# Low Pass Filter with Limited Attenuation   Whatsapp  Our analysis will now continue with the low-pass filter of Fig. 2, which has limited attenuation at the high-frequency end. That is, the output will not drop to zero as the frequency becomes relatively high. The filter is similar in construction to Fig. 1, but note that now $mathbf{V}_{o}$ includes the capacitive element. Fig. 1: High-pass filter with attenuated output. Fig. 2: Low-pass filter with limited attenuation
At $f=0 \mathrm{~Hz}$, the capacitor can assume its open-circuit equivalence, and $\mathbf{V}_{o}=\mathbf{V}_{i}$. At high frequencies the capacitor can be approximated by a short-circuit equivalence, and
﻿$$\mathbf{V}_{o}=\frac{R_{2}}{R_{1}+R_{2}} \mathbf{V}_{i}$$
A plot of $V_{o}$ versus frequency is provided in Fig. 3(a). A sketch of $A_{V}=V_{o} / V_{i}$ will appear as shown in Fig. 3(b).  Fig. 3: Low-pass filter with limited attenuation
An equation for $\mathbf{V}_{o}$ in terms of $\mathbf{V}_{i}$ can be derived by first applying the voltage divider rule:
$$\mathbf{V}_{o}=\frac{\left(R_{2}-j X_{C}\right) \mathbf{V}_{i}}{R_{1}+R_{2}-j X_{C}}$$
and
\begin{aligned} \mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}} &=\frac{R_{2}-j X_{C}}{R_{1}+R_{2}-j X_{C}}=\frac{R_{2} / X_{C}-j}{\left(R_{1}+R_{2}\right) / X_{C}-j} \\ &=\frac{(j)\left(R_{2} X_{C}-j\right)}{(j)\left(\left(R_{1}+R_{2}\right) / X_{C}-j\right)} \\ &=\frac{j\left(R_{2} / X_{C}\right)+1}{j\left(\left(R_{1}+R_{2}\right) / X_{C}\right)+1}=\frac{1+j 2 \pi f R_{2} C}{1+j 2 \pi f\left(R_{1}+R_{2}\right) C} \end{aligned}
so that
$$\bbox[10px,border:1px solid grey]{\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{1+j\left(f / f_{1}\right)}{1+j\left(f / f_{c}\right)}} \tag{1}$$
with
$$\quad f_{1}=\frac{1}{2 \pi R_{2} C} \text{ and } f_{c}=\frac{1}{2 \pi\left(R_{1}+R_{2}\right) C}$$
The denominator of Eq. (1) is simply the denominator of the low-pass function. The numerator, however, is new and must be investigated. Applying Logarithmic equation :
$$A_{V_{\mathrm{dB}}}=20 \log _{10} \frac{V_{o}}{V_{i}}=20 \log _{10} \sqrt{1+\left(f / f_{1}\right)^{2}}+20 \log _{10} \frac{1}{\sqrt{1+\left(f / f_{c}\right)^{2}}}$$
For $f \gg f_{1},\left(f / f_{1}\right)^{2} \gg 1$, and the first term becomes
$$20 \log _{10} \sqrt{\left(f \mid f_{1}\right)^{2}}=20 \log _{10}\left(\left(f \mid f_{1}\right)^{2}\right)^{1 / 2}=\left.20 \log _{10}\left(f \mid f_{1}\right)\right|_{f \gg f_{1}}$$
which defines the idealized Bode asymptote for the numerator of Eq. (1).
At $f=f_{1}, 20 \log _{10} 1=0 \mathrm{~dB}$, and
at $f=2 f_{1}, 20 \log _{10} 2=6 \mathrm{~dB}$.
For frequencies much less than $f_{1},\left(f / f_{1}\right)^{2} \ll 1$, and the first term of the Log equation expansion becomes $20 \log _{10} \sqrt{1}=20 \log _{10} 1=0 \mathrm{~dB}$, which establishes the low-frequency asymptote.
The full idealized Bode response for the numerator of Eq. (1) is provided in Fig. 4. Fig. 4: Idealized and actual Bode response for the magnitude of $\left(1+j\left(f / f_{1}\right)\right)$.
We are now in a position to determine $A_{\text {vdB }}$ by plotting the asymptote for each function of Eq. (1) on the same frequency axis, as shown in Fig. 5. Note that $f_{c}$ must be less than $f_{1}$ since the denominator of $f_{1}$ includes only $R_{2}$, whereas the denominator of $f_{c}$ includes both $R_{2}$ and $R_{1}$. Fig. 5: $A_{V_{\mathrm{dB}}}$ versus frequency for the low-pass filter with limited attenuation of Fig. 2.
Since $R_{2} /\left(R_{1}+R_{2}\right)$ will always be less than 1 , we can use an earlier development to obtain an equation for the drop in $\mathrm{dB}$ below the 0-$\mathrm{dB}$ axis at high frequencies. That is,
\begin{aligned}20 \log _{10} R_{2} /\left(R_{1}+R_{2}\right) &=20 \log _{10} 1 /\left(\left(R_{1}+R_{2}\right) / R_{2}\right) \\ &=20 \log _{10} 1 -20 \log _{10}\left(\left(R_{1}+R_{2}\right) / R_{2}\right)\end{aligned}
and
$$\bbox[10px,border:1px solid grey]{20 \log _{10} \frac{R_{2}}{R_{1}+R_{2}}=-20 \log _{10} \frac{R_{1}+R_{2}}{R_{2}}} \tag{2}$$
as shown in Fig. $5$.
In region 1 of Fig. 5, both asymptotes are at 0 $\mathrm{~dB}$, resulting in a net Bode asymptote at 0 $\mathrm{~dB}$ for the region. At $f=f_{c}$, one asymptote maintains its 0-$d B$ level, whereas the other is dropping by $6 \mathrm{~dB} /$ octave. The sum of the two is the 6-$\mathrm{dB}$ drop per octave shown for the region. In region 3 the -6-$d \mathrm{~B}$ /octave asymptote is balanced by the +6-$\mathrm{dB}$ /octave asymptote, establishing a level asymptote at the negative $\mathrm{dB}$ level attained by the $f_{c}$ asymptote at $f=f_{1}$. The $\mathrm{dB}$ level of the horizontal asymptote in region 3 can be determined using Eq. (2) or by simply substituting $f=f_{1}$ into the asymptotic expression defined by $f_{c}$.
The full idealized Bode envelope is now defined, permitting a sketch of the actual response by simply shifting $3 \mathrm{~dB}$ in the right direction at each corner frequency, as shown in Fig. 5.
The phase angle associated with $\mathbf{A}_{V}$ can be determined directly from Eq. (1). That is,
$$\bbox[10px,border:1px solid grey]{\theta=\tan ^{-1} f / f_{1}-\tan ^{-1} f / f_{c}} \tag{3}$$
A full plot of $\theta$ versus frequency can be obtained by simply substituting various key frequencies into Eq. (3) and plotting the result on a log scale.
The first term of Eq. (3) defines the phase angle established by the numerator of Eq. (1). The asymptotic plot established by the numerator is provided in Fig. 6. Note the phase angle of $45^{\circ}$ at $f=$ $f_{1}$ and the straight-line asymptote between $f_{1} / 10$ and $10 f_{1}$. Fig. 6: Phase angle for $\left(1+j\left(f \mid f_{1}\right)\right)$.
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. $7$. Fig. 7: Phase angle for the low-pass filter of Fig. 2.
The asymptotes of Fig. $7$ clearly indicate that the phase angle will be $0^{\circ}$ in the low-frequency range and $0^{\circ}\left(90^{\circ}-90^{\circ}=0^{\circ}\right)$ in the high-frequency range. In region 2 the phase plot drops below $0^{\circ}$ due to the impact of the $f_{c}$ asymptote. In region 4 the phase angle increases since the asymptote due to $f_{c}$ remains fixed at $-90^{\circ}$, whereas that due to f1 is increasing.
In the midrange the plot due to f1 is balancing the continued negative drop due to the $f_{c}$ asymptote, resulting in the leveling response indicated. Due to the equal and opposite slopes of the asymptotes in the mid-region, the angles of $f_{1}$ and $f_{c}$ will be the same, but note that they are less than $45^{circ}$. The maximum negative angle will occur between $f_{1}$ and $f_{c}$. The remaining points on the curve of Fig. $7$ can be determined by simply substituting specific frequencies into Eq. (1). However, it is also useful to know that the most dramatic (the magnitude also goes through its greatest changes (such as at $f_{1}$ and $f_{c}$ ).

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250