Sum in two variables for response of NON-LTI discrete time system

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ALESSIO MANIÀ
ALESSIO MANIÀ 2022-4-5
评论： Paul 2022-4-5
Hi everyone.
I'm quite new to MATLAB and I'm trying to write a code that's able to perform the sum written below, to compute the response of a NON-LTI discrete-time system (so it can't be used the function conv()).
I cannot figure how to define the two functions and how to iterate on them to get the correct result.
This is one of the many tries I made, defining two anonymus functions: the result is always zero, which is of course wrong!
delta = 0.01;
nmax= 30;
n = -nmax:1:nmax;
kmax=30;
k=-kmax:1:kmax;
%====================================================================
x = @(n) (1/3).^(n).*heaviside(n+delta); %Input signal
h = @(n, k) (1/2).^(n-k).*heaviside(n+delta); %Impulse response
%====================================================================
yn = 3*(0.5).^n.*heaviside(n+delta); %Expected response
ni=1:length(n);
for ki=1:length(k)
xa = x(ki);
ha = h(ni, ki);
y(ni)=sum(xa.*ha); %Response (WRONG!)
end
%====================================================================
stem(n,yn,'k*');
hold on
stem(n, y, 'r:');
hold off
If anyone has an idea to make this code work, I'd be very grateful!
Thanks in advandce.

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Paul
Paul 2022-4-5
编辑：Paul 2022-4-5
Hi @ALESSIO MANIÀ
1. The loop should be over n, beause each time through the loop it should be computing y(ni), i.e., a single element of y.
for ni = 1:length(n)
2. Internal to the loop, it should compute xa and ha for all of the values of k.
3. But ha should only be computed for the loop value of n
xa = x(k);
ha = h(n(ni),k)
4. I assume the code uses delta to deal with heaviside(0) = 1/2. FWIW, the value of heaviside(0) can be controlled with sympref, though I think the code is fine as is.
  2 个评论
ALESSIO MANIÀ
ALESSIO MANIÀ 2022-4-5
Thanks a lot @Paul, your answer solves all the issues I was trying to manage perfectly!
And I really appreciated also the tip about the heaviside function, which as you've seen I tried to manage in a not very elegant way!
Paul
Paul 2022-4-5
Ran in:
Glad to help. Actually, I thought that was pretty clever how you managed heaviside(). And that's a subtle problem with heaviside() that's easy to miss. For that reason, I avoid using heaviside() in discrete-time problems and instead just define my own function for the discrete-time unit step.
The loop approach very clearly shows the algorithm, but if you're interested it can be replaced with a single line of code using "implicit expansion," which is discussed in the documentation
nmax= 30;
n = -nmax:1:nmax;
kmax=30;
k=-kmax:1:kmax;
%====================================================================
dunitstep = @(n) double(n >= 0); %Discrete unit step
x = @(n) (1/3).^(n).*dunitstep(n); %Input signal
h = @(n, k) (1/2).^(n-k).*dunitstep(n); %Impulse response
%====================================================================
yn = 3*(0.5).^n.*dunitstep(n); %Expected response
y = sum(x(k).*h(n(:),k),2); %Computed response
%====================================================================
stem(n,yn,'k*');
hold on
stem(n, y, 'r:');
hold off

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