Out of bounds, don't know why and unit step function. Additionally what is unit-ramp function?

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cdburnet 2015-4-26

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https://ww2.mathworks.cn/matlabcentral/answers/213672-out-of-bounds-don-t-know-why-and-unit-step-function-additionally-what-is-unit-ramp-function

回答： Philip Caplan 2015-4-27

So I'm receiving the following error and I don't know why: "Attempted to access ap1(2); index out of bounds because numel(ap1)=1." I could swear that everything should work...

Outside of that, I'm supposed to find the unit-step response and unit-ramp response. Now I think I can achieve that with the heaviside function but there are a variety of ways. From my understanding a unit-step function basically changes input into binary, except for 0. So all positive numbers will be 1s, all negatives will be 0s, and any zeros will be 0.5.

In the problem I've been provided, I just have "u" provide to me to represent the unit-step and unit-ramp functions. However, I wasn't given a timespan. I've omitted it just to try to troubleshoot this error.

t1=linspace(0,10,100);
  function dap1=P1(t1,ap1)
      dap1=zeros(3,1);
      dap1=[-5 -25 -5; 1 0 0; 0 1 0].*[ap1(1);ap1(2);ap1(3)]+[1;0;0] %*heaviside(t1)
  end
[~,ap1]=ode45(@P1,t1,1);
y1=[0 25 5].*[ap1(1);ap1(2);ap1(3)

In past problems this was achieved via a for loop as shown below with zb and zb(i). To me that's strange because in the situation with the for loop I interpret this as assuming that the unit step function is always one.

%% DOWN HERE for the loop I'm referring to VVVV

 function Pother
        L=2; %variables
        g=9.8; %variables
        thet0=0.2; %variables
        dthet0=0; %variables
        tspan2=linspace(0,10,100); %time variation
        zb=[10,.2]; %function b(t) where b(t)=10u(t) and 0.2u(t)
%d2thet/dt2=(-g*sin(thet)-b(t)*cos(thet))/L
 L=2; %variables
        g=9.8; %variables
        thet0=0.2; %variables
        dthet0=0; %variables
        tspan2=linspace(0,10,100); %time variation
        zb=[10,.2]; %function b(t)
        function Tprimea=TA(t,za,zb,L) %Defining a matrix Tprimea under TA
            Tprimea=(zeros(size(za)));
            Tprimea(1)=za(2);
            Tprimea(2)=(-g*sin(za(1))-zb*cos(za(1)))/L;
        end
      for i=1:2 %The i changes the function b(t): represented here by zb
        [~,za]=ode45(@TA,tspan2,[thet0;dthet0],[],zb(i),L); %ode solver
        subplot(1,2,i) %This changes the placement of each new graph
        plot(tspan2,za(:,1)) %time vs theta
      end
    end

Additionally, what is unit-ramp function? I'm beginning to wrap my head around the step function but I am completely lost with the ramp.

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Answer 1

Philip Caplan 2015-4-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/213672-out-of-bounds-don-t-know-why-and-unit-step-function-additionally-what-is-unit-ramp-function#answer_176887

在 MATLAB Online 中打开

When you called "ode45", you were only passing in a scalar value for the initial condition. However, it looks like your "odefun" expects a 3x1 solution vector at each time step and returns a 3x1 "dydt" vector. I don't know the details of the ODEs you are solving but the modified code below executes without errors:

function testODE()
t1=linspace(0,10,100);
[~,ap1] = ode45(@P1,t1,ones(3,1));
y1 = [0 25 5]*[ap1(1);ap1(2);ap1(3)]
end
function dap1 = P1(t1,ap1)
  dap1 = [-5 -25 -5; 1 0 0; 0 1 0]*[ap1(1);ap1(2);ap1(3)]+[1;0;0]; %*heaviside(t1)
end