Your approach is correct. The reason the freqz plot did not look correct is that you were passing the filtered signal to it, not the filter coefficients themselves. This approach creates the filter separately, then uses the filter function to filter the signal (since that is essentially what the original ‘yn’ did). The only significant difference is that, and passing the filter coefficients to freqz.
Try this —
fs = 200; %sampling frequency
Ts = 1/200; %sampling time
t = 0:Ts:1;
x = sin(2*pi*2*t) + sin(2*pi*10*t) + sin(2*pi*90*t); %signal
figure(1)
plot(t,x); %unfiltered signal
title('Signal with 2, 10, and 90 Hz');
xlabel('time (s)');
%Difference equation of Hanning Moving Average filter:
yn = zeros(size(t));
ind = 3:length(t);
y = [0.25 0.5 0.25]; % Separate Vector Of Filter Coefficients
yn = filter(y, sum(y), x); % Filter Signal
%Attempt at creating the moving average equation.
figure(2)
plot(t,yn)
title("Hanning Moving Average")
%Me just plotting the filtered Hanning Moving Average plot with
%freqz():
figure(10)
freqz(y,1, 2^16, fs)
The freqz plot now looks correct, displaying the Bode plot of a lowpass filter with a cutoff of about 50 Hz.
In order to see that in detail, after the freqz plot, temporaily add:
set(subplot(2,1,1),'XLim',[0 60])
set(subplot(2,1,2),'XLim',[0 60])
in order to see the -6 dB frequency (at 50 Hz), denoting the filter stopband.
