High Pass RC 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 If we substitute which we recognize as the cutoff frequency of earlier sections, we obtain 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: providing an equation for the magnitude and phase of the high-pass filter in terms of the frequency levels. Using Eq. (3), and, substituting the magnitude component of Eq. (2), and Recognizing that $ \log _{10} \sqrt{x}=\log _{10} x^{1 / 2}=\frac{1}{2} \log _{10} x $, we have For frequencies where $ f \ll f_{c} $ or $(f_{c} / f)^{2} \gg 1 $, and but resulting in However, logarithms are such that and substituting $ b=f_{c} / f $, we have 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}$. 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} $ ). The actual response will approach an asymptote (straight-line segment) defined by $ A_{V_{\mathrm{d} B}}=0 \mathrm{~dB} $ since at high frequencies with 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, At $ f=2 f_{c} $, At $f=f_{c} / 2$, 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 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} $, 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 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), for a difference of $ 90^{\circ}-84.29^{\circ} \cong 5.7^{\circ} $ from the idealized response.
Substituting $ f=10 f_{c} $, In summary, therefore,