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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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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, \(\sin u / u\).

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.

Establish the integral $$ \int_{0}^{\infty} \sin a t \cos b t d t=\frac{a^{2}}{a^{2}-b^{2}} $$ by first multiplying the integrand by \(e^{-s t}\) and allowing \(s \rightarrow 0\) after integration. This, or a similar artifice, is of necessary in evaluating Fourier transform integrals. A Fourier transform so evaluated is then described as a limiting form.

The voltage gain of a multistage low-pass a mplifier is accurately approximated by the gaussian function $$ |\breve{G}(\omega)|=G_{0} e^{-(\omega / 2 \pi \Delta \omega)^{2}} . $$ Show that \(\Delta \omega\) is the rms bandwidth of the amplifier. Show also that the so-called \(u p p e r\) halfpower frequency \(\nu_{0}\) [the frequency at which \(\left(|\breve{G}| / G_{0}\right)^{2}=\frac{1}{2}\) ] is related to \(\Delta \omega\) by $$ \nu_{0} \approx 0.6 \Delta \omega, $$ where \(\nu_{0}\) is an ordinary (cyclic) frequency. Show then that the uncertainty relation (12.2.7) implies that $$ \nu_{0} \Delta t \approx \frac{1}{3} $$ when \(\Delta t\), the duration of a narrow pulse whose frequency spectrum is limited by the amplifier, has its minimum permitted value. This form of the uncertainty relation expresses reasonably well the relation between the rise- time of an amplifier (see Prob. 11.4.3 for a definition of risetime) and its bandwidth expressed by its upper half-power frequency.

Prove the similarity relation that if \(F(\omega)\) is the Fourier transform of \(f(t)\) then \((1 / a) F(\omega / a)\) is the Fourier transform of \(f(a t)\), where \(a\) is a real, positive constant.
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