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Please help me to solve this problem below.

I want to run this coding based on the time given by the system.

Given,

A = ( { {E^(I [Beta]1 + I [Beta]3) Cos[[Beta]2t], E^(I [Beta]1 - I [Beta]3) Sin[[Beta]2t]}, {-E^(-I [Beta]1 + I [Beta]3) Sin[[Beta]2t], E^(-I [Beta]1 - I [Beta]3) Cos[[Beta]2t]} } ) /. {[Beta]1 -> 0, [Beta]3 -> 0} [Beta]2 = Pi; For[i = 1, i < 11, i++; t++, t = t + TimeUsed[]; Print[A]].

My problem are: 1. I cannot run this coding since I only want β2 influence by time. I want to look the changes of β2 only on time(For example start with t=0s until t=10s). 2. It is true that I used command TimeUsed[] for the time?(I want the time come from the computer). 3. How can I plot the graph from this command TimeUsed[]. I can't make the graph.

Thank you so much for helping me.

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    $\begingroup$ A = {{Exp[I* β1 + I* β3] Cos[β2* t], Exp[I* β1 - I* β3] Sin[β2* t]}, {-Exp[-I* β1 + I* β3] Sin[β2* t], Exp[-I* β1 - I* β3] Cos[β2* t]}}; β1 = 0; β2 = Pi; β3 = 0; t = 0; tu = Table[ t++; t = t + TimeUsed[]; Print[A]; t, {i, 1, 10}] and then ListPlot[tu] but you need to explain more clearly what you mean by "only β2 dependent on time" $\endgroup$
    – Bill
    Commented Jun 5, 2016 at 17:56
  • $\begingroup$ Tq Bill, what I mean by 'only β2 dependent on time' is β2 is varies with time. For example, 1st case, β2=Pi. 2nd case, β2=2Pi and in both cases, the value of β1 and β3 remain constant. that mean the value β1 and β3 not varies with time. $\endgroup$
    – munirah
    Commented Jun 14, 2016 at 3:14
  • $\begingroup$ I did the coding suggest by u but why the value of identity, I cannot be evaluate? i get the output as: {{-0.8053078857111251 + 0.*I, 0.5928568201610549 + 0.*I}, {-0.5928568201610549 + 0.*I, -0.8053078857111251 + 0.*I}} $\endgroup$
    – munirah
    Commented Jun 14, 2016 at 3:23

1 Answer 1

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I suppose I know what you mean. If you actually want β2 t changed while β2 keeps the constant Pi, I guess the following two codes can solve your problem:

β1 = 0.;
β2 = Pi;
β3 = 0.;
A={{Exp[I β1 + I β3] Cos[β2 t], Exp[I β1 - I β3] Sin[β2 t]}, {-Exp[-I β1 + I β3] Sin[β2 t], Exp[-I β1 - I β3] Cos[β2 t]}};
  1. Using Animate (or Manipulate) while setting the setting bar's running speed the same as your computer time's running speed:

    Animate[Evaluate@MatrixForm@Chop@A, {t, 0, 1}, DefaultDuration -> 1]
    
  2. Using Dynamic:

    Dynamic[With[{t = FractionalPart@SessionTime[]}, Evaluate@Chop@A], UpdateInterval -> .1]
    
  3. I suppose you will need a plot of A as a function of t, so the following code will solve this problem:

    Grid[MapIndexed[Plot[#, {t, 0, 1}, AxesLabel -> {"t", "Value at " <> ToString /@ #2}] &, A, {2}]]
    

I will further update this post if you can make clearer what you want to do.

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  • $\begingroup$ Simply change the DefaultDuration will do the job. In the second solution, use FractionalPart[SessionTime[]/t0] $\endgroup$
    – Wjx
    Commented Jun 14, 2016 at 3:21
  • $\begingroup$ Tqvm Wjx for replying. Sorry, this solution is given by you? , It is mean DefaultDuration will run the coding? In second solution why add /t0 at the Session time $\endgroup$
    – munirah
    Commented Jun 14, 2016 at 3:30
  • $\begingroup$ Thank you Wjx. I mention wrong person. I have question;1. do the coding and run it. but the value of Identity cannot be valued. For example, I do the Dynamic and get the result as: Dynamic[With[{t = FractionalPart[SessionTime[]]}, Evaluate[MatrixForm[A]]], ImageSizeCache -> {261., {10., 17.}}, UpdateInterval -> 0.1]. 2. I also wonder about the t interval; In 1(using animate), the Default Duration is increased by 1. it is mean by 1 second? How can I fix the duration? For example, I want to fix from t=0s until t=10s. $\endgroup$
    – munirah
    Commented Jun 14, 2016 at 3:36
  • $\begingroup$ I suppose that's my solution...... DefaultDuration measures the time the dynamic will run for a time by default. /10 means that the time it costs to change fractional part from 0 to 1 will increase to 10 times the original. $\endgroup$
    – Wjx
    Commented Jun 14, 2016 at 3:38
  • $\begingroup$ I'm very sorry again Wjx, I mention wrong person for the solution. I read the wrong name. $\endgroup$
    – munirah
    Commented Jun 14, 2016 at 3:39

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