Laplace from symbolic to tf tooling.

1 回表示 (過去 30 日間)
am
am 2021 年 2 月 1 日
コメント済み: am 2021 年 2 月 1 日
Hello, I want to use the function stepinfo().
I understood from different posts that "Control System Toolbox objects and Symbolic Math Toolbox objects are inherently incompatible" (@Star Strider). But in all the responses I saw in the forum, the transfer function was already available.
How do I get the the transfer function from a non laplacian function?
The faulty code:
syms t
k = 3
T2 = 10
c = 2
time_vector = [1:1:50];
y =@(t) k*(1.-exp(-t/T2))*c
laplace(y)
results = y(time_vector);
laplace(y)
Error: Check for missing argument or incorrect argument data type in call to function 'laplace'.

回答 (1 件)

Walter Roberson
Walter Roberson 2021 年 2 月 1 日
syms s t
k = 3
k = 3
T2 = 10
T2 = 10
c = 2
c = 2
time_vector = [1:1:50];
y = @(t) k*(1.-exp(-t/T2))*c
y = function_handle with value:
@(t)k*(1. -exp(-t/T2))*c
Y = laplace(y(t), t, s)
Y = 
[n, d] = numden(Y)
n = 
6
d = 
yTF = tf(sym2poly(n), sym2poly(d))
yTF = 6 ---------- 10 s^2 + s Continuous-time transfer function.
step(yTF, time_vector)
  4 件のコメント
am
am 2021 年 2 月 1 日
編集済み: am 2021 年 2 月 1 日
Thank you, I missed you just needed to install it.
I want to see the stepinfo(), the raise time, settling time etc.
Now I do not have errors, I do have a plot but no step responses. How can I visualise the stepinfo()?
syms s t
k = 3
T2 = 10
c = 2
time_vector = [1:1:50];
y = @(t) k*(1.-exp(-t/T2))*c
results = y(time_vector);
plot(time_vector,results)
hold on
grid on
xlabel('Time')
ylabel('Temperature')
Y = laplace(y(t),t, s)
[n,d] = numden(Y)
yTF = tf(sym2poly(n), sym2poly(d))
stepinfo(yTF)
am
am 2021 年 2 月 1 日
Sorry to recomment, but I am unsure you get a pling if one simply edits :).
How can I visualise the stepinfo() please?

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