Chapter 6: Problem 53
A transfer function is given by $$ H(f)=\frac{100}{1+j(f / 1000)} $$ Sketch the asymptotic magnitude and phase Bode plots to scale. What is the value of the half-power frequency?
Short Answer
Expert verified
The half-power frequency is 1000 Hz. Sketch the magnitude at 40 dB, decreasing by 20 dB/decade, and the phase from 0° to -90°.
Step by step solution
01
Identify the Form of the Transfer Function
The given transfer function is \( H(f)=\frac{100}{1+j(f / 1000)} \). This is a first-order low-pass filter characterized by a DC gain of 100 and a pole at \( f = 1000 \) Hz. The standard form of a first-order low-pass filter transfer function is \( H(f) = \frac{K}{1 + j(f / f_c)} \), where \( f_c \) is the cutoff frequency. Here, \( f_c = 1000 \) Hz.
02
Sketch the Asymptotic Magnitude Bode Plot
The magnitude of the transfer function \( |H(f)| \) at low frequencies (\( f \ll 1000 \) Hz) approximately equals the DC gain, 100, or 40 dB in logarithmic scale (since \( 20 \log_{10} 100 = 40 \)). At high frequencies (\( f \gg 1000 \) Hz), the magnitude decreases at a rate of -20 dB/decade. The magnitude plot intersects the line at 40 dB at \( f = 1000 \) Hz and decreases beyond this frequency.
03
Sketch the Asymptotic Phase Bode Plot
At low frequencies (\( f \ll 1000 \) Hz), the phase angle of the transfer function, \( \phi(f) = \arg(H(f)) \), is approximately 0 degrees. As the frequency approaches the cutoff frequency \( f = 1000 \) Hz, the phase begins to lag. At \( f = f_c = 1000 \) Hz, the phase angle is -45 degrees. For \( f \gg 1000 \) Hz, the phase approaches -90 degrees.
04
Determine the Half-Power Frequency
The half-power frequency, also known as the cutoff frequency, is defined as the frequency at which the magnitude of the transfer function is \( |H(f)| = \frac{|H(0)|}{\sqrt{2}} \). In decibels, this corresponds to a 3 dB reduction from the DC gain. For this low-pass filter, the half-power frequency is the same as the cutoff frequency, \( f_c = 1000 \) Hz.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transfer Function
A transfer function is a mathematical description that connects the input and output of a system in the frequency domain. It is particularly useful in analyzing and designing systems, such as filters in electronics. The transfer function, given as \[H(f)=\frac{100}{1+j(f / 1000)}\] in this exercise, tells us how different frequencies are altered as they pass through the system.
In the form of \[H(f) = \frac{K}{1 + j(f / f_c)}\] it specifies:
In the form of \[H(f) = \frac{K}{1 + j(f / f_c)}\] it specifies:
- The DC gain, represented by \(K\), which is 100 in this case.
- The cutoff frequency, \(f_c\), where we observe a behavior change, which is 1000 Hz.
Low-Pass Filter
Low-pass filters are essential components in signal processing that allow signals with a frequency lower than a designated cutoff frequency to pass through, while attenuating signals with frequencies higher than the cutoff. The transfer function provided in this exercise,\[H(f)=\frac{100}{1+j(f / 1000)}\], represents a first-order low-pass filter.
This filter effectively "smooths out" signals by eliminating high-frequency noise, making it invaluable for applications where clean signals are required. The operation of a low-pass filter can be visualized in Bode plots, which show how the magnitude and phase of the output signal change with frequency.
In summary, low-pass filters are crucial for processing real-world signals, focusing on removing noise and preserving crucial signal components below a certain frequency threshold.
This filter effectively "smooths out" signals by eliminating high-frequency noise, making it invaluable for applications where clean signals are required. The operation of a low-pass filter can be visualized in Bode plots, which show how the magnitude and phase of the output signal change with frequency.
In summary, low-pass filters are crucial for processing real-world signals, focusing on removing noise and preserving crucial signal components below a certain frequency threshold.
Cutoff Frequency
The cutoff frequency of a filter is the point at which the filter starts to effectively "cut off" higher frequencies. For the transfer function \[H(f)=\frac{100}{1+j(f / 1000)}\], the cutoff frequency is given as \(f_c = 1000\) Hz. This frequency marks where the filter's output power is exactly half the power of the input signal. In Bode plots, the cutoff frequency is prominently shown where there is a noticeable change in how the system responds to frequency changes.
Here are some key roles of cutoff frequency:
Here are some key roles of cutoff frequency:
- Sets the threshold for high-frequency attenuation.
- Defines -3 dB point, indicating the frequency where the output is muted by half in power.
- Plays a vital role in designing filters for optimized performance.
Asymptotic Magnitude
The asymptotic magnitude of a Bode plot helps in approximating the behavior of a filter at various frequencies. For the given transfer function, \(H(f)=\frac{100}{1+j(f / 1000)}\), we calculate it as follows:
The asymptotic approach allows engineers to quickly assess how a filter influences different frequency components, though it doesn’t capture all detailed nuances of the actual frequency response curve.
- At low frequencies (when \(f \ll 1000\) Hz): The magnitude is about 100, or 40 dB.
- As frequency increases beyond \(f_c = 1000\) Hz, the magnitude decreases at a rate of -20 dB/decade.
The asymptotic approach allows engineers to quickly assess how a filter influences different frequency components, though it doesn’t capture all detailed nuances of the actual frequency response curve.
Phase Angle
The phase angle in a Bode plot measures the time displacement between the input and output signals of a system, represented in degrees. In the discussed example, the phase angle for the transfer function \(H(f)=\frac{100}{1+j(f / 1000)}\), evolves as follows:
Overall, comprehending phase angle changes aids in designing systems that respond correctly under varying operating conditions, ensuring precise performance and reliability.
- At low frequencies, the phase angle is nearly 0 degrees, indicating minimal lag.
- As frequency approaches 1000 Hz, the phase shifts, reaching -45 degrees at the cutoff frequency.
- Beyond 1000 Hz, the phase shift intensifies to -90 degrees.
Overall, comprehending phase angle changes aids in designing systems that respond correctly under varying operating conditions, ensuring precise performance and reliability.