Advice using function titles with abs command and plotting with function command

Hi all, continuing my code for a fourier series assignment and have hit a brick wall surrounding the function handles and the abs command. Since my last question i have changed the layout of the code and created a single function from the piecewise function i originally had. However, adding this in seperately it was plotting fine (apart from a few data cursor errors when moving the plot(error updating pointdatatip), but then adding the second fplot line stops the first from plotting. Additionally, the errors i am recieving are exclusively related to the first fplot line.

My code is below:

clear
ft=@f;
w=2;
T=(2*pi)/w;

hold on
fplot(ft,[-pi,pi])

%evaluating C0
c0=(1/T)*integral(ft,-T/2,0)+(1/T)*integral(ft,0,T/2);

%evaluating Cn values up to n=10 from n=1
for n = 1:10
    f3=@(t) ((4+t)/2).*exp(-1j*n*w*t);
    f4=@(t) ((2-t).*cos(2*t)).*exp(-1j*n*w*t);
    c(n,1)=(1/T)*integral(f3,-T/2,0)+(1/T)*integral(f4,0,T/2);
end 

%turning values n=1:10 to conjugates
cc=conj(c);

%series addition for k=-10:10
k=(1:10).';
s=@(t) c0+sum(c(k).*exp(1j*k*w*t),1)+sum(cc(k).*exp(1j*-1*k*w*t),1);

fplot(s,[-pi,pi],'r')

%evaluating the average energy of the signal
f5=@(t) abs((4+t)/2).^2;
f6=@(t) abs((2-t).*cos(2*t)).^2;
E=(1/T)*integral(f5,-T/2,0)+(1/T)*integral(f6,0,T/2);

Above and below this point i would prefer to use the abs(ft).^2 above, however, this works just fine

I am not sure how to implement the line abs(ft-s) below, as they might need scalar values or arrays?

z1=0;
z2=0;
x=0;
while z2<1/10
    s2=@(t) c0+sum(c(x).*exp(1j*x*w*t),1)+sum(cc(x).*exp(1j*-1*x*w*t),1);
    x=+1;
    z1=ft-s2;
    z2=abs(z1);
end


function y=f(t)

if (-pi<t)&(t<0)
    y=(4+t)/2;
elseif (0<=t)&(t<pi)
    y=(2-t).*cos(2*t);
end
end

Thanks in advance for any help given

[Jan]

Please explain "it will not work" with any details.

Is it intented, that y is undefined in f(), when it is < -pi or > pi?

[Jai Harnas]

Yes, so y being the output, the interval for the given equation is [-pi,pi]

edit: so i just changed it so y is defined for <-pi and >pi and now both plot on the same graph. However, now this error occurs:

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your
function to return an output with the same size and shape as the input arguments.

[Jan]

Where does this warning occur?

It is still not clear to me, what your problem is.

[Jai Harnas]

The warning is exclusive to the first fplot line and im not sure how to fix it

Answer 1

[Walter Roberson]

    s2=@(t) c0+sum(c(x).*exp(1j*x*w*t),1)+sum(cc(x).*exp(1j*-1*x*w*t),1);
    x=+1;
    z1=ft-s2;

s2 is a function handle. You need to invoke it with a parameter.

   x=+1;

You are assigning positive 1 to x, not adding 1 to x.

[Jai Harnas]

Thank you, i have changed the x=+1; to x=x+1.

ft and s2 are both function handles in this scenario, so a better question from me would be how i subtract two function handles to give me a scalar which i can then apply the absolute command to to fit my while loop. If it can be done at all?

[Walter Roberson]

No, you cannot take the difference between function handles. A function handle is a pointer to a block of memory where the struct() is that holds the information about the function handle. If there were a meaning for subtraction it would mean how far apart the data structures are in memory.

You can invoke a function handle on a value to get a numeric result. However, you would need a definite t to invoke the handle on, and based on what you have posted, you do not have a definite t.

[Jai Harnas]

Okay thank you, thats what i was wondering, is there anyway to overcome this or would i have to re-write my code in another way with a definite value for t?

Answer 2

[Jan]

If f the values t == pi and t == -pi are excluded also. Is this wanted?

Do you want f to accept vectors for t? Then:

function y = f(t)
y        = zeros(size(t));
index    = (-pi<t) & (t<0);
y(index) = (4 + t(index)) / 2;
index    = (0<=t) & (t<pi)
y(index) = (2 - t) .* cos(2 * t);
end

The IF condition must be a scalar, so if pi<t is converted implicitly to if all(pi < t).

[Jai Harnas]

Accepting scalars for what i am trying to do throughout this code is fine. Implementing this code however, still gives the error telling me that the function behaves unexpectedly

[Jan]

@Jai Harnas: Please do not rephrase the error message in your word, but post a copy of the complete message. Then it woulöd start to get clear, in which part of the code the problem is.

"Accepting scalars for what i am trying to do throughout this code is fine." - I do niot think so. You get a corresponding error message. In you original code you provide the interval [-pi, pi], but for t==pi and t==-pi the function f does not reply an output.