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A bandpass filter has a center, or resonant, frequeng of \(80 \mathrm{krad} / \mathrm{s}\) and a quality factor of 8 . Find the bard width, the upper cutoff frequency, and the lowercult off frequency. Express all answers in kilohertz.

Short Answer

Expert verified
Bandwidth is approximately 1.592 kHz, upper cutoff is 13.525 kHz, lower cutoff is 11.937 kHz.

Step by step solution

01

Understand the Definitions

The center frequency \( \omega_0 \) of a bandpass filter is given as \( 80\ \text{krad/s} \), and the quality factor \( Q \) is 8. We need to convert the frequency from radians per second to hertz by dividing by \( 2\pi \).
02

Calculate Bandwidth

The bandwidth \( BW \) of a bandpass filter is given by the formula \( BW = \frac{\omega_0}{Q} \). Given \( \omega_0 = 80\ \text{krad/s} = 80000\ \text{rad/s} \), substitute the values: \( BW = \frac{80000}{8} = 10000\ \text{rad/s} \).
03

Convert Bandwidth to Hertz

Convert the bandwidth to hertz by dividing by \( 2\pi \): \( BW_{Hz} = \frac{10000}{2\pi} \approx 1591.55\ \text{Hz} \). To convert Hz to kHz, divide by 1000: \( BW_{kHz} \approx 1.592\ \text{kHz} \).
04

Calculate Upper and Lower Cutoff Frequencies

The upper and lower cutoff frequencies are symmetrical around \( \omega_0 \). Use \( \omega_u = \omega_0 + \frac{BW}{2} \) and \( \omega_l = \omega_0 - \frac{BW}{2} \). Substituting the values: \( \omega_u = 80000 + \frac{10000}{2} = 85000\ \text{rad/s} \) and \( \omega_l = 80000 - \frac{10000}{2} = 75000\ \text{rad/s} \).
05

Convert Upper and Lower Cutoff Frequencies to Hertz

Convert the upper and lower cutoff frequencies to hertz by dividing by \( 2\pi \): \( f_u = \frac{85000}{2\pi} \approx 13524.89\ \text{Hz} \) and \( f_l = \frac{75000}{2\pi} \approx 11936.62\ \text{Hz} \). Convert these to kHz by dividing by 1000: \( f_u \approx 13.525\ \text{kHz} \) and \( f_l \approx 11.937\ \text{kHz} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quality Factor
The quality factor, often denoted by the letter \( Q \), is a measure of how "selective" a bandpass filter is with respect to its center frequency. It indicates the sharpness of the peak of the frequency response curve. A higher \( Q \) means a narrower bandwidth, allowing the filter to be more selective.
This is particularly useful in applications such as radio tuning or signal processing where you need to isolate a narrow range of frequencies.
  • In mathematical terms, \( Q \) is defined as the ratio of the resonant frequency \( \omega_0 \) to the bandwidth \( BW \).
  • The formula is \( Q = \frac{\omega_0}{BW} \).
  • For a given center frequency, a higher \( Q \) results in a smaller \( BW \), meaning the filter allows only a smaller range of frequencies around \( \omega_0 \) to pass through.

Understanding the quality factor helps in designing filters that can effectively differentiate between closely spaced frequency signals.
Cutoff Frequency
The cutoff frequencies of a bandpass filter are crucial points that define where the filter transitions from passing a signal to attenuating it. These frequencies are typically defined as the points where the output signal is reduced to half its power, corresponding to a drop of approximately 3 dB.
For a bandpass filter, there are two cutoff frequencies: the lower cutoff frequency \( low\omega_l \) and the upper cutoff frequency \( \omega_u \). Both are symmetrically placed around the center frequency \( \omega_0 \).
  • The lower cutoff frequency \( \omega_l \) is calculated using \( \omega_l = \omega_0 - \frac{BW}{2} \).
  • The upper cutoff frequency \( \omega_u \) is calculated using \( \omega_u = \omega_0 + \frac{BW}{2} \).
  • The difference between the upper and lower cutoff frequencies gives the filter's bandwidth.

In practice, knowing the cutoff frequencies helps in understanding the filter's range of operation, ensuring that only the desired frequencies are emphasized while unwanted frequencies are suppressed.
Resonant Frequency
The resonant frequency, often denoted as \( \omega_0 \), is the center frequency of a bandpass filter. It's the frequency at which the filter allows signals to pass through with maximum amplitude, typically leading to the least attenuation.
This frequency is crucial because it determines the midpoint of the frequencies that the filter will pass.
  • Mathematically, the resonant frequency is neither the upper nor the lower cutoff; rather, it is the frequency at the center of the filter's passband.
  • A change in the resonant frequency will shift the passband of the filter, thereby altering which signals can pass through unimpaired.
  • In electronic design, this frequency is often set based on the application needs, such as in musical instruments, radio stations, or communication devices.

Understanding the resonant frequency is key to designing filters that achieve the desired frequency response, ensuring optimal performance of electronic systems.

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Most popular questions from this chapter

Use a 25 nF capacitor to design a low-pass passice filter with a cutoff frequency of \(160 \mathrm{krad} / \mathrm{s}\) a) Specify the cutoff frequency in herla b) Specify the value of the filter resistor. c) Assume the cutoff frequency cannot increaset more than \(8 \%\). What is the smallest value load resistance that can be connected across the output terminals of the filter? d) If the resistor found in (c) is connected acrow the output terminals, what is the magnitude \(H(j \omega)\) when \(\omega=0 ?\)

Using a \(25 \mathrm{mH}\) inductor, design a high-pass, \(R L\) passive filter with a cutoff frequency of \(160 \mathrm{krad} / \mathrm{s}\) a) Specify the value of the resistance. b) Assume the filter is connected to a pure resistive load. The cutoff frequency is not to drop below \(150 \mathrm{krad} / \mathrm{s} .\) What is the smallest load resistor that can be connected across the output termi- nals of the filter?

Use a 20 nF capacitor to design a series \(R L C\) bard ter frequency of the filter is \(20 \mathrm{kHz}\), and the quality factor is 5 a) Specify the values of \(R\) and \(L\) b) What is the lower cutoff frequency in kilohert? c) What is the upper cutoff frequency in kilone d) What is the bandwidth of the filter in \(k\) .

Use a \(25 \mathrm{mH}\) inductor to design a low-pass, \(R L\), passive filter with a cutoff frequency of \(2.5 \mathrm{kHz}\) a) Specify the value of the resistor. b) \(\mathrm{A}\) load having a resistance of \(750 \mathrm{S}\) is connected across the output terminals of the filter. What is the corner, or cutoff, frequency of the loaded filter in hertz?

Given the following voltage transfer function: $$\begin{aligned} H(s) &=\frac{V_{o}}{V_{i}} \\ &=\frac{4 \times 10^{6}}{s^{2}+500 s+4 \times 10^{6}} \end{aligned}$$ a) At what frequencies (in radians per second) is the ratio of \(V_{o} / V_{i}\) equal to unity? b) At what frequency is the ratio maximum? c) What is the maximum value of the ratio?

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