/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 A band-limited voltage signal is... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 64

A band-limited voltage signal is found to have the complex Fourier coefficients presented in the table below. Calculate the average power that the signal would supply a \(4-\Omega\) resistor $$\begin{array}{ccc} \hline n \omega_{0} & \left|c_{n}\right| & \theta_{n} \\ \hline & & \\ 0 & 10.0 & 0^{\circ} \\ \omega & 8.5 & 15^{\circ} \\ 2 \omega & 4.2 & 30^{\circ} \\ 3 \omega & 2.1 & 45^{\circ} \\ 4 \omega & 0.5 & 60^{\circ} \\ 5 \omega & 0.2 & 75^{\circ} \\ \hline \end{array}$$

Short Answer

Expert verified
The average power is approximately 24.32 W.

Step by step solution

01

Understand the Problem

We need to calculate the average power supplied by a band-limited voltage signal through a \(4-\Omega\) resistor. The Fourier coefficients \(|c_n|\) for different harmonics \(n\omega_0\) are given.
02

Calculate Power for Each Harmonic

For each harmonic, the power from a single component in a resistive load \(R\) is given by \(P_n = \frac{|c_n|^2}{2R}\). We will calculate this for each harmonic using \(R = 4\) ohms.
03

Power Calculation for \(n=0\)

For \(n=0\), \[ P_0 = \frac{|c_0|^2}{2 \times 4} = \frac{10.0^2}{8} = 12.5 \text{ W} \]
04

Power Calculation for \(n=1\)

For \(n=1\), \[ P_1 = \frac{|c_1|^2}{8} = \frac{8.5^2}{8} = 9.03125 \text{ W} \]
05

Power Calculation for \(n=2\)

For \(n=2\), \[ P_2 = \frac{|c_2|^2}{8} = \frac{4.2^2}{8} = 2.205 \text{ W} \]
06

Power Calculation for \(n=3\)

For \(n=3\), \[ P_3 = \frac{|c_3|^2}{8} = \frac{2.1^2}{8} = 0.55125 \text{ W} \]
07

Power Calculation for \(n=4\)

For \(n=4\), \[ P_4 = \frac{|c_4|^2}{8} = \frac{0.5^2}{8} = 0.03125 \text{ W} \]
08

Power Calculation for \(n=5\)

For \(n=5\), \[ P_5 = \frac{|c_5|^2}{8} = \frac{0.2^2}{8} = 0.005 \text{ W} \]
09

Sum All Powers to Find Average Power

The total average power is the sum of the powers from all harmonics: \[ P_{avg} = P_0 + P_1 + P_2 + P_3 + P_4 + P_5 = 12.5 + 9.03125 + 2.205 + 0.55125 + 0.03125 + 0.005 = 24.32375 \text{ W} \]
10

Convert to Final Result

Since the power is often presented with sensible rounding: The average power provided by the signal to the resistor is approximately \(24.32\text{ W}\).

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Key Concepts

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

Understanding Average Power
Average power in this context is the mean amount of energy a voltage signal supplies to a load over time. It is crucial for evaluating how much useful work a system can perform. In the exercise, we want to determine how this band-limited signal influences a resistor, a component that dissipates power in the form of heat. Calculating average power involves taking into account all frequency components, or harmonics. The equation for average power from each harmonic is given by:
  • For each harmonic: \(P_n = \frac{|c_n|^2}{2R}\)
where \(|c_n|\) is the magnitude of the Fourier coefficient for each harmonic, and \(R\) is the resistance. By summing up these individual powers from each harmonic, you get the total average power supplied to the resistive load.
Basics of Harmonic Analysis
Harmonic analysis is the process of determining the different frequency components present in a signal. These components are called harmonics and represent integer multiples of a base frequency. In our example, the term \(n\omega_0\) represents the frequency of each harmonic component where \(n\) is an integer and \(\omega_0\) is the base, or fundamental, frequency.
  • The 0th harmonic is essentially the DC component, or the average value of the signal.
  • Higher harmonics are critical for shaping the waveform, contributing to its unique characteristics.
Fourier series is a powerful tool for breaking down a complex, periodic signal into a sum of simpler sinusoidal components. Each harmonic has its amplitude and phase, which helps in reconstructing the original signal from these components.
Resistive Load Power Considerations
When a signal is applied to a resistive load, the resistance determines how much power is dissipated as heat. In our scenario, the resistor value is \(4 \Omega\). Power dissipation in a resistor is calculated using the basic formula derived from Ohm's Law:
  • Power through resistance, \(P = \frac{V^2}{R}\)
where \(V\) is voltage across the resistor. For each harmonic, the power calculation is simplified by directly using the Fourier coefficient specific to that harmonic. This results in power contributions from each harmonic component, organized by their respective Fourier amplitude values \(|c_n|\). The calculated power for each harmonic accounts precisely to how energy is used within the load.
Understanding Complex Fourier Coefficients
Complex Fourier coefficients \(c_n\) are crucial in harmonic analysis as they encapsulate both the magnitude and phase information of each harmonic component. They express how a signal alternates and vibrates at each frequency. Each coefficient has two parts:
  • Amplitude \(|c_n|\) which determines the intensity of the harmonic.
  • Phase \(\theta_n\) which indicates the shift relative to the origin.
These components are represented in the form \(|c_n|e^{j\theta_n}\), where \(j\) is the imaginary unit. The magnitude \(|c_n|\) is essential in power calculations because it directly influences the power each harmonic contributes. In practical terms, you can work with these coefficients easily, using them to approximate any periodic function by summing harmonics; this provides a clear picture of how a signal contributes to system behavior.

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Most popular questions from this chapter

The voltage across the terminals of a circuit is \\[\begin{aligned} v(t)=& 30+20 \cos \left(60 \pi t+45^{\circ}\right) \\ &+10 \cos \left(60 \pi t-45^{\circ}\right) \mathrm{V} \end{aligned}\\] If the current entering the terminal at higher potential is \\[\begin{aligned} i(t)=& 6+4 \cos \left(60 \pi t+10^{\circ}\right) \\ &-2 \cos \left(120 \pi t-60^{\circ}\right) \mathrm{A} \end{aligned}\\] find: (a) the rms value of the voltage, (b) the rms value of the current, (c) the average power absorbed by the circuit.

Obtain the exponential Fourier series for \(f(t)=t\) \(-1

Determine the period of these periodic functions: (a) \(f_{1}(t)=4 \sin 5 t+3 \sin 6 t\) (b) \(f_{2}(t)=12+5 \cos 2 t+2 \cos \left(4 t+45^{\circ}\right)\) (c) \(f_{3}(t)=4 \sin ^{2} 600 \pi t\) (d) \(f_{4}(t)=e^{j 10 t}\)

Find the quadrature (cosine and sine) form of the Fourier series \\[f(t)=2+\sum_{n=1}^{\infty} \frac{10}{n^{3}+1} \cos \left(2 n t+\frac{n \pi}{4}\right)\\]

Evaluate each of the following functions and see if it is periodic. If periodic, find its period. (a) \(f(t)=\cos \pi t+2 \cos 3 \pi t+3 \cos 5 \pi t\) (b) \(y(t)=\sin t+4 \cos 2 \pi t\) (c) \(g(t)=\sin 3 t \cos 4 t\) (d) \(h(t)=\cos ^{2} t\) (e) \(z(t)=4.2 \sin \left(0.4 \pi t+10^{\circ}\right)\) \(+0.8 \sin \left(0.6 \pi t+50^{\circ}\right)\) (f) \(p(t)=10\) \((\mathrm{g}) q(t)=e^{-\pi t}\)
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