Symbolic Calculation to Numeric value for limit of an integral

Hi, I've got a problem trying to evaluate numerically a symbolic expression, which by the way should represent the power of a signal.

I run my script and as a result I get the symbolic formula I am trying to solve; when I apply "vpa" I obtain the same result, instead with "eval" or "feval" there will occur errors.

Here the functions I use, potenza and fmodulsquared:

function [y] = potenza (signal)

hMod = @fmodulsquared;
hModsign = @(t) hMod(signal(t));
syms D;
resultIntegral = (int(  hModsign, -D/2 , +D/2  ))./D;
y = limit(resultIntegral,D,+Inf); 
end

function z = fmodulsquared(x)
z = abs(x).^2 ;
end

Now, running this script -which I use to test the function potenza - :

signal = @(t) -3*(sin((t-5/2)/3));

power = potenza(signal);
fprintf("La potenza del segnale è: %s \n", power);
vpa(power)

I get :

La potenza del segnale è: limit(int(9*abs(sin(t/3 - 5/6))^2, t, -D/2, D/2)/D, D, Inf) 
 
ans =
 
limit(int(9*abs(sin(t/3 - 5/6))^2, t, -D/2, D/2)/D, D, Inf)
 

using eval instead of vpa I get:

Error using evalin
Unrecognized function or variable 't'.

Error in sym/eval (line 14)
s =
evalin('caller',vectorize(map2mat(char(x))));

[darova]

Please sohw signal

power = potenza(signal);

[Nicola Scrof]

I am sorry, I forgot it when I pasted the script, now you can see it. Thank you very much.

Answer 1

[darova]

Look at your signal

t = linspace(0,100);
signal = @(t) -3*(sin((t-5/2)/3));
area(t,signal(t))
axis equal

It never ends! Integral is area under the curve

[Nicola Scrof]

That's true, but thanks to the Symbolic Math Toolbox I can get a symbolic answer integrating it, indeed if I apply the integral to the sinusoidal function, as I do here:

syms D;
signal = @(t) -3*(sin((t-5/2)/3));
int(signal, -D/2, +D/2);

I get the following:

ans =
 
18*sin(D/6)*sin(5/6)

Then, I can divide ans by D and apply the limit as following:

>> signalMed = ans/D
 
signalMed =
 
(18*sin(D/6)*sin(5/6))/D
 
>> limit(signalMed, D, +Inf)
 
ans =
 
0

So troubles arrive when I square the module of the sinusoidal function, through fmodulsquared

[darova]

If the value square (all positive) the answer is inf

[Nicola Scrof]

Thank you for your help. I have found the solution, which is to divide by D before integrating! In this way, I get the result I wanted, as shown in the picture .