Let's use e(-x) and e(-2x), since e(+20) is so big.
subplot(311), plot(x,y,'-r.',x,h,'-g.');
grid on; legend('y','h'); xlabel('x'); title('y(x),h(x)')
subplot(312); plot(xc,q,'-b.');
xlabel('xc'), title('conv(y,h)'); grid on
Try it.
There is no graphical way that I know of to illustrate the convolution function. You can understand the value of the convolution integral at one particular overlap value, i.e. you can attempt to illustrate q(xc) at a specific value of xc, by plotting x and a backwards, translated-by-xc version of h(x). This is not enough, however. The value of q(xc) is the sum of the area under the curve y(x)*h(xc-x). Therefore I have also plotted that curve, in black. In the code and plot below, "hft" stands for "h, flipped and translated".
For xc=2:
hft=[hflip(end-4:end),zeros(1,N-4-1)];
subplot(313), plot(x,y,'-r.',x,hft,'-g.',x,y.*hft,'-k.')
grid on; xlabel('x'), title('y(x), h(2-x), y(x).*h(2-x)'); legend('y','hft','y.*hft');
The sum of all the values on the black curve in the bottom plot should equal the value of q at xc=2 in the middle plot.
fprintf('q(xc=2)=%.2f, sum(y.*hft)=%.2f.\n',q(5),sum(y.*hft));
q(xc=2)=0.32, sum(y.*hft)=0.32.
Does it? Yes. Multiply both values by deltax, if you want to get area under the curve, instead of simple sum.
To get the next point, i.e. q(xc=2.5) on the blue curve in the middle plot, shift the green curve on the bottom plot one point to the right, i.e. put the green peak at 2.5, instead of 2. Then recalculate the black curve on the bottom plot, and add up the black values on the recalculated bottom plot. The shift again to get the next point on the middle plot. And so on, until you get get all the blue points on the middle plot.
