Question 9.33: The “inversion theorem” for Fourier transforms states that φ...

The “inversion theorem” for Fourier transforms states that

\tilde{\phi}(z)=\int_{-\infty}^{\infty} \tilde{\Phi}(k) e^{i k z} d k \quad \Longleftrightarrow \quad \tilde{\Phi}(k)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \tilde{\phi}(z) e^{-i k z} d z .                           (9.198)

\text { Use this to determine } \tilde{A}(k) \text {, in Eq. 9.20, in terms of } f(z, 0) \text { and } \dot{f}(z, 0) .

\tilde{f}(z, t)=\int_{-\infty}^{\infty} \tilde{A}(k) e^{i(k z-\omega t)} d k                          (9.20)

\left[\text { Answer: }(1 / 2 \pi) \int_{-\infty}^{\infty}[f(z, 0)+(i / \omega) \dot{f}(z, 0)] e^{-i k z} d z\right]
The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
\tilde{f}(z, 0)=\int_{-\infty}^{\infty} \tilde{A}(k) e^{i k z} d k \Rightarrow \tilde{f}(z, 0)^{*}=\int_{-\infty}^{\infty} \tilde{A}(k)^{*} e^{-i k z} d k . \text { Let } l \equiv-k ; \text { then } \tilde{f}(z, 0)^{*}=

\int_{\infty}^{-\infty} \tilde{A}(-l)^{*} e^{i l z}(-d l)=\int_{-\infty}^{\infty} \tilde{A}(-l)^{*} e^{i l z} d l=\int_{-\infty}^{\infty} \tilde{A}(-k)^{*} e^{i k z} d k \text { (renaming the dummy variable } l \rightarrow k ).

f(z, 0)=\operatorname{Re}[\tilde{f}(z, 0)]=\frac{1}{2}\left[\tilde{f}(z, 0)+\tilde{f}(z, 0)^{*}\right]=\int_{-\infty}^{\infty} \frac{1}{2}\left[\tilde{A}(k)+\tilde{A}(-k)^{*}\right] e^{i k z} d k . Therefore

\frac{1}{2}\left[\tilde{A}(k)+\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(z, 0) e^{-i k z} d z.

\text { Meanwhile, } \dot{\tilde{f}}(z, t)=\int_{-\infty}^{\infty} \tilde{A}(k)(-i \omega) e^{i(k z-\omega t)} d k \Rightarrow \dot{\tilde{f}}(z, 0)=\int_{-\infty}^{\infty}[-i \omega \tilde{A}(k)] e^{i k z} d k.

(Note that \omega=|k| v, here, so it does not come outside the integral.)

\dot{\tilde{f}}(z, 0)^{*}=\int_{-\infty}^{\infty}\left[i \omega \tilde{A}(k)^{*}\right] e^{-i k z} d k=\int_{-\infty}^{\infty}\left[i|k| v \tilde{A}(k)^{*}\right] e^{-i k z} d k=\int_{\infty}^{-\infty}\left[i|l| v \tilde{A}(-l)^{*}\right] e^{i l z}(-d l)

=\int_{-\infty}^{\infty}\left[i|k| v \tilde{A}(-k)^{*}\right] e^{i k z} d k=\int_{-\infty}^{\infty}\left[i \omega \tilde{A}(-k)^{*}\right] e^{i k z} d k .

\dot{f}(z, 0)=\operatorname{Re}[\dot{\tilde{f}}(z, 0)]=\frac{1}{2}\left[\dot{\tilde{f}}(z, 0)+\dot{\tilde{f}}(z, 0)^{*}\right]=\int_{-\infty}^{\infty} \frac{1}{2}\left[-i \omega \tilde{A}(k)+i \omega \tilde{A}(-k)^{*}\right] e^{i k z} d k .

\frac{-i \omega}{2}\left[\tilde{A}(k)-\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \dot{f}(z, 0) e^{-i k z} d z, \text { or } \frac{1}{2}\left[\tilde{A}(k)-\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[\frac{i}{\omega} \dot{f}(z, 0)\right] e^{-i k z} d z.

Adding these two results, we get \tilde{A}(k)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[f(z, 0)+\frac{i}{\omega} \dot{f}(z, 0)\right] e^{-i k z} d z  .   qed

Related Answered Questions