Low Pass RC Filter

Facebook
Whatsapp
Twitter
LinkedIn
For the low-pass filter of Fig. 1,
$$ \begin{aligned} \mathbf{A}_{v} &=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\frac{-j X_{C}}{R-j X_{C}}=\frac{1}{\frac{R}{-j X_{C}}}+1 \\ &= \frac{1}{1+j \frac{R}{X_{C}}}=\frac{1}{1+j \frac{R}{\frac{1}{2 \pi f C}}}=1+j \frac{1}{\frac{f}{\frac{1}{2 \pi R C}}} \end{aligned} $$
and
$$ \bbox[10px,border:1px solid grey]{\mathbf{A}_{V}=\frac{1}{1+j\left(f / f_{c}\right)} } \tag{1}$$
with
$$ \bbox[10px,border:1px solid grey]{f_{c}=\frac{1}{2 \pi R C}} \tag{2}$$
as defined earlier.
Low-pass filter
Fig. 1: Low-pass filter.
Note that now the sign of the imaginary component in the denominator is positive and $ f_{c} $ appears in the denominator of the frequency ratio rather than in the numerator, as in the case of $ f_{c} $ for the high-pass filter.
In terms of magnitude and phase,
$$\bbox[10px,border:1px solid grey]{\mathbf{A}_{V}=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}=\mathbf{A}_{V} \angle \theta=\frac{1}{\sqrt{1+\left(f / f_{c}\right)^{2}}} \angle-\tan ^{-1}\left(f / f_{c}\right)} \tag{3}$$
An analysis similar to that performed for the high-pass filter will result in
$$\bbox[10px,border:1px solid grey]{A_{V_{\mathrm{dB}}}=-20 \log _{10} \frac{f}{f_{c}} f \gg f_{c}} \tag{4}$$
Note in particular that the equation is exact only for frequencies much greater than $ f_{c} $, but a plot of Eq. (4) does provide an asymptote that performs the same function as the asymptote derived for the high-pass filter.
In addition, note that it is exactly the same as Eq. (2), except for the minus sign, which suggests that the resulting Bode plot will have a negative slope [recall the positive slope for Eq. (2)] for increasing frequencies beyond $ f_{c} $.
Bode plot for the high-frequency region of a low-pass R-C filter
Fig. 2: Bode plot for the high-frequency region of a low-pass R-C filter.
A plot of Eq. (4) appears in Fig. $ 2 $ for $ f_{c}=1 \mathrm{kHz} $. Note the 6- $ \mathrm{dB} $ drop at $ f=2 f_{c} $ and the 20- $ \mathrm{dB} $ drop at $ f=10 f_{c}$.
At $ f \gg f_{c} $, the phase angle $ \theta=-\tan ^{-1}\left(f / f_{c}\right) $ approaches $ -90^{\circ} $, whereas at $ f \ll f_{c}, \theta=-\tan ^{-1}\left(f / f_{c}\right) $ approaches $ 0^{\circ} $. At $ f=f_{c}, \theta= $ $ -\tan ^{-1} 1=-45^{\circ} $, establishing the plot of Fig. 3. Note again the $ 45^{\circ} $ change in phase angle for each tenfold increase in frequency.
Phase plot for a low-pass R-C filter.
Fig. 3: Phase plot for a low-pass R-C filter.
Even though the preceding analysis has been limited solely to the R-C combination, the results obtained will have an impact on networks that are a great deal more complicated. One good example is the high and low-frequency response of a standard transistor configuration. Some capacitive elements in a practical transistor network will affect the low-frequency response, and others will affect the high-frequency response.
In the absence of the capacitive elements, the frequency response of a transistor would ideally stay level at the mid-band value. However, the coupling capacitors at low frequencies and the bypass and parasitic capacitors at high frequencies will define a bandwidth for numerous transistor configurations.
In the low-frequency region, specific capacitors and resistors will form an R-C combination that will define a low cutoff frequency. There are then other elements and capacitors forming a second R-C combination that will define a high cutoff frequency. Once the cutoff frequencies are known, the -3-dB points are set, and the bandwidth of the system can be determined.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024 realnfo.com
Privacy Policy