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Chapter 6: Problem 15

Consider a function \(f(t)\) that is zero for negative \(t\) and equals \(\exp (-t / 2 \tau)\) for \(t \geqq 0 .\) Find its Fourier coefficients \(A(\omega)\) and \(B(\omega)\) in the continuous superposition $$ f(t)=\int_{0}^{\infty}[A(\omega) \sin \omega t+B(\omega) \cos \omega t] d \omega $$

Short Answer

Expert verified
\(A(\omega) = \frac{\omega}{(1/(2\tau))^2 + \omega^2}\), \(B(\omega) = \frac{1/(2\tau)}{(1/(2\tau))^2 + \omega^2}\).

Step by step solution

01

Understanding the Function

The function given is \( f(t) = \exp(-t / 2\tau) \) for \( t \geq 0 \) and zero otherwise. This is essentially an exponentially decaying function defined only for non-negative \( t \). Our goal is to express this function as a continuous superposition of sine and cosine functions as described.
02

Establish the Fourier Series Representation

We need to find the coefficients \( A(\omega) \) and \( B(\omega) \) such that the integral representation of \( f(t) \) holds for all \( t \geq 0 \):\[ f(t) = \int_{0}^{\infty} [A(\omega) \sin(\omega t) + B(\omega) \cos(\omega t)] d\omega.\]
03

Identify the Role of Sine and Cosine Functions

The sine and cosine functions form an orthogonal basis over the interval from 0 to infinity. This means the function \( f(t) \) can be expressed in terms of sine and cosine to find the coefficients \( A(\omega) \) and \( B(\omega) \).
04

Determine Coefficient \( B(\omega) \)

Since \( f(t) \) is non-zero only for \( t \geq 0 \) and is of exponential form, it predominantly resembles a cosine transformation. We compute \( B(\omega) \) using the formula:\[ B(\omega) = \int_{0}^{\infty} f(t) \cos(\omega t) \cdot dt = \int_{0}^{\infty} \exp(-t / 2\tau) \cos(\omega t) \cdot dt.\]Evaluate this integral to get:\[ B(\omega) = \frac{1}{(1/(2\tau))^2 + \omega^2} \cdot \frac{1}{2\tau}.\]
05

Derive Coefficient \( A(\omega) \)

For \( A(\omega) \), we compute it similarly by considering the transformation involving sine:\[ A(\omega) = \int_{0}^{\infty} f(t) \sin(\omega t) \cdot dt = \int_{0}^{\infty} \exp(-t / 2\tau) \sin(\omega t) \cdot dt.\]This integral turns out to be:\[ A(\omega) = \frac{\omega}{(1/(2\tau))^2 + \omega^2}.\]
06

Compile the Final Expressions for the Coefficients

Based on integrations performed in Steps 4 and 5, the coefficients are: \[ A(\omega) = \frac{\omega}{(1/(2\tau))^2 + \omega^2} \] \[ B(\omega) = \frac{1/(2\tau)}{(1/(2\tau))^2 + \omega^2} \]These coefficients satisfy the integral representation.

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Key Concepts

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

Exponential Function
The exponential function is a key player in many mathematical scenarios, especially when it comes to describing decay processes. In our exercise, the function given is an exponential one:
\( f(t) = \exp(-t / 2\tau) \) for \( t \geq 0 \).
This function starts from a peak value at \( t = 0 \) and decays towards zero as \( t \) increases.
  • Exponential functions are expressed as powers of the constant \( e \), which is approximately equal to 2.718.
  • In this context, the exponent \(-t/2\tau\) indicates an exponential decay rate determined by the parameter \( \tau \).
  • These functions are often used to model natural processes such as radioactive decay or cooling.
Given that the function is zero for negative time values, it emphasizes processes starting from a defined point, useful in practical boundary conditions.
Orthogonal Functions
Orthogonal functions play a significant role in simplifying complex problems and breaking them into manageable parts. In our Fourier Transform context:
  • Sine and cosine functions constitute orthogonal bases over specific intervals, meaning they are independent and can represent any signal in those intervals without interfering with each other.
  • When functions are orthogonal, the integral of their product over a certain range is zero unless the functions are identical.
These properties are instrumental in Fourier analysis, allowing us to decompose a complicated function into simpler sine and cosine components.
In the exercise, this orthogonality facilitates finding the Fourier coefficients \( A(\omega) \) and \( B(\omega) \) that describe the function \( f(t) \) as a sum of sines and cosines.
Sine and Cosine Transform
The sine and cosine transforms are variations of the Fourier Transform designed to decompose functions into their frequency components.
By focusing on either just sine or cosine terms, these transforms can effectively capture specific type behaviors in function representation.
  • For an exponentially decaying function like ours, the cosine transform usually plays a more prominent role. This is due to its nature of handling oscillations while decaying from the starting point.
  • In the exercise, we derive the cosine coefficient \( B(\omega) \) using an integral representation that suits the decay model. The sine coefficient \( A(\omega) \) complements it, capturing the oscillatory nature.
Thus, we arrive at the continuous superposition involving both sine and cosine terms, which together reconstruct the original function accurately.
Integral Representation
Integral representation serves as a foundational tool in transforming functions from time-domain to frequency-domain.
In this process, such as with the Fourier Transform:
  • We aim to express the given function in terms of integrals involving sine and cosine components over continuous frequencies.
  • The task involves computing integrals of the form \( \int_{0}^{\infty} f(t) g(t) \cdot dt \), where \( g(t) \) can be sine or cosine, ensuring all variations in \( f(t) \) are accounted for.
With the exercise, this methodical integration yields coefficients \( A(\omega) \) and \( B(\omega) \) which make the transformation into the frequency domain possible, offering a powerful method to analyze the behavior of the function in terms of its frequency content.

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Most popular questions from this chapter

Suppose \(\psi(t)\) is zero outside of the interval from \(t_{1}\) to \(t_{2}\), which has duration \(t_{2}-t_{1}=\Delta t\) and central value \(\frac{1}{2}\left(t_{1}+t_{2}\right)=t_{0}\). Suppose \(\psi(t)\) is equal to cos \(\omega_{0}\left(t-t_{0}\right)\) within that interval. (a) Show that \(\psi(t)\) can be Fourier-analyzed as follows: $$ \begin{aligned} \psi(t) &=\int_{0}^{\infty} B(\omega) \cos \omega\left(t-t_{0}\right) \\ \pi B(\omega) &=\frac{\sin \left[\left(\omega_{0}+\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}+\omega}+\frac{\sin \left[\left(\omega_{0}-\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}-\omega} \end{aligned} $$ (b) Show that if \(\Delta t\) is much shorter than the period of any frequency that we can measure or are interested in, then \(\pi B(\omega)\) has the constant value \(\Delta t\). (c) Show that if \(\Delta t\) contains many oscillations, i.e., if \(\omega_{0} \Delta t \geqslant 1\), then, for \(\omega\) sufficiently near \(\omega_{0}, B(\omega)\) is given essentially by the second term only: $$ \pi B(\omega) \approx \frac{\sin \left[\left(\omega_{D}-\omega\right) \frac{1}{2} \Delta t\right]}{\omega_{0}-\omega}, \quad\left|\omega_{0}-\omega\right| \ll\left|\omega_{0}+\omega\right| $$ (d) Sketch \(\psi(t)\) and \(B(\omega)\) for part \((c)\). This problem can help us to understand collision broadening of spectral lines. An undisturbed atom emitting almost monochromatic visible light has a mean decay time of about \(10^{-8} \mathrm{sec}\), and thus the Fourier spectrum of its radiation has a bandwidth \(\Delta \nu\) of about \(10^{8}\) cps. If the atoms are in a gas- discharge-tube light source, then it turns out that the bandwidth of the emitted light (called "linewidth" in optics) is about \(10^{9} \mathrm{cps}\), rather than \(10^{8} \mathrm{cps}\). Part of the reason for this "line broadening" is the fact that the atoms do not radiate in a free and undisturbed manner; they collide. A collision results in a sudden change in amplitude or phase constant or both. That is similar to the situation illustrated by the truncated harmonic oscillator, A given atom may spend most of its time "unexcited." Occasionally it is excited into an oscillatory motion of the optical (valence) electrons (we are speaking classically; a more accurate picture requires quantum mechanics). The atom begins to oscillate as a damped harmonic oscillation with decay time of order \(10^{-8}\) sec. However, within a time \(\Delta t\) of about \(10^{-9} \mathrm{sec}\) (in a typical gas- tube light source), it has a collision that truncates the oscillation in some random way. If one adds the light from many such sources, the bandwidth \(\Delta v\) will be given by \(\Delta p \approx(1 / \Delta t) \approx 10^{9} \mathrm{cps}\).

This experiment requires a piano. Trill two adjacent notes (a halftone apart). First pick two notes near the top of the keyboard. Trill slowly, then as fast as you can. Estimate the trill frequency. Can you still easily make out the two notes of the trill? Now trill two adjacent notes near the bottom of the keyboard, first very slowly, then gradually more rapidly. Is there a speed at which the two notes blend into a messy, indistinguishable mixture? Estimate the frequency where things get messy. Then do arithmetic and decide how good your ear and brain are at recognizing two separate maxima in the Fourier analysis, even when the frequency widths of the peaks (at balf-maximum intensity) are not small compared to the frequency spacing between the maxima.

In Prob. \(2.31\) you derived the dispersion law for sawtooth shallow-water standing waves, obtaining the result \(v_{\mathrm{p}} \approx\) \(1.1 \sqrt{g h} .\) For sinusoidal shallow-water waves the result turns out to be \(v_{\varphi}=\sqrt{g h}\). Thus shallow-water waves are nondispersive. (The phase velocity does not depend on the wavelength.) Instead of standing waves we now consider shallow-water traveling wave packets. Since the waves are nondispersive, a single "solitary wave" or "tidal wave" will propagate without changing its shape (approximately). Such waves, called tsunami, can be excited by undersea earthquakes in the ocean. The average water depth in the deep ocean is about 5 kilometers: \(h=5 \times 10^{5} \mathrm{~cm}\). Tidal waves of horizontal length much longer than \(5 \mathrm{~km}\) are therefore "shallow- water" waves. Tsunami waves propagate in the deep ocean at a velocity $$ v=\sqrt{g h}=\sqrt{(980) 5 \times 10^{5}}=2.2 \times 10^{4}=220 \text { meter } / \text { see } $$ \(\simeq 495\) miles/hour, which is somewhat slower than a typical jet airplane. How long does it take such a tidal wave to propagate from Alaska to Hawaii? In 1883 the volcano Krakatoa blew up, creating the world's biggest explosion. (Krakatoa is located in Sunda Strait, between Sumatra and Java. An account of the explosion can be found in any encyclopedia.) Huge tidal waves and atmospheric waves were created. Recently it has been discovered that there are air traveling waves with velocity about \(220 \mathrm{~m} / \mathrm{sec}\). (Recall that ordinary sound velocity at \(0^{\circ} \mathrm{C}\) is \(332 \mathrm{~m} / \mathrm{sec}\). On the average the air is colder than that, so the velocity is less than that.) The existence of these air waves probably explains how the tidal water waves from Krakatoa appeared on the far sides of land masses that should have blocked the water waves. Apparently the tidal waves "jumped over" the land masses by coupling to the air waves having the same velocity (and same excitation time). See the article by F. Press and D. Harkrider, "Air-Sea Waves from the Explosion of Krakatoa" Science \(154,1325(\) Dec. 9,1966\()\). In the experiment make your own shallow-water tidal waves as follows: Take a square pan a foot or two long. Fill it with water to a depth of about \(\frac{1}{2}\) or \(1 \mathrm{~cm}\). Give the pan a quick nudge (or lift one end and drop it suddenly). You will create two traveling wave packets, one at the near end and one at the far end, traveling in opposite directions. Follow the bigger of the packets. Measure the velocity by timing the wave for as many pan lengths as you can (probably about four). A stopwatch helps. Alternatively, you can count out loud as the packet hits the walls, memorize the "musical tempo," and finally measure the tempo with an ordinary watch. How well do your results agree with \(v=\sqrt{g h}\). As the depth of the water increases, you will finally get to the point where the waves are not shallow-water waves. Then the dispersion relation gradually goes over to the deep-water gravitational-wave dispersion relation \(\omega^{2}=g k\), i.e.r $$ v_{w}=\lambda \nu=\sqrt{\frac{g \lambda}{2 \pi}} $$ (We shall derive this relation in Chap. \(7 .)\) Thus the wave packet will spread out and not maintain its shape. For sufficiently shallow water (less than \(1 \mathrm{~cm}\), roughly), the shape is maintained fairly well for several feet. Finally, make a traveling tidal wave in your bathtub by suddenly pushing the entire end of the tub water with a board. Measure the down and back time and thus measure the velocity. Is it \(\sqrt{g h}\). Notice the breakers!

Suppose \(f(t)\) is zero except in the interval from \(t=t_{1}\) to \(t=t_{2}\) of duration \(\Delta t=t_{2}-t_{1}\) and centered at \(t_{0}=\frac{1}{2}\left(t_{1}+t_{2}\right)\). Suppose that in this interval \(f(t)\) makes exactly one sinusoidal oscillation at angular frequency \(\omega_{0}\), starting and ending with value zero at \(t_{1}\) and \(t_{2}\left(\right.\) i.e., \(\left.\Delta t=T_{0}=2 \pi / \omega_{0}\right)\). Find the Fourier coefficients \(A(\omega)\) and \(B(\omega)\) in the continuous superposition $$ f(t)=\int_{0}^{\infty}\left[A(\omega) \sin \omega\left(t-t_{0}\right)+B(\omega) \cos \omega\left(t-t_{0}\right)\right] d \omega $$ Make a rough plot of the Fourier coefficients versus \(\omega\) and a sketch of \(f(t)\).

Assume the dispersion law is given by a single resonance, and neglect damping; i.e., assume $$ c^{2} k^{2}=\omega^{2}\left(1+\frac{\omega_{p}{ }^{2}}{\omega_{0}^{2}-\omega^{2}}\right), \quad \omega_{p}^{2}=\frac{4 \pi N e^{2}}{m} $$ where \(N\) is the number of resonating electrons per unit volume. (a) Sketch the square of the index of refraction, \(n^{2}\), versus \(\omega\), for \(0 \leqq \omega<\infty\). The important features are the value and slope at \(\omega=0\), at \(\omega\) slightly less than \(\omega_{0}\) and slightly greater than \(\omega_{0}\), at \(\omega=\sqrt{\omega_{0}^{2}+\omega_{p}^{2}}\), and at infinity. How do you interpret the region where \(n^{2}\) is negative? The region near \(\omega_{0}\) ? (b) Derive the following formula for the square of the group velocity: $$ \left(\frac{v_{g}}{c}\right)^{2}=\frac{1+\frac{\omega_{p}^{2}}{\omega_{0}^{2}-\omega^{2}}}{\left[1+\frac{\omega_{p}^{2} \omega_{0}^{2}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}}\right]^{2}} $$ Sketch \(\left(v_{g} / c\right)^{2}\) versus \(\omega .\) Show that \(\left(v_{d} / c\right)^{2}\) is always less than unity, as required by the theory of relativity. Show that \(v_{0}{ }^{2}\) is negative in the same frequency region where \(n^{2}\) is negative. For what frequency is the group velocity greatest? What is the group velocity at that frequency?
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