# Sketching the Bode Response

In the previous section we found that normalized functions of the form appearing in Fig. $1$ had the Bode envelope and the $\mathrm{dB}$ response indicated in the same figure. In this section we introduce additional functions and their responses that can be used in conjunction with those of Fig. 1 to determine the dB response of more sophisticated systems in a systematic, time-saving, and accurate manner.
Fig. 1: dB response of (a) low-pass filter and (b) high-pass filter.
Fig. 2: High-pass filter with attenuated output.
As an avenue toward introducing an additional function that appears quite frequently, let us examine the high-pass filter of Fig. $2$ which has a high-frequency output less than the full applied voltage. Before developing a mathematical expression for $\mathbf{A}_{V}=\mathbf{V}_{o} / \mathbf{V}_{i}$, let us first make a rough sketch of the expected response.
At $f=0 \mathrm{~Hz}$, the capacitor will assume its open-circuit equivalence, and $V_{o}=0 \mathrm{~V}$. At very high frequencies, the capacitor can assume its short-circuit equivalence, and
$$V_{o}=\frac{R_{2}}{R_{1}+R_{2}} V_{i}=\frac{4 \mathrm{k} \Omega}{1 \mathrm{k} \Omega+4 \mathrm{k} \Omega} V_{i}=0.8 V_{i}$$
The resistance to be employed in the equation for cutoff frequency can be determined by simply determining the Thevenin resistance "seen" by the capacitor.
Fig. 3: Determining RTh for the equation for cutoff frequency.
Setting $V_{i}=0 \mathrm{~V}$ and solving for $R_{T h}$ (for the capacitor C) will result in the network of Fig. 3, where it is quite clear that
$$R_{Th} = R_1 + R_2 = 1 kΩ + 4 kΩ = 5 kΩ$$
Therefore,
$$f_{c}=\frac{1}{2 \pi R_{T h} C}=\frac{1}{2 \pi(5 \mathrm{k} \Omega)(1 \mathrm{nF})}=31.83 \mathrm{kHz}$$
A sketch of $V_{o}$ versus frequency is provided in Fig. 4(a). A normalized plot using $V_{i}$ as the normalizing quantity will result in the response of Fig. 4(b). If the maximum value of $A_{V}$ is used in the normalization process, the response of Fig. 4(c) will be obtained.
Fig. 4: Finding the normalized plot for the gain of the high-pass filter of Fig. 2 with attenuated output
For all the plots obtained in the previous section, $V_{i}$ was the maximum value, and the ratio $V_{o} / V_{i}$ had a maximum value of 1 . For many situations, this will not be the case, and we must be aware of which ratio is being plotted versus frequency. The $\mathrm{dB}$ response curves for the plots of Figs. 4(b) and 4(c) can both be obtained quite directly using the foundation established by the conclusions depicted in Fig. 1, but we must be aware of what to expect and how they will differ. In Fig. 4(b) we are comparing the output level to the input voltage.
In Fig. 4(c) we are plotting $A_{V}$ versus the maximum value of $A_{V}$. On most data sheets and for the majority of the investigative techniques commonly employed, the normalized plot of Fig. 2(c) is used because it establishes $0 \mathrm{~dB}$ as an asymptote for the $\mathrm{dB}$ plot. To ensure that the impact of using either Fig. 4(b) or Fig. 4(c) in a frequency plot is understood, the analysis of the filter of Fig. $2$ will include the resulting $\mathrm{dB}$ plot for both normalized curves.
For the network of Fig. 2:
$$\mathbf{V}_{o}=\frac{R_{2} \mathbf{V}_{i}}{R_{1}+R_{2}-j X_{C}}=R_{2}\left[\frac{1}{R_{1}+R_{2}-j X_{C}}\right] \mathbf{V}_{i}$$
Dividing the top and bottom of the equation by $R_{1}+R_{2}$ results in
$$\mathbf{V}_{o}=\frac{R_{2}}{R_{1}+R_{2}}\left[\frac{1}{1-j \frac{X_{C}}{R_{1}+R_{2}}}\right]$$
but
\begin{aligned}-j \frac{X_{C}}{R_{1}+R_{2}} &=-j \frac{1}{\omega\left(R_{1}+R_{2}\right) C}=-j \frac{1}{2 \pi f\left(R_{1}+R_{2}\right) C} \\&=-j \frac{f_{c}}{f} \text { with } f_{c}=\frac{1}{2 \pi R_{T h} C} \text { and } R_{T h}=R_{1}+R_{2}\end{aligned}
so that
$$\mathbf{V}_{o}=\frac{R_{2}}{R_{1}+R_{2}}\left[\frac{1}{1-j\left(f_{c} / f\right)}\right] \mathbf{V}_{i}$$
If we divide both sides by $\mathbf{V}_{i}$, we obtain
$$\bbox[10px,border:1px solid grey]{\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{R_{2}}{R_{1}+R_{2}}\left[\frac{1}{1-j\left(f_{c} / f\right)}\right]} \tag{1}$$
from which the magnitude plot of Fig. 4(b) can be obtained. If we divide both sides by
$$\mathbf{A}_{V_{\max }}=R_{2} /\left(R_{1}+R_{2}\right),$$
we have
$$\bbox[10px,border:1px solid grey]{\mathbf{A}_{V}^{\prime}=\frac{\mathbf{A}_{V}}{\mathbf{A}_{V_{\max }}}=\frac{1}{1-j\left(f_{c} / f\right)}} \tag{2}$$
from which the magnitude plot of Fig. 4(c) can be obtained. Based on the past section, a $\mathrm{dB}$ plot of the magnitude of $A_{V}^{\prime}=$ $A \sqrt{ } A_{V_{\max }}$ is now quite direct using Fig. 1(b). The plot appears in Fig. $5$.
Fig. 5: dB plot for A′v for the high-pass filter of Fig. 2.
For the gain $A_{V}=V_{o} / V_{i}$, we can apply :
$$20 \log _{10} a b=20 \log _{10} a+20 \log _{10} b$$
where
$$20 \log _{10}\left\{\frac{R_{2}}{R_{1}+R_{2}}\left[\frac{1}{1-j\left(f_{c} l f\right)}\right]\right\}$$
$$=20 \log _{10} \frac{R_{2}}{R_{1}+R_{2}}+20 \log _{10} \frac{1}{\sqrt{1+\left(f_{c} / f\right)^{2}}}$$
The second term will result in the same plot of Fig. 5, but the first term must be added to the second to obtain the total $\mathrm{dB}$ response. Since $R_{2} /\left(R_{1}+R_{2}\right)$ must always be less than 1 , we can rewrite the first term as
$$20 \log _{10} \frac{R_{2}}{R_{1}+R_{2}}=20 \log _{10} \frac{1}{\frac{R_{1}+R_{2}}{R_{2}}}=20 \log _{10} 1-20 \log _{10} \frac{R_{1}+R_{2}}{R_{2}}$$
and
$$20 \log _{10} \frac{R_{2}}{R_{1}+R_{2}}=-20 \log _{10} \frac{R_{1}+R_{2}}{R_{2}}$$
providing the drop in $\mathrm{dB}$ from the $0$-$\mathrm{dB}$ level for the plot. Adding one log plot to the other at each frequency, will result in the plot of Fig. 6.
Fig. 6: Obtaining a dB plot of $Av_{dB}$
For the network of Fig. 2, the gain
$$\mathbf{A}_{v}=\mathbf{V}_{o} / \mathbf{V}_{i}$$
can also be found in the following manner:
$$\mathbf{V}_{o} =\frac{R_{2} \mathbf{V}_{i}}{R_{1}+R_{2}-j X_{C}}$$
\begin{aligned} \mathbf{A}_{V}&=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}} =\frac{R_{2}}{R_{1}+R_{2}-j X_{C}}\\ &=\frac{j R_{2}}{j\left(R_{1}+R_{2}\right)+X_{C}}=\frac{j R_{2} / X_{C}}{j\left(R_{1}+R_{2}\right) / X_{C}+1} \\ &=\frac{j \omega R_{2} C}{1+j \omega\left(R_{1}+R_{2}\right) C}=\frac{j 2 \pi f R_{2} C}{1+j 2 \pi f\left(R_{1}+R_{2}\right) C}\end{aligned}
and
$$\bbox[10px,border:1px solid grey]{\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{j\left(f / f_{1}\right)}{1+j\left(f / f_{c}\right)}} \tag{3}$$
with
$$f_{1}=\frac{1}{2 \pi R_{2} C} \quad and \quad f_{c}=\frac{1}{2 \pi\left(R_{1}+R_{2}\right) C}$$
The bottom of Eq. (3) is a match of the denominator of the lowpass function of Fig. 1(a). The numerator, however, is a new function that will define a unique Bode asymptote that will prove useful for a variety of network configurations. Applying Equation:
$$20 \log _{10} \frac{V_{o}}{V_{i}}=20 \log _{10}\left[\frac{f}{f_{1}}\right]\left[\frac{1}{\sqrt{1+\left(f / f_{c}\right)^{2}}}\right]$$ $$=20 \log _{10}\left(f / f_{1}\right)+20 \log _{10} \frac{1}{\sqrt{1+\left(f / f_{c}\right)^{2}}}$$
Let us now consider specific frequencies for the first term.
At $f=f_{1}$ :
$$20 \log _{10} \frac{f}{f_{1}}=20 \log _{10} 1=0 \mathrm{~dB}$$
At $f=2 f_{1}$ :
$$20 \log _{10} \frac{f}{f_{1}}=20 \log _{10} 2=+6 \mathrm{~dB}$$
At $f=\frac{1}{2} f_{1}$
$$20 \log _{10} \frac{f}{f_{1}}=20 \log _{10} 0.5=-6 \mathrm{~dB}$$
Fig. 7: dB plot of f/f1.
A $dB$ plot of $20 \log_10( f/f1)$ is provided in Fig. 7. Note that the asymptote passes through the 0-dB line at f = f1 and has a positive slope of +6 dB/octave (or 20 dB/decade) for frequencies above and below f1 for increasing values of f.
If we examine the original function $Av$, we find that the phase angle associated with $j f/f1 = f/f1 \angle 90°$ is fixed at $90°$, resulting in a phase angle for $Av$ of $90° - tan^{-1}( f/fc) = tan^{-1}( fc/f )$.
Now that we have a plot of the dB response for the magnitude of the function f/f1, we can plot the dB response of the magnitude of Av using a procedure outlined by Fig. 8.
Fig. 8: Plot of $Av_{dB}$ for the network of Fig. 2
Solving for f1 and fc:
$$f_{1}=\frac{1}{2 \pi R_{2} C}=\frac{1}{2 \pi(4 \mathrm{k} \Omega)(1 \mathrm{nF})}=39.79 \mathrm{kHz}$$
with
$$f_{c}=\frac{1}{2 \pi\left(R_{1}+R_{2}\right) C}=\frac{1}{2 \pi(5 \mathrm{k} \Omega)(1 \mathrm{nF})}=31.83 \mathrm{kHz}$$
For this development the straight-line asymptotes for each term resulting from the application of Log equation will be drawn on the same frequency axis to permit an examination of the impact of one line section on the other. For clarity, the frequency spectrum of Fig. $8$ has been divided into two regions. In region 1 we have a 0-d B asymptote and one increasing at $6 \mathrm{~dB} /$ octave for increasing frequencies. The sum of the two as defined by the Log equation is simply the 6-dB/octave asymptote shown in the figure.
In region 2 one asymptote is increasing at $6 \mathrm{~dB}$, and the other is decreasing at $-6 \mathrm{~dB} /$ octave for increasing frequencies. The net effect is that one cancels the other for the region greater than $f=f_{c}$, leaving a horizontal asymptote beginning at $f=f_{c}$. A careful sketch of the asymptotes on a $\log$ scale will reveal that the horizontal asymptote is at $-1.94 \mathrm{~dB}$, as obtained earlier for the same function. The horizontal level can also be determined by simply plugging $f=f_{c}$ into the Bode plot defined by $f\left(f_{1}\right.$; that is,
\begin{aligned}20 \log \frac{f}{f_{1}} &=20 \log _{10} \frac{f_{c}}{f_{1}}=20 \log _{10} \frac{31.83 \mathrm{kHz}}{39.79 \mathrm{kHz}} \\&=20 \log _{10} 0.799=-1.94 \mathrm{~dB}\end{aligned}
The actual response can then be drawn using the asymptotes and the known differences at $f=f_{c}(-3 \mathrm{~dB})$ and at $f=0.5 f_{c}$ or $2 f_{c}(-1 \mathrm{~dB})$.
In summary, therefore, the same $\mathrm{dB}$ response for $\mathbf{A}_{V}=\mathbf{V}_{o} / \mathbf{V}_{i}$ can be obtained by isolating the maximum value or defining the gain in a different form. The latter approach permitted the introduction of a new function for our catalog of idealized Bode plots that will prove useful in the future.