How does the step function work?
2 ビュー (過去 30 日間)
古いコメントを表示
Hello, I am trying to figure out how the 'step' function work. I have a transfer function:
num = [-5.21];
den = [1 6 73];
G = tf(num, den);
Getting the inverse laplace of this transfer function manually, I get:
y(t)=((-5.21)/8)*exp(-3t)*sin(8t)
When I use the 'step' function the final steady state in the graph is -0.0714, when the final steady state of the inverse laplace approaches 0. There are great differences too in the graphs as the one worked manually it shows that the oscillation first peaks at -0.39 while matlab shows the first oscillation first peaks at -0.09. Where did I go wrong? I would greatly appreciate an answer. Thanks!
Matlab:

Desmos:

0 件のコメント
採用された回答
Star Strider
2022 年 4 月 10 日
You forgot to actually use the Heaviside unit step function as an input!
num = [-5.21];
den = [1 6 73];
G = tf(num, den)
syms s t
H = num / poly2sym(den,s) * laplace(heaviside(t))
h = ilaplace(H)
figure
hfp = fplot(h, [0 2]);
grid
EndVal = hfp.YData(end)
So the step result is correct!
.
2 件のコメント
その他の回答 (0 件)
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

