# Use of syms, piecewise & conv commands

8 vues (au cours des 30 derniers jours)
Ben le 11 Sep 2023
Modifié(e) : Paul le 13 Sep 2023
I'm trying to define the values of x(t) at various values of t to plot the input h(t) and x(t) vs the output y(t) after performing convolution. I am getting the a error when trying to run my code and am hoping one of the experts on here may be able to assist. I'm still quite new to MATLAB, I've looked at the associated help documetation, however the error code points me to the inbuilt conv() function which is way over my head. My code is below, along with the error message I am receiving.
Thanks in advance for any help.
Code
t = [0:1:10] % define range of t values
t = 1×11
0 1 2 3 4 5 6 7 8 9 10
h = (exp(-(t-2)).*heaviside(t-2)); % define function h(t)
%x = t .* (t >= 0) .* (t <= 1); % define function x(t) (commented out to attempt piecewise command)
syms x(t)
x(t) = piecewise((0 <= t) & (t <= 1), t, 0)
x(t) = y = conv(h, x); % carry out convolution
Error using indexing
Invalid argument at position 2. Symbolic function indexing evaluates the function at the input arguments. To perform colon indexing call FORMULA before indexing.

Error in conv (line 32)
c = conv2(a(:),b(:));
figure;
plot(t, h, 'r', 'LineWidth', 1, 'DisplayName', 'h(t) = e^{-(t-2)} * u(t-2)'); % plotting h(t)
hold on; % holf to allow multiple plots on one axes
plot(t, x, 'b', 'LineWidth', 1, 'DisplayName', 'x(t) = t (0 <= t <= 1)'); % plotting x(t)
plot(t, y, 'g', 'LineWidth', 1, 'DisplayName', 'y(t) = x(t) * h(t)'); % plotting y(t)
title('Plot of h(t), x(t), and y(t)'); % add title
legend('Location', 'northeast'); % define legend location
hold off; % hold off to allow plotting of figure
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### Réponse acceptée

Paul le 12 Sep 2023
Modifié(e) : Paul le 12 Sep 2023
Hi Ben,
First you have to decide if you want to solve the problem for a closed form expression using symbolic tools, like syms and piecewise and int, or if you want to approximate the solution numerically using ordinary numeric arrays and conv (you could also go for a numeric solution using integral). The code as posted is mixing numeric and symbolic in ways that can't be done.
Considering that you already have x(t) defined symbolically, and I'm sure you can define h(t) symbolically, I'd encourage you to use int to come up with the expression for the convolution x(t)*h(t) (unless you're required to use conv?).
If you try that and run into problems, feel free to respond with the updated code and explain what problem you're having.
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Paul le 12 Sep 2023
Modifié(e) : Paul le 13 Sep 2023
Hi Ben,
The solution you've posted should be reconsidered for a couple of reasons. The use of the 'same' option results in a partial solution, at best. Also, if using 'same', the output values of conv do not correspond to the values in t. Finally, assuming that the goal is approximate the convolution integral, the output of conv has to be multiplied by the sampling period.
First, let's get the exact solution.
Define the signals
syms t real
syms h(t) x(t)
h(t) = exp(-(t-2))*heaviside(t-2);
x(t) = t*rectangularPulse(0,1,t);
Plot them
figure
fplot([x(t),h(t)],[0 10]),grid
xlabel('t') Compute the convolution integral, taking advantage of the fact that h(t) = x(t) = 0 for t < 0
syms tau real
syms y(t)
y(t) = int(h(tau)*x(t-tau),tau,0,t)
y(t) = Plot the result
figure
fplot(y(t),[0 20]),grid
xlabel('t')
hold on
Now take samples of x(t) and h(t) and use conv with the default option 'full' and multiply by the sampling period
t = [0:0.01:10]; % define range of t values
T = t(2); % sampling period
%h = (exp(-(t-2)).*heaviside(t-2)); % define function h(t)
h = exp(-(t-2)).*(t>=2); % define function h(t)
x = t .* (t >= 0) .* (t <= 1); % define function x(t)
y = conv(h,x)*T; % carry out convolution
y has more samples than both x and h, so we have to create an associated time vector
ty = (0:numel(y)-1)*T;
Add to plot (show only a subset of the points in so we can clearly see the overlay).
plot(ty,y,'o','MarkerIndices',1:10:numel(y))
That looks like a pretty good approximation. But, if you zoom in tight you'll see it's not exact, though it could be improved with a smaller sampling period. Also, if you zoom in around t = 10 seconds, you'll see the effect of cutting off the samples of h(t) at t = 10. You could extend the t vector to a larger value, but there will always be a price to pay for using a finite number of samples of h to approximate a signal that lasts forever.
If we use the 'same' option, we get an output that has the same number of elements as the first argument to conv, which in this case is h, which in this problem has the same number of elements as t.
ysame = conv(h,x,'same')*T;
[numel(ysame) numel(h) numel(t)]
ans = 1×3
1001 1001 1001
But the conv output with 'same' doesn't correspond to the values in t. Rather, ysame is the 'central' part of the convolution (unfortunately the doc page conv doesn't define 'central'), which in this case is basically a 10 second interval in the middle of 0 <= t < = 20, which is 5 <= t <= 15
plot(t+5,ysame,'kx','MarkerIndices',1:20:numel(ysame))
legend('h(t)*x(t)','conv(full)','conv(same)') Finally, the problem with heaviside is that, with the default sympref, it evaluates to 1/2 when the input evaluates to 0
heaviside(0)
ans = 0.5000
Typically, that's not what we want when using discrete-time processing to approximate continuous time results (though sometimes it may be).

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