Read the passage given below and answer the question One very common use of filters is bandwidth limiting. Analog filter implementation consists of two categories: passive and active. The active filters are further classified as high‐pass, low‐pass, band‐pass, band‐reject and all‐pass. Butterworth and Chebyshev are commonly used practical filters. The key characteristic of Butterworth filter is that it has a flat passband and stopband. The practical limit for most RC active filters is close to 30 kHz. The Chebyshev response is referred to as an equiripple response because passband is characterised by a series of ripples that have equal maximum levels and equal minimum levels besides exhibiting flat stpoband.
The characteristic equation for the output voltage of an All‐pass filter is given by:
The passage introduces the concept of filters, which are widely used for bandwidth limiting in electronic circuits. Filters are broadly categorized as passive and active filters. Active filters, often implemented using components like operational amplifiers (op-amps), capacitors, and resistors, are further classified based on their frequency response characteristics:
The question specifically asks about the characteristic equation for the output voltage of an All-pass filter. While the passage mentions All-pass filters as a type of active filter, it does not provide the specific equation. The characteristic equation, often represented as the transfer function (\(V_o / V_{in}\)), describes how the filter modifies the input signal to produce the output signal.
An All-pass filter's main function is to alter the phase relationship between frequencies in a signal while keeping the amplitude response relatively constant across the frequency spectrum. A common implementation uses an operational amplifier (op-amp) in a specific configuration with resistors and capacitors. The transfer function of such a circuit can be derived using circuit analysis techniques.
A typical transfer function for an All-pass filter circuit has the form:
\(\frac{V_o}{V_{in}} = \frac{1 - sRC}{1 + sRC}\)
where \(s\) is the complex frequency variable (\(s = j\omega\)) and \(\omega = 2\pi f\) is the angular frequency. Substituting \(s = j2\pi f\), the transfer function becomes:
\(\frac{V_o}{V_{in}} = \frac{1 - j2\pi fRC}{1 + j2\pi fRC}\)
This means \(V_o = V_{in} \left(\frac{1 - j2\pi fRC}{1 + j2\pi fRC}\right)\). Let's examine the structure of the provided options and see if any match this form or a variation that simplifies to this form.
The options provided are expressions for \(V_o\) in terms of \(V_{in}\) and contain the term \(j2\pi fRC\). Let's look at Option 3:
\(\rm V_o = V_{in}\left(-1+\frac{2}{j2\pi fRC+1}\right)\)
We can manipulate the term in the parenthesis:
\(-1+\frac{2}{j2\pi fRC+1} = \frac{-(j2\pi fRC+1) + 2}{j2\pi fRC+1}\)
\(= \frac{-j2\pi fRC - 1 + 2}{j2\pi fRC+1}\)
\(= \frac{1 - j2\pi fRC}{1 + j2\pi fRC}\)
This result matches the standard form of the transfer function for an All-pass filter, \(\frac{V_o}{V_{in}} = \frac{1 - j2\pi fRC}{1 + j2\pi fRC}\). Therefore, multiplying by \(V_{in}\), we get:
\(\rm V_o = V_{in}\left(\frac{1 - j2\pi fRC}{1 + j2\pi fRC}\right)\)
This confirms that Option 3 represents the characteristic equation for the output voltage of a typical All-pass filter circuit configuration.
Let's briefly look at why other options are incorrect:
Thus, Option 3 is the correct characteristic equation for the output voltage of an All-pass filter among the given choices.
| Filter Type | Frequency Response | Key Characteristic |
|---|---|---|
| Low-Pass | Passes low frequencies, attenuates high frequencies | Roll-off above cutoff frequency |
| High-Pass | Passes high frequencies, attenuates low frequencies | Roll-off below cutoff frequency |
| Band-Pass | Passes a specific range of frequencies | Center frequency and bandwidth |
| Band-Reject (Notch) | Attenuates a specific range of frequencies | Center frequency and notch width |
| All-Pass | Passes all frequencies with equal gain | Phase shift varies with frequency |
All-pass filters are not used for shaping the magnitude of the frequency spectrum like other filter types. Their primary utility comes from their ability to modify the phase response of a signal without affecting its amplitude. This property makes them valuable in various applications:
The characteristic equation \(V_o = V_{in}\left(\frac{1 - j2\pi fRC}{1 + j2\pi fRC}\right)\) shows that the magnitude of the transfer function is \(\left|\frac{1 - j2\pi fRC}{1 + j2\pi fRC}\right| = \frac{\sqrt{1^2 + (-2\pi fRC)^2}}{\sqrt{1^2 + (2\pi fRC)^2}} = 1\), confirming that the gain is unity (or constant) for all frequencies \(f\). The phase shift, however, is \(\arg\left(\frac{1 - j2\pi fRC}{1 + j2\pi fRC}\right) = \arg(1 - j2\pi fRC) - \arg(1 + j2\pi fRC) = -\arctan(2\pi fRC) - \arctan(2\pi fRC) = -2\arctan(2\pi fRC)\), which clearly varies with frequency \(f\).
In choke input filter circuit, the first element is _______.
In the frequency response graph of an amplifier the 3 dB point refers to :
Which of the following statements is NOT correct for a Chebyshev Filter?
Which of the following statements is NOT correct about Butterworth filter?
FIR filters
1. are non-recursive
2. use feedback
3. are recursive
4. do not adopt any feedback
Select the correct choice.