Compute convolution y[n]=x[n]*h[n]: x[n]={2,0,1,-1,3}; h[n]={1,2,0,1}
60 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
My approach:
x=[2 0 1 -1 3];
h=[1 2 0 1];
% therefore
y=conv(x,h)
y =
2 4 1 3 1 7 -1 3
4 Kommentare
Walter Roberson
am 2 Dez. 2023
It would depend on whether that u[n-1] is the unit step function or not.
If it is then x[n] would be 0 for n < 0, and 1 for n >= 0 -- an infinite stream of 1's. And h[n] would be 0 for n <= 1, and 0.9^n for n > 1 -- an infinite stream of non-negative numbers. You cannot express that as a finite convolution sequence.
Let us see what it would turn out like for continuous functions:
sympref('heavisideatorigin', 1)
syms n integer
x(n) = heaviside(n)
h(n) = (sym(9)/sym(10))^n * heaviside(n-1)
syms t tau
C(t) = int(x(tau) * h(t-tau), tau, 0, t)
[C(-1), C(0), C(1)]
assume(sign(t-1) == 1)
simplify(C)
Antworten (1)
Image Analyst
am 2 Dez. 2023
@Li Hui Chew, yes your approach is correct. Is that all you wanted - confirmation of your approach?
0 Kommentare
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!