0
$\begingroup$

Plotting Extended Zone Scheme:

i want to know plot this -3pi to 3pi how can iContourPlot[ Cos[k] == Cos[5.12 Sqrt[e]] + 5*Sinc[5.12 Sqrt[e]], {k, 0, π}, {e, 0.1, 1.52}, FrameLabel -> {"k[\!\(\*SuperscriptBox[\(nm\), \(-1\)]\)]", "Energy[eV]"}, PlotLabel -> Cos[k] == Cos[5.12 Sqrt[e]] + 5*Sinc[5.12 Sqrt[e]]]

i want to know plot this -3pi to 3pi how can i

$\endgroup$
  • 1
    $\begingroup$ The code you've posted does not correspond to the image. Could you tell us which parameter range do you want to alter and where are you stuck? $\endgroup$ – Kuba May 29 '17 at 12:22
  • $\begingroup$ i want plot k range [-3pi to 3pi] vs e $\endgroup$ – kim May 29 '17 at 13:00
  • $\begingroup$ who anser k range [0 to 3pi] as ContourPlot[ Cos[k] + I Floor[k/[Pi]] == Cos[5.12 Sqrt[e]] + 5 Sinc[5.12 Sqrt[e]] + I Floor[5.12 Sqrt[e]/[Pi]], {k, 0, 3 [Pi]}, {e, 0, 3.5}, GridLines -> {({#1 [Pi], Dashed} &) /@ Range[3], None}, FrameLabel -> {SequenceForm[k, " [\!(*SuperscriptBox[(nm), (-1)])]"], "Energy [eV]"}, PlotLabel -> Cos[k] == Cos[5.12 Sqrt[e]] + 5 Sinc[5.12 Sqrt[e]]] $\endgroup$ – kim May 29 '17 at 13:01
1
$\begingroup$

Maybe you can use RegionFunction:

ContourPlot[
    Cos[k] == Cos[5.12 Sqrt[e]] + 5 Sinc[5.12 Sqrt[e]],
    {k,0,3 Pi}, {e,0.1,3.5},
    RegionFunction -> (
        (IntervalMemberQ[Interval[Pi^2 (Quotient[#,Pi]+{0,1})^2/5.12^2], #2])&
    )
]

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.