Convolution of two signal

How can I write the convolution code of the function below?

u(t)=sin(2*pi*f*t)
g(t)=exp(-t/tau)

Analytically or with con(,) ?

[Walter Roberson]

If you have the symbolic toolbox then it might be easiest to use the laplace transform:

 laplace(u convolved with g) = laplace(u) times laplace(g)

and take the inverse laplace of both sides...

[Tayfun Çelebi]

thanks. I have the symbolic toolbox

Answer 1

[Image Analyst]

Try this:

numPoints = 1000; % Whatever.
f = .10; % Whatever.
tau = 2; % Whatever.
t = linspace(0, 4*pi); % Whatever.
u = sin(2*pi*f*t);
g = exp(-t/tau);
out = conv(u, g, 'full');
% Plot u
subplot(3, 1, 1);
plot(t, u, 'b-', 'LineWidth', 2);
grid on;
xlabel('t', 'FontSize', 12);
ylabel('u', 'FontSize', 12);
% Plot g
subplot(3, 1, 2);
plot(t, g, 'b-', 'LineWidth', 2);
grid on;
xlabel('t', 'FontSize', 12);
ylabel('g', 'FontSize', 12);
% Plot out
subplot(3, 1, 3);
% plot(t, out, 'b-', 'LineWidth', 2);  % Would need to compute the new t if you want this.
plot(out, 'b-', 'LineWidth', 2);  % Just plot vs. index.
grid on;
xlabel('t', 'FontSize', 12);
ylabel('out', 'FontSize', 12);

% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0.04, 1, 0.96]);

[Tayfun Çelebi]

Thanks for your effort. But I have some questions in my mind.

why did you assign the tau and f as 0.1, 2?

i tried a lot of numbers in the code below for tau and f but i was not sure about any response of convolution. Furthermore, when i convolve the system analytically, can i get same answer like your asnwer ?

t0=-1;
tf=10;
N=50; %5000 100 is tried
dt=(tf-t0)/N;
t=t0:dt:tf;
freq=1; % 0.1 10 5 is tried
tau=3;  % 0.1 100 10 1 is tried 
%--------------------------------------------------------------------------
%Creating Functions
%--------------------------------------------------------------------------
u=0.5*sin(2*pi*freq*t);
g=10*exp(-t/tau);

%--------------------------------------------------------------------------
%Convolve with conv function code
%--------------------------------------------------------------------------
c=conv(g,u);

[Image Analyst]

If you don't like the tau and f I chose arbitrarily, then substitute the values you want.

Note that the output signal is longer than the input signal. That's how convolution works, because the signals start to overlap before the center of the moving signal is over the fixed signal (when the moving signal is to the left of the stationary signal), and there will still be overlap on the right end because the moving signal can keep moving until the left element of the moving signal is over the right element of the stationary signal. So the times on the output signal will be before and after the stationary signal.

You should get the same signal if you do it analytically and evaluate the analytical signal at the quantized t values of the digitial signal.

[Tayfun Çelebi]

As you know, This is continious signal so we convolve the two signals with integral operator. I tried the integral operator but at the end of operation, the response of 'conv' and integral solutuion is different with each other. What is the differences ? Codes below,

%syms t tau
%ans_intgrl=int(exp(-tau)*sin(t-tau),tau,0,t)

t_1 = 0:0.0635:12.5664;
ans_intgrl=4/5*(exp(-t_1/2)-cos(t_1)+1/2*sin(t_1));

% response of analytical convolution
subplot(2,2,4);
plot(ans_intgrl, 'b-', 'LineWidth', 2);
xlabel('t', 'FontSize', 15);
ylabel('Y out', 'FontSize', 15);
grid on;
title('Convolution of Two Signals without conv function');
% 

[Image Analyst]

Why do you think that formula is the convolution?

[Tayfun Çelebi]

Because the convolution formula for Continious time is that. I took it from the book and definition of convolution :

https://en.wikipedia.org/wiki/Convolution

[Walter Roberson]

convolution is an integral from -infinity to +infinity, not from 0 to t.

[Tayfun Çelebi]

I agree with you but ı would like to convolve the system only in the range of 0 and t bacause

at the begining of code , I assumed that the time is in range of o and t.

[Walter Roberson]

Convolution is defined over infinity. However notice that the definition involves integral over tau of f(tau) * g(t-tau). That t is what you will change over 0 to some value.

Effectively for each time in your range you have to calculate the integral.

Sometimes you get lucky and the integral has a closed form solution that makes it efficient to calculate at different values, but when it does not then you just have to evaluate a series of numeric integrals.

[Tayfun Çelebi]

thanks for your valuable answer. so,Could you help me to fix the codes ? I indicated the wrong row below.

%syms t tau
%ans_intgrl=int(exp(-tau)*sin(t-tau),tau,0,t)
t_1 = 0:0.0635:12.5664;    % this row is fixed !!!!!!!!!!!!!!!!!!!!
ans_intgrl=4/5*(exp(-t_1/2)-cos(t_1)+1/2*sin(t_1));
% response of analytical convolution
subplot(2,2,4);
plot(ans_intgrl, 'b-', 'LineWidth', 2);
xlabel('t', 'FontSize', 15);
ylabel('Y out', 'FontSize', 15);
grid on;
title('Convolution of Two Signals without conv function');

[Walter Roberson]

Hmmmm... when i use the convolution integral, I get a result that would be infinite except when sin(f*pi*t) is 0.

[Tayfun Çelebi]

Actually,i am confused about the problem. I hope someone can help us.

[Walter Roberson]

The laplace formulation I mentioned earlier works quickly... but I do not know what the "right" answer is, so I cannot verify the laplace version.

I have been trying Wolfram Alpha's Convolve function, but it gives up very easily I find.

[Csanad Levente Balogh]

Hi!

In the examples above, the results of the convolution are plotted over the index. How can I calculate the time axis for the convolution results? Also, the amplitude is bigger than it supposed to be, isn't there some scaling that should be applied?

[Image Analyst]

If you use 'same', the t axis will be the same. If you use 'full', the t axis will be bigger on both ends by half a window width. You can calculate what that time would be. If you use 'valid', it will of course be less, coming in by half a window width on either end.

The amplitude is what it is and it is correct. If you want to scale it, you're free to do so. If you make the sum of the kernel window, g, elements 1, then the mean of the output signal will be the same as the mean of the input signal.

[Paul]

The following code shows how to compute the result three different ways, assuming that both functions to be convolved are zero for t < 0 (as suggested in the original question):

syms t f tau s w
syms u(t,f) g(t,tau) yconvint(t,f,tau)

u(t,f) = sin(2*sym(pi)*f*t)*heaviside(t);
g(t,tau) = exp(-t/tau)*heaviside(t);

% time domain convolution integral
yconvint(t,f,tau) = int(g(w,tau)*u(t-w,f),w,0,t);
yconvint(t,f,tau) = simplify(rewrite(yconvint(t,f,tau),'sincos'));

% inverse Laplace
ylaplace(t,f,tau) = ilaplace(laplace(u(t,f))*laplace(g(t,tau)));

% discretized convolution
% assume values for f and tau
fval = 1; tauval = 0.5;

% time vector
dt = 0.02;
tvec = 0:.02:10;

ufunc = matlabFunction(u(t,f),'Vars',[t f]);
gfunc = matlabFunction(g(t,tau),'Vars',[t tau]);

yconvdis = conv(ufunc(tvec,fval),gfunc(tvec,tauval));

plot(tvec,double(yconvint(tvec,fval,tauval)),tvec,double(ylaplace(tvec,fval,tauval)),'o',tvec,yconvdis(1:numel(tvec))*dt,'x'),grid

The discretized convolution needs to be scaled by dt to match the true, continuous convolution.