Question

I need to model an electric rectifier and to plot both in and out signals with MATLAB. The rectifier is made of a RC circuit which charge as fast as the voltage increases but discharges way slower so the out signal is more or less flat. It’s supposed to look like that:

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I tried to code it on MATLAB and I got this (my circuit rectifies negative voltage but same principle):
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To get the same figure as the one from wikipedia I tried to compute the intersection between the downing exp curve (red) and the rising sinus curve (blue) so I just had to add a sin curve and a downing exp curve at right intervals to get the out signal.
Here is my code:

f=@(x)sin(2*pi*250000*x+pi/2);%oscillateur de référence
f1=@(x)sin(2*pi*250000*x);
g=@(x)exp(-x*10^4);%décharge du détecteur de crête
h=@(x)f(x)-g(x);%intersection des deux fonctions

format long;
inter=fzero(h,[3.82*10^-6,3.90*10^-6]);

y1=g(0:10^-12:inter);
y2=f(inter:10^-12:4*10^-6);
y3=sin(2*pi*250000*(0:10^-12:1*10^-6));

y=-[y3 y1 y2 y1 y2];

y4=-f1(linspace(0,8*10^-6,length(y)));

x=linspace(0,10*10^-6,length(y));%abscisse

plot(x,y,x,y4);

But why is there a gap between the curves on my figure ?

Answer 1

You really don’t need to find the intersection points. You can reproduce the same curve with a series of nested max() calls and logical operations. Here’s an example:

f=@(x)sin(2*pi*250000*x);
discharge=1e-6; %# quarter period when discharge begins
g=@(x)exp(-rem(x-discharge,(1e-5)/2.5)*10^5); %#modulo over the period to repeat. 
h=@(x)max(f(x).*(x<discharge),max(f(x),g(x)).*(x>=discharge)); %# the rectified signal

y=linspace(0,1e-5,1e4);
plot(y,f(y),y,h(y))