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I´m reading the book Numerical Simulation of Optical Wave Propagation by Jason Schmidt.

In section 2.1-2.2, the author performs the FFT of a centered and a shifted Gaussian pulse and makes the statement that the time shifted signal has a higher maximum frequency than the unshifted one. For that reason more samples are required in getting the shifted signal spectrum.

Why does the time shifted image have a higher highest frequency? I don´t understand.

How much I should increase my sampling rate for a given time shift?

Thanks for your attention. Carolina Rickenstorff

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  • $\begingroup$ To be clear, it is being stated that a time-shift results in a frequency shift? The time-shift property of the Fourier transform states that it results in a phase shift in the Fourier domain. $\endgroup$
    – Envidia
    May 13, 2021 at 20:57
  • $\begingroup$ It'd help to know the larger context. The frequency of the frequency (rather of spectrum as if it were the signal) will be greater, i.e. spectrum will be oscillatory. This is only a phase shift and does not change the absolute value of the spectrum, nor required sampling rate. $\endgroup$ May 13, 2021 at 21:23
  • $\begingroup$ There's no pulse compression going on, right? It's just that the pulse is shifted? $\endgroup$
    – TimWescott
    May 13, 2021 at 23:36

3 Answers 3

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I was able to take a look at the book. You are right it says: "The highest significant frequency in the shifted Gaussian signal is higher than that in the centered Gaussian". I think this is misleading.

Looking at Figs 2.2 and 2.3 I can only think that the author is saying that in Fig 2.3 the phase is "significant" across the whole spectrum, while in Fig 2.2 phase is zero everywhere.

However, the magnitude curves are identical. If magnitude is zero then phase is irrelevant. (i.e. atan (small/small)). The required sample interval is governed by the magnitude spectrum. That is the required sample interval is not affected by a simple time shift, as stated by the above commenters. If this wasn't the case, then real-data analog-to-digital conversion would be pretty difficult!

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In the linear medium, electromagnetic waves (light, IR including) with different frequencies do not interact with each other. For a tight spectral composition ($Δω \ll ω_0$) of the pulse, the ratio of constituent waves' intensities does not change on propagation, and the spectral composition does not change either. But the shape of the pulse envelope, as well as the Fourier transform of this envelope, can change and does change.

I reproduce a typical (as you can encounter in EE courses) derivation of equations for the pulse envelope shape evolution that somewhat loosely mixes the square root of light intensity with the lightwave amplitude. I believe you can find the rigorous derivation of propagation equations for the dispersive medium in the book you cite; if it is not there, see the other sources, for example, IEEE's Journal of Lightwave Technology or Elsevier's Optical Fiber Technology Journal, maybe the past century articles.

When the group velocity $v_g$ (and a group velocity refractive index $n_g = {c / v_g}$) is wavelength-dependent in the medium, the chromatic dispersion phenomena change the pulse shapes on propagation. Even without chromatic dispersion, the modal dispersion leads to pulse broadening in optical fibers.

A Gaussian pulse propagating in optical fibers (or other media) can be made chirped, and in fact the design of optical communication systems uses the chirped pulses to mitigate the pulse broadening effect. Let the initial amplitude of a chirped pulse is $$ A(0,t) = A_0\exp\left(-{\frac {1+iC} {2}}\left({t \over T_0}\right)^2\right) $$ where $C$ is the chirp coefficient. The Fourier transform of the chirped pulse is $$ A(0,ω) = A_0{\sqrt{\left({\frac {2πT_0^2} {1+iC}}\right)}}\exp{\left(-{\frac {ω^2T_0^2} {2(1+iC)} }\right)} $$ For the pulse propagating in a $z$ direction in the dispersive medium the frequency components of the pulse can be written down as $$ A(z,ω) = A(0,ω)\exp(iβz) $$

For a narrow frequency pulse ($Δω \ll ω_0$) the propagation constant $β(ω)$ can be expanded in a Taylor series $$ β(ω) = \bar n (ω)ω/c \sim β_0 + β_1(Δω) + {1 \over 2}β_2(Δω)^2 + {1 \over 6}β_3(Δω)^3 $$ where $β_1 = {1 / v_g(ω_0)}$ and $β_2$ is proportional to the dispersion parameter D.

For the carrier wavelength different from zero-dispersion wavelength we can neglect the $β_3$ term contribution, and write down the time domain propagation equation in the form $$ A(z,t')={1 \over 2π}\int_{-∞}^{∞}A(0,ω)\exp(-iωt' + {i \over 2}β_2 ω^2 z)dω = \\ {\frac {A_0 T_0} {\sqrt{\left(T_0^2 - iβ_2 (1+iC)z\right)}}}\exp{\left(-{\frac {(1+iC){t'}^2} {2(T_0^2-iβ_2 z (1+iC))}}\right)} $$ where the substitute is made: $t' = t - βz$, which takes out the explicit dependence on carrier's group velocity $v_g$ (via $β_1$) from the equation.

Under the conditions stated, the Gaussian pulse envelope keeps the Gaussian shape on propagation even in the presence of dispersion, with the pulse width changing according to $$ T = T_0{\sqrt{\left((1 + {Cβ_2 z \over T_0^2})^2 + ({β_2 z \over T_0^2})^2\right)}} \tag {eq. T(z)} $$ . The parameters $β_2,T_0$ can be combined into a parameter called dispersion length $L_D = T_0^2/β_2$. The unchirped pulse width unconditionally increases as $\sqrt{\left(1+(z/L_D)^2\right)}$ on propagation, while the chirped pulse width, if $β_2 C < 0$ holds, starts its evolution with shrinking, reaching its minimum value at a distance of $L_D·C/(1+C^2)$.

You cannot get rid of pulse broadening entirely for all times in the dispersive medium. But with a specially prepared chirped pulse, you can delay the moment when the broadening process impairs your pulse critically, and even reverse the broadening process replacing it with shrinking for cable lengths on the order of $L_D⋅C/(1+C^2)$. For a longer optical path, the pulse width will inevitably start to increase roughly proportional to the propagation distance, but in the meantime you should certainly increase the spatial resolution when sampling the contracted pulse envelope.

You can deduce the formula for the corrected spatial resolution using the equation $\text T(z)$ for the pulse width evolution. In other words, higher spatial (and temporal) frequencies can appear in the spatial (and temporal) Fourier transform of the pulse envelope as the pulse propagates in space and time in the dispersive medium (or waveguide) -- providing the condition $β_2 C < 0$ holds -- and consequently tightlier spaced readouts are required to sample the advanced signal when the current pulse width is shorter than the original width.

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Clarifying for signal processing, I looked into this from VVT's answer and it appears this has nothing to do with FFT of some time shifted waveform but rather the physical consequence of doing so in the application.

From Dispersion wiki:

enter image description here

Time evolution of a short pulse in a hypothetical dispersive medium (k=w^2) showing that the longer wavelength components travel faster than the shorter wavelengths (positive GVD), resulting in chirping and pulse broadening.

Reading further:

if a pulse travels through a material with negative group-velocity dispersion, shorter wavelength components travel faster than the longer ones, and the pulse becomes negatively chirped, or down-chirped, decreasing in frequency with time.

So a greater sampling frequency is required for a time shifted Gaussian pulse propagated through optical media since it'd have lost less frequency. (correct me if I'm wrong)

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