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:
Matlab graph of the TF
Desmos:
Response of the TF

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Star Strider
Star Strider 2022년 4월 10일

2 개 추천

You forgot to actually use the Heaviside unit step function as an input!
num = [-5.21];
den = [1 6 73];
G = tf(num, den)
G = -5.21 -------------- s^2 + 6 s + 73 Continuous-time transfer function.
syms s t
H = num / poly2sym(den,s) * laplace(heaviside(t))
H = 
h = ilaplace(H)
h = 
figure
hfp = fplot(h, [0 2]);
grid
EndVal = hfp.YData(end)
EndVal = -0.0716
So the step result is correct!
.

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Arcadius
Arcadius 2022년 4월 10일
That was right, thank you :)
Star Strider
Star Strider 2022년 4월 10일
As always, my pleasure!

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