DTFT and LSI response of signal processing

Verify the symmetry of the real signal

Assume

Establish the evenodd function of sequence parity decomposition, calculate the even part and odd part of x(n), and draw the original sequence and component sequence graph; find the DTFT of the component sequence, draw the graph, and verify the symmetry properties of the real signal DTFT .
Sequence parity decomposition uses
xe(n)=[x(n)+x(-n)]*1/2;
xo(n)=[x(n)-x(-n)]*1/2;
experimental code as follows:

t= -3:1:12;
N=16;
x=cos(t*pi/2);
xf=fliplr(x);
xe=(x+xf)*(1/2);%even
xo=(x-xf)*(1/2);%odd
figure;
subplot(3,1,1);
stem(t,x);
title("original");
subplot(3,1,2);
stem(t,xe);
title("even");
subplot(3,1,3);
stem(t,xo);
title("odd");
 
[Xe,w]=freqz(xe,1,512,'whole');
[Xo,w]=freqz(xo,1,512,'whole');
figure;
subplot(2,1,1);
stem(w,Xe);
title("even");
subplot(2,1,2);
stem(w,Xo);
title("odd");

The original sequence and the odd and even sequence are as follows:

The Fourier transform of the parity sequence is as follows:

You can see that the real part of the sequence has symmetry

The amplitude and phase response of the filter

A third-order filter is described by the following difference equation:

Draw the amplitude and phase response of the filter, and explain the type of filter (low pass, band pass, high pass)

b = [0.0181,0.0543,0.0543,0.0181];
a = [1.0000,-1.7600,1.1829,-0.2781];
[h,t]=impz(b,a);
k = [0:500];
w = (pi/500)*k;
t = t';
h = h';
H = h * ( exp(-j*pi/500) ).^(t'*k);
magH = abs(H);
angH = angle(H);
figure;
subplot(2,1,1);
plot(w/pi,magH);
title('Magnitude part');
subplot(2,1,2);
plot(w/pi,angH);
title('Angle part');

The result is as shown

It can be seen that this is a low pass filter