Single value after integration instead of matrix

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)

[Torsten]

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]

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]

What is z ? As written, MATLAB expects a single, previously defined value for it.

[Özgür Alaydin]

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]

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]

Dear Torsten

Thanks a lot.

This is exactly what i want. I have learnt one more command (numel).

Regards.

Answer 1

[Özgür Alaydin]

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