How can I apply filter with loop-based function instaed of using filter( ) : built in MATLAB function?

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Krittanai il 17 Set 2022
Modificato: Jan il 18 Set 2022
Hello
I'm trying to plot the generated signal x against filtered x with loop-based function, but I don't know how to do it. Can somebody instuct me how to do it? Here is my attached code. Thank you for your help in advance!
%% Applying the filter to data
% create time sequence n for 501 samples from 0 to 500
n = linspace(0, 500, 500+1);
% assume a signal x
x = 0.7*sin(0.02*pi*n) + sin(0.1*n)+ 0.1*sin(rand()*n);
plot(n,x)
%% Example : HPF
%{
Difference Equation
y[n] = 0.5x[n] + 0.4x[n-1] + 0.1x[n-2]
in a form of [a]y = [b]x
%}
a_1 = 1; b_1 = [0.5 0.4 0.1];
% apply filter
y_hpf1 = filter(b_1,a_1,x);
% plot the generated signal x against filtered x
figure();
plot(n,x);
title("x with HPF")
hold();
plot(n,y_hpf1);
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Risposte (1)

Jan il 17 Set 2022
Modificato: Jan il 17 Set 2022
On the other hand: "y[n] = 0.5x[n] + 0.4x[n-1] + 0.1x[n-2]" is easy to implement and the question sounds like a homework assignement. Start with pre-allocating the output y, which has the same size as the input x:
y = zeros(size(x));
You cannot apply the formula to y(1), because there is no x(0) and x(-1). So use y(1) = 0.5 * x(1) and y(2) = 0.5 * x(2) + 0.4 * x(1). Then create a for loop to calculate the remaining elements.
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Krittanai il 18 Set 2022
Also, I have to use "y[n] = 0.5x[n] + 0.4x[n-1] + 0.1x[n-2]" & "y[n] = 0.5x[n] - 0.4x[n-1] - 0.1x[n-2]" as the problem states.
Jan il 18 Set 2022
Modificato: Jan il 18 Set 2022
@Krittanai: "but I can't" - please mention, what this means. It is easier to solve a problem than to guess, what the problem is. So if you get an error message, read it and if you do not understand it, post a copy of the complete copy here. If the result differ from your expectations, explain the difference.
The simple method myFilter assumes, that the filter parameters a and b have the correct dimensions:
a = [2, 0, 0]; % Not just a = [2]
b = [0.5 0.4 0.1];
You find a version on the linked page, which cares for scalar a also (Link)
"I have to use "y[n] = 0.5x[n] + 0.4x[n-1] + 0.1x[n-2]" " - then simply convert this to Matlab:
for k = 3:length(x)
y(k) = 0.5 * x(k) + 0.4 * x(k-1) + 0.1 * x(k-2);
end
I've inserted * for the multiplication and converted [ and ] to ( and ). You see, this is not magic.
I've mentioned before, that the first 2 elements of y need extra care.

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