Chapter 6: Problem 21
Draw the circuit diagram of a first-order \(R C\) lowpass filter and give the expression for the half-power frequency in terms of the circuit components. Sketch the magnitude and phase of the transfer function versus frequency.
Short Answer
Expert verified
An RC low-pass filter has a cutoff frequency \( f_c = \frac{1}{2\pi RC} \). The magnitude decreases and phase shifts negative with increasing frequency.
Step by step solution
01
Understanding the Circuit
A first-order RC low-pass filter consists of a resistor (R) and a capacitor (C) connected in series. The input voltage (
Vin
) is applied across the series combination, and the output voltage (
Vout
) is measured across the capacitor. This configuration allows only low frequencies to pass through while attenuating high frequencies.
02
Determine the Transfer Function
The transfer function (H(s)) of the RC low-pass filter is the ratio of the output voltage to the input voltage, given as:\[ H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{1}{sRC + 1} \]where \( s = j\omega \) and \( \omega \) is the angular frequency.
03
Calculate the Half-Power Frequency
The half-power frequency, or cutoff frequency, \(f_c\), occurs where the output power is half of the input power. This frequency is often characterized by a magnitude of \(\frac{1}{\sqrt{2}}\) or 0.707 times the maximum magnitude. It can be calculated as:\[ f_c = \frac{1}{2\pi RC} \]
04
Sketch the Magnitude Response
The magnitude |\(H(j\omega)\)| of the transfer function as a function of frequency \(\omega\) is expressed by:\[ |H(j\omega)| = \frac{1}{\sqrt{(\omega RC)^2 + 1}} \]This forms a curve that indicates attenuation at higher frequencies. At \(\omega = 0\), the magnitude is 1, and it decreases as \(\omega\) increases, initially dropping at the cutoff frequency \(f_c\).
05
Sketch the Phase Response
The phase \(\phi(\omega)\) of the transfer function as a function of frequency \(\omega\) is given by:\[ \phi(\omega) = -\tan^{-1}(\omega RC) \]At \(\omega = 0\), the phase is 0 degrees. As \(\omega\) increases, the phase shift becomes more negative, reaching -45 degrees at the cutoff frequency \(f_c\), and continues to approach -90 degrees as \(\omega\) goes to infinity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transfer Function
In an RC low-pass filter, understanding the transfer function is essential as it characterizes the filter's mathematical behavior. The transfer function, often denoted as \(H(s)\), is a complex representation that relates input and output voltages. Specifically, it's defined by the ratio \(H(s) = \frac{V_{out}(s)}{V_{in}(s)}\). For the RC low-pass filter, the expression simplifies to \(H(s) = \frac{1}{sRC + 1}\), where \(s = j\omega\) and \(\omega\) refers to angular frequency.
- \(V_{out}(s)\) is the Laplace transform of the output voltage.
- \(V_{in}(s)\) is the Laplace transform of the input voltage.
- \(RC\) represents the time constant of the filter.
Cutoff Frequency
The cutoff frequency, also known as the half-power frequency, is vital in filter design as it marks the threshold where the filter starts to attenuate incoming frequencies. For an RC low-pass filter, the cutoff frequency \(f_c\) is given by the formula \(f_c = \frac{1}{2\pi RC}\). At this frequency:
- The output power is half of the input power, represented by a magnitude of \(\frac{1}{\sqrt{2}}\) or 0.707 of the maximum input voltage.
- The phase of the output signal is shifted by -45 degrees.
Frequency Response
The frequency response of an RC low-pass filter offers insight into how different frequencies are processed. It's divided into two main characteristics: magnitude response and phase response.