How does the step function work?

2 views (last 30 days)
Arcadius
Arcadius on 10 Apr 2022
Commented: Star Strider on 10 Apr 2022
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

Accepted Answer

Star Strider
Star Strider on 10 Apr 2022
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!
.
  2 Comments
Arcadius
Arcadius on 10 Apr 2022
That was right, thank you :)
Star Strider
Star Strider on 10 Apr 2022
As always, my pleasure!

Sign in to comment.

More Answers (0)

Categories

Find more on Mathematics in Help Center and File Exchange

Products


Release

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!