Single value after integration instead of matrix

2022 2 24
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更新时间：2022 3 5
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Dear all,
I have a code as given below.
I want to integratethis function and in the end i want to have numeric matrix. But when i run the code i am getting single value. What i want is having a one-dimensional matrix, each row should be calculated from the integration.
When i run the code i am getting "0" as a result.
Normally z is step function with a intensity 0.228. Middle of z is 0 it gives a function like ;
In the end i assign z -> alfa*sin(wt) and want to integrate over t. I should have a matrix which should give sligtly different shape then given above (bended side trough the middle).
Please help....
Vb1=0.228; dz=1E-11;
Ltot = 20e-9;
z=-Ltot/2:dz:Ltot/2;
L = 10e-9; alfa=5e-9; w=1E+14; T=2*pi/w;
t= 0:T/(length(z)-1):T;
z =z + alfa.*sin(w.*t);
Vb =@(t) ((z<-L/2).*(Vb1) + (z>-L/2).*(Vb.*(z./(2*Ltot)+0.25)) + (z>L/2).*(Vb1));
V0 = (1/T).*integral(Vb,0,T)

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Torsten
Torsten 2022-2-24
  1. The definition z =@(z,t) (z + alfa.*sin(w.*t)); will nowhere be used.
  2. Vb =@(z,t) ((z<-L/2).*(Vb1) + (z>-L/2).*0 + (z>L/2).*(Vb1)); can't be used as a function handle for "integral" since "integral" needs a function with only one input argument, not two.
  3. The spacings z=-Ltot/2:dz:Ltot/2; and t= 0:T/(length(z)-1):T; won't have any effect.
We still don't know what the function is you are trying to integrate.
Özgür Alaydin
Özgür Alaydin 2022-2-24
Dear Torsten,
We can change the function as Vb =@(t) ((z<-L/2).*(Vb1) + (z>-L/2).*(Vb.*(z./(2*Ltot)+0.25)) + (z>L/2).*(Vb1));
Vb =@(t) ((z<-L/2).*(Vb1) + (z>-L/2).*(Vb.*(z./(2*Ltot)+0.25)) + (z>L/2).*(Vb1));
The previous one is the simplest one.
As you ask, when i change the function, it is givinf error too.
Torsten
Torsten 2022-2-24
What is z ? As written, MATLAB expects a single, previously defined value for it.
Özgür Alaydin
Özgür Alaydin 2022-2-24
编辑：Özgür Alaydin 2022-2-24
z is as written. It is matrix. i assign z as z+sin(wt).
So the function Vb is z and t dependent. I need to integrate for t but V0 must be matrix same size with z.
Torsten
Torsten 2022-2-24
编辑：Torsten 2022-2-24
If you set z = z + alfa.*sin(w.*t), z is an array of values - the integration variable t has disappeared.
So Vb does no longer depend on t.
Maybe you want this:
Vb1=0.228; dz=1E-11;
Ltot = 20e-9;
z=-Ltot/2:dz:Ltot/2;
L = 10e-9; alfa=5e-9; w=1E+14; T=2*pi/w;
for i=1:numel(z)
fz =@(t) z(i) + alfa.*sin(w.*t);
Vb = @(t) ((fz(t)<-L/2).*(Vb1) + (fz(t)>-L/2).*(Vb1.*(fz(t)./(2*Ltot)+0.25)) + (fz(t)>L/2).*(Vb1));
V0(i) = (1/T).*integral(Vb,0,T,'ArrayValued',true)
end
Özgür Alaydin
Özgür Alaydin 2022-2-25
Dear Torsten
Thanks a lot.
This is exactly what i want. I have learnt one more command (numel).
Regards.

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Özgür Alaydin
Özgür Alaydin 2022-3-5

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Vb1=0.228; dz=1E-11;
Ltot = 20e-9;
z=-Ltot/2:dz:Ltot/2;
L = 10e-9; alfa=5e-9; w=1E+14; T=2*pi/w;
for i=1:numel(z)
fz =@(t) z(i) + alfa.*sin(w.*t);
Vb = @(t) ((fz(t)<-L/2).*(Vb1) + (fz(t)>-L/2).*(Vb1.*(fz(t)./(2*Ltot)+0.25)) + (fz(t)>L/2).*(Vb1));
V0(i) = (1/T).*integral(Vb,0,T,'ArrayValued',true)
end

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