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Perform the same computation as in Sec. \(24.3,\) but for the current as specified by $$i(t)=5 e^{-1.25 t} \sin 2 \pi t \quad \text { for } 0 \leq t \leq T / 2$$ $$i(t)=0\quad\quad\quad\quad\quad\quad \text { for } T / 2 < t \leq T$$ where \(T=1\) s. Use five-point Gauss quadrature to estimate the integral. .

Short Answer

Expert verified
To estimate the integral of the given current function using five-point Gauss quadrature, first change the limits of integration to \(-1 \leq x \leq 1\) using substitution \(t = \frac{T}{4}(x + 1)\) and \(\text{d}t = \frac{T}{4}\text{d}x\). Transform the function \(i(t)\) accordingly and integrate over the intervals \(0 \leq t \leq T/2\) (using Gauss quadrature) and \(T/2 < t \leq T\) (integral will be zero). Add the two integrals to find the final result: \(I = \int_{-1}^{1}i(t)\frac{T}{4}\text{d}x\).

Step by step solution

01

Define the Functions

(The given function is defined in two separate time intervals, so we will define it accordingly and the variables being used are also specified.) \(i(t) = \begin{cases} 5e^{-1.25t}\sin(2\pi t), & 0 \leq t \leq T/2 \\ 0, & T/2 < t \leq T \end{cases}\) where \(T = 1\text{s}\)
02

Change the limits of integration

(As we are using Gauss quadrature, we need to transform the limits of integration from the given limits to the reference interval \((−1, 1)\) using the following substitutions:) Given time interval \(0 \leq t \leq T/2 \Rightarrow -1 \leq x \leq 1\) Substitutions: \(t = \frac{T}{4}(x + 1)\) and \(\text{d}t = \frac{T}{4}\text{d}x\)
03

Change the functions

(Transform the function \(i(t)\) using the substitutions for \(t\) and \(\text{d}t\).) \(i(t) = 5e^{-1.25(\frac{T}{4}(x + 1))}\sin(2\pi (\frac{T}{4}(x + 1)))\) and \(\text{d}t = \frac{T}{4}\text{d}x\)
04

Integrate

(Integrate the function \(i(t)\) over the specified intervals using Gauss quadrature with five points.) The first interval \(0 \leq t \leq T/2\): Integrate the function over the reference interval \((-1, 1)\) using Gauss quadrature with \(5\) points, \(I_1 = \int_{-1}^{1}i(t)\frac{T}{4}\text{d}x\). The second interval \(T/2 < t \leq T\): Since the function is \(0\) in this interval, the integral would be \(0\), i.e., \(I_2 = 0\).
05

Sum the integrals

(Add both integrals to get the final result.) \(I = I_1 + I_2 = \int_{-1}^{1}i(t)\frac{T}{4}\text{d}x\) This is the final integral value estimated using the five-point Gauss quadrature method.

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Key Concepts

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

Numerical Integration
Numerical integration is a mathematical tool used to approximate the value of an integral, especially when an analytical solution is difficult or impossible to find. In simple terms, it's a method to calculate the area under a curve when we can't do it exactly by hand. One powerful technique for numerical integration is **Gauss Quadrature**.

Gauss Quadrature is efficient because it uses specific points and weights to provide an accurate integral approximation. These points, known as **Gauss points**, are predetermined roots of orthogonal polynomials. By focusing calculations only at these strategically chosen points, Gauss Quadrature often requires fewer evaluations to achieve high accuracy compared to other methods, like the Trapezoidal or Simpson's rule.
  • Ideal for polynomial integrands.
  • Requires transformation of the integration interval to \((-1, 1)\), which simplifies the computation.
  • In our example, a five-point Gauss quadrature is used, meaning it evaluates the integral at five different points within the interval.
Understanding numerical integration is crucial because it allows us to handle complex, real-world problems in engineering, physics, and beyond, where exact solutions are elusive.
Piecewise Functions
Piecewise functions are a type of function defined by different expressions based on the input value. Simply put, these functions 'switch' expressions depending on the interval of the input variable. This feature makes them perfectly suited for modeling situations where conditions change.

In the given exercise, we're dealing with a piecewise function to represent the current \(i(t)\). The function is defined differently over two intervals:
  • For \(0 \leq t \leq T/2\), the current is described by \(5e^{-1.25t}\sin(2\pi t)\).
  • For \(T/2 < t \leq T\), the current is \(0\).
This switch effectively models scenarios like signals that turn off after a certain period.

Working with piecewise functions requires recognizing and correctly applying the change in conditions or expressions at defined points. These functions are commonly found in fields like control systems and electrical engineering, as they can model phenomena such as switches and pulses.
Signal Processing
Signal processing involves analyzing, manipulating, and interpreting signals—functions that carry information. A signal can be anything that conveys information, such as sound waves, electrical currents, or radio waves.

In our scenario, the function \(i(t)\) represents an electrical current that varies over time and is piecewise-defined. Such functions are typical in signal processing, where the signal might start strong and then fade out or stop suddenly, like the current in our example which zeroes out after half a second.
  • Important for understanding and designing systems that use or generate signals, like telecommunications and audio processing.
  • Applications include filtering, encoding, and even compressing signals for efficient transmission and storage.
Decomposing and understanding piecewise integrated signals, as seen with \(i(t)\), helps us grasp the fundamental nature of the signals we're dealing with, a central task in signal processing, ensuring effective communication and functionality in modern technology. The use of integration, such as with Gauss Quadrature, aids in quantifying and understanding these signals accurately.

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Most popular questions from this chapter

Using the following data, calculate the work done by stretching a spring that has a spring constant of \(k=300 \mathrm{N} / \mathrm{m}\) to \(x=0.35 \mathrm{m}:\) $$\begin{array}{l|llllllll} F, 10^{3} \mathrm{N} & 0 & 0.01 & 0.028 & 0.046 & 0.063 & 0.082 & 0.11 & 0.13 \\\ \hline x, \mathrm{m} & 0 & 0.05 & 0.10 & 0.15 & 0.20 & 0.25 & 0.30 & 0.35 \end{array}$$

Suppose that the current through a resistor is described by the function $$i(t)=(60-t)^{2}+(60-t) \sin (\sqrt{t})$$ and the resistance is a function of the current, $$R=12 i+2 i^{2 / 3}$$ Compute the average voltage over \(t=0\) to 60 using the multiplesegment Simpson's 1/3 rule.

Fully developed flow of a Bingham plastic fluid moving through a 12 -in diameter pipe has the given velocity profile. The flow of a Bingham fluid does not shear the center core, producing plug flow in the region around the centerline. $$\begin{array}{l|ccccccc} \text { Radius, } r, \text { in } & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Velocily, } v, & \text { if } / \mathrm{s} & 5.00 & 5.00 & 4.62 & 4.01 & 3.42 & 1.69 & 0.00 \end{array}$$ Find the total volume flow rate \(Q\) using the relationship \(Q=\int_{r_{1}}^{r_{2}} 2 \pi r v d r+v_{c} A_{c},\) where \(r\) is the radial axis of the pipe. \(R\) is the radius of the pipe, \(v\) is the velocity, \(v_{c}\) is the velocity at the core, and \(A_{\mathrm{c}}\) is the cross-sectional area of the plug. Solve the problem using two different approaches. (a) Fit a polynomial curve to the noncore data and integrate. (b) Use multiple-application Simpson's 1/3 rule to integrate. (c) Find the percent error using the integral of the polynomial fit as the more correct value.

A jet fighter's position on an aircraft carrier's runway was timed during landing: $$\begin{array}{l|ccccccc} t, s & 0 & 0.52 & 1.04 & 1.75 & 2.37 & 3.25 & 3.83 \\ \hline x_{i} \mathrm{m} & 153 & 185 & 210 & 249 & 261 & 271 & 273 \end{array}$$ where \(x\) is the distance from the end of the carrier. Estimate (a) velocity \((d x / c t r)\) and \((b)\) acceleration \((d v / d t)\) using numerical differentiation.

As in Sec. \(24.2,\) compute \(F\) using the trapezoidal rule and Simpson's 1/3 and Simpson's 3/8 rules but use the following force. Divide the mast into 5 -ft intervals. $$F=\int_{0}^{30} \frac{250 z}{6+z} e^{-z / 10} d z$$

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