integral of two added function can't be implemented

im trying to answer q5 here

%Q4 answer 
clc
clear
u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
h1= @(t) u(t)-2.*u(t-1)+u(t-2)

h1 = function_handle with value:
    @(t)u(t)-2.*u(t-1)+u(t-2)

h2= @(t) h(t) + h1(t)

h2 = function_handle with value:
    @(t)h(t)+h1(t)

fplot(h2,[-3,3])
grid on 
ylim([-1 2])

% Q5 answer 
clc
clear
t=@(t) t

t = function_handle with value:
    @(t)t

u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
h1= @(t) u(t)-2.*u(t-1)+u(t-2)

h1 = function_handle with value:
    @(t)u(t)-2.*u(t-1)+u(t-2)

h2= @(t) h(t) + h1(t)

h2 = function_handle with value:
    @(t)h(t)+h1(t)

i=integral(h2,-inf,t)

Error using integral
Limits of integration must be double or single scalars.

fplot(i,[-3,3])
grid on 
ylim([-1 2])

Answer 1

[Paul]

Hi Faisal,

If the plot for the answer to Q4 is corret, then does the integration really need to start from -inf?

However, I suggest revisiting Q4, that solution does not look correct. Why would the output of the system be the sum of the impulse response and the input?

[Faisal Al-Wazir]

ok i think i didn't understood the qustion

i did some changes to q4

can you tell me if it's correct now ?

clc
clear
u = @(t) double(t>=0);
x= @(t) u(t)-2.*u(t-1)+u(t-2)

x = function_handle with value:
    @(t)u(t)-2.*u(t-1)+u(t-2)

h1 = @(t) exp(-t).*(x(t)-x(t-1));
%h2= @(t) h(t) + h1(t)
fplot(h1,[-5,5])
grid on 
ylim([-1 2])

[Paul]

If h1(t) is supposed to be the answer to Q4, then no, it's not correct.

Also, the expression for h in the Question isn't correct. The u(t-1) should be u(t-2).

How are you trying to solve this problem?

[Faisal Al-Wazir]

yes my bad so if i change u(t-1) to u(t-2) well it be right ?

if no, then what can i do ?

clc
clear
u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
x= @(t) u(t)-2.*u(t-1)+u(t-2)

x = function_handle with value:
    @(t)u(t)-2.*u(t-1)+u(t-2)

h1 = @(t) exp(-t).*(x(t)-x(t-2));
y1=@(t)  (u(t)-2.*u(t-1)+u(t-2)).* (exp(-t).*(u(t)-u(t-1))) %maybe this one is right 

y1 = function_handle with value:
    @(t)(u(t)-2.*u(t-1)+u(t-2)).*(exp(-t).*(u(t)-u(t-1)))

fplot(h1,[-3,3])
grid on 
ylim([-1 2])

[Paul]

No, just changing the u(t-1) to u(t-2) in the definition of h will not be enough. Typical approaches to solving this problem would be to use convolution or the Laplace transform.

The convolution integral can be computed numerically using integral, in closed form using int, or approximated via the convolution sum via conv.

The Laplace transform solution can be optained using laplace and ilaplace.

[Faisal Al-Wazir]

thanks paul but when i use conv function it gives an empty plot

clc
clear
u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
x= @(t) u(t)-2.*u(t-1)+u(t-2)

x = function_handle with value:
    @(t)u(t)-2.*u(t-1)+u(t-2)

h1= @(t) conv(h,x)

h1 = function_handle with value:
    @(t)conv(h,x)

  
fplot(h1,[-3,3])
grid on 
ylim([-1 2])

Warning: Error updating FunctionLine.

The following error was reported evaluating the function in FunctionLine update: Input arguments to function include colon operator. To input the colon character, use ':' instead.

[Paul]

Recheck the doc page for conv. The inputs are vectors of numbers, not anonymous functions as you've tried above. So you have to evaluate the functions at the values of t of interest. Also, conv computes the convolution sum, it needs to be scaled appropriately to approximate a convolution integral. Here's an example.

u = @(t) double(t >= 0);             % unit step
h = @(t) exp(-2*t).*(u(t) - u(t-1)); % impulse response
x = @(t) t.*(u(t) - u(t-2));         % input
y = @(t,dt) conv(h(t),x(t))*dt;      % output, scaled by dt to convert from sum to integral
tval = 0:.001:5;
dt = tval(2);
yval = y(tval,dt);
yval = yval(1:numel(tval)); % only retain as many points as in t
plot(tval,yval)

[Faisal Al-Wazir]

thank you so much paul

but for q5 the same method doesn't work

%%
clc
clear
u = @(t) double(t >= 0);             % unit step
h = @(t) exp(-2*t).*(u(t) - u(t-1)); % impulse response
x = @(t) t.*(u(t) - u(t-2));         % input
y = @(t,dt) conv(h(t),x(t))*dt;      % output, scaled by dt to convert from sum to integral
tval = 0:.001:5;
dt = tval(2);
yval = y(tval,dt);
yval = yval(1:numel(tval)); % only retain as many points as in t
plot(tval,yval)

%%
%q5
y1=@(t) int(y,-inf,tval)

y1 = function_handle with value:
    @(t)int(y,-inf,tval)

yval1 = y1(tval,dt);

Error using solution
Too many input arguments.

yval1 = yval1(1:numel(tval));
plot(tval,yval1)

[Paul]

That's because int is used to find a closed-form expression for an integral based on a symbolic expression of the integrand. But you don't have any symbolic expressions, just a sequence of numbers that are (approximate) samples of y at values of t. In this case, consider using cumtrapz.

[Faisal Al-Wazir]

well i tried doing what you said but with no luck

and i still don't know how to add the limits (-inf to t)

%%
clc
clear
u = @(t) double(t >= 0);             % unit step
h = @(t) exp(-2*t).*(u(t) - u(t-1)); % impulse response
x = @(t) t.*(u(t) - u(t-2));         % input
y = @(t,dt) conv(h(t),x(t))*dt;      % output, scaled by dt to convert from sum to integral
tval = 0:.001:5;
dt = tval(2);
yval = y(tval,dt);
yval = yval(1:numel(tval)); % only retain as many points as in t
plot(tval,yval)

%%
%q5
y1= @(t) cumtrapz(y)

y1 = function_handle with value:
    @(t)cumtrapz(y)

yval1 = y1(tval,dt);

Error using solution
Too many input arguments.

yval1 = yval1(1:numel(tval));
plot(tval,yval1)

[Paul]

No need to use a function handle. For example, suppose I want to integrate y(tau)*dtau from 0 to t, with y(t) equal to t^2. We know that the result should be t^3/3. Let's try it

tval = 0:.001:3;
yval = tval.^2;
plot(tval,cumtrapz(tval,yval))

Compare to closed form expression

plot(tval,tval.^3/3)

[Faisal Al-Wazir]

so the final answer is this right ?

but the limits are not -inf to t

%%
clc
clear
u = @(t) double(t >= 0);             % unit step
h = @(t) exp(-2*t).*(u(t) - u(t-1)); % impulse response
x = @(t) t.*(u(t) - u(t-2));         % input
y = @(t,dt) conv(h(t),x(t))*dt;      % output, scaled by dt to convert from sum to integral
tval = 0:.001:5;
dt = tval(2);
yval = y(tval,dt);
yval = yval(1:numel(tval)); % only retain as many points as in t
plot(tval,yval)

%%
%q5
tval = 0:.001:3; 
yval = conv(h(tval),x(tval))*dt;
yval = yval(1:numel(tval))

yval = 1×3001

1.0e+00 * 0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0003 0.0003 0.0003 0.0003 0.0004 0.0004 0.0004

plot(tval,cumtrapz(tval,yval))

[Paul]

Looks ok for this example, but you'll need to adapt it to the actual problem. Not sure why you recomputed tval and yval for Q5 from Q4, but it doesn't matter I suppose.

"but the limits are not -inf to t"

Yes, the effective limits are 0 to t when using cumtrapz for this problem.

From first principles, what is y(t) for t < 0? You should be able to answer this question based only on the form of h(t) and x(t). If you know the answer to this question, then you'll know the answer about the limits of integration.

[Faisal Al-Wazir]

well the issue i don't have the answer

[Paul]

The impulse response is 0 for t < 0. The input to the system is 0 for t < 0. Consequently, would you expect the output of the system to be non-zero for any t < 0? If so, why? If not, then does it matter if the lower limit of integration on y(t) is -inf or 0?

[Faisal Al-Wazir]

greetings paul

i reviewed the code that we were discussing and i think the values you were using are differnt then mine

%%
%q4
clc
clear
u = @(t) double(t >= 0);             % unit step
h = @(t) exp(-t).*(u(t) - u(t-2)); % impulse response
x = @(t) u(t-1)-2*u(t-2)+u(t-3);         % input
y = @(t,dt) conv(h(t),x(t))*dt;      % output, scaled by dt to convert from sum to integral
tval = 0:.001:5;
dt = tval(2);
yval = y(tval,dt);
yval = yval(1:numel(tval)); % only retain as many points as in t
plot(tval,yval)

%%
%q5
tval = 0:.001:3;
yval = conv(h(tval),x(tval))*dt;
yval = yval(1:numel(tval));
plot(tval,cumtrapz(tval,yval))

[Paul]

Yes, as I stated previously, I was showing an example that would have to be modified for the specific problem at hand.

Your current code changes the definition of x(t) from the original Q4. For Q5, consider extending tval to a longer time.