How to convolve signal and transfer function

Hello,

I have a input signal given by

    Vin(t) = u(t)-u(t-3)    

and a transfer function in time domain given by

    h(t) = ( exp^(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 )

I want to be able to convolve these two equations but i am unsure of how to do so. I have read through the documentation for conv() but cannot find a way to do this.

Any help would be greatly appreciated.

Thanks

Answer 1

[Star Strider]

Try this —

t = linspace(0, 10)

t = 1×100

0 0.1010 0.2020 0.3030 0.4040 0.5051 0.6061 0.7071 0.8081 0.9091 1.0101 1.1111 1.2121 1.3131 1.4141 1.5152 1.6162 1.7172 1.8182 1.9192 2.0202 2.1212 2.2222 2.3232 2.4242 2.5253 2.6263 2.7273 2.8283 2.9293

u = @(t) heaviside(t);
Vin = @(t) u(t)-u(t-3);
h = @(t) ( exp(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 );

y = conv(Vin(t), h(t), 'same');

figure
plot(t, y)
grid
xlabel('t')
ylabel('y(t)')

.

[Tawseef Patel]

Thank you for your reply. The only thing I wanted to ask as a follow up is, if I take the input signal into the frequency domain via laplace and same with the transfer function I get a graph similar to the one attached.

syms s t

% input
vint = heaviside(t) - heaviside(t-3);
vins = laplace(vint); 

% transfer function
ht = ( exp(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 );
hs = laplace(ht);

% output
ws = (vins*hs) ;
wt = ilaplace(ws);

fplot(wt, [-1,10]); hold on; fplot(vint, [-1 10])

which does not resemble the graph that you have attached in your response. Any clue as to why that may be? They should both be returning the same output.

Thanks

[Star Strider]

The reult using the laplace transform is correct.

However a similar result can occur with conv with the 'full’ (default) option —

t = linspace(0, 10);
u = @(t) heaviside(t);
Vin = @(t) u(t)-u(t-3);
h = @(t) ( exp(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 );

y = conv(Vin(t), h(t));

figure
plot(t,y(1:numel(t)))
grid
xlabel('t')
ylabel('y(t)')

.

[Tawseef Patel]

Works! Thank you so much.

[Star Strider]

As always, my pleasure!

.

[Paul]

A few thoughts to consider.

syms s t
% input
vint(t) = heaviside(t) - heaviside(t-3);
vins = laplace(vint); 

Keep in mind that Matlab assumes that t covers all of real line. Because ht is (I think) supposed to be the impulse response, it's best to ensure that ht(t) = 0 for t < 0.

% transfer function
ht(t) = (( exp(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 )) * heaviside(t);
hs = laplace(ht);
% output
ws = (vins*hs) ;
wt(t) = ilaplace(ws);

As shown in the plot, wt is not zero for t<0, so probably best to enforce that as well

wt(t) = wt(t) * heaviside(t);
figure; hold on
fplot(wt,   [-1,10]);   % output
%fplot(vint, [-1 10]);   % input
%fplot(ht,   [-1 10]);   % impulse response

Note that wt is now zero for t < 0.

Can also compute the exact output using the convolution integral

syms tau real
wt1(t) = int(vint(tau)*ht(t - tau),tau,-inf,inf);
fplot(wt1, [-1 10],'o')

The conv() function computes the convolution sum, which can approximate the convolution integral when scaled by the time step in the data. Also, need to be careful when using heaviside() to generate samples of a continuous signal because of the default value of heaviside(0)

t = linspace(0, 10);
dt = t(2) - t(1);
%u = @(t) heaviside(t); % by default, heavside(0) = 1/2, but at t = 0, we should have u(0) = 1 by definition of unit step
u = @(t) (t>=0);
Vin = @(t) u(t)-u(t-3);
h = @(t) ( exp(-2*t)/6 ) + ( 3/16*exp(-t) ) - ( exp(-5*t)/48 );
y = conv(Vin(t), h(t))*dt;  % must scale by dt
plot(t,y(1:numel(t)),'-x')
legend('wt - Laplace','wt - convolution integral','wt - convolution sum')