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Chapter 12: Problem 3

Show that when \(\psi(t)\) is expressed by (12.1.3), $$ \overline{|\psi(t)|^{2}}=\sum_{-\infty}^{\infty} \breve{A}_{n} \check{A}_{n}^{*}=\sum_{-\infty}^{\infty}\left|\check{A}_{n}\right|^{2} $$ where the bar denotes a time average taken over the fundamental period \(T_{1}\). This relation is very similar to (1.8.17), except for a factor of \(\frac{1}{2}\). Account for the difference.

Short Answer

Expert verified
Answer: The factor of \(\frac{1}{2}\) in expression (1.8.17) is caused by the definition of the time average in that expression, which is for amplitude square, not for the time function \(\psi(t)\) itself. In our case, the time average was directly defined for the time function square magnitude. This accounts for the difference in the factor of \(\frac{1}{2}\) between our result and expression (1.8.17).

Step by step solution

01

Given Expression for \(\psi(t)\)

The given expression for \(\psi(t)\) is represented by Equation (12.1.3): $$ \psi(t) = \sum_{-\infty}^{\infty} \check{A}_n e^{jn\omega_1 t} $$ where \(\check{A}_n\) is the Fourier series coefficient and \(\omega_1\) is the fundamental frequency.
02

Expression for \(|\psi(t)|^2\)

To find the square magnitude of \(\psi(t)\), we need to multiply \(\psi(t)\) by its complex conjugate: $$ |\psi(t)|^2 = \psi(t) \psi^*(t) = \left( \sum_{-\infty}^{\infty} \check{A}_n e^{jn\omega_1 t} \right) \left( \sum_{-\infty}^{\infty} \check{A}_n^* e^{-jn\omega_1 t} \right) $$
03

Time average of \(|\psi(t)|^2\)

Now we need to take the time average over one fundamental period \(T_1\). The time average is defined as: $$ \overline{|\psi(t)|^{2}} = \frac{1}{T_1} \int_{0}^{T_1} |\psi(t)|^2 dt $$ Substitute the expression for \(|\psi(t)|^2\) from Step 2: $$ \overline{|\psi(t)|^{2}} = \frac{1}{T_1} \int_{0}^{T_1} \left( \sum_{-\infty}^{\infty} \check{A}_n e^{jn\omega_1 t} \right) \left( \sum_{-\infty}^{\infty} \check{A}_n^* e^{-jn\omega_1 t} \right) dt $$
04

Distribute the summations and integrate

Distribute the summations and integrate each term separately: $$ \overline{|\psi(t)|^{2}} = \frac{1}{T_1} \sum_{-\infty}^{\infty} \sum_{-\infty}^{\infty} \int_{0}^{T_1} \check{A}_n \check{A}_m^* e^{j(n-m)\omega_1 t} dt $$ We know that for any integers \(n\) and \(m\): $$ \int_{0}^{T_1} e^{j(n-m)\omega_1 t} dt = \begin{cases} T_1, & \mbox{if } n = m \\ 0, & \mbox{if } n \neq m \end{cases} $$ Applying this to our expression, we get: $$ \overline{|\psi(t)|^{2}} = \sum_{-\infty}^{\infty} \check{A}_n \check{A}_n^* = \sum_{-\infty}^{\infty} |\check{A}_n|^2 $$
05

Comparing with expression (1.8.17)

Now, we need to compare the obtained result with expression (1.8.17). We see that the difference is a factor of \(\frac{1}{2}\). The factor of \(\frac{1}{2}\) in expression (1.8.17) is caused by the definition of the time average in that expression, as it is for amplitude square, not for the time function \(\psi(t)\) itself. In our case, the time average was directly defined for the time function square magnitude. This accounts for the difference in the factor of \(\frac{1}{2}\) between our result and expression (1.8.17).

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Key Concepts

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

Complex Conjugate
In mathematics and physics, a complex conjugate refers to the pair of complex numbers where one is the inverse sign of the imaginary component of the other. If we have a complex number represented as \( z = a + bi \), where \( a \) and \( b \) are real numbers, then its complex conjugate is \( z^* = a - bi \). Complex conjugates are fundamental in simplifying expressions that involve complex numbers, especially in Fourier series.

When you multiply a complex number by its conjugate, it yields a real number. This property is used to find the magnitude squared of complex signals or functions. For the function \( \psi(t) \) in the original exercise, multiplying \( \psi(t) \) by its complex conjugate \( \psi^*(t) \) results in the expression \( |\psi(t)|^2 \).

This operation is often used to eliminate the imaginary part and focus on the amplitude of signals, which is crucial when analyzing periodic functions in the context of Fourier series and signal processing.
Time Average
The concept of a time average is widely used when dealing with cyclic or periodic events, such as signals in electrical engineering and other disciplines. A time average refers to the average value of a function over a certain period. This is particularly useful in understanding and analyzing the behaviors of systems that oscillate over time.

For a periodic function \( f(t) \) with period \( T \), the time average \( \overline{f(t)} \) is calculated as:
\[ \overline{f(t)} = \frac{1}{T} \int_0^T f(t) \, dt \]
This formula averages the function over one complete cycle. In the context of the Fourier series, the time average allows us to simplify expressions by factoring out elements that contribute negligibly over a full period, thus focusing on the primary components.

In our exercise, the time average is applied to the square of the magnitude of \( \psi(t) \), helping us link the original function to its frequency components, encapsulated in the Fourier coefficients.
Fundamental Frequency
Fundamental frequency, often denoted as \( f_0 \), is the lowest frequency in a periodic waveform and defines the repetition rate of the waveform. It is crucial in breaking down complex signals into simpler sinusoidal components using Fourier series. Each periodic function can be represented as a sum of sine and cosine terms (harmonics), each with frequencies that are integer multiples of the fundamental frequency.

For a function \( \psi(t) \), the Fourier representation includes a fundamental frequency \( \omega_1 \) given by:
\( \omega_1 = \frac{2\pi}{T_1} \), where \( T_1 \) is the fundamental period.

Higher multiples of this fundamental frequency make up the harmonics in the signal. These create a richer and more complex representation of the original function. Understanding the fundamental frequency is essential in signal processing and helps to analyze and reconstruct signals accurately.

In the original exercise, the coefficients \( \check{A}_n \) are associated with these harmonic frequencies, allowing us to express \( \psi(t) \) as a sum of multiple frequency components. This decomposition is central to many applications, including audio signal processing, telecommunications, and vibration analysis.

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Most popular questions from this chapter

Show that the Fraunhofer diffraction integral for a one-dimensional aperture, such as a single slit (Sec. 10.4) or a grating (Secs. \(10.7\) and 10.8) may be expressed in the form $$ \psi(s)=\breve{C} \int_{-\infty}^{\infty} \breve{T}(x) e^{i \theta x} d x $$ where \(s \equiv \kappa\left(\sin \theta_{o}+\sin \theta_{o}\right)\) and \(\breve{T}(x)\) is the aperture transmission function (9.4.19), possibly \(\dagger\) For further details concerning holography and other applications of Fourier analysis to optical problems, see M. Born and E. Wolf, "Principles of Optics," 3 ded., sec. \(8.10\), Pergamon Press, New York, 1965; A. Papoulis, "Systems and Transforms with Applications in Optics," McGraw-Hill Book Company, New York, 1968; and J. W. Goodman, "Introduction to Fourier Optics," McGraw-Hill Book Company, New York, \(1968 .\) complex, that expresses the wave- transmission properties of the aperture. Hence show that the diffracted-wave amplitude, except for a constant factor, is the Fourier transform of the aperture transmission function. It is thus no accident that the Fourier transform of a rectangular pulse (Prob. \(12.22\) ) is functionally the same as the Fraunhofer diffraction-amplitude function for a single slit, \(\sin u / u\).

Apply the Kramers-Kronig relations (12.7.15) and (12.7.16) to the two parts of the (delay) transfer function of Prob. 12.6.1, T) \((\omega)=\cos \omega t_{2}-j \sin \omega t_{2}\), to verify that one part implies the other, and vice versa.

What limitations must be put on the mathematical behavior of a transfer function at an infinite frequency if the Kramers-Kronig integrals are to converge? Answer: \(T_{r}(\infty)\) remains finite, \(T_{i}(\infty)=0\).

The uncertainty relation (12.2.7) relates a waveform duration in the time domain to the bandwidth of its energy spectrum in the frequency domain. In the case of a traveling wave of finite duration, what can be said about the waveform extent \(\Delta x\) in the space domain and the spread \(\Delta \kappa\) of the wave numbers in the wave-number domain? In quantum mechanics the energy \(E\) and momentum \(p\) of a particle are related to frequency and wave number by de Broglie's relations \(E=\hbar \omega\) and \(p=\hbar \kappa\), where \(\hbar\) is Planck's constant divided by \(2 \pi\). On the basis of these relations and the uncertainty relation, obtain Heisenberg's uncertainty principle \(\Delta E \Delta t \geq \frac{1}{2} \hbar\) and \(\Delta p \Delta x \geq \frac{1}{2} \hbar\). This principle is an inescapable part of a wave description of nature.

A pulse described by the Dirac delta function \(\delta(t)\) has negligible duration, infinite height at \(t=0\), but unit area. Find the amplitude spectrum of \(\delta(t)\) as a limiting case by setting the area of the rectangular pulse in Prob. 1222 equal to unity and then letting \(\tau \rightarrow 0 .\) Answer: \(F(\omega)=1\).
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