# High-Pass R-C Filter

Let us start by reexamining the high-pass filter of Fig. 1. The high-pass filter was chosen as our starting point because the frequencies of primary interest are at the low end of the frequency spectrum. The voltage gain of the system is given by
$$\begin{split} \mathbf{A}_{V}&=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{R}{R-j X_{C}}\\ &=\frac{1}{1-j \frac{X_{C}}{R}}=\frac{1}{1-j \frac{1}{2 \pi f C R}}\\ &=\frac{1}{1-j\left(\frac{1}{2 \pi R C}\right) \frac{1}{f}} \end{split}$$
If we substitute
$$f_{c}=\frac{1}{2 \pi R C}$$
which we recognize as the cutoff frequency of earlier sections, we obtain
$$\mathbf{A}_{v}=\frac{1}{1-j\left(f_{c} / f \right)} \tag{1}$$
We will find in the analysis to follow that the ability to reformat the gain to one having the general characteristics of Eq. (1) is critical to the application of the Bode technique. Different configurations will result in variations of the format of Eq. (1), but the desired similarities will become obvious as we progress through the material.
In magnitude and phase form:
$$\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=A_{V} \angle \theta=\frac{1}{\sqrt{1+\left(f_{c} / f \right)^{2}}} \angle tan ^{-1}\left(f_{c} / f\right) \tag{2}$$
providing an equation for the magnitude and phase of the high-pass filter in terms of the frequency levels.
Using Eq. (3),
$$A_{V_{\mathrm{dB}}}=20 \log _{10} A_{V} \tag{3}$$
and, substituting the magnitude component of Eq. (2),
$$A_{v_{\mathrm{dB}}}=20 \log _{10} \frac{1}{\sqrt{1+\left(f_{c} / f\right)^{2}}}=20 \log _{10} 1 - 20 \log _{10} \sqrt{1+\left(f_{c} / f\right)^{2}}$$
and
$$\mathbf{A}_{V_{\mathrm{dB}}}=-20 \log _{10} \sqrt{1+\left(\frac{f_{c}}{f} \right)^{2}}$$
Recognizing that $\log _{10} \sqrt{x}=\log _{10} x^{1 / 2}=\frac{1}{2} \log _{10} x$, we have
\begin{aligned} A_{V_{\mathrm{dB}}} &=-\frac{1}{2}(20) \log _{10}\left[1+\left(\frac{f_{c}}{f}\right)^{2}\right] \\ &=-10 \log _{10}\left[1+\left(\frac{f_{c}}{f}\right)^{2}\right] \end{aligned}
For frequencies where $f \ll f_{c}$ or $(f_{c} / f)^{2} \gg 1$,
$$1+\left(\frac{f_{c}}{f} \right)^{2} \cong\left(\frac{f_{c}}{f} \right)^{2}$$
and
$$A_{V_{\mathrm{AB}}}=-10 \log _{10}\left(\frac{f_{c}}{f} \right)^{2}$$
but
$$\log _{10} x^{2}=2 \log _{10} x$$
resulting in
$$A_{V_{\mathrm{dB}}}=-20 \log _{10} \frac{f_{c}}{f}$$
However, logarithms are such that
$$-\log _{10} b=+\log _{10} \frac{1}{b}$$
and substituting $b=f_{c} / f$, we have
$$A_{V_{\mathrm{dB}}}=+20 \log _{10} \frac{f}{f_{c}} \quad f \ll f_{c} \tag{4}$$
To plot Eq. (4), consider the following levels of increasing frequency:
For $f = fc/10$, $f/fc = 0.1$ and $20 \log_10 0.1 = -20 dB$
For$f = fc/4, f/fc = 0.25$ and $20 \log_10 0.25 = -12 dB$
For $f = fc/2, f/fc = 0.51$ and $20 \log_10 0.5 = -6 dB$
For $f = fc, f/fc = 1$ and $20 \log_10 1 = 0 dB$
Note from the above equations that as the frequency of interest approaches $f_{c}$, the $\mathrm{dB}$ gain becomes less negative and approaches the final normalized value of $0 \mathrm{~dB}$. The positive sign in front of Eq. (4) can therefore be interpreted as an indication that the $\mathrm{dB}$ gain will have a positive slope with an increase in frequency. A plot of these points on a log scale will result in the straight-line segment of Fig. $1$ to the left of $f_{c}$.
Fig. 1: Idealized Bode plot for the low-frequency region.
For the future, note that the resulting plot is a straight line intersecting the $0 dB$ line at $f_{c}$. It increases to the right at a rate of $+6 \mathrm{~dB}$ per octave or $+20 \mathrm{~dB}$ per decade. In other words, once $f_{c}$ is determined, find $f_{c} / 2$, and a plot point exists at $-6 \mathrm{~dB}$ (or find $f_{c} / 10$, and a plot point exists at $-20 \mathrm{~dB}$ ).
Bode plots are straight-line segments because the dB change per decade or octave is constant.
The actual response will approach an asymptote (straight-line segment) defined by $A_{V_{\mathrm{d} B}}=0 \mathrm{~dB}$ since at high frequencies
$$f \gg f_{c} \text { and } f_{c} / f \cong 0$$
with
$$A_{V_{\mathrm{dB}}}=20 \log _{10} \frac{1}{\sqrt{1+\left(f_{c} / f \right)^{2}}}=20 \log _{10} \frac{1}{\sqrt{1+0}}=20 \log _{10} 1=0 \mathrm{~dB}$$
The two asymptotes defined above will intersect at $f_{c}$, as shown in Fig. 1, forming an envelope for the actual frequency response. At $f=f_{c}$, the cutoff frequency,
$$A_{V_{\mathrm{dB}}}=20 \log _{10} \frac{1}{\sqrt{1+\left(f_{c} / f \right)^{2}}}=20 \log _{10} \frac{1}{\sqrt{1+1}}=20 \log _{10} \frac{1}{\sqrt{2}} =-3 \mathrm{~dB}$$
At $f=2 f_{c}$,
$$A_{V_{\mathrm{dB}}}=-20 \log _{10} \sqrt{1+\left(\frac{f_{c}}{2 f_{c}} \right)^{2}}=-20 \log _{10} \sqrt{1+\left(\frac{1}{2} \right)^{2}} =-20 \log _{10} \sqrt{1.25}=-\mathbf{1} \mathbf{d B}$$
At $f=f_{c} / 2$,
\begin{aligned}A_{V_{\mathrm{dB}}} &=-20 \log _{10} \sqrt{1+\left(\frac{f_{c}}{f_{c} / 2}\right)^{2}}=-20 \log _{10} \sqrt{1+(2)^{2}} \\&=-20 \log _{10} \sqrt{5} \\&=-7 \mathbf{d B}\end{aligned}
separating the idealized Bode plot from the actual response by $7 \mathrm{~dB}- 6 \mathrm{~dB}=1 \mathrm{~dB}$, as shown in Fig. 1.
Reviewing the above, at $f=f_{c}$, the actual response curve is $3 \mathrm{~dB}$ down from the idealized Bode plot, whereas at $f=2 f_{c}$ and $f_{c} / 2$, the actual response curve is 1 dB down from the asymptotic response. The phase response can also be sketched using straight-line asymptotes by considering a few critical points in the frequency spectrum. Equation (2) specifies the phase response (the angle by which $\mathbf{V}_{o}$ leads $\mathbf{V}_{i}$ ) by
$$\theta=\tan ^{-1} \frac{f_{c}}{f} \tag{5}$$
For frequencies well below $f_{c}\left(f \ll f_{c}\right), \theta=\tan ^{-1}\left(f_{c} / f\right)$ approaches $90^{\circ}$ and for frequencies well above $f_{c}\left(f \gg f_{c}\right), \theta=\tan ^{-1}\left(f_{c} l f\right)$ will approach $0^{\circ}$, as discovered in earlier sections of the chapter.
At $f=f_{c}$,
$$\theta=\tan ^{-1}\left(f_{c} / f\right)=\tan ^{-1} 1=45^{\circ}$$
Defining} $f \ll f_{c}$ for $f=f_{c} / 10$ (and less) and $f \gg f_{c}$ for $f=10 f_{c}$ (and more), we can define
an asymptote at $\theta=90^{\circ}$ for $f \ll f_{c} / 10$, an asymptote at $\theta=0^{\circ}$ for $f \gg 10 f_{c}$, and an asymptote from $f_{c} / 10$ to $10 f_{c}$ that passes through $\theta=45^{\circ}$ at $f=f_{c}$.
The asymptotes defined above all appear in Fig. 2. Again, the Bode plot for Eq.(5) is a straight line because the change in phase angle will be $45^{\circ}$ for every tenfold change in frequency.
Substituting $f=f_{c} / 10$ into Eq. (5),
$$\theta=\tan ^{-1}\left(\frac{f_{c}}{f_{c} / 10}\right)=\tan ^{-1} 10=84.29^{\circ}$$
Fig. 2: Phase response for a high-pass R-C filter.
for a difference of $90^{\circ}-84.29^{\circ} \cong 5.7^{\circ}$ from the idealized response.
Substituting $f=10 f_{c}$,
$$\theta=\tan ^{-1}\left(\frac{f_{c}}{10 f_{c}}\right)=\tan ^{-1} \frac{1}{10} \cong \mathbf{5 . 7}$$
In summary, therefore,
at $f=f_{c}, \theta=45^{\circ}$, whereas at $f=f_{c} / 10$ and $10 f_{c}$, the difference between the actual phase response and the asymptotic plot is $5.7^{\circ}$.
Example 1:
a. Sketch $A_{V_{\mathrm{dB}}}$ versus frequency for the high-pass $R-C$ filter of Fig.$3$.
b. Determine the decibel level at $f=1 \mathrm{kHz}$.
c. Sketch the phase response versus frequency on a log scale.
Fig. 3: Example 1.
Solution:
a.
$$f_{c}=\frac{1}{2 \pi R C}=\frac{1}{(2 \pi)(1 \mathrm{k} \Omega)(0.1 \mu \mathrm{F})}=1591.55 \mathrm{~Hz}$$
The frequency $f_{c}$ is identified on the log scale as shown in Fig. 4. A straight line is then drawn from $f_{c}$ with a slope that will intersect$-20 \mathrm{~dB}$ at $f_{c} / 10=159.15 \mathrm{~Hz}$ or $-6 \mathrm{~dB}$ at $f_{c} / 2=795.77 \mathrm{~Hz}$.
A second asymptote is drawn from $f_{c}$ to higher frequencies at $0 \mathrm{~dB}$.The actual response curve can then be drawn through the $-3$-dB level at $f_{c}$ approaching the two asymptotes of Fig. 4. Note the plot at $f=2 f_{c}$ and $0.5 f_{c}$.
Fig. 4: Frequency response for the high-pass filter of Fig. 3.
Note that in the solution to part (a), there is no need to employ Eq. (4) or to perform any extensive mathematical manipulations.
b.
$$\left|A_{v \mathrm{aB}}\right|=20 \log _{10} \frac{1}{\sqrt{1+\left(\frac{f_{c}}{f}\right)^{2}}}=20 \log _{10} \frac{1}{\sqrt{1+\left(\frac{1591.55 \mathrm{~Hz}}{1000}\right)^{2}}}$$
$$=20 \log _{10} \frac{1}{\sqrt{1+(1.592)^{2}}}=20 \log _{10} 0.5318=-5.49 \mathrm{~dB}$$
as verified by Fig. $4$.
c. See Fig. 5.
Fig. 5: Phase plot for the high-pass R-C filter
.
Note that $\theta=45^{\circ}$ at $f=f_{c}=1591.55 \mathrm{~Hz}$, and the difference between the straight-line segment and the actual response is $5.7^{\circ}$ at $f=f_{c} / 10=159.2 \mathrm{~Hz}$ and $f=10 f_{c}=15,923.6 \mathrm{~Hz}$
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