Embarrassing Laplace question using ilaplace

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Robert
Robert el 16 de En. de 2017
Respondida: Star Strider el 20 de En. de 2017
I always get stuck using ilaplace And then 6 months goes by and i get stuck again because i don't use it often enough
Im just trying to plot the step response of a first order system using the step command which works fine as always. Just for kicks i want to take the same function take the inverse laplace function and plot it against t manually They should yield the EXACT same answer but yet it doesn't.
Can anyone see my mistake?
clear all
clc
syms R C S
R = linspace(1,10000,5);
C = .0001;
for r = 1:5
Vout_in_t = ilaplace(5/(R(r)*C*S+1))
ezplot(Vout_in_t,[0,2])
hold on
my = tf([1], [1 1/(R(r)*C)])
figure
opt = stepDataOptions
opt.StepAmplitude = 5
step(my,opt)
hold on
end

Respuestas (2)

Jyotish Robin
Jyotish Robin el 20 de En. de 2017
Editada: Jyotish Robin el 20 de En. de 2017
Hi Robert,
From the given code , what I understood is that the transfer function you are trying to use is the one below:
S/(1+RCS)
In that case , you have to rewrite the line which defines your transfer function as:
>>my = tf([1 0], [(R(r)*C) 1])
Hope this helps!

Star Strider
Star Strider el 20 de En. de 2017
If you want to see the response to the step input, you need to convolve it with the Laplace transform of the unit step, (1/s).
See if this does what you want:
syms R C s
R = linspace(1,10000,5);
C = .0001;
for r = 1:5
Vout_in_t = ilaplace(5/(R(r)*C*s+1) * 1/s)
ezplot(Vout_in_t,[0,2])
hold on
my = tf([1], [1 1/(R(r)*C)])
figure
opt = stepDataOptions
opt.StepAmplitude = 5
step(my,opt)
end
hold off

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