% EE3032-2 Winter 2019-20 W3D4 example

% In class yesterday, we calculated the segment of the output y(t) on 0<=t<=1
% given the input x(t) and the impulse response h(t).

% Today in MATLAB we will:
% 1. graph x(t) and h(t)
% 2. have MATLAB compute the convolution y(t) and graph it
% 3. graph our computed segment of y(t) and confirm it matches.

dt = 0.001; % Time step for plotting and Riemann integral approximation
t = -0.5:dt:2.5;
h = zeros(size(t)); % impulse response value vector, initially all 0s
r1 = 0<=t & t<=1; % Boolean vector that is true in region 1: 0<=t<=1
h(r1) = 1-t(r1); % In region 1: apply the given line equation to calculate h values.

x = zeros(size(t));
x(r1) = t(r1); % x has the same region 1 as h, but different values
r2 = 1<=t & t<=2; % region 2 where x(t) = 1
x(r2) = 1;

yCalc = zeros(size(t)); % using calculus, we found y(t) in region 1
yCalc(r1) = t(r1).^2/2 - t(r1).^3/6;

y = dt*conv(x,h); % conv is digital convolution; multiplying by dt gives Riemann approximation to continuous-time convolution
t2 = (t(1)*2):dt:(t(end)*2); % apply the width property to find the time support for the convolution result
plot(t,h, t,x, t2,y, t,yCalc)
legend('h(t)', 'x(t)', 'y(t)', 'yCalcR1(t)')