How can I create an array with a for loop?

Hi I’m trying to create an array of values that are output from a function. The function goes from -20 to 50 but I can’t create an array that large so I’m confused on what to do. The attached picture is what I have so far and it’s not working.
% code
clear all;
close all;
A = zeros(1:70);
for m = 1:70;
n = -50:20;
X = power(0.8,n)
A(m) = X
end
stem(n,arrayofA(n))

1 commentaire

Do not use the accepted answer: a loop is NOT required:
n = -50:20;
A = power(0.8,n);

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Réponses (2)

Torsten
Torsten le 12 Mar 2018
A = zeros(1,71);
n = -50:20;
A = power(0.8,n)

1 commentaire

Jos (10584)
Jos (10584) le 12 Mar 2018
This is preferred over the solution with a for-loop. However, there is no need to initialise A with zeros first, since you overwrite it at the third line.

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Birdman
Birdman le 12 Mar 2018
Modifié(e) : Birdman le 12 Mar 2018
Try this:
A=zeros(1,70);n=-50:20;
for m=1:numel(n)
A(m)=power(0.8,n(m));
end
stem(n,A)

5 commentaires

Paige Coffeen
Paige Coffeen le 12 Mar 2018
This one looks like it worked! Can you explain what is the m=1:numel(n) function doing? That seems to be the connecting factor I needed.
Thank you!!
Birdman
Birdman le 12 Mar 2018
It is the condition for the loop to be operated. m starts at 1, incremented by 1 and the loop is over when m is equal to the number of elements in n vector, which is 71. At each step, the statement in for loop is executed and the result is stored in A array.
Jos (10584)
Jos (10584) le 12 Mar 2018
As Torsten showed, A = power(0.8,n) would do. No need for a for-loop
y = zeros(1,12);
for i = 1:12
y(i+1) = y(i) + ((5 - (4/50)*y(i)));
end
y
y = 1×13
0 5.0000 9.6000 13.8320 17.7254 21.3074 24.6028 27.6346 30.4238 32.9899 35.3507 37.5227 39.5209
x = zeros(1,10);
for i = 10:-1:2
x(i-1) = x(i) + ((5 + (4/50)*x(i)));
end
x
x = 1×10
62.4378 53.1831 44.6140 36.6796 29.3330 22.5306 16.2320 10.4000 5.0000 0
Good Day,
May I ask how can I insert the start and stop variable on above code? for example I want to start in 17 to 30 or 15 to 2 with decrementing step.
Thanks and Regards,
Dennis
Walter Roberson
Walter Roberson le 24 Août 2021
You should ask your own question for that.

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