the denominator of the high-pass function of Fig. 1(b). The numerator, however, is new and must be investigated.
$$\bbox[10px,border:1px solid grey]{\log _{10} a b=\log _{10} a+\log _{10} b} \tag{A}$$
Applying Eq. (A):
$$A_{V_{\mathrm{dB}}}=20 \log _{10} \frac{V_{o}}{V_{i}}=20 \log _{10} \sqrt{1+\left(f_{1} / f\right)^{2}}+20 \log _{10} \frac{1}{\sqrt{1+\left(f_{c} / f\right)^{2}}}$$
For $ f \ll f_{1},\left(f_{1} / f\right)^{2} \gg 1 $, and the first term becomes
$$20 \log _{10} \sqrt{\left(f_{1} / f\right)^{2}}=\left.20 \log _{10}\left(f_{1} / f\right)\right|_{f \ll f_{1}}$$
which defines the idealized Bode asymptote for the numerator of Eq. (1).
At $ f=f_{1}, \quad 20 \log _{10} 1=0 \mathrm{~dB} $
At $ f=0.5 f_{1}, \quad 20 \log _{10} 2=6 \mathrm{~dB} $
At $ f=0.1 f_{1}, \quad 20 \log _{10} 10=20 \mathrm{~dB} $
For frequencies greater than $ f_{1}, f_{1} / f \ll 1 $ and $ 20 \log _{10} 1=0 \mathrm{~dB} $, which establishes the high-frequency asymptote. The full idealized Bode plot for the numerator of Eq. (1) is provided in Fig. $ 4 $.
Fig. 4: Idealized and actual Bode response for the magnitude of ($1 - j (f1/f)$).
We are now in a position to determine $ A_{V_{\mathrm{dB}}} $ by plotting the asymptotes for each function of Eq. (1) on the same frequency axis, as shown in Fig. 5. Note that $ f_{c} $ must be more than $ f_{1} $ since $ R_{1} \| R_{2} $ must be less than $ R_{1} $.
Fig. 5: $Av_{dB}$ versus frequency for the high-pass filter with limited attenuation of Fig. 1.
When determining the linearized Bode response, let us first examine region 2 , where one function is $ 0 \mathrm{~dB} $ and the other is dropping at $ 6 \mathrm{~dB} $ /octave for decreasing frequencies. The result is a decreasing asymptote from $ f_{c} $ to $ f_{1} $.
At the intersection of the resultant of region 2 with $ f_{1} $, we enter region 1 , where the asymptotes have opposite slopes and cancel the effect of each other. The resulting level at $ f_{1} $ is determined by $ -20 \log _{10}\left(R_{1}+R_{2}\right) / R_{2} $ as found in earlier sections. The drop can also be determined by simply substituting $ f=f_{1} $ into the asymptotic equation defined for $ f_{c} $.
In region 3 both are at $ 0 \mathrm{~dB} $, resultasymptotic equation defined for $ f_{c} $. In region 3 both are at $ 0 \mathrm{~dB} $, resulting in a 0 -dB asymptote for the region. The resulting asymptotic and actual responses both appear in Fig. 5.
The phase angle associated with $ A_{V} $ can be determined directly from Eq. (1); that is,
$$\bbox[10px,border:1px solid grey]{\theta=-\tan ^{-1} \frac{f_{1}}{f}+\tan ^{-1} \frac{f_{c}}{f}} \tag{2}$$
A full plot of $ \theta $ versus frequency can be obtained by simply substituting various key frequencies into Eq. (2) and plotting the result on a log scale. The first term of Eq. (2) defines the phase angle established by the numerator of Eq. (1). The asymptotic plot resulting from the numerator is provided in Fig. 6. Note the leading phase angle of $ 45^{\circ} $ at $ f=f_{1} $ and the straight-line asymptote from $ f_{1} / 10 $ to $ 10 f_{1} $.
Fig. 6: Phase angle for $ \left(1-j\left(f_{1} / f\right)\right) $.
Now that we have an asymptotic plot for the phase angle of the numerator, we can plot the full phase response by sketching the asymptotes for both functions of Eq. (1) on the same graph, as shown in Fig. $ 7 $.The asymptotes of Fig. $ 1 $ clearly indicate that the phase angle will be $ 90^{\circ} $ in the low-frequency range and $ 0^{\circ}\left(90^{\circ}-90^{\circ}=0^{\circ}\right) $ in the high-frequency range.
Fig. 7: Phase response for the high-pass filter of Fig. $ 1 $.
In region 2 the phase angle is increasing above $ 0^{\circ} $ because one angle is fixed at $ 90^{\circ} $ and the other is becoming less negative. In region 4 one is $ 0^{\circ} $ and the other is decreasing, resulting in a decreasing $ \theta $ for this region.
In region 3 the positive angle is always greater than the negative angle, resulting in a positive angle for the but opposite, the angles at $ f_{c} $ and $ f_{1} $ are the same. Figure $ 7 $ reveals that the angle at $ f_{c} $ and $ f_{1} $ will be less than $ 45^{\circ} $. The maximum angle will occur between $ f_{c} $ and $ f_{1} $, as shown in the figure. Note again that the greatest change in $ \theta $ occurs at the corner frequencies, matching the regions of greatest change in the $ \mathrm{dB} $ plot.
Example 1: For the filter of Fig. 8:
a. Sketch the curve of $ A_{V_{\mathrm{dB}}} $ versus frequency using a log scale.
b. Sketch the curve of $ \theta $ versus frequency using a log scale.
Fig. 8: Example 1.
Solution:
a. For the break frequencies:
$$ f_{1}=\frac{1}{2 \pi R_{1} C}=\frac{1}{2 \pi(9.1 \mathrm{k} \Omega)(0.47 \mu \mathrm{F})}=\mathbf{3 7 . 2} \mathbf{H z} $$
$$ f_{c}=\frac{1}{2 \pi\left(\frac{R_{1} R_{2}}{R_{1}+R_{2}}\right) C}=\frac{1}{2 \pi(0.9 \mathrm{k} \Omega)(0.47 \mu \mathrm{F})}=\mathbf{3 7 6 . 2 5} \mathbf{~ H z} $$
The maximum low-level attentuation is
$$\begin{aligned}-20 \log _{10} \frac{R_{1}+R_{2}}{R_{2}} &=-20 \log _{10} \frac{9.1 \mathrm{k} \Omega+1 \mathrm{k} \Omega}{1 \mathrm{k} \Omega} \\&=-20 \log _{10} 10.1=-\mathbf{2 0 . 0 9} \mathbf{d B}\end{aligned}$$
The resulting plot appears in Fig. $ 9 $.
Fig. 9: $Av_{dB}$ versus frequency for the filter of Fig. 8.
b. For the break frequencies:
At $f=f_{1}=37.2 \mathrm{~Hz}$
$$\begin{aligned}\theta &=-\tan ^{-1} \frac{f_{1}}{f}+\tan ^{-1} \frac{f_{c}}{f} \\ &=-\tan ^{-1} 1+\tan ^{-1} \frac{376.25 \mathrm{~Hz}}{37.2 \mathrm{~Hz}} \\ &=-45^{\circ}+84.35^{\circ} \\ &=\mathbf{3 9 . 3 5}^ {\circ} \end{aligned}$$
At $f=f_{c}=376.26 \mathrm{~Hz}$
$$\begin{aligned} \theta &=-\tan ^{-1} \frac{37.2 \mathrm{~Hz}}{376.26 \mathrm{~Hz}}+\tan ^{-1} 1\\ &=-5.65^{\circ}+45^{\circ} \\ &=\mathbf{3 9 . 3 5}^ {\circ} \end{aligned}$$
At a frequency midway between $ f_{c} $ and $ f_{1} $ on a log scale, for example, $ 120 \mathrm{~Hz} $ :
$$\begin{aligned} \theta &=-\tan ^{-1} \frac{37.2 \mathrm{~Hz}}{120 \mathrm{~Hz}}+\tan ^{-1} \frac{376.26 \mathrm{~Hz}}{120 \mathrm{~Hz}} \\&=-17.22^{\circ}+72.31^{\circ} \\ &=\mathbf{5 5 . 0 9 ^ { \circ }}\end{aligned}$$
The resulting phase plot appears in Fig. 10.
Fig. 10: v (the phase angle associated with Av) versus frequency for the filter of Fig. 8.