Frequency response of discrete PT1

Question

MOC 2020-6-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/546416-frequency-response-of-discrete-pt1

回答： MOC 2020-6-11

I want to calculate the frequency response of a discrete PT1 system with T=T1 and K=1. (the code is at the bottom of the question)

To validate the magnitude response, I want to use an sinusoidal input signal with linearly increasing frequency from 0 to fs/2.

The magnitude response should be the envelope of the transient response.

However, this is only the case if I change the line

u = sin(2*pi*f.*t);

to

u = sin(pi*f.*t);

Why? I assume that I have any kind of faulty reasoning here, but I can't find it.

Thanks in advance.

clear
%% Input (linear frequency sweep)
Ts = .1e-3;
duration = 30;
t = (0:Ts:duration);
fs = 1/Ts;
f = (fs/2)*t/duration;
u = sin(2*pi*f.*t); 
%% PT1-system continuous
T1 = 1e-3;
G = tf(1,[T1 1]);
%% PT1 discretized
H = c2d(G,Ts,'zoh');
%% PT1 trasient response
x0 = 0;
x = lsim(H,u,[],x0,'zoh');
%% PT1 freq. response
[h,~] = freqz(H.Numerator{1:end},H.Denominator{1:end},f,fs);
%% Plot
figure(1),clf(1),hold on
plot(u),plot(x),plot(abs(h))

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Answer 1

MOC 2020-6-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/546416-frequency-response-of-discrete-pt1#answer_449633

在 MATLAB Online 中打开

Found the error by myself now.

I did not use the instantaneous frequency for the calculation of the sweep signal, as stated here or here.

The corrected code is:

clear
%% Input (linear frequency sweep)
Ts = .1e-3;
duration = 30;
t = (0:Ts:duration);
fs = 1/Ts;
f = (fs/2)*t/duration;
finst = 1/2*(fs/2)/duration*t;
u = sin(2*pi*finst.*t);
%% PT1-system continuous
T1 = 1e-3;
G = tf(1,[T1 1]);
%% PT1 discretized
H = c2d(G,Ts,'zoh');
x0 = 0;
%% PT1 trasient response
x = lsim(H,u,[],x0,'zoh');
%% PT1 freq. response
[h,~] = freqz(H.Numerator{1:end},H.Denominator{1:end},f,fs);
%% Plot
figure(1),clf(1),hold on
plot(u),plot(x),plot(abs(h))