Matlab Opens smithplot in new figure

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Nils
Nils am 28 Jul. 2023
Kommentiert: dpb am 28 Jul. 2023
Hi all,
I want to design an Application with the new App Designer that uses a smith chart. But when I would like to plot the smith chart on the already Defined UI.Axes Object, but when I try to plot it, it opens in a new Figure. I already tried to use the axes command, but it doesn't help. How can I fix this behavior
Thank you very much
d = dipole;
freq = linspace(60e6,90e6,200);
s = sparameters(d,freq);
axes(app.UIAxes)
hg = smithplot(s);

Akzeptierte Antwort

Kevin Holly
Kevin Holly am 28 Jul. 2023
For the smithplot function, you need to change the Parent to the app.UIAxes
d = dipole;
freq = linspace(60e6,90e6,200);
s = sparameters(d,freq);
hg = smithplot(s,'Parent',app.UIAxes);

Weitere Antworten (1)

dpb
dpb am 28 Jul. 2023
Bearbeitet: dpb am 28 Jul. 2023
d = dipole;
freq = linspace(60e6,90e6,200);
s = sparameters(d,freq);
Error using matlab.internal.lang.capability.Capability.require
Support for Java user interfaces is required, which is not available on this platform.

Error in waitbar (line 46)
Capability.require(Capability.Swing);

Error in em.EmStructures/analyze

Error in em.PortAnalysis/impedance (line 164)
status = analyze(obj,freq);

Error in em.SharedPortAnalysis/protectedsparameters (line 98)
Z = impedance(obj,freq);

Error in em.PortAnalysis/sparameters (line 43)
S = protectedsparameters(obj,freq,z0);
hUIF=uifigure; % make a uifigure, not figure
app.UIAxes=axes(hUIF); % put the axes inside it; is regular figure axes otherwise
hg = smithplot(app.UIAxes,s); % put the plot in this axes; otherwise it defaults to plain
Well, can't do that in the available environment, but the key is you have to tell it to use your app axes; by default everything goes to the command window graphics environment; just because the code is in the app, that doesn't change that behavior, the plotting functions aren't aware of that context. Whether they could be made to be so, I don't; interesting Q? for possible enhancement, maybe???
  2 Kommentare
Les Beckham
Les Beckham am 28 Jul. 2023
I believe that "uaxes" should be "uiaxes" in the next to last line of your code example.
dpb
dpb am 28 Jul. 2023
It died on something in the instrumentation toolbox stuff before it got there so didn't notice.
It is a typo, indeed, but I intended it to be just axes() with the uifigure as parent; that's what I've always done; I see that uiaxes() will create the uifigure so that's better way to go...thanks.

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