DTFT on for filter

Question

Tu Nguyen 2022-2-15

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1650390-dtft-on-for-filter

评论： Paul 2022-2-16

I need to find DTFT of h1 and h2 on MATLAB, but the graph and what I calculation is very different on h2. If anyone knows, please help me check

h1 = [.5 .5];

h2 = [0.5 -0.5];

omega = -4*pi:0.01:4*pi;

fz1 = freqz(h1,1,omega);

fz2 = freqz(h2,1,omega);

figure (7);

subplot(2,1,1);

plot(omega/pi, abs(fz1));

grid on

title('Magnitude response of h1[n]');

hold on

subplot(2,1,2);

plot(omega/pi,abs(fz2));

grid on

title('Magnitude response of h2[n]');

hold off

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Answer 1

Paul 2022-2-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1650390-dtft-on-for-filter#answer_897110

在 MATLAB Online 中打开

What exactly is not matching for h2? Did you do something different for h1? Looks like freqz returns the DTFT of h2 as it should. Well, at least the magnitude, I didn't check the phase.

h1 = [.5 .5];

h2 = [0.5 -0.5];

omega = -4*pi:0.01:4*pi;

fz1 = freqz(h1,1,omega);

fz2 = freqz(h2,1,omega);

figure;

subplot(2,1,1);

plot(omega/pi, abs(fz1));

grid on

title('Magnitude response of h1[n]');

subplot(2,1,2);

hold on

plot(omega/pi,abs(fz2));

grid on

title('Magnitude response of h2[n]');

plot(omega/pi,abs(h2(1)+h2(2)*exp(-1j*omega)),'ro','MarkerIndices',1:20:numel(omega))

hold off

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Tu Nguyen 2022-2-16

Because when I check the point of h2 at pi/2, it should result in imaginbary number, nut on the graph it return 0.717. That's why I think I did something wrong on the h2 DTFT

Paul 2022-2-16

在 MATLAB Online 中打开

The DTFT of h2 at omega = pi/2 is

h2 = [0.5 -0.5];
fz2 = h2(1) + h2(2)*exp(-1j*pi/2)
fz2 = 0.5000 + 0.5000i

Its magnitude, which is what is being plotted using the abs() function, is

abs(fz2)
ans = 0.7071

which is indeed 0.7071, and is shown on the plot above at the x-axis value of 1/2 (because the independent variable for those plots is omega/pi). Feel free to show the manual calculations if they are yielding a different result.

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