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Chapter 12: Problem 3

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.

Short Answer

Expert verified
Answer: Yes, the real and imaginary parts of the transfer function are interconnected according to the Kramers-Kronig relations. By applying the Kramers-Kronig relations to the real and imaginary parts, we verified that one part implies the other, and vice versa.

Step by step solution

01

Identify real and imaginary parts of the transfer function

First, we need to identify the real and imaginary parts of the transfer function T(ω). The transfer function is given as: T(ω) = \(\cos(ωt_2) - j \sin(ωt_2)\) Here, the real part is \(\cos(ωt_2)\), denoted as R(ω), and the imaginary part is \(-\sin(ωt_2)\), denoted as I(ω).
02

Rewrite the Kramers-Kronig relations

Now, we need to rewrite the Kramers-Kronig relations (12.7.15) and (12.7.16) in terms of R(ω) and I(ω). The relations are given as: Relation (12.7.15): R(ω) = \(\frac{1}{\pi}\,\text{P}\int_{-\infty}^{\infty}\frac{I(u)}{u-\omega}du\) Relation (12.7.16): I(ω) = -\( \frac{1}{\pi}\,\text{P}\int_{-\infty}^{\infty}\frac{R(u)}{u-\omega}du\)
03

Apply Kramers-Kronig relation (12.7.15) to I(ω)

With the expressions for R(ω) and I(ω), we will now apply the Kramers-Kronig relation (12.7.15) to I(ω) and check if we obtain R(ω). Calculating R(ω) using I(ω) and relation (12.7.15): R(ω) = \(\frac{1}{\pi}\,\text{P}\int_{-\infty}^{\infty}\frac{-\sin(ut_2)}{u-\omega}du\) It can be observed that the integral is a well-known result for the Hilbert transform of a sine function, and therefore we have: R(ω) = \(\cos(ωt_2)\) We can see that we obtain the correct real part R(ω) by applying the Kramers-Kronig relation (12.7.15) to I(ω).
04

Apply Kramers-Kronig relation (12.7.16) to R(ω)

Now, we will apply the Kramers-Kronig relation (12.7.16) to R(ω) and check if we obtain I(ω). Calculating I(ω) using R(ω) and relation (12.7.16): I(ω) = -\( \frac{1}{\pi}\,\text{P}\int_{-\infty}^{\infty}\frac{\cos(ut_2)}{u-\omega}du\) Again, this integral is a known result for the Hilbert transform of a cosine function, so we have: I(ω) = \(-\sin(ωt_2)\) We can see that we obtain the correct imaginary part I(ω) by applying the Kramers-Kronig relation (12.7.16) to R(ω).
05

Conclusion

By applying the Kramers-Kronig relations (12.7.15) and (12.7.16) to the real and imaginary parts of the given transfer function, we verified that one part implies the other, and vice versa. This shows the interconnection between the two parts of the transfer function.

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Most popular questions from this chapter

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.

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.

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, \(\sin u / u\).

Show that the Fourier transform of \(f(t)=e^{-(t / \tau)^{2}}\) is \(e^{-(\omega \tau / 2 \pi)^{2}}\). Show that the duration of \(f(t)\), defined by \((12.24)\), is \(\tau / 2\) and that the width \(\Delta \omega\) of its energy spectrum, defined by \((12.2 .5)\), is \(1 / \tau\) and hence that the product \(\Delta l \Delta \omega\) has precisely the minimum value \(\frac{1}{2}\) permitted by the uncertainty relation (12.2.7).

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.
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