[EMERGENCY NEED CRITICAL HELP] Numerically determine the Voltage

DO i differentiate i(t) first, then substitute into the eq for v(t) ? please help

see attached pic for the full details.

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency

Answer 1

[Jim Riggs]

You are given an equation for i(t). All you need to do is integrate it directly, but your instructions are to perform the integration numerically, and not use the int function.

First, set the problem constants:

 Q0 = 0;
 C = .003;

To perform numerical integration, you need to define the discrete time parameters;

 Tstart = 0;   % integration start time
 Tstop = 7;    % integration stop time
 dt = .001;   % this is the time step
 nstep = (Tstop-Tstart)/dt + 1;  % This is the number of integration steps

Next, you might want to initialize the data vectors:

 clear time current volt dv

 time = zeros(1,nstep);
 current = zeros(1,nstep);
 volt = zeros(1,nstep);
 dv = zeros(1,nstep);   % this is the differential voltage which is being integrated

Now perform the integration calculation loop:

 for i=1:nstep
     time(i) = (i-1)*dt;
     current(i) = 0.3 + 0.1*exp(-5*time(i)) * sin(25*pi*time(i));  % this is the current function, i(t)
     dv(i) = current(i)*dt + Q0;  % this is the quantity inside the integral

% numerical integration is simply summing
   if i ==1
       volt(i) = 1/C * dv(i);
   else
       volt(i) = volt(i-1) + 1/C * dv(i);
   end
end

Time step, dt

In general, the smaller the time step (dt) the more accurate the answer will be. Note that the current function has a frequency of 12.5 Hz (25 * Pi * t), so this is one of the determining factors regarding what size time step you should use. The number of integration samples should be at least 10 per cycle. This would equate to a time step of 125 per second, or 0.008 seconds. dt should be smaller than this. I chose to use 1000 Hz, dt=.001.

It's always a good idea to try different values to see how it effects the answer.