Light Propagation
Part 2: Temporal Behavior

ECE 5358/6358                              han q le - copyrighted
Use solely for students registered for UH ECE 6358/5368 during courses - DO NOT distributed (copyrighted materials)

Segment 1

1. Wave carrier and modulation: basic concepts

1.1 Introduction

Hardly do we ever deal with strictly monochromatic (single-frequency), continuous wave (CW) in applications. In fact, such a lightwave has very limited appications such as illumination, interference, metrology, or sometimes, just to deliver raw power... 

More important applications, such as optical communications or optical signal processing use light as signals to carry information. A signal is made by modulating a wave (referrred to as carrier wave). The modulation can be on the wave amplitude, frequency, or phase. Such a light signal has behavior that is not as simple as that of a monochromatic, continuous wave.

A spatial analogy is plane wave. It’s the simplest of all waves and can be a useful approximation in many problems. But do we always use plane waves? Focusing a laser beam onto a target, whether in an optical disk, a laser printer, or into a waveguide or fiber requires considering waves with spatial profiles more complicated than that of a plane wave. We see that Gaussian beams have properties more complex than those of a plane wave for example.

This topic is the temporal analogy of the spatial beam that we study. Below is a slide about the analogy: a laser beam needs to be focused into a small spot and be made into a very short pulse for optical communication.

A plane wave cannot describe a tightly focused beam. A monochromatic continuous wave cannot describe a short pulse. What we deal with here is called a wave packet.

What is there in a wave packet or light pulse that we need to learn? Isn’t it just as simple as a continuous wave but with finite duration? Here is something for us to think about. Look at this demo:
Link to section 2.2.1

1.2 Review - carrier and modulation (skip if you are familiar)

Link to 1100

1.2.1 Carrier wave basic illustration

Carrier is a pure single frequency wave, like this (e^(i (k z - ω t))) that we have seen before. For simplicity, we even make the wave one dimension. Such a wave is a pure harmonic, because it has only one spatial frequency k, and one temporal frequency ω.

Manipulate[ tmax=2 π/ω ;
Animate[Plot[{Cos[v z-t],Cos[v k z-ω t+ϕ]},{z,-20,20}, PlotRange->{{-20,20},{-1,1}},
	       AspectRatio-> 0.2   , PlotStyle->{{RGBColor[1,0.,0], Thickness[0.005]},{RGBColor[0.,0.,1], Thickness[0.005]}} 
,Filling-> {{1-> {Axis,Hue[0,1,1,0.25]}},{2-> {Axis, Hue[0.5,1,1,0.4]}}}	 ]
,{t,0,tmax}, AnimationRunning->False]
,{k,1,2}, {ω,1,2}, {ϕ,0,2 π} ,{{v,1},-1,1,2}]

Vary various parameters, the red wave is the reference, the blue wave is the wave with variable k, ω, direction, and phase. Imagine where one can find wave like that.

1.2.2 Phase velocity

You can see in the illustration above that the wave is moving from left to right or right to left. How fast does it move? what speed? how do we determine the speed?

Animate[Show[Plot[Cos[ z-t],{z,-20,20}, PlotRange->{{-20,20},{-1,1}},
	       AspectRatio-> 0.2   , PlotStyle->{{RGBColor[0.,0.,1], Thickness[0.005]}} 
,Filling-> {1->  {Axis, Hue[0.5,1,1,0.4]}}	 ]
,Graphics[ {Red,Disk[{-18+ t ,Cos[-18]},{0.8,0.16}]}]  ]
,{t,0,40}, AnimationRunning->False]

You can see in the illustration above that the wave is moving from left to right. We can just follow the red dot. There is nothing special about the red dot we pick. Any dot on the wave is OK. We note that the distance the red dot travel in one period is one wavelength. Since the function is periodic, the value is the same when translated by one period. This is most obvious answer we have learned about the speed of a harmonic (Sine or Cosine) wave:
                             v= (λ/T)=(λ/(1/f)) =λ f =((2 π)/k)  (ω/(2 π)) = (ω/k)       (1.2.1)

But is this always correct? What if the wave is not purely Sine or Cosine?
There is a conceptually more fundamental definition. Let's look at the wave itself in terms of the expression 
                                       (e^(i Φ[z;t]))        where Φ[z;t] is the phase function.
As an example of the Gaussian beam we have studied before (if not, it’s OK to skip this example):
                        (( w (Subscript[k, 1]))/(Abs[Q[z]] i))(e^(-(Subscript[k, 1]) (ρ^2) Im[(1/(2Q[z]))]  )) (e^(i  ( (Subscript[k, 1])(z+ (ρ^2) Re[(1/(2 Q[z]))])- Arg[Q[z]]-ω t)  ))
                        where  Q[z]=z-i  (Subscript[z, 0]); (Subscript[z, 0]) =(((Subscript[k, 1])(w^(2 )))/2) = ((π (w^(2 )))/λ)  
Notice that the wave is separated into 2 parts:
                             (e^(-(Subscript[k, 1]) (ρ^2) Im[(1/(2Q[z]))]  ))    is the amplitude.
                             (e^(i  ( (Subscript[k, 1])(z+ (ρ^2) Re[(1/(2 Q[z]))])- Arg[Q[z]]-ω t)  ))  has the form  (e^(i Φ[z;t]))
where:                        Φ[z,ρ;t]=(Subscript[k, 1])(z+ (ρ^2) Re[(1/(2 Q[z]))])- Arg[Q[z]]-ω t     
is the phase.

We applied this concept spatially to determine the phasefront as given by the surface of constant phase:

We know that the phasefronts move, and clearly the velocity of a point of this wave is given by how far a distance Δ(OverVector[x]) it moves for a given time Δt while the phase remains the same, i. e. for Φ[z,ρ;t]=constant. We don’t care for amplitude (which is getting smaller as the wave spread). For simplicity, let consider only the one-dimension case with z as the space variable:
                         Δ Φ[z,t] =(∂_z) Φ[z,t]Δz+(∂_t) Φ[z,t]Δt =0             (1.2.2)
Thus, the phasefront velocity is:   v=((Δz)/(Δt))=-(((∂_t) Φ[z,t])/((∂_z) Φ[z,t]))                           (1.2.3)
In general:                                 v=-(((∂_t) Φ[z,t])/(|∇ Φ[z,t]|))(OverHat[n])                          (1.2.4)
where (OverHat[n]) is the normal vector of the phasefront gradient.
A key aspect of this definition and is different from the illustration with the red dot above is that it is the constant phasefront and NOT constant amplitude that determines the velocity of the wave.

Let’s apply the concept to the pure harmonic wave above: (e^(i (k z - ω t) ))
The phase term is:                Φ[z;t]=k z - ω t    
                                                   v=-(((∂_t)(k z - ω t))/((∂_z)(k z - ω t)))=(ω/k)          (1.2.5)
 which is exactly the same as (1.2.1) without invoking the concepts of period and wavelength.

Thus, for the carrier wave above, (ω/k) is the wave velocity. It is also called the phase velocity, because it is the velocity that if we move at which, the phase of the carrier would appear to be constant.

1.2.3 Instantaneous frequency

Let’s consider this example:

Play[Sin[880 2Pi  t  ],{t,0,2}]

Now, let’s listen to this:

Play[Sin[220 2Pi  t (1+3 t) ],{t,0,2}]

What is the difference between the 2 signals? (see Link to 1100 for more about frequency modulation)

What is the frequency of the second signal?

Let’s think fundamentally: what is a frequency of a wave if it is not always purely single frequency like the chirp signal above?
See the demonstration below. If we sit at a fixed location, we will see the wave amplitude oscillate in time like this

ω1=2 Pi   ;
Animate[Show[Plot[Cos[ 2 π z-ω1 t],{z,-1.2,1.2}, PlotRange->{{-1.5,1.5},{-1.2,1.2}},
	       PlotStyle->{{RGBColor[0.,0.,1], Thickness[0.005]}} 
,Filling-> {1->  {Axis, Hue[0.5,1,1,0.4]}}	 ]
,Graphics[ {Red,Disk[{0  ,Cos[ω1 t]},0.1]}]  ]
,{t,0,1}, AnimationRunning->False]

ω1=2 Pi  ;ω2=3.7 Pi  ;
Animate[GraphicsRow[{
Show[Plot[Cos[ 2 π z-ω1 t],{z,-1.2,1.2}, PlotRange->{{-1.5,1.5},{-1.2,1.2}},
	       PlotStyle->{{RGBColor[0.,0.,1,0.1], Thickness[0.005]}} 
,Filling-> {1->  {Axis, Hue[0.5,1,1,0.1]}}	 ]
,Graphics[ {Red,Disk[{0  ,Cos[ω1 t]},0.1]}]  ]
,Show[Plot[Cos[ 2 π z-ω2 t],{z,-1.2,1.2}, PlotRange->{{-1.5,1.5},{-1.2,1.2}},
	       PlotStyle->{{RGBColor[0.,0.,1,0.1], Thickness[0.005]}} 
,Filling-> {1->  {Axis, Hue[0.5,1,1,0.1]}}	 ]
,Graphics[ {Red,Disk[{0  ,Cos[ω2 t]},0.1]}]  ]
}] 
,{t,0,1}, AnimationRunning->False]

The frequency is the number of cycles per unit time. If the ball moves up down real fast, it's high frequency. If it moves slowly, it's low frequency.

We have a concept: instantaneous frequency. It is the speed of the phase change vs. time:
ω(t)=|((∂ Φ[t])/(∂t))|

Let's apply the concept to  (e^(i (k z - ω t) ))
The phase term is:                Φ[z;t]=k z - ω t    
                                                   |((∂ (k z - ω t))/(∂t))|=ω          (1.2.6)

Now we can look at this chirp signal:

Play[Sin[220 2Pi  t (1+3 t) ],{t,0,2}]

And we can plot the instantaneous frequency:

t=. ;
(∂_t)(220 2Pi  t (1+3 t) )

Plot[(1320 π t+440 π (3 t+1))/(2 π ),{t,0,2}, Frame-> True, GridLines->Automatic, FrameLabel->{"Time (sec)","Instantaneous f (Hz)"}, LabelStyle->{24}]

So we can see that we do get the right answer.

2. Envelope and wave packet

2.1 Discussion of formulation

We have seen in AM modulation:

AM[t_]:=(1+ 0.23 Cos[2 π *0.0231 t+0.873]
                        +0.39 Cos[2 π *0.085  t+1.33]
                         +0.275 Cos[2 π *0.11 t-1.9])  ;
Manipulate[
Plot[{Cos[2 π (x-t)]*AM[(x-t)] ,AM[(x-t)], - AM[(x-t)]}
, {x,0,50},
PlotPoints-> 200, PlotStyle-> {{RGBColor[1,0,0]}, {RGBColor[0,0,1]},{RGBColor[0,0,1]}}, AspectRatio-> 0.3],{t,0,50,1}, DisplayAllSteps-> True]

The blue part is called the “envelop” that is imposed on the “red part” carrier.
Or this:

f=1500 ;
Sound[Play[(1+Cos[20 t (1+0.5 t)])*Cos[2 π f  t],{t,0,7}] ]

Plot[1+Cos[20 t (1+0.5 t)],{t,0,7}, AspectRatio->0.25, PlotPoints->300,Filling-> Axis]

Often, many applications use a short light pulse, such as in optical communication. This is what we call a wave packet. The short light pulse can be of any shape. A common shape is:

Plot[0.25(1+Clip[4 (t+1)])*(1+Clip[4*(-t+1)]),{t,-3,3}, PlotStyle->{Thick, Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}]

But for studying purpose, a very useful shape is Gaussian:

Plot[Exp[-t^2],{t,-3,3}, PlotStyle->{Thick, Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}]

Suppose we have a signal function S[t].

How do we formulate this envelope S[t] on a lightwave (e^(i (k z -ω t)))?
Is it just                                     S[t] (e^(i (k z -ω t))) ?

For example, we have (e^(-(t^2))) as a pulse envelope in time. If we plot in time, the signal would look like this:

ω=10*Pi ; k=10*Pi; z=0 ;
Plot[Exp[-t^2] Cos[k z -ω t] ,{t,-3,3}, PlotStyle->{Thick, Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> All]

But what about space? let t=0, and we plot vs. z, this is what it looks like:

ω=10*Pi ; k=10*Pi;  t=0 ;
Plot[Exp[-t^2] Cos[k z -ω t] ,{z,-3,3}, PlotStyle->{Green}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> All]

Clearly, this does not make sense. A pulse must have finite extend in space as well if we take a snap shot of the pulse at time =0. If we let the signal be: Exp[-z^2] Cos[k z -ω t]:

ω=10*Pi ; k=10 π; t=0 ;
Animate[Plot[Exp[-(z^2)] Cos[k z -ω t] ,{z,-2.5,2.5}, PlotStyle->{Thick, Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> {-1,1}, PlotPoints->100]
,{t,0,1}]

The problem here we see is that the pulse doesnot move!
We can make the pulse move by making the envelope moving along with the carrier:

ω=10*Pi ; k=10  π; t=0 ;
Animate[Plot[Exp[-(  z-(ω/k) t)^2] Cos[k z -ω t] ,{z,-3,3}, PlotStyle->{ Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> {-1,1}, PlotPoints->200]
,{t,-1,1}]

So, this is our light pulse formulation:  
                                             E[z;t]=(e^(-(((((z-(ω/k) t)/w)))^2)))  (e^(-i (k z -ω t)))                            (2.1.1)
Or:                                        E[z;t]=(e^(-((((ζ/w)))^2)))  (e^(-i k ζ))  where  ζ=z-(ω/k) t

Is it correct?

2.2 Frequency domain concept - note on frequency domain vs time domain (e. g. FDTD)

If we use the wave equation in LectSect 1, it is obvious that it satisfies the wave equation with the propagation velocity:                 
                                                        (Subscript[v, p])=(ω/k)               (2.2.1a)
which is defined as the phase velocity, which is also given by:
                                                   (Subscript[v, p])=(ω/(k[ω]))=(c/(n[ω]))      (2.2.1b)

But we recall a discussion in LectSet 1 that the wave equation is not necessarily correct for any media with matter and for non-single-frequency light field. 
                                     (∇^2)E  -((μ ϵ)/(c^2))((∂^2)/(∂ (t^2)))E =0                              1(4.1.10)   (2.2.2)
To really think about it, let’s consider a numerical approach that allows us to solve the time-dependent equation above. A very powerful numerical method is known as FDTD (finite-difference time-domain) algorithm. Here is a 1-D illustration:

2.2.1 One-dimensional FDTD illustration: incident on a dielectric

Below is a simple Mathematica code for 1-D time-domain numerical solution (FDTD) for a light pulse incident on an dielectric.

S=Table[(E^(-I 2 π (i-300)/50.))   Exp[-(((i-400))^2)/150.^2],{i,1,3000}] ;
ListPlot[Re[S], Joined-> True, PlotRange-> {{0,800},All}]

Demo FDTD calculation for pulse propagation in air and dielectric

Compare the behavior of the light pulse in air or vacuum (left) and in the dielectric (right).

Is it correct? We know that it does NOT look correct! What is meant “not correct” here is not about numerical accuracy. The result is numerically accurate, but it is not correct with respect to physics.
Thus, it is important to remember that even if we can solve the time-dependent equation above, because it is correct only for purely harmonic wave but not generally for any wave, it is not necessarily most helpful for us to study the behavior of modulated light.

So, what does physically correct simulation look like?

Demo:  group velocity 1

Why the FDTD result above is not correct? Let’s take a look of the wave equation in Lect Set 1. We modify Eq. 1(4.1.4c) with zero current and space charge:

(∇^2)E -(µ/(c^2))((∂^2)/(∂ (t^2)))(E + 4 π P) = -4 π ∇ (∇. P)       1(4.1.4c-modified) (2.2.3)
We see that the only way (2.2.3) can become (2.2.2) for the FDTD approach is: 
                                                      P[t]=χ E[t]                     (2.2.4)
which means that the induced polarization of atoms and molecules is strictly instantaneous.

2.2.2 Frequency domain

We know that strictly instantaneous induced polarization is not correct. As discussed in LectSet1, atoms and molecules in a medium has a time-dependent response to the EM field. The induced polarization is an accumulative effect that depends on the history of the field:
                                           P[t]=(∫_(-∞)^t)(OverBar[(OverBar[α])])[t-τ] · E[τ]dτ         1(4.1.12)
Let’s express the fields by their FT 
                     ∫P[ω](e^(-i ω t))dω=(∫_(-∞)^t)(OverBar[(OverBar[α])])[t-τ] · (∫E[ω](e^(-i ω τ))dω)dτ      
                                                =∫(∫_(-∞)^t)(OverBar[(OverBar[α])])[t-τ] · (e^(-i ω τ))dτ  E[ω]dω    
                                                =∫ (e^(-i ω t))(∫_(-∞)^t)(OverBar[(OverBar[α])])[t-τ] · (e^(i ω (t-τ)))dτ E[ω]dω    
                                                =∫(e^(-i ω t)) ((∫_0^∞)  (OverBar[(OverBar[α])])[u] · (e^(i ω u))du )E[ω]dω  
                                                =∫(OverBar[(OverBar[χ])])[ω]·E[ω](e^(-i ω t)) dω
Hence:                             P[ω]=(OverBar[(OverBar[χ])])[ω]·E[ω]                            1(4.1.14)

This means that in the frequency domain, we can write:
                                       E[ω]+4π P[ω]=(1+4 π (OverBar[(OverBar[χ])])[ω])· E[ω]                                                 
                                                                 = (OverBar[(OverBar[ϵ])])[ω]· E[ω]            (2.2.5)
Then, (2.2.3) will yield the wave equation (2.2.2) (assuming ∇. P=0) for the component ω of the wave:
                                     (∇^2)E[ω]  -((μ  )/(c^2))(OverBar[(OverBar[ϵ])])[ω] ·((∂^2)/(∂ (t^2)))E[ω] =0                       1(4.1.10)   (2.2.2)
                                     (∇^2)E[ω]  +((ω^2)/(c^2))μ (OverBar[(OverBar[ϵ])])[ω] · E[ω] =0                       1(4.1.10)   (2.2.6)
which is the Helmholtz equation. 
Thus, for time-dependent behavior of light wave (not counting the harmonic oscillation, of course), it is more convenient to study in the frequency domain (FD).
The reason we employ frequency domain approach is not just a matter of convenience. It gives us physical insights that time domain sometimes cannot.

Before we engage in frequency domain discussion, here is a note on advanced FDTD for optics applications:
- include material dispersion (frequency dependent dielectric -> Time-response model)
- any dispersion in general (e. g. structural)
- nonlinear optics
They are done by including all quantities in wave equations, not just E and H but also polarization field P, including non-linear P. However this means much larger array size for computation and more time consuming, which reduces its attractiveness.

2.3 Spectral decomposition and linear superposition for a light pulse or wave packet

The starting point for the frequency domain is the Helmholz equation, which is simplified to 1D in this case:
                              ((∂^2)/(∂ (z^2)))E+ ((ω^(2 ))/(c^2))μ ϵ E=0:                     (2.3.1)
The solution is (e^(i (k z - ω t))) 
where                      k=(ω/c) (Sqrt[ϵ[ω] μ])=(ω/c)n[ω]              (2.3.2)
From these solution (e^(i (k z - ω t))), how do we construct a finite pulse of our interest such as below?

Figure: Carrier, envelope, and pulse

ω=10*Pi ; k=10  π; t=0 ;
Row[{Plot[{Cos[k z -ω t],Exp[-(  z-(ω/k) t)^2]} ,{z,-3,3}, PlotStyle->{{Thickness[0.002],Hue[0.24,1,0.8,0.8]},{Thickness[0.01],Dashed, Blue}}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> {-1,1}, PlotPoints->200]
,Plot[Exp[-(  z-(ω/k) t)^2] Cos[k z -ω t] ,{z,-3,3}, PlotStyle->{ Red}, Filling-> {1-> {Axis,Hue[0.2,1,1,0.3]}}, PlotRange-> {-1,1}, PlotPoints->200]}]

Review demo: Linear superposition of discrete plane wave light beam

Demo: Linear superposition of discrete waves

2.3.1 Review: signal spectrum and power spectral density

Consider a signal: S[t], what is its power or energy spectrum (or spectral density)?
                                               P[ω]=(Underscript[Lim, T-> ∞])(1/T)(((|(∫_(-T/2)^(T/2))(e^(i ω t))S[t]dt|))^2) (if power signal)     (2.3.4a)
                                               E[ω]=(((|(∫_(-∞)^∞)(e^(i ω t))S[t]dt|))^2)      (if energy signal)            (2.3.4b)
The spectral component of the signal is:
                                                    a[ω]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(i ω t))S[t]dt                            (2.3.5)
And by FT theorem:
                                                    S[t]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)a[ω](e^(-i ω t))dt                           (2.3.6)
which is a linear combination of harmonic (e^(-i ω t)) with spectrum a[ω]

If the signal has a carrier (e^(-i (Subscript[ω, 0]) t )) with an envelope S[t]:    (e^(-i (Subscript[ω, 0]) t ))S[t] , then:
                                                  a[ω]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(i ω t))(e^(-i (Subscript[ω, 0]) t ))S[t]dt                     (2.3.7)
                                                         =(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(i (ω-(Subscript[ω, 0])) t))S[t]dt         
                                                  a[Ω]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(i  Ω t))S[t]dt                               (2.3.8a)
and for the inverse:
                                                  S[t]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(-i  Ω t))a[Ω]dt                              (2.3.8b)    
where:                                                Ω≡ω-(Subscript[ω, 0])                                                 (2.3.8c)

In other words, spectral component is simply the FT of the envelope.

Review concept: power spectral density and spectrum. Link to soundwave demo

Applying the concept in the demo of soundwave above to lightwave, we have the concept of optical spectrum.

Instruments to measure light spectrum.

(Si raman laser)
Diode laser spectra

If we have a cw laser signal that can be approximated as a pure single-frequency light wave:
                                                 E=(Subscript[E, 0]) (e^(i (k z-ω t)))
and we put the signal through a spectrum analyzer, what is its spectrum? (power spectral density). If we chop this light with a very fast chopper, what does its spectrum look like?

2.3.2 Review example: spectral decomposition of a square pulse

Consider a pure wave  (e^(i (k z -(Subscript[ω, 0]) t))) from a perfect laser. Its spectrum is δ[ω-(Subscript[ω, 0])]. Suppose we have a shutter that can open and close really fast and let the chopped off laser light pulse to pass through an optical spectrum analyzer. What kind of spectrum will we get?

Consider the following square pulse:

a=1  ; n=1.45  ; ω=2 π*14 ;
Plot[ Re[(E^(-I  ω  t)) (UnitStep[t+a/2] -UnitStep[t-a/2])/a],{t,-1,1},PlotStyle-> {{RGBColor[1,0,0], Thickness[0.005]}},ImageSize-> {600,250},
AspectRatio->0.3 ]

What is its energy spectrum? (or energy spectral density)?

ω= . ;
(1/(2 π))(∫_(-τ/2)^(τ/2))(E^(I ω t))(E^(-I  ω0  t))\[DifferentialD]t

ω0=100. ;  τ=1  ;
LogPlot[(((((sin((1/2) τ (ω-ω0)))/(π (ω-ω0)))))^2),{ω,50,150}, PlotRange-> All, GridLines-> Automatic, PlotRange-> {10^-6,1}]

Thus, the spectrum would be a Sinc^2 function and look like the above. It is no longer a delta function, which means that it is no longer strictly monochromatic: it has other spectral components.

Per Fourier transform theorem, we can reconstruct a square pulse:
                                         (∫_(-∞)^∞)((sin((1/2) τ (ω-ω0)))/(π (ω-ω0)))(e^(i (k[ω] z -ω t)))dω

FourierTransform[((sin((1/2) τ Ω))/(π Ω)), Ω, t]

τ=1 ;ω=2 π*14 ;
Plot[Re[(E^(-I  ω  t))(((1/2) (Sqrt[(π/2)]) Sign[t+(τ/2)]-(1/2) (Sqrt[(π/2)]) Sign[t-(τ/2)])/π)],{t,-1,1}, PlotStyle-> {Thick,Red}, Filling-> Axis]

This expression:           (∫_(-∞)^∞)((sin((1/2) τ (ω-ω0)))/(π (ω-ω0)))(e^(i (k[ω] z -ω t)))dω
is the linear superposition of each component (e^(i (k[ω] z -ω t))) with amplitude: ((sin((1/2) τ (ω-ω0)))/(π (ω-ω0)))

2.3.3 Spectral decomposition of a Gaussian pulse

Now we can apply this concept to our Gaussian pulse:
Let z=0. The signal is:         (e^(-(((t/τ))^2))) (e^(-i (Subscript[ω, 0]) t))                         (2.3.9)
The spectral component is:       a[ω]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(-(((t/τ))^2)))(e^(i (ω-(Subscript[ω, 0])) t))dt     (2.3.10a)
For simplicity, we define:           Ω≡ω-(Subscript[ω, 0])                    (2.3.10b)

τ=. ; t=. ; Ω=. ;
(1/(Sqrt[2 π]))(∫_(-∞)^∞)(E^(-(((t/τ))^2)))(E^(I Ω t))\[DifferentialD]t

Hence:                        a(Ω)=(1/(Sqrt[2 ]))τ (e^(-(1/4) (τ^2) (Ω^2)))      (2.3.11)
The E field is:
                           E[z;t]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)a[Ω]  (e^(i (k[ω] z - ω t))) dω          (2.3.12)
which is the linear superposition of every spectral component wave (e^(i (k[ω] z - ω t))), which satisfies the Helmholtz eq. with an associated amplitude   a(Ω)=(1/(Sqrt[2 ]))τ (e^(-(1/4) (τ^2) (Ω^2))). Hence:
                           E[z;t]=(1/(2 (Sqrt[π])))τ(e^(-i  (Subscript[ω, 0]) t))(∫_(-∞)^∞)(e^(-(1/4) (τ^2) (Ω^2))) (e^(i (k[ω]z - Ω t))) dΩ     (2.3.13)

where we replace ω=(Subscript[ω, 0])+Ω  so that we can take the carrier wave (e^(-i  (Subscript[ω, 0]) t)) outside the integral. 
What prevents us from applying straightforward inverse Fourier transform of (2.3.13) ?
It is the unknown dispersion relation k[ω]: This is the difference between what we are doing here, vs. the FDTD illustration in 2.2

We cannot integrate unless we know what k[ω] is. For simplicity we consider the case of NO dispersion and let k[ω] simple be:
                             k[ω]=(ω/c) n[ω]=(ω/c) n                      (2.3.14)
                             k[ω]z=(Subscript[ω, 0])(z/c) n+Ω(( z)/c) n                 (2.3.15)
In other words, we assume the refractive index be just a constant vs, ω. Thus, we expect that this result should give us the same results as the FDTD in Section 2.2.
The the inverse Fourier transform can be easily applied:
               E[z;t]=(1/(2 (Sqrt[π])))((τe)^(i  (Subscript[ω, 0]) ((z/c)n - t)))(∫_(-∞)^∞)(e^(-(1/4) (τ^2) (Ω^2))) (e^(i  Ω ((z/c)n - t))) dΩ     (2.3.16)

ω= .; n=.  ; t=.  ; z=.  ; τ= .  ; 
(1/(2 (Sqrt[π])))τ(E^(I  (Subscript[ω, 0]) ((z/c)n- t)))(∫_(-∞)^∞)(E^(-(1/4) (τ^2) (Ω^2))) (E^(I  Ω ((z/c)n - t))) \[DifferentialD]Ω

Hence:
                          E(z,t)=(e^(i (Subscript[ω, 0]) (((z n)/c)-t)))(e^(-((((c t-z n))^2)/((c^2) (τ^2)))))=(e^(i (Subscript[ω, 0]) (((z n)/c)-t)))(e^(-((((((z n)/c)-t))^2)/( (τ^2)))))           (2.3.15)

Another way to do it:

(1/(2 (Sqrt[π])))τ (E^(I  (Subscript[ω, 0]) ((z/c)n- t))) FourierTransform[(E^(-(1/4) (τ^2) (Ω^2))), Ω, ζ]

If we look at (2.3.15) and our “guess” solution in (2.1.1), they are the same!  But this is ONLY because we make a highly simplistic assumption. The guessing method cannot help us in realistic case. The spectral linear superposition method can. It is not this particular result, it is the approach.

Although we use Gaussian pulse throughout, the results are of general validity and can be applied to other type of pulse shapes. See example below for raised cosine pulse shape. In HW, we will deal with other shapes.

Example: Raised cosine pulse

2.4 Summary of key ideas

- For applications such as optical fiber communication, we don’t just deal with continuous carrier wave; we deal with modulated light wave
- The most common form of modulation in OFC is intensity modulation (amplitude modulation) often in the form of a short pulse. A stream of signal is made up of light packets (light pulses)
- The propagation behavior of lightwave packets is important in fiber transmission.

3. A note about dispersion: linear and quadratic dispersion

Are the above colorful phenomena of the same underlying physics as these:

The answer is NO!
They may look the same, but the physics are very different. We have learned about the soap film. It is caused by interference. We will learn about diffraction, which gives many insects their beautiful color. But the latters are not the same as the one we are studying right now.

What we study is the “rain bow” effects. It is the dispersion of light by materials, which is water in the case of rainbow.

Some very important practical applications:

Refractive index (phase velocity) vs. frequency:

Atoms and molecules in materials are not the only cause of dispersion. Structures such as waveguides also have dispersion. Waveguide dispersion revisited. 
Later on, we will see that fiber dispersion, although very small, is quite significant because of its length that is used for optical communications. It is NOT just the dispersion coefficient, but also the length (how long the light pulse travel).

In the above example, we perform the superposition by taking k[ω]=((n ω)/c) with the assumption that the index of refraction is a constant of frequency. This is what we did to integrate: 
                 (1/(2 (Sqrt[π])))(∫_(-∞)^∞)(((e^(-(1/4) (τ^2) (Ω^2))) τ) (e^(i (k z-ω t)))/.k->((n ω)/c))dω     
Except for vacuum or highly rarified media such as air, low density gases, k(ω) is never that simple in most media. It can be a complicated function of ω. But we can simplify it somewhat by expand k(ω) around the center frequency (Subscript[ω, 0]):
          k(ω)= k((Subscript[ω, 0]))+((dk)/(dω))(ω-(Subscript[ω, 0])) +(1/(2!))(((d^2)k)/(d(ω^2)))(((ω-(Subscript[ω, 0])))^2)+ ...     (3.1.1)
As we know, the inverse of  ((dk)/(dω)) is called the group velocity (GV):
                              (Subscript[v, g])=(((((dk)/(dω))))^(-1))                               (3.1.2)
The second order term (((d^2)k)/(d(ω^2)))is called the 2nd-order dispersion term, and the term 2π(((d^2)k)/(d(ω^2))) is called the dispersion coefficient. There are other definitions and names for this term, such as group velocity dispersion (abbreviated as GVD, which can also expressed in different ways and different units: (d/(df))((1/(Subscript[v, g])))=2π(((d^2)k)/(d(ω^2))), or (d/(dλ))((1/(Subscript[v, g])))=-(c/((Subscript[λ, 0])^2)) (d/(df))((1/(Subscript[v, g]))). But for calculation, all must be converted to the appropriate expression and unit), but all is proportional to the second order derivatives as we can see later.
Depending on applications, in some cases, even 3rd order terms must be considered.

We will examine the role of these terms in the temporal behavior of the propagation of light.

4. Pulse propagation in media: group velocity

4.1 Overview

Demo: Linear superposition of discrete waves - part 2 - with group velocity

4.2 Group velocity

To appreciate the meaning of the group velocity, we recall in the above example that if n[ω] is constant, then:
                      (1/(Subscript[v, g]))=((dk)/(dω))=(d/(dω))(ω/c)n =(n/c)          (4.2.1)
Then, the GV is c/n which is the same as phase velocity   (Subscript[v, p])=(ω/(k[ω]))=(c/(n[ω])) in (4.2.1)
The phase carrier and the envelope move together as the first example above. If there is no dispersion term, it's still easy to see that the pulse maintains its shape. From Eq.  (2.3.6) and (2.3.7)
                           E[z;t]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)a[Ω]  (e^(i (k[ω] z - ω t))) dω          (2.3.12)

Substitute using the above:     k(ω)= k((Subscript[ω, 0]))+((dk)/(dω))(ω-(Subscript[ω, 0])) +(1/(2!))(((d^2)k)/(d(ω^2)))(((ω-(Subscript[ω, 0])))^2)+ ...     (3.1.1)
but we will keep only the 1st order:

k=. ; ω= .;  z=. ; t=  .;
(E^(I (k z - ω t))) /. k->(Subscript[k, 0])+(ω-(Subscript[ω, 0]))/(Subscript[v, g])

(E^(I (z (((ω-(Subscript[ω, 0]))/(Subscript[v, g]))+(Subscript[k, 0]))-t ω))) /. ω-(Subscript[ω, 0])-> Ω

(E^(I (z ((Ω/(Subscript[v, g]))+(Subscript[k, 0]))-t ω))) /. ω->  Ω+(Subscript[ω, 0])

(E^(I Expand[(z ((Ω/(Subscript[v, g]))+(Subscript[k, 0]))-t ((Subscript[ω, 0])+Ω))]))

which can be written:     (e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0])))) (e^(i   Ω ((z/(Subscript[v, g]))-t)))            (4.2.2)

The first factor has no Ω dependence and can be taken outside the integral in (2.3.12):
                   E[z;t]=(1/(Sqrt[2 π]))(e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0]))))(∫_(-∞)^∞)a[Ω]  (e^(i   Ω ((z/(Subscript[v, g]))-t))) dΩ       (4.2.3)
But recall from the above that:
                      a[Ω]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(i  Ω t))S[t]dt                               (2.3.8a)
and for the inverse:
                       S[t]=(1/(Sqrt[2 π]))(∫_(-∞)^∞)(e^(-i  Ω t))a[Ω]dt                              (2.3.8b)    
Hence:         E[z;t]=(e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0]))))S[t-(z/(Subscript[v, g]))]                              (4.2.4)
In other words, the pulse simply moves with 2 different speeds:
     - the carrier (e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0])))) moves with phase velocity ((Subscript[ω, 0])/(Subscript[k, 0]))
     - the envelope moves with group velocity (Subscript[v, g])=(((((dk)/(dω))))^(-1))

For a Gaussian pulse:
                   E[z;t]=(e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0]))))(e^(-(((((z/(Subscript[v, g]))-t ))^2)/(τ^2))))                              (4.2.5)   
We can explicitly perform Fourier transform if we wish, and get exactly the same result:

Ω=. ; τ= .;  z=. ; t= .; 
(1/(2 (Sqrt[π])))τ(∫_(-∞)^∞)(E^(-(1/4) (τ^2) (Ω^2)))  (E^(I   Ω ((z/(Subscript[v, g]))-t))) \[DifferentialD]Ω

which gives the same as above:
                    E[z;t]= (e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0]))))(e^(-(((((z/(Subscript[v, g]))-t ))^2)/(τ^2))))       (4.2.5)

It is obvious that (e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0])))) is the carrier term, moving at the phase velocity (Subscript[v, p])=(ω/(k[ω]))=(c/(n[ω])). The second term (e^(-(((((z/(Subscript[v, g]))-t ))^2)/(τ^2)))) is obviously the Gaussian envelope moving at the group velocity (Subscript[v, g]).

Note that E[z;t]= (e^(i ((Subscript[k, 0]) z-t (Subscript[ω, 0]))))(e^(-(((((z/(Subscript[v, g]))-t ))^2)/(τ^2)))) cannot satisfy the “typical” wave equation: 
                                              (∇^2)E  -((μ ϵ)/(c^2))((∂^2)/(∂ (t^2)))E =0 
because the 2 terms have different propagation speeds as mentioned. This is an example why the generic wave equation as we often see and discussed in LectSet 1 does not apply because of the frequency-dependence response (and causality in time-response to impulse stimulus) of the medium.

When we study more into material dispersion properties, we will see Kramers-Kronig relations implicitly contains the time-dependence and causality nature of medium response. An explicit example is the Lorentz oscillator model in which the “damping” time of the electron appears in the dispersion of the dielectric constant.

Demo: Group velocity concept illustration 1

Demo: Group velocity concept illustration 2 - carrier and envelope

Notice that the envelope travels slower than the carrier. That  is because the envelope moves at the group velocity which is smaller than the carrier or phase velocity.

The most important implication is that we can only transmit information at the speed of GV, not at the phase velocity. This is the same point often made when, near a sharp dispersion point of an absorption feature for example, where the phase index can be less than 1, the phase velocity can be larger than the speed of light in vacuum c, but that is not in any violation of Relativity, because the GV is always < c.