How to Design a lowpass filter by cascading 10 sections of the first-order IIR lowpass filter

I design all of this filter but i can't understand how to make filter cascading 10 section.

out teacher said you can conv the transfer function 10 time but it didn't work. any idea?

%% low-pass IIR cascade K = 10 and compare with first-order low pass 
clc; clear all; close all;
w = 0:pi/255:pi;

% For Tenth-Order Low-Pass IIR | Transfer Function
num1 = [1 1];
den1 = [1 0.31];
y = (num1) / (den1) ;
for i=1:9
    y = conv (y,y)
end
h1 = freqz(y, 1, w);
g1 = 20*log10(abs(h1));
% PLOT CODE For Tenth-Order
figure (2);
plot(w/pi,g1);grid minor; axis ([0 1 -100 5]);
xlabel("\omega /\pi");ylabel("Gain in dB");
title("LOW-PASS Cascade K=10 IIR Filter");

i had problem in conv part that its result is incorrect.

Answer 1

[Mathieu NOE]

hello

yes , conv is doing what your professor is asking , but this is the right way to use it (on the num and den part of your IIR filter)

%% low-pass IIR cascade K = 10 and compare with first-order low pass 
clc; clear all; close all;
w = 0:pi/255:pi;
% For Tenth-Order Low-Pass IIR | Transfer Function
num1 = [1 1]/1.5267;  % the /1.5267 is to make static gain = 1 (0 dB) (my suggestion)
den1 = [1 0.31];
% y = (num1) / (den1) ;

num = num1; % init
den = den1; % init

h1 = freqz(num, den, w);
g1 = 20*log10(abs(h1));
% PLOT CODE For first-Order
figure (1);
plot(w/pi,g1);grid minor; axis ([0 1 -100 5]);
xlabel("\omega /\pi");ylabel("Gain in dB");
title("LOW-PASS Cascade K=1 IIR Filter");

for i=1:9
    num = conv (num,num1);
    den = conv (den,den1);
end
h1 = freqz(num, den, w);
g1 = 20*log10(abs(h1));
% PLOT CODE For Tenth-Order
figure (2);
plot(w/pi,g1);grid minor; axis ([0 1 -100 5]);
xlabel("\omega /\pi");ylabel("Gain in dB");
title("LOW-PASS Cascade K=10 IIR Filter");

[HADIMARGO]

tnq very much.

according to this code i could solve that.

and i had to add line 8 and 9.