Moving average without using movmean() or conv()

43 vues (au cours des 30 derniers jours)
FS
FS le 10 Oct 2020
Commenté : Rena Berman le 7 Mai 2021
Write a function that performs the moving average of any 1-dimensional input data for any non-zero and positive order M. Make sure to add M-1 zeros at the beginning of the data so that your output data will be the same length as your input data. You must not use built in functions like movmean() or conv().
y[n] = 1/M Σx[n − k]
lower limit k= 0 upper limit M −1
I have this so far:
function
t1=0:0.001:1;t2=1.001:0.001:2;t3=2.001:0.001:3;
t=[t1 t2 t3];
x1=10*ones(1,length(t1));
x2=15*ones(1,length(t2));
x3=20*ones(1,length(t3));
x=[x1 x2 x3];
M=20;
for i=1:length(t)-M
y1(i)=(1/M)*sum(i:i+M-1);
end
y=[zeros(1,M) y1];
figure
subplot(211)
plot(t,x,'b')
ylabel('Signal')
axis([0 3 0 30])
subplot(212)
plot(t,y,'r')
ylabel('Moving avg')
xlabel('Time (s)')
axis([0 3 0 inf])
  11 commentaires
Rena Berman
Rena Berman le 7 Mai 2021

(Answers Dev) Restored edit

Rena Berman
Rena Berman le 7 Mai 2021
(Answers Dev) Restored edit

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Réponse acceptée

Image Analyst
Image Analyst le 10 Oct 2020
I'm attaching a fully manual 2-D convolution. It should be easy for you to adapt it to 1-D
  1 commentaire
Image Analyst
Image Analyst le 20 Oct 2020
I see you're editing the original question to follow my advice in my comment above:
"
You need an outer loop over n but not like you have. You need to use M, not k
for n=1:length(x)-(2*M+1) % M is the window width
Then the inner loop needs to be over k where k goes from 0 to M-1, just like your instructions said. Then you need to sum y and divide by M, again, like the instructions said, not like what you did which was to divide by (2*k+1). If you sum a sequence of M elements in y, then divide by M, you get the average value in that segment of y.
"
So, did that help you? Are we done yet and you can Accept this answer? Or do you still have questions?

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