MATLAB: How can i integrate numerically by mantaining some parameters (need to differentiate later)

integrationnumericalparameter

Hi guys! I'm having problems with MATLAB, I'm not such an expert and have not used it for a bit, and I'm trying to evaluate an integral of a function with more variables. The variables are x and d (Rs,Rc have finite values that are given before). This is a piece of the function:

fun =@(x,d) 1/(((d+Rs+Rc)^2*sin(x)^2-Rs^2)^2);

This has to be integrated between 0 and pi for x. The result has to be differentiated in d after.

I= int(fun,x,0,pi); (i know this is wrond but I don't know how to write it right)
D=diff(I,d);

I get a problem of "Undefined function or variable 'd'." and can't do the integral. I don't know how to avoid the problem. i tried to solve it symbolically, but the function is too long and the program can't solve it. I don't know how to solve the problem with my actual knowledge. The final thing I need is the value of the differential as function of d.

Thanks in advance!

EDIT: If you may understand better, there it is the full expression I need to integrate-differentiate:

fun= @(x,d,Rs,Rc) 1/(((d+Rs+Rc)^2*sin(x)^2-Rs^2)^2)*( -((d+Rs+Rc)^2*cos(x)^2+(d+Rs+Rc)^2-Rs^2)/(((d+Rs+Rc)^2-Rs^2)^(1/2))+((d+2*Rs+Rc)*(d+Rc)*((d+Rs+Rc)^2*cos(x)^2+3*Rc*(d+Rs+Rc)*cos(x)+(d+Rs+Rc)^2-Rs^2)+Rc*(d+Rs+Rc)*cos(x)*(2*Rc^2+6*Rc*(d+Rs+Rc)*cos(x)+3*(d+Rs+Rc)^2*cos(x)^2))/((Rc^2+2*Rc*(d+Rs+Rc)*cos(x)+(d+Rs+Rc)^2-Rs^2)^(3/2)));

 fun= @(x,d,Rs,Rc) 1/(((d+Rs+Rc)^2*sin(x)^2-Rs^2)^2)*( -((d+Rs+Rc)^2*cos(x)^2+(d+Rs+Rc)^2-Rs^2)/(((d+Rs+Rc)^2-Rs^2)^(1/2))+((d+2*Rs+Rc)*(d+Rc)*((d+Rs+Rc)^2*cos(x)^2+3*Rc*(d+Rs+Rc)*cos(x)+(d+Rs+Rc)^2-Rs^2)+Rc*(d+Rs+Rc)*cos(x)*(2*Rc^2+6*Rc*(d+Rs+Rc)*cos(x)+3*(d+Rs+Rc)^2*cos(x)^2))/((Rc^2+2*Rc*(d+Rs+Rc)*cos(x)+(d+Rs+Rc)^2-Rs^2)^(3/2)));

Best Answer

Manually differentiate "fun" with respect to d.

If the result is "dfun(x,d)", then

D = @(d) integral(@(x)dfun(x,d),0,pi,'ArrayValued',true)

should work.

Best wishes

Torsten.

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MATLAB: 5G NR Downlink Carrier Waveform Generation

Hi Javier,

You can provide any values. But the best guide to run different simulations is the Transmission bandwidth configurations provided in the standard spec of TS 38.104 Section 5.3.2.

For example to configure a value of Channel Bandwidth of 30 MHz of 15 kHz SCS and 50 MHz bandwidth of 60 kHz SCS, make the following updates to carrier structure and bwp structure:

% Carriers Configuration
carriers = [];
carriers(1).SubcarrierSpacing = 15; % 15 kHz
carriers(1).NRB = 160;  % 160 resource blocks comprise 30 MHz bandwidth
carriers(1).RBStart = 0;
carriers(2).SubcarrierSpacing = 60; % 60 kHz
carriers(2).NRB = 65;    % 65 resource blocks comprise to 50 MHz bandwidth
carriers(2).RBStart = 0;
% Bandwidth parts configuration (Occupying complete carrier for both numerologies)
bwp = [];
bwp(1).SubcarrierSpacing = carriers(1).SubcarrierSpacing;          % BWP Subcarrier Spacing same as carrier subcarrier spacing 1
bwp(1).CyclicPrefix = 'Normal';         % BWP Cyclic prefix for 15 kHz
bwp(1).NRB = carriers(1).NRB;                        % Size of BWP

bwp(1).RBOffset = 0;                   % Position of BWP in SCS carrier

bwp(2).SubcarrierSpacing = carriers(2).SubcarrierSpacing;          % BWP Subcarrier Spacing
bwp(2).CyclicPrefix = 'Normal';         % BWP Cyclic prefix for 60 kHz
bwp(2).NRB = carriers(2).NRB;                        % Size of BWP
bwp(2).RBOffset = 0;                   % Position of BWP in SCS carrier

This will generate the output waveform for both the numerologies.

Hope this helps.

Regards,

Sriram

MATLAB: FSOLVE requires all values returned by functions to be of data type double.

If you want to solve a system of equations symbolically or using symbolic variables, use the solve function instead of fsolve. The fsolve function requires that the output of the function you specify as the first input be double precision numbers, not symbolic expressions.

If you must use fsolve with symbolic variables inside the function, you must convert those symbolic expressions into numbers using double before returning the numbers (not the symbolic expressions) from the function.

As a very simple example, put all this code into a function example373677.m and run it:

function y = example373677
y = fsolve(@mysubfun, 3);
 
function z = mysubfun(q)
x = sym('x');
r = x.^3-2*x-5;
z = double(subs(r, q));

Of course, with solve this is easier.

syms x
y = solve(x.^3-2*x-5)

You may need to use vpa to approximate y numerically.

vpa(y)

One of the solutions returned in y should match the result you received from running the function.

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