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I am trying to plot the function as the figure below. However, I can't think of a way to do it.

Maybe it is easy to you but I am just a newbie. Also I have searched the help section but I couldn't find the result.

Hope some could guide me to plot this. Thank you.

enter image description here

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One way of doing this is by defining a function which relates $\frac{\omega_2}{\omega_{UG}}$ to $\frac{\omega_1}{\omega_{UG}}$. One simple implementation can be written as

 fun[xin_?NumberQ,phase_?NumberQ]:=
 Quiet@FindRoot[phase==90-(ArcTan[1/z]+ArcTan[1/xin])*180/\[Pi],{z,2}][[1,2]]

Here we are setting up a function, for a particular phase, which will numerically solve for $f\left(\frac{\omega_2}{\omega_{UG}},\theta\right)=\frac{\omega_2}{\omega_{UG}}$.

We can plot our function directly with

Plot[{fun[x,50],fun[x,60],fun[x,65],fun[x,70]},{x,0,10},
PlotRange->{Automatic,{0,10}},
PlotLegends->{"50\[Degree]","60\[Degree]","65\[Degree]","70\[Degree]"},
AxesLabel->{"Subscript[\[Omega], 2]/Subscript[\[Omega], UG]",
"Subscript[\[Omega], 1]/Subscript[\[Omega], UG]"}]

enter image description here

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ContourPlot[Evaluate[90 \[Degree] - (ArcTan[1/x ] + ArcTan[1/y ]) == #*
Pi/180 & /@ {50, 60, 65, 70}], {x, 0, 10}, {y, 0, 10}, PlotLegends -> {"50\[Degree]", "60\[Degree]", "65\[Degree]", "70\[Degree]"}]

Click here for a detailed description ContourPlot.

enter image description here

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