Why I don't get a curve plot by adding term2 to phi?

Hey,

I have an issue where I am stucked for several days. I have the following code:

to run it you need to have the attached matlab file (matlab) on your computer. The issue I am encountering arise when I add 'term2' to 'phi' expression. Before adding it when I plot real and imaginary part of 'exp(-1i*phi)' I have a nice curve which seems reasonable. However, when I add 'term2' to the expression of 'phi' I end up with a plot which is mostly like a scatter thing rather than a curve which should not happen. I want to use 'exp(-1i*phi)' in another expression and having such plot when adding 'term2' to 'phi' is affecting the result badly. I am not sure if this issue originates from somewhere in the code or something else. I have attached the figures I get for real and imaginary part of 'exp(-1i*phi)' in this two situations. The two above ones is for having 'term2' and the two below ones is when removing 'term2' from 'phi' expression. I really appreciate your comments.

L= 0.00435001414404276;

c= 2.99792458E8;

sigma = 2.7E6;

center = 24075189.2672552;

numGaussians = 9;

CombF = c/(793.3863E-9);

Start = CombF+700E6;

End = CombF+900E6;

nu = linspace(Start-60E6, End+90E6, 12000);

gaussEqn1 = zeros(1, length(nu));

for k = 1 : numGaussians

if k==1

b = Start;

else

b = Start+(center * (k-1));

end

thisGaussian1 = 1*exp(-((nu-b).^2)/(2.*(sigma.^2)));

% Add into accumulator array:

gaussEqn1 = gaussEqn1 + thisGaussian1;

dcomb1 = gaussEqn1;

end

load matlab.mat dconst dY Real_nalpha

d = dY - dconst + dcomb1;

dnu =nu(2)-nu(1);

Dif=diff(Real_nalpha)/dnu;

nu = nu(1:length(nu)-1);

Real_nalpha = Real_nalpha(1:length(Real_nalpha)-1);

d = d(1:length(d)-1);

term2=(nu).*Dif;

phi=(4*pi.*(Real_nalpha+term2).*nu*L)/c;

plot(nu,real(exp(-1i*phi)))

hold on;

plot(nu,imag(exp(-1i*phi)))

which shows real part of 'exp(-1i*phi)' when adding term2 which shows imaginary art of 'exp(-1i*phi)' when adding term2

which shows real part of 'exp(-1i*phi)' when removing term2

which shows imaginary art of 'exp(-1i*phi)' when removing term2

I am not sure why I have the two above curves for the real and imaginary part of 'exp(-1i*phi)' when adding term2 in the expression for 'phi'.

Answer 1

[William Rose]

@Shaily_T,

Please post a question in the simplest possible form that will allow others to assist you.

In this case, you can make a matlab.mat file with the five variables you need. I did that, and I attached it. You could do likewise.

Then you can illustrate your unexpected result as follows:

load matlab.mat   % load c, L, nu, Real_nalpha, term2
phi1=(4*pi.*(Real_nalpha).*nu*L)/c;         % compute phi1
phi2=(4*pi.*(Real_nalpha+term2).*nu*L)/c;   % compute phi2

Plot real and imaginary parts:

subplot(2,2,1); plot(nu,real(exp(-1i*phi1))); ylabel('Real'); title('Without term2')
subplot(2,2,3); plot(nu,imag(exp(-1i*phi1))); ylabel('Imag')
subplot(2,2,2); plot(nu,real(exp(-1i*phi2))); ylabel('Real'); title('With term2')
subplot(2,2,4); plot(nu,imag(exp(-1i*phi2))); ylabel('Imag')

You are suprised that the plots on the right are different from the plots on the left. They are different because term2 is a 1x11999 vector whose values are not constant. You used many lines of code to calculate term2. It is not surprising that term2 might vary. That variation makes phi2 vary, which produces the observed result.

If you are suprised that term2 changes, then review the formulas you used to calculate term2. Why would you NOT expect term2 to vary?

[Shaily_T]

@William Rose

Hi, Thanks for your response and guidance on how to post the question in the best way.

So, you think the code itself and the element wise multiplication don't appear to have an issue? I think variation of "term2" is fine but a variation which result in a curve similar to "exp(-1i*phi1)" with some changes and not something similar to the plot of "exp(-1i*phi2)" which is a scatter like plot. The reason I don't expect to see this kind of plot is I am using "exp(-1i*phi2)" in another equation and I have data for the result of that equation. When I use "exp(-1i*phi2)" in that equation the result doesn't look similar to the experimental data I have and this originates from the issue of getting kind of scatter plot for "exp(-1i*phi2)".

[William Rose]

@Shaily_T,

You're welcome. Thank you for being a good sport. I admit I get cranky sometimes.

The code that generates term2 is complicated. I don't know if it is doing what you want. But even a very simple term2 can produce plots like the ones you have obtained. Consider the following:

load matlab.mat   % load c, L, nu, Real_nalpha, term2

Instead of using term2 from the file, I will make a very simple term2, a straight line:

term2=1:length(Real_nalpha);
phi1=(4*pi.*(Real_nalpha).*nu*L)/c;         % compute phi1
phi2=(4*pi.*(Real_nalpha+term2).*nu*L)/c;   % compute phi2

Plot real and imaginary parts:

subplot(2,2,1); plot(nu,real(exp(-1i*phi1))); ylabel('Real'); title('Without term2')
subplot(2,2,3); plot(nu,imag(exp(-1i*phi1))); ylabel('Imag')
subplot(2,2,2); plot(nu,real(exp(-1i*phi2))); ylabel('Real'); title('With term2')
subplot(2,2,4); plot(nu,imag(exp(-1i*phi2))); ylabel('Imag')

Interesting. In the example above, term2=[1,2,3,...] This demonstrates that the addition of a simple, linearly increasing, term2 can make your plots look totally different.

[Shaily_T]

@William Rose No worries! I appreciate your time!

Yes, you are right. Indeed it is a nonlinear phenomena and really complex. With all those lines I've written in the first posted code I am trying to calculate "term2" in the same way as the obtained equation for "phi". I was thinking maybe there is an issue with the way I am writing the code and maybe it doesn't completely reflect the equation. Because of the data I have for the final quantity which uses "exp(-1i*phi)", I am sure the "exp(-1i*phi)" should not look like the scatter like plots I see now. I am sure there is not a problem with the calculation of "Real_nalpha". But I cannot figure out where is the issue.

[William Rose]

@Shaily_T

You want to plot where

and term2 is supposed to be .

Are you confident that the equation above (which involves nu squared in the second term) is correct?

The K-K relation usualy specifies the imaginary part, given the real part, or vice versa, but I do not see the imaginary number i in your equation.

You said you trust your calculation for n_r., and I assume your vector ν is correct.

MOre later, must go.

[William Rose]

@Shaily_T,

[edit: Added a - sign in front of s*n_i in equaiton below. The - sign indicates attenuation when s is >0.]

I am guessing that your equations have to do with the phase ϕof a light wave after it travels a distance 2L through a medium with index of refraction n(nu). For a non-dispersive medium (i.e. n=constant), n is real, and a wave that starts with amplitude 1 will, after a distance 2L, be represented by , where , and s is a real number>0,. If the index of refraction, n, is a function of nu, then n will be complex. The real and imaginary parts of n(nu) are related by the K-K equations, also known as the HIlbert transform. Now the wave is represented by , where . Now it is clear that the imaginary part causes attenuation, because , where the first term, involving , is a real number between 0 and 1 and indicates attenuation.

You wrote that term2 is derived from the real part, using the K-K relationship. But if that is true, then it should ne the imaginary part of n. But according to your equaiton for the phase, term2 is not the imaginary part of n. Is that a possible error?

I am out of my league here. Maybe my understanding of phase angles and index of refraction is wrong.

[William Rose]

@Shaily_T,

You said "I am sure there is not a problem with the calculation of "Real_nalpha"." You provided Real_nalpha in the file matlab.mat. I renamed it nRe, for brevity. You also provided a formula for term2, and you said the formula comes from the K-K relations. Therefore I computed nIm from nRe, using the formula you gave:

nIm=nu(2:end).*diff(nRe)./diff(nu);

This calculation is much easier than the many lines of code you used to compute term2. The resulting vector nIm matches the vector term2 in the file matlab.mat. See plot below. This confirms that term2 and nRe are related by the formula which you gave.

Notice that the horizontal axis is frequency in MHz, relative to the base frequency of the range shown. The base frequency is 378 THz, which corresponds to about 0.8 um wavelength in vacuum, i.e. near infrared, i.e. slightly longer wavelength than visible light. The values for term2=nIm range from -5000 to +5000. These correspond to attenuation factors in the equation . Values of nIm=-5000 and nIm=+5000 are very large amplification and attenuation factors, respectively. Are you confident in the equation you gave for term2? Could there be a constant that you forgot?