iir filter for loop code
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c1 = 8
c2 = 2
c3 = 7
Can you guys help me to find the first seven impulse response of the IIR filter with filter coefficient b0 = 0.05 * c1, b1 = 0.03 * c2, b2 = 0.02 * c3, a1 = 0.5, a2 = 0.5 using for loop code in matlab, this picture may also help
thanks <3
回答（1 个）
Mathieu NOE
2021-10-25
helo
here you are
clc
clearvars
c1 = 8;
c2 = 2;
c3 = 7;
b0 = 0.05 * c1;
b1 = 0.03 * c2;
b2 = 0.02 * c3;
a1 = 0.5;
a2 = 0.5;
% manual for loop coding
x = [1; zeros(6,1)];
samples = length(x);
y(1) = b0*x(1) + 0 + 0 + 0 + 0;
y(2) = b0*x(2) + b1*x(1) + 0 + a1*y(1) + 0;
for k = 3:samples
y(k) = b0*x(k) + b1*x(k-1) + b2*x(k-2) + a1*y(k-1) + a2*y(k-2);
end
figure(1)
plot(y)
10 个评论
Robert Manalo
2021-10-25
Mathieu NOE
2021-10-25
the output is y
you can see the values if you type y in the workspace
Robert Manalo
2021-10-25
Robert Manalo
2021-10-28
Mathieu NOE
2021-10-28
hello
see bleow
c1 = 8;
c2 = 2;
c3 = 7;
b0 = 0.05 * c1;
b1 = 0.03 * c2;
b2 = 0.02 * c3;
a1 = 0.5;
a2 = 0.5;
%% filter numerator / denominator in vector form
B = [b0 b1 b2];
A = [1 a1 a2];
%% manual for loop coding
x = [1; zeros(6,1)];
y = myfilter(x,B,A);
figure(1)
plot(y)
%%%%%%%%%%%%%%%%%%%
function y = myfilter(x,B,A)
samples = length(x);
b0 = B(1);
b1 = B(2);
b2 = B(3);
a1 = A(2);
a2 = A(3);
y(1) = b0*x(1) + 0 + 0 + 0 + 0;
y(2) = b0*x(2) + b1*x(1) + 0 + a1*y(1) + 0;
for k = 3:samples
y(k) = b0*x(k) + b1*x(k-1) + b2*x(k-2) + a1*y(k-1) + a2*y(k-2);
end
end
Kennedy Allen Aday
2021-10-28
got it bro! <3 thank you so much
Mathieu NOE
2021-10-28
do you accept my answer this time ?
tx
Robert Manalo
2021-10-30
Johnny Dela Vega
2021-10-31
@Mathieu NOE Hey, can I ask you one question? I have a code for the same problem but was rejected. Can you tell me what is wrong with my code?
function y=IIR_Filter(b,a,x)
c=[4 8 1 2 3];
b0=0.05*c(4);
b1=0.03*c(3);
b2=0.02*c(2);
b=[b0 b1 b2];
a=[0.5 0.5];
x=[1 0 0 0 0 0 0];
N=length(a)-1;
M=length(b)-1;
y=zeros(1,length(x));
for n=1:length(x)
y1=0;
for k=1:N
if n-k>=1
y1=y1+a(k+1)*y(n-k);
end
end
y2=0;
for k=0:M
if n-k>=1
y2=y2+b(k+1)*x(n-k);
end
end
y(n)=-y1+y2;
end
Mathieu NOE
2021-11-19
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2021-10-25
2021-11-19
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