Thread: Mathematica - 2D Discontinuous plotting

Dear MHF members,

I need to plot a figure as in the following figure.


I use the code

Code:

Plot[{t, t - 2 \[Pi]}, {t, 0, 8 \[Pi]}, PlotStyle -> {Red, Blue}, 
 Ticks -> {Table[j \[Pi], {j, 0, 8}], Table[j \[Pi], {j, -2,  8}]}]

to get the following image.


But I don't know how to remove the segments $\displaystyle (\pi,2\pi)$, $\displaystyle (3\pi,4\pi)$, $\displaystyle (5\pi,6\pi)$ and $\displaystyle (7\pi,8\pi)$ from the $\displaystyle x$-axis and remove the same parts of the functions $\displaystyle t$ and $\displaystyle t-2\pi$.

Thanks.
bkarpuz

Google suggests something like this might work:

Code:

p1 := Plot[ 2x , {x, -1, 1}];
p2 := Plot[2x, {x, 5, 10}];
Show[p1, p2];

Originally Posted by SpringFan25

Google suggests something like this might work:

Code:

p1 := Plot[ 2x , {x, -1, 1}];
p2 := Plot[2x, {x, 5, 10}];
Show[p1, p2];

Thanks SpringFan25.

The following works for me.
But still don't know how to remove the intervals from the $\displaystyle x$-axis.

Code:

Show[Table[{Plot[t, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Red], 
   Plot[t-2 \[Pi], {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Blue], 
   Plot[0, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> {Black, Thick}]}, {j, 0, 3}], 
 Ticks -> {Table[j \[Pi], {j, 0, 7}], Table[j \[Pi], {j, -2, 7}]}, 
 PlotRange -> All]

this might be useful, it show code doing what you want. I ahve no idea how it works though!

http://www.st-andrews.ac.uk/~pl10/c/djmpark/Assets/C3527823/BrokenAxisGraph.pdf

Well, a really cheesy way to get the desired output would be something like this:

Code:

Show[Table[{Plot[t, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Red], 
   Plot[t - 2 \[Pi], {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Blue], 
   Plot[0, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> {Black}]}, {j, 0, 3}], 
 Ticks -> {Table[j \[Pi], {j, 0, 7}], Table[j \[Pi], {j, -2, 7}]}, 
 PlotRange -> {{0, 7 \[Pi]}, {-2 \[Pi], 7 \[Pi]}}, AspectRatio -> 7/9,
 Epilog -> 
  Inset[Show[
    Table[Plot[0, {x, j \[Pi], j \[Pi] + \[Pi]}, 
      PlotRange -> {{0, 7 \[Pi]}, {-2 \[Pi], 7 \[Pi]}}, Axes -> False,
       PlotStyle -> {{Thickness[.006], White, EdgeForm[]}}, 
      AspectRatio -> 7/9], {j, 1, 7, 2}]], {0, 0}, {0, 0}, {7 \[Pi], 
    Automatic}]
 ]

I.e., draw white lines over the parts of the axis you don't want. If you use it to produce a vector graphic to include elsewhere, this might be a problem, though.

Perhaps you could try to play around with the Inset[] function; I have a feeling that it should be possible to create a table of plots for each interval, and then add them to a plot with not axes using Inset. The positioning might be a bit tricky, and you might have to pay attention to AspectRatio.

Originally Posted by Zarathustra

Well, a really cheesy way to get the desired output would be something like this:

Code:

Show[Table[{Plot[t, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Red], 
   Plot[t - 2 \[Pi], {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> Blue], 
   Plot[0, {t, 2 j \[Pi], (2 j + 1) \[Pi]}, 
    PlotStyle -> {Black}]}, {j, 0, 3}], 
 Ticks -> {Table[j \[Pi], {j, 0, 7}], Table[j \[Pi], {j, -2, 7}]}, 
 PlotRange -> {{0, 7 \[Pi]}, {-2 \[Pi], 7 \[Pi]}}, AspectRatio -> 7/9,
 Epilog -> 
  Inset[Show[
    Table[Plot[0, {x, j \[Pi], j \[Pi] + \[Pi]}, 
      PlotRange -> {{0, 7 \[Pi]}, {-2 \[Pi], 7 \[Pi]}}, Axes -> False,
       PlotStyle -> {{Thickness[.006], White, EdgeForm[]}}, 
      AspectRatio -> 7/9], {j, 1, 7, 2}]], {0, 0}, {0, 0}, {7 \[Pi], 
    Automatic}]
 ]

I.e., draw white lines over the parts of the axis you don't want. If you use it to produce a vector graphic to include elsewhere, this might be a problem, though.

Perhaps you could try to play around with the Inset[] function; I have a feeling that it should be possible to create a table of plots for each interval, and then add them to a plot with not axes using Inset. The positioning might be a bit tricky, and you might have to pay attention to AspectRatio.

Actually, this is what I was looking for. I really appreciate your help Zarathustra. I will read and try to understand it. Many many thanks.
bkarpuz